Solution Manual = Linear circuits Analysis - 2e, Decarlo
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Transcript of Solution Manual = Linear circuits Analysis - 2e, Decarlo
PROBLEM SOLUTIONS CHAPTER 1.
Solution 1.1. (a) Charge on one electron: -1.6019×10-19 C. This means that charge on 1013 electrons
is: -1.6019×10-6 C. Net charge on sphere is: 1.6019×10-6 C (POSITIVE).
Solution 1.2. (a) 1 atom ≡ -4.646×10-18 C. By proportionality, 64g ≡ NA atoms.
3.1g ≡ ? atoms ⇒ 3.1g ≡3.1NA
64 atoms.
Total Charge = −4.646 ×10−18 C
atom×
3.1× 6.023 ×1023
64atoms =−1.3554 ×105 C
(b) Total charge per atom is -4.646×10-18 C. Total charge per electron is –1.6019×10-19 C. Therefore,there are 29 electrons per atom of copper.
(c) 0.91 A ≡ 0.91 C/s. i =∆Q
∆t ⇒ ∆t =
∆Q
i=
1.36 ×105
0.91=1.49 ×105 sec .
(d) We know there are 3.1NA
64= 2.9174 ×1022 atoms in the penny. Removing 1 electron from
0.05 ×3.1NA
64 atoms means removing 0.05 ×
3.1NA
64 electrons. Therefore,
Net charge = 0.05 ×3.1NA
64×1.6019 ×10−19 = 234C
Solution 1.3 (a) 7.573 ×1017 × −1.6019 ×10−19( ) = −0.1213C
(b) Current =0.1213
10−3 = 121.3A flowing from right to left.
(c) Again, use proportionality:
10 A =x ×1.6019 ×10−19
60sec ⇒ x =
10 × 60
1.6019 ×10−19 = 3.75 ×1021
(d) i t( ) =dq
dt= 1− e−5t A. This is an exponential evolution with an initial value of 0, a final value of 1,
and a time-constant of 1/5 (signal reaches ~63% of it’s final value in one time-constant).
0.2 time in sec
i(t) 1
(e) Current is the slope of the charge waveform. Therefore, by inspection:
Solution 1.4 (a) 6.023×1023×(-1.6019×10-19) = –9.65×104 C.
(b) Current flows from right to left (opposite electrons), and:
I =9.65 ×104
10−3 = 9.65 ×107 A
(c) Using proportionality:
5A =x × 1.6019 ×10−19
60sec ⇒ x =
5 × 60
1.6019 × 10−19 =1.87 ×1021
(d) i t( ) =dq
dt= 1+ 0.5πcos πt( ) ⇒ i 1sec( ) = 1− 1.57 = −0.57A. Current flows from left to right.
Solution 1.5 (a) i t( ) = 1− 4e−2 t + 3e−3t t ≥ 0. Then
q t( ) = i t( )dt0
t
∫ = 1 − 4e−2τ + 3e−3τ( )d0
t
∫ τ = τ]−∞t − 4 e−2τdτ
0
t
∫ + 3 e−3τdτ0
t
∫
= t − 4 −0.5e−2τ[ ]0
t+ 3 −0.333e−3τ[ ]0
t= t + 2e−2t − e−3t −1
(b) By inspection:
(c) q t( ) = 120cos 120πt( ) . Hence
i t( ) =dq
dt= −120π× 120sin 120πt( ) =−14400πsin 120πt( ) A
Solution 1.6. (a) i(t) = 1 − cos(πt) A. Hence
q( t) = i(τ)dτ0
t
∫ = 1 − cos(πτ)( )dτ0
t
∫ = t −1
πsin(πτ)
0
t= t −
1
πsin(πt) C
(b) Charge is integral of current. Graphically, the charge at time t is the area under the current curve up to
time t: (note the quadratic nature between 2 and 4 seconds)
Solution 1.7
Again, Q is the running area under the current curve. Between 0 and 3 seconds, current decreases linearly
until zero. So, Qtot = 7.5 C. From 0 to 6: Qtot = 7.5 + Q3_6 = 7.5 -1/1×0.5 + -1/1×0.5 + -1×1 = 5.5 C,
where the curve from 3 to 6 was divided into two triangular sections and one rectangular one.
Solution 1.8 Charge is the area under the current curve. Thus, Q = 0.1*4 – 0.1*2 = 0.2 C.
Solution 1.9 Calculate the change in energy for the electron: ∆E = Q ∆V = 3.218×10-15.
Equate this to kinetic energy:
3.218 ×10−15 =1
2mv2 ⇒ v = 8.4 ×107m / s
where the mass of an electron, 9.1066×10-31 has been substituted.
Solution 1.10 P = VI. Hence I = P/V = 2×103/120 = 16.6667 A
PROBLEM Solution 1.11 (a) It is necessary to integrate the i(t) curve to obtain q(t). We do this
interval by interval:
(i) 0 ≤ t ≤ 1 ms, q( t) = 0 + dτ0
t
∫ = t µC
(ii) 1 ms ≤ t ≤ 2 ms, q( t) = 1 − 2 dτ1
t
∫ = 3 − 2t µC
(iii) 2 ms ≤ t ≤ 3 ms, q( t) = −1+ dτ2
t
∫ = −3 + t µC
(iv) 3 ms ≤ t ≤ 5 ms, q( t) = 0 + 8 − 2τ( )dτ3
t
∫ = 8t − t2 −15 µC
(v) 5 ms ≤ t, q( t) = 0 µC
0 1 2 3 4 5 6-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Cha
rge
in m
icro
C
TextEnd
Time in ms
(b) Voltage is the ratio of the power and current curve. In this case, the division can be done graphically
by inspection. Note that the ratio of a quadratic function and a linear function is a linear function:
0 1 2 3 4 5 6-1
-0.5
0
0.5
1
1.5
2
Vol
tage
in V
TextEnd
Time in ms
Solution 1.12 (a) VA = P/I = 20/4 = 5 V
(b) PB = VI = 2×7 = 14 W
(c) VC = P/I = -3W/3A = -1V
(d) ID = P/V = -27W/3V = -9A
(e) IE = P/V = 2/1 = 2A
(f) PF = VI = -4×5 = -20W
In all of the above, note that the direction of the current flow relative to the polarity of the voltage across a
device determines whether power is delivered or absorbed. Power is absorbed when current flows from the
positive terminal of the device to the negative one.
Solution 1.13 (a) By inspection: Circuit Element (CE) 1 absorbs –5W, and CE 2 absorbs 6W.
(b) Compute power absorbed by all elements including independent sources:
I3A: -15
CE1: -5
V3V: -12
CE2: +6
V5V: 10
I2A: 16
----------
Sum: 0 (Verifies conservation of power.)
Solution 1.14 (a) Compute power absorbed:
I5A: -85
CE1: 98
V3V: 33
CE2: 16
V7V: -42
I2A: -20
-------------
Sum: 0
(b) Add all terms:
I-source: Pabsorbed = −3 1− e−t( ) = −3 + 3e−t watts
V-source: Pabsorbed = −2 3e−t −1( ) = −6e−t + 2 watts
CE1: Pabsorbed = 3e−t × 3 1− e−t( ) = 9e−t − 9e−2t watts
CE2: Pabsorbed = 3e−t −1( ) 3e−t −1( ) = 9e−2t − 6e−t +1 watts
Simple algebraic manipulation of the the sum of all the above terms reveals that the result is zero.
Solution 1.15 (a) When IL = 1, P = VLIL = (16-4)×1 = 12 W. When IL = 2, P = VLIL = (16-16)×1 = 0.
(b) P = (16-4IL2)IL. Differentiate this w.r.t. IL and set to zero: 16 – 12IL
2 = 0. Therefore, IL = 1.155A.
Solution 1.16 (a) When IL = 2, P = (16-4)×2 = 24W. When IL = 3, P = (16 - 9)×3 = 21 W.
(b) Maximum occurs in the interval from 0 to 4: P = (16 - IL2) IL
Differentiate w.r.t. IL and set to zero: 16 – 3IL2 = 0.
Therefore, IL = 2.31 A.
Solution 1.17 (a) Power is the product of the current and voltage. We can compute the product
graphically:
0 0.5 1 1.5 2 2.5 30
2
4
6
8
10P
ower
in W
atts
TextEnd
Time in s(b)
W t( ) = p t( )dt0
t
∫ = 10 −10e−7τ( )dτ0
t
∫ = 10τ]0t − − 10
7 e−7τ[ ]0
t=10t +
10
7e−7t −
10
7
This can be used as an aid to plot the work function:
0 0.5 1 1.5 20
1
2
3
4
5
6
7
8
9
Time in s
Ene
rgy
in J
TextEnd
Solution 1.18 (a) Since, i t( ) = 115 − 23t mA ,
q 7( ) = i τ( )dτ0
7
∫ = 115t −23t2
2
0
7
×10−3 = 0.2415 C
(b) Energy is the integral of power:
E = p t( )dt0
7
∫ = v t( ) × i t( )dt0
7
∫ = 25 i t( )dt0
7
∫
= 25 × 0.2415 = 6.0375 C
Solution 1.19 (a) ∆t = 100oF, Rate of temp. increase is 2.5 Wh/oF per gallon:
Energy = 2.5Wh/oF/gallon×100 oF×30gallons = 7500 Wh = 2.7×107J.
(b) Heater generates P = 120×10 = 1200 W. We want 7500 Wh. Therefore, the total number of hours
needed is 7500Wh/1200W = 6.25 h.
Solution 1.20 First compute the change in temperature required, in oF:
∆t = 80-25 = 55oC = 55×9/5 oF= 99oF
Next, compute the energy spent every hour, which means on 40 gallons of water:
E = 2.5 Wh/oF/gallon×99oF×40gallons = 9900 Wh
Since the heater is not 100 % efficient, we spend more energy than is actually needed to heat the water:
E_spent = 9900 Wh/0.9 = 11000 Wh
So, far, this was the energy spent every hour. Over six hours, the total energy spent is:
E6h = 11000×6 = 66,000 Wh
Finally, the total energy spent per month is Em = 66,000×30 = 1980 kWh
and the bill is 1980 kWh×0.14$/kWh = $277.2
Solution 1.21
Energy = 120 W × 6 h = 720 Wh = 0.72 kWh
Therefore, cost per day = 0.72 kWh × 8 = 5.76 cents, and cost per month is 5.76×31 = $1.785.
Solution 1.22
We need to compute the difference between the inner diameter of the tube and the outer one in order to get
the cross-sectional area:
area = πRout2 −πRin
2 = π0.0032 − π0.00182 =1.81 ×10−3 m2
Then, R = 1.7×10-5×(12/1.81) = 11.3 mΩ.
Solution 1.23 L = 20 m, W = 0.015m, H = 0.001 m. Thus, A = W×H, and R = 5.1× copper×L/A
= 0.116 Ω.
Solution 1.24. (a) 500 ft, 20 gauge wire: 10.35 Ω/1000 ft from table 1.3. This implies that
R = 5.175 Ω.
(b) 55 ft, 20 gauge, nickel wire:
R = 5.1×10.35
1000× 55 = 2.9 Ω
(c) Rtot = 2.9 + 5.175 = 8.08 Ω.
Solution 1.25. R T( ) = R 20( ) 1 +α T − 20( )[ ] . Substituting at T = –10 yields:
21 = R 20( ) 1+ 0.0039 −30( )[ ] or R 20( ) = 23.78 Ω
Evaluating at T = +10 yields,
R 10( ) = 23.78 + 23.78 × 0.0039 × −10( ) or R 10( ) = 22.85Ω
Solution 1.26. For tungsten, we know that α = 0.0045. Therefore:
R 150( ) = R 20( ) 1 +α T − 20( )[ ] = 200 1 + 0.0045 150 − 20( )[ ]= 317Ω
Rate of change of resistance is (317-200)/(150-20) = 0.9 Ω/oC.
Solution 1.27. Plug numbers directly into the same formula as problem 1.26:
0.0022 = 0.002 + 0.002×0.0039(T-20)
Rearrange to obtain: T = 45.64o C.
Solution 1.28. (a) Power in a wire: P = I2R. Rearranging, we can express the current as
I = PR .
Substitute given P and R to obtain I = 0.707 mA.
(b) Use the same formula for current obtained above to get 50 A.
Solution 1.29. Use formula for power: P = V2/R. Rearranging, R = V2/P = 96 Ω.
Solution 1.30 (a) I = V/R = 12 A, out of the positive terminal of the battery.
(b) Up through the resistor.
(c) Absorbed power by resistor: P = V2/R = 14.4 W. Same power is delivered by source.
(d) From table 1.2 and 1.3, 1000 feet of 18 AWG aluminum wire has resistance:
»R1000ft = 1.6*6.51R1000ft = 1.0416e+01
By proportionality, 1000 × 0.1 = L ×10.416 . Hence,
»L = 100/10.416L = 9.6006e+00 meters.
Solution 1.31 (a) V = 10 V.
(b) P = V2/R, which means that R = V2/P = 100/25 = 4 Ω.
(c) I = V/R = 10/4 = 2.5 A. Current flow is downwards through resistor.
(d) Up through resistor.
(e) P = V2/R10 = 100/10 = 10 W. Hence, I10 = V/R10 = 1 A. Without applying material from a future
chapter, a legitimate way to obtain Isource is to apply conservation of power first and then compute Isource
from the power formula. Hence, Psource =10 + 25 = 35 watts. Using material from a later chapter, in
particular KCL, we may conclude that, Isrc = 2.5 + 1 = 3.5 A. Thus, Psource = VIsource = 10×3.5 = 35 W.
This approach indicates that power is conserved.
Solution 1.32 (a) From 0 to 1 s, i(t) = 10-3t. Thus, i2R = 10−6t2R is the power absorbed during this
interval. Integrating this expression for the power from 0 to 1 s gives us the total energy used:
10−6t3R
30
1
= 500010−6
3= 0.001667 J.
Finally, we need to multiply this by 2 to account for the interval from 1 to 2 seconds. Thus, the total energy
spent is 3.33 mJ.
(b) The same charge that got transported in one direction during the interval from 0 to 1 is being
transported back in the interval from 1 to 2 (by symmetry). Therefore, total charge transfer is zero.
Solution 1.33. (a) 60 W + 120 W = 180 W.
(b) P = IV þ I = P/V = 180/12 = 15 A.
(c) P = Energy/Time þ Time = 1.2 MJ/180 W = 6.67×104 sec = 1.85 h.
Solution 1.34. P = I2R. Therefore, 325 = 25×(5+4+2R). Solving for R, yields R = 2 Ω.
Solution 1.35. (a) Use definition of power and substitute given power:
V2 = P × R = 98 × 2 = 14 V
Similarly, I3 =P
R=
12
3= 2 A, V4 = P × R = 16 × 4 = 8 V, I5 =
768.8
5= 12.4 A, and
V6 = 486 × 6 = 54 V.
(b) Ptot = Pdissipated = 98 + 12 + 16 + 768.8 + 486 = 1380.8 W.
(c) Vin = V2 + V6 = 68 V. Iin + I3 = I5 + I4 and I4 =V4
4. Thus, Iin = I5 + I4 – I3 = 12.4 A.
Solution 1.36. (a) Sources are the top, right-most, and bottom left. The reason is that current flows out
of the positive terminal of the device.
(b) The 32/16 element is a 2 Ω resistor. The 54.5/18.167 element is a 3Ω resistor. The 13/2.167 element
is a 6 Ω resistor. The 93/2.833 element is a 32.827 Ω resistor. The 24/5 element is a 4.8 Ω resistor.
Solution 1.37. Power: 12 = Ix2R, which means that R = 12/Ix
2. Now, analyze the loop: 16 = Ix(R+4).
Substitute the value of R into this expression: 16 = Ix12
Ix2 + 4Ix . Hence: Ix
2 − 4Ix + 3 = 0 .
This equation has two solutions: one is at Ix = 1 A or R = 12 Ω. The other is at Ix = 3 A or R = 4/3 Ω.
Solution 1.38. (a) Conservation of power:
16Ix = 4Ix2 +12 + 10 − 6
Hence
0 = Ix2 − 4Ix + 4 = Ix − 2( )2
Thus, Ix = 2 A.
(b) 32Ix = 4Ix2 + 28 ⇒ Ix
2 −8Ix + 7 = Ix − 7( ) Ix −1( ) = 0 . Hence, Ix = 7A or Ix = 1A .
Solution 1.39. (a)
(i) AA: I = 36/12 = 3A
BB: I = 24/12 = 2A
CC: I = 14.4/12 = 1.2A
(ii) Sum = 6.2A
(iii) P = VI = 6.2×12 = 74.4W. This is equal to the sum of the powers absorbed by the bulbs.
(iv) R = V/I
AA: R = 12/3 = 4Ω
BB: R = 12/2 = 6Ω
CC: R = 12/1.2 = 10Ω
(b) Each AA bulb draws 3 A. Thus, up to five bulbs can be connected without blowing the fuse (5×3=15).
So, 6 or more would blow the fuse.
(c) Similar analysis suggests that 13 or more bulbs would blow the fuse. Intuitively, the bulbs draw less
current, so more of them can be used.
Solution 1.40. (a)
p(t) = i2 t( )R = 20cos 2πt( )[ ]2 ×10 = 40001 + cos 4πt( )
2= 2000 1 + cos 4πt( )( ) W
(b)
W t( ) = p t( )dt0
t
∫ = 2000t + 2000 cos 4πt( )dt0
t
∫ = 2000t +2000sin 4πt( )
4π J
0 0.5 1 1.5 2 2.5 30
1000
2000
3000
4000
5000
6000
Ene
rgy
in J
TextEnd
Time in s
Solution. 1.41. When the switch is closed, a constant current of 5/10000 = 0.5 mA flows through the
circuit. When the switch is open, no current flows. So, 50% of the time, a 0.5 mA current flows, and the
other 50% no current flows. The average current is therefore 0.25 mA.
Solution. 1.42 When the switch is at A, the current is 5/5000 = 1 mA. When the switch is at B, the
current is 5/10000 = 0.5 mA. Now, the switch is at position A 20% of the time (1ms out of a 5ms period,
after which the events repeat). So, the average current is 0.2*1 mA + 0.8*0.5 mA = 0.6 mA.
Solution. 1.43 The current in the load resistor is 2 A. So, the power is 22×RL = 8 W.
Solution 1.44. Vin = IinR1 è Iout = µVin/R2 = µIinR1/R2.
Solution 1.45 (a) I1 = Vin/R1. Hence, Vout = αVinR2/R1.
(b)
Vout
Vin=
αR2
R1=
100 ×10
R1= 5 ⇒ R1 = 200 Ω
(c)
Power − gain =
α2Vin2
R12 R2
Vin2
R1
=α2 R2
R1= 500
Solution 1.46 (a) V1 = 200 mA × 5 Ω = 1 V implies V2 = 0.8×8 = 6.4 V. Hence
Vout = 5×6.4 = 32 V and Iout = 32/64 = 0.5 A.
(b) Current Gain = 0.5/0.2 = 2.5.
(c) Power values for the 5, 8, and 64 Ω resistors are, respectively, P5 = 0.2 W, P8 = 5.12 W, P64 = 16 W.
Solution 1.47 (a) I1 = 5A è I2 = 3×5/3 = 5 A è Iout = 25 A, Vout = 50 V.
(b) Voltage Gain = 5.
(c) Pin = 5×5×2 = 50 W, P1 = 5×5×3 = 75W, P2 = 25×50 = 1250W.
Solution 1.48 I1 = Vin/10 = 0.1 A, VR = 10×(Vin/10)×R = R; Vout = 5R×10 = 50R =50RVin
Vout/Vin = 50R. If we want Vout/Vin to be 150, R has to be 3 Ω.
SOLUTIONS CHAPTER 2
SOLUTION 2.1. Using KCL at the center node of each circuit:
(a) I3 = I2 − I1 = −1 − 2 = −3A
(b) I3 = I1 + I2 − I4 = 2 −1 − 0.5 = 0.5A
SOLUTION 2.2. KCL at the bottom node gives I1 = −7 − 8 = −15A , and at the right node
I4 =−6 − 8 = −14 . From these, KCL at the top node gives I3 = I4 − 5 = −19A, and finally at the central
node gives I2 = 6 + I3 − 7 = −20 A,
SOLUTION 2.3. Use a gaussian surface on the top triangle. Performing KCL around this surface yields
1A − 2A + 3A + 4 A − 5A = I = 1A.
SOLUTION 2.4. Use a gaussian surface around the bottom rectangle. KCL yields
I1 = 2A +10A + 3A = 15A .
SOLUTION 2.5. Using KVL, V1 = 55V −15V + 105V −100V − 30V = 15V .
SOLUTION 2.6. Using KVL, Vx = 5V −1V −1V −1V +1V −1V = 2V .
SOLUTION 2.7. Using KVL once again.
v1 = 7 + 6 + 5 = 18V
v2 = 6 + 7 −8 = 5V
v3 = −5 − 6 = −11V
v4 = 8 − 7 =1V
SOLUTION 2.8. KVL is used to find the voltage across each current source, and KCL to find the current
through each voltage source.
I3V = 6A − 7A = −1A
I4V = I3V + 8A = 7A
I5V =−8A − 6A = −14A
V7 A = 4V + 3V = 7V
V8A = −4V + 5V = 1V
V6 A = V8 A − 3V = −2V
Chap 2 Probs P2 - 2 © R. A. DeCarlo, P. M. Lin
SOLUTION 2.9. Using the same method as before, the current and voltages are found through and across
each sources.
I5V = 9 +8 − 7 = 10 ⇒ P = 50W
I4V = −6 − I5V = −16 ⇒ P =−64W
I2V = 6 − 7 = −1A ⇒ P = −2W
I3V = − I2V − 9 = −8A ⇒ P =−24W
V8A = 4 − 5 = −1V ⇒ P =−8W
V9 A = 3 + V8 A = 2V ⇒ P = 18W
V7 A = 2 − V9 A = 0 ⇒ P = 0W
V6 A = 5 − V7A = 5 ⇒ P = 30W
Summing all the power give 0W, hence conservation of power.
SOLUTION 2.10. Doing KVL around the right loop does not balance out. Changing 8V to 5V would fix
this.
SOLUTION 2.11. Using KVL to determine the voltages, and KCL to determine the currents:
Vy = 8V
Vx = Vy − 4 = 4V
Ia = 4 A
Iy = 4 −14 + 2Ia = −2A
Ix = Ia − Iy = 6A
SOLUTION 2.12. First Vin = I2 ⋅ 8Ω = 24V . Then I1 = Vin / 3Ω = 8A and I3 = 12A − I1 − I2 =1A.
Therefore RL = Vin / I3 = 24Ω ⇒ P = I3 ⋅ Vin = 24W
SOLUTION 2.13. (a) First, from current division, get
I1 =1/ 3
1/ 3 +1/ 6 + 1/ RL
⋅ 12 − aI1( ) ⇒ I1 =
12 / 3
1 + a( ) / 3 +1/ 6 +1/ RL
.
(b) Using the previous equation and solving for 1/ RL = 12/ 3I1( ) −1/ 6 − 1+ a( ) / 3 = 0.5S or RL = 2Ω .
The power P = I32 RL =
1/ RL
1/ 3 +1/ 6 +1/ RL
⋅ 12 − aI1( )
⋅ RL = 18W
SOLUTION 2.14. For the power delivered by the source to be 60W, the voltage across it should be
V = P / 2A = 30V . Therefore the current through the 20Ω must be I20Ω = 30 / 20 = 1.5A , and by KCL the
current through IRL= 2 − I20Ω = 0.5A . From this, RL = V / IRL
= 60Ω .
Chap 2 Probs P2 - 3 © R. A. DeCarlo, P. M. Lin
SOLUTION 2.15. Writing KVL around the loop 25V − 4I −15V − 5I − I = 0 ⇒ I =1A, and
P5Ω = I 2R5Ω = 5W
SOLUTION 2.16. The total power supplied by the source is P = 50V ⋅ 0.5A = 25W . The power absorbed
by the resistor is P60Ω = 0.5A( )2 ⋅ 60Ω = 15W . Therefore by conservation of power, the power absorbed by
X is 10W.
SOLUTION 2.17. (a) As this loop is open, no current flows through it, so IR is 0A. The output voltage is
VOUT = −2V + 3V − 2V = −1V by KVL.
(b) Writing out the KVL equation around the loop 3 − 2 − IRR − 2 − IR2R − IRR = 0 ⇒ −1 = IR 4R.
Therefore IR =−1/ 4R and VOUT = IR R = −1/ 4V .
SOLUTION 2.18. Writing out KVL around the loop 60 − 30I − 30 − 20 + 60 − 40I = 0 ⇒ I = 1A . From
ohm's law R = V / I = 30Ω.
SOLUTION 2.19. (a) Using Ohms law Iin = V2 / 20 +12( ) = 0.75A, and V1 =12 ⋅ Iin = 9V . To find R, write
KCL and get VR = 30 − V2 = 6V . Therefore using Ohms law again, R = VR / IIN = 8Ω .
(b) Writing KVL around the loop, 30 = aV1 + IinR + Iin 20 + V1, and substituting Iin = V1 / 12,
V1 = 30 / R + 32( ) /12 + a[ ] is obtained. Next substitute back V1 =12Iin and solve for
R =30
12I IN
− a
⋅12 − 32 = 40Ω
SOLUTION 2.20. (a) i. Using R = Vxy / Ibat the value of each resistors starting with the top one are 2.7Ω,
0.6Ω, and 0.25Ω. Using the same relationship, the resistance for the motor is 1.25Ω.
ii. Using P = Vxy2 / R the power dissipated by each resistor is 16.875W, 3.75W, 1.5625W, and for the
motor 7.8125W.
iii. The relative efficiency is = 7.8125/ (12 ⋅2.5) ⋅100 = 26 %
(b) i. Performing voltage division across each resistor
VAB = 0
VBC =12 ⋅ RBC / (RBC + RCD + Rmotor ) = 3.43V
VCD = 12 ⋅ RCD / (RBC + RCD + Rmotor ) =1.43V
Vmotor = 12 ⋅ Rmotor / (RBC + RCD + Rmotor ) = 7.14V
Chap 2 Probs P2 - 4 © R. A. DeCarlo, P. M. Lin
ii. Ibat = 12 / (RBC + RCD + Rmotor ) = 5.71A
iii. The relative efficiency is = (Vmotor2 / Rmotor)/ (12 ⋅5.71) ⋅100 = 59.5 %
(c) i. Repeating the steps from (b), the voltages across the first two resistance are 0, then across the other
and the motor 2V, and 10V
ii. Ibat = 12 / (RCD + Rmotor) = 8A
iii. And the relative efficiency is = (Vmotor2 / Rmotor)/ (12 ⋅8) ⋅100 = 83.3 %
(d) What is the largest equivalent resistance of the motor that will draw 30A? R = 12/ 30A = 0.4Ω.
SOLUTION 2.21. (a) Observe that i =− IO , thus v = ki3 = −kI03 .
(b) Using KVL and previous equation, vx = (R1 + R2)IO + VO + kIO3 .
(c) The power is = IOvx = (R1 + R2)IO2 + VOIO + kIO
4
SOLUTION 2.22. I100Ω =0.04
100= 0.02 A. Therefore V300Ω = 0.02 × (100 + 200) = 6 V. By KCL,
I150Ω = 0.02 +6
300= 0.04 A. Req, seen by the source, is 300 Ω. Therefore Vs = 0.04 × 300 = 12 V.
SOLUTION 2.23. Using KCL IR = 5 − 3 = 2A , and KVL VR = 10 + 6 = 16V . Thus
R = VR / IR =16 / 2 = 8Ω.
SOLUTION 2.24. Using KCL, KVL, along with Ohm’s law,
I5Ω = 6 − 7 = −1A
I4V = 8 − I5Ω = 9A
I2Ω = 8 + 7 =15A
V6 A = 4 + 5I5Ω = −1V
V8A = −4 + 2I2Ω = 26V
V7 A = V8 A − 5I5Ω = 31V
Now, the power delivered or absorbed by each element is calculated:
P6A = I6A ⋅V6 A = −6W
P7A = I7A ⋅V7 A = 217W
P8A = I8A ⋅ V8A = 208W
P4V = I4V ⋅ 4 = 36W
P5Ω = I5Ω2 ⋅5 = 5W
P2Ω = I2Ω2 ⋅ 2 = 450W
Chap 2 Probs P2 - 5 © R. A. DeCarlo, P. M. Lin
Note that for passive elements, when the power is positive it is absorbed, while for independent sources it is
generated when the power is positive.
SOLUTION 2.25. Note that I1 = 6A . Thus by KCL
I3 = 6 − 0.5I1 = 3A
I2 = 2 + 0.2I1 = 3.2A
I4 = 8 − 0.3I1 = 6.2A
And finally using KVL
V2 = 8A ⋅1+ 4I4 + 3I3 = 41.8V
V1 = 2I2 −3I3 = −2.6V
SOLUTION 2.26. (a) Using KCL,
I4 = 5 − 4 =1A
I3 = I4 − 2 = −1A
I2 = 3 − 2 = 1A
I1 = − I2 − 5 = −6A
(b) Using KVL and Ohm’s law,
VIV
VIV
VIV
VIV
44
1010
1212
183
44
33
22
11
==−==
==−==
(c)
WVVVP
WVVP
WVVVP
WVVAP
A
A
A
A
60)(5
24)(4
40)(2
90)(3
1345
434
3212
123
=−+==−−=
−=−−==−=
Chap 2 Probs P2 - 6 © R. A. DeCarlo, P. M. Lin
SOLUTION 2.27. Write KVL around the outside loop, 40 = 500Ix + (400 + 200)i . And write KCL
equation i = Ix − 2I x . Solving yields Ix = −0.4A . The dependent source delivers 2I x ⋅(−600i) = 192W , and
the independent 40I x = −16W . Finally the resistors absorb 500Ix2 + 400i2 + 200i2 =176W verifying the
conservation of energy since the source generate 192W-16W=176W.
SOLUTION 2.28. By voltage division V2 =[((90 ||180) + 60)||40]
[((90||180) + 60)||40] +160⋅
60
60 + (90||180)⋅ Vs = 1/14 ⋅Vs .
Therefore Vs = 14V2 = 280V .
SOLUTION 2.29. By voltage division
vx = 9V ⋅18 + 3
(18 + 3) + 6= 7V
SOLUTION 2.30. By voltage division we get the following two equations in order to solve for the two
unknowns.
V2 = V1 ⋅R1
R1 + R2
V1 =100V ⋅R1 + R2
R1 + R2 + 60
Solving yields R1 = 40Ω , and R2 = 100Ω .
SOLUTION 2.31. Dividing 1400 in four gives 350. If we only need 1/4 and 2/4, the resistor string can be
made of three resistances: 350Ω, 350Ω, and 700Ω.
SOLUTION 2.32. Using voltage division, at t=0 vR = 15 ⋅2R
3R= 10V , and t = 5 s vR = 10V , and at t = 10
the voltage goes back to 0V.
Chap 2 Probs P2 - 7 © R. A. DeCarlo, P. M. Lin
0 5 10 150
1
2
3
4
5
6
7
8
9
10
time (sec)
Volts
SOLUTION 2.33. By voltage division
Vb =Rb
Ra + RbVin and Vd =
Rd
Rc + RdVin
By KVL, if Vout = 0, then
0 = Vout = Vb − Vd =Rb
Ra + Rb−
Rd
Rc + Rd
Vin
For arbitrary Vin , this requires that Rb
Ra + Rb=
Rd
Rc + Rd or equivalently that Rb Rc = Ra Rd .
SOLUTION 2.34. First Geq = 1m +1.5m + 2m + 3m = 7.5mS . By current division
I2 =100mA ⋅1.5m
Geq
= 20mA , P = 100mA ⋅ I2 /1.5mS =1.33W .
SOLUTION 2.35. By current division, for I1 to be 2A then 160 + R = 300 ||600 for an even split. Thus
R = 40Ω .
SOLUTION 2.36. By current division, i1 = 0.4A ⋅1/10
1/10 + 1/ 40= 0.32 A . Therefore using KVL
vd =10i1 − 0.25i1 = 3.12V .
SOLUTION 2.37. (a) Req = (8k | |2k) + (9k ||1k) = 2.5kΩ
Chap 2 Probs P2 - 8 © R. A. DeCarlo, P. M. Lin
(b) Req = 2k ||[(2k | |2k) + (2k | |2 k)] = 1kΩ
SOLUTION 2.38. (a) Req = 2 + 15 +10 + 10 + 40 + 30 + 20 + 8 =135Ω .
(b) Four of the resistors are shorted, thus Req = 2 + 15 +10 + 8 = 35Ω .
(c) Lumping the series resistance together Req = 8 + [50 ||(50||25)] + 2 = 22.5Ω
SOLUTION 2.39. (a) Req = [2R + (4R | |4 R)]||[2 R + (4 R | |4 R)] = 2R
(b) Req = 2R | |2 R + (4R | |4 R | |4 R | |4 R) = 2R
SOLUTION 2.40. (a) First Req = 150 + [375||(250 + 500)] = 400Ω . Next Iin = 14V / Req = 35mA . The
power delivered by the source is then = 14Iin = 0.49W .
(b) Req = 150 + [375||(250 + 500)||1k] = 350Ω, and Iin = 14 / Req = 40mA . The power delivered by the
source is = 14Iin = 0.56W .
As the equivalent resistance decreases, more of it gets dissipated by it.
SOLUTION 2.41. Req1 > Req2. Without going into a detailed analysis using methods of Chapter 3, we
present the following intuitive argument. First note that the points a and b represent points on an
unbalanced bridge circuit meaning that the voltage between a and b would not be zero. Also note that when
two resistors are placed in parallel, the equivalent resistance becomes smaller than either resistance. The
addition of the resistor R in circuit 2 essentially creates an internal parallel resistance resulting in an Req2
lower than Req1.
SOLUTION 2.42. Req1 = Req2. As was the case in the previous problem, this is a balanced bridge circuit.
Hence no voltage appears between a and b making the additional resistor irrelevent.
SOLUTION 2.43. (a) Rin = [(20||20) +10]||(1/ 0.12)||(1/ 0.08) = 4Ω.
(b) Rin = 6R ||[(R || R ||0.5R) + 0.75R + (2R | |2 R)] = 1.5R
SOLUTION 2.44. (a) The infinite resistance are essentially open circuits, thus
Req = 1 + 2 + 3 + 4 + 2 + 4 + 3 + 2 +1 = 22Ω
(b) 0 resistances are short circuits. Labeling one branch x and the other y, it can be seen that the circuit is a
set of 3 resistor strings in parallel to each other between x and y, then added in series to the two 1 Ohm
resistor. Thus Req = [(2 + 3)||(4 + 2 + 4)||(2 + 3)]+1 +1 = 4Ω .
Chap 2 Probs P2 - 9 © R. A. DeCarlo, P. M. Lin
(c) Writing out Req = 1 + [Rx ||(2 + 3 + 4 + 2 + 4)] + 3 + 2 +1 = 7 + [Rx ||15] , and solving for Rx = 3.75Ω.
(d) No, it requires methods to be covered in the next chapter.
SOLUTION 2.45. Using the formulas for parallel resistances, the circuit of figure 2.45 reduces to
(a) RAC = (2 +1) / /6( ) + 8[ ] / /10 = 5 Ω
(b) RAB cannot be calculated by series parallel formulas, but RBC can be done.
RBC = (8 +10) / /6( ) + 2[ ] / /1 = 0.86667 Ω
SOLUTION 2.46. (a) Req = 300 + (R ||5.6k), thus R = 800Ω.
(b) Req = R + (R ||1.2K) , the following quadratic equation must be solved R2 +1.4k ⋅ R −1.2M = 0 . This
yields R = 600Ω .
(c) Req = 500 + 300 + (800||400 || R). Solving for R yields 800Ω .
SOLUTION 2.47.
(a) Using the fact that the resistance seen into terminal a-b is the same as that seen in terminal c-d, we can
obtain the following relationship. eqeq RRRR ||+= . This produces a quadratic equation whose solution is
Req = 1.618R .
(b) Using the previous argument )5(||105 eqeq RR ++= . Solving for Ω= 18.11eqR .
SOLUTION 2.48. By current division Ix =1/ 18k
1/18k +1/ 9k
⋅
1
6k + (9k ||18k)1
6k + (9k ||18k)+ 1
4k
⋅36m = 3mA
Chap 2 Probs P2 - 10 © R. A. DeCarlo, P. M. Lin
SOLUTION 2.49. The 500 Ω resistor has no effect on the current entering the circuit to its right.
0.15 =30
R+
30
600=
30
R+ 0.05
Hence R = 30/0.1 = 300 Ω.
SOLUTION 2.50. (a) First, express the total current as I =120
0.5 + (20 ||30||40 || RL2 ). Next, find RL2 that
will cause I to be 15A. Thus RL2 = 40Ω or less will cause the fuse to blow as this will cause the current to
be 15A or more.
(b) Repeating the previous procedure, RL2 = 20Ω .
(c) RL2 = 120Ω .
SOLUTION 2.51. At time 0, all switches are open and Vout =260
260 + 40⋅ 220 =190.7V .
Then at t = 5s, switch one closes and
Vout =260||260
(260 ||260) + 40⋅ 220 = 168.2V .
At t = 10s,
Vout =130 ||260||260
(130 ||260||260) + 40⋅ 220 =136.2V .
Finally at t =15 s,
Vout =65||130 ||260||260
(65||130 ||260||260) + 40⋅ 220 = 98.5V .
0 2 4 6 8 10 12 14 16 18 2080
100
120
140
160
180
200
Time (sec)
Volts
Chap 2 Probs P2 - 11 © R. A. DeCarlo, P. M. Lin
SOLUTION 2.52. (a) Lumping the two sources together and the resistors into an equivalent resistor gives
i1( t) =9cos(2 t) − 3cos(2t)
7k + 9k + 8k + (2k | |3k | |6 k)= 0.24cos(2t)mA .
(b) By current division i2 (t) =1/ 6k
1/ 2k +1/ 3k +1/ 6k
⋅ i1(t) = 40cos(2 t) A .
SOLUTION 2.53. (a) Starting with,
Req1 = 5||(10 + 10) = 4Ω
Req 2 = 10||(6 + Req1) = 5ΩReq 3 = 5 + Req2 = 10Ω
(b) Using the values R just obtained,
Va = 100 ⋅Req 2
Req3
= 50V
Vb = Va ⋅Req1
6 + Req1
= 20V
Vc = Vb ⋅ 1010 +10
= 10V
(c) Finally,
Iin =100
Req3
=10A
Id =Va
6 + Req1
= 5A
Ie = Vb
10 +10= 1A
.
SOLUTION 2.54. (a) Circuit a: Using voltage division,
vout (t) = vin (t) ⋅300||(20 + 30 + 50)
[300 ||(20 +30 + 50)] + 5
⋅
30
30 + 20 + 50
= 33.75sin(377t)V , and Ohm’s law
iout(t) = vout( t)/ 30 = 1.125sin(377t )A . The instantaneous power is then
P(t) = iout (t) ⋅ vout (t) = 37.969sin 2(377t)W .
Circuit b: By current division
iout(t) = iin (t) ⋅1/ (20 + 30 + 50)
1/ (20 + 30 + 50) +1/ 300 +1/ (50 +100)
= 60sin(377t)A , and from Ohm’s law
vout (t) = 50 ⋅ iout(t) = 3000sin(377t)V . The instantaneous power is P(t) = 180sin2 (377t)kW .
(b) No, since the current source forces the amount of current in the circuit.
Chap 2 Probs P2 - 12 © R. A. DeCarlo, P. M. Lin
SOLUTION 2.55. (a) Noting that i2 = v1 /10 = 6A, then we can write KCL at the top left node,
isource = i2 + v1 / 6 + (v1 − 5i2 )/ 5 = 22A. Thus P = 60 ⋅ 22 = 1.32kW .
(b) First, determine the current through each resistor:
i2 = 60 /10 = 6A
i2.5Ω =60
2.5 + (5||5)
= 12A
i5Ω = 1/ 2 ⋅i2.5Ω = 6A
Then calculate the power absorbed by each resistor:
P10Ω = 10i22 = 360W
P2.5Ω = 2.5i2.5Ω2 = 360W
P5Ω = 5i5Ω2 = 180W
SOLUTION 2.56. From Ohm’s law I1 = 100m / 200 = 0.5mA . By current division
IRL =20k
20k + 200
⋅150I1 = 75.257mA , and PRL = 200IRL
2 = 1.103W .
SOLUTION 2.57. First, using voltage division, Vx = Vs
2
2 +1
= (2 / 3)Vs . Then using KCL and the
previous equation, Is = (Vs / 3) − Vx = −(1/ 3)Vs . Finally using Ohm’s law Req = Vs / Is =−3Ω.
SOLUTION 2.58. Observing the following relationship, V1 = Vin , the following nodal equation can written:
Iin = Vin / 3 + Vin / 6 − 2Vin = −1.5Vin .
SOLUTION 2.59. Step 1. From voltage division
V1 =18
18 + 4 + 2Vs = 0.75Vs and Vin =
22
24Vs =
11
12Vs
Hence
Pin =Vin
2
22=
11×11
22 ×144Vs
2 =11
288Vs
2
Step 2. For the load, by current division
Chap 2 Probs P2 - 13 © R. A. DeCarlo, P. M. Lin
I2Ω =6
6 + 2AV1 =
3
4A
3
4Vs
=
9A
16Vs
Therefore
P2Ω = 2 × I2Ω2 = 2
81A2
256Vs
2 =81A2
128Vs
2
Step 3. P2Ω = 10 × Pin implies that
81A2
128Vs
2 =1011
288Vs
2
Hence
A =128 ×110
81 × 288= 0.7769
SOLUTION 2.60. By voltage division V1 =6
6 + 2Vin = (3/ 4)Vin . By current division, and substituting the
previous equation I2 =3
3 + 64V1 = Vin . Using voltage division and Ohm’s law, and substituting the previous
equation,
Vout = 4.5I2
1010 + 5
= 3Vin = 30V
Iout = 4.5I2 / (10 + 5) = 0.3Vin = 3A
Finally, from the previous equations | Vout / Vin |= 3 .
SOLUTION 2.61. Writing out KCL when the switch is closed, ibat = 150A +Vbat − 0.04ibat
240
. Solving
gives ibat = 150.02A and Vout ≈ 6V . When the switch is open Vout = Vbat
240
240 + 0.04
≈12V . Therefore,
the reason for the radio stopping is insufficient supply voltage.
SOLUTION 2.62. (a) Using the following relationship P = V 2 / R, the resistance of each headlight on low
beam is R = V 2 / P = 4.11Ω .
(b) Using the same relationship R = 2.22Ω .
(c) By voltage division, Vout =14.7240
240 + 0.04= 14.698V .
(d) Using voltage division, Vout = 14.7240 ||4.11||4.11
(240||4.11||4.11) + 0.04= 14.417V
Chap 2 Probs P2 - 14 © R. A. DeCarlo, P. M. Lin
(e) Using voltage division, Vout =14.7240||2.22||2.22
(240 ||2.22||2.22) + 0.04= 14.186V
SOLUTION 2.63. By voltage division
11.96 =15
15 + R012 =
180
15 + R0
Therefore
R0 =180 −15 ×11.96
11.96= 0.050167 Ω
SOLUTION 2.64. (a) Using KVL Vt = 102 − 0.05 ⋅80 = 98 V.
(b) Using KVL Vt = 102 + 0.05 ⋅ 50 =104.5 V.
(c) P = Vt ⋅50 = 5.225. kW
SOLUTION 2.65. Minimum load means the minimum load resistance that the system can handle.
»MaxPwr = 0.8*50e6
MaxPwr = 40000000
»Vs = 750e3;
»Iline = MaxPwr/Vs
Iline = 5.3333e+01
»Rmin = Vs/Iline
Rmin = 1.4062e+04, i.e., Rmin = 14.062 kΩ.
SOLUTION 2.66. (a) Using the following general form for a non-ideal voltage source: vout = −Rsiout + Vs ,
one sees that for zero current vout = Vs = 40V . The slope of the line is −40
1000= −Rs = −0.04Ω , thus
Rs = 0.04Ω .
(b) This curve represents a resistor’s I-v characteristic, thus the slope 60
11
3
= R = 45Ω .
(c) The general form for a non-ideal current source is iout = −1
Rs
Vout + I s . When the voltage is zero,
iout = Is = 5A . From the slope of the line, −4000
5, Rs = 4000 / 5 = 800 Ω.
Chap 2 Probs P2 - 15 © R. A. DeCarlo, P. M. Lin
SOLUTION 2.67. Using the following formula: T
n
nI
Cn
= 1, solve for T, and get 0.625 hrs, or 37.5 min.
SOLUTION 2.68. Using the same equation as before and solving for Cn = nIT
n
1/
, with n=10, and
T=55/60 hrs, the capacity obtained is 20 Ah.
SOLUTION 2.69. C20 = 50 Ah
(a) In eq. 9, solving for I with n=20, and T=10, I=4.2A
(b) Calculate the capacity for n=10 and T=10, this yields 42 Ah.
SOLUTION 2.70. (a) Using a sequence of voltage division,
V1 = 50mV4850
= 48mV
V2 = 50V1
195
200= 2.34V
Vload = 2.5V2 = 5.85V
And the power is RRL= Vload
2 / RL = 2.278W .
(b) Following is the graph, and the script used to generate it.
0 10 20 30 40 50 60 700
1
2
3
4
5
Pow
er in
Wat
ts
0 10 20 30 40 50 60 700
200
400
600
800
Resistance in Ohms
Cur
rent
in m
A
%Script for Question 70 in chapter 2
RL=8:1:64;
V2=2.34;
IL=2.5*V2 ./ RL;
%Note the use of the ".*" which means that the division
%is performed for each value of RL.
Chap 2 Probs P2 - 16 © R. A. DeCarlo, P. M. Lin
PL=RL .* (IL .^ 2);
%Plot the Power versus RL
subplot(2,1,1);
plot(RL,PL);
ylabel('Power in Watts');
%Plot the Current versus IL
subplot(2,1,2);
plot(RL,1000.*IL);
xlabel('Resistance in Ohms');
ylabel('Current in mA');
%The use of subplot lets you subdivide the graphing
%window in two halfs.
SOLUTION 2.71. (a) Using the following script:
%Script for problem 2.71
R1=15; R2=4; R3=9; R4=2; R5=8;
R6=18;
Ra= R4+R5;
Ga= 1/Ra;
Gb= Ga+1/R1;
Rb= 1/Gb;
Rc= 1/(1/R6+1/R3)+Rb;
Gc= 1/Rc;
Geq= Gc+1/4;
Req= 1/Geq;
Irc= 20*Gc/Geq;
Vrb= Irc*Rb;
Vout= Irc*(Ga/Gb)*8;
Req
Vout
So (a) Req = 3Ω , and (b) Vout = 24V
SOLUTION 2.72. Using the following script:
%Script for problem 2.72
R1=1e3; R2=2.2e3; R3=2e3; R4=5e3; R5=3e3; R6=R5;
R7=3.2e3; R8=1.2e3; R9=1.6e3;
Chap 2 Probs P2 - 17 © R. A. DeCarlo, P. M. Lin
Ga=1/R7+1/(R8+R9);
Ra=1/Ga;
Gb=Ga+1/R6;
Rb=1/Gb;
Gc=1/R4+1/(R5+Rb);
Rc=1/Gc;
Gd=1/R2+1/(R3+Rc);
Rd=1/Gd;
Geq=1/R1+Gd;
Req=1/Geq
%Going through the same step to find Vout
Id=200e-3*(Gd/(Geq));
Ic=Id*((1/(R3+Rc))/Gd);
Ib=Ic*((1/(R5+Rb))/Gc);
Ia=Ib*Ga/Gb;
Iout=Ia*((1/(R8+R9))/Ga)
Vout=Iout*R9
The following values are obtained:
Req =
5.9121e+02
Iout =
5.5431e-03
Vout =
8.8689e+00
R1
a b c d
R9 R7 R6 R4 R2
SOLUTION 2.73. Using the following script:
%Script for problem 2.73
R1=20; R2=40; R3=60; R4=30; R5=10; R6=135;
R7=150; R8=300; R9=130; R10=200; R11=50;
Ga=1/R10+1/R11;
Ra=1/Ga;
Rb=Ra+R9+(1/(1/R7+1/R8));
Chap 2 Probs P2 - 18 © R. A. DeCarlo, P. M. Lin
Gb=1/Rb;
Gc=Gb+1/R6;
Rc=1/Gc;
Rd=Rc+R5+(1/(1/R3+1/R4));
Gd=1/Rd;
Ge=Gd+1/R2;
Re=1/Ge;
Rin=R1+Re
Ie=10/Rin;
Id=Ie*Gd/Ge;
I1=Id*(1/R6)/Gc
Ib=Id*Gb/Gc;
Vout=Ib*Ra
The following values are obtained:
Rin = 50ΩVout = 0.667V
I1 = 33.3mA
R1 e d
R2
R11
R10
R9 R8 R7
R6
R5
R3
R4 a c b
SOLUTION 2.74. An identical procedure to the one followed in the previous problem will yield the
following values:
Rin = 50.53ΩIout = 133.8mA
PROBLEM SOLUTIONS CHAPTER 3.
Solution 3.1. Select the bottom node as the reference node, and write a node equation at the positive
terminal of the V1 resistor:
V1 − V0
3R+
V1
6R+
V1 − 4V0
6R= 0
⇒ 2V1 − 2V0 + V1 +V1 − 4V0 = 0
⇒ 4V1 = 6V0
⇒ V1 = 1.5V0
Solution 3.2 Write a node equation at the top node:
0.6 −Vx
100−
2Vx
100−
Vx
50= 0
⇒ −Vx − 2Vx − 2Vx = −60
⇒ −5Vx = −60
⇒ Vx = 12V
Solution 3.3
0.6 −Vx
100+
25Vx
100−
Vx
50−
Vx − 0.2Vx
40= 0
⇒ −3Vx
100+
25Vx
100−
8Vx
400= −0.6
⇒ Vx = −0.6 × 400
80⇒ Vx = −3V
Solution 3.4 (a)
It is evident from the figure that Vc = 20. We need to write two equations in Va and Vb and put them in
matrix form. In this case, we can write the matrix equation by inspection. Note that the resistors are
identified by conductance values.
15m −5m
−5m 35m
Va
Vb
=
0.5
0.5
(b) Solve the matrix equation by inverting the left-most matrix:
Va
Vb
=
1525 − 25
35m 5m
5m 15m
0.5
0.5
=1
0.5
35 5
5 15
0.5
0.5
=40
20
(c) Vx = Vab = Va −Vb = 20V , Vda = -Va = -40, Vdb = -Vb = -20.
(d) Pi = 0.5×40 = 20W, Pv = 20×(20-20) = 0. Pdiss = 40×40×10m + 20×20×5m + 20×20×5m = 20 W.
Power delivered equals dissipated power.
Solution 3.6 Write two nodal equations:
Vs1 − V1
3000= Is3 +
V1 −V2
6000V2
30000=
Vs2 − V2
12000+
V1 − V2
6000
Rewrite equations as:
2Vs1 − 2V1 = 6000Is3 +V1 −V2
2V2 = 5Vs2 − 5V2 +10V1 −10V2
Cast into a matrix equation
−3 1
−10 17
V1
V2
=
6000Is3 − 2Vs1
5Vs2
Solving the matrix equation yields:
V1
V2
=
181.46
124.39
Power absorbed by the 6k resistor is (V1-V2)2/R = 0.5429W .
Similarly, Ps1 = (Vs1-V1)/3000×Vs1 = 4.7W, Ps2 = (Vs2 −V2) /12000 ×Vs2 = −0.32W
Ps3 = Is3 × (Vs2 − V1) = −1.21W
Solution 3.7 (a) Again, the matrix equation can be written by inspection:
G1 + G2 + G4 −G4
−G4 G3 + G4 + Gs
VB
VC
=
50G1
50G3
(b) Substituting the values of conductances and inverting the above matrix equation yields:
VB = 34.0132V
VC = 33.6842V
(c) Power delivered is 80.7566W . Using the Principle of Conservation of Power:
Pdel = P1 + P2 + P3 + P4 + P5
or,
Pdel = 50 ×VA −VB
20+
VA −VC20
= 80.7566W
(d) In this part, we take the above matrix equation and solve it for each value of Gs. If we do this, we can
get a feel for the behavior of VB and VC w.r.t. changes in Gs. The following plot is the voltage difference
between the two nodes as a function of Gs, and hence as a function of temperature.
As can be seen, in this figure, the voltage difference between B and C does not change linearly with Rs.
Since this resistance itself changes linearly with temperature, this means that VB-VC does not change
proportionally to temperature.
Solution 3.8 The answer is:
G1 + G2 −G1 −G2 0
−G1 G1 + G3 + G4 + G7 −G4 −G7
−G2 −G4 G4 + G5 + G6 + G2 −G6
0 −G7 −G6 G6 + G8 + G7
V1
V2
V3
V4
=
− Is1
0
0
Is2
Solution 3.9 Write the matrix equation by inspection:
4 /100 −1/100 −1/100
−1/100 4 /100 −1/100
−1/100 −1/100 4 /100
V1
V2
V3
=
Vs1100Vs1100Is2
Solving the equation in MATLAB, we get: V1 = 7, V2 = 7, V3 = 11, and P = 0.6.
P = Vs1 ×VS1 − V1
100+
VS1 −V2100
= 10 ×
310
+3
10
= 0.6W
Solution 3.10 (a)
Nodal equation for A:
VA −Vs110
+VA −VB
10+
VA −VC10
= 0
(b) At node B:
VB10
+VB −VA
10− Is2 = 0
(c) At node C:
Is2 − Is3 +VC − VA
10= 0
(d) Manipulate algebraically to cast as the following matrix equation:
3 −1 −1
−1 2 0
−1 0 1
VA
VB
VC
=Vs1
10Is2
10 Is3 − Is2( )
VA
VB
VC
=
−13 13
1313
−43 13
(e) P = (-10-13.333)/10×(-10) = 23.33W.
Solution 3.12
We are required to write the equations in matrix form. First, write a node equation at VA and Vout:
VA − 5 +VA
5+
VA −Vout
10= 0
Vout −VA
10− 7.5VA +
Vout
10= 0
Now group the coefficients for VA and Vout, and write the matrix equation:
1+1/5 +1/10 −1/10
−1/10 + 7.5 2/10
VA
Vout
=
5
0
VA
Vout
=
1
−37
where the matrix inversion was performed in MATLAB. The ratio of the output voltage to the input voltage
is -37/5.
Solution 3.13 (a) Nodes A and B are already labeled:
VA − 9( )0.1 + VA − 0.5VB( )0.2 + VA − VB( )0.3 = 0
VB −VA( )0.3 + VB − 9( )0.5 + 0.4VB = 0
This can be rearranged Into:
0.1+ 0.2 + 0.3( )VA − 0.3VB − 0.2 × 0.5( )VB = 0.9
−0.3VA + 0.3 + 0.5 + 0.4( )VB = 4.5
The matrix equation can now be easily obtained:
0.6 −(0.5)(0.2) − 0.3
−0.3 1.2
VA
VB
=
0.9
4.5
VA
VB
=
4.8
4.95
(b) Iin = -(VA-Vs)0.1 - (VB-Vs)0.5 = 2.445A
(c) Ps = VSIin = 22.005W, Pdep = 0.5 ×Vout × (0.5Vout −VA) × 0.2 = −1.151W .
(d) P = V2/R = (4.95)×(4.95)×0.4 = 9.801W.
Solution 3.14 (a) We write node equations at VA and VB:
−Is1 +VA
20k+ gm1VA +
VA −VB
10k= 0
VB − VA
10k− gm2 VA − VB( ) +
VB
2.5k+ Is2 − gm1VA = 0
Rearranging, we have:
1
20k+ gm1 +
1
10k
VA −
VB
10k= Is1
−1
10k− gm2 − gm1
VA +
1
10k+ gm2 +
1
2.5k
VB = −Is2
(b)
1/20 k + gm1 +1/10k −1/10k
−1/10k − gm2 − gm1 1/10k + gm2 +1/2.5k
VA
VB
=
Is1
− Is2
(c) The above matrix is inserted into MATLAB, with all the values substituted, to obtain:
VA
VB
=
9.722
5.972
(d) Vo = VA-VB = 3.75V
(e) P1 = VAIs1 = 0.0292W, Pgm1 = -Vogm1VA = -0.008W, Pgm2 = VBgm2Vo = 0.0112W, P2 = -VBIs2 = -
0.0119W.
Solution 3.15 I1 = 0.4. Write nodal equations at A and B
VA100
+VA − VB
20+ 0.03 VA − VB( ) = 0.4
VB − VA20
+VB40
+VB − 80VB
4040
= 0
Rearranging and casting into matrix form:
1/100 +1/20 + 0.03 −1/20 − 0.03
−1/20 1/20 +1/40 −1/40
VA
VB
=
0.4
0
VA
VB
=
40
40
It is obvious then, that Vx, the voltage between A and B, is zero.
Solution 3.16
1) VA = 3000ix = 3000 ×VA −VB
9000=
VA − VB3
Equation at node B:
VB − VA9000
+VB
6000+
VB −VD18000
= 0 is equivalent to:
2) −2VA + 6VB − VD = 0
Equation at node D:
VD − VB18000
+VD
9000+ IS = 0 which can be rewritten as:
3) 3VD −VB = −360 Solving the system formed by equations (1), (2) and (3) we obtain:
VA = 9V , VB = −18V , VD = −126V .
Solution 3.17 (a) Choose E as the reference node
VA = 2ix
At node B
6 = (VB-VC)/3 + (VB-VA)/2
Or 6 = 5/6VB –1/3VC –VA/2
At node C, VC = 2iy
At node D, VD = -12V
iy = (VD-VA)/2 = VD/2 –ix
From here on, the solution involves algebraic manipulations to solve the system of equations. MATLAB or
hand analysis can be performed to obtain:
VA = 48V, VB = 12V, VC = -60V.
(b) P6A = 6(12 = 72W
I12V = 30-8 = 22A ( P12V = 264W.
P2i x= 2ix × (−iz − iY ) = 2 × 24 × (+18 + 30) = 2304W
P2i y= 2iy × (−ix +
vc − vD6
) = 2 × (−30)× (−24 − 8) =1920W
(c) P3 Ω = ix × (VB −VC ) =1728W = 1728W
P6 Ω = 6×8×8 = 384W
P2 Ωy = 2×iy×iy = 1800W
P2 Ω = (VB-2ix) (VB-2ix)/2 = 648W.
(c) VD = -12V, VA = 2ix, VC = 2iz.
Substitute the above VA and VC into the node equation for node B:
iz = (VB-2ix)/2 = VB/2 –ix and
ix = (VB-2iz)/3 = VB/3 –2/3iz
Substitute iz into ix to obtain: ix = 0, VA = 0. Then, VB can be deduced to be 12.
Finally, VC = 12V.
Now, compute the powers.
P6A = 72W , P3 Ω = 0, P6 Ω = (VC-VD)2/6 = 96W, P2 Ωy = 72W, P2 Ωz = 72W.
P2ix = 0W , P2iy = 12 × 4 = 48W
Solution 3.18 The three node equations at A, C, and D are:
−0.8 − 0.3 = 0.015VA + 0.02VA − 0.02VC
Vc = 440
−0.8 + 2.5 = −0.005VD + 0.025VC − 0.025VD
As can be seen, these really reduce to only two equations in two unknowns. These can be solved rather
easily either by hand or by MATLAB to obtain: VA = 220V, VD = 310V . Note that all of these voltages
are already referenced to node B (i.e. VA = VAB, etc).
Solution 3.19 (a) Supernode is BC (50 V source).
(b) Only one node equation needs to be written:
P12V =12 × (VA − VD
2+
VC − VD6
) = 12 × (6 + 4) =120W,
VB90
−VA90
+VB10
+VC10
−VA10
+VC90
= 0
with the constraint that VC −VB = 50.
(c) The constraint equation can be substituted into the B node equation to obtain
VB = 125V. Thus, VC = 175V, and ix = (VC-VA)/10 = -12.5A.
(d) (VA-VB)/90 = 1.94A è P300 = 300×(1.94+12.5) = 4332W.
VC/90 = 1.94 è P50 = -528W.
P50 = 50 × (ix +VC90
) = −528W
Solution 3.20 (a) VB = VA – 440 and VC = VA – 460.
(b) Supernode is one including A, B, and C.
(c) VC − 40( )0.15 + 0.05VB + 0.25VA − 25 + 0.2 VA − 40( ) = 0
This can easily be rearranged to get VA = 200V.
(d) power Ps = VA × I = 200 × 25 = 5000W = 5KW
Solution 3.22 (a) VC = Vs2 = 6V
(b) Ix = 0.01VA
(c) Supernode at A,B, encompassing the controlled source. So, we have one equation:
Is1 = 0.01VA + 0.0125VB + 0.1VB − 0.1VC
(d) Substitute the constraint equations, VA –VB = 20Ix = 0.2VA , (equivalently: VA =VB0.8
) into the above
equation: VA =VB0.8
Is1 = 0.01VB
0.8+ 0.0125VB + 0.1VB − 0.1VC
⇒ VB = 6.4V
⇒ VA = 8V
(e) Ix = 0.08A.
(f) P0.0125Ω = VB2 / R = 0.512W
(g) P = Is1 ×VA = 0.2 × 8 =1.6W
Solution 3.23 (a) VC = Vs2 = 50V.
(b) ix = VA/100.
(c) Supernode A,B:
Is1 = 0.01VA + 0.05VB + 0.05VB − 0.05Vs2 + 0.09VA − 0.09Vs2
VA = VB + 300ix = VB + 3VA
(d) Solving the above two equations yields: VA = -90V, VB = 180V.
(e) ix =VA100
=−90100
= −0.9A
(f) (VB-VC) (VB-VC)/R = 845W.
(g) PS1= IS1
× VA = 2 × (−90) = −180W
Solution 3.24 (a) VB − VC = 3Vx = 3VB; VC = −2VB
(b) Supernode at B and C, encompassing controlled source.
(c)
Is2 = (VB10
−VA10
) +VB10
+VC10
+ (VC10
−VA10
); VC = −2VB; 10Is2= −2(VB +VA)
equivalently: 2VA + 2VB = −10
(d)
(0.1VA − 0.1Vs1) + (0.1VA − 0.1VB) + (0.1VA − 0.1VC ) = 0; 0.3VA + 0.1VB = 0.1VS1; equivalently:
3VA − VB = −10
(e) Again, any method can be used to simplify and solve the system of two equations. The solution is:
VA = -2.5V, VB = -2.5V.
Solution 3.25 (a) VA = Vs1 = 16V.
(b) Supernode at C and D, encompassing controlled voltage source.
(c) Is2 = (0.75mVD − 0.75mVB ) + (1mVC −1mVA)
(d) VC = 4VB + VD
(e)
0.75mVD + (0.75mVB − 0.75mVD) + (0.25mVB − 0.25mVA) = 0
or 1mVB − 0.25mVA = 0
(f) We now have three equations in VB, VC, and VD. These can be solved using any method. By inspection,
we can immediately deduce VB from VA using the last equation: VB = 4V.
The remaining two equations can be solved to obtain: VC = 20V and VD = 4V.
Solution 3.26 (a) The supernode is the combination of A, C, and the controlled voltage source.
(b) Write node equations starting at the supernode:
(G2VA − G2Vin ) + (G3VA ) + (G4VA − G4VB) + (G6VC ) + (G5VC − G5VB ) = 0
⇒ G2 + G3 + G4( )VA + −G4 − G5( )VB + G6 + G5( )VC = G2Vin
and
(2G6VC ) + (G4VB − G4VA ) + (G5VB − G5VC ) + (G1VB − G1Vin ) = 0
⇒ −G4( )VA + G1 + G4 + G5( )VB + 2G6 − G5( )VC = G1Vin
and
VA −VC = 3Vx, VA −VC = 3(Vin − VA ), 4VA − VC = 3Vin
In matrix form:
0.8
−0.4
4
−0.5
0.6
0
0.2
0.1
−1
VA
VB
VC
=6
6
180
(c) The above system of equations can be solved to obtain: VA = 38.75V, VB = 40V, VC = -25V.
(d) Iin = (G7Vin )+ (G2Vin – G2VA) + (G1Vin – G1VB) = 5A.
è Req = 12Ω and P = 300W.
P = Vin × Iin = 60 × 5 = 300W
(e) Iout = VCG6 = -2.5A è P = 62.5W.
Solution 3.27 (a) Supernode is A,B encompassing controlled voltage source.
(b)
(VA – Vs1) + 0.4VB + (0.2VB – 0.2VC) = 0
è VA + 0.6VB – 0.2VC = Vs1
(c) VA – VB = IB = 0.4 VB è VA = 1.4VB.
(d) Is2 = 0.2VB + (0.2VC – 0.2VB) = 0.2VC.
(e)
In matrix form:
1
1
0
0.6
−1.4
0
−0.2
0
0.2
VA
VB
VC
=8
0
2
The solution is: VC = 10V,VA = 7V ,VB = 5V .
(f) i = (VA-Vs1)1S = -1 è Pccvs = (VA −VB ) × (VA − Vs1 ) ×1S = −2W
Pvccs = (Vs1-VC)(0.2VB) = -2W.
Solution 3.28 (a) Supernode at A,C, CCVS.
(b) Node equation at supernode:
Is + 0.25mVA = G1VA + (G2VA –G2VB) + (G5VC) + (G4VC – G4VB)
è Is = (G1 + G2 – 0.25m)VA + (–G2 – G4)VB + (G4+G5)VC
Constraint:
VA – VC = 104ix = 104G3VB
è 0 = VA – 104G3VB – VC
At node B:
G3VB + (G2VB – G2VA) + (G4VB – G4VC) = 0
è –G2VA + (G3+G2+G4) – G4VC = 0
è (c) Matrix equation:
G1 + G2 − 0.25m −G2 − G4 G4 + G5
1 −104G3 −1
−G2 G3 + G2 + G4 −G4
VA
VB
VC
=Is
0
0
;
0
1000
−0.2
−1
−1000
1.1
1
−1000
−0.8
VA
VB
VC
=2
0
0
(d) Substitute the values of conductances and solve the above matrix equation in MATLAB to obtain:
VA = -38V, VB = -20V, VC = -18V.
(e) P = V × I = (104G3VB) × −Is + VAG1 + G2(VA − VB)[ ] = (−20) × −2m −1.9m − 3.6m[ ] = 0.15W
Solution 3.29
Loop equation: Vin = 2kI1 + 500(I1 + 20m)
è Vin = 2500I1 + 10 è I1 = 20mA.
Pvin = 20m×60 = 1.2W.
PI = 20m×(500I1 + 500×20m) = 0.4W.
P2k = I1×I1×R = 0.8W.
P500 = (I1 + 20m)2R = 0.8W.
total power absorbed by resistors: PR = 0.8 + 0.8 = 1.6
total power delivered by sources: Ps =1.2 + 0.4 W=1.6W
Conservation of power is verified.
Solution 3.30
Loop equation: 100(I1 − 0.5) + 200I1 + 500 × (I1 + 20m) = 0
è I1 = 0.05A.
P0.5A = I × V100Ω; V100Ω = 100 × (0.5 − 0.05); where V100Ω is the voltage on the 100Ω resistor.
P0.5A = 0.5(0.5×100 – 0.05×100) = 22.5W.
P20m = 20m(I1+20m)500 = 0.7W.
Solution 3.31
Loop equation: 3.3 = 50I1 + (50m + I1)100 + (I1 – 30m)40 + (I1 – 50m)60
è I1 = 0.01A.
The power delivered by the independent voltage source:
P = I1×3.3 = 0.033W.
Solution 3.32
Loop equation: 50 = 300I1 + (I1 – 0.4I1)500
50 = (300 + 500 – 200)I1 è I1 = 0.0833A.
Power absorbed by the 500Ω resistor.
P500 = (I1 – 0.4I1)2500 = 1.25W.
Solution 3.33
Loop equation: 1000(I1 − Is) + 4000I1 + 5000(I1 − gmVx ) = 0→10000I1 − 2Vx = 50
and
Vx = 1000(Is − I1) →1000I1 + VX = 50 .
Solve the above two equations in I1 and Vx to obtain: I1 = 12.5mA, Vx = 37.5V.
Thus, Req = Vx/Is = 750_,
P = Ivccs ×Vvccs = gm ×Vx × 5000 × (gmVx − I1)=0.1875W
Solution 3.34
Loop equation: Vin = 2Iin + 14Iin – 10V1
V1 = 2Iin
After replacing V1 in the loop equation we obtain:
è Vin = – 4Iin è
R1eg =VinIin
= −4Ω
Solution 3.35
Loop equation: Vs = 500I1 + 100(I1 + 0.5) + 400(I1 − 0.001Vx ) +100(I1 + 0.005Vy )
Vx = 500I1,Vy = 400I1 - 400×0.001Vx = 400I1 – 200I1 = 200I1
After replacing Vx and Vy in the loop equation we obtain:
Vs – 50 = 1000I1 è I1 = 0.1A
Vy = 200I1 = 20V è P400ohm = Vy2 /400Ω= 1W
Req = Vs/I1 = 150/0.1 = 1500Ω.
Solution 3.36
Select clockwise loop current I1 in the left loop. Select anti clockwise loop current I2 in the right loop.
The two mesh equations are:
12 = I1 +10(I1 + I2)
and 10(I2 + I1) + 2I2 + 12 = 0
The two simultaneous equations can be solved easily to obtain: I1 = 0.75A, I2 = +0.375A.
P10ohm = (I1 + I2)2/10 = 0.127W.
Battery 1 supplies more current. (I1 > I2)
Solution 3.37
(a) The equation for the left loop is:
660 = I1R + 1.296(I1 + I2)+ 590 + I1R
The equation for the right loop is:
660 = (0.3 – R) I2+ 1.296 (I1 + I2) + 590 + (0.3 – R) I2
Simplifying the two equations:
70 = 1.596I1 + 1.296I2
70 = 1.296I1 + 1.596I2
The solution of these two equations is: I1 = I2 = 24.2A.
(b) I1 + I2 = 48.4 , voltage across locomotive = 590 + 48.4×1.296 è power = 31592W.
(c) Because the locomotive is 1/3 distance from either station it follows that
R = 1/3×0.3 = 0.1Ω. The two equations become:
70 = I1(2R + 1.296) + 1.296I2
70 = 1.296I1 + (1.296 + 0.6 – 2R)I2
The solution of these two equations is: I1 = 32.64A, I2 = 16.32A.
Current in locomotive motor I1 + I2 = 48.96A.
Voltage across locomotive 590 + (I1 + I2) × 1.296
It follows that:
è P = (I1 + I2)(590 + 49×1.296) = 31993W.
Solution 3.38
(a)
(b) The three loop equations are:
660 – 590 = 0.1I1 + 1.296 (I1 – I2) + 0.1I1
0 = 1.296 ( I2 − I1) + 0.2I2 + 1.296 ( I2 − I3)
–70 = 1.296( I3 − I2) + 0.2I3
These three equations can be solved using any method to obtain:
I1 = 46.8A, I2 = 0, I3 = – 46.8A.
(c) Motor currents are 46.8A each.
(d) Ps = VI = 660×46.8 = 30.9kW. Each source supplies 30.9kW.
Solution 3.39 (a) Define three meshes with three mesh currents. The first, I1 , is a clockwise current
around the first mesh. The second, I2 , is a clockwise current around the middle loop of the circuit
(through the 10mS, 5ms, and 5ms conductances). The third, I3 , is a counterclockwise current through the
right-most loop containing the voltage source.
*current names shown above.
(b) I1 = 0.5A
(I2
10m−
I110m
) +I25m
+ (I25m
+I35m
) = 0.
(I3
25m+
I35m
) +I25m
= 20
These are two equations in two unknown currents. After grouping the terms, it can be verified that:
I2 = 0.1A, I3 = 0.
(c) Vx = 20V
Vad = (0.5 – I2)/10m = 40V
Vbd = 20V
(d) P0.5 = Va × 0.5 = 40×0.5 = 20W
P20V = 0W
Presistors = 2× I2 2/5m + (0.5 – I2)2/10m = 20W
The conservation of power is verified.
Solution 3.40 (a) We can either write down the equations or evaluate the matrix by inspection:
90 ×(I2 − 4.8m) + 10kI2 + 50 = 0
90kI3 + 10 × (I3 − 4.8m) = 50
OR
100k 0
0 100k
I2
I3
=
90k × 4.8m − 50
10k × 4.8m + 50
(b) The solution of the above equation is: I2 = 3.82mA, I3 = 0.98mA.
(c) Current source: P = 4.8m×[(4.8m – I2)90k + (4.8m – I3)10k] = 0.61W.
Voltage source: P = 50(I3 – I2) = – 142mW.
Solution 3.41 (a) By inspection:
112k −90k −10k
−90k 100k 0
−10k 0 100k
I1
I2
I3
=180
−60
60
(b) Using MATLAB:
I1 = 4.4mA, I2 = 3.36mA, I3 = 1.04mA
→ ix = − I2 = −3.36mA
(c) P180 = 180×4.4m = 0.792W, P60 = 60( I3 − I2) = -0.139W.
Solution 3.42 The matrix equation is:
8 −6 3
−6 8 −2
3 −2 4
I1I2
I3
=14
0
6
whose solution is: I1 = 4A, I2 = 3A, I3 = 0A
v = ( I1 + I3 + I2) ×2 = 2V
Solution 3.43 (a) First, note that two mesh currents are needed. Two clockwise currents are defined: I1
in the middle loop, and I2 in the right-most loop:
Middle loop equation:
100I1 – 100Is1 + 20Ix + 80I1 – 80I2 = 0, where Ix = Is1 – I1
and
Right-most loop equation:
80I2 – 80I1 + 10I2 + Vs2 = 0
These can easily be cast into the following matrix equation:
160
−80
−80
90
I1I2
=
16
−6
(b) The solution of this equation is: I1 = 0.12A, I2 = 0.04A.
(c) VA = 100(Is1 – I1) = 8V and VB = 80x(I1 − I2) = 6.4V .
(d) Ps1 = Is1VA = 1.6W.
(e) P0.0125S = (I1 − I2)2/0.0125 = 0.512W.
Solution 3.44 (a) Create two clockwise mesh currents in the top loop (I2) and the bottom-left loop (I1) .
The bottom-right loop has an independent current source. Writing the loop equations:
Vs1 = 200 (I1 – I2) + 200 (I1 + Is2)
200 (I2 – I1) + 100I2 + 300Ix + 200 (I2 + Is2) = 0, where Ix = I1 – I2
(b) Solving, we get: I1 = -0.1A, I2 = -0.7A, Ix = 0.6A.
(c) VB = (I1 + Is2)200 = 130V.
(d) Pvs1 = I1Vs1 = –25W, Pis2 = (VB + (Is2 + I2)200)Is2 = 105W, P300ix = (−I2)(300Ix) = 126W.
Solution 3.45 (a) Create two clockwise mesh currents in the top loop (I1) and the middle loop (I2) (all
resistor loop):
Top loop equation:
0.5vx = 500 (I1 - I2,) +500I1 where vx = –500I1
and
Middle loop equation:
600 (I2 – Is1) + 500 (I2 – I1) + 900 (I2 + Is2) = 0
(b) Solving, we get: I1 = 0.015A, I2 = 0.0375A, vx = -7.5V.
(c) Pis1 = Is1[0.5vx + (Is1 – I2)600] = 109.7W
P0.5vx = 0.5vx(I1 – Is1) = 1.63W
Pis2 = (I2 + Is2)×900×Is2 = 53.2W
Solution 3.46 Write the mesh equations in terms of R’s and then substitute the values from the matrix:
Mesh 1 equation:
v1 = R1 ( i1 − i3) + R2 (i1 – i2)-25i2From this equation, and the first row of the matrix equation, we can deduce that
R1 = 5 Ωand
R2 + 25 = 40 ⇒ R2 = 15Ω.
Similarly:
Mesh 3 equation: R1(i3 − i1) + i3R4 + R3(i3 − i2) =0
From which we can deduce:
R3 = 25 Ω and R4 = 5 Ω.
Solution 3.47
Modified loop 1 equation:
Vs1 = 3MI1 + v + 2MI1
Constraint equation:
– I1 + I2 = Is3
Modified loop 2 equation:
v = 2MI2 + Vs2 + 8MI2
Or in matrix form:
5M 0 1
−1 1 0
0 −10M 1
I1
I2
v
=Vs1
Is3
Vs2
Solving: I1 = -1.1 A, I2 = -0.95 A.
The power Ps3 = Is3 xv= 7.58 W
Solution 3.50 I2 = 2A, I3 = –7A
Loop 1 equation:
Vs = 3I1 + 3(I1 − I2) + 6(I1 + I3 − I2) + 2vy + 2(I1 + I3) =14I1 + 2vy − 74 ⇒ 14I1 + 2VY = 88
vy = 3(I1 – I2)= 3I1 − 6
Solving the above system, we obtain: I1 = 5A è vy = 9, v can be found from the loop 3 equation
v + (I1 + I3 − I2).6 + 2v y + 2(I3 + I1) = 0 . Solving, we obtain v =10V
Finally, Pvs = Vs×I1 = 70W.
Solution 3.51
Mesh 1 equation:
9kI1 + 3k (I1 – I4) + 6k (I1 – I3) + 12k (I1 – I2) = 0
where we have used the fact that Ix = I1 – I4 (and I4 = – 4mA)
Mesh 3 equation:
6k (I3 – I1) – v2 +2kI3 + v = 0
Mesh 2 equation:
2.4kI2 + 12k (I2 – I1) – v = 0
Constraint equations:
I2 – I3 = 0.5mA
I3 – I4 = 0.5Ix = 0.5I1 – −0.51I4 = 0.51I1 + 2mA ; 0.51I1 − I3 = −2mA
The above five equations need to be put into matrix form:
30k −12k −6k 0 0
−6k 0 8k 1 −1
−12k 14.4k 0 −1 0
0 1 −1 0 0
0.5 0 −1 0 0
I1I2
I3
v
v2
=
−12
0
0
0.5m
2m
The solution is:
I1
I2
I3
I1I2
I3
v
v2
=
−0.002A
−0.0025A
−0.003A
−12V
−24V
Ix = I1 − I4 = 2mA
Power delivered by 0.5mA current source: P1 = 0.5mxv = −6mW
Power delivered by the dependent current source: P2 = 0.5Ix ⋅ v2 = 1m × (−24) = −24mW
Power delivered by 4mA current source: P3 = 4m × (3k ⋅ Ix − V2) = 120mW
Solution 3.52
Write the following equations:
Mesh 2 equation:
100 (I2 – I1) + 150 (I2 – I4 – I6) + v = 0
Mesh 4 equation:
3vx + 30 (I4 – I3) + 150 (I4 - I2 + I6) = 0
Mesh 5 equation:
800I5 – 3vx + 10( I5 − I1 + I6) =0
Mesh 6 equation:
250I6 +150(I6 + I4 − I2) +10(I6 + I5 − I1) = 0
Constraint equation:
I1 I2 I3
I4
I5
I6
Is2= −I2 + I3
Substituting the values of I1 = Is1 = 1.15 and I3 = Is3 = 0.95 , and noting that
vx = (I1 – I6 – I5)10
we can write:
250 −150 0 −150 1
−150 180 −30 120 0
0 0 840 40 0
−150 150 10 410 0
1 0 0 0 0
I2
I4
I5
I6
v
=
100I130I3 − 30I1
40I1
10I1I3 − IS2
I2
I4
I5
I6
v
=
0.65
0.45
0.05
0.1
35
Finally, vx = 10V and v is as given above.
Solution 3.53 (a)
Replace the voltage source by current sources:
At node 1
is1 = (V2 − V1) /12k
At node 2
(V2 − V1) /12k + (V2 − V3) /90k + (V2 − V4) /10k = 0
At node 3
is2 = (V3 – V2) /90k + V3/10k
At node 4
is2 + (V4 −V2) /10k + V4/90k = 0
V1 = 180
V4 – V3 = 60
(b) In matrix form:
−1/12k 1/12k 0 0 −1 0
−1/12k 1/12k +1/90 k +1/10k −1/90k −1/10k 0 0
0 −1/90 k 1/90k +1/10k 0 0 −1
0 −1/10k 0 1/90 k +1/10k 0 1
1 0 0 0 0 0
0 0 −1 1 0 0
V1
V2
V3
V4
is1is2
=
0
0
0
0
18060
The solution from MATLAB is
180.0000
127.2000
33.6000
93.6000
-0.0044
0.00232
(c) Power delivered by S1 is:
Ps1 = Vs1 × (−Is1) = 0.792W
Power delivered by S2 is:
Ps2 = Vs2 × (−Is2) = −0.139W
Solution 3.54 (a) Replace the 100 ohm resistor, the controlled voltage source, and vs2 by current source.
Then write the node equations:
Is1 = ix + I1 + 0.09 (V1 – V3)
I1 = V2/20 + 0.05 (V2 – V3)
– I2 = 0.05 (V3 – V2) + 0.09 (V3 – V1)
V3 = Vs2
V1 – V2 = 300ix = 300V1/100
In matrix form:
0.1
0
−0.09
0
2
0
0.1
−0.05
0
1
−0.09
−0.05
0.14
1
0
1
−1
0
0
0
0
0
1
0
0
V1
V2
V3
I1I2
=
2
0
0
50
0
(b) Using MATLAB to solve the above system the solution is:
V1 = -90.0000V
V2 = 180.0000V
V3 = 50.0000V
I1 = 15.5000A
I2 = -6.1000A
(c) Power delivered by the current source is
Ps1 = Is1 × V1 = −180W
Power delivered by the voltage source is:
Ps2 = Vs2 × (−I2) = 305W
Solution 3.55 Modify the circuit so that it looks like the following:
The modified node equations are:
Is1 = - Iy + Ix
Ib = Iy + 0.2 (VC – VB)
Is2 = 0.2 (VC – VB) + Ix
The equations describing the constitutive relationships of elements in the original network are:
VA – VB = Ib
Ix = 0.2Vb
Ib = 0.4Vb
Iy = 8 −VA
These can be cast into a matrix equation and solved easily to obtain the same result as previously arrived
at.
In matrix form:
0
0
0
1
0
0
1
0
−0.2
−0.2
−1
0.2
0.4
0
0
0.2
0.2
0
0
0
0
1
0
1
0
−1
0
0
−1
1
0
0
0
0
1
0
−1
0
−1
0
−1
0
−1
0
0
0
0
0
0
VA
VB
VC
Ix
Iy
Ib
Is1
=
0
0
2
0
0
0
8
VA
VB
VC
Ix
Iy
Ib
Is1
=
7V
5V
10V
1A
1A
2A
0A
Solution 3.56 Modify the circuit as follows:
The modified node equations are:
At node A: Is = G1VA + G2(VA −VB ) + Ia
At node B: G2(VA −VB ) = Ix + G4(VB −VC )
At node C: 0.25mVA + Ia = G4(VC − VB) + G5VC
The equations describing the constitutive relationships of elements in the original network are:
Ix = G3VB
VA – VC = 104Ix
These can be cast into a matrix equation that can be solved in MATLAB.
In Matrix form:
0.25m
−0.2m
−0.25m
0
1
−0.2m
1m
−0.8m
−0.1m
0
0
−0.8m
1m
0
−1
1
0
−1
0
0
0
1
0
1
−104
VA
VB
VC
Ia
Ix
=
2m
0
0
0
0
The solution is:
VA
VB
VC
Ia
Ix
=
−38V
−20V
−18V
0.0075A
−0.002A
We observe that we have obtained the same results as in problem 3.28.
Solution 3.57 Replace dependent source by i35 (from 3 to 5). Also, replace voltage source by i10 (from 1
to 0). Now, write the modified noted equations. The reference node is O:VO = 0V :
At node 1: i10 = (V6 – V1) + (V2 – V1)
At node 2: 2 = (V2 – V1) + (V2 – V3)
At node 3: i35 = (V4 – V3) + (V2 – V3)
At node 4: 2 = 2 + (V4 – V3) + V4
At node 5: i35 = (V5 – V6) + V5
At node 6: 2 = (V5 – V6) – V6
Constraints:
V3 – V5 = 15vx = 15V4
V1 = 5
The following matrix equation is obtained:
−2 1 0 0 0 1 −1 0
−1 2 −1 0 0 0 0 0
0 1 −2 1 0 0 0 −1
0 0 −1 2 0 0 0 0
0 0 0 0 2 −1 0 −1
0 0 0 0 1 −2 0 0
0 0 1 −15 −1 0 0 0
1 0 0 0 0 0 0 0
V1
V 2
V 3
V 4
V 5
V 6
i10
i35
=
0
2
0
0
0
2
0
5
The solution of this equation is obtained from MATLAB:
V1 =5.0000V
V2 =3.3571V
V3 =0.2857V
V4 = -0.1429V
V5 =1.8571V
V6 = -0.0714V
i10 = -6.7143A
i35 = 3.7857A
The power delivered by the dependent voltage source connected between nodes 3 and 5:
P35 =15vx (−i35) = 15 ×V4 × (−i35) = 8.115W
The power delivered by the current source connected between nodes 2 and 4:
P24 = (2A) × (V2 −V4 ) = 7W
The power delivered by the current source connected between nodes 4 and 6:
P46 = (2A) × (V4 −V6) = −0.143W
The power delivered by the voltage source connected between nodes 1 and 0:
P10 = 5V × (I12 + I16) = 5x (V1 −V2) + (V1 − V6)[ ] = 33.57W
Solution 3.59 Using the appropriate element stamps for each element of the circuit, we obtain the
following system:
0.15 + 0.2 −0.15 −0.2 0
−0.15 0.15 + 0.05 0 −1
−0.2 0 0.25 + 0.2 1
0 −1 1 0
VA
VB
VC
Ix
=
−8 − 3
3
25
440
Solution 3.60
1/20 k +1/10k + gm1 −1/10k
−1/10k − gm1 − gm2 gm2 +1/10k +1/2.5 k
V1
V2
=
Is1
−Is2
The solution is the same as that of problem 3.14.
Solution 3.62 (a) Because RT (T ) can be approximated by a straight line between (250Ω,0oC) and
(80Ω,50oC) it follows that:
RT (T )= – 3.4T + 250
(b) For T = 25oC , RT = 165Ω
(c) The voltage across the RT + RL series combination can be obtained from voltage division:
VT ,L =RT + RL
RT + 2RL + R⋅12 = 4.7857V
This is the same as the voltage across Rx because the meter is at zero deflection. Thus,
RxRx + R
⋅12 = 4.7857. It follows that Rx =165.84Ω.
(d) We first denote the nodes:
A - the node common to R,Rx and the voltmeter;
B - the node common to R,R and the voltage source;
C - the node common to RL,RT and RL;
D - the node common to Rx,RL and the voltage source.
The reference node is D:VD = 0. It follows that VB =12V .
We also have: vout = VA − VC
The node equations are:
12
0
At node A:VA −VC
Rm+
VA −12R
+VARx
= 0
Equivalently: VA(RRx + RmRx + RmR) - VC ⋅ RRx = 12RmRx
VA × 4199.86 − VC × 41.46 =19900.8 (1)
At node C:VC −12R + RL
+VC
RT + RL+
VC − VARm
= 0
Equivalently: (VC −12) × 0.004 +VC
RT + 2.5+ (VC −VA) ⋅10−4 = 0
VC (0.004 ⋅ RT + 0.01 +1+ 0.00025 + RT ×10−4) −
−VA(RT ×10−4 + 0.00025) = 0.48 × (RT + 2.5)
The last equation can be rewritten as:
VC (0.0041RT +1.01025) - VA(RT ×10−4 + 0.00025) = 0.48(RT + 2.5) (2)
From (1) and (2), we obtain:
(0.0041RT +1.01025) ×−19900.8 + VA × 4199.86
41.46−
− VA × (RT ×10−4 + 0.00025) = 0.48(RT + 2.5)
Equivalently: VA × (RT × 0.415 +102.337) = 2.448 × RT + 486.12
It follows that VA =2.448 × RT + 486.120.415 × RT +102.337
From the equation at node A:
vout = VA − VC = −Rm ×VA −12
R+
VARx
=
2.448 × RT + 486.120.415 × RT +102.337
×(−100.3) + 480
At T = 0oC: RT = 250Ω. It follows that vout = −54.4415V
At T = 50oc: RT = 80Ω. It follows that vout = 80V
(e): The formula has been derived at part d):
T RT vout
0oC 250Ω −54.4415V
5oC 233Ω −52.4136V
10oC 216Ω −50.2368V
15oC 199Ω −47.8938V
20oC 182Ω −45.3650V
25oC 165Ω −42.6273V
30oC 148Ω −39.6537V
Solution 3.63
Place a source Vin between C and D, and calculate the current drawn from the source as below:
Loop 1 equation:
I1R1 + (I1 − I2)R2 + (I1 − I3)R3 = 0
Equivalently:
I1(R1 + R2 + R3) − I2R2 − I3R3 = 0
Loop 2 equation:
(I2 − I1)R2 + I2R4 −1 = 0
Equivalently:
−I1R2 + I2(R2 + R4) =1
Loop 3 equation:
1 + I3R5 + (I3 − I1)R3 = 0
Equivalently:
−R3I1 + I3(R3 + R5) = −1
We obtain the following system of equations:
30I1 − 4I2 − 6I3 = 0
−4I1 + 6I2 =1
−6I1 +14I3 = −1
⇒ I1 = 0.0096A , I2 = 0.1731A , I3 = −0.0673A;
Iin = I2 − I3 = 0.2404 A
Reg,CD =VinIin
=1
0.2404= 4.16Ω
Solution 3.64
The node equation at node A is:
VAG1 + (VA − VB)G2 + (VA −VC )G3 = 0
Equivalently:
(G1 + G2 + G3)VA −VBG2 − VCG3 = 0
The supernode is identified by a Gaussian surface enclosing the controlled voltage source. The supernode
equation is:
G2(−VA + VB) − 6 + G4VC + G3(VC − VA ) = 0
Equivalently, we have:
−VA(G2 + G3) + G2VB +VC (G3 + G4 ) = 6
One way of obtaining the solution to the problem is:
We multiply the above two equations by 30.
−30(G2 + G3)VA + 30G2VB + 30(G3 + G4 )VC =180
and
30(G1 + G2 + G3)VA − 30G2VB − 30G3VC = 0
By equating the coefficient of the above two equations with the coefficients of the first and second given
equations, we obtain:
30G2 = 30 ⇒ G2 = 0.1S
30G3 = 2 ⇒ G3 = 0.067S
30(G1 + G2 + G3) = 11⇒ G1 = 0.2S
30(G3 + G4 ) = 32 ⇒ G4 = 0.87S
can be obtained as follows:
VC −VB = VX = (VC − VA )
Equivalently:
VA −VB + (1− )VC = 0
By comparing with the third given equation ⇒ = 3.
Solution of 3.66
(a)
At node A: (VA-VC)/2 + (VA-VB)/2 + (VA-VD)/2 = 14
At node B: (VB-VA)/2 + (VB-VC)/2 + (VB-VD)/2 = 7
At node C: (VC-VA)/2 + (VC-VB)/2 + (VC-VD)/2 + 2VC = 0
At node D: (VD-VA)/2 + (VD-VB)/2 + (VD-VC)/2 + 0.5VD = 0
These can be solved in MATLAB to obtain:
22.0000
18.5000
7.5000
12.0000
(b) Mesh analysis would result in the same voltages
The loops and their current loops are:
A,C,14A:I1A,C,D:I2
A,B,D:I3
B,C,D:I4
C,D,reference node:I5
B,D,7 A:I6
I1 = 14A,I6 = 7A
Loop ACDequation: 2(I2 − I1) + 2(I2 − I3) + 2(I2 + I4 + I5) = 0
Loop BCD equation: 2(I3 + I6 + I4 ) + 2(I2 + I4 + I5) + 2I4 = 0
Loop ABD equation: 2(I3 + I4 + I6) + 2(I3 − I2) + 2I3 = 0
Loop CDref node equation: 2(I5 + I2 + I4 ) + 2(I5 − I6) + 0.5(I5 + I1) = 0
In matrix form:
6
2
−2
2
−2
2
6
0
2
6
2
2
2
2
0
4.5
I2
I3
I4
I5
=
28
−14
−14
7
;
I2
I3
I4
I5
=
6.75
1.75
−5.5
1
VA = 2(I1 − I2) + 0.5(I1 + I5) = 22V
VB = 2(I4 + I3 + I6) + 2(I6 − I5) = 18.5V
VC = 0.5(I1 + I5) = 7.5V
VD = 2(I6 − I5) =12V
(c) Mesh analysis requires more work.
(d) The removal of the top resistor will result in more node equations than loop equations. The addition of
a resistor between node A and reference node will result in more loop equations than node equations.
Thevenin Probs, 7/11/01 - P4.1 - @R.A. Decarlo & P. M. Lin
PROBLEM SOLUTIONS CHAPTER 4
SOLUTION 4.1. First, find Vout / Vs for each circuit. Then solve for R knowing
Vout = P ⋅10 =±14.142V .
(a) Writing KCL at the inverting terminal, 1 /1k(v− − vs) =1/ R(Vout − v− ) ⇒ Vout / Vs = −R / 1k , since
the inverting terminal is a virtual short. Solving for R = −Vout ⋅1k / Vs = 2.828kΩ .
(b) Writing KCL at the inverting terminal, Vs /1.5k = (Vout − Vs)/ R ⇒ Vout / Vs = R / 1.5k +1, solving
for R = 1.5k(Vout / Vs −1) = 2.743kΩ.
(c) From (a) Vout / Vs = −12k / R , thus R = −12k ⋅Vs / Vout = 4.243kΩ.
(d) This is the same circuit as (b) except the output voltage is taken across two resistors. Thus
Vout =P
1010 + 6( ) = 22.627V . Using the general form from (b), R = 400 Vout / Vs −1( ) = 1.410kΩ
SOLUTION 4.2. (a) First, find the voltage at the non-inverting terminal as v+ = 1/ 2 ⋅Vs . Then write KCL
at the inverting terminal, and make use of the virtual short property,
(Vs / 2)/10 k = (Vout − Vs / 2) / 30k ⇒ Vout / Vs = 30k(1/ 20k +1/ 60k) = 2 .
(b) Relating the output of the amplifier to the output of the circuit, Vout = Vamp(500 / 800) . Then writing
KCL at the inverting terminal, Vs / 400 = (Vamp − Vs) /1.2k ⇒ Vamp / Vs = 1.2k / 400 +1 = 4. Therefore
Vout / Vs = (Vamp / Vs ) ⋅ (Vout / Vamp ) = 2.5.
(c) Note that since no current goes into the non-inverting terminal of the op-amp, the voltage at that node is
–Vs. KCL at the inverting terminal, −Vs / 4k = (Vout + Vs )/ 20k ⇒ Vout / Vs = −6.
SOLUTION 4.3. Write KCL for both terminals,
(V− −Vi ) /1k = (Vo − V− ) / 2k
V− /1k = (Vo − V− )/ 3k
Solving and doing the appropriate substitutions, Vo / Vi = −8 .
SOLUTION 4.4. This is essentially the basic inverting configuration, which is defined as
Vo / Vi = −2k / 1k = −2 .
Thevenin Probs, 7/11/01 - P4.2 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.5. (a) By voltage division VL = 1V ⋅100
200= 0.5V . Using Ohm’s law
Is = IL =1
100 +100= 5mA .
(b) No current flows in the input terminal of an ideal op-amp, thus Is = 0A and VL = 1V . From Ohm’s
law Ia = IL = VL /100 = 10mA .
SOLUTION 4.6. (a) Using voltage division,
V1 = Vs32||(8 + 24)
[32||(8 + 24)] + 8
=
23
Vs
Vout = V124
24 + 8
= 0.5Vs
(b) By voltage division,
V1 = Vs32
32 + 40
= 0.8Vs
Vout = V124
24 + 8
= 0.6Vs
(c) Using voltage division, V1 = Vs32
32 +8
= 0.8Vs , as no current enters the non-inverting terminal of the
op-amp. Due to the virtual short property, Vout = V124
24 + 8
= 0.6V1 . This is indeed the same results as
(b), which should be expected because of the isolation provided by the ideal buffers.
SOLUTION 4.7. Write KCL at the inverting terminal,
−Vs1 /1k − Vs2 / 2k = Vout / 4k ⇒ Vout = −4Vs1 − 2Vs2 = 40mV .
SOLUTION 4.8. (a) The voltage at the non-inverting terminal is V+ = 3 / 2V , KCL at the inverting
terminal gives (1.5− 2.5) /10k = (Vout −1.5)/ 30k ⇒ Vout = −1.5V . The power is
P = Vout2 / 500 = 4.5mW .
(b) The voltage at the non-inverting terminal is 3V this time, thus KCL
(3 − 2.5)/ 10k = (Vout − 3) / 30k ⇒ Vout = 4.5V . The power is P = Vout2 / 500 = 40.5mW .
SOLUTION 4.9. (a) Define the point between the two op-amp as Vint. Observe that the first op-amp is in
the basic inverting configuration, and the second the non-inverting configuration. By inspection,
Thevenin Probs, 7/11/01 - P4.3 - @R.A. Decarlo & P. M. Lin
Vint / Vs = −R1 / 1k
Vout / Vint = (1+ R2 /1k )
Cascading the two stages, (Vint / Vs)(Vout / Vint ) = Vout / Vs =− R1 /1k(1+ R2 /1k) . Solving for
R1 = 20 ⋅1k / (1+ R2 /1k) = 5kΩ. The power absorbed is P = (20 ⋅0.5)2 / 8 =12.5W .
(b) Using the same equations as (a), solve for R2 = (20 ⋅1k / 2k −1)1k = 9kΩ .
(c) Rewriting the equation obtained in (a), R12 +1kR1 − 20M = 0, and solving the quadratic equation yields
R1 = R2 = 4kΩ .
SOLUTION 4.10. This is a cascade of two non-inverting configuration op-amp of the form
Vo / Vs = (1+10k /10k ) for each. Therefore 2 ⋅ 2 = 4.
SOLUTION 4.11. This system is made up of a non-inverting stage with a gain of 1+10k/10k, a voltage
divider of gain 8k/(2k+8k), and a second non-inverting stage of gain 1+10k/10k. The product of all three
yields Vout / Vin = (2)(0.8)(2) = 3.2.
SOLUTION 4.12. (a) By inspection, the gain of the first stage is –1. Then write KCL for the second stage
Vs1 / 2R − Vs2 / R = Vout / 2R ⇒ Vout = Vs1 − 2Vs2 = 10V .
(b) The first stage gain is –0.5, thus Vout = 2R(0.5Vs1) / 2R − 2R(Vs2)/ 0.5R =−7.5V , using the same
procedure as in (a).
SOLUTION 4.13. (a) This is a cascade of a summing amplifier with the following transfer characteristic,
Vo = −4Vs1 − 2Vs2 , and an inverting stage of gain –1.5. Thus Vout = 1.5(4Vs1 + 2Vs2 ) = 2.25V .
(b) Notice that the only difference is the gain of the inverting stage, which is now –2. Therefore
Vout = 2(4Vs1 + 2Vs2) = 3V .
SOLUTION 4.14. This circuit is a cascade of two summing amplifier where the output of the first is an
input of the second stage. The transfer function of the first stage is Vo = −2RVs1 / 2R − 2RVs2 / R , which
is substituted in the transfer function of the second stage to obtain
Vout = −R[−2RVs1 / 2R − 2RVs2 / R]/ R − RVs3 / R = Vs1 + 2Vs2 − Vs3 =−2V .
SOLUTION 4.15. Writing KCL at the inverting node, −V1 / R1 − V2 / R2 − V3 / R3 = Vout / Rf , and
solving for Vout = −Rf
R1V1 +
Rf
R2V2 +
Rf
R3V3
.
Thevenin Probs, 7/11/01 - P4.4 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.16. Referring figure P4.15, the value of the resistance must satisfy the following
constraints: R1 = R2 = R3 = 3R
Rf = R
These will yield the inverted average. If polarity is a concern, a second inverting stage should be added
with a unity gain, i.e. both R’s equal.
SOLUTION 4.17. Using the topology of 4.12 the following parameters are chosen,
Ga1 = 3, Ga2 = 5, Gb1 = 2, Gb2 = 4
For the time being assume G f = 1. Now we calculate δ = (1+ 3 + 5) − (2 + 4) = 3, this sets Gg = 3.
(a) The requirement for G f = 10µS sets the scaling factor K = 10µ / 1 =10µ . This then yields the
following set of parameters,
Ga1 = 30µS, Ga2 = 50 µS, Gb1 = 20µS, Gb2 = 40µS, G f = 10µS, Gg = 30µS
(b) The requirement for G f = 2µS , sets the scaling constant to 2uS. So the following parameters are
obtained:
Ga1 = 6µS, Ga2 =10 µS, Gb1 = 4 µS, Gb2 = 8µS
Furthermore for Gg = 12µS , ∆G = 6µS in order to make the incident conductance equal at both terminal.
(c) Using the starting values from (a), one could choose a scaling constant of 5 µS. This will yield the
following resistances:
Ra1 = 66.67kΩ, Ra2 = 40kΩ, Rb1 = 100kΩ, Rb2 = 50kΩ, R f = 200kΩ, Rg = 66.67kΩ
These are all reasonable values for circuit implementation.
SOLUTION 4.18. (a) Choosing the following initial values:
Ga1 = 3S, Ga2 = 5S, Gb1 = 11S, Gb2 = 4S, G f =1S
Thevenin Probs, 7/11/01 - P4.5 - @R.A. Decarlo & P. M. Lin
then calculate δ = (1+ 3 + 5) − (11+ 4) =−6 . Thus Gg = 1S , and ∆G =1 + 6 = 7S . Scaling everything by
1µS, yield this final set of parameters, which meet the requirements.
Ga1 = 3µS, Ga2 = 5µS, Gb1 =11µS, Gb2 = 4µS, G f = 1µS, Gg =1µS, ∆G = 7uS
(b) The set of parameters remains unchanged, except for ∆G which now becomes 6uS in order to
maintain the equal termination conductance requirement due to Gg = 0S .
(c) Scale the initial parameters of (a) by 5uS, and get the following set of resistances:
Ra1 = 66.67kΩ, Ra2 = 40kΩ, Rb1 = 18.18kΩ, Rb2 = 50kΩ, Rf = 200kΩ, Rg = 200kΩ, ∆R = 28.57kΩ
SOLUTION 4.19. (a) Choosing the following initial set of parameters:
Ra1 = 1/ (4S) = 0.25Ω, Ra2 = 1/ (2S) = 0.5Ω, Rb1 = 1/ (5S) = 1/ 5Ω, Rb2 =1/ (4S) = 0.25Ω, R f = 1Ω
and δ = (1+ 4 + 2) − (5 + 4) = −2 , thus choose Rg =1/ (1S) = 1Ω and ∆R = 1/ (1+ 2) = 1/ 3Ω . To meet
the Rf = 50kΩ requirement, all the parameters must be scaled by 50k, which gives
Ra1 = 12.5kΩ, Ra2 = 25kΩ, Rb1 = 10kΩ, Rb2 =12.5kΩ, R f = 50kΩ, Rg = 50kΩ, ∆R = 16.67kΩ
(b) Same as (a) with a 100k scaling constant:
Ra1 = 25kΩ, Ra2 = 50kΩ, Rb1 = 20kΩ, Rb2 = 25kΩ, Rf = 100kΩ, Rg = 100kΩ, ∆R = 33.33kΩ
SOLUTION 4.20. (a) When the op-amp is in its active region vout / vs = −5 . Thus it will operate in its
active region when −3 ≤ vs ≤ 3, and will saturate at 15V when vs ≤ −3 , and at –15V when vs ≥ 3. SPICE
yield the following plot:
Thevenin Probs, 7/11/01 - P4.6 - @R.A. Decarlo & P. M. Lin
(b) Using SPICE the following plot is obtained:
Thevenin Probs, 7/11/01 - P4.7 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.21. The first stage is in a summing configuration, thus its output is, assuming it's in the
active region of operation, -15 V which means it is just about to saturate. The second stage is in the
inverting configuration with a gain of -1.5, which means that the overall output will be saturated at 15V.
SOLUTION 4.22. When vin −80kvin +1.5
100k
> 0, or vin > 6 the output of the comparator saturates at
–15 V, when it is vin < 6 , it will saturate at 15 V. The following plot is obtained from SPICE.
SOLUTION 4.23. When vin −10kvin + 20
110k
> 0 , or vin > 2 the output of the comparator will be
saturated at -15V. Otherwise when it is < 2V the output saturates at 15V. In SPICE:
Thevenin Probs, 7/11/01 - P4.8 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.24. Based on the same reasoning as the previous questions,
The output will be +Vsat, when vin < vref 1 −R1 + R2
R2
= −
R1
R2vref , and –Vsat for
vin > vref 1 −R1 + R2
R2
= −
R1
R2vref .
SOLUTION 4.25. Using the previously derived relationship, and the topology of figure P4.24, set
vref = −1.5V , and R1 = 2kΩ and R2 = 3kΩ . Set the power supplies to the Op-amp to +/– 10V to satisfy
the Vsat requirement. Also the input to the inverting and non-inverting terminal are reversed for fig. P4.24.
Verifying in SPICE we obtain the following,
Thevenin Probs, 7/11/01 - P4.9 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.26. The design that fulfills the requirement is the same as for P4.25, with the input to the op
amp reversed. The following is obtained from SPICE,
Thevenin Probs, 7/11/01 - P4.10 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.27. First, for the comparator to give +Vsat for the lower voltages, the inputs to the op amp
in the topology of P4.24 must be interchanged. Then the components are chosen to satisfy the following
relationship, vswitch = vref 1−R1 + R2
R2
=−
R1
R2vref . Choose vref = −1.5V , and R1 = 2k and R2 = 1k.
Verifying in SPICE,
Thevenin Probs, 7/11/01 - P4.11 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.28. Write KCL at the inverting terminal, noting that the no current flows into it:
(V− − vin )/ R = (vout − V− )/ R . Use the following relationship vout = A(V+ − V− ) = −AV− . Solving
using the previous two equations yields vout / vin = −A
A + 2.
SOLUTION 4.29. (a) By inspection the voltage gain for the ideal case is –1. When A=1000, the gain
becomes –0.998, thus 0.2%.
(b) Repeating the method of P4.28, and setting vout / vin = −ARf
Rf + R1 + AR1to –1 and solving for
Rf =10.417kΩ.
(c) Solving the previous equation when the gain is –1, Rf / R1 = (A + 1)/ ( A −1).
SOLUTION 4.30. (a)The first part was obtained in P4.29. Rearranging the equation yields
vout / vin = −Rf
R1
1
1 + 1+ Rf / R1( ) / A
.
Thevenin Probs, 7/11/01 - P4.12 - @R.A. Decarlo & P. M. Lin
(b) The error is caused by (1+ Rf / R1) / A in the denominator, and may be defined, in percent, as
100 −1
1+ 1+ Rf / R1( ) / A
⋅100. Thus for the conditions listed in the problem, it will always be less than
2.05%. With A = 10000 it will be less than 0.21%.
SOLUTION 4.31. (a) Substituting the non-ideal model, and writing KCL at the inverting terminal,
(V− − vin )/ R1 + V− / Rin = (vout − V− ) / Rf is obtained. Now observe the following dependencies,
iout = vout / RL , and vout =− AV− − (iout + (vout − V− )/ R f )Rout . Using these three equations, substitute
the second into the third and then solve for vout / vinusing the last two. This yields
V− = vout + vinRf
R11/
Rf
R1+
Rf
Rin+1
= vout
−1 −Rout
RL−
Rout
R f
A − Rout
Rf
and
vout / vin = −Rf
R1
1
1+
1 + RoutRf
+ RoutRL
1+Rf
R1+
Rf
Rin
A −Rout
Rf
A gain of –9.988
(b) For an ideal op-amp the gain is −Rf / R1=-10.
(c) The error is about 0.1175%.
SOLUTION 4.32. The gain is –9.883, and the error 1.16%
SOLUTION 4.33. This derivation was performed in P4.31.
Thevenin Probs, 7/11/01 - P4.13 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.34. Assume that the sliding contact is at the bottom of Rp. Then, writing KCL at the
inverting terminal yields vin / Ro = (vout − vin )/ Rp . This implies vout / vin = 1 + Rp / Ro . When the
slider is at the top, it is evident that vout = vin . Therefore 1 ≤ vout / vin ≤1 +Rp
Ro.
SOLUTION 4.35. Writing KCL at inverting input, and making use of voltage division,
−vin / R1 = βvout / Rf[ ] where β is the fraction of vout that appears across Rf. Hence, vout
vin=−
Rf
βR1.
When the slider is at the top β= 1 and vout
vin=−
Rf
R1. When the slider is at the bottom, the fraction of vout
appearing across Rf is β=Rf / / R0
Rf / / R0 + Rp=
Rf R0
Rf + R0×
1Rf R0
R f + R0+ Rp
=Rf R0
R f R0 + Rp(R f + R0). Hence
1
β=
Rf R0 + Rp(Rf + R0 )
R f R0=1 +
Rp
R0+
Rp
R f. It follows that
vout
vin=−
Rf
βR1=−
Rf
R11 +
Rp
R0+
Rp
Rf
.
Therefore the range of achievable voltage gain is
−Rf
R1≥
vout
vin≥ −
Rf
R11 +
Rp
R0+
Rp
Rf
SOLUTION 4.36. Using the basic non-inverting configuration of figure 4.10 characterized by
vout / vin = 1+Rf
R1
, i.e., µ = 1+
Rf
R1
.
SOLUTION 4.37. At first glance, one might use two inverting configurations, figure 4.5, in cascade.
However, such would not have infinite input resistance. To circumvent this problem we add a buffer
amplifier as per figure 4.7 at the front end of a cascade of two inverting configurations. The resulting
overall gain is µ =R f1
R11
Rf 2
R12
. Indeed, such a configuration can achieve theoretically any gain greater
than zero.
Thevenin Probs, 7/11/01 - P4.14 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.38. Using a single inverting amplifier configuration, figure 4.5, preceded by a buffer stage
of figure 4.7. The gain is µ = −Rf
R1.
SOLUTION 4.39. By KVL for figure P4.39a, Vo = −i1Rf . Thus to achieve Vo = −i1rm in figure P4.39b,
we set Rf = rm .
SOLUTION 4.40. Writing KCL at the inverting node of the ideal op amp yields IL = Vi / Ra , which is
indeed independent of the load resistor which has no effect on the load current.
SOLUTION 4.41. The current through the LED is IL = 10R1
10k
/ 3.8k , so for (a) it is 1.32mA and for
(b) 2.11mA.
SOLUTION 4.42. Applying KCL at the inverting terminal, IL = vin / R1 . Again, ideally, RL does not
affect IL.
SOLUTION 4.43. (a) Defining a temporary voltage Vo at the output of the op-amp, we can write KCL at
the inverting and non-inverting terminal:
(V− − 2)/ 1k = (Vo − V− ) / 2k
V− /100 + (V− − Vo) / 200 = Iout
Substituting the first equation into the second and simplifying causes Vo to drop out and Iout = 20mA .
(b) The answer remains the same as the value of the load resistance was not used for finding the load
current.
SOLUTION 4.44. Using the same approach as for the previous question, but with resistor labels instead,
the following equations are obtained from KCL:
V− = R2Vs + R1VoR2 + R1
Iout = V−R2 + R1R1R2
−
VoR2
Substituting the first into the second yields Iout = Vs / R1 .
Thevenin Probs, 7/11/01 - P4.15 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.45. (a) Vs = 5 V, (b) Iout = 10mA sets R1 = Vs / Iout = 500 Ω. (c) From KVL and Ohm's
law, Is = (Vs + RL Iout ) / αR1 . We require Is < 0.5 mA. This means that in the worst case, RL = 500 Ω,
(Vs + RLIout )
0.5 ×10−3R1=
5 + 500 × 0.01
0.25= 40 <α .
(d) From KVL and Ohm's,
Vo = −RLIout − (Iout + (RL Iout )/ R1)R2 ≥ −20 V
Hence
R2 ≤20 − RL Iout
(Iout + (RL Iout ) / R1)=
20 − 5
0.01 + 5/ 500= 750
Hence one design is to pick R2 = 750 Ω and α = 40 which impliesαR2 = 30 kΩ.
SOLUTION 4.46. The exact same design as P4.45 can be used with the isolation buffer of figure 4.7
placed at the input of it in order to provide the infinite input resistance needed by P4.46b.
SOLUTION 4.47. The general expression for this summing circuit is
Vout = −Rf
RoVo −
Rf
R1V1 −
Rf
R2V2 −
R f
R3.
(a) Using the expression above | Vout |=| −1 − 0 − 0 − 8 | E = 9E .
(b) | Vout |=| −0 − 2 − 4 − 0 | E = 6E .
(c) It has to be a linear combination of 8, 4, 2, 1, thus [1 1 0 1] would yield 13E.
(d) With the same approach, [0 1 1 1].
SOLUTION 4.48. For this implementation we add an extra R-2R branch along with an extra summing
input to the op amp. From the theory developed in Example 4.9 the total resistance seen by the source is
2R. For the total power supplied by the source to be less than 0.02 W, we require R ≥E2
2 0.02( ) = 2.5 kΩ.
Thevenin Probs, 7/11/01 - P4.16 - @R.A. Decarlo & P. M. Lin
SOLUTION 4.49. The same steps as in the previous questions are repeated. Because the resistance seen
by the source is unchanged no matter how many branches are added to the R-2R network,
R ≥E2
2 0.01( ) = 5 kΩ.
SOLUTION 4.50. (a) If the input is 3vmax / 8, then the first comparator will give –Vsat, keeping S2 down.
The next comparator will output +Vsat, causing S1 to go up. After subtraction, the input to the last
comparator is vmax / 8 yielding +Vsat at its output since its input is slightly above the reference input
level. Thus the logic output values are [0 1 1].
(b) Putting in 6vmax / 8, will cause +Vsat and S2 to go up. The input to the second comparator will be
2vmax / 8, which will cause +Vsat and S1 to go up. The input to the last comparator will be 0, thus it will
output –Vsat. The corresponding logic output is [1 1 0].
SOLUTION 4.51. Simply add a subtractor and switch to the last comparator, followed by an additional
comparator. The reference level to the new (additional) comparator will be vmax− /16 , and its output will be
the new least significant bit.
SOLUTION 4.52. (a) Writing the node equation for figure P4.52c,
Vout
RL' +
(Vout − V1)
10k=
A(0 − V1) − Vout
Rout
which implies that
Vout / V1 =
1
10k−
A
Rout
1
RL' + 1
10k+ 1
Rout
For figure P4.52a, the corresponding node equation is
Vout
RL=
A(0 − Vin ) − Vout
Rout
which leads to
Thevenin Probs, 7/11/01 - P4.17 - @R.A. Decarlo & P. M. Lin
Vout / Vin =−
A
Rout
1RL
+ 1Rout
Note that 1
RL=
1
RL1 +
1
10k, which when substituted into the later equation make both of them
approximately the same since the 1/10k term in the numerator of Vout / V1 has a negligible contribution.
(b) Writing the node equation for figure P4.52d, yields
Vout
RL' +
(Vout − V2)
10k=
A(0 − V1) − Vout
Rout
Hence
V1 = V2100|| Rin
(100 || Rin ) +10k
≈
V2
101
Solving produces Vout / V2 =
1
10k−
A
101Rout
1
RL1 + 1
10k+ 1
Rout
. Note that as in (a) the 1/10k term in the numerator is
negligible; after eliminating this negligible term, one sees that Vout
V2 is 101 time smaller than
Vout
Vin.
SOLUTION 4.53. (a) Using the equation just derived, after substituting in the values, the gain is –980.392
(b) From the previous equation, Vout / V1 = −980.382 ; write KCL at the non-inverting terminal to obtain,
Vin − V1
10k=
V1
Rin+
V1 − Vout
10k; substitute Vout = −980.382V1; solve for V1 / Vin , and then multiply both gains
to obtain (Vout / V1)(V1 / Vin ) = Vout / Vin = −0.9979 .
(c) They only differ by about 0.01%, thus they are very similar.
SOLUTION 4.54. Writing out the transfer equation, Vout =R2
R1Vs2 −
R2
R1Vs1 , thus R2 / R1 = 4 . Using
R2 = 100kΩ , R1 = 25kΩ . As expected SPICE shows to noticeable difference in outputs when the source
resistances are varied.
SOLUTION 4.55. Due to the ideal nature of the op-amp, the voltage VRb= Vs2 − Vs1 . By KVL
Thevenin Probs, 7/11/01 - P4.18 - @R.A. Decarlo & P. M. Lin
V2 = Vs2 + Ra(Vs2 − Vs1)/ Rb
V1 = Vs1 − Ra(Vs2 − Vs1) / Rb
Next, V1 − V2 = (Vs1 − Vs2)(1+ 2Ra / Rb) .
SOLUTION 4.56. (a) Noticing that the final stage is a summing op-amp in which Vout =R2
R1V1 −
R2
R1V2 .
From the previous question, Vout =R2
R1(V1 −V2) =
R2
R1(1+ 2Ra / Rb)(Vs1 − Vs2) . Thus
α =R2
R1(1+ 2Ra / Rb). The gain α can be varied by adjusting the single resistance Rb.
(b) Picking the set of values below will satisfy the requirement:
R2 = 100kΩ, R1 =100kΩ, Ra = 20kΩ, Rb = 10kΩ .
(c) Doing the SPICE simulation using the parameters from (b) yield 5 V at the output for
Vs1 −Vs2 = 2 −1 V. Setting Rb arbitrarily to 20 kΩ, the output now becomes 3 V, which agrees with the
relationships developed earlier.
PROBLEM SOLUTIONS CHAPTER 5.
Solution 5.1. (a) Vs = 10 V, P = 20 W and P = Vs×Is implies Is = 2 A.
(b) Rin = Vs/Is = 10/2 = 5 Ω
(c) By the linearity/proportionality property: Vs
new
Vsold =
Isnew
Isold which implies
210
=Isnew
2 implies
Isnew = 0.4 A.
(d) Pnew = Vsnew × Is
new = 2 × 0.4 = 0.8 watts. Observe that
Pnew
Pold =0.820
≠Vs
new
Vsold =
210
It follows that the proportionality property does not hold for power calculations.
Solution 5.2 First note that the ratio IR/VS is constant. With the given values of voltage and current, this
ratio is:
IR/VS = 0.25/25 = 0.01
Power dissipated in the resistor is
P = IR2R = 2.5 è IR
2 = 2.5/R = 0.25 è IR = 0.5
Since IR is always 0.01×VS, it follows that VS = 50V.
Solution 5.3 Label the resistances R1, R2, and so on in the manner shown in Example 5.11. In this
problem, we have R1 to R10 (the last being the 2 Ohm resistance at the voltage source). First, assume that
V1 (the voltage across R1) is 1V. Then evaluate the rest of the currents and voltages until you deduce the
resulting VS. It should be noted that the equivalent resistance looking into R3, R5, R7, and R9 is always
2Ω.
V1 =1 ⇒ I1 =V14
= 0.25 ⇒ I2 = 0.25 ⇒ V2 = I2 × 2 = 0.5 V
V3 = V1 + V2 = 1.5 è I3 =V33
= 0.5 è I4 = I3 + I2 = 0.75 è V4 = I4 × 4 = 3
V5 = V3 + V4 = 4.5 è I5 =V53
= 1.5 è I6 = I5 + I4 = 2.25 è V6 = I5 × 4 = 9
V7 = V6 + V5 =13.5 è I7 =V73
= 4.5 è I8 = I7 + I6 = 6.75 è V8 = I8 × 4 = 27
V9 = V8 + V7 = 40.5 è I9 =V93
= 13.5 è I10 = I9 + I8 = 20.25 è V10 = I10 × 2 = 40.5
VS = V9 + V10 = 40.5 + 40.5 = 81 V
Thus, an 81 V input produces a 1 V output è Vout = (1/81)×VS = 2 V.
Solution 5.4 Label the resistances R1 to R10 progressively from right to left just like in the previous
problem. Then, assume Iout = 1 and proceed as follows:
Iout = 1 è V1 = Iout × 4 = 4 è I2 = V14
= 1
I3 = I1 + I2 = 2 è V3 = I3 × 4 = 8 è V4 = V3 + V1 = 12 è I4 =V43
= 4
I5 = I4 + I3 = 6 è V5 = I5 × 4 = 24 è V6 = V5 + V4 = 36 è I6 =V63
= 12
I7 = I6 + I5 = 18 è V7 = I7 × 4 = 72 è V8 = V7 + V6 = 108 è I8 =V83
= 36
I9 = I8 + I7 =54 è V9 = I9 × 4 = 216 è V10 = V9 + V8 =324 è I10 =V103
= 108
IS = I10 + I9 = 162
è Iout/IS = 1/162 è Iout = (1/162)×40.5 = 0.25A.
Solution 5.5 (a) MATLAB code given in problem.
(b) Subsitute to obtain V1 = 10V.
(c) Req = VS/IS = 11.6667Ω.
(d) First, define r1 = 1:0.25:10;
then create an outermost loop around the code of part (a) as: for j=1:length(r1)
then, in the statement defining R, do R = [R1(j), R2, R3, R4, R5, R6, R7, R8]’;
Finally, replace the last statement with Vs(j) = V(n) + V(n-1); end;
The following is the resulting plot:
Solution 5.6 (a)
The following code can be used:
n = 9;
v = zeros(n,1);
i = zeros(n,1);
r = [r1 r2 r3 r4 r5 r6 r7 r8 r9]’;
i(1) = 1;
v(1) = i(1)*r(1);
i(2) = i(1);
for k=2:2:n-2
v(k) = r(k)*i(k);
v(k+1) = v(k)+v(k-1);
i(k+1) = v(k+1)/r(k+1);
i(k+2) = i(k+1) + i(k);
end;
v(8) = i(8)*r(8);
v(9) = v(8) + v(7);
i(9) = v(9)/r(9);
Is = i(9) + i(8);
It follows that Is =16.9877A.
(b) By the proportionality property: I1new
I1old =
Isnew
Isold → I1
new =1
16.9877× 200mA =11.77mA
(c) Req = v(9)/Is = 38.15Ω.
Solution 5.7 Va = 12V, iB = 60m
By inspection:
Vout_a = 300/900×12 = 4V
Vout_b = (300||600)×60m = 12
_ Vout = 4 + 12 = 16V.
*SOLUTION 5.8. Part 1: Set the 3 A current source to zero. This generates an open circuit in place of
the current source eliminating the effect of the series 0.1 S resistor. The equivalent circuit is:
By voltage division,
VL12V =
1
0.25 + 0.2 + 0.051
0.25 + 0.2 + 0.05+
1
0.1
×12 =2
2 +10×12 = 2 V
Part 2: Set the 12 V source to zero. This generates a short circuit in place of the voltage source which
shorts out the effect of the 0.5 S resistor. The equivalent circuit is:
Note that the 0.1 S resistor in series with the 3 A source is redundant to the calculation of VL. Hence, by
Ohm's law,
VL3A =
1
0.25 + 0.2 + 0.05 + 0.1× 3 =
3
0.6= 5 V
Therefore by superposition,
VL = VL12V + VL
3A = 2 + 5 = 7 V
Solution 5.9 Replace the dependent source by an independent voltage source VS:
In the following analysis, we have to always compute Va because It defines the constraint on Vs. So, when
only the 88V source Is active, VA Is the result of voltage division between the 60||30 Ω resistor and the
120||30 Ω resistor. So,
Va_1 = 40V
And, since deactivated VS, Vout_1 = 0.
Now, due to the 55V source, we have
VS
Now, the 120 and 60 Ω resistors are in parallel, and the same can be said about the 30 and 30 Ω resistors.
Thus, another voltage divider gives:
Va_2 = – 15V and Vout_2 = 0V.
Finally, when VS1 is active, the left part of the circuit consists only of resistances, so Va_3 = 0. Vout is given
by another divider formula:
Vout_3 = 90/100×VS
Now add all contributions:
Va = 40 –15 + 0 = 25V
Vout = 0 + 0 + 0.9VS, where VS = 2Va.
Vout = 0.9×2×25 = 45V.
Finally, P = V2/R = 22.5W.
Solution 5.10
Due to 3A source:
iout = 1A by current division between the two paths. So, vout_1 = 2 V.
Due to 1A source:
iout = 2/3A again by current division. So, vout_2 = 4/3V.
Due to 1vV source:
iout = 14/6 = 7/3A (by Ohm’s Law). So, vout_3 = 14/3V.
Finally, vout = 6/3 + 4/3 + 14/3 = 8V, and the power delivered by the source is 8×1 = 8 W.
Solution 5.11
Due to 22 V source:
Req = 900||225 = 180. Now, by voltage divider:
Vout_1 = 0.5×22 = 11 V.
Due to the 20 V source:
Req = 180||225 = 100 Ω. So, again, by voltage division:
Vout_2 = 900/(900+100)×20 = 18 V.
Finally, due to current source:
We have three resistances in parallel with a resistance equal to 90 Ω. So, Vout_3 = 0.1×90 = 9 V.
Vout = 11 + 18 + 9 = 38V and P = 38×38/900 = 1.6 W.
Solution 5.12
Find contribution to Vout :
First, note that no current flows through R1 - R3 because of the virtual ground property of the op-amp.
Thus, this circuit is identical to the inverting amplifier studied in Chapter 4. So,
0
0
0_ VR
RV f
out −=
Similarly, when each of the other sources is activated, the circuit will be an inverting amplifier. So,
Vout _1 = −R f
R1V1, Vout _2 = −
Rf
R2V2, Vout _3 = −
Rf
R3V3
Vout = −RfV0R0
+V1R1
+V2R2
+V3R3
Solution 5.13 Due to the 4 V source, the circuit looks like an inverting amplifier:
So, Vout_1 = -30/10×(-4) = 12 (from the results of Chapter 4).
Again, note here that no current flows through the two resistances connected to the + terminal of the op
amp. Since no current flows through them, then no voltage develops across them. So, the + terminal can be
assumed to be connected to ground, and this is why we say that the circuit looks like that of the inverting
amplifier.
Now, due to the 6 V source:
The voltage at the + terminal is established by a resistive divider between the two 10K resistors. So, this
voltage is 3V. Thus, the voltage at the negative terminal is also 3V. We can now use KVL on the inverting
side of the op-amp to get:
Vout_2 = 3 + 30k×0.3m = 12V
So, Vout = 12 + 12 = 24V, and P = (24)×(24)/500 = 1.15W.
Solution 5.14
(a) When VS2 is deactivated. The circuit looks like two inverting amplifiers in cascade.
Thus, by inspection, V1 = –2VS1 and Vout_1 = –3V1 = 6VS1 = 3V.
(b) Similarly, when VS1 is zero, V1 is zero because the first inverting amplifier has zero input. Thus, the
circuit consists just of the second inverting amplifier:
Vout_2 = – 3/2VS2 = – 4.5
(c) Vout = 3 – 4.5 = – 1.5V.
*SOLUTION 5.15. For Vs1 and Vs2 , the situation reduces to the analysis of two inverting amplifiers in
cascade. For Vs3 , the situation is simply a single stage inverting amplifier. Note that because of the
virtual ground at the inverting terminal of the op amp, when Vs1 and Vs2 are zero, they have no
contribution to the input of the second stage.
(a) With Vs2 and Vs3 set to zero,
Vouts1 =
−R
R
−2R
2R
Vs1 =Vs1 = 5 V
(b) With Vs1 and Vs3 set to zero,
Vouts2 =
−R
R
−2R
R
Vs2 = 2Vs2 = 2 × (−2.5) = −5 V
(c) With Vs1 and Vs2 set to zero,
Vouts3 =
−RR
Vs3 = −Vs3 = −2 V
(d) By superposition,
Vout = Vouts1 + Vout
s2 + Vouts2 = −2 V
Solution 5.16
If the op-amp were ideal, we would get:
Vout = – 4VS1 – 2VS2 = – 26
This is clearly beyond the linear range of operation of the op-amp. In other words, the amplifier responds
in a non-linear manner to this level of input. Hence, superposition, which relies on linearity, cannot be
used.
Solution 5.17
We know by linearity that
Vout = aIs1 + bVs2
Substitute the first measurement to obtain:
5 = 0 + b×10 è b = 0.5
Now, take the second measurement:
1 = a×10 + 0 è a = 0.1
So,
Vout = 0.1Is1 + 0.5Vs2
At 20A, 20V: Vout = 12V
Solution 5.18
Again Vout = aIs1 + bVs2
Substitute the two measurements to obtain:
5a + 10b = 15
2a + 5b = 10
These two simultaneous equations in a and b can easily be solved to obtain:
a = – 5 and b = 4
Therefore, at 1A and 5V,
Vout = – 5 + 20 = 15V
Solution 5.19 (a)
Again,
Vout = aIs1 + bVs2
Substituting the result of the first measurement into this equation yields:
a×4×cos(2t) + 0 = –2cos(2t) è a = – 0.5
Now, substitute the second measurement:
0 + 10b = 55 è b = 5.5
Therefore, for the given input current and voltage:
Vout = –cos(2t) –55cos(2t) = –56cos(2t) V
(b) Vout = 2cos(5t) + 110cos(5t) = 112cos(5t) V
Solution 5.20 (a)
First of all,
Iload = aVa + bIb
Now, substitute the two measurements into this equation:
7a + 3b = 1
9a + b = 3
Solving these two equations for the unknowns a and b, we get
a = 0.4 and b = – 0.6
(b) Iload = 0.4×15 –0.6×9 = 0.6A
Solution 5.21 (a)
We know that the output is going to be a linear combination of the three inputs:
Vout = aIs1 + bVs2 + cVs3
Now, substitute the three measurements into this relationship:
50ma – 2b + 5c = – 13
0 + 3b + 5c = 2
0 + 2b + 4c = 0
These equations can be written in matrix form and solved as follows:
0.05 −2 5
0 3 5
0 2 4
a
b
c
=−13
2
0
⇒
a
b
c
=100
4
−2
(b) Substitute the given values of the input sources to obtain:
Vout = 100×1 – 4×40 + 2×10 = – 40V
Solution 5.22
Vout = AIs1 + BVs2 + CVs3
Substituting the measurements into this equation results in a system of three equations and three
unknowns. This system can be written in matrix form by inspection:
30 ⋅10−3 2 −1
−20 ⋅10−3 4 2
−10 ⋅10−3 −3 1
A
B
C
=11.5
27
−14
⇒
A
B
C
=150
5.5
4
This can be solved to obtain:
Solution 5.23 (a)
The coefficient matrix is inverted, and both sides of the nodal equation are multiplied by it to obtain:
Va
Vb
=
43.0108 0.233
43.0108 −1.4337
0.02Vs1 − 0.00125Vs2
Vs2
Expanding the first row of the above equation gives:
Va = 0.8602Vs1 – 0.0538Vs2 + 0.233Vs2
This is exactly in the form required, where A = 0.8602 and B = 0.1792
(b) For this part, we expand the second row of the equation:
Vb = 0.8602Vs1 – 0.0538Vs2 – 1.4337Vs2
Again, this is in the desired form, where A = 0.8602 and B = – 1.4875
(c) Vab = Va – Vb = 0 + 1.667 Vs2
Thus, A = 0 and B = 1.667
Solution 5.24 (a)
Again, we invert the coefficient matrix to obtain
I1I2
I3
v1
v2
=
0.0022 0.0022 0.0022 0.6296 −0.0741
0.0022 0.0022 0.0022 −0.7037 0.2593
0.0022 0.0022 0.0022 −0.3704 −0.0741
−0.2222 0.7778 −0.2222 137.037 −92.5926
−0.3333 −0.3333 0.6667 −11.1111 77.7778
Vs1
0
0
Is2
0
Expanding the first row of this equation:
I1 = 0.0022Vs1 + 0.6296Is2
So, A = 0.0022, B = 0.6296
(b) Similarly, expanding the third row:
I3 = 0.0022Vs1 + -0.3704Is2
So, A = 0.0022, B = -0.3704
(c) By the same procedure, A = -0.3333 and B = -11.111
Solution 5.25
Invert the coefficient matrix and mujltiply both sides of the equation in the problem by this inverse matrix
to obtain
V1
V2
V3
Ia
Ib
Ic
Id
=
0 0 0 −3 2 −1 −4
0 0 0 −3 1 −1 −4
0 0 0 12 −3 3 16
0 0 0 −4 1 −1 −5
0 −1 0 −16.6 9.2 −5.2 −21.8
1 1 0 −42.1 −16.2 11.7 53.8
1 1 1 26.1 −12.2 7.7 32.8
Is1
0
0
0
Vs2
0
0
The second to last row can be expanded to get Ic = Is1 – 16.2Vs2 è A = 1, B = –16.2
Solution 3.26 (a)
First compute the response due to V1:
By voltage divider:
Vout_1 = 6/(2+6+1)×V1 = 2/3V1
Then, due to I2:
I2 flows through the 20 ohm resistor in series with 6||3. Thus,
Vout_2 = –2I2
Therefore, Vout = 2/3V1 – 2I2
(b) Vout = 8cos(10t) – 4
*SOLUTION 5.27. (a) By linearity Vout = A Vs1 + B Is2 .
To find A, let Is2 = 0. The circuit becomes a ladder network as follows.
Let VoutA = 1 V. Then
»I1 = 1/420 + 1/70
I1 = 1.6667e-02
»V400 = I1*40 + 1
V400 = 1.6667e+00
»I400 = V400/400
I400 = 4.1667e-03
»I2 = I1 + I400
I2 = 2.0833e-02
»Vs1 = I2*20 + V400
Vs1 = 2.0833e+00
»A = 1/Vs1
A = 4.8000e-01
To find B, let Vs1 = 0. The circuit becomes a ladder network as follows.
Again assume that VoutB = 1 V. Then
»I1 = 1/420 + 1/70
I1 = 1.6667e-02
»V400 = I1*40 + 1
V400 = 1.6667e+00
»I400 = V400/400
I400 = 4.1667e-03
»I20 = V400/20
I20 = 8.3333e-02
»Is2 = I20 + I400 + I1
Is2 = 1.0417e-01
»B = 1/Is2
B = 9.6000e+00Hence by linearity Vout = 0.48 Vs1 + 9.6 Is2 .
(b)
»Vout = A*20 + B* 0.5
Vout = 1.4400e+01
(c) Doubling resistances does not change voltage ratios hence A is the same. However, the doubling also
doubles the voltage to current ratio. Hence, B is doubled. It follows that if all resistances are
doubled, thenVout = 0.48 Vs1 + 19.2 Is2
Solution 5.28 (a) (b)
We solve parts (a) and (b) at the same time. First, we find the responses to Vs1:
Equivalent resistance across Vo2: 60||30 = 20
Now, by voltage divider: Vo2_1 = 20/140Vs1
This voltage now divides between Vo1 and the 10 ohm resistance:
Vo1_1 = 20/30Vo2_1 = 2/3×1/7Vs1 = 2/21Vs1
Now, compute the responses due to Is2:
Equivalent resistance across Vo2: 60||120 = 40Ω
By voltage division: Vo2_2 = 40/50Vo1_2
Where Vo1_2 = Is2×(50||20) = 14.286Is2
Now, add the contributions:
Vo1 = 0.0952Vs1 + 14.286Is2
Vo2 = 0.1429Vs1 + 11.4288Is2
Solution 5.29 (a)
Find contribution due to v1:
The parallel combination results in a 12 ohm resistance in series with the remaining two. Thus, by voltage
division:
vout1 = 24/48v1 = 0.5v1
Now, due to v2
The equivalent resistance across v in this figure is 36||(24+12) = 18, which means that by voltage divider:
v = 0.5v2
Similarly, by another voltage division application
vout2 = 24 / 36 × v= 0.66×0.5v2 = 0.333v2
Now, due to i3
Define Req1 = 24 + 36||18 = 36 Ω. This resistance is in parallel with the 12 ohm resistance to introduce
an equivalent of 9 Ω. The total voltage that develops across this 9 ohm resistance is –9is3. This voltage
divides between vout and the 36||18 resistance:
vout3 = 24/36v1 = -9×24/36is3 = -6is3
Finally, due to i4:
A similar analysis of this resistive network can reveal that vout4 = 6is4. Thus
vout = 0.5v1 + 0.333v2 – 6is3 + 6i s4
(b) For this part, note that scaling resistance values does not affect voltage ratios. This can be evident from
the application of any voltage divider formula. On the other hand, scaling resistances does affect current-
to-voltage or voltage-to-current ratios. This is by definition of a resistance! So, in the above equation,
doubling the resistances does not affect the first two terms, but doubles the second two terms.
Solution 5.30 (a)
Define two clockwise mesh currents: I1 in the bottom left loop and I2 in the top loop. The bottom right
loop has a current source, so it will not be considered:
Va – (I1-I2)3 – I2 – (I1 – ib) = 0
and
6I2 – I2 + (I2 – I1)3 = 0
Solving these two equations for the two currents gives:
I1 = 8 A and I2 = 3 A
The power delivered by the dependent source is: P = ix × I ix where I ix is the current leaving the ' +'
terminal of the dependent voltage: P = ix × (I2 − I1) = I2 × (I2 − I1) = −15W
è vout = I1 + ib = 8 + 26 = 34V
(b) Now, we express vout = Ava + Bib
Turning off ib, we still have two loops, in which we can define the same mesh currents as above to obtain:
va – 3(I1 – I2) – I2 – I1 = 0
I2 – 3(I2 – I1) – 6I2 = 0
Solving for the two currents, we get I1 = 4/13va which sets up vout across the 1 ohm resistor:
Vout1 = 4/13va
Now, turn off the voltage source:
Write a node equation at vout:
bout
b
voutoutout
iv
ivvv out
13
9
0316
2
6
=?
=−−
++
So, vout = 4/13va + 9/13ib
(c) Substitute into the above equation: vout = 35V
Solution 5.31 By inspection:
where the leftmost current is 0.25 with a resistance of 32Ω. Similarly, the downward current is 0.75 and its
resistance is 32Ω. This reduces to:
Solution 5.32
The circuit can be transformed as follows:
Write two nodal equations at 1 and 2:
05166.9
92
02
92
1
2212
211
=−+++−
=−+−+−
mk
vk
vm
kvv
kvv
mk
vm
These two equations can be solved using any method to obtain:
v1 = 11.2V and v2 = 2.4V
The power absorbed by the 9.6kΩ resistor is:
P =V2
2
9.6 ×103 = 0.6mW
Solution 5.33
Then
Output voltage is 5V
(b) P = 1.25W
(c) For a given resistance, doubling the voltage increases the current by two times. So, the current is
doubled. It follows that Voutnew = 2 × Vout
old = 10V
Solution 5.34
This circuit is easy enough to solve by inspection. Vs = 28V.
Solution 5.35
Now, write two node equations:
mv
vvv
mvvv
80125
032.0500500
12500500500
21
12
121
=++−
=+−
These two equations can be solved to get v1 = 2.8V, v2 = -0.4V.
Solution 5.36
Now, we can write the node equations:
02.01.011005050100
100100
095.020
4.02.025
100255050
2122
1121
=−−++−+−
=−+++−+−
vvvv
vvvv
Solving these two equations yields:
v1 = 25V and v2 = 20V
*SOLUTION 5.37. After a source transformation on the 30 V independent source and one on the 9Vx
dependent source we obtain the circuit below.
Transforming the two dependent voltage sources and combining yields the following circuit.
Writing a single node equation for Vx yields
7.5 =Vx
4+
Vx
2+
Vx
2= 1.25Vx
Hence, Vx = 6 V.
Solution 5.38
Replace the dependent source with a temporary independent source. When doing the analysis, always
compute Va in order to keep track of the constraint on the dependent source.
When Vt is not active, vout is obtained from a voltage divider between the 60||30 combination and the 15
ohm resistor:
vout1 = 4/7Vs
Similarly va1 = 3/7Vs
Now, compute the response due to the temporary source.
Straightforward voltage division also applies here to get:
va2 = -2/7Vt
and vout2 = -5/7Vt
So, vout = 4/7Vs – 5/7Vt
va = 3/7Vs – 2/7Vt, where Vt = µva è va = 3/7Vs – 2µ/7va
Rearranging,
sa Vv27
3
+=
Then,
vout =4
7Vs −
5µ7
va
vout =4 − µ7 + 2µ
Vs
Solution 5.39 Replace the dependent source by a temporary independent source:
When Vt is shorted, the result is a ladder network. The input resistance looking each of the vertical
branches is R. Label these vertical branches V1, Va, V2 from left to right. It follows that
V1 = R/3RVs
_ Va_1 = R/2R × R/3RVs = 1/6Vs
Also, Vout_1 = 0
Now, short VS and turn on Vt.
Again, the result is a ladder network (note the symmetry in the above figure). Thus, we can write by
inspection:
Va_2 = 1/6Vt
Vout_2 = Vt
Adding the contributions:
Vout = 0 + Vt = Vt = µVa where we have substituted the constraint on Vt
and Va = 1/6Vs + 1/6Vt = 1/6Vs + 1/6 µVa
è Va = [1/(6 – µ)]Vs
Substitute this Va into the expression for Vout:
sout VV−
=6
Solution 5.40
The first step is to replace the dependent source with a temporary independent source. Then, superposition
can proceed as usual.
Now, let’s short the temporary source, Vt. Again, this network is a ladder network, like the one in the
previous problem. However, now, the equivalent resistance looking into each of the vertical branches (from
left to right) is different. Now, it is
Req = 20R||(R+4R) = 20R×5R/25R = 4R
Now, again, define the voltages across these three vertical branches (from left to right) as V1, Va, V2. It
follows by voltage division that
V1 = 4/9Vs
è Va_1 = 4/5 × 4/9 × Vs
è Va_1 = 16/45 Vs
It should be noted that Vout = 0.
Now, short the input source and find the response due to Vt. Again, in this case, the circuit is identical to
the case when Vs was active (note the symmetry in the above figure). Therefore,
Va_2 = 16/45 Vt
Vout_2 = Vt
Adding the two contributions:
( ) sa
asa
VV
VVV
−=?
+=
1645
145
16
45
16
andsout
atout
VV
VVV
164516−
=
==
Solution 5.41
First, replace the controlled source by a temporary independent source:
Now, do a source transformation on the Vt source:
The equivalent resistance seen by the current source is 99Ω. Therefore:
Vout = -99/100Vt = -0.99Vt
Now, Vt = µVg and IL = -0.99Vt/990k where IL is the current through R2
è VR2 = ILR2 = (100k/990k)×(-0.99Vt) = -0.1Vt
By KVL
Vg = VR2 + Vs = -0.1Vt + Vs = -0.1µVg + Vs
Rearranging
sg VV11.0
1
+=
Finally, and substituting:
91.8
1.011
99.0
−=
+−=
s
out
s
out
VV
VV
Solution 5.42
Ra
Rbb
xaa
xbb
V
V
in
out
1
10
10
10
1++=
++=
At R = 0
20
201
0
0
=?
==
b
b
V
V
in
out
At infinite R
11
1
1
8
8
ab
a
b
V
V
in
out
=?
==
Finally, at R = 10
4
5.0
101018020
1
1
1
1
=?=?
=++
b
a
aa
Substituting these values for R = 2 yields
1425.01
2420 =↔+↔+=
in
out
V
V
SOLUTION 5.43. From the chapter
Vout
Vin=
b0 + b1µa0 + a1µ
=b0 + b1µ1+ a1µ
Equivalently,
−µVout
Vina1 + b0 + µb1 =
Vout
Vin
Plugging in the data yields three equations in three unknowns which in matrix form are:
0 1 0
−154 1 1
−168 1 2
a1
b0
b1
=264
154
84
»A=[0 1 0;-154 1 1;-168 1 2]
A =
0 1 0
-154 1 1
-168 1 2
»y = [264 154 84]'
y =
264
154
84
»Coefs = A\y
Coefs =
2.8571e-01
2.6400e+02
-6.6000e+01
Therefore, a1 = 0.28571, b0 = 264, and b1 = –66. Making a1 = 1, will yield a different set of answers.
When m = ∞, Vout/Vin = b1/a1 = –231.
Thevenin Probs, 7/24/01 - P6.1 - @R.A. Decarlo & P. M. Lin
PROBLEM SOLUTIONS CHAPTER 6
SOLUTION 6.1. (a) Voc is found by removing RL and doing voltage division.
VVVOC 28100600
600
300)700||600(
)700||600(63 =
+×
+=
RTH is found by setting the source to zero and by calculating the equivalent resistance seen looking back
between the A and B terminal.
Ω=+= 200600||]100)600||300[(THR
(b) Using
LL IRP ⋅=
the power for each resistance may be found by substituting the appropriate RL in the following equation.
L
LTH
OC RRR
VP ⋅
+
=2
For 50 Ω, 200 Ω, and 800 Ω , the power obtained is 627.2 mW, 980.0 mW, and 627.2 mW respectively.
The use of Thevenin equivalent does reduce the effort in obtaining the answer.
SOLUTION 6.2. (a) To find RTH, open circuit the current source and short-circuit the voltage source. The
resulting resistance seen from terminal A-B is 1 kΩ. Using superposition, the contribution of the current
and voltage source at the open circuit output may be summed as 30 V (2 k/4 k) + 10 mA (2 k/4 k) (2 kΩ).
VOC is then 25 V and ISC = VOC/RTH is 25 mA.
(b) Following is a plot of
L
LTH
OC RRR
VP ⋅
+
=2
for RL from 100 Ω to 4 k Ω.
Thevenin Probs, 7/24/01 - P6.2 - @R.A. Decarlo & P. M. Lin
SOLUTION 6.3. (a) Turning off the two sources
RTH = (60 + 60)||40 = 30 Ω,
and using superposition
VOC = 6V40
40 + 60 + 60
+ 0.1A
60
40 + 60 + 60
40 = 3 V.
ISC is the obtained as 100 mA.
(b) Using
L
LTH
OC RRR
VP ⋅
+
=2
a load of 90 Ω will absorb 56.25 mW.
(c) It absorbs 75 mW; hence the 30 Ω resistor absorbs more power.
SOLUTION 6.4. As both resistor divider ration are the same (3/6), the voltage at A and B is the same
resulting in a VOC of 0 V.
RTH = (3K | |6 K) + (9K ||18K) = 8 kΩThe relation VOC/ISC cannot be used in this situation.
SOLUTION 6.5. Using superposition
VOC = 2020 ×103
20 ×103 + 5 ×103
−105 ×103
20 × 103 + 5 ×103
+ 20 ⋅ 20sin(50t) = 14 + 400sin(50t) V,
and
Thevenin Probs, 7/24/01 - P6.3 - @R.A. Decarlo & P. M. Lin
RTH = (20k | |5k) + 20k = 24 kΩ.
ISC can then be found using the VOC/RTH relationship as 0.58+20sin(50t) mA.
SOLUTION 6.6. Once again using superposition
VKmAKKKK
KmA
KKK
KVVOC 2042546
584
454
486
672 =⋅+⋅
+++
++= ,
and
Ω=+++= KKKKKKRTH 1024]6||)84[(
SOLUTION 6.7. Using source transformation, (a) is obtained from the original circuit. Then combining
in series the resistors and voltage sources, and retransforming them (b) is obtained. Finally adding the two
currents and transforming back the circuit to its Thevenin form (c) is obtained.
From (c),
WRIP
mAkkI
kR
VV
LL
L
TH
oc
15.0
5)66/(60
6
60
2 ==
=+=Ω=
=
Thevenin Probs, 7/24/01 - P6.4 - @R.A. Decarlo & P. M. Lin
SOLUTION 6.8. First, each source in series with 2R, can be replaced by an up going current source of
RVx 2/ in parallel with 2R. Then starting from the left, the two 2R in parallel are combined and then
retransformed to a voltage source of 2/oV in series with 2R once added to the series resistance. Repeating
the previous steps,
RRVVVVRRVRVRVRV
RVRRRVRVRVRVVV
RRVVVRRVRVRV
RVRRRVRVRVV
RRVVRRVRV
RVRRRVRV
oo
oo
oo
oo
oo
oo
+++++?+++++?+++
++++?+++?++
+++?+?+
2/4/8/16/||2/4/8/16/
2/||2||2||4/8/16/22/4/8/
2/4/8/||2/4/8/
2/||2||2||4/8/22/4/
2/4/||2/4/
2/||2||2||4/22/
321321
32121
2121
211
11
1
Thus 2/4/8/16/ 321 VVVVV ooc +++= and RRTH 2= .
SOLUTION 6.9. First we find RTH
RTH = 2K + [ 6K | |3K( ) + 15K ||10K( )]| |24 K = 8KΩ
Use nodal analysis to solve for VOC. At node a
1mA =1/ 6K(Va − Vc ) +1/ 3K(Va − Vd ) + 1/ 24K(Va) or
(1/ 6K +1/ 3K +1/ 24K)Va −1/ 6K(Vc) −1/ 3K(Vd ) = 1m
doing the same for the supernode, and the equation inside the supernode yields the following two
equations:
(1/ 6K +1/ 3K)Va − (1/ 6K +1/15 K)Vc − (1/ 3K +1/10 K)Vd = 0, and
Vc − Vd = 30
using these three equations, it is now possible to solve for the three unknowns VA, VC, and VD. Using
MATLAB they are respectively 4.5V, 22.8750V, and -7.1250V. VOC being VA, 4.5V.
a
c d
SOLUTION 6.10.
RTH = [(18K | |9K) + 66K]| |36K = 24K
Thevenin Probs, 7/24/01 - P6.5 - @R.A. Decarlo & P. M. Lin
By superposition, noting that all resistances are in k-ohms,
VOC =66+(9||18)
66+(9||18)+36⋅36 ⋅ 2.5 +
(66 + 36)||9
[(66 + 36)||9] +18⋅
36
36 + 66⋅18 −
36
36 + (9||18) + 6630 = 52 V
When a 2 kΩ is connected, the current IL becomes VOC/(RTH+2k) = 2mA; thus the power absorbed is 8
mW.
SOLUTION 6.11. (a) Introduce a test voltage source at the output, and write the nodal equations in matrix
form:
(1/100 +1/ 50) −1/ 50 −1/100
−1/ 100 −1/100 (1/ 100 +1/ 200 +1/100)
−1/ 50 (1/ 50 +1/100 +1/100) −1/100
Vtest
VC
VD
=itest
0
Vs / 100
Solving we obtain, Vtest = 100itest + 2Vs / 3. From eq. 6.10 RTH =100 Ω, voc = 2Vs / 3.
C D
(b) To obtained the power the following equation is used, P =voc
RTH + RL
2
⋅ RL .
Thevenin Probs, 7/24/01 - P6.6 - @R.A. Decarlo & P. M. Lin
0 20 40 60 80 100 120 140 160 180 2000.3
0.4
0.5
0.6
0.7
0.8
0.9
1
1.1
1.2
Load resistance in Ohms
Po
we
r/V
s in
mW
SOLUTION 6.12.
VOC = 0, as no independent sources are present. Writing the following nodal equation where vx is the
voltage across both ports,
ix = ((vx − iz ) − ix ) + (vx − iz )/ 2, RTH can be found as vx / ix =2.5 +
1.5,
SOLUTION 6.13.
Defining vi and ii as the voltage across and current into the input ports, writing the nodal equation at the
input: ii + gmvx = 1/ 200K(vi − vx ) . We can also get the following equation vx = 200K(gmvx + ii) . Using
the previous two equations we can solve for RTH = vi / ii =−200K 1+200K ⋅ gm
1 + 200K ⋅ gm
+1
1 + 200K ⋅ gm
.
This yields a gm of 10 µS.
SOLUTION 6.14.
ISC is null as no independent source are present. To find RTH vi and ii are defined as the voltage across and
current into the input ports. Writing the nodal equation we get:
ii = Vx / 1.8K + (1/ 200)(Vx − 3 / 4Vx) , and vi = Vx −300 ⋅ ii . Solving RTH = vi / ii =−600Ω .
SOLUTION 6.15.
First, write out the equation around loop 1:
Thevenin Probs, 7/24/01 - P6.7 - @R.A. Decarlo & P. M. Lin
Vx = i1 ⋅1200 − i2 ⋅100 − i3 ⋅300 . Then substituting the following relationships,
i1 = Ix , i2 = 2I x , and i3 = 0.01Vx + i2 = 0.01Vx + 2Ix , and solving for Vx / Ix = RTH = 100Ω . VOC = 0 as no
independent sources are present.
1
2
3
SOLUTION 6.16.
Introduce a test source at the output terminals, and write out the nodal equations in matrix for the top node,
and the supernode comprised of the current controlled voltage source (ccvs). vb is the node left of ccvs, and
vc the node to the right.
1/ 30 +1/ 20 −1/ 20 −1/ 30
−1/ 20 −1/ 30 + 0.4 1/ 20 − 0.4 2 / 30
1/ 6 −1 5/ 6
vtest
vb
vc
=itest
0
0
Solving, we obtain vtest =−90itest , thus RTH =−90 Ω.
b c
SOLUTION 6.17.
(a) Turn off independent source. Introduce a test source and write loop equation:
vx = 6ix − 4i1 +10ix . Note that i1 = ix . Now solve for vx / ix = RTH = 12Ω .
Thevenin Probs, 7/24/01 - P6.8 - @R.A. Decarlo & P. M. Lin
Short the input and write the loop and nodal equations:
3A = i1 + ISC
10i1 = 4i1 + 6ISC
Solving yields ISC = 1.5A, and VOC = ISC RTH = 18V .
(b) In MATLAB the following plot is generated: P =VOC
RTH + RL
2
⋅ RL for 1 ≤ RL ≤ 24Ω.
0 5 10 15 20 251
2
3
4
5
6
7
Resistance in Ohms
Pow
er
in W
atts
Maximum power is absorbed by 12 Ω load.
SOLUTION 6.18.
To find thevenin resistance, introduce a test source and write the following equations:
vs = 20i1 + 40i1 + 40i1 =100i1, and is = vs /100 + i1. Solving for vs / is = RTH = 50Ω .
Next, use the following nodal equation; 0.2A = i1 + ISC , and loop equation 20i1 + 40i1 = 40ISC . Solving
using these two equations yields ISC = 0.12A , and consequently VOC = RTHISC = 6V .
SOLUTION 6.19.
(a) Introduce a test source, vs, and get the following two equations: vs −15is =−Vx , and
kVx + Vx / 3 + Vx / 5 + is = 0. Solving yields vs / is = RTH =15 +1
k +8 /15
, or 95/6 Ω for k = 2/3.
Next, write the following nodal equation kVx + Vx / 3 + (Vx −1) / 5 = 0 and observe that VOC = 1V − Vx .
Thus solving yields VOC = 1−1/ 5
k + 8
15
= 5/ 6V for k = 2/3, and ISC = VOC / RTH =1/19 A .
Thevenin Probs, 7/24/01 - P6.9 - @R.A. Decarlo & P. M. Lin
(b) Solving the previously obtained equation for k when VOC = 0, yields k = -1/3, and consequently
RTH = 20 Ω.
SOLUTION 6.20.
Introduce a test source, vt, and get the following equations: it + ix = ix ⋅Vs − 300ix
300⇒ Vs = 300it + 300ix ,
and vt = Vs − 300ix +100it . Solving yields vt / it = RTH = 400Ω .
Observe how no current flows through the 300 resistor in parallel with the dependent source. Thus VOC
will always be 0 V and is independent of VS.
SOLUTION 6.21. (a) For this part, consider the modified circuit below.
Step 1: Applying KCL to node A, we have
1
400VA −100( ) +
VA
800+ 0.1 − Is = 0
Multiplying through by 800 yields
3VA = 800Is +120
Step 2. Computing Vs, we have
Vs = VA − 2000ix = VA −2000
400100 − VA( ) = 6VA − 500
Hence
Vs = 6800
3Is + 40
− 500 =1600Is − 260 = RthIs + Voc
Therefore Rth = 1.6 kΩ and Voc = −260 V.
(b) By linearity, Voc = −130 V.
Thevenin Probs, 7/24/01 - P6.10 - @R.A. Decarlo & P. M. Lin
SOLUTION 6.22.
Introduce a test source, vt,, and get the following two equations: vt = (30m − V1 / 100)400 − V1 = 12 − 5V1 ,
and it = 0.06V1 + 30m − V1 / 100 = 30m + 0.05V1 . Solving obtain vt = −100it +15. Thus RTH = −100Ω ,
and ISC = VOC / RTH =−150mA .
SOLUTION 6.23. Insert Itest as per text. Hence
1
40+
1
120
VC −
16
40= Itest
Solving, we obtain
VC = 30Itest +12
By KVL,
Vtest = VC + 30 × 2i1 + Itest( ) = VC +60
4016 − VC( ) + 30Itest =−0.5VC + 30Itest + 24
Substituting for VC yields
Vtest = 15Itest +18
Hence, Voc =18 V and Rth = 15 Ω. Thus i =18
75 + 15= 0.2 A. Further, P75 = 75(0.2)2 = 3 W.
C
SOLUTION 6.24.
(a) and (b)
Thevenin Probs, 7/24/01 - P6.11 - @R.A. Decarlo & P. M. Lin
vs
-
+
-
+
v 2
1
(c) Writing KVL around loop 1, loop 2, and finally relating v to vx ,
vs = 10(0.02 − I1) + v − 20I1v =−5I2 −15(I2 − 0.02)
3vx = I1 − I2 = 3[10(0.02 − I1)] = 0.6 − 30I1
(d) Rewriting these in matrix form,
−30 0 1
0 20 1
31 −1 0
I1
I2
v
=1
0
0
vs +
−0.2
0.3
0.6
(e), (f), and (g). Using MATLAB,
»A=[-30 0 1;0 20 1; 31 -1 0];
»b1 = [1 0 0]';
»b2 = [-.2 .3 .6]';
»I1 = inv(A)*b1
I1 =
-1.5385e-03
-4.7692e-02
9.5385e-01
»I2 = inv(A)*b2
I2 =
1.9231e-02
-3.8462e-03
3.7692e-01
»Rth=-1/I1(1)
Rth = 650
Note: Isc = I2(1) = 0.01923 A
»Voc= -Rth*I2(1)
Thevenin Probs, 7/24/01 - P6.12 - @R.A. Decarlo & P. M. Lin
Voc =
1.2500e+01
(h)
SOLUTION 6.25.
(a)
(b) Write four nodal equations,
is = (VA − VC ) / 2k
is + 1m = (VC − VD )/ 6k + (VC − VE )/ 3k
VE / 2k = (VC − VD) / 6k − VD /15k
VE / 2k = VE /10k + (VE − VC )/ 3k
(c)
1/ 2k −1/ 2k 0 0
0 1/ 2k −1/ 6k −1/ 3k
0 1/ 6k −(1/ 6k + 1/ 15k) −1/ 2k
0 −1/ 3k 0 −1/ 15k
VA
VC
VD
VE
=
1
1
0
0
is +
0
1m
0
0
(d) Solving in MATLAB VA = 5.818kis + 3.8182 , thus
RTH = 5.818kΩVOC = 3.8182V
(e) This only changes VOC = 3818 ⋅8m = 30.54V .
SOLUTION 6.26. For this circuit, no current flows though the 20 Ω resistor. Therefore VAB = VCB .
Further, from the examples in the chapter, VCB = 4Vs . Hence, Voc = VAB = VCB = 4Vs . Also, shorting
terminals A and B, yields Isc =VCB
20=
4Vs
20=
Vs
5. It follows that Rth =
Voc
Isc= 20 Ω. Note that the
Thevenin equivalent to the left of C-B is a voltage source of value 4Vs. Therefore the Thevenin equivalent
to the left of A-B is 20 Ω in series with 4Vs.
SOLUTION 6.27. (a) From the previous problem VCB = 4Vs . Thus by voltage division,
VOC = VAB = VCB
180
180 + 20
= 3.6Vs . Next find ISC = VCB / 20 = Vs / 5, and then RTH = VOC / ISC =18Ω .
Thevenin Probs, 7/24/01 - P6.13 - @R.A. Decarlo & P. M. Lin
(b) This changes the voltage division at the output, thus VAB / Vs = 4180||162
(180||162) + 20
= 3.24 .
SOLUTION 6.28. (a) Writing the following two KCL equations,
vtest / 4k = (Vo − vtest ) /12 k
Itest = (vtest − Vo ) /15k
where Vo is the voltage at the output of the op-amp. Doing the appropriate substitution get
vtest = −15k / 3 ⋅ Itest , thus
RTH = −5kΩVOC = 0V
Since no independent source exist right of A-B
(b) Applying Ohm’s law Is = 1/ (10k − 5k )Vs = 0.2mVs
SOLUTION 6.29. (a) Adding a test source at terminal A-B, and noting that the voltage at the output of the
op-amp is Vo = −5 / 2Vs . Write KCL at terminal A,
vtest / 900 = (Vo − v test )/ 100 + itest
vtest = 90itest − 2.25Vs
Where one sees by inspection that
RTH = 90ΩVOC = −2.25Vs
ISC = VOC / RTH = −0.025Vs
(b) Noticing the virtual short to ground provided by the ideal op-amp, RTH = 20kΩ , and VOC = 0V since no
independent sources are present right of the input terminal.
SOLUTION 6.30. The output voltage of the ideal op amp is −2.5Vs1 − 2Vs2 which drives a voltage divider
circuit. Hence
Voc = VAB = 0.9(−2.5Vs1 − 2Vs2 ) =−2.25Vs1 −1.8Vs2
Further,
Isc =−2.5Vs1 − 2Vs2
100= −0.0225Vs1 − 0.018Vs2
Finally,
Rth =Voc
Isc=
0.9(−2.5Vs1 − 2Vs2)(−2.5Vs1 − 2Vs2)
100
= 90 Ω
Thevenin Probs, 7/24/01 - P6.14 - @R.A. Decarlo & P. M. Lin
Equivalently if one sets Vs1 and Vs2 to zero, then the ouput terminal of the op amp goes to ground. Hence
Rth = 100 / /900 = 90 Ω
SOLUTION 6.31. Define the node at the output of the op-amp as Vo. Note how the circuit left of this
node is a general summing circuit as per text. Thus, Vo = 2Vs2 − 4Vs1. Hence we simply replace the op
amp circuit to the left of the 20 Ω resistor by an ideal voltage source of value V0. Hence
Voc =80
80 + 20V0 + 2I = 0.8V0 + 2
−V0
100
= 0.78V0 = 1.56Vs2 − 3.12Vs1
Alternately, one can introduce a test source at the output terminal and write out a set of equations using
KVL,
Vo = 2Vs2 − 4Vs1 = −20I − 80(I − itest )
vtest = 2I + 80(itest − I)
Solving yields vtest = 17.6itest + 0.78(2Vs2 − 4Vs1) , and by inspection,
RTH =17.6ΩVOC = 0.78(2Vs2 − 4Vs1) = 56Vs2 − 3.12Vs1
SOLUTION 6.32. Upon inspection when the op-amp is in active range, for inputs from –3V to 3V,
RTH =1kΩVOC = 0V
However when the input exceeds 3V, the output of the op-amp will saturate at –15V, and
Vs = −15V + 6kIs . Therefore from eq. 6.10,
RTH = 6kΩVOC = −15V
When the input is less than –3V, Vs = 15 + 6kIs , thus
RTH = 6kΩVOC = 15V
SOLUTION 6.33. (a) The op-amp configuration has a gain of –4. So when the input is between –3V and
3V it is operating in its active region, thus
Thevenin Probs, 7/24/01 - P6.15 - @R.A. Decarlo & P. M. Lin
RTH = 4kΩVOC = 0V
When the input is greater than 3V the output saturates at –12V and
Vs = 20kIs −12
RTH = 20kΩVOC = −12V
When the input is less than –3V the output saturates at 12V,
Vs = 20kIs +12
RTH = 20kΩVOC = 12V
(b) When the input is in the active range,
RTH =16kΩVOC = −4Vs
When it is greater than 3V,
RTH = 0kΩVOC = −12V
And when less than –3V,
RTH = 0kΩVOC = 12V
The last two obtained using figure 6.28.
The maximum power is when the output is in saturation, P = VOC( )2/ 28k = 6mW .
SOLUTION 6.34. From the table the following two equations can be written:
6 = 2RTH + voc
12 = 8RTH + voc
Putting in matrix form and solving,
1 1
8 1
RTH
voc
=
6
12
−1/ 6 1/ 6
4 / 3 −1/ 3
6
12
=
RTH
voc
Thus RTH =1 kΩ since current was in mA, and voc = 4 V.
SOLUTION 6.35. (a) From Ohm’s law, IAB = VAB / RL . Thus 0.2uA and 0.1uA.
(b) Note how using this topology VAB = voc − IABRTH , thus
Thevenin Probs, 7/24/01 - P6.16 - @R.A. Decarlo & P. M. Lin
1 −0.2u
1 −0.1u
voc
RTH
=
0.4
1
−1 2
−10M 10M
0.4
1
=
voc
RTH
Thus, voc = 1.6V , and RTH = 6MΩ .
SOLUTION 6.36. (a) Although the text describes finding Rth from a measurement or calculation of both
Voc and Isc, measurement of Isc is often impractical. Hence the procedure outlined in this problem
provides a more practical means of determining the Thevenin equivalent.
Since the internal meter resistance is 10 MΩ,
IAB (µA) =VAB
10+
VAB
RL
Hence, the completed table is:
RL(M Ω) vAB (V) IAB (µA)
2 0.4 0.24
10 1 0.2
(b) The terminal relationship assuming a Thevenin equivalent is given by
VAB = Voc − RthIAB
In matrix form with the data from the table
1 −0.24
1 −0.2
Voc
Rth
=
0.4
1
Hence Voc = 4 V and Rth = 15 MΩ. Note that since we used µA and V, Rth is in MΩ.
SOLUTION 6.37. (a) For this scenario, the circuit is essentially a voltage source with a resistance R in
series with the circuit under test, in parallel with a voltmeter measuring the voltage division between the
Thevenin Probs, 7/24/01 - P6.17 - @R.A. Decarlo & P. M. Lin
later two. Therefore if replacing R = 0 by R = R2 causes the voltage measured by the voltage meter to
drop by half, then by voltage division, Eo / 2 = Eo
RTH
R2 + RTH
, and RTH = R2 .
(b) Using the same reasoning and voltage division,
Eo = Vs
RTH
RTH + Rs
Eo / 2 = Vs
RTH
RTH + Rs + R2
Therefore RTH + Rs = R2 or equivalently RTH = R2 − Rs .
(c) Again by voltage division,
Eo = VsRTH || Rm
(RTH || Rm ) + Rs
Eo / 2 = VsRTH || Rm
(RTH || Rm ) + Rs + R2
Therefore (RTH || Rm ) + Rs = R2 or RTH Rm
RTH + Rm= R2 − Rs . Solving for RTH =
R2 Rm − RsRm
Rm + Rs − R2.
SOLUTION TO 6.38. (a) The voltmeter measures the voltage division between the two resistors involved
thus,
Eo = Voc
Eo / 2 = Voc
R2
RTH + R2
and RTH = R2 .
(b) Now
Eo = Voc
Rm
Rm + RTH
Eo / 2 = Voc
Rm || R2
(Rm || R2) + RTH
From the division of the former by the later R2 = RTH || Rm . And from the former
Voc = (Rm + RTH )Eo / Rm = (1+ RTH / Rm )Eo .
SOLUTION TO 6.39. Using the relation developed in problem 6.37 RTH =R2Rm − Rs Rm
Rm + Rs − R2
= 5kΩ
SOLUTION TO 6.40. Using the equation developed in problem 6.38,
Thevenin Probs, 7/24/01 - P6.18 - @R.A. Decarlo & P. M. Lin
RTH =R2 Rm
Rm − R2
= 4MΩ
Voc = (1+ RTH / Rm )Eo = 20V
SOLUTION TO 6.41. Writing the line equation in the following general form,
i = vab / RTH − isc
i = 2vab − 4
Thus RTH = 0.5Ω , and isc = 4A .
SOLUTION TO 6.42. (a) In this range the curve appears to be varying linearly between (0V,0mA) and
(0.2V, 0.1mA) pair, thus writing the line equation i = (0.1m / 0.2)v yields RTH = 2kΩ ⇒ Voc = 0V .
(b) Writing the line equation of the following form, i = v / RTH − isc , into a matrix equation,
0.2 −1
0.7 −1
1/ RTH
isc
=
0.1m
10.1m
and solving gives
RTH = 50Ωisc = 3.9mA
voc = 0.195V
(c) First make a guess as to which region of the curve, N will operate in. Assuming that it will operate in
the A-B region, then by KVL i(t) = (vs (t) + vb − voc )/ (R + RTH ) = (50sin(1000t) − 0.095) / 550. It can be
seen that this guess is wrong as the range of i(t) is not in the appropriate region. Assuming the 0-A region,
by KVL i(t) = (vs (t) + vb − voc )/ (R + RTH ) = (50sin(1000t)m + 0.1)/ 2500 = 0.02sin(1000t) + 0.04mA .
Note that the highest current is 0.06mA, thus still in the appropriate region of operation.
(d) Repeating the procedure above and guessing the region A-B, by KVL
i(t) = (vs (t) + vb − voc )/ (R + RTH ) = (200sin(1000t)m + 305m) /100 = 2sin(1000t) + 3.05mA . The
maximum and minimum current are 5.05 and 1.05 mA respectively, thus the assumption made was correct.
SOLUTION TO 6.43. (a) By the power transfer theorem RL = RTH . For circuit (a) RL = 80 ||240 = 60Ω,
and circuit (b) RL = (900||180) + 50 = 200Ω.
(b) Finding isc for each circuit: (a) isc = 36 / (80 + 240||60) ⋅ 240 / 300 = 225mA and (b)
isc = 60mA900 ||180
(900||180) + 50 + 200
= 22.5mA . The power is now obtained from P = isc / 2( )2
RL . Thus
759mW for (a) and 25.3mW for (b).
Thevenin Probs, 7/24/01 - P6.19 - @R.A. Decarlo & P. M. Lin
SOLUTION TO 6.44. Finding the Thevenin equivalent,
RTH = 30||15( ) +10[ ] ||80 = 16Ω
Voc = −32V
The value for the load resistance is 16Ω, and the power is P =Voc
2
2
/ RL =16W .
SOLUTION TO 6.45. (a) Find the Thevenin equivalent,
RTH =12k | |6k + 8k =12kΩ = RL
Voc = 8k(2mA) + 246
6 +12
= 24V
The power is P = Voc / 2( )2/ RL = 12mW .
(b) The maximum power will be transferred to the load, when the value of its load is closest to 12kΩ. Thus
the power is P =Voc
RTH +10k
2
⋅10k =11.9mW .
(c) Same reasoning as (b) the power is P =Voc
RTH +15k
2
⋅15k =11.9mW .
SOLUTION TO 6.46. (a) The Thevenin equivalent to the left of RL has Rth = 12 + 20 / /180 = 30 Ω and
Voc =1 ×18 +180
180 + 2040 = 54 V. Therefore, for maximum power transfer
GL + 2GL + 3GL = 6GL =6
RL=
1
30
Hence RL =180 Ω.
(b) For this part, let VL denote the voltage (top to bottom) across the load. With RL =180 Ω, then the
parallel combination equals Rth and hence VL = 27 V. It follows that
P180 =(27)2
180= 4.05 watt.
Since the terminal voltages are the same, the absorbed power is inversely proportional to the resistance.
Hence P90 = 8.1 watt, and P60 = 12.15 watt.
Thevenin Probs, 7/24/01 - P6.20 - @R.A. Decarlo & P. M. Lin
SOLUTION 6.47. To find the Thevenin equivalent introduce a test voltage source at the output, and write
KCL at the two node in the circuit. By inspection the following matrix expression is obtained:
(1/ 200 +1/ 400) −1/ 400
−1/ 400 0.0015 +1/1k +1/ 400
VA
Vtest
=
0.1
Itest
Solving Vtest = 240Itest + 8. And RTH = 240 Ω, Voc = 8 V. By voltage division, the voltage across the load
resistor is 4V, and the power delivered to it is 66.7mW.
VA Vtest
SOLUTION 6.48. Using KCL get ix =1mA , which independent of what is connected to the output. Thus
isc = 10ix =10mA
RTH = 3kΩ
The power is then P = 10mA / 2( )2 ⋅3k = 75mW .
SOLUTION 6.49. Performing a source transformation on (a), and combining the elements will simplify to
one current source going up of 2/3A in parallel with a 10Ω resistor. This is essentially the Norton
equivalent of the circuit,
isc = 2/ 3A
RTH =10Ω
For (b), combine the voltage sources and resistor in series. The circuit obtained is one voltage source of
5V in series with a 45Ω resistor. This is the Thevenin equivalent,
voc = 5V
RTH = 45Ω
(a) The value of the load resistor is simply the thevenin resistance obtained above.
(b) Using Ohm’s law for (a) VL = isc(10 ||10) = 10/ 3V , and voltage division for (b) VL = voc(1/ 2) = 2.5V
(c) Using the following formula, P = VL2 / RL , (a) absorbs 1.1W and (b) 139mW, thus (a) absorbs more
power.
Thevenin Probs, 7/24/01 - P6.21 - @R.A. Decarlo & P. M. Lin
SOLUTION 6.50. (a) Note that the circuit left of the terminal is already in its Thevenin form. The load
RL = R | |( R + 300)
R2 + R(300 − 2RL ) − 300RL = 0
Solving, R = 71.6Ω . By voltage division, the voltage across the load is 5V. The power absorbed isP = Voc / 2( )2
/ RL = 416.7mW .
(b) The following script can be used to plot the power absorbed by the load versus R:
%Script for problem 6.50b
R=0:2:400;
%Calculate Load resistance
RL= 1./((1./R)+1./(R+300));
%Calculate the power
P=(10.*(RL./(RL+60))).^2./RL;
%Plot the power versus R
plot(R,P);
ylabel('Power in Watts');
xlabel('Resistance in Ohms')
SOLUTION 6.51. First, find the Thevenin equivalent by writing out the transfer equation vab = 200i + 40.
Thus RTH = 200Ω⇒ Voc = 40V . The maximum power will then be P = (Voc / 2)2 / 200 = 2W .
SOLUTION 6.52. The assumption that all controlling voltages or currents for dependent sources within
Ni are assumed to be in Ni, implies that the nodal equation matrix of figure P6.52a has the partitioned
form:
G11 G12 0
G21 G22 G23
0 G32 G33
VN1
Vm
VN 2
=IN1
Im
IN 2
(*)
where VN1 is the vector of UNKNOWN and INDEPENDENT node voltages internal to N1 and VN2 is
the vector of UNKNOWN and INDEPENDENT node voltages internal to N2. The right side of the
Thevenin Probs, 7/24/01 - P6.22 - @R.A. Decarlo & P. M. Lin
equation consists of (effective) currents injected into the appropriate node. However, IN1 depends only on
sources in N1 and IN2 depends only on sources in N2.
At this point we must presume that the matrix equation (*) has a unique solution, i.e., the determinant
of the coefficient matrix is non-zero. Hence we can calculate VN1, Vm, and VN2 uniquely. As such, by
considering the first row of (*), we can assert that VN1 satisfies
G11VN1 = IN1 − G12Vm (**)
Note that we are not claiming that we can solve for VN1 from this equation.
Replacing N2 by a voltage source of value Vm results in the network of figure P6.52b. For this
network, the nodal equations are
G11VN1 = IN1 − G12Vm (***)
where Gij is the same as in (*). Clearly, this is the same as equation (**). Again we presume there is a
unique solution to this equation, i.e., the determinant of G11 is non-zero. If so, we can solve for VN1
uniquely and the result is the same as that obtained by solving (*).
This theorem can be extended to RLCM networks (to be studied in later chapters) or even
nonlinear networks under appropriate conditions.
To emphasize the subtlety of this result and the need for unique solvability in each network,
consider the following circuit.
The resulting nodal equations are:
Thevenin Probs, 7/24/01 - P6.23 - @R.A. Decarlo & P. M. Lin
2 −1 0 0
−1 0.5 −0.5 0
0 −0.5 2.5 −1
0 0 −1 2
V1
V2
V3
V4
=
−2
0
0
4
There exists a unique solution and from MATLAB, we find
Vm = V3 = 2 V
To apply the voltage source substitution, we replace N2 by a voltage source of 2 V and obtain thefollowing circuit.
The nodal equations here are
2 −1
−1 0.5
V1
V2
=
−2
1
Observe that the coefficient matrix has a zero determinant. Thus there is either no solution or many
solutions, i.e., no unique solution. This demonstrates that unique solvability of the larger network does not
imply the unique solvability of the smaller derived network.
SOLUTION 6.53. The assumption that all controlling voltages or currents for dependent sources within
Ni are assumed to be in Ni, implies that the loop equation matrix of figure P6.53a has the partitioned form:
R11 R12 0
R21 R22 R23
0 R32 R33
IN1
Im
IN 2
=EN1
Em
EN 2
(*)
Thevenin Probs, 7/24/01 - P6.24 - @R.A. Decarlo & P. M. Lin
where IN1 is the vector of unknown and independent loop currents internal to N1 and IN2 is the vector of
unknown and independent loop currents internal to N2 and Im a single independent loop current common
to N1 and N2. The right side of the equation represents the net contribution of voltage sources present in
the appropriate loop. However, EN1 depends only on N1 and EN2 depends only on N2.
At this point we must presume that the matrix equation (*) has a unique solution, i.e., the determinant
of the coefficient matrix is non-zero. Hence we can calculate IN1, Im, and IN2 uniquely. As such, by
considering the first row of (*), we can assert that IN1 satisfies
R11IN1 = EN1 − R12Im (**)
Note that we are not claiming that we can solve for IN1 from this equation.
Replacing N2 by a current source of value Im results in the network of figure P6.53b. For this
network, the loop equations are
R11IN1 = EN1 − R12Im (***)
where Rij is the same as in (*). Clearly, this is the same as equation (**). Again we presume there is a
unique solution to this equation, i.e., the determinant of R11 is non-zero. If so, we can solve for IN1
uniquely and the result is the same as that obtained by solving (*). For some subtlety in the proof refer to
the solution of 6.52.
This theorem can be extended to RLCM networks (to be studied in later chapters) or even
nonlinear networks under appropriate conditions.
SOLUTION 6.54. (a) The thevenin equivalent to the left of terminal A-B is
RTH = [(30||60) + 20]||10 = 8Ω , and using Ohm’s law along with voltage division
VOC = 15 −1015V
15 + 60
30
30 + 30
= 14V
(b) Doing the same for the circuit right of terminal A-B. RTH = [(30||15) + 10]||20 =10Ω and
VOC = 7.5 − 207.5V
15 +15
30
30 + 30
= 5V .
(c) Using superposition, VAB = 1410
18
+ 5
8
18
= 10V .
(d) (e) (vCB − 15) / 30 + vCB / 60 = (10 − vCB )/ 20 Hence, vCB = 10V .
Thevenin Probs, 7/24/01 - P6.25 - @R.A. Decarlo & P. M. Lin
SOLUTION TO P6.55. The proof is based on superposition. Let us consider the figure below where N1
and N2 are differently named but identical networks.
We first compute the contribution to ia from the independent sources in N1 with those of N2 deactivated.
Let this current be ia1 . The contribution to ia from the independent sources in N2 with those of N1
deactivated is ia2 . But because N1 and N2 are identical, ia
1 = −ia2. Hence by superposition
ia = ia1 + ia
2 = 0. By the current source substitution theorem we can replace the lines by current sources of
value 0-amp. This defines an open circuit and the connecting line can be replaced by an open circuit.
From the given network we also note by KVL that Vx + Vy = 0 which implies that Vx = –Vy . On the other
hand, since the networks are identical, Vx = Vy . Thus we conclude that Vx = Vy = 0. Thus we can replace
Vx and Vy by a voltage source of 0-volt (voltage source substitution theorem) which is the definition of a
short circuit.
SOLUTION 6.56. Label the potential between each line starting from the top as Vx1, Vx2 on the left, and Vy1
and Vy2 on the right. Now by superposition and linearity notice that
Vx1 = −Vy1
Vx2 = −Vy2
because the independent source is negative on the right side. Additionally, from KVL,
Vx1 = Vy1
Vx2 = Vy2
The only way all these condition can be met, is if all the voltages are 0 V, or short circuited.
Thevenin Probs, 7/24/01 - P6.26 - @R.A. Decarlo & P. M. Lin
SOLUTION 6.57. (a) Using the results from P6.55, no current flows between the two halves. So the right
hand side circuit may be analyzed as if it was stand alone. By voltage division then,
va =3 + 2
3 + 2 +1
18 = 15V .
(b) From the results of P6.56, all the lines crossing the symmetry line are shorted together. Consequently,
by voltage division, va =6 ||3
(6 ||3) +1
(−18) = −12V .
SOLUTION 6.58. Note how this circuit is the same as in P6.57: it is just redrawn with the neighboring
resistors added in parallel or in series. Using superposition, we can solve for va when the sources [vs1 vs2]
are [18 18], and then [18 –18]. By linearity adding the two contribution will be equivalent to solving for
[36 0] directly, since adding the source contributions [18 18]+[18 –18] =[36 0]. The contributions of
15V–12V were obtained in P5.57; thus va = 3V .
SOLUTION 6.59. Yes since
45 27[ ] = 36 36[ ] + 9 −9[ ] = 2 18 18[ ] +1/ 2 18 −18[ ] = 2(15) + (12) / 2 = 24V .
SOLUTION 6.60. For this proof we attach an arbitrary network N to each of the networks N1 and N2 in
figure P6.60 as shown below.
N may have internal independent sources, but we consider N1 and N2 external excitations to N and
we assume no violation of KVL in the attachment. Choose node 3 as a reference node. Then
V13 = Va − Vc and V23 = Vb − Vc for both figures. Hence N1 and N2 provide the identical external
Thevenin Probs, 7/24/01 - P6.27 - @R.A. Decarlo & P. M. Lin
excitations to N and hence all currents and voltages in N remain unchanged.
The extension of this result to 4 external nodes is shown in the figure below. The verification is
the same as above. The extension of course to n-terminals is clear.
SOLUTION 6.61. Using the E-shift theorem, remove the 9 V source from each branch and add it to the
4V source, and notice by inspection that VOC = −4 + 9 = 5 V.
SOLUTION 6.62. (a) Writing a KCL equation for each node in N2:
I1 = Ia
I2 = Ia − Ia = 0
I3 = Ia − Ia = 0
I4 =− Ia
Do the same for N1:
I1 = Ia
I2 = 0
I3 = 0
I4 =− Ia
This shows that the two have identical outcomes.
(b)
Thevenin Probs, 7/24/01 - P6.28 - @R.A. Decarlo & P. M. Lin
SOLUTION 6.63. Using the I-shift theorem, this circuit is essentially a –2A source in parallel with a
series combination of resistors, and a 5A source. Thus isc = 5 − 2 = 3A .
SOLUTION 6.64. (a) This can be done by inspection. An equal source is connected between A-C and C-
B; thus by the I-shift theorem, it is equivalent to the same source just connected between A-B.
(b) In figure 6.64c the VCCS is replaced by a resistor using the Ohm’s law relationship
100Ω = V1 / (0.01V1).
(c) (d) By voltage division V1 = Vs
100||900
(100 ||900) + 10
, and by Ohm’s law
Vout = Vs
100||900(100 ||900) +10
0.01 20k | |5k( ) = 36Vs
Vout / Vs = 36
L&C Probs, 11/15/01 P:7-1 © R. A. DeCarlo, P. M. Lin
PROBLEM SOLUTIONS CHAPTER 7
SOLUTION 7.1. Given the coil has 48 turns and 12 turns/cm, we know that the length of the coil is 4 cm.Since the length of the coil is greater than 0.4 times its diameter, the formula given in the question can beused:
L =4 ×10−5 × 0.02( )2 × 48( )2
18 × 0.02( ) + 40 × 0.04( ) =18.81 H
SOLUTION 7.2. Part 1. Applying (7.1)The voltage vL(t) can be computed using the inductor v-i relationship:
vL (t) = LdiL (t)
dt
The calculations for vL(t) for t = 0s to 5s are summarized in the following table:
Time Interval d/dt (iin(t)) vL(t)0s< t ≤1s 2 As-1 1V1s< t ≤3s -2 As-1 -1V3s< t ≤4s 2 As-1 1V4s< t ≤5s 0 As-1 0V
Below is the plot of VL vs time.
Part 2. Applying (7.4)
wL (t) =12
(0.5 iin2( t))
In the time interval 0s < t ≤ 1s, iin (t) = 2t . Thus
wL (t) =12
0.5iin2 ( t)( ) = t2
In the time interval 1s < t ≤ 3s, iin (t) = 4 − 2t . Thus
L&C Probs, 11/15/01 P:7-2 © R. A. DeCarlo, P. M. Lin
wL (t) =12
0.5iin2 ( t)( ) =
14
(4 − 2t)2 = t2 − 4t + 4
In a similar way, wL (t) can be computed in the remaining intervals.The calculations for wL(t) for t = 0s to 5s are summarized in the following table:
Time Interval iin(t) wL(t)0s< t ≤1s 2t t2
1s< t ≤3s 4-2t t2-4t+43s< t ≤4s -8+2t t2-8t+164s< t ≤5s 0 0
Below is the plot of wL vs t.
SOLUTION 7.3. Applying (7.2)
iL ( t) = iL (0) +1L
vL ( )d0
t∫
It is assumed that iL(0) = 0 A. Using the preceding formula and the fact that vin(t) = t in the time interval0s < t ≤ 2s, the current is
iL ( t) =1
0.5d = t2
0
t∫ in the time interval 0s < t ≤ 2s
In the time interval 2s < t ≤ 3s vin (t) = 2 . Hence
iL ( t) =1
0.5vL ( )d
0
t∫ = iL (2) +
10.5
2dt2
t∫ = 4 + 4 2
t = 4t − 4
In a similar way iL ( t) can be computed in the remaining intervals.
The calculations for iL(t) for t = 0s to 6s are summarized in the following table:
L&C Probs, 11/15/01 P:7-3 © R. A. DeCarlo, P. M. Lin
Time Interval vin(t), V iL(t), A0s< t ≤2s t t2
2s< t ≤3s 2 4t-43s< t ≤5s -2 20-4t5s< t ≤6s 0 0
Below is the plot of iL vs t.
SOLUTION 7.4. Part 1. Using (7.1), we have
vin (t) = 0.2 ×10−3 ddt
is( t)[ ] = 0.2 ×10−3 ×1000cos(1000 t) = 0.2cos(1000t)mV
For the 2mH inductor
iout ( t) = iout (0) +1L
10vin0
t
∫ ( )d =1
2 ×10−3 2cos0
t
∫ (1000 )d = sin(1000t)mA
Below is a sketch of iout vs t.
Part 2. Instantaneous power delivered by the dependent source is given by
pL ( t) = vL (t) × iL (t) = 10vin ( t) × iout ( t) = 2cos(1000t) × sin(1000t) = sin(2000t) W
L&C Probs, 11/15/01 P:7-4 © R. A. DeCarlo, P. M. Lin
Part 3. The energy stored in the 2mH inductor is given by
WL (t) =12
LiL2(t) =
12
Liout2( t) = sin2(1000t)nJ
Below is a sketch of WL vs t
SOLUTION 7.5. Part 1 For the excitation in Figure P7.5b,
i1(t) = i1(0) +1
0.5vin
0
t
∫ ( )d , i2(t) = i2(0) +1
0.25vin
0
t
∫ ( )d = 2i1( t)
It is assumed that i1(0) = i2(0) = 0 A . In the interval 0s < t ≤ 1s, vin (t) = −10V . Hence, in this interval
i1(t) =1
0.5vin ( )d
0
t∫ =
10.5
(−10)d0
t∫ = −20t
i2( t) =1
0.25vin ( )d
0
t∫ =
10.25
(−10)d0
t∫ = −40t
In the interval 1s < t ≤ 2s, vin (t) = −5V . Hence, in this interval,
i1(t) = i1(1)+1
0.5vin( )d
1
t∫ = −20 +
10.5
(−5)d1
t∫ = −10t −10
i2( t) = i2(1)+1
0.25vin( )d
1
t∫ = −40 +
10.25
(−5)d1
t∫ = −20t − 20
In a similar fashion i1(t) and i2(t) can be computed in other intervals.
The calculations for i1(t) and i2(t) for t = 0s to 5s are summarized in the following table:
L&C Probs, 11/15/01 P:7-5 © R. A. DeCarlo, P. M. Lin
Time Interval vin(t), V i1(t), A i2(t), A0s< t ≤1s -10 -20t -40t1s< t ≤2s -5 -10-10t -20-20t2s< t ≤3s 0 -30 -603s< t ≤4s 5 -60+10t -120+20t4s< t ≤5s 10 100+20t 200+40t
Below are the plots of i1 vs t and i2 vs t.
Part 2 For the excitation in Figure P7.5c,
i1(t) = i1(0) +1
0.5vin
0
t
∫ ( )d , i2(t) = i2(0) +1
0.25vin
0
t
∫ ( )d = 2i1( t)
It is assumed that i1(0) = i2(0) = 0 A. In the interval 0s < t ≤ 1s, vin (t) =10t . Hence, in this interval
i1(t) = i1(0) +1
0.5vin( )d
0
t∫ =
10.5
(10 )d0
t∫ = 10t2
and
i2( t) = i2(0) +1
0.25vin( )d
0
t∫ =
10.25
(10 )d0
t∫ = 20t2
In the interval 1s < t ≤ 3s, vin (t) =10t − 20 . Hence, in this interval
i1(t) = i1(1)+1
0.5vin( )d
1
t∫ =10 +
10.5
(10 − 20)d1
t∫ =10t2 − 40t + 40
i2( t) = i2(1)+1
0.25vin( )d
1
t∫ = 20 +
10.25
(10 − 20)d1
t∫ = 20t2 − 80t + 80
The calculations for i1(t) and i2(t) for t = 0s to 3s are summarized in the following table:
L&C Probs, 11/15/01 P:7-6 © R. A. DeCarlo, P. M. Lin
Time Interval vin(t), V i1(t), A i2(t), A0s< t ≤1s 10t 10t2 20t2
1s< t ≤3s 10t-20 10t2-40t+40 20t2-80t+80
Below are the plots of i1 vs t and i2 vs t.
SOLUTION 7.6. For the circuit in Figure P7.6, the parallel combination of the 0.75mH and 1.5mHinductors can be replaced by an inductor with the inductance of (0.75||1.5)mH. The v-i relationship forthis inductor is
iin (t) = iin (0) +1
[(0.75) (1.5)] ×10−3vs( )d =
1
0.5 ×10−3 vs( )d0
t
∫ A0
t∫
The series combination of the 0.8mH and 0.6mH can be replaced by an (0.8mH + 0.6mH) inductor. Thev-i relationship for this inductor gives:
vout (t) = (0.8 + 0.6) ×10−3 ddt
6iin (t)[ ] =0.8 + 0.6
0.5× 6
ddt
vs0
t
∫ ( )d
= 16.8vs( t) V
Below is a plot of vout(t) vs t.
L&C Probs, 11/15/01 P:7-7 © R. A. DeCarlo, P. M. Lin
Assume iL(0) =0 for all inductors.
iout ( t) = 6iin (t) =6
0.5 ×10−3 vs0
t
∫ ( )d =12 ×103 vs0
t
∫ ( )d A
In the interval 0s < t ≤ 1s, vs(t) = 1mV . Hence,
iin (t) =1
0.5 ×10−3 vs( )d0
t∫ = 2t(A), iout ( t) = 6iin (t) = 12t(V )
The power p(t) on the same interval is computed as:
p( t) = 6iin ( t) × vout (t) = (12t) × (16.8×10−3) = 0.202t(W )
In the interval 1s < t ≤ 3s, vs(t) = 2mV . Hence,
iin (t) = iin (1) +1
0.5 ×10−3 vs( )d1
t∫ = 2 + 2 × 2 1
t = 4t − 2(A)
iout ( t) = 6iin (t) = 24t −12(A)and
p( t) = 6iin ( t) × vout (t) = (24 t −12) × (16.8× 2 ×10−3) = 0.806t− 0.403(W )
In a similar fashion, iout(t) and p(t) can be computed for the remaining intervals.
The calculations for iout(t) and the instantaneous power delivered by the dependent source, p(t), for t = 0sto 6s are summarized in the following table:
Time Interval vs(t), V vout(t), V iout(t), A p(t), W0s< t ≤1s 1×10-3 16.8×10-3 12t 0.202t
1s< t ≤3s 2×10-3 33.6×10-3 -12+24t 0.806t-0.403
3s< t ≤5s -2×10-3 -33.6×10-3 132-24t 0.806t-4.435
5s< t ≤6s 0 0 12 0
L&C Probs, 11/15/01 P:7-8 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.7. For 0 ≤ t < 2 s, using the inductor v-i relationship, we have
iin (t) =1
0.5vin ( )d
0
t
∫ =2
0.25π−cos(0.25π )[ ]0
t =8π
1− cos(0.25πt)[ ] A
The associated energy stored as a function of t for this time interval is
W0.5(t) =16
π 2 1− cos(0.25πt)[ ]2 J
The energy for the second inductor remains zero over this interval.
For 2 s ≤ t, we have
iin (t) = i0.5(t) + i0.25( t) = i0.5( t) +1
0.25vin ( ) d
2
t
∫
=8π
1− cos(0.25πt)[ ] +4
0.25π−cos(0.25π )[ ]2
t =8π
1− cos(0.25πt)[ ] −16π
cos(0.25πt)
=8π
−24π
cos(0.25πt) A
Here the current
i0.5( t) =8π
1− cos(0.25πt)[ ] A
in which case the energy stored over the interval [2,t] is
W0.5(2,t) =12
0.5 i0.52 (t) − i0.5
2 (2)( ) =12
0.5 i0.52 (t) −
64
π 2
J
Further
i0.25( t) = −16π
cos(0.25πt) A
in which case the energy stored over the interval [2,t] is
W0.25(2,t) =12
0.25 i0.252 (t) − i0.25
2 (2)( ) =12
0.25i0.252 ( t) J
SOLUTION 7.8. Let the 5mH inductor be L1 and the 20mH inductor be L2.For 0 ≤ t < 3ms,
iL1(t) = iL1(0) +1L1
vs0
t
∫ ( )d =1
5 ×10−3 12cos0
t
∫ (500 )d = 4.8sin(500t) mA
L&C Probs, 11/15/01 P:7-9 © R. A. DeCarlo, P. M. Lin
iL 2(t) = iL2(0) +1L2
vs0
t
∫ ( )d =1
20 ×10−3 12cos0
t
∫ (500 )d = 1.2sin(500t) mA
For t ≥ 3ms,
iL1(t) = 4.8sin(500t) mA as L1 is still subjected to the same voltage.
on the other hand,
iL 2(t) = iL2(3− ms) +1
L2vL2
3+ ms
t
∫ ( )d = iL 2(3− ms) +1
L20
3+ ms
t
∫ d
= iL 2(3− ms)
=1.2sin(500 × 3 ×10−3) =1.197mA
For 0≤ t < 3ms, the energies stored in the inductors are given as follows:
WL1(t) =1
2L1 iL1(t)[ ]2 = 57.6sin 2(500t) nJ,
WL2( t) =1
2L2 iL2( t)[ ]2 = 14.4sin 2(500t) nJ
For t ≥ 3ms, the energies stored in the inductors are given as follows:
WL1(t) =1
2L1 iL1(t)[ ]2 = 57.6sin 2(500t) nJ,
WL2( t) =1
2L2 iL2( t)[ ]2 = 14.328 nJ
SOLUTION 7.9. Given the dielectric parameters and the dimensions of the capacitor, the capacitance of thepaper capacitor is given by
C = r 0Ad
= 3× 8.854 ×10−12 ×0.04 × 0.8
10−4 = 8.5nF
SOLUTION 7.10. . Part 1. Applying the capacitor v-i relationship:
iC ( t) = Cddt
vC ( t)( ) =1 ×100 ×1000 × (− sin(1000t)) = −0.1sin(1000t) A
Part 2. Applying the capacitor v-i relationship:
L&C Probs, 11/15/01 P:7-10 © R. A. DeCarlo, P. M. Lin
iC ( t) = Cd
dtvC ( t)( )
10 ×10−3 cos(1000t) = Cddt
sin(1000t)( )
10 ×10−3cos(1000t) =1000C cos(1000t)Therefore, C=10µF.
SOLUTION 7.11. Applying the capacitor v-i relationship for C1 and C2:
iC1(t) = (2mF)ddt
vin (t)( ), iC 2(t) = (6mF)ddt
vin ( t)( ) = 3iC1( t),
In the interval –1s < t ≤ 0s, vin(t) = 5t + 5. Hence, in this time interval
iC1(t) = 2 ×10−3 ×d(vin (t))
dt= 10−2 A
and
iC 2(t) = 6 ×10−3 ×d(vin ( t))
dt= 3×10−2 A
Using KCL, iin(t) can be computed as
iin (t) = iC1(t) + iC 2(t) = 4 ×10−2 A
In the interval 0s < t ≤ 1s, vin(t) = 5. Thus
iC1(t) = iC2( t) = 0 A
Using KCL, iin (t) = iC1(t) + iC 2(t) = 0 A.In a similar fashion iC1(t), iC2(t) and iin(t) can be computed for the remaining intervals.
The calculations for iC1(t), iC2(t) and iin(t) for t = -1s to 6s are summarized in the following table:
Time Interval d/dt (vin(t)), Vs-1 iC1(t), mA iC2(t), mA iin(t), mA-1s< t ≤0s 5 10 30 400s< t ≤1s 0 0 0 01s< t ≤3s -5 -10 -30 -403s< t ≤4s 20 40 120 1604s< t ≤5s 0 0 0 05s< t ≤6s -15 -30 -90 -120
Below are the plots of iC1(t), iC2(t) and iin(t).
L&C Probs, 11/15/01 P:7-11 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.12. Applying (7.6)
vC (2) = vC (0) +1C
iC ( )d0
2
∫ = 4V +1C
d0
2
∫ = 5V
vC (3) = vC (2) +1C
iC ( )d2
3
∫ = vC (2) +1C
2d2
3
∫ = 5V +1V = 6V
Applying (7.11), the energies stored in the capacitor over the intervals [0,2] and [2,3] are given by
WC[0,2] =12
C vC2 (2) − vC
2 (0)[ ] = 9 J
WC[2,3] =12
C vC2 (3) − vC
2 (2)[ ] =11 J
SOLUTION 7.13. Part 1 Applying (7.6)
L&C Probs, 11/15/01 P:7-12 © R. A. DeCarlo, P. M. Lin
vC1(t) = vC1(0) +1
C1iin ( )d
0
t
∫ = 1
0.25iin ( )d
0
t
∫
vC 2( t) = vC2(0) +1
C2iin ( )d
0
t
∫ = 1
0.1iin ( )d
0
t
∫
In the interval 0s < t ≤ 1s, iin (t) = 2t . Hence, in this interval
vC1(t) =1
0.25iin ( )d =
10.25
( 2)o
t∫
0
t= 4 ×106 t2(V )
and
vC 2( t) =1
0.1iin ( )d =
10.1
( 2)o
t∫
0
t=10 ×106t2(V )
In the interval 1s < t ≤ 3s, iin (t) = 2 ×10−3 (A). Thus
vC1(t) = vC1(1) +1
0.25× iin ( )d =
1
t∫ 4 +
10.25
× (2 ×10−3 )1
t= 8 ×103t − 4(V )
and
vC 2( t) = vC2(1) +1
0.1× iin ( )d =
1
t∫ 10 +
10.1
× (2 ×10−3 )1
t= 20 ×103t −10(V )
In a similar fashion, vC1(t) and vC2(t) can be computed for the remaining intervals.
The calculations for vC1(t) and vC2(t) for t = 0s to 8ms are summarized in the following table:
Time Interval iin(t), A vC1(t), V vC2(t), V0s< t ≤1ms 2t 4×106 t2 10×106 t2
1ms< t ≤3ms 2×10-3 -4 + 8×103 t -10 + 20×103 t
3ms< t ≤5ms -2×10-3 44 - 8×103 t 110 - 20×103 t5ms< t ≤8ms 0 4 10
Below are the plots of vC1(t) and vC2(t).
L&C Probs, 11/15/01 P:7-13 © R. A. DeCarlo, P. M. Lin
Part 2 Applying (7.12),
WC1(t) =12
C1vC12(t) , WC 2(t) =
12
C2vC 22( t)
The expressions for WC1(t) and WC2(t) for t = 0ms to 8ms are listed in the following table:
Time Interval WC1(t), µJ WC2(t), µJ
0s< t ≤1ms 0.125 × (4 ×106 t2)2 0.05 × (10×106 t2)2
1ms< t ≤3ms 0.125 × (-4 + 8×103 t)2 0.05 × (-10 + 20×103 t)2
3ms< t ≤5ms 0.125 × (44 – 8×103 t)2 0.05 × (110 – 20×103 t)2
5ms< t ≤8ms 0.125 × (4)2 0.05 × (10)2
Below are the plots of WC1(t) and WC2(t).
Part 3 Since the current iC1(t) stays constant at 0A after t = 5ms,
vC1(∞) = vC1(5) = 4V
vC 2(∞) = vC 2(5) = 10V
SOLUTION 7.14. Part 1 Applying (7.6),
L&C Probs, 11/15/01 P:7-14 © R. A. DeCarlo, P. M. Lin
vin (t) = vin (0) +1C
iin ( )d0
t
∫ = 1
0.5 ×10−3 iin ( )d0
t
∫
In this part, we use the current excitation signal described in Figure P7.14b. In the interval 0s < t ≤ 1s,iin (t) = −10(mA). Thus, in this interval
vin (t) =1
0.5 ×10−3 (−10 ×10−3)do
t∫ = −20t(V )
In the interval 1s < t ≤ 2s, iin (t) = −5(mA). Hence, in this interval
vin (t) = vin (1)+1
0.5 ×10−3 (−5 ×10−3)d1
t∫ = −20 + (−10t +10) = −10t −10(V )
In a similar fashion, vin(t) can be computed for the remaining intervals. The calculations for vin(t) for t = 0sto 5s are summarized in the following table:
Time Interval iin(t), mA vin(t), V0s< t ≤1s -10 -20t1s< t ≤2s -5 -10 -10t2s< t ≤3s 0 -303s< t ≤4s 5 -60+10t4s< t ≤5s 10 -100+20t
Below is the plot of vin(t) vs. time.
Part 2. In this part, we use the current excitation signal described in Figure P7.14c. In the interval 0s < t≤ 1s, iin (t) = 10(mA) . Thus, in this interval
vin (t) =1
0.5 ×10−3 (10×10−3 × )do
t∫ = 10t2(V)
In the interval 1s < t ≤ 3s, iin (t) = 10t − 20(mA) . Hence, in this interval
vin (t) = vin (1)+1
0.5 ×10−3 10−3 × (10 − 20)do
t∫ = 10 +10(10 2 − 40 )1
t =10t2 − 40t + 40 (V)
L&C Probs, 11/15/01 P:7-15 © R. A. DeCarlo, P. M. Lin
The calculations for vin(t) for t = 0s to 3s are summarized in the following table:
Time Interval iin(t), mA vin(t), V0s< t ≤1s 10t 10t2
1s< t ≤3s 10t-20 10t2 - 40t + 40
Below is the plot of vin(t) vs. time.
SOLUTION 7.15. Part 1 Using the capacitor v-i relationship,
iin (t) = C1ddt
(vs(t)) = 20 × 6 ×1500 × cos(1500t) = 0.18cos(1500t) A
Then, we can find vout by applying (7.6),
vout (t) = vout (0) +1
0.5m2iin ( )d
0
t
∫
= 10 +2 × 0.18
0.5m ×1500sin(1500t)
= 10 + 0.48sin(1500t) V
Part 2 The instantaneous power delivered by the independent source is given by
p( t) = 2iin ( t) ×vout (t) = 0.36cos(1500t) × [10+ 0.48sin(1500t)] W
Part 3 Applying (7.11), the energy stored in the capacitor over the interval [0,t] is given by
WC1[0, t] =12
C1 vs2(t) − vs
2(0)[ ] = 360sin2(1500t) J
L&C Probs, 11/15/01 P:7-16 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.16. Part 1 From the solution of problem 7.15,
vout (t) = vout (0) +1
0.5m2iin ( )d
0
t
∫
= vout (0) +1
0.5m2 × 20
dd
vs( )( )d0
t
∫
= vout (0) + 0.08dd
vs( )( )d0
t
∫
= vout (0) + 0.08vs( )
If we assume vout(0) = 0V, then
vout (t) = 0.08 vs( )
The following is a plot of vout(t) vs time.
Part 2 The instantaneous power delivered by the dependent source is given by
p( t) = vout ( t) × 2iin (t) = ( vout (0) + 0.08 vs( t)) × 2 × 20d
dtvs( t)( )
= 40( vout (0) + 0.08 vs( t)) d
dtvs(t)( ) W
If we assume vout(0) = 0V, then
p( t) = 3.2 vs( t) ddt
vs(t)( ) W
The calculations for d/dt (vS(t)) and p(t) for t = 0s to 6ms are summarized in the following table:
L&C Probs, 11/15/01 P:7-17 © R. A. DeCarlo, P. M. Lin
Time Interval d/dt (vS(t)), Vs-1 p(t), W0ms< t ≤1ms 2 × 10
312.8 × t
1ms< t ≤3ms -2 × 103
12.8 × (t-2)
3ms< t ≤4ms 2 × 103
12.8 × (t-4)4ms< t ≤6ms 0 0
The following is a plot of p(t) vs time.
Part 3 The energy stored in the 0.5-mF capacitor is given by
WC (t) =12
× 0.5m × vout2(t) = 0.25( vout (0) + 0.08 vs( t))2 mJ
If we assume vout(0)=0V, then
WC (t) = 1.6(vs(t))2 J
In the interval 0s < t ≤ 1ms, vs(t) = 2 ×103t(V ). Thus, in this interval
WC (t) = 6.4t2J
In the interval 1s < t ≤ 3ms, vs(t) = 2 ×103 × (2 − t)(V ). Hence, in this interval
WC (t) = 6.4 × (2 − t)2J
In a similar way WC(t) can be computed for the interval 3ms < t ≤ 4ms.
The following is a plot of WC(t) vs time.
L&C Probs, 11/15/01 P:7-18 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.17. Part 1 From the solution of problem 7.16,
vout (t) = vout (0) + 0.08 vs( )
If we assume vout(0)=0V, then
vout (t) = 0.08 vs( )
The following is a plot of vout(t) vs time.
Part 2 The instantaneous power delivered by the dependent source is given by.
p( t) = 40( vout (0) + 0.08 vs(t)) ddt
vs( t)( ) W
If we assume vout(0)=0V, then
p( t) = 3.2 vs( t) ddt
vs(t)( ) W
The calculations for d/dt (vS(t)) and p(t) for t = 0s to 6ms are summarized in the following table:
L&C Probs, 11/15/01 P:7-19 © R. A. DeCarlo, P. M. Lin
Time Interval d/dt (vS(t)), Vs-1 p(t), W0ms< t ≤1ms 2 × 103 12.8 × t1ms< t ≤2ms 0 02ms< t ≤4ms -2 × 103 12.8 × (t-3)4ms< t ≤5ms 0 05ms< t ≤6ms 2 × 103 12.8 × (t-6)
The following is a plot of p(t) vs time.
Part 3 The energy stored in the 0.5-mF capacitor is given by
WC (t) = 0.25( vout (0) + 0.08 vs(t))2 mJ
If we assume vout(0) = 0V, then
WC (t) = 1.6(vs(t))2 J
In the interval 0ms < t ≤ 1ms, vs(t) = 2 ×103t(V ). In this interval
WC (t) = 6.4t2J
In the interval 1ms < t ≤ 2ms, vs(t) = 2(V ). In this interval
WC (t) = 6.4( J )
In a similar fashion WC(t) can be computed for the remaining intervals.
The following is a plot of WC(t) vs time.
L&C Probs, 11/15/01 P:7-20 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.18. Part 1 Applying the capacitor v-i relationship for the equivalent capacitor of the seriescombination of 0.3 mF and 0.6 mF capacitors
vin (t) = vin (0) +1
0.3 ×10−3 0.6 ×10−3is( )d
0
t
∫
= (0) +1
0.3×10−3 0.6 ×10−360 ×10−3 sin(100 )d
0
t
∫
= 3 − 3cos(100t) V
Therefore,
iout ( t) = (0.2 ×10−3 + 0.8 ×10−3)ddt
(10vin (t)) = 3sin(100t) V
Below is a sketch of iout(t) vs time.
Part 2 The instantaneous power delivered by the dependent source is given by
p( t) = iout ( t) ×10vin (t) = 90sin(100t) × [1− cos(100t)] W
L&C Probs, 11/15/01 P:7-21 © R. A. DeCarlo, P. M. Lin
Part 3 The instantaneous energy stored in the 0.8mF capacitor is given by
WC (t) =12
CvC2( t) =
12
(0.8mF) 10vin ( t)[ ]2 = 0.361− cos(100t)[ ]2 J
SOLUTION 7.19.
Part 1 Since Q = CV, the charge that resides on each plate of the capacitor = 10µF × 100V = 1mC
Part 2 Since V = Q/C, the required voltage = 1mC/5µF = 200V
Part 3 Since V = Q/C, the required voltage = 50µC/1µF = 50V
Part 4 The energy required = 0.5 × 10µF × (100V)2 = 0.05J
SOLUTION 7.20. When 0s≤ t <2s, vC(t) = 25V. Conservation of charge requires that
q1(2−) + q2(2−) = q1(2+ ) + q2(2+ ). Since q2(2− ) = 0 C it follows that
25 V × 150 mF = vC(2+)(150 mF + 100 mF)
Hence
vC (2+ ) =25V ×150mF + 0V ×100mF
150mF +100mF= 15 V
Thus, for t > 2s, vC(t) = 15 V.
SOLUTION 7.21. For this solution consider the figure below in which C1 and C2 are labeled.
There are two cases to consider: (i) t < 2 ms and (ii) t > 2 ms.
Case 1. t < 2 ms. Here, since the current source is zero for t < 0, and C1 is uncharged at t = 0,
vC (t) =1C1
is( )d−∞
t
∫ = vC (0) +1C1
is( )d0
t
∫ =1C1
is( )d0
t
∫
=−12
15 ×10−3 × 500e−500[ ]0
t=1.6 1− e−500t( ) V
L&C Probs, 11/15/01 P:7-22 © R. A. DeCarlo, P. M. Lin
Note that vC (2− ms) =1.0114 V. Hence the energy stored over [0, 2 ms] is
WC1(0 < t < 2ms) = 0.0192 1− e−500t( )2 J
andWC2(0 < t < 2ms) = 0
Case 2. 2 ms < t. At t = 2 ms the switch closes, forcing a discontinuity in the capacitor voltages. Tocalculate the capacitor voltages at 2
+ ms, we use conservation of charge. Here, the relevant equation is:
qC1(2+ ms) + qC 2(2+ ms) = qC1(2− ms) + qC2(2− ms) + is( )d
2− ms
2+ ms
∫
Note that since vC 2(2− ms) = 0, qC 2(2− ms) = C2vC 2(2− ms) = 0 and the integral of the boundedcontinuous function is(t) over an infinitesimal interval is zero, this equation reduces to
qC1(2+ ms) + qC 2(2+ ms) = qC1(2− ms)
or equivalently, since for t > 2 ms, vC1(t) = vC 2(t) = vC (t) ,
C1vC (2+ ms) + C2vC (2+ ms) = C1vC (2− ms)Therefore
vC (2+ ms) =C1
C1 + C2vC (2− ms) =
1515 + 25
×1.0114 = 0.37927 V
and it follows that
vC (t) = vC (2+ ms) +1
Ceqis( ) d
2ms
t
∫ = 0.379 +1
C1 + C2is( )d
2ms
t
∫
= 0.379 +−12
40 ×10−3 × 500e−500[ ]0.002
t= 0.379 + 0.6 0.36788 − e−500t( ) V
Hence the energy stored in the two capacitors over the interval [2+ ms, t] is
WCeq(2+ ms < t) =
12
CeqvC2 ( t) −
12
CeqvC2 (2+ ms) J
whereas the instantaneous stored energy, i.e., the energy stored over (-∞,t > 2 ms] is given by
WCeq(t) =
12
CeqvC2 ( t)
What happens between 2-ms and 2
+ms is beyond the scope of the material in this chapter. Please
refer to problem 51 in chapter 8 for an explanation.
L&C Probs, 11/15/01 P:7-23 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.22. Part 1 Applying the inductor v-i relationship,
vL (t) = Ld
dtiin (t)
= 2.5mH × −200te−10t + 20e−10 t( )= 0.05e−10 t (−10t +1)
Applying the capacitor v-i relationship, we have:
vC (t) = vC (0) +1C
iin ( )d0
t
∫
=20
1mte−10
0
t
∫
= 20 ×103 −0.1 e−10 − 0.01e−10[ ]0
t
= 2 ×103 − te−10 t − 0.1e−10t + 0.1[ ] V
and
vin (t) = vL ( t) + vC ( t)
= 0.05e−10t (−10t +1)+ 2000 − te−10t − 0.1e−10t + 0.1[ ]= −2000.5 × te−10 t −199.95 × e−10t + 200(V)
The sketches of vL(t), vC(t) and vin(t) are shown below.
L&C Probs, 11/15/01 P:7-24 © R. A. DeCarlo, P. M. Lin
Part 2 The energy stored in the inductor is given by
WL (t) =12
L[iin (t)]2 = 0.5t2e−20 t J
The sketch of WL(t) vs. time is shown below.
Part 3 The energy stored in the capacitor is given by
WC (t) =12
C vC (t)[ ]2 = 2 −te−10t = 0.1e−10 t + 0.1[ ]2kJ
The sketch of WC(t) vs time is shown below.
L&C Probs, 11/15/01 P:7-25 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.23. Applying the inductor v-i relationship we have that
vL (t) = 0.25d[is( t)]
dt= 4cos(4 t) (V )
Applying the capacitor v-i relationship it follows that
iC ( t) = 0.25d[2v2( t)]
dt= 0.25 × 2 × 4 × 4 × [− sin(4t)] = −8sin(4 t) A
SOLUTION 7.24.
Part 1. By KCL is( t) = iC1(t) + iC 2(t). Applying the v-i relationship for capacitor C1 and C2 we have:
iS (t) = (10mF + 20mF)d
dtvs(t)
= 30 ×10−3 × (12te−5t − 30t2e−5t )
= 0.36te−5t − 0.9t2e−5t A
Part 2. Applying the v-i relationship for the 20mH inductor, it follows that
vout (t) = (20mH )d
dt[12is( t)]
= 0.24 × 4.5t2e−5t − 3.6te−5t + 0.36e−5t( )= 1.08t2e−5t − 0.864te−5t + 0.0864e−5t (V )
Part 3 The energy stored in the 20mH inductor for t > 0 is given by
WL (t) =12
(20mH )[12iS( t)]2
= 1.44 0.36te−5t − 0.9t2e−5t( )2J
= 1.44 × t2e−10 t (0.36 − 0.9t)2 J
SOLUTION 7.25. We denote by vC(t) the voltage across the capacitors C1 and C2 and by iL(t) the currentthrough the inductors L1 and L2. The equivalent capacitance of the parallel combination of C1 and C2 is(C1 + C2) and thus:
iS (t) = (C1 + C2)dvC (t)
dt
Using the v-i relationship for the capacitor C2 it follows that
iC 2(t) = C2dvC (t)
dt
By replacing dvC ( t)
dt, it follows that
L&C Probs, 11/15/01 P:7-26 © R. A. DeCarlo, P. M. Lin
iC 2(t) =C2
C1 + C2× iS(t) =
23
is(t)
The equivalent inductance of the series combination of L1 and L2 is (L1 + L2) and thus:
9iC 2( t) = (L1 + L2)diL (t)
dt
Using the v-I relationship for the inductor L2 it follows that
vout (t) = vL2(t) = L2di2( t)
dt
By replacing diL (t)
dt, it follows that
vout (t) =L2
L1 + L2× 9iC 2( t) =
23
9iC 2( t) = 6iC 2(t) = 4is(t)
SOLUTION 7.26. Using the v-i relationship for the capacitor we can write:
vC 2( t) =1
C2iC ( )d
−∞
t
∫
vC1(t) =1C1
iC ( )d−∞
t
∫
where iC(t) is the current through the capacitors C1 and C2. Since C1 and C2 are connected in series:
vin (t) = vC1( t) + vC 2(t) =1C1
+1
C2
iC ( )d
−∞
t
∫
It follows that iC ( )d−∞
t
∫ = vin (t)C1C2
C1 + C2
. Hence
vC 2( t) =1
C2iC ( )d
−∞
t
∫ =1
C2×
C1C2C1 + C2
vin (t) =C1
C1 + C2vin ( t)
Since L1 and L2 are combined in parallel it follows that
vout (t) =1
L1+
1
L2
−1d(AvC 2( t))
dt
=L1L2
L1 + L2× A ×
dvC 2(t)dt
=L1L2
L1 + L2× A ×
C1C1 + C2
dvin ( t)dt
L&C Probs, 11/15/01 P:7-27 © R. A. DeCarlo, P. M. Lin
where the last equality follows by replacing vC2(t) with C1
C1 + C2vin (t)
SOLUTION 7.27. Observe first that the 0.3H and 0.6H parallel inductances combine to make a 0.2Hinductance. Also the 0.4H and 1.2H parallel inductances combine to make a 0.3H inductance. Finally, the0.2H and 0.3H inductances are combined in series and the equivalent inductance is 0.5H. Shortly, all theabove steps can be written as: Leq = (0.3H || 0.6H) + (0.4H || 1.2H) = 0.5H.
SOLUTION 7.28. Observe first that the 1mH and 5mH inductors combine to make a 6mH inductance.This inductance combines in parallel with the 3mH inductance to make a 2mH inductance. The next stepis to combine in series the 2mH inductance with the 10mH inductance. The equivalent inductance is12mH. This inductance is combined in parallel with the 36mH inductance and the result is 9mH. Finally,the 9mH inductance is combined in series with the 4mH inductance and the result is 13mH. Shortly, allthe above steps cam be written as: Leq = [(5mH + 1mH) || 3mH + 10mH] || 36mH + 4mH = 13mH.
SOLUTION 7.29. Step 1. The parallel combination of the 0.6mH and 1.2mH inductors is equivalent to a0.4mH inductor.Step 2. The series combination of the 2.4mH and 0.4mH inductors is equivalent to a 2.8mH inductor.Step 3. The parallel combination of the 2.8mH and 7mH inductors is equivalent to a 2mH inductor.
SOLUTION 7.30. The three inductors can be arranged in the seven fashions as shown below.
(a) (b) (c)
(d) (e)
(f) (g)
Leqa = 1mH
Leqb = 1mH + 1mH = 2mH
Leqc = 1mH || 1mH = 0.5mH
Leqd = 1mH + 1mH + 1mH = 3mH
Leqe = 1mH + (1mH || 1mH) = 1.5mH
Leqf = 1mH || (1mH + 1mH) = 0.667mH
Leqg = 1mH || 1mH || 1mH) = 0.333mH
L&C Probs, 11/15/01 P:7-28 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.31. For Fig. P7.31a, Leq = 2 + 15 + 10 + 10 + 40 +30 + 20 + 8 = 135mH
The circuit in Fig. P7.31b is equivalent to the following circuit.
Therefore, Leq = 2 + 15 + 10 + 8 = 35mH
The circuit in Fig. P7.31c is equivalent to the following circuit.
Therefore, Leq = 2 + [(15+10) || (10+40) || (20+30)] + 8 = 22.5mH
SOLUTION 7.32. When the switch is open, the circuit in Fig. P7.32 can be rearranged as the following.
Therefore, Leq = 8L || 8L = 4L
When the switch is closed, the circuit in Fig. P7.32 can be rearranged as the following.
Therefore, Leq = 4L || 4L + 4L || 4L = 4L
L&C Probs, 11/15/01 P:7-29 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.33. Leq1=Leq2 Without going into a detailed analysis, we present the following intuitiveargument. Note that the points a and b represent points on a balanced bridge circuit meaning that thevoltage between a and b would be zero. Therefore, no current will flow through the additional inductanceL. Therefore the presence of L does not affect the equivalent inductance value.
SOLUTION 7.34. Leq1>Leq2 Without going into a detailed analysis, we present the following intuitiveargument. Note that the points a and b represent points on an unbalanced bridge circuit, meaning that thevoltage between a and b would not be zero. Also note that when two inductors are placed in parallel, theequivalent inductance becomes smaller than either inductance. The addition of the inductor L in circuit 2essentially creates an internal parallel inductance resulting in an Leq2 lower than Leq1.
SOLUTION 7.35. First we add an iin label to the circuit as shown below.
From KVL and the derivative definition of the capacitor
vin (t) = vL1( t) + vL 2(t) = L1diin (t)
dt+ L2
diin ( t)dt
= L1 + L2( ) diin ( t)dt
Equivalently,diin ( t)
dt=
1L1 + L2( ) vin (t)
It follows that
vLk(t) = Lk
diin (t)dt
=Lk
L1 + L2vin ( t)
which is the required voltage division formula.
SOLUTION 7.36. Leq = (11mH) || (19.25mH) + 3mH = 10mH. Applying the voltage division formula,
vL1(t) = vin ( t) ×3mHLeq
= 60te−t mV
and
vL2( t) = vin (t) − vL1( t) =140te−t mV
L&C Probs, 11/15/01 P:7-30 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.37. First consider the circuit below which contains the additional label of vin(t).
From KCL and the integral definition of the inductor,
iin (t) = iL1(t) + iL 2(t) =1L1
vin ( )d0
t
∫ +1L2
vin ( ) d0
t
∫
Equivalently
vin ( )d0
t
∫ =1
1
L1+
1
L2
iin (t)
Therefore
iLk(t) =
1Lk
vin ( )d0
t
∫ =
1
Lk1
L1+
1
L2
iin ( t) =
L1L2
LkL1 + L2
iin (t) for k = 1,2
SOLUTION 7.38. In the circuit illustrated in Fig. 7.38,
Leq = (12mH + 27mH) || (130mH) = 30mH
Applying the current division formula,
iL1(t) =12 + 27
12 + 27 +130iin (t) = 0.231e−t 2
mA
iL 2(t) =130
12 + 27 +130iin ( t) = 0.769e−t 2
mA
Also,
vin (t) = Leqddt
iin ( t) = −60te−t 2 V
The instantaneous energy stored in the 130-mH is given by
WL1(t) =12
L1 iL1(t)[ ]2 = 0.8e−2t 2 nJ
L&C Probs, 11/15/01 P:7-31 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.39. Using the inductor v-i relationship,
vin (t) = Leqddt
iin ( t) = 30mH ×ddt
iin ( t)
The calculations for vin(t) for t = -2s to 7s are summarized in the following table:
Time Interval d/dt (iin(t)), mAs-1 vin(t), mV-2s< t <=0s 100 30s< t <=2s -200 -62s< t <=3s 0 03s< t <=4s 200 64s< t <=6s 100 36s< t <=7s -200 -6
From the solutions to problem 7.38,
iL1(t) = 0.231 iin ( t)
iL 2(t) = 0.769 iin (t)
Below are the plots for vin(t), iL1(t) and iL2(t).
L&C Probs, 11/15/01 P:7-32 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.40. Part 1 Using the current division formula,
iL 2(t) =18
18 + 6iin (t) = 90cos(300 t) mA
Part 2 Leq=1.5mH + 6mH || 18mH = 6mH
vin (t) = Leqddt
iin (t) = -0.679sin(300 t) V
Part 3 Instantaneous power delivered by the source is given by
p( t) = vin (t) × iin(t) = [-0.679sin(300 t)]× [120cos(300 t)] mW
= -40.7sin(600 t) mW
Below is a plot of p(t)
SOLUTION 7.41. Part 1 Leq = 0.3 || 0.9 + 0.4 || 0.4 = 0.425 H
For computing iin(t) we need to apply
iin (t) = iin (t0) +1
Leqvin ( )d
t0
t
∫
to each interval [0, 1], [1, 2], …, [n, n+1], …
The initial condition for the interval [n, n+1] for n even is:
iin (n) = iin (0) +1
Leqvin ( )d
0
n
∫
We assume that iin(0) = 0 and it follows that iin(n) = 0 for n even, since
L&C Probs, 11/15/01 P:7-33 © R. A. DeCarlo, P. M. Lin
vin ( )d = 00
n
∫for n even. The initial condition for the interval [n, n+1] for n odd is:
iin (t) = iin (n −1) +1
Leqvin ( )d
n−1
n
∫ =1
0.425(16 ) n−1
n = 37.6 A
For the interval [n, n+1] with n even:
iin (t) = iin (n) +1
Leqvin ( )d
n
t
∫ = 0 +1
0.42516 n
t = 37.6( t − n) A
For the interval [n, n+1] with n odd:
iin (t) = iin (n) +1
Leqvin ( )d
n
t
∫ = 37.6 +1
0.425(−16 ) n
t = 37.6( n +1− t) A
For the interval [n, n+1] with n even:
iin (t) = iin (n) +1
Leqvin ( )d
n
t
∫ = 0 +1
0.425(16 ) n
t = 37.6( t − n) A
Hence,
iin (t) =37.6(t − 2n), 2n < t ≤ 2n +1
37.6(2n + 2 − t), 2n +1< t ≤ 2n + 2
where n is a non-negative integer. Below is a sketch for the input current vs. time.
Part 2 The total energy stored in the set of four inductors is given by
L&C Probs, 11/15/01 P:7-34 © R. A. DeCarlo, P. M. Lin
W (t) =1
2Leq iin (t)[ ]2
= 300(t − 2n)2 J for 2n < t ≤ 2n +1
300(2n + 2 − t)2 J for 2n +1< t ≤ 2n + 2
where n is a non-negative integer.
Below is a sketch of W(t) vs. t.
Part 3 By the current division formula, the current through the 0.9H inductor is given by
iL 0.9H (t) =0.3
0.3 + 0.9iin (t)
=9.4( t − 2n) A for 2n < t ≤ 2n +1
9.4(2n + 2 − t) A for 2n +1< t ≤ 2n + 2
where n is a non-negative integer.
Below is a sketch of the iL0.9H vs. time.
L&C Probs, 11/15/01 P:7-35 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.42.. a)Step 1. The parallel combination of the 1µF and 2µF capacitances is equivalent to a 3µF capacitance.
Step 2. The series capacitors 1.5µF and 3µF combine to make a 1µF capacitance.
Step 3. Finally, the 2µF capacitance is combined in parallel with the 1µF capacitance that was the result of
Step 2 to make a 3µF capacitance.Shortly, the above steps can be written as:
Ceq = [(1µF + 2µF) || 1.5µF] + 2µF] = 3µF
b) Proceeding in a similar fashion as in part a:
Ceq = 30mF + [9mF || (9mF + 18mF) || 5.4mF] = 33mF
SOLUTION 7.43. a)Step 1. Combine the parallel capacitances of 1µF and 2µF to make a 3µF capacitance.
Step 2. Combine the series capacitances of 1.5µF and 3µF to make 1µF capacitance.
Step 3. Combine the parallel capacitances of 2.5µF and 1µF (the result of Step 2) to make a 3.5µFcapacitance.Shortly, the above steps can be written in a condensed form as:
Ceq = 2.5µF + [1.5µF || (1µF + 2µF) = 3.5µF
b) Proceeding in a similar fashion as in part a:
Ceq = (1mF || 2 mF) + (1.2mF || 2mF || 1.5mF) + (4mF || 2.6667mF || 3.2mF || 1.6mF) = 2.1667mF
SOLUTION 7.44.
Part 1 Ceq = ((( 30 || 60 + 40 ) || 30 + 40 ) || 30 + 40 ) || 30 + 40 = 60mF
Part 2 The value of vin(t) is given by
vin (t) = vin (0) +1C
iin ( )d0
t
∫ = 0 +1
0.06200sin(20 )d
0
t
∫ = 166.7 −166.7cos(20t) mV
SOLUTION 7.45. When the switch is open, the circuit in figure P7.45 can be represented by thefollowing.
L&C Probs, 11/15/01 P:7-36 © R. A. DeCarlo, P. M. Lin
Therefore, Ceq= 6.4C
When the switch is closed, the circuit in figure P7.45 can be represented by the following.
Therefore, Ceq= ( 4 + 4 ) || ( 16+ 16 ) = 6.4C
SOLUTION 7.46. Ceq1<Ceq2 Without going into a detailed analysis, we present the following intuitiveargument. Note that the points a and b represent points on an unbalanced bridge circuit, meaning that thevoltage between a and b would not be zero. Also note that when two capacitors are placed in parallel, theequivalent capacitance becomes bigger than either capacitance. The addition of the capacitor C in circuit 2essentially creates an internal parallel capacitance resulting in a Ceq2 higher than Ceq1.
SOLUTION 7.47. The three capacitors can be arranged in the seven fashions as shown below.
For configuration (a), Ceq = 1 || 1 || 1 = 0.3333µF
For configuration (b), Ceq = 1 || (1 + 1) = 0.6667µF
For configuration (c), Ceq = 1 || 1 + 1 = 1.5µF
For configuration (d), Ceq = 1 + 1 + 1 = 3µF
For configuration (e), Ceq = 0 + 1 + 1 = 2µF
For configuration (f), Ceq = 1 || 1 + 0 = 0.5µF
For configuration (g), Ceq = 1 + 0 + 0 = 1µF
L&C Probs, 11/15/01 P:7-37 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.48. As Q = CV for any capacitor, C=Q/V. The equivalent capacitance of the network isgiven by
Ceq =1C2V
||1C3V
||1C5V
= 0.1 F
SOLUTION 7.49. First consider the circuit below which contains the additional label of iin(t).
From KVL and the integral definition of the capacitor,
vin (t) = vC1( t) + vC 2(t) =1C1
iin ( )d +0
t
∫ 1C2
iin ( )d0
t
∫Equivalently,
iin ( )d0
t
∫ =1
1
C1+
1
C2
vin (t)
Therefore,
vCk (t) =1
Ckiin ( )d
0
t
∫ =
1
Ck1
C1+
1
C2
vin (t) =
C1C2
CkC1 + C2
vin (t)
Hence,
vC1(t) =C2
C1 + C2vin ( t)
vC 2( t) =C1
C1 + C2vin (t)
which is the required voltage division formula.
SOLUTION 7.50. By the voltage division formula,
vC 3(t) =6 ||6
6|| 6 + 9vin (t) = 21(1− e−20t ) V
SOLUTION 7.51. Part 1 By the voltage division formula,
L&C Probs, 11/15/01 P:7-38 © R. A. DeCarlo, P. M. Lin
vC1(t) =0.05 + 0.15
0.05 + 0.15 + 0.1vin ( t) = 6.667sin(120 t) V
vC 2( t) = vin (t) − vC1( t) = 3.333sin(120 t) V
Part 2 Let the 0.1µF capacitor be C1 and the 0.15µF capacitor be C2. The energies stored in the 0.1µF
and the 0.15µF capacitors are given by
WC1[0, t] =12
C1 vC12(t) − vC1
2(0)[ ] = 2.222sin 2(120 t) J
WC2[0,t] =12
C2 vC 22( t) − vC 2
2(0)[ ] = 0.8333sin2(120 t) J
SOLUTION 7.52. Part 1 As both the terminals of 0.08F are tied to the voltage source vin(t),
vC1(t) = vin ( t) =100e−2 t V
By the voltage division formula,
vC 2( t) =0.06
0.03 + 0.06vin (t) = 66.67e−2t V
Part 2 Let the 0.03-F capacitor be C2. The energy stored in the 0.03-F capacitor over the interval [0,t] isgiven by
WC2[0,t] =12
C2 vC 22( t) − vC 2
2(0)[ ] = 66.67 e−4 t −1[ ] J
SOLUTION 7.53. First consider the circuit below which contains the additional label of vin(t).
From KCL and the derivative definition of the capacitor
iin (t) = iC1(t) + iC 2(t) = C1dvin (t)
dt+ C2
dvin ( t)dt
= C1 + C2( ) dvin ( t)dt
Equivalently,dvin ( t)
dt=
1C1 + C2( ) iin (t)
It follows that
iCk(t) = Ck
dvin (t)dt
=Ck
C1 + C2iin ( t)
which is the required current division formula.
L&C Probs, 11/15/01 P:7-39 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.54. Part 1 By the current division formula,
iC1(t) =0.08
0.08 + 0.03 0.06iin ( t) = 80e−2 t A
iC 2(t) =0.03 0.06
0.08 + 0.03 0.06iin (t) = 20e−2t A
Part 2 Let the 0.03-F capacitor be C2.
vC 2( t) = vC2(0) +1
C2iC 2( )d
0
t
∫ = 333.3(1− e−2t ) V
The energy stored in the 0.03-F capacitor over the interval [0,t] is given by
WC2(0, t) =12
C2 vC 22( t) − vC 2
2(0)[ ] =1666.5(1− e−2t )2 J
SOLUTION 7.55. Part 1 By the current division formula,
iC 2(t) =2
1 + 2is(t) = 20sin(250t) mA
Part 2 By the voltage division formula,
vout (t) =2
1+ 2× 9ic2( t) = 0.12sin(250t) V
Part 3 The current in the 2mH inductor is given by
iL 2(t) = iL2(0) +1L2
vout ( )d0
t
∫ = −0.24 cos(250t) −1[ ] A
The energy stored in the 2-mH inductor is given by
WL2( t) =12
L2iL 22( t) = 57.6 cos(250t) −1[ ]2 J
SOLUTION 7.56. Part 1 The dual network N* is shown below.
Part 2 Since the equivalent conductance seen by Is* is 500S, the equivalent resistance seen by Is* is equalto 0.002Ω.
L&C Probs, 11/15/01 P:7-40 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.57. Part 1 The dual network N* is shown below.
Part 2 S*eq = 350S I*out = 8A.
SOLUTION 7.58. First we draw the graph, given below, of the circuit in figure P7.58. We construct thegraph associated with dual network graph for N* from the graph of N. The dual network graph is givenby the dashed lines.
This circuit and its dual have the branch characteristics given in the following table:
ORIGINAL NETWORK DUAL NETWORK
V1 = 40 V I1* = 40 A
V2 = 500 I2 I2* = 500 V2
*
I3 = 2 I2 V3* = 2 V2
*
V4 = 400I4 I4* = 400V4
*
V5 = 200 I5 I5* = 200 V5
*
Finally replace the branches by the elements given in the table above. This produces the dual networkbelow.
SOLUTION 7.59. The dual for the circuit in figure P7.59 is shown below.
L&C Probs, 11/15/01 P:7-41 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.60. First we draw the graph, given below, of the circuit in figure P7.60.
This circuit and its dual have the branch characteristics given in the following table:
ORIGINAL NETWORK DUAL NETWORK
V1 = 5 I1 I1* = 5V1
*
I2 = 8 A V2* = 8 V
I3 = 6 A V3* = 6 V
V4 = 4 V I4* = 4 A
V5 = 2 I5 I5* = 2V5
*
I6 = 7 A V6* = 7 V
Now we construct the graph associated with dual network graph for N* from the graph of N. The dualnetwork graph is given by the dashed lines.
L&C Probs, 11/15/01 P:7-42 © R. A. DeCarlo, P. M. Lin
The graph of the dual network then is pulled out and flipped vertically to produce the graph topology of thedual network, N*.
Finally replace the branches by the elements given in the table above. This produces the dual networkbelow.
SOLUTION 7.61. Part 1 We first redraw the circuit to eliminate branch crossing. The resultantschematic is shown below.
L&C Probs, 11/15/01 P:7-43 © R. A. DeCarlo, P. M. Lin
Then we draw the graph, given below, of the above circuit. We construct the graph associated with dualnetwork graph for N* from the graph of N. The dual network graph is given by the dashed lines.
The resultant dual network given by the above graph is shown below.
Part 2 The equivalent resistance Req* seen by the current source in the dual circuit is equal to 1/Req = 1/αΩ.
SOLUTION 7.62. Part 1 We first redraw the circuit to eliminate branch crossing. The resultantschematic is shown below.
L&C Probs, 11/15/01 P:7-44 © R. A. DeCarlo, P. M. Lin
Then we draw the graph, given below, of the above circuit. We construct the graph associated with dualnetwork graph for N* from the graph of N. The dual network graph is given by the dashed lines.
The resultant dual network given by the above graph is shown below.
Part 2 If we label the conductances in the dual network by their corresponding resistance values, we havethe circuit below.
L&C Probs, 11/15/01 P:7-45 © R. A. DeCarlo, P. M. Lin
By comparing to the original network, we conclude that Req* = Req.
Part 3 Since the general relationship between Req and Req* is given by Req= 1/Req*. We can write
Req= 1/Req* =1/Req
Solving the above equation, we know that Req = 1 Ω.
Part 4 To retain the special properties of parts (b) and (c), we require one resistance value must be areciprocal of the other one.
SOLUTION 7.63. The dual network is shown below.
SOLUTION 7.64. The dual network is shown below.
From the answer of problem 7.55,
iC2(t) = 20 sin (250t) mA, and vout(t) = 0.12 sin (250t) V
Therefore, vC2*(t) = 20 sin (250t) mV and iout* (t) = 0.12 sin (250t) A
The energy stored in the 2mF capacitor is given by
WC2* ( t) =
12
C2* vout
* (t)[ ]2=
12
L2 iL 2(t)[ ]2 = 57.6[cos(250t) −1]2 J
SOLUTION 7.65. The dual network is shown below.
L&C Probs, 11/15/01 P:7-46 © R. A. DeCarlo, P. M. Lin
The equivalent inductance of the dual circuit (Leq*) = (1+2) || 1.5 + 2.5 = 3.5µH.
SOLUTION 7.66. By voltage division
v1( t) =
1
0.41
0.4+
1
0.2
vs(t) =0.2
0.2 + 0.4vs(t) =
13
vs( t)
Note that the 0.4 F resistor in parallel with the dependent voltage source has no effect on i1(t) and hence isredundant. By the definition of a capacitor,
i1(t) = 0.2dv1( t)
dt=
115
×dvs( t)
dt
Finally, since the last capacitor is initially uncharged,
vout (t) =1
0.5i1( )d
0
t
∫ =215
dvs( )d
d0
t
∫ =2
15vs( t)
A sketch is given below.
SOLUTION 7.67. Following the method outlined in page 269 of the text, we require
vC (t0) −1C
i0( )d ≥140
1/200
∫With i0(τ) = 2A, vC(0) = 20 V, it follows that
20 −14 ≥1C
× 2A ×1
200s
C ≥ 1.667mFTherefore, the standard capacitor value of 1.8mF should be chosen for this application.
L&C Probs, 11/15/01 P:7-47 © R. A. DeCarlo, P. M. Lin
SOLUTION 7.68. From the relationships given in Problem 7.67, we have
vL (t) =ddt
L(t)iL( t)[ ] = I0ddt
L(t)[ ] = I0L1 41T1
−t
T12
Below is a plot of vL(t) vs t.
From the plot, we notice that when t=T1, the value of vL(t) changes from positive to negative. This meansthat vL(t) will change sign when a car is inside the loop. Therefore, we can make a circuit to monitor thevoltage vL(t) and whenever a negative voltage is detected, the left turn signal should be initiated during thenext traffic light change.
1st Order Circuit Probs 11/26/01 P8-1 © R. A. DeCarlo, P. M. Lin
SOLUTIONS PROBLEMS CHAPTER 8
SOLUTION 8.1. (a) By KCL, CdvC (t)
dt= −
vC (t)R
or dvC ( t)
dt+
vC (t)RC
= 0 . Using 8.12
vC (t) = vC (0)e−t / =10e−t V where = RC =1s . Plotting this from 0 to 5 sec
»t = 0:.05:5;
»vc = 10*exp(-t);
»plot(t,vc)
»grid
»xlabel('time in s')
»ylabel('vc(t) in V')
0 1 2 3 4 50
2
4
6
8
10
time in s
vc(t)
in V
TextEnd
(b) The solution has the same general form as for (a),iC ( t) = iC (0+)e−t / = −10
5 ×103 e−t = −2e−t mA =
−vC (t)R
.
1st Order Circuit Probs 11/26/01 P8-2 © R. A. DeCarlo, P. M. Lin
0 1 2 3 4 5-2
-1.5
-1
-0.5
0
time in s
ic(t)
in m
A
TextEnd
(c) By linearity, if vC(0) is cut in half, all resulting responses are cut in half. If vC(0) is doubled, then all
resulting responses are doubled. Alternately, one view this as a simple change of the initial condition
with the same conclusion reached from linearity.
SOLUTION 8.2. (a) From inspection of the general form, 8.12, 0.1/ = 1= 0.1/ RC ⇒ C = 0.1/R = 5
µF.
(b) Since τ = RC = 0.1, vC (t) = 10e−10 t V.
SOLUTION 8.3. (a) The general solution form is vC (t) = vC (0)e−t / = vC (0)e−t /ReqC
. Using the
given data, take the following ratio, vC (0.001)vC (0.002)
=18.3946.7668
= 2.7183 =vC (0)e−(0.001)/
vC (0)e−(0.002)/ = e0.001/ . Hence,
»K = 18.394/6.7668
K =2.7183
»tau = 1e-3/log(K)
tau =0.0010
»C = 5e-6;
»Req = tau/C
Req =200.0008
»% Req = R*4e3/(R+4e3)
1st Order Circuit Probs 11/26/01 P8-3 © R. A. DeCarlo, P. M. Lin
»R = Req*4e3/(4e3-Req)
R =210.5272
»vC0 = 6.7668/exp(-0.002/tau)
vC0= 49.9999 V
(b)
» % vC (t) = 50e−1000 t V
»t = 0:tau/100:5*tau;
»vc = vC0*exp(-t/tau);
»plot(t,vc)
»grid
»xlabel('Time in ms')
»ylabel('vC(t) in V')
SOLUTION 8.4. After one time constant the stored voltage, 8 V, decays to 8/e = 2.943 V. From the
graph, the time at which the output voltage is 2.94 V is approximately 0.19 s. Thus = 0.19 s, and R =
τ/C = 190 Ω.
SOLUTION 8.5. (a) The general form of the inductor current is iL ( t) = iL (0)e−t / = 0.15e−t / A where
= L / R = 2 ×10−3 s. Plotting iL ( t) = 0.15e−500 t A from 0 to 10 msec yields:
»t = 0:.01e-3:10e-3;
»iL = 0.15*exp(-t/2e-3);
»plot(t,iL)
»grid
1st Order Circuit Probs 11/26/01 P8-4 © R. A. DeCarlo, P. M. Lin
»xlabel('Time in s')
»ylabel('Inductor Current')
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.010
0.02
0.04
0.06
0.08
0.1
0.12
0.14
0.16
Time in s
Indu
ctor
Cur
rent
TextEnd
(b) Here vL (t) = −RiL (t) = −22.5e−500 t V.
0 0.001 0.002 0.003 0.004 0.005 0.006 0.007 0.008 0.009 0.01-25
-20
-15
-10
-5
0
Time in s
Indu
ctor
Vol
tage
in V
TextEnd
(c) Using linearity, for iL(0) = 50 mA, then vL (t) =−22.5
3e−500t = −7.5e−500t V and for iL(0) = 250
mA, then vL (t) = −22.5250150
e−500t = −37.5e−500t V.
SOLUTION 8.6. Since = L / R, we can solve for L = 5 mH. Then solving iL ( t) = iL (0)e−t / for iL(0)
at t = 4 µsec yields iL(0) =15 mA.
1st Order Circuit Probs 11/26/01 P8-5 © R. A. DeCarlo, P. M. Lin
SOLUTION 8.7. (a) We desire to solve iL ( t) = iL (0)e−t / for iL(0) and R in = 0.08 /(R +103).
Using the following ratio, iL (0.05 ms)iL (0.15 ms)
=9.1971.2447
=iL (0)e−(0.05m)/
iL (0)e−(0.15m)/ = e0.1×10−3 / = 7.3889. Hence
»K = 9.197/1.2447
K =7.3889
»tau = 0.1e-3/log(K)
tau =5.0000e-05
»L = 0.08;
»Req = L/tau
Req =1.6000e+03
»R = Req - 1000
R = 599.9862
»iL0 = 9.197e-3/exp(-0.05e-3/tau)
iL0 = 0.0250 A
(b) = 80m /(R +1000) = 50 sec , iL ( t) = 0.025e−t /50×10−6 A.
SOLUTION 8.8. By Ohm’s law, vR(0+) = −(4k ||16k)iL (0+) = −32 V. The time constant
= L /(4 k ||6k) = 25 sec , i.e.,
»Req = (4e3*16e3/(4e3+16e3))
Req =3200
»L = 0.08;
1st Order Circuit Probs 11/26/01 P8-6 © R. A. DeCarlo, P. M. Lin
»tau = L/Req
tau =2.5000e-05
Using the general equation, vR( t) = vR (0+)e−t / = −32e−t /25 V. Equivalently,
vR( t) = −ReqiL (t) = −Req 0.01e−4×104 t V.
SOLUTION 8.9. (a) Note that the Thevenin resistance seen by the capacitor is Rth:
»R1 = 360+60*120/(60+120)
R1 =400
»Rth = 400*1200/1600
Rth =300
Hence, vC (t) = vC (0)e−t / = 80e−t /0.15 V where = RTHC = 300 × 0.5 ×10−3 = 0.15 s.
0 0.2 0.4 0.6 0.8 10
10
20
30
40
50
60
70
80
Cap
acito
r V
olta
ge (
V)
TextEnd
Time in s
(b) Here iC (0+) = −vC (0+ ) / RTH = −0.2667 . Therefore iC ( t) = iC (0+)e−t / = −0.2667e−t /0.15 A.
Equivalently, iC ( t) =−vC (t)
Rth= −0.2667e−t /0.15 A.
(c) By voltage division, vR(0+) = vC (0)60||120
(60||120) +160 + 200
= 8 V; for t > 0 vR( t) = 8e−t /0.15 V.
1st Order Circuit Probs 11/26/01 P8-7 © R. A. DeCarlo, P. M. Lin
SOLUTION 8.10. First, find the Thevenin equivalent seen at the left of the inductor. Introducing a test
source in place of the inductor we obtain the following KCL equation at that node.
itest = vtest /1k + vtest − 200vtest1k
200 . Let vtest = 1 V. Then
»itest = 1e-3 + (1 - 200*1e-3)/200
itest =0.0050
»Rth = 1/itest
Rth =200
Thus Rth = 200 Ω, = L / Rth = 0.25 ms, and iL ( t) = 0.025e−4000 t A. Next from 8.13b find
vL (t) = −(RTH × iL (0))e−4000 t = −5e−4000 t V, and from Ohm’s law ix (t) = −5e−4000 t mA.
SOLUTION 8.11. For all parts it is necessary to find the Thevenin equivalent resistance seen by the
capacitor. To this end we apply an external test current to the remainder of the circuit to obtain:
vtest = R1itest + R2(itest + itest )
Thus
Rth = vtest / itest = R1 + R2(1+ ) = 120 + 70(1+ )
(a) With α = 4, Rth = 470 Ω, = RthC = 0.1175 s, and vC (t) = vC (0)e−t / = 50e−8.51t V.
0 0.1 0.2 0.3 0.4 0.5 0.6 0.70
10
20
30
40
50
Time in s
Cap
acito
r V
olta
ge (
V)
TextEnd
(b) With α = -4, Rth = −90 Ω, = RthC = −0.0225 s, and vC (t) = 50e44.44t V.
1st Order Circuit Probs 11/26/01 P8-8 © R. A. DeCarlo, P. M. Lin
0 0.02 0.04 0.06 0.08 0.1 0.120
1000
2000
3000
4000
5000
6000
7000
8000
Time in s
Cap
acito
r V
olta
ge (
V)
TextEnd
Note how this is not a stable design as V increases exponentially without bound.
(c) From the general equation developed at the beginning, Rth = 120 + 70(1+ ) > 0 requires that
> −2.7143.
SOLUTION 8.12. Find the Thevenin resistance left of the inductor. Forcing a test current source into
the output node,
vtest = R1itest + R2 itest − R1itest( ) = 100 + 50(1− 100)( )itest = 150 − 5000( )itest
and
Rth = vtest / itest =150 − 5000 Ω
(a) Using the above equation, Rth = −350 Ω and = L / Rth = −7.1429 ×10-5 s. Hence,
iL ( t) = 0.1e14000t A, an unbounded response due to the presence of the negative equivalent resistance.
(b) Rth = 100 Ω, = L / Rth = 2.5 ×10−4 s, vR(0+) = −R2(1− R1)iL(0) = 0 , but more importantly
vR2(t) = −R2(1− R1)iL( t) = 0 × iL ( t) = 0 .
(c) R1 + R2 1− R1( ) = 150 − 5000( ) > 0 implies < 0.03.
SOLUTION 8.13. Over a long period of time the inductor L, is seen as a short circuit. Thus at time 0-,
the current through the inductor is, by current division, Is/2. As such, iL ( t) = 0.5Is e−Rt /2L A. A sketch
will reveal an exponentially decreasing current from an initial 0.5Is A.
1st Order Circuit Probs 11/26/01 P8-9 © R. A. DeCarlo, P. M. Lin
SOLUTION 8.14. This is similar to problem 8.13. Here the current turns off at time zero instead of a
switch opening. By current division iL (0+) = Is /2 . The difference between this problem and problem
8.13 is that the Thevenin resistance seen by the inductor is different. Here, RTH = 2R ||0.5R = 0.4R . So
for t > 0, iL = (Is /2)e−R tht /L = 0.5Ise−Rt /2.5L A. A sketch of this function plotted with respect to this
new time constant will be identical to the one in problem 8.13.
Define the time constant of problem 8.13 as τold. The slower decay in the plot below represents
the fall of the inductor current for problem 8.14 relative to that of problem 8.13 which is the faster
decaying curve in the plot below.
1st Order Circuit Probs 11/26/01 P8-10 © R. A. DeCarlo, P. M. Lin
SOLUTION 8.15. Over a long period of constant applied voltage, a capacitor looks like an open circuit.
By voltage division, vC (0+ ) =34
Vs and τ = 3RC. Hence
vC (t) = vC (0+)e−t / = 0.75Vse−t /(3RC ) V
SOLUTION 8.16. Same problem as 8.15 for t < 0. For t > 0 the effective resistance changes.
Rth = 3R / /R = 0.75R and τ = 0.75RC. Thus, vC (t) = 0.75Vse−t /(0.75RC) V. Same behavior as in the
previous problem, except for a faster decay than in problem 8.15 due to a smaller effective resistance.
Note how the decreased resistance affects the RC and RL circuit differently.
SOLUTION 8.17. For t < 0 the inductor looks like a short circuit. Let R1 =1333/ /800 = 500 Ω. The
current supplied by the source is Is =12
100 + 500= 0.02 A. By current division,
iL (0+) = 0.02 ⋅1333
800 +1333= 0.0125 A
For t > 0, the switch is opened and the inductor sees only the 800 Ω resistor. Hence, = L / R = 25 µsec
and
1st Order Circuit Probs 11/26/01 P8-11 © R. A. DeCarlo, P. M. Lin
iL ( t) = 12.5e−40000 t mA
0 20 40 60 80 1000
2
4
6
8
10
12
14
Time in micro seconds
Indu
ctor
cur
rent
in m
A
TextEnd
SOLUTION 8.18. (a) For t < 0 the applied voltage is constant and at t = 0, the capacitor is like an open
circuit. By voltage division, vC (0+ ) = 3025k ||6.25k
(25k ||6.25k) +1k= 25 V. For 0 < t < 1 ms, the source is off
and the capacitor discharges through three resistors in parallel; thus
= RthC = 833.33 × 0.6 ×10−6 = 0.5 ms and vC (t) = 25e−2000 t V.
(b) From continuity vC (0.001)= 25e−2000(0.001) = 3.383. For t > 1ms, the capacitor keeps on
discharging through only one resistance, the 25 kΩ resistor; thus the new time constant is τnew = 15 ms,
and vC (t) = 3.383e−(t−0.001)/0.015 V.
(c) PlotvC (t) = 25e−2000 t u(t) − u( t − 0.001)[ ] + 3.383e−(t−0.001)/0.015u(t − 0.001) V
»t = 0:.01:12;
»vc = 25*exp(-2*t) .*(u(t)-u(t-1))+3.3834*exp(-(t-1)/15) .*u(t-1);
»plot(t,vc)
»grid
»xlabel('Time in milli-secs')
»ylabel('Capacitor Voltage in V')
1st Order Circuit Probs 11/26/01 P8-12 © R. A. DeCarlo, P. M. Lin
0 2 4 6 8 10 120
5
10
15
20
25
Time in milli-secs
Cap
acito
r V
olta
ge in
V
TextEnd
SOLUTION 8.19. (a) From Ohm’s law iL (0+) =54
60 + 30= 0.6 A. For t > 0, the Thevenin resistance
seen by the inductor is Rth = (60 + 30) || 720 = 80 Ω and τ = L/Rth = 1/160 s. Thus
iL ( t) = iL (0+ )e−t / = 0.6e−160t A. From Ohm’s law and current division
v(t) = −60 ×720
90 + 720× iL( t) = −32e−160 t V
(b) From continuity property, iL (0.01±) = 0.6−160(0.01) = 121.14 mA. For t > 10 ms, the Thevenin
resistance seen by the inductor is Rth = (690 + 30) || 720 = 360 Ω and τnew = L/Rth = 1/720 s. Hence,
iL ( t) = 121.14e−720(t−0.01) mA for t > 10 ms. From Ohm’s law and current division
v(t) = −690 ×720
720 + 720× iL (t) = −41.793e−720(t−0.01)u(t − 0.01) V
Therefore
iL ( t) = 0.6e−160 t u( t) − u( t − 0.01)[ ] + 0.12114e−720(t−0.01)u(t − 0.01) A
1st Order Circuit Probs 11/26/01 P8-13 © R. A. DeCarlo, P. M. Lin
0 5 10 150
0.1
0.2
0.3
0.4
0.5
0.6
0.7
Time in milli-seconds
Indu
ctor
Cur
rent
in A
TextEnd
SOLUTION 8.20. For both circuits we first compute the Thevenin resistance seen to the right of the
capacitor for 0 ≤ t ≤ 60 ms. If we excite the circuit to the right of the capacitor over this time interval,
then itest =vtest200
+(1− 0.25)vtest
100= 0.0125vtest . Let Rth1 =
10.0125
= 80 Ω.
(a) For t < 0, the capacitor acts as an open circuit. Using voltage division, vC (0+ ) =80
80 +133.380 = 30
V. For 0 ≤ t ≤ 60 ms, the time constant is 1 = Rth1C = 40 ms, and
vC (t) = vC (0+ )e−t = 30e−t (Rth1C ) = 30e−25t V
From continuity, vC (60+ ms) = 6.694 V. The new Thevenin resistance is Rth2 = 200 Ω. Thus for t >
60 ms, the time constant is 2 = 200C =100 ms, and vC (t) = 6.694e−10(t−0.06) V. The resulting
capacitor voltage is plotted below.
1st Order Circuit Probs 11/26/01 P8-14 © R. A. DeCarlo, P. M. Lin
0 50 100 150 200 2500
5
10
15
20
25
30
time in ms
Cap
acito
r V
olta
ge (
V)
TextEnd
(b) It is the same circuit as above for t < 0; thus vC (0+ ) = 30 V. However the Thevenin resistances seen
by the capacitor are different because there is no switch to disconnect the independent voltage source and
its series resistance. First for 0 ≤ t ≤ 60 ms, the Thevenin resistance to the right remains as
Rth1 =1
0.0125= 80 Ω. However, for 0 ≤ t ≤ 60 ms, the Thevenin resistance seen by the capacitor
changes to Rth3 = Rth1 //133.3 = 50 Ω. Then new time constant is 3 = Rth3C = 25 ms and for 0 ≤ t ≤60 ms
vC (t) = 30e−40t V
From continuity, vC (60+ ms) = 2.72 V. The new Thevenin resistance is Rth4 = 200/ /133.3 = 80 Ω.
Thus for t > 60 ms, the time constant is 4 = 80C = 40 ms, and vC (t) = 2.72e−25(t−0.06) V. The
resulting capacitor voltage is plotted below.
1st Order Circuit Probs 11/26/01 P8-15 © R. A. DeCarlo, P. M. Lin
0 50 100 150 200 2500
5
10
15
20
25
30
time in ms
Cap
acito
r V
olta
ge (
V)
TextEnd
(c) For t < 60 ms, the voltage decays faster in (b) due to the smaller time constant. Similarly, for t > 60
ms.
SOLUTION 8.21. Following, are the switching times with the time constants associated with them.
t = 0 ⇒ Rth = 20 kΩ ⇒ = 20 ms
t = 5 ms ⇒ Rth = 4 kΩ ⇒ = 4 ms
t = 7.5 ms ⇒ Rth = 800 Ω ⇒ = 0.8 ms
It follows that with t in ms,
vC (t) = 10e−0.050t u( t) − u(t − 5)[ ] + 7.788e−0.250( t−5) u(t − 5) − u(t − 7.5)[ ] + 4.169e−1.25(t−7.5)u(t − 7.5) V
1st Order Circuit Probs 11/26/01 P8-16 © R. A. DeCarlo, P. M. Lin
0 2 4 6 8 10 120
1
2
3
4
5
6
7
8
9
10
time in ms
Cap
acito
r V
olta
ge (
V)
TextEnd
*SOLUTION TO 8.22. This solution is done in MATLAB.
% Define switching times, inductance, and Thevenin equivalent resistances.
t1= 12e-6;t2=18e-6;t3=21e-6;L= 0.1;rth1= 800;rth2= 8000;rth3=1600;rth4= 32000;
% Define time constants for each of the four time intervals.
tau1= L/rth1tau2= L/rth2tau3= L/rth3tau4= L/rth4
tau1 = 1.2500e-04tau2 = 1.2500e-05tau3 = 6.2500e-05tau4 = 3.1250e-06
% Compute initial inductor currents for each of the four time intervals.
il1= 100e-3;il2=il1*exp(-t1/tau1)
1st Order Circuit Probs 11/26/01 P8-17 © R. A. DeCarlo, P. M. Lin
il3=il2*exp(-(t2-t1)/tau2)il4=il3*exp(-(t3-t2)/tau3)
il2 = 9.0846e-02il3 = 5.6214e-02il4 = 5.3580e-02
% Determine inductor currents for each of the four time intervals. Plot.
t = 0:0.5e-7:36e-6;seg1= il1*exp(-t/tau1) .*(ustep(t)-ustep(t-t1));seg2 =il2*exp(-(t-t1)/tau2) .*(ustep(t-t1)-ustep(t-t2));seg3=il3*exp(-(t-t2)/tau3) .*(ustep(t-t2)-ustep(t-t3));seg4= il4*exp(-(t-t3)/tau4) .* ustep(t-t3);iL=seg1 + seg2 +seg3 + seg4;
plot(t,iL)grid
SOLUTION 8.23.
For circuits with a forced voltage, equation 8.19c is used as a general solution,
vC (t) = vC (∞) + vC to+( ) − vC ∞( )[ ]e−
t
RTHC
.
(a) At time zero the voltage is 0 V. As time approaches infinity, the capacitor looks like an open withvoltage 10 V. The Thevenin resistance is 10 kΩ. Thus for t > 0.
1st Order Circuit Probs 11/26/01 P8-18 © R. A. DeCarlo, P. M. Lin
vC (t) = 10 + −10[ ]e−
t
2
=10(1− e−0.5t ) V.
(b) With vin(t) = 0 and vC (0+ ) = 5 V, vC (t) = 5e−
t
2
= 5e−0.5t V.
Part (a) Part (b)
(c) From linearity, vC (t) = 10(1− e−0.5t ) + 5e−0.5t = 10 − 5e−0.5t V. Using Ohm’s law,
iC ( t) =vin (t)
104 −vC ( t)
104 . Thus iC ( t) = 1− 1− 0.5e−0.5t( ) = 0.5e−0.5t mA.
0 2 4 6 8 100
0.1
0.2
0.3
0.4
0.5
time in s
Cap
acito
r C
urre
nt (
mA
)
TextEnd
(d) This is the same as (a), under the condition that the input is 1.5 times larger. Hence by linearity,
vC (t) = 1.5 ×10(1− e−0.5t ) = 15(1− e−0.5t ) V.
(e) By linearity, ANSWER = –2x(ANSWER to (b)) + 3x(ANSWER to (a)):
1st Order Circuit Probs 11/26/01 P8-19 © R. A. DeCarlo, P. M. Lin
vC (t) = −2 × 5e−0.5t + 3×10(1− e−0.5t ) =15(1− e−0.5t ) = 30 − 40e−0.5t V
SOLUTION 8.24.(a) At t = 0-, the capacitor looks like an open circuit; therefore, by voltage division and the continuity
property, vC (0− ) = vC (0+ ) =3R4R
Vs1 = 0.75Vs1 . Similarly, at t = ∞, vC (∞) =3R4R
Vs2 = 0.75Vs2 . The
circuit time constant is = (3R / /R)C = 0.75RC . Hence
vC (t) = 0.75Vs2 + 0.75 Vs1 −Vs2[ ]e−
t
0.75RC
.
(b) A sketch will show an exponentially varying voltage from 0.75Vs1 converging to 0.75Vs2 with thecomputed time constant.(c) The response to the initial condition when the inputs are set to zero, zero-input response, is
vC (t) = Vs1e−
t
0.75RC
. The zero order response, the response with 0V initial condition to a forced
voltage, is vC (t) = Vs2 −Vs2e−
t
0.75RC
.
SOLUTION 8.25.For RL circuits with a forced current, equation 8.19b is used as a general solution:
iL ( t) = iL (∞) + iL to+( ) − iL ∞( )[ ]e−R th
L(t−t o)
.
Since Rth = R = 100 Ω and L = 0.4 H, we have = 4 ms and
iL ( t) = iL (∞) + iL to+( ) − iL ∞( )[ ]e−250(t−to )
(a) Here, iL (0) = 0 and as time approach infinity, the inductor becomes a short and
iL (∞) =10 100 = 0.1 A. Thus iL ( t) = 0.11− e−250 t( ) A.
(b) Here iL (0) = −50 mA and because the input is zero, iL (∞) = 0 . Thus, iL ( t) = −0.05e−250t A. Plotsfor parts (a) and (b) appear below.
1st Order Circuit Probs 11/26/01 P8-20 © R. A. DeCarlo, P. M. Lin
0 5 10 15 20-0.06
-0.04
-0.02
0
0.02
0.04
0.06
0.08
0.1
Time in ms
Indu
ctor
Cur
rent
(A
)
TextEnd
(c) By linearity iL ( t) = 0.11− e−250 t( ) − 0.05e−250t = 0.1− 0.15e−250t A. Further, by KVL and Ohm's
law, vL (t) = vin ( t) −100iL ( t) implies vL (t) = 10 −10 +15e−250t = 15e−250t V.
0 5 10 15 200
5
10
15
Time in ms
Indu
ctor
Vol
tage
(V
)
TextEnd
(d) Observe that the new initial condition is –0.5 times the old one and that the new input voltage is 1.5times the old one. Therefore, by linearity,
iL ( t) = 1.5 × 0.11− e−250 t( ) + (−0.5) × −0.05e−250t( ) = 0.15 − 0.125e−250 t
and thus
vL (t) = 15 −15 +12.5e−250t = 12.5e−250t V
The plot is similar to part (c) with initial point 12.5 instead of 15.
SOLUTION 8.26. For this problem Rth = 2R / /0.5R = 0.4R in which case = L Rth = L 0.4R .
1st Order Circuit Probs 11/26/01 P8-21 © R. A. DeCarlo, P. M. Lin
(a) At t = 0, the inductor looks like a short circuit. Hence by current division, iL (0−) = iL (0+ ) = 0.5Is1 .A similar argument yields iL (∞) = 0.5Is2 . Using the general form of the solution,
iL ( t) = 0.5Is2 + 0.5 Is1 − Is2[ ]e− 0.4Rt
L .(b) A sketch will show an exponentially varying current from 0.5Is1 A converging to 0.5Is2 .(c) The response to the initial condition when the inputs are set to zero, zero-input response, is
iL ( t) = 0.5Is1e− 0.4Rt
L . The zero state response, the response with no initial condition, to the input
Is2u(t), is iL ( t) = Is2 1− e− 0.4Rt
L
.
SOLUTION 8.27. For this problem, vC (0) = 20 −10e−0.4t( )t=0
=10 V and
vC (∞) =10Is = 20 =t→∞lim 20 −10e−0.4t( ) . Hence Is = 2 A. Further,
iC ( t) = CdvC (t)
dt= 4Ce−0.4t = 0.4e−0.4t which implies that C = 0.1 F. Since
=1/0.4 = 2.5 = (10+ R)C = 0.1(10 + R) , it follows that R = 15 Ω.
SOLUTION 8.28.(a) The Thevenin resistance for this configuration is Rth = 1000/ /1000 = 500 Ω and = RthC = 0.25 s.
Hence vC (t) = vC (0+ )e−t =15e−4 t V is the zero-input response.
(b) Using a source transformation and voltage division, vC (∞) = 3 V. Thus vC (t) = 3 1− e−4 t( ) V.
(c) Here vC (∞) = 4 V, thus vC (t) = 4 1− e−4 t( ) V.
(d) This is the superposition of parts (b) and (c), i.e., vC (t) = 7 1− e−4t( )(e) The complete response is the superposition of parts (d) and (a), i.e., vC (t) = 7 + 8e−4 t V.(f) From linearity,
vC (t) = 0.5 × 7 1− e−4t( ) + 2 ×15e−4 t = 3.5 + 26.5e−4 t
SOLUTION 8.29. Solution done in MATLAB
%Problem 8.29
%RTH= (60||120)+120%tau=L/RTH
RTH=1/(1/60+1/120)+120;tau=0.2/RTH;
%Using superposition at t<0il01=36/120/2;il02=(72/180)*(60/180);il0= il01+il02;
1st Order Circuit Probs 11/26/01 P8-22 © R. A. DeCarlo, P. M. Lin
% At t>0 36volts is off thusilinf=il02;
t=0:5*tau/1000:5*tau;ilt=ilinf+(il0-ilinf)*exp(-t/tau);
plot(t,1000*ilt);xlabel('Time in seconds');ylabel('Current in mA');
0 1 2 3 4 5 6 7x 10-3
120
140
160
180
200
220
240
260
280
300
Time in seconds
Curr
ent i
n m
A
*SOLUTION 8.30.
» % Rth = 120 + 120\\60 = 160 kohm» % tau = Rth*C»tau = 160e3*0.5e-6tau = 0.0800» % Initial capacitor voltage is computed by» % voltage division and superposition»vc0 = 24+24vc0 = 48» % At t = ∞, capacitor looks like an open circuit. Hence»vcinf=24;»t = 0:1e-3:5*tau;»vct = vcinf+(vc0-vcinf)*exp(-t/tau);»plot(t*1000,vct)»grid»xlabel('Time in msec')»ylabel('Capacitor voltage (V)')
1st Order Circuit Probs 11/26/01 P8-23 © R. A. DeCarlo, P. M. Lin
0 50 100 150 200 250 300 350 40020
25
30
35
40
45
50
Time in msec
Cap
acito
r vo
ltage
(V
)
*SOLUTION 8.31A
»vc0 =0;»% Consider 0 ≤ t ≤ 0.5»vcinf = 50;»% Rth = 600\\300 = 200 ohms»% tau1 = Rth*C»tau1 = 300e-6*200»vc0 = 0;»t = 0:5e-3:1;»vct = (vcinf+(vc0-vcinf)*exp(-t/tau1)) .* (ustep(t)-ustep(t-.5));»»% Consider 0.5 ≤ t ≤ 1»tau2 = tau1»vc5 = (vcinf+(vc0-vcinf)*exp(-.5/tau1))»vcinf2 = 80;»vct2 = (vcinf2+(vc5-vcinf2)*exp(-(t-0.5)/tau1)) .* ustep(t-0.5);»vca = vct+vct2;»plot(t,vca)»grid»xlabel('Time in sec')»ylabel('Capacitor Voltage (V)')
1st Order Circuit Probs 11/26/01 P8-24 © R. A. DeCarlo, P. M. Lin
0 0.2 0.4 0.6 0.8 10
10
20
30
40
50
60
70
80
Time in sec
Cap
acito
r V
olta
ge (
V)
*SOLUTION 8.31B
»vc0 =0;»% for 0 ≤ t ≤ 0.5»vcinf = 75;»tau1 = 300e-6*300»vc0 = 0;»t = 0:5e-3:1;»vct = (vcinf+(vc0-vcinf)*exp(-t/tau1)) .* (ustep(t)-ustep(t-.5));»% for 0.5 ≤ t ≤ 1»tau2 = 300e-6*200»vc5 = (vcinf+(vc0-vcinf)*exp(-.5/tau1))»vcinf2 = 80;»vct2 = (vcinf2+(vc5-vcinf2)*exp(-(t-0.5)/tau1)) .* ustep(t-0.5);»vcb = vct+vct2;»plot(t,vcb)»grid»xlabel('Time in sec')»ylabel('Capacitor Voltage (V)')»pause»plot(t,vca,t,vcb,'b')»grid
1st Order Circuit Probs 11/26/01 P8-25 © R. A. DeCarlo, P. M. Lin
0 0.2 0.4 0.6 0.8 10
10
20
30
40
50
60
70
80
Time in sec
Cap
acito
r V
olta
ge (
V)
Comparison of the two responses.
0 0.2 0.4 0.6 0.8 10
10
20
30
40
50
60
70
80
SOLUTION 8.32.
%Problem 8.32
%Consider t < 0%vin=-20V, thusvc0=(8/10)*(-20);
%For 0<t<20msRTH=1/(1/2e3+1/8e3);tau1=5e-6*RTH;
1st Order Circuit Probs 11/26/01 P8-26 © R. A. DeCarlo, P. M. Lin
vcinf=(8/10)*20;t=20e-3;vc20ms=vcinf+(vc0-vcinf)*exp(-t/tau1);
%For 20ms<tRTH=1/(1/2e3+1/8e3+1/1.6e3);tau2=5e-6*RTH;
%By superpositionvcinf2=20*0.5+20*0.4;t=0:(40e-3)/1000:40e-3;
vct= (ustep(t)-ustep(t-20e-3)).*(vcinf+(vc0-vcinf).*exp(-t./tau1)) ...+ ustep(t-20e-3).*(vcinf2+(vc20ms-vcinf2).*exp(-(t-0.02)./tau2));
plot(t,vct);grid;xlabel('time in seconds');ylabel('Volts');
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04-20
-15
-10
-5
0
5
10
15
20
time in seconds
Volts
SOLUTION 8.33.(a)%Problem 8.33%(a)
%at t<0 only one source is contributing thus,il0=24/60*0.5;%For t>0RTH=60+1/(1/30+1/60);tau=16e-3/RTH;
%As t goes to infinity, by superposition,ilinf=24/(60)*0.5+24/80*1/3;t=0:5*tau/1000:5*tau;ilt=ilinf+(il0-ilinf).*exp(-t/tau);figure(1);plot(t,1000.*ilt);grid;xlabel('time in seconds');ylabel('Current in mA');
1st Order Circuit Probs 11/26/01 P8-27 © R. A. DeCarlo, P. M. Lin
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1x 10-3
200
210
220
230
240
250
260
270
280
290
300
time in seconds
Curr
ent i
n m
A
(b) Using 8.23, The time constant remains the same as in (a). In order to find the inductor voltage, wemust do so indirectly by solving for vL (t) = VA − 60IL . At t=0+,
iL (0+) = 200mA
VA = 246090
+ 243090
− iL (0+ )(60|| 30) = 20V
vL (0+ ) = x( t0+ ) = VA − 60iL (0+ ) = 8V
For t > 0, X e = vL (∞) = 0V , Thus vL (t) = 8e−5000 tV .
(c) By linearity,
iL ( t) = 600 − 200e−5000 tmA
vL (t) = 16e−5000 tV
VA
SOLUTION 8.34.
(a) Since the voltage has been constant for a long time, the capacitor acts as an open circuit. Thus by
voltage division and continuity, vC (0− ) = vC (0+ ) = 0.75Vo .(b) RTH = 6R ||18R || 3R =1.8R .
(c) For that period of time the switch is closed, vC (t) = 0.9Vo − 0.15Voe−t /1.8RC .
(d) Using the previous equation, and by continuity, vC (T− ) = vC (T + ) = 0.9Vo − 0.15Voe−T /1.8RC .(e) The time constant remains the same as the only difference is the source turning off.
(f) For t > T, vC (t) = vC (T + )e−(t−T )/1.8RC .(g)
1st Order Circuit Probs 11/26/01 P8-28 © R. A. DeCarlo, P. M. Lin
0 2 4 6 8 10 12 14 16 180
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
Normalized time in terms of tau
Norm
aliz
ed V
olts
in te
rms
of V
o
SOLUTION 8.35.
(a) For t < 0, The switch is closed, the current source off, and the voltage source has been providing a
constant voltage for a prolonged period of time. Thus, vC (0− ) = −50uV .
(b) Since voltage in continuous across a capacitor, vC (0+ ) = vC (0− ) = −50uV .(c) The thevenin resistance seen by the capacitor is RTH = 200||200 =100Ω , thus = RTHC = 0.2s .
(d) As t goes to infinity the capacitor voltage goes to 16V, thus vC (t) = 16 + (−50u −16)e−5tV .
(e) Again using the continuous property of a capacitor, vC (0.5+ ) = vC (0.5− ) = 14.687V .(f) For t > 0.5s, the switch is open, thus RTH = 200Ω . = RTHC = 0.4s .
(g) As t goes to infinity, the capacitor voltage goes to 32V. vC (t) = 32 + (14.687− 32)e−2.5(t−0.5)V .(h)
0 0.5 1 1.5 2 2.5-5
0
5
10
15
20
25
30
35
Time in seconds
Vo
lts
SOLUTION 8.36.
(a) vC (0+ ) = −5V(b) Doing so in matlab.%Problem 8.36b
%Initial conditionvc0=-5;%From 0<t<80usRTH=300e3;tau=RTH*(1/3)*1e-9;
1st Order Circuit Probs 11/26/01 P8-29 © R. A. DeCarlo, P. M. Lin
%So the initial condition on Voutvout0=(10+5)/300e3*60e3-5; %-2Vvoutinf1=10;t=80e-6;vout80us= 10+(vout0-10)*exp(-t/tau);
%For t > 80us%tau stays the samevoutinf2=-5;
t=0:160e-6/1000:160e-6;vout= (ustep(t)-ustep(t-80e-6)).*(10+(vout0-10).*exp(-t./tau)) ...+ustep(t-80e-6).*(-5+(vout80us+5).*exp(-(t-80e-6)./tau));plot(t,vout);xlabel('time in seconds');ylabel('Volts');grid;%Notice that Vout is vin-vct times a constant 60/300 plus vct%Thus Vout=vin(60/300)+vct(1-60/300)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6x 10-4
-2
-1
0
1
2
3
4
5
time in seconds
Vo
lts
(c)
1st Order Circuit Probs 11/26/01 P8-30 © R. A. DeCarlo, P. M. Lin
(d)
SOLUTION 8.37.
1st Order Circuit Probs 11/26/01 P8-31 © R. A. DeCarlo, P. M. Lin
If the source voltage has been –10 V for a long time then the switch is open and vC (0+ ) = −10V . Thetime constant with the switch open is = 5us . At t > 0, the input voltage changes to 20V. It then followsthat vC (∞) = 20V , and
vC (t) = 20 − 30e−t /5uV . Using the elapsed time formula, 8.24, we wish to find when the switch closes.
ta − 0 = 5us ⋅ln−10 − 200 − 20
= 2.03us . At that time the input voltage is still 20V, and the switch closes.
The time constant is now = 98ns , and vC (∞) = 0.39V . Note that because the voltage converges to avalue greater than zero, this time interval will be from 2.03us to 5us when the input changes back to–10V, thus
vC (t) = 0.39 − 0.39e−t /98nV .vC (5us) = 0.39V . At t > 5us, the voltage changes to –10 V, so
vC (t) = −0.2 + (0.39 + 0.2)e−t /98nV .Using the elapsed formula, we get tb = 0.1us for the voltage to go down to 0V again and cause the switchto open again. At this point the time constant becomes the original value again and
vC (t) = −10 +10e−t /5uV for t > (0.1+5) us.
SOLUTION 8.38.
(a) Introduce a test current source at the output and write KVL,
itest = vtest /200 + (vtest − 6V ) /200 + (6V − vtest ) / 400. Solving for itest = vtest3
400
− 0.015 . This
implies the following,RTH = 400 /3ΩiSC = 15mA
vOC = RTH iSC = 2V
(b) Using the general form, vC (t) = 2 − 8e−15000 t .
SOLUTION 8.39.
(a) Introduce a test voltage and solve for KVL,vtest = 5kitest +101(vtest +1− 40itest) −1 + 40itest
vtest = −10itest −1
vOC = −1V
RTH = −10Ω(b) The complete response is vC (t) = −1 + et V. Note that the voltage goes to infinity as t goes to infinitybecause of the negative time constant.
SOLUTION 8.40.
Compute the thevenin equivalent seen by the inductor at t > 0. Using KCL write,itest = ⋅100 ⋅ itest + (vtest −100itest + 25) /50 . Then one obtains the following,
1st Order Circuit Probs 11/26/01 P8-32 © R. A. DeCarlo, P. M. Lin
itest = vtest /125 +1/5 A
RTH = 125ΩiSC = −1/5 A
vOC = −25VA t < 0, the applied voltage has been –25V for a long time. Using the previously obtained thevenin
equivalent and linearity, iL (0−) = 1/5 A . iL (∞) = iSC = −1/5 A , So iL ( t) = 0.2 + 0.4e−6250 tA .
SOLUTION 8.41.(a) Introducing a test source and using KCL,
itest =vtest −1.5vs
40+
vtest
100
0.2 =vs
40+
1.5vs − vtest
40
Solving for the test source current in terms of the test voltage, itest =vtest50
− 0.12 A. Thus the thevenin
equivalent is,Rth = 50Ωvoc = isc * Rth = 6V
(b) vC (t) = 6 − 6e−200 t V.
(c) From linearity all the currents increase by the same ratio, thus vC (t) = 15 −15e−200 t V.(d)
(e)
1st Order Circuit Probs 11/26/01 P8-33 © R. A. DeCarlo, P. M. Lin
SOLUTION 8.42.
%Problem 8.42C= 1e-6;vc0=0;
%For 0 < t < 5msRth= 20e3;tau1= Rth*C;vcinf1= 50e-3*Rth;vc5ms= vcinf1+(vc0-vcinf1)*exp(-5e-3/tau1);
%For 5ms < t < 7.5msRth= 4e3;tau2= Rth*C;vcinf2= 50e-3*Rth;vc75ms= vcinf2+(vc5ms-vcinf2)*exp(-(7.5e-3-5e-3)/tau2);
%For t > 7.5msRth=800;tau3= Rth*C;vcinf3= 50e-3*Rth;t=0:12e-3/1000:12e-3;vct= (ustep(t)-ustep(t-5e-3)).*(vcinf1+(vc0-vcinf1).*exp(-t/tau1)) ...+ (ustep(t-5e-3)-ustep(t-7.5e-3)).*(vcinf2+(vc5ms-vcinf2).*exp(-(t-5e-3)/tau2)) ...+ (ustep(t-7.5e-3)).*(vcinf3+(vc75ms-vcinf3).*exp(-(t-7.5e-3)/tau3));plot(t,vct);grid;
1st Order Circuit Probs 11/26/01 P8-34 © R. A. DeCarlo, P. M. Lin
xlabel('time in seconds');ylabel('Volts');
0 0.002 0.004 0.006 0.008 0.01 0.0120
50
100
150
200
250
time in seconds
Vol
ts
SOLUTION 8.43.
(a) From the thevenin resistance RTH = 800Ω , 1 =125 s .(b) From the thevenin resistance RTH = 8kΩ , 2 = 12.5 s .(c) From the thevenin resistance RTH = 1.6kΩ , 3 = 62.5 s .(d) From the thevenin resistance RTH = 32kΩ, 4 = 3.125 s .(e) 0 mA.(f) In MATLAB:%Problem 8.43f
tau1= 125e-6;tau2= 12.5e-6;tau3= 62.5e-6;tau4= 3.125e-6;vs= 100;
il0= 0;ilinf1= vs/800;ilinf2= vs/8e3;ilinf3= vs/1.6e3;ilinf4= vs/32e3;
il12us= ilinf1+(il0-ilinf1)*exp(-12e-6/tau1);il18us= ilinf2+(il12us-ilinf2)*exp(-(18e-6-12e-6)/tau2);il21us= ilinf3+(il18us-ilinf3)*exp(-(21e-6-18e-6)/tau3);
t= 0:36e-6/1000:36e-6;ilt= (ustep(t)-ustep(t-12e-6)).*(ilinf1+(il0-ilinf1).*exp(-t/tau1))+ ...(ustep(t-12e-6)-ustep(t-18e-6)).*(ilinf2+(il12us-ilinf2).*exp(-(t-12e-6)/tau2))+ ...(ustep(t-18e-6)-ustep(t-21e-6)).*(ilinf3+(il18us-ilinf3).*exp(-(t-18e-6)/tau3))+ ...(ustep(t-21e-6)).*(ilinf4+(il21us-ilinf4).*exp(-(t-21e-6)/tau4));
plot(t,1000*ilt);grid;xlabel('time in seconds');ylabel('Current in mA');
1st Order Circuit Probs 11/26/01 P8-35 © R. A. DeCarlo, P. M. Lin
0 0.5 1 1.5 2 2.5 3 3.5 4x 10-5
0
5
10
15
time in secondsC
urr
ent i
n m
A
SOLUTION 8.44.
The first stage is a differentiator, and from 8.25, the output of the first op-amp is
= −RCdvin (t)
dt= 0.25RCe−0.25t . The second stage is an integrator and using 8.26,
vout (t) =−1
2RC(0.25RCe−0.25 d )
0
t∫ = 0.5 e−0.25[ ]0
t= 0.5e−0.25t
SOLUTION 8.45.
(a) First note the following relationships,
vout (t) = vC 2(t) =1
C2iC 2( )d
0
t∫
vin (t) = vC1( t)
iC 2(t) = −vin (t) / R − C1dvC1(t)
dt= − vin ( t)
R− C1
dvin (t)dt
Doing the appropriate substitution, and solving,
vout (t) = −1
C2Rvin ( )d
0
t∫ −
C1C2
vin ( t) +C1C2
vin (0).
(b) 4
C2Re−0.25t −1( ) +
C1C2
1− e−0.25t( ) V.
(c) −1C2R
sin( t) +C1C2
1− cos( t)( ) V.
SOLUTION 8.46.
These are two integrator in cascade. Using 8.26, the output of the first stage is
−1
RCvin ( )d = 2 cos(50 )[ ]0
t mV = 2cos(50 t) − 20
t∫ mV. Using the same equation again,
vout (t) = −1
RC(2cos(50 )d = −10 −
250
sin(50 )
0
t
mV = 2cos(50t)mV0
t∫
1st Order Circuit Probs 11/26/01 P8-36 © R. A. DeCarlo, P. M. Lin
*SOLUTION 8.47. The following solution is done in MATLAB
c= 1e-6;rf= 10e6;rs=1e6;tau=c*rf;vgain= -10/1;vofinal= -3*vgain;
% Part (a)
voinit= 0;t= 0: tau/100: tau;vout= vofinal + (voinit -vofinal).*exp(-t/tau);
% Time at which output voltage reaches saturation is tsat
tsat= tau*log((0 - vofinal)/(15- vofinal))
plot (t,vout)gridxlabel('Time in secs')ylabel('Output Voltage in volts')
tsat = 6.9315e+00
% Part (b)
voinit=-5;vout= vofinal + (voinit -vofinal).*exp(-t/tau);tsat= tau*log((-5 - vofinal)/(15- vofinal))plot (t,vout)gridxlabel('Time in secs')
1st Order Circuit Probs 11/26/01 P8-37 © R. A. DeCarlo, P. M. Lin
ylabel('Output Voltage in volts')
tsat = 8.4730e+00
(c) vC(0-) = 0. Observe saturation at about 6.4 seconds.
V(IVM)
Time (s)Leaky Integrator-Transient-7
(V)
+0.000e+000
+5.000
+10.000
+15.000
+0.000e+000 +2.000 +4.000 +6.000 +8.000 +10.000
vC(0-) = -5 V. Observe that the time of saturation is 8.198 seconds.
1st Order Circuit Probs 11/26/01 P8-38 © R. A. DeCarlo, P. M. Lin
V(IVM)
Time (s)Prb Sol 8.47-Transient-16
(V)
-5.000
+0.000e+000
+5.000
+10.000
+15.000
+0.000e+000 +2.000 +4.000 +6.000 +8.000 +10.000
Note that in both cases the time of saturation is much lower than in the MATLAB computations which
assume an ideal op amp. In the Burr Brown model used by the SPICE simulation of this circuit, the
input resistance is 2 MΩ which is comparable with the external input resistance. Hence the assumption
of an infinite input resistance is not valid for the SPICE simulation and causes the discrepancy in the
time of saturation. However, if the external input resistance is changed to 10 kΩ and the feedback
resistance to 100 kΩ with a corresponding change in the capacitor to 100 µF, one obtains results
comparable to the MATLAB computations.
SOLUTION 8.48.Since the op-amps do not load the first stage of the circuit, we can find its transfer function for the op-
amp stage as vout (t)
v + (t)= (1+ K) .
(a) From the problem statement, we know that the overall function, is a scaled integrator. As the op-amp
stage only provides gain it is logical to assume that the R-C stage will perform the integration of the input
times some constant, G. With this in mind we have
vout (t) = G(1+ K) 10sin( )d =−10G(1+ K)
cos( t) −1( )0
t∫ , where G must be negative.
(b) Using the same reasoning, vout (t) = G(1+ K)d(10sin( t))
dt= 10G(1+ K) cos( t)where G is
positive.
For low frequency (a) yields a big output, while (b) a small one. For high frequency the reverse happens.
1st Order Circuit Probs 11/26/01 P8-39 © R. A. DeCarlo, P. M. Lin
SOLUTION 8.49.
For (b) the integral i-v relationship is
v(t) = vCeq(0+ ) +1
Ceqi(
0
t∫ )d = vC1(0+ ) + vC 2(0+ ) +
C1 + C2C1C2
i( )d0
t∫ . Repeating the same for (a),
vC1(t) = vC1(0+ ) +1
C1i( )d
0
t∫
vC 2( t) = vC2(0+ ) +1
C2i( )d
0
t∫
By KVL the two capacitor voltage can be added together, thus give the same relationship as for (b).
SOLUTION 8.50.
First calculate Ceq = C1 ||C2 = 0.2F . Then find the initial voltage
vCeq(0+ ) = vC1(0+ ) + vC 2(0+ ) = 30V . The final voltage will be 12V, and the time constant is
= Req ⋅Ceq = 0.4s . Thus vout (t) = 12 + (30 −12)e−2.5tV .
*SOLUTION 8.51. (a) After the switch closes, we have the circuit shown below.
From Chapter 7, Ceq = 0.5 F and vC (0+ ) = vC1(0+ ) − vC 2(0+ ) = 2 − 0 = 2 V. Hence,
iR (t) = iR(0+ )e−t / = iR (0+ )e−t /RC eq =
vC (0+ )R
e−t /RCeq = 4e−4 tu( t) A
(b) For this part we apply the integral definition of the capacitor. Specifically,
1st Order Circuit Probs 11/26/01 P8-40 © R. A. DeCarlo, P. M. Lin
vC1(t) = vC1(0+ ) +1
C1iC1( ) d
0
t
∫ = 2 − iR( ) d0
t
∫ = 2 − 4 e−4 d0
t
∫ = 1+ e−4 t V
and
vC 2( t) = vC2(0+ ) +1
C2iC 2( ) d
0
t
∫ = 0 + iR ( )d0
t
∫ = 4 e−4 d0
t
∫ =1− e−4 t V
(c) The energy stored at t = 0+ for each capacitor is:
WC1(0+ ) = 0.5C1vC12 (0+ ) = 2 J
and
WC2(0+ ) = 0.5C2vC 22 (0+ ) = 0 J
Further at t = ∞,
WC1(∞) = 0.5C1vC12 (∞) = 0.5 J
and
WC2(∞) = 0.5C2vC 22 (∞) = 0.5 J
Computing total instantaneous stored energies, we have
WCtot(0+ ) = 2 J and WCtot(∞) = 1 J
Hence the decrease in stored energy from 0+ to ∞ is 1 J.
(d) Computing the energy dissipated in the resistor over [0+, ∞) is
WR(0,∞) = R iR2 ( )d
0
∞
∫ = RvC
2 (0+ )
R2
e−2t /RCeq d
0
∞
∫ =vC
2 (0+ )R
×RCeq
−2× e
−2 t /RCeq ]0
∞
=CeqvC
2 (0+ )
2= 1 J
(e) From the expressions developed in part (d), the dissipated energy is independent of the value of R.
R only affects the rate at which energy is dissipated. Clearly, the energy stored at 0 is 2 J while
the energy dissipated over [0,∞) is 1 J and the remaining energy at t = ∞ is 1 J. Hence
conservation of energy is verified.
1st Order Circuit Probs 11/26/01 P8-41 © R. A. DeCarlo, P. M. Lin
SOLUTION 8.52. (a) Using the relations developed in P8.49,
Ceq = 0.2F
vCeq =1− 0.5 = 0.5V
iR (t) = iR(0+ )e−t / = iR (0+ )e−t /RC eq =
vCeq(0+ )
Re
−t /RC eq = e−10 tu(t)
(b) For this part we apply the integral definition of the capacitor. Specifically,
vC1(t) = vC1(0+ ) +1
C1iC1( ) d
0
t
∫ =1− iR( ) d0
t
∫ =1− e−10 d0
t
∫ =1
10(9 + e−10 t ) V
and
vC 2( t) = vC2(0+ ) +1
C2iC 2( ) d
0
t
∫ = 0.5 + 4 iR ( )d0
t
∫ = 0.5 + 4 e−10 d0
t
∫ =1
10(9 − 4e−10t ) V
(c) The energy stored at t = 0+ for each capacitor is:
WC1(0+ ) = 0.5C1vC12 (0+ ) = 0.5 J
and
WC2(0+ ) = 0.5C2vC 22 (0+ ) = 31.25mJ
Further at t = ∞,
WC1(∞) = 0.5C1vC12 (∞) = 405mJ
and
WC2(∞) = 0.5C2vC 22 (∞) =101.25mJ
Computing total instantaneous stored energies, we have
WCtot(0+ ) = 531.25mJ and WCtot(∞) = 506.25mJ
Hence the decrease in stored energy from 0+ to ∞ is 25 mJ.
(d) Computing the energy dissipated in the resistor over [0+, ∞) is
1st Order Circuit Probs 11/26/01 P8-42 © R. A. DeCarlo, P. M. Lin
WR(0,∞) = R iR2 ( )d
0
∞
∫ = RvCeq
2 (0+ )
R2
e
−2 t /RCeq d0
∞
∫ =vCeq
2 (0+ )
R×
RCeq
−2× e
−2 t /RCeq ]0
∞
=CeqvCeq
2 (0+ )
2= 25mJ
(e) From the expressions developed in part (d), the dissipated energy is independent of the value of R.
R only affects the rate at which energy is dissipated. Clearly, the energy stored at 0 is 531.25 mJ
while the energy dissipated over [0,∞) is 506.25 mJ and the remaining energy at t = ∞ is 25 mJ.
Hence conservation of energy is verified.
SOLUTION 8.53. As all the switches are open initially, the initial current through the inductors is 0A.
For 0 < t < 50 ms, iL ( t) = 54.54 − 54.54e−20000 t mV. At t > 50 ms, the equivalent inductance is 10 mH,
the initial current through the 110 mH inductance is 54.54 mA, and through the 11 mH inductance 0 A.
So assuming the current splits equally between the two branches in steady state,
iL1 = 27.27 + (54.54 − 27.27)e−220000 t
iL 2 = 27.27 − 27.27e−220000t
SOLUTION 8.54. (a) Charges will distribute in order to achieve equal voltage by KVL. Since q=CV,
vR(0−) = 0V , due to equal capacitance the charges will distribute half and half, vR(0+ ) = 0.5V .
(b) The equivalent capacitance is 2 F, thus vR( t) = 0.5e−0.5t .
*SOLUTION 8.55. (a) Writing a node equation at v we have for all t,
4dvdt
+v4
+ 4ddt
v − vs( ) +v − vs( )
2= 0 (*)
Equivalently,
8dvdt
= −3v4
+ 4dvsdt
+vs2
(**)
Grouping terms and dividing by 8 yields when t > 0,
dvdt
= −332
v +116
(***)
1st Order Circuit Probs 11/26/01 P8-43 © R. A. DeCarlo, P. M. Lin
Notice that vs = 1 for t > 0 and for t > 0, dvs( t)
dt=
du(t)dt
= 0.
(b) By inspection v(0-) = 0; both capacitors are uncharged at 0
-. Recall from part (a) that KCL at the
node for v yields (*) which is equivalent to (**). Since conservation of charge follows by
integrating (*) or equivalently integrating (**) we have
8dvd
0−
0+
∫ d = −34
v
0−
0+
∫ d + 4dvsd
0−
0+
∫ d + 0.5 vs
0−
0+
∫ d
Since the integral of a finite integrand over an infinitesimal interval is zero, we have equivalently,
8dvd
0−
0+
∫ d = 0 + 4dvsd
0−
0+
∫ d + 0
Evaluating these integrals we obtain
8 v(0+ ) − v(0−)( ) = 4 vs(0+ ) − vs(0− )( ) = 4u(0+ ) = 4
(c) Since v(0-) = 0, v(0
+) = 0.5 V. Since v satisfies (***), i.e.,
dvdt
= −332
v +116
≡ −1
v + F (***)
from equation 8.17,
v(t) = F + v(0+ ) − F[ ]e−t / =23
−16
e−3t /32
u(t) V
Using v(0-) would have led to an incorrect answer.
SOLUTION 8.56. (a)
x(t) = K1e−t / + K2
dx( t)
dt= −
K1 e−t /
doing the substitution, −K1 e−t / = −
1(K1e−t / + K2) + F . In order to satisfy the equality, K2 = F .
1st Order Circuit Probs 11/26/01 P8-44 © R. A. DeCarlo, P. M. Lin
(b) x(t0+ ) = K1e
−t 0 / + F , and K1 = x( t0+ ) − F[ ]et 0 / .
(c) x(t) = K1e−t / + K2 = x( t0+ ) − F[ ]e−(t−t0 )/ + F .
SOLUTION 8.57. (a) From the graph, the initial and final values are 0 and 80 V respectively. That sets
the following constraint, 100R2
R1 + R2= 80V . From vC ( ) = 80 − 80e−1 = 50.57V . Looking at the Graph
= 5ms . Thus R1 ||R2 ⋅C = 5ms . Solving, R1 = 6250Ω and R2 = 25kΩ .
(b) Using the same equalities, R1 = 0.25R2 = 2kΩ , and C =2k || 8K
= 3.125uF .
SOLUTION 8.58. (a) From the graph, the initial and final values are 0 and 100 mA respectively. Thus
R1 = 200 /100m = 2kΩ. From iL ( ) = 100 −100e−1mA = 63.21mA , the graph shows a = 20ms . Thus
L /(R1 || R2) = 20ms, and R2 = 0.25mΩ.
(b) R1 stays the same, L = 20ms(R1 || R2) = 20H .
SOLUTION 8.59. This question is done in matlab
%Problem 8.59
tau1= 20*1;
tau2= 1;
%For 0 < t < ta
vo= 0;
vinfa= 10;
%Using the elapsed time formula,
ta=tau1*log((0-10)/(9-10));
%For ta < t < tb
vinfb=0;
tb=tau2*log((9-0)/(1-0));
%For tb < t < tc
vinfc=vinfa;
tc=tau1*log((1-10)/(9-10));
1st Order Circuit Probs 11/26/01 P8-45 © R. A. DeCarlo, P. M. Lin
%Next switching is just a repeat of ta < t < tb
t1=ta;
t2=ta+tb;
t3=t2+tc;
t4=t3+tb;
t5=t4+tc;
t=0:t5/1000:t5-1/1000;
vt= (ustep(t)-ustep(t-t1)).*(10-10.*exp(-t/tau1))+ ...
(ustep(t-t1)-ustep(t-t2)).*(9.*exp(-(t-t1)/tau2))+ ...
(ustep(t-t2)-ustep(t-t3)).*(10-9.*exp(-(t-t2)/tau1))+ ...
(ustep(t-t3)-ustep(t-t4)).*(9.*exp(-(t-t3)/tau2))+ ...
(ustep(t-t4)-ustep(t-t5)).*(10-9.*exp(-(t-t4)/tau1));
Frequency= 1/(tb+tc)
plot(t,vt);
grid;
xlabel('time in seconds');
ylabel('Volts');
(b) The frequency is 0.0217 Hz.
0 20 40 60 80 100 120 1400
1
2
3
4
5
6
7
8
9
time in seconds
Volts
SOLUTION 8.60. When the switch is in position A, =18.18ms . In position B it is = 99.5us . Using
the elapsed time formula, find ta, when the output voltage reaches 90 V.
1st Order Circuit Probs 11/26/01 P8-46 © R. A. DeCarlo, P. M. Lin
ta =18.18msln60 −136.3690 −136.36
= 9.07ms . At this point the switch goes to B, and the elapsed time until the
voltage reaches 60 V is tb = 99.5usln90 − 59.4560 − 59.45
= 0.4ms . Adding both time, F =105.6Hz .
SOLUTION 8.61. (a) The circuit can be rearranged in a series of one Vsolar V voltage source, one Lstore
inductor, and one Rstore+Rsolar resistor.
(b) iL ( t) =Vsolar
Rsolar + Rstore1− e−(Rstore+Rsolar )t /L[ ]A .
(c) In this time period the circuit reduces to an Lsotre inductor in series with a Rstore+R1 resistor.
(d)
iL (T1−) =
Vsolar
Rsolar + Rstore
iL ( t) = iT1
− e−(Rstore+ R1)(t−T 1)/L A.
(e) The two elements in series are an Lstore inductor and a resistor Req = Rstore + (R1 || R2).
(f)
iL (T2−) = i
T1− e−(Rstore+ R1)(T2−T 1)/Lsotre A
iL ( t) = iT2
− e−(Re q)(t−T 2)/LstoreA
(g)
PVsolar = VsolariL (t) =V 2
solar
Rsolar + Rstore1− e−(Rstore+Rsolar)t /L[ ]W
PRsolar = RsolariL2( t)W
PRstore = RstoreiL2( t)W
PLstore = LstoreiL ( t)diL (t)
dtW
(h) WL (0,t) =12
LiL2(t)J
SOLUTION 8.62. The light turns off when the current through it goes down to 0.5 mA. This
corresponds to ib = 10uA , and a voltage across the capacitor of vC = ib(R1 + 2k) + 0.5 = 0.7V . The time
constant of this circuit is = (R1 + 2k) || 5k ⋅1000uF = 4s. The final voltage across the capacitor is by
voltage division, 0.1V. Thus using the elapsed time formula t1 = 4ln1.5 − 0.10.7 − 0.1
= 3.39s .
1st Order Circuit Probs 11/26/01 P8-47 © R. A. DeCarlo, P. M. Lin
SOLUTION 8.63. (a) Since RC = 10-3
s, from equation 8.25, va( t) = −RCdvs(t)
dt= cos(1000t) V.
Hence vb( t) = −RCdva( t)
dt== − sin(1000t) V, and vout (t) = −
R f
Rvb(t) = sin(1000t) = vs( t) V.
(b) With the switch moved to position B, there is no source in the circuit. But the output at the
switching instant is sin(1000t) V which coincides with vs(t). Hence, the input to the first amp remains
the same and the circuit continues to produce vout (t) = sin(1000t) V, i.e., the circuit becomes an
oscillator.
*SOLUTION 8.64. Before attacking the problem proper, consider driving an ideal unity gain integrator
with the square wave of figure P8.64b. If we start the integration when the square wave goes positive,
then we have a triangular waveform as follows:
On the other hand, if we start the integration when the square wave goes negative, we get the following
waveform
One concludes that without some further physical assumptions, there is no unique solution to this
problem.
Physically speaking all capacitors have a leakage resistance. Hence, in modeling the capacitor we
put a very large resistance in parallel with an ideal C, producing a nearly ideal leaky integrator circuit.
The leaky integrator circuit has a first order response. Hence over time, when the circuit reaches steady
state, the dc level of the resulting output waveform will be proportional to the average level of
1st Order Circuit Probs 11/26/01 P8-48 © R. A. DeCarlo, P. M. Lin
the square wave which is zero in this case.
Comments: Actually the proportionality constant in the above statement is the overall dc gain of the
integrator-inverter. See the formulas given in P22.16. Adding up the two formulas, we have
Output(t)max + Output(t)min = K (Vmax + Vmin)
and K = H(0), i.e., the dc gain of the first order low pass system. This leads to
Average of output = (dc gain) (average of input levels)
See the analysis in example 8.7 and later an exact analysis is given in problems 22.15 and 22.16. In
other words, one would expect that the output of our (leaky) integrator in steady state to be given by the
waveform below.
Now we can start to solve the problem. The first part is to design a (leaky) integrator circuit to
produce a triangular waveform of value 2 V peak-to-peak. For this we consider the following figure
which consists of the leaky integrator followed by an inverter.
1st Order Circuit Probs 11/26/01 P8-49 © R. A. DeCarlo, P. M. Lin
To handle this analysis, recall that i ≅ ∆q/∆t in which case ∆q = C∆v. Hence, to have a peak-to-peak
voltage at v2(t) and v3(t) of 2 V, we require that
∆v =iin∆t
C=
9R1
×50 ×10−6
C= 2
Hence R1C = 2.25×10-4. If we choose R1 = 10 kΩ, then C = 22.5 nF. At this point the waveform of
v3(t) is given below.
In order to complete the design, we must raise the portion of the curve with positive slope by 1 V and
lower the portion with negative slope by –1 V. This can be done by adding one-ninth of vin(t) to v3(t).
This can be done by using the following circuit. In this circuit, there is a voltage-divider at the non-
1st Order Circuit Probs 11/26/01 P8-50 © R. A. DeCarlo, P. M. Lin
inverting terminal of the second op amp. Here V+ equals one-eighteenth of vin(t). However the gain of
the non-inverting portion is 2; therefore one-ninth of the input is added to v3(t) as desired.
CHAPTER 9 PROBLEM SOLUTIONS
SOLUTION TO PROBLEM 9.1. If we can compute expressions for K and q that are real,then these quantities exist by construction. Consider that A, B, K and q must satisfy thefollowing relationship:
K cos(ωt + θ) = K cos(θ)( )cos(ωt) + −K sin(θ)( )sin(ωt) ≡ Acos(ωt) + Bsin(ωt)
Therefore K cos(θ) = A and −K sin(θ) = B . Consequently,
K cos(θ)( )2 + −Ksin(θ)( )2 = K2 = A2 + B2
in which case K = A2 + B2 . Further,
K sin(θ)
K cos(θ)= tan(θ) =
−B
Ain which case
θ = tan−1 −B
A
with due regard to quadrant.
SOLUTION TO PROBLEM 9.2. For the inductor,
WL (t) = 12 L ⋅ iL
2(t) =1
2L Vo
C
Lsin
1
LCt
2
=CV0
2
2sin2 1
LCt
and for the capacitor,
WC (t) = 12 C ⋅vC
2 (t) = 12 C Vo cos
1
LCt
2
=CV0
2
2cos2 1
LCt
.
Hence,
WC + WL = 12 C ⋅ vC
2 (t) + 12 L ⋅ iL
2(t) =CV0
2
2sin2 1
LCt
+ cos2 1
LCt
=
CV02
2
SOLUTION TO PROBLEM 9.3. Since x(t) = (K1 + K2t)e−αt ,
x' (t) = − α ⋅K1e−αt + K2e−αt − α ⋅t ⋅ K2e−αt
and
x' ' (t) = α2 ⋅ K1e−αt − α ⋅K2e−αt − α ⋅K2e−αt + α2 ⋅ t ⋅ K2e−αt
Substituting into the differential equation, we have
α2 ⋅ K1e−αt − 2α ⋅K2e−αt + α2 ⋅ t ⋅ K2e−αt
+ 2α − α ⋅K1e−αt + K2e−α t − α ⋅t ⋅ K2e−αt[ ] + α2 K1e−αt + K2te−αt[ ] = 0
This means that the solution form satisfies the differential equation.
SOLUTION TO PROBLEM 9.4.
(a) Suppose x(T ) = 0 at some T. Then K1es1T = −K2es2T . Since esiT > 0 whenever si
is real and T is finite, K1 & K2 must have opposite signs.
(b) For this we solve for T and show there can only be one solution. Since
K1es1T = −K2es2T and esiT > 0 ,
K1
−K2=
es2T
es1T implies ln
K1
−K2
= ln
es2T
es1T
= s2 − s1( )T
Hence the unique solution is given by
T = ln
K1
−K2
s2 − s1( )
provided s2 ≠ s1 which is the case for distinct roots.
SOLUTION TO PROBLEM 9.5. Suppose x(T ) = 0 at some T >0. This is true if and only if
K1es1T = −K2Tes1T (*)
Since es1T > 0 and T > 0, (*) is true if and only if K1 = −K2T which is true if and only ifK1 & K2 have opposite signs.
SOLUTION 9.6. (a) Denote one period of oscillation by T. Then by definition9950T = 2π . Hence, T = 0.63148 ms. The time constant of decay is 1 ms. Therefore,NT = N × 0 . 6 3 1 4 8 = 1. Hence N =1.5836 cycles.
(b) Here observe that the time constant of decay is 1/σ s. Hence NT = N2πωd
=1
σ. One
concludes that N =ωd
2πσ.
SOLUTION 9.7. The differential equation for the capacitor voltage is
d2vC (t)
dt2 +1
LCvC (t) =
1
LCVs
For t > 0, the characteristic equation is s2 +1
LC= 0 . Hence from table 9.2, the solution
for either the inductor current or capacitor voltage has the general form
vC (t) = Acos(ωdt) + Bsin(ωdt) + XF
where ωd =1
LC. From table 9.2, XF =
Vs
LC
1
LC
= Vs. Further,
vC (0+ ) = A + XF = A + Vs = 0Hence A = −Vs . Also,
vC' (0+ ) = ωd B =
1
LCB =
iL (0+ )
C= 0
Hence B = 0. Therefore
vC (t) = −Vs cos(ωdt) +Vs = Vs 1 − cost
LC
V
To obtain the expression for iL(t) (= iC(t)), we can either repeat the above derivation ordifferentiate and multiply by C. We choose the latter. Therefore
iL (t) =CVs
LCsin
t
LC
=Vs
L
C
sint
LC
A
SOLUTION TO PROBLEM 9.8. Essentially this is example 9.7, case 1, with literals and R =∞. Clearly the circuit is a driven parallel LC circuit having characteristic equation
s2 +1
LC= s +
j
LC
s −j
LC
= s + jωd( ) s − jωd( ) = 0
Thus we obtain
iL (t) = Acos(ωdt) + Bsin(ωdt) + XF = Acos1
LCt
+Bsin1
LCt
+XF
Here XF = IS is the value of the current when the inductor is shorted and the capacitor isopen. Applying the initial conditions,
iL (0+) = A + Is = 0 ⇒ A = − IsFurther
iL' (0+) =
vL (0+)
L=
vC (0+)
L= 0 =
1
LCB
because capacitor voltage is continuous and because
iL' (0+) =
d
dtAcos
1
LCt
+ B sin1
LCt
+ XF
t=0
=B
LC
Hence B = 0. Therefore,
iL (t) = Is 1− cos(1
LCt)
Rather than repeat the above derivation,
vC (t) = vL (t) = LdiL (t)
dt= LIs
d
dt1 − cos
1
LCt
= Is
L
Csin
1
LCt
SOLUTION TO PROBLEM 9.9. Observe that the circuits of figures (a) and (b) are dualcircuits. Hence the numerical value of vout(t) and iout(t) are the same for the sameexcitation. Since the circuit is linear, when the excitation is doubled, the response isdoubled (given zero initial conditions) by linearity. Therefore, iout (t) = 2g(t) .
SOLUTION TO PROBLEM 9.10.(a) The initial conditions are:
vc (0−) = 0 = vc( 0+)iL (0−) =10 / 0.5 = 20 = iL( 0+)
(b)
WL (0)=1
2L ⋅iL
2(0)= 0.5 J
WC (0) = 0
(c) Maximum value of vC when all energy is in capacitor must satisfy 1
2C ⋅ vC,max
2 = 0.5
or equivalently vC,max = 1000 V.
(d) From the text development, the parallel LC circuit has the differential equation,
d2vC
dt2 +1
LCvC = 0
The solution form is:
vC (t) = A ⋅cos1
LCt
+B ⋅sin1
LCt
The initial conditions are:vC (0)= A = 0
and
vC ' ( 0+) =1
LCB =
iC (0+)
C= −
iL( 0+)
CHence
B = −1000Thus
vC (t) = −1000 ⋅ sin1
LCt
V
SOLUTION TO PROBLEM 9.11.The switch has been closed for a long time which means that the inductor acts like a short
and the capacitor like an open. Hence, at t = 0-, vL = 0 and iC = 0. Hence Is divides
equally between the resistors, i.e.,
iL (0−) = iL( 0+) = Is / 2 and vC (0−) = vC ( 0+) = 0
For t ≥ 0, the differential equation is
d2vC
dt2 +1
LCvC = 0
with corresponding response
vC (t) = A ⋅cos1
LCt
+B ⋅sin1
LCt
Evaluating at the initial conditions,
vC (0)= A = 0and
vC ' ( 0 )= B ⋅1
LC=
iC (0+)
C=
−iL (0+)
Cin which case
B = −IS
2
L
CTherefore,
vC (t) = −Is
2
L
Csin
1
LCt
SOLUTION TO PROBLEM 9.12.
Natural frequency is 1
LC= 5000 ⋅ 2π . Hence,
»C = 100e-9;»L = 1/((10e3 *pi)^2 *C)L = 1.0132e-02
By voltage divider,vC (0−) = vC ( 0+) = 20 mV
Current through L,iL (0−) = iL( 0+) = 0
Voltage across capacitor satisfies,
vC (t) = Acos1
LCt
+ B sin1
LCt
Using the ICs,
vC (0+) = A = 20 mVand
vC ' ( 0 )= B ⋅1
LC=
iC (0+)
C= −
iL( 0+)
C= 0
HencevC (t) = 20 ⋅cos 10,000πt( ) mV
SOLUTION TO PROBLEM 9.13. By definition
ω =1
LC= 2π ⋅40
in which case»C = 0.1e-3;»w = 2*pi*40;»L = 1/(w^2 *C)L = 1.5831e-01 (rad/s).
Observe thatiL (0−) = iL( 0+) = 1A
From the given circuit, the capacitor is never connected to a source. Therefore,vC (0−) = vC ( 0+) = 0. Also, since
d2iLdt 2 +
1
LCiL = 0
it follows thatiL (t) = Acos(ωt) + Bsin(ωt)
From the initial conditionsiL (0)= A = 1
and
iL' (0+ ) = Bω =vC (0+ )
L= 0
iL (t) = cos(80πt) A
SOLUTION TO PROBLEM 9.14.(a) From the continuity property, the capacitor voltage and inductor current remain thesame,
vC (0−) = vC ( 0+) = 5 ViL (0−) = iL( 0+) = 1
However, iC (0−) = 0 but iC (0+ ) = −iL (0+) = −1 A and vL (0−) = 0 but vL (0+) = 5 V.These values change to maintain satisfaction of KVL and KCL.
(b)
vC (t) = Acos1
LCt
+ B sin1
LCt
wherevC (0+) = 5 = A
vC ' ( 0+) = B1
LC=
iC (0+)
C= −
1
C ⇒ B = −
π2
vC (t) = 5cos(2πt) −π2
sin(2πt) V
Alternately, from equation 9.4,
vC (t) = K cos(2πt + θ)
and from equation 9.17b K = A2 + B2 and θ = tan−1 −B
A
in which case
»A = 5; B = -pi/2;»K = sqrt(A^2 + B^2)K = 5.2409e+00»theta = atan2(-B,A)theta = 3.0440e-01»thetadeg = theta*180/pithetadeg = 1.7441e+01
Hence, K = 5.24 and θ = 17o.
(c)
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 2-6
-4
-2
0
2
4
6C
apac
itor
Vol
tage
(V
)
TextEnd
Time in s
SOLUTION TO PROBLEM 9.15. As before, ω =1
LC= 2π ⋅40 in which case L = 0.158
H.Now,
vC (0−) = vc (0+) =100
125⋅ 25 = 20 mV
andiL (0−) = iL( 0+) = 10 mA
The solution form is:iL (t) = Acos(80πt) + Bsin(80πt)
whereiL (0+) = A = 10 mA
and
iL' (0+) =vL( 0+)
L=
vC (0+)
L= 80πB ⇒ B = 0.50265 mA
SOLUTION TO PROBLEM 9.16.
(a) Note: vC (0−) = vC ( 0+) = −1 V and iL (0)= 5 A. The characteristic equation is:
s2 +1
RCs +
1
LC= s2 + 5s + 4 = (s +1)(s + 4 )= 0
As stated, the circuit is overdamped. Hence,
vC (t) = K1es1t + K2es2 t = K1e−t + K2e−4t
vC (0+) = K1 + K2 = −1
vC ' ( 0+) = s1K1 + s2K2 = −K1 − 4K2 =iC (0+)
C= 0
⇒ K1 =−s2
s2 − s1= −1.3333 and K2 =
s1s2 − s1
= 0.3333 .
Finally,
vC (t) = −1.3333e−t + 0.3333e−4t VUsing MATLAB to plot:
»t = 0:.02:4.5;»vc = -1.33333*exp(-t) + 0.33333*exp(-4*t);»plot(t,vc)»grid»ylabel('Capacitor Voltage in V')»xlabel('Time in s')
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-1
-0.9
-0.8
-0.7
-0.6
-0.5
-0.4
-0.3
-0.2
-0.1
0
Cap
acito
r V
olta
ge in
V
TextEnd
Time in s
(b)K1 + K2 =1
vC ' ( 0+) = s1K1 + s2K2 = −K1 − 4K2 =iC (0+)
C=
−10
C= −10
Hence,»A = [1 1;-1 -4]A = 1 1 -1 -4»b = [1;-10]b = 1 -10»K = A\bK = -2 3
yielding
vC (t) = −2e−t + 3e−4t V
0 0.5 1 1.5 2 2.5 3 3.5 4 4.5-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Cap
acito
r V
olta
ge in
V
TextEnd
Time in s
Obviously, there is only one zero crossing.
SOLUTION TO PROBLEM 9.17.(a) First,
vC (0−) = vC ( 0+) =100
30030 =10 V
iL (0−) = iL( 0+) = 0.1 A
Clearly, iC (0−) = 0 and vL (0−) = 0 . However,
iL (0+) + iC (0+) +vC (0+)
66.667= 0.1 + iC (0+) + 0.15 = 0
Hence, iC (0+ ) = −0.25 A. Further,vL (0+) = 10 V.
(b) The characteristic equation is:
s2 +1
RCs +
1
LC= 0
From MATLAB,»R = 66.667;C = 25e-6; L = 0.5;»b = 1/(R*C)b = 6.0000e+02»c = 1/(L*C)c = 80000»si = roots([1 b c])si = -3.9999e+02 -2.0000e+02
We take the roots to be: s1 = −400 and s2 = −200 .
(c) Overdamped response implies,
vC (t) = K1es1t + K2es2 t = K1e−400t + K2e−200t
(d)vC (0+) = K1 + K2 =10
vC ' ( 0+) = s1K1 + s2K2 = −400K1 − 200K2 =iC (0+)
C= −104
»A = [1 1;-400 -200];»b = [10;-1e4];»K = A\bK = 40 -30
Finally,
vC (t) = 40e−400 t −30e−200t V(e)»t=0:0.01e-3:25e-3;»vc = 40*exp(-400*t) - 30*exp(-200*t);»plot(t,vc)»grid»ylabel('Capacitor Voltage in V')»xlabel('Time in s')
0 0.005 0.01 0.015 0.02 0.025-6
-4
-2
0
2
4
6
8
10
Cap
acito
r V
olta
ge in
V
TextEnd
Time in s
SOLUTION TO PROBLEM 9.18.(a)
iL (0−) = iL( 0+) = 0vC (0−) = vC (0+) = 5 V
At t = 0+, the circuit is a series RLC with R = 12.5 Ω, L = 2.5 H, and C=0.1 F. Theresulting characteristic polynomial is:
s2 +RL
s +1
LC= s2 + 5s + 4 = (s + 4)(s +1) = 0
Hence, s1,s2 = −4,−1 and the form of the response is:
vC (t) = K1e−4 t + K2e−t
At t = 0+,vC (0+) = K1 + K2 = 5
and
vC '(0) = −4K1 − K2 =iC (0+)
C=
iL (0+)C
= 0
Solve for K1 and K2 we obtain:»A = [1 1; -4 -1];»b = [5;0];
»K = A\bK = -1.6667e+00 6.6667e+00
Hence,
vC (t) = −1.66667e−4 t + 6.66667e−t V
(b)iL (0+) = iL(0−) = 0
Since, the stable (passive) circuit contains no source for t > 0, all initial energy isabsorbed by the resistor. Hence limt→∞ iL ( t) = 0 , i.e., iL (∞) = 0 .
iL ( t) = iC (t) = CdvCdt
=23
e−4 t −23
e−t A
SOLUTION TO PROBLEM 9.19. This circuit is a series RLC in which case the
characteristic equation is always: s2 +RL
s +1
LC= 0
(a) For this time period, R = 2 kΩ in which case the characteristic equation is found andsolved in MATLAB as follows:
»R = 2e3; L = 0.1; C = 0.1e-6;»b = R/Lb = 20000»c = 1/(L*C)c = 100000000»s12 = roots([1 b c])s12 = -10000 -10000
The roots are repeated and: s1, s2 = –10,000. The form of the response is:
iL (t) = (K1 + K2t)e−10000t
To find K1 and K2:
iL (0+) = K1 = 2.5and
iL '(0+) = −25000 + K2 =vL (0+)
L=
vC (0+) − RiL (0+)L
=6 − 2000 × 2.5
0.1= −49940
⇒ K2 = −24940
Thus
iL (t) = (2.5− 24940t)e−10000 t 0 ≤ t ≤ 0.1msec
(b) After the switch closes, R = 1 kΩ,
iL (0.1×10−3) = (2.5 − 24940 × 0.1×10−3)e−1 = 0.0022073and
iL '(10−3) = −25000e−1 − 24940e−1 + 24940e−1 = −9197
The new characteristic equation is computed as follows:
»R = 1e3; L = 0.1; C = 0.1e-6;b = R/Lb = 10000»c = 1/(L*C)c = 100000000»s12new = roots([1 b c])s12new = -5.0000e+03 + 8.6603e+03i -5.0000e+03 - 8.6603e+03i
Hence s1,s2 = −5000 ± j8660.25. The form of the new solution is:
iL ( t') = e−5000 t' Acos( t ') + Bsin( t')[ ] A
where t'= t − 0.1×10−3 and = 8660.25 rad/sec. Observe that
iL ( t'= 0) = 0.0022073 = Aand
iL '( t'= 0) = −5000A + 8660.25B = −9197From MATLAB,»w = imag(s12new(1));»sig = real(s12new(1));»B = (-9197 -sig*2.2073e-3)/wB = -1.0607e+00Hence for t > 0.1 ms,
iL ( t) = e−5000(t−0.1ms) 0.0022073cos ( t − 0.1ms)( ) −1.0607sin ( t − 0.1ms)( )[ ] A
(c)
0 0.5 1 1.5 2 2.5 3 3.5 4x 10-3
-1
-0.5
0
0.5
1
1.5
2
2.5
iL(t
), in
A
TextEnd
time in s
K1 = 2.5;K2 = -24940;t = 0:0.01e-3:4e-3;A = 0.0022073;B = (-9197 -sig*2.2073e-3)/wB = -1.0607e+00iL = (K1 + K2*t) .* exp(-10000*t) .* (u(t)-u(t-1e-4)) ...+ exp(-5000*(t - 1e-4)) .* (A*cos(w*(t - 1e-4))+B*sin(w*(t - 1e-4))) ....* u(t -1e-4);plot(t,iL)gridiL = (K1 + K2*t) .* exp(-10000*t) .* (u(t)-u(t-1e-4))
(d)
»K1 = 2.5;K2 = -24940;»t = 0.1e-3;»iL = (K1 + K2*t) .* exp(-10000*t)iL = 2.2073e-03» % The energy stored in the inductor over [0,0.1ms] is in J:»WL = 0.5*0.1*(iL^2 - 2.5^2)WL = -3.1250e-01» % The energy stored in the capacitor over [0,0.1ms] first» % requires computation of vL and then vC.»vL = 0.1*(K2*exp(-10000*t) - 10000*(K1 + K2*t) .* exp(-10000*t))vL = -9.1970e+02»vC = vL + 2000*iLvC =
-9.1528e+02» % The energy stored in the capacitor over [0,0.1ms] is in J:»WC = 0.5*0.1e-6*(vC^2 - 6^2)WC = 4.1885e-02
» % To compute energy dissipated in resistor, we make» % use of conservation of energy: WR + WC + WL = 0
»WR = -WL - WCWR = 2.7061e-01
SOLUTION TO PROBLEM 9.20. For this problem, iL (0) = 8 A, vC (0) = 20 V. For the
parallel RLC, the characteristic equation s2 +1
RCs +
1LC
= 0 is solved as follows:
»R = 20; C = 0.01e-3; L = 25e-3;»si = roots([1 1/(R*C) 1/(L*C)])si = -4.0000e+03 -1.0000e+03»s1 = si(1); s2 = si(2);
Hence
vC (t) = K1e−4000 t + K2e−1000t
To compute the constants,
vC (0) = K1 + K2 = 20and
vC '(0) = −4000K1 −1000K2 =iC (0+)
C= −
iL (0+) +vC (0+)
RC
»iL0 = 8; vC0 = 20;»vCprime = -(iL0 +vC0/R)/CvCprime = -9.0000e+05»A = [1 1;-4000 -1000];»b = [20;vCprime];»K = A\bK = 2.9333e+02 -2.7333e+02
Hence,
vC (t) = 293.33e−4000t − 273.33e−1000 t VAlso,
iL ( t) = −vC (t)
R− CvC
' ( t) = −2.933e−4000t +10.933e−1000 t A
where»K1 = K(1); K2 = K(2);»KR1 = K1/R; KR2 = K2/R;»KCp1 = -4000*K1*C; KCp2 = -1000*K2*C;»KL1 = -KR1-KCp1KL1 = -2.9333e+00»KL2 = -KR2 - KCp2KL2 = 1.0933e+01»t = 0:0.01e-3:4e-3;»iL = KL1*exp(-4000*t) + KL2*exp(-1000*t);»plot(t,iL)»grid»xlabel('Time in s')»ylabel('Inductor Current in A')»
0 0.5 1 1.5 2 2.5 3 3.5 4x 10-3
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1
2
3
4
5
6
7
8
9
Time in s
Indu
ctor
Cur
rent
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TextEnd
SOLUTION TO PROBLEM 9.21.(a) The circuit is critically damped and the characteristic polynomial is:
s2 +1
RCs +
1LC
= (s + 20)2 = s2 + 40s + 400 = 0
»R = 2; C = 1/80;»L = 1/(400*C)L = 2.0000e-01
(b) vC (0) = 10 V and vC '(0+ ) = −800 =iC (0+ )
C=
−iL (0+ ) − vC (0+ ) / 2C
.
»vC0 = 10; vCp0 = -20*10 -600vCp0 = -800»iL0 = C*800-vC0/2iL0 = 5Hence: iL(0-) = iL(0+) = 5 A.
(c)By simple KCL,
iL ( t) =−180
dvCdt
−vC2
= 5e−20t +150 ⋅ t ⋅e−20t A
SOLUTION TO PROBLEM 9.22.(a) The characteristic equation for the series RLC
s2 +Req
Ls +
1
LC= 0
For critically damped response, want (Req / L)2 − 4 / (LC ) = 0 . Solving yields Req = 4
Ω. Hence,
Req =5R
5 + Rimplies that R = 20 Ω.(b) Solving for the resulting roots implies that»si = roots([1 Req/L 1/(L*C)])si = -50 -50»s1 = 50;
Hence
vC (t) = (K1 + K2t)e−50t
From the initial conditions, vC (0) = K1 = 5 and
vC '(0) = −250 + K2 =iC (0+)
C=
iL (0+)C
=−5
0.01= −500
Hence, K2 = –250 and
vC (t) = (5 − 250t)e−50t V
Set to zero and solve for t = 0.02. See MATLAB plot.
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.05-1
0
1
2
3
4
5
SOLUTION TO PROBLEM 9.23.(a) The circuit is a series RLC. Hence
»R = 40; C = 0.25e-3; L = 0.1;si = roots([1 R/L 1/(L*C)])si = -200 -200»s1 = si(1);
Since the circuit is critically damped,
iL ( t) = (K1 + K2t)e−200 t AUsing initial conditions,
iL (0) = K1 = 1
iL '(0) = s1K1 + K2 = −200K1 + K2 =vL (0+)
L=
−vC (0+) − 40iL (0+)
L= −350
HenceK2 = −150
Thus
iL ( t) = (1−150t)e−200 t A
Now using MATLAB,»R/Lans = 400»1/(L*C)ans = 40000
»y = dsolve('D2y + 400*Dy + 40000*y = 0, y(0) = 1,Dy(0) = -350')y =exp(-200*t)-150*exp(-200*t)*t
This answer coincides with the analytical solution.
(b) As above, the form of the solution is
vC (t) = (K1 + K2t)e−50t
Applying initial conditions,
vC (0) = K1 = 5
vC '(0) = s1K1 + K2 = −200K1 + K2 =iC (0+)
C=
iL (0+)
C= 4000
Hence, K2 = 5000 and
vC (t) = (5 + 5000t)e−200 t V
In MATLAB,»y = dsolve('D2y + 400*Dy + 40000*y = 0, y(0) = 5,Dy(0) = 4000')y =5*exp(-200*t)+5000*exp(-200*t)*t
which verifies the analytical solution. Plotting we obtain,
0 0.005 0.01 0.015 0.02 0.025 0.03 0.035 0.04 0.045 0.050
2
4
6
8
10
12
Observe, there is no zero crossing as per problem 9.5.
SOLUTION TO PROBLEM 9.24.Source has been on for a long time and turns off at t = 0. Hence
iL (0−) = iL (0+) = 0
vC (0−) = vC (0+) = 0.1× 40 = 4 V
At t ≥ 0, we have a series RLC circuit:
»R = 40; C = 4e-3; L = 2;si = roots([1 R/L 1/(L*C)])si = -1.0000e+01 + 5.0000e+00i -1.0000e+01 - 5.0000e+00i»wd = imag(si(1))wd = 5»sig = -real(si(1))sig = 10
Hence
vC (t) = e−10 t Acos(5 t) + B sin(5t)( )Applying initial conditions,
vC (0) = A = 4
vC '(0) = −10A + 5B =iC (0+)
C=
iL (0+)C
= 0
Hence, B = 8. The final form is:
vC (t) = e−10 t 4cos(5t) + 8sin(5 t)( ) V
Similarly,
iL ( t) = e−10 t A cos(5t) + Bsin(5t)( ) AApplying initial conditions,
iL (0) = A = 0
iL '(0) = −10A + 5B =vL (0+)
L=
−vC (0+) − 40iL (0+)L
= −2
Thus
iL ( t) = −0.4e−10 t sin(5t) A
SOLUTION TO PROBLEM 9.25.
At t = 0-, the capacitor is an open circuit and the inductor is a short circuit. The
resulting circuit is a simple resistive network. The first step in the solution is to solve thisnetwork for the initial conditions on the capacitor and inductor. Specifically, solve forthe capacitor voltage (i.e. the voltage across the series connection of the 6 Ω resistor andthe independent voltage source) and the inductor current (i.e. the current flowing throughthe 4 Ω resistor).
Verify that iL (0−) = iL (0+) =1 A and vC (0−) =vC (0+) =12 V.
When the switch opens, the branch containing the independent voltage source iseliminated. So, we end up with a series RLC circuit. The equivalent resistance is
Req = 4 + 24 / / 4 8= 20 Ω.
»R = 20; C = 0.01; L = 2;si = roots([1 R/L 1/(L*C)])si = -5.0000e+00 + 5.0000e+00i -5.0000e+00 - 5.0000e+00i
»sig = -real(si(1))sig = 5»wd = imag(si(1))wd = 5
iL ( t) = e−5t A cos(5t) + Bsin(5t)[ ] AApplying ICs,
iL (0) = A =1
iL '(0) = −5A + 5B =vL (0+)
L=
−8
2= −4
Hence B = 0.2. It follows that
iL (t) = e−5t cos(5t) +1
5sin(5t)
A
SOLUTION TO PROBLEM 9.26. The series RLC circuit has characteristic equation
s2 +RL
s +1
LC= s2 +
10L
s +1
LC= s2 + 2 s + 2 + d
2 = 0
From the given response, = 10 =102L
which implies that L = 0.5 H. Further,
1LC
=2C
= 2 + d2 =102 + 10 3( )2
= 400
Hence, C = 5 mF.
Now, from given response, iL (0) = 0 and iL '(0) = 500 3 =vL (0)
L= 2vL(0) .
HencevL (0) = 250 3 V. In addition, vC (0) = −10iL (0) − vL (0) = −250 3 V and
vC '(0) = iL (0) / C = 0
To find the capacitor voltage we have
vC (t) = e−10 t Acos(10 3t) + B sin(10 3t)( )
It follows that vC (0) = −250 3 = A and
vC '(0) = 0 = 2500 3 +10 3 BB = –250. Therefore
vC (t) = −e−10 t 250 3 cos(10 3t) + 250sin(10 3t)( )
SOLUTION TO PROBLEM 9.27. (a) From the given information,
vC (t) = Ae− t cos( t + ) V
where = 2 f = 2 / T =2
0.5 ×10−3 = 4000π rad/s. f = 2 kHz.
The time constant of the exponential term is 1/σ which is 2 ms from the figure.
Hence σ = 500 . The characteristic equation of the parallel RLC is:
s2 +1
RCs +
1LC
= s2 +1
104Cs +
1LC
= s2 + 2 s + 2 + d2 = 0
Since 1
104C= 2 = 1000, C = 0.1 µF. Further,
1LC
=107
L= 2 + d
2 =106 +16π2106
implies that
»L = 1e7/(1e6 + 16*pi^2 *1e6) or L = 63 mH.
(b) From the figure and the above calculations, vC(0) = 10 V and
vC (t) = 10e−500 t cos(4000π t) V,Hence
vC '(t) = −5000e−500t cos(4000π t) − 40000πe−500t sin(4000π t)
implying that vC' (0) = −5000 . Thus
iL (0) = −vC (0) /10 4 − CvC' (0) = −0.001 + 5000 × 0.1×10−6 = −0.5 ×10−3 A
SOLUTION TO PROBLEM 9.28. As given vC (t) = Ae−3t cos(4 t + ) and R = 10 Ω. Thecharacteristic polynomial of the parallel RLC circuit is:
s2 +1
RCs +
1LC
= s2 +1
10Cs +
1LC
= s2 + 2 s + 2 + d2 = 0
Hence 1/(RC) = 6 implies C = 1/60 F. Further d =
4 × 60
L− 36
2= 4 . Hence L = 2.4
H.To change R to obtain a cricially damped circuit,
60R
2
=4
LC=100
Hence R2 = 36 or R = 6 Ω. It follows that 2σ = 10 or σ = 5. The form of the response is:
vC (t) = (K1 + K2t)e−5t
SOLUTION TO PROBLEM 9.29. For all cases, vC(0-)= vC(0+) = 0 and iL(0-)= iL(0+) =20/20 = 1 A. Further for all cases the circuit is a parallel RLC with characteristicequation:
s2 +1
RCs +
1LC
= s2 + 2 s + 2 + d2 = 0
(a)»L = 2e-3; C = 5e-6;»c = 1/(L*C);»R = 10;»b = 1/(R*C);»si = roots([1 b c])si = -10000 -10000»% Solution is cricitally damped.
Thus
vC (t) = (K1 + K2t)es1t = (K1 + K2t)e−104 t V
From ICs,vC (0) = K1 = 0
vC '(0) = s1K1 + K2 = K2 =iC (0+)
C=
−iL (0+)
C= −2 ×105
Hence
vC (t) = −2 ×105te−104 t V
(b)»R = 100;»si = roots([1 1/(R*C) 1/(L*C)])
si = -1.0000e+03 + 9.9499e+03i -1.0000e+03 - 9.9499e+03i
vC (t) = e−1000 t Acos(9950 t) + B sin(9950t)( ) VFrom ICs.
vC (0+) = A = 0
vC '(0) = −1000A + 9950B =iC (0+)
C=
−iL (0+)C
= −2 ×105
in which case B = –20.1. Thus
vC (t) = −20.1e−1000t sin(9950t) V
(c)»R = 87/17;si = roots([1 1/(R*C) 1/(L*C)])si = -3.6328e+04 -2.7527e+03
We define s1,s2 = –2753, –36327. Thus
vC (t) = K1es1t + K2es2t = K1e−2753t + K2e−36327 t VFrom the IC's
vC (0) = K1 + K2 = 0
vC '(0) = s1K1 + s2K2 = −2 ×105
»A = [1 1;si(2) si(1)];»b = [0; -2e5];»K = A\bK = -5.9568e+00 5.9568e+00
Therefore
vC (t) = −5.9568 e−2753t − e−36327t( ) V
SOLUTION TO PROBLEM 9.30. For all cases, vC(0-) = vC(0+) = 0 and iL(0-) = iL(0+) =10/100 = 0.1 A. Further for all cases the circuit is a parallel RLC with characteristicequation:
s2 +1
RCs +
1LC
= s2 + 2 s + 2 + d2 = 0
(a)»R = 50; C = 0.04e-3; L = 0.625;»si = roots([1 1/(R*C) 1/(L*C)])si = -400 -100
Define the two roots as:s1 = –100s2 = –400
Hence,
vC (t) = K1e−100 t + K2e−400t V
From, the initial conditionsvC(0+) = 0 = K1 + K2
and
vC '(0) = s1K1 + s2K2 = −100K1 − 400K2 =−iL (0+)
C= −2500
»A = [1 1;si(2) si(1)];»b = [0;-2500];»K = A\bK = -8.3333e+00 8.3333e+00
Therefore
vC (t) = −8.3333 e−100 t − e−400t( ) V
(b)»L = 0.4;»si = roots([1 1/(R*C) 1/(L*C)])si = -2.5000e+02 -2.5000e+02
Thus
vC (t) = (K1 + K2t)e−250t VFrom IC's,
vC (0) = K1 = 0
vC '(0) = s1K1 + K2 = K2 =iC (0+)
C=
−iL (0+)
C= −2500
Therefore
vC (t) = −2500te−250 t V
(c)»L = 0.2;»si = roots([1 1/(R*C) 1/(L*C)])si = -2.5000e+02 + 2.5000e+02i -2.5000e+02 - 2.5000e+02i
vC (t) = e−250 t A cos(250t) + Bsin(250t)[ ] VFrom ICs.
vC (0+) = A = 0
vC '(0) = −250A + 250B = 250B =iC (0+)
C=
−iL(0+)C
= −2500
in which case B = –10. Thus
vC (t) = −10e−250t sin(250t) V
SOLUTION TO PROBLEM 9.31.(a) The indicated behavior occurs when the resistance causes the circuit to be criticallydamped, i.e.,
1RC
2
−4
LC=
4R
2
−16 = 0
Thus R = 1 Ω and
R =1 = R0 + et−5 = 0.8 + et−5
So»t = 5+log(1-0.8)t = 3.3906e+00 (years)
(b) Here, we have a series case: the indicated behavior occurs when the resistancecauses the circuit to be critically damped, i.e.,
RL
2
−4
LC= R2 −144 = 0
Thus R = 12 Ω and
R =12 =R0
1 + et−5 =15
1 + et−5
»t = 5 + log((15-12)/12)t = 3.6137e+00 (years)
SOLUTION TO PROBLEM 9.32. Step 1: Since the step functions are 0 from t = ∞ up to t =0-,
vC(0-) = vC(0+) = 0, iL(0-) = iL(0+) = 0
Step 2: At t = 0+, we have
iC (0+) = 2 −vC (0+)
20− iL (0+) = 2 A
andvL (0+) = vC (0+) − 5iL (0+) − 50 = −50 V
SOLUTION TO PROBLEM 9.33. Since the step function is 0 from t = ∞ up to t = 0-,
vC(0-) = vC(0+) = 0, iL(0-) = iL(0+) =0
Since the circuit is a parallel RLC
s2 +1
RCs +
1LC
= s2 +100s +1600 = 0
»R = 4; L = 0.25; C = 2.5e-3;»si = roots([1 1/(R*C) 1/(L*C)])si = -80 -20
Hence s1,s2 = -20, -80. The general form is:
vC (t) = K1e−80 t + K2e−20t + X f V
When the capacitor is open and the inductor is a short, Xf = 0. Thus,
vC (t) = K1e−80 t + K2e−20t VFrom the ICs
vC(0+) = 0 = K1 + K2
and
vC '(0) = −80K1 − 20K2 =iC (0+)
C=
−iL (0+) − vC (0+) 8 + (20 − vC (0+)) 8
C
=2.5
2.5 ×10−3 =1000
»A = [1 1;-80 -20];»b = [0; 1000];»K = A\bK =
-1.6667e+01 1.6667e+01Hence,
vC (t) = −16.667 e−80 t − e−20t[ ] V
SOLUTION TO PROBLEM 9.34. From the continuity property and the fact that at t = 0-,the capacitor looks like an open and the inductor looks like a short at t = 0-,
iL(0-) = iL(0+) = 1AvC(0-) = vC(0+) = 0
Since the circuit is a parallel RLC
s2 +1
RCs +
1LC
= s2 + 500s + 40000 = 0
»R = 40; C = 0.05e-3;L = 0.5;»si = roots([1 1/(R*C) 1/(L*C)])si = -400 -100
Define the two roots as:s1 = –100s2 = –400
Hence, the general form is:
iL ( t) = K1e−100 t + K2e−400t + X f
When the capacitor is open and the inductor is a short, Xf = –1 A. Thus,
iL ( t) = K1e−100 t + K2e−400t −1
From, the initial conditionsiL (0+) = K1 + K2 −1 =1
and
iL '(0) = −100K1 − 400K2 =vL (0+)
L=
vC (0+)L
= 0
»A = [1 1;-100 -400];»b = [2;0];»K = A\bK = 2.6667e+00
-6.6667e-01
Hence,
iL ( t) =83
e−100 t −23
e−400t −1 A
SOLUTION TO PROBLEM 9.35.
(a) From the problem specs and the continuity property,
iL(0-) = iL(0+) = 0.008 AvC(0-) = vC(0+) = 2
At t = ∞, the inductor looks like a short and the capacitor looks like an open; hence iL(∞)
= 0 and vC(∞) = 400×0.006 = 2.4 V. The circuit is a series RLC with characteristic
polynomial
s2 +RL
s +1
LC= 0
»R = 400; C = 6.6667e-6; L = 0.2;si = roots([1 R/L 1/(L*C)])s1 = si(1); s2 = si(2);si = -1.5000e+03 -5.0000e+02
Hence
vC (t) = K1e−1500 t + K2e−500t + 2.4 VUsing initial conditions
vC(0) = K1 + K2 + 2.4 = 2
vC '(0) = −1500K1 − 500K2 =iC (0+)
C=
iL (0+)C
=8 ×10−3
6.6667 ×10−6 =1200
»A = [1 1;-1500 -500];»b = [2-2.4;1200]b = -4.0000e-01 1.2000e+03»K = A\bK = -1.0000e+00 6.0000e-01
Hence
vC (t) = −e−1500 t + 0.6e−500t + 2.4 V
(b) vL(t) is going to have the same form as vC(t) above except that vL(∞) = 0 since theinductor is a short at t = ∞. Alternately however, we have
vL (t) = LdiL (t)
dt= L
diC (t)dt
= LCd2vC (t)
dt2 = −3e−1500t + 0.2e−500t V
SOLUTION TO PROBLEM 9.36.
At t = 0, inductor is a short circuit and the capacitor is an open circuit. Since the currentsource is 0 at t = 0-, and the continuity property,
iL(0-) = iL(0+) = –1 AvC(0-) = vC(0+) = 65 V
For positive time, we have a series RLC circuit with characteristic equation
s2 +RL
s +1
LC= 0
»R = 65; C = 0.1e-3; L = 0.1;si = roots([1 R/L 1/(L*C)])s1 = si(1); s2 = si(2);si = -4.0000e+02 -2.5000e+02
At t = ∞, vC(∞) = 0.6*65 = 39 V. Hence
vC (t) = K1e−400 t + K2e−250t + 39 V
Using initial conditions,vC(0) = K1 + K2 + 39 = 65
vC '(0) = −400K1 − 250K2 =iC (0+)
C=
iL (0+)C
=−1
10−4 = −104
»A = [1 1;s1 s2];»b = [65-39; -1e4];K=A\bK = 2.3333e+01 2.6667e+00
Hence,
vC (t) = 23.333e−400t + 2.6667e−250 t + 39 V
SOLUTION 9.37.
(a) Rth = 200//50 + R = (40 + R). The characteristic equation is:
s2 +1
RthCs +
1
LC
= 0
Critically damped means that both roots are the same, so the discriminant is zero, i.e.
1
RthC
2
−4
LC= 0
Equivalently,
Rth = 40 + R = 0.5L
C= 50 Ω
Thus R = 10 Ω.(b) Short the inductor and open the capacitor. Because the capacitor is in parallel
with the shorted inductor at t = 0-, vC(0+) = vC(0
-) = 0. The Thevenin equivalent
resistance seen by the LC-parallel combination is Rth = 50 Ω from part (a). A simple
calculation indicates that Voc = 0.8*50 = 40 V. Therefore, iL(0+) = iL(0
-) = 40/50 = 0.8
A. To find vR(0+) we use the following equivalent circuit:
Hence,
vR(0+ ) = −40 ×10
10 + 40= −8 V.
To compute the derivative of vR at 0+, consider that
d
dtvR(t)( ) =
d
dt10 × iR(t)( ) = 10
d
dt
−40 − vC (t)
50
= −0.2
d
dtvC (t)( ) = −0.2
iC( t)
C
HencedvR(t)
dt t=0+
= −0.2iC (0+ )
C= −4000iC (0+ )
But
iC (0+ ) =−40
50− iL( 0+ ) = −0.8 − 0.8 = −1.6 A
Therefore,dvR(t)
dt t=0+
= −4000iC (0+ ) = 6400 V/s
(c) Since the circuit is critically damped the roots of the characteristic equation are
s1,2 = −1
2RthC= −200
According to table 9.2 for t ≥ 0,
vR(t) = K1 + K2t( )e−200 t + XF
It follows from the circuit and this equation that
XF = vR(∞) = −8 V
K1 = vR(0+) − XF = 0
dvR(t)
dt t=0+
= −200K1 + K2 = K2 = 6400
Therefore
vR(t) = 6400te−200t − 8( )u(t) V
A plot of the waveform is given below
SOLUTION TO PROBLEM 9.38. For this problem we first compute the Theveninequivalent of the circuit to the left of the capacitor for t > 0. Consider
Now observe
Is = I1 + I2 =Vs −12
500+
Vs − kv1500
=Vs −12
500+
Vs − k 12 −Vs( )500
It follows that
Vs =5002 + k
Is +
1 + k2 + k
12 = Rth Is +Voc
The parallel LC is now driven by this Thevenin equivalent.
»L = 0.01; C = 1e-6;»% Critical damping means (1/(Rth*C))2 - 1/(L*C) = 0»x = sqrt(4/(L*C));»Rthcrit = 1/(C*x)Rthcrit = 50»kcrit = (500 - 2*50)/50kcrit = 8»% For parallel circuit, larger Rth means less damping»% Hence, smaller Rth means overdamped. Smaller Rth»% means larger k. Therefore k > 8 is the ranger for»% overdamped response.
For the critically damped response we have R = Rth = 50 Ω; hence
»R = 50; C = 1e-6; L = 0.01;si = roots([1 1/(R*C) 1/(L*C)])si =
-10000 -10000
in which case
vC (t) = (K1 + K2t)e−50t + X f
At t = ∞, vC (∞) = vL (∞) = 0 in which case X f = 0. From the initial conditions,
vC (0) = K1 = 0
vC '(0) = s1K1 + K2 = K2 =iC (0+)
C=
Voc
Rthcrit− iL (0+)
CHence
K2 =10.8 50
10−6 = 2.16 ×105
Therefore
vC (t) = 2.16 ×105e−50 t V
SOLUTION TO PROBLEM 9.39. (a) The series RLC leads to a characteristic equation ofthe form
s2 +RL
s +1
LC= s2 +
240.2
s +1
0.2C= s2 +120s +
5C
= 0
For a critically damped response, 1202 =20C
. Hence, C = 1.3889 mF.
(b)»C = 20/120^2C = 1.3889e-03»L = 0.2; R = 24;»si = roots([1 R/L 1/(L*C)])si = -60 -60Hence
iL (t) = (K1 + K2t)e−60t + 0.4 A
where iL(∞) = 0.4 because at t = ∞, the capacitor looks like an open and the inductor likea short. Hence all current from the source flows through the inductor.
Using the initial conditions,
iL(0) = K1 + 0.4 = 0 ⇒ K1 = – 0.4
iL '(0) = s1K1 + K2 = 60 × 0.4 + K2 =vL (0+)
L=
−vC (0+) − 24 iL(0+) − 0.4( )L
= 48
Hence, K2 = 24 and
iL ( t) = (−0.4 + 24t)e−60 t + 0.4 A
SOLUTION TO PROBLEM 9.40. This problem differs from 39 in the initial conditioncalculation. Specifically,
iL(0-) = iL(0+) = –0.4vC(0-) = vC(0+) = 0
Again
iL (t) = (K1 + K2t)e−60t + 0.4 Aand
iL(0+) = K1 + 0.4 = – 0.4 ⇒ K1 = – 0.8
iL '(0) = 60 × 0.8 + K2 =vL (0+)
L=
−vC (0+) − 24 iL (0+) − 0.4( )L
= 96
Hence,K2 = 48
and
iL ( t) = (−0.8 + 48t)e−60t + 0.4 A
SOLUTION TO PROBLEM 9.41.(a) At 0-, the capacitor is an open circuit and inductor is a short circuit. So,
vC(0-) = vC(0+) = 5 ViL(0-) = iL(0+) = 5×10-3 – 4×10-3 = 1 mA
Now, at 0+, replace the capacitor by a 5 V voltage source and the inductor by a 1 mAcurrent source. Also, the original independent current source is turned off. Solve theresulting circuit to obtain.
iC + 5×10-3 – 1×10-3 = 0
⇒ iC(0+) = –4 mA
vL(0+) = 0
(b) Since the circuit is a parallel RLC, the characteristic polynomial is
s2 +1
RCs +
1LC
= s2 +1
104Cs +
1LC
= s2 + 2 s + 2 + d2 = 0
»R = 1e3; C = 0.5e-6; L = 0.184;si = roots([1 1/(R*C) 1/(L*C)])si = -1.0000e+03 + 3.1416e+03i -1.0000e+03 - 3.1416e+03i
»wd = imag(si(1))wd = 3.1416e+03
»sig = -real(si(1))sig = 1000
Also, at t = ∞, iL(∞) = 5×10-3 A. Therefore
iL ( t) = e−1000 t Acos(π ×103t) + Bsin(π ×103t)[ ] + 5 ×10−3 A
Using initial conditions
iL (0) = A + 5 ×10−3 =10−3 ⇒ A = −4 ×10−3
and
iL '(0) = −1000A + π ×103B = 4 + π ×103B =vL (0+)
L= 0
Hence,»B = -4e-3/piB = -1.2732e-03
Finally,
iL ( t) = −e−1000 t 4 c o s (π ×103t) +1.2732sin( π ×103t)[ ] + 5 mA
SOLUTION TO PROBLEM 9.42. The response here coincides with that of problem 41 up totime t = 2s. At this point we need the new initial conditions on the circuit at t = 2+.However, at t = 2, e−1000 t = 0 for all practical purposes. Hence, iL (2) = 5 ×10−3 A.
Differentiating the expression for iL(t) and evaluating at t = 2 yields zero by inspection.
This follows because Ke−2000 = 0 for K in the range of 1 to 104. This can also be seen
from the circuit because at t = 2 s, the capacitor has charged to 5 V, making
iL' (2+) =
vL (2+)L
= 0.
To find steady state current, solve the circuit with the new current source value and withthe capacitor and inductor as open and short circuits, respectively:
iL (∞) = 5 ×10−3 + 4 ×10−3 = 9 ×10−3
iL ( t) = −e−1000(t−2) A cos π ×103(t − 2)( ) + B sin π ×103( t − 2)( )[ ] + 9 ×10−3 A
Using the new initial conditions
iL (2+) = A + 9 ×10−3 = 5 ×10−3 ⇒ A = −4 ×10−3
and
iL '(2+) = −1000A + π ×103B = 4 + π ×103B =vL (0+)
L= 0
Hence B = –1.2732×10-3
and for t ≥ 2s,
iL ( t) = −e−1000(t−2) 4cos π ×103( t − 2)( ) +1.2732sin π ×103(t − 2)( )[ ] + 9 mA
SOLUTION TO PROBLEM 9.43. At t = 0-, iL(0-) = 10/20 = 0.5 A and vC(0-) = –5 V by theusual considerations. At t = 0+, we have a parallel RLC circuit. Hence
s2 +1
RCs +
1LC
= s2 +1
104Cs +
1LC
= s2 + 2 s + 2 + d2 = 0
»R = 10; C = 0.05e-3; L = 0.01;si = roots([1 1/(R*C) 1/(L*C)])si = -1.0000e+03 + 1.0000e+03i -1.0000e+03 - 1.0000e+03i»wd = imag(si(1))wd = 1.0000e+03»sig = -real(si(1))sig = 1000
Also, at t = ∞, vC(∞) = –5 V. Therefore
vC (t) = e−1000 t A cos(103t) + Bsin(10 3t)[ ]− 5 V
Using initial conditions
vC (0) = A − 5 = −5 ⇒ A = 0and
vC '(0) = −1000A +1000B =103B =iC (0+)
C=
−0.5
0.05 ×10−3 = −104
Hence, B =10 and
vC (t) = −10e−1000t sin(103t) − 5 VFurther
vL (t) = −5 − vC ( t) = 10e−1000 t sin(103t) V
SOLUTION TO PROBLEM 9.44. (a) The circuit is a driven series RLC. Hence
s2 +RL
s +1
LC= s2 + 2 s + 2 + d
2 = 0
»R = 400; C = 0.5e-6; L = 0.2;si = roots([1 R/L 1/(L*C)])si = -1.0000e+03 + 3.0000e+03i -1.0000e+03 - 3.0000e+03i»wd = imag(si(1))sig = -real(si(1))wd = 3000sig = 1000
Also, at t = ∞, vC(∞) = 400×0.006 = 2.4 V. Therefore
vC (t) = e−1000 t A cos(3000 t) + B sin(3000t)[ ] + 2.4 V
Using initial conditions
vC (0) = A + 2.4 = 2 ⇒ A = −0.4and
vC '(0) = −1000A + 3000B = 400 + 3000B =iC (0+)
C=
iL (0+)C
= 1.6 ×104
Hence, B = 5.2 and
vC (t) = e−1000 t −0.4cos(3000t) + 5.2sin(3000t)[ ] + 2.4 V
(b) Consistent with underdamped circuit behavior and because the inductor behaves as ashort and the capacitor as an open at t = ∞ (iL(∞) = 0),
iL ( t) = e−1000 t Acos(3000t) + Bsin(3000t)[ ] V
Using the initial conditions,
iL (0+) = A = 0.008 AFurther,
iL '(0) = −1000A + 3000B = −8 + 3000B =vL (0+)
L=
400(0.006 − 0.008) − 20.2
= −14
Hence, B = −0.002 and
iL ( t) = e−1000 t 0.008cos(3000t) − 0.002sin(3000 t)[ ] VFinally
vL (t) = vR(t) − vC (t) = 400 0.006 − iL (t)[ ] − vC (t) = 2.4 − 400iL (t) − vC ( t)
which implies that
vL (t) = e−1000 t 3.6cos(3000t) −13.2sin(3000t)[ ] V
SOLUTION TO PROBLEM 9.45. This circuit is the same series RLC as problem 44. Note
that at t = ∞, vC(∞) = –2.4 V. Hence
vC (t) = e−1000 t A cos(3000 t) + B sin(3000t)[ ]− 2.4 V
Now, the initial conditions are:
vC(0-) = 2.4 = vC(0+) V, iL(0-) = 0 = iL(0+)Thus
vC (0) = A − 2.4 = 2 . 4 ⇒ A = 4.8Further,
vC '(0) = −1000A + 3000B = −4800 + 3000B =iC (0+)
C= 0
Hence, B =1.6 and
vC (t) = e−1000 t 4.8cos(3000t) +1.6sin(3000 t)[ ]− 2.4 V
SOLUTION TO PROBLEM 9.46. For all three cases, assuming iL is pointing downward,
vC (0−) = vC (0+) = 0 and iL (0−) = iL (0+) = 0.1 AAt t = ∞,
vC (∞) = 0 and iL (∞) = −0.2 AFurther
iC (0+) = −iL (0+) +−10 − vC (0+)
50= −0.1− 0.2 = −0.3 A
Lastly, all three cases are for a parallel RLC whose characteristic equation is:
s2 +1
RCs +
1LC
= 0
(a)»R = 50; C = 0.04e-3; L = 0.625;si = roots([1 1/(R*C) 1/(L*C)])s1 = si(1); s2 = si(2);si = -400 -100
Hence,
vC (t) = K1e−100 t + K2e−400t V
From, the initial conditionsvC(0+) = 0 = K1 + K2
and
vC '(0) = s1K1 + s2K2 = −100K1 − 400K2 =iC (0+)
C=
−0.3
0.04 ×10−3 = −7500
»A = [1 1;si(2) si(1)];»b = [0;-7500];»K = A\bK = -25 25
Therefore
vC (t) = −25 e−100t − e−400t( ) V
(b)»R = 50; C = 0.04e-3; L = 0.4;»si = roots([1 1/(R*C) 1/(L*C)])s1 = si(1); s2 = si(2);si = -2.5000e+02 -2.5000e+02
Thus
vC (t) = (K1 + K2t)e−250t VFrom IC's,
vC (0) = K1 = 0
vC '(0) = s1K1 + K2 = K2 =iC (0+)
C= −7500
Therefore
vC (t) = −7500te−250 t V
(c)»L = 0.2;»si = roots([1 1/(R*C) 1/(L*C)])si = -2.5000e+02 + 2.5000e+02i -2.5000e+02 - 2.5000e+02i
vC (t) = e−250 t A cos(250t) + Bsin(250t)[ ] VFrom ICs.
vC (0+) = A = 0
vC '(0) = −250A + 250B = 250B =iC (0+)
C= −7500
in which case B = –30. Thus
vC (t) = −30e−250t sin(250t) V
SOLUTION TO PROBLEM 9.47.
At t = 0-, he capacitor is open and the inductor is a short. This together with thecontinuity property implies vC(0-) = vC(0+) = 10 V and iL(0-) = iL(0+) = 1 A byinspection.
Now, for t = 0+, vin = 0, replace capacitor and inductor with a voltage source and acurrent source, respectively (values are those of the initial conditions). Solve for initialcapacitor current and initial inductor voltage to obtain:
vL(0+) = -10 ViC(0+) = iL(0+) – iR1 – iR2 = – 1 A
Notice that the resulting circuit is an undriven parallel RLC circuit with Req = 10//10 = 5Ω.»R = 5; C = 0.01; L = 4/3;si = roots([1 1/(R*C) 1/(L*C)])s1 = si(1); s2 = si(2);si = -15 -5
Hence,
vC (t) = K1e−15t + K2e−5t V
From, the initial conditionsvC(0+) = 10 = K1 + K2
and
vC '(0) = s1K1 + s2K2 = −15K1 − 5K2 =iC (0+)
C=
−10.01
= −100
»A = [1 1;-15 -5];»b= [10; -100];»K = A\bK = 5.0000e+00 5.0000e+00
vC (t) = 5e−15t + 5e−5t V
Verify with dsolve function in matlab: dsolve('D2y+20*Dy+75*y=0,y(0)=10,Dy(0)=-100')
»dsolve('D2y+20*Dy+75*y=0,y(0)=10,Dy(0)=-100')ans =5*exp(-5*t)+5*exp(-15*t)
0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10
1
2
3
4
5
6
7
8
9
10
SOLUTION TO PROBLEM 9.48.
For t = 0-, there is no source present nor has there been a non-zero excitation. Hence,
vC(0-) = vC(0+) = 0 and iL(0-) = iL(0+) = 0
At t = 0+, replace capacitor and inductor by 0-valued voltage and current sources toobtain:
vL(0+) = 10 V, iC(0+) = 1 A
For t > 0, we have a driven parallel RLC circuit with vC(∞) = 10 V. Thus
»R = 5; C = 0.01; L = 4/3;si = roots([1 1/(R*C) 1/(L*C)])s1 = si(1); s2 = si(2);si = -15 -5
and
vC (t) = K1e−15t + K2e−5t +10 V
From, the initial conditionsvC(0+) = 0 = K1 + K2 +10
and
vC '(0) = s1K1 + s2K2 = −15K1 − 5K2 =iC (0+)
C=
10.01
=100
»A = [1 1;-15 -5];»b= [-10; 100];»K = A\bK = -5.0000e+00 -5.0000e+00
in which case
vC (t) = −5e−15t − 5e−5t +10 V
To compute iL, note that iL(∞) = 1 A,
iL ( t) = K1e−15t + K2e−5t +1 A
From, the initial conditionsiL(0+) = 0 = K1 + K2 +1
and
iL '(0) = s1K1 + s2K2 = −15K1 − 5K2 =vL (0+)
L=
104 / 3
= 7.5
»A = [1 1;-15 -5];»b= [-1; 7.5];»K = A\bK = -2.5000e-01 -7.5000e-01
iL ( t) = −0.25e−15t − 0.75e−5t +1 A
SOLUTION TO PROBLEM 9.49.
Input to this circuit is a superposition of the inputs in problems 9.47 and 9.48. So, theoutput of the circuit here is a superposition of the output of the circuit in problems 9.47and 9.48:
vC (t) = 10e−15t +10e−5t −10 V
For t > 0, by linearity this is the difference of the zero-input circuit response (i.e., due tothe IC's as per problem 47) and the zero-state (zero ICs) as per problem 48.
SOLUTION TO PROBLEM 9.50.(a)
vC(0-) = –60 V, iL(0-) = –0.1 A(b) By continuity property,
vC(0+) = –60, iL(0+) = –0.1(c)Replace capacitor by voltage source of value –60 V and inductor by current source ofvalue –0.1 A.
vL(0+) + vC(0+) = 60⇒ vL(0+) = 120
andiC(0+) + iR1(0+) – 1 – iL(0+) = 0
⇒ iC(0+) = 1
(d) Req = 120//600 = 100 Ω.»R = 100; C = 1e-3; L = 2;si = roots([1 1/(R*C) 1/(L*C)])si = -5.0000e+00 + 2.1794e+01i -5.0000e+00 - 2.1794e+01i»sig = -real(si(1)); wd = imag(si(1));
(e) Note that if the excitation of 60 V had remained forever, then iL(∞) would be 0.1 A.Therefore for the interval 0 < t < 1,
iL ( t) = e−5t A cos(21.794 t) + B sin(21.794t)[ ] + 0.1 A
(f)iL(0+) = A + 0.1 = –0.1
⇒ A = –0.2
iL' (0+) = −5A + 21.794B =1 + 21.794B =
vL (0+)L
= 60
Hence, B = 2.707 and
iL ( t) = e−5t −0.2cos(21.794t) + 2.707sin(21.794t)[ ] + 0.1 A
(g) For t > 1, the forcing function is zero iL(∞) = 0. Thus,
iL ( t) = e−5(t−1) A cos 21.794(t −1)( ) + Bsin 21.794(t −1)( )[ ] A
From part (f) of the exp(-5t) term, we can guess that iL(1) approximates 0.1 and also that
vC(1) approximates its steady state value of 60 V. These may be off by percent or two,
but are good enough for our engineering calculations. It follows that vL(1+)approximates –60 V.
iL (1+) = A = 0.1and
iL' (1+) = −5A + 21.794B = −0.5 + 21.794B =
vL (1+)L
= −30
Here B = –1.35. Hence
iL ( t) = e−5(t−1) 0.1cos 21.794( t −1)( ) −1.35sin 21.794( t −1)( )[ ] A
*SOLUTION 9.51. To find the initial conditions, use the following equivalent circuit at t =
0–.
By inspection iL (0+ ) = iL( 0− ) = 1 A and vC (0+ ) = vC( 0− ) = 5 V.To find the characteristic roots, set independent source to zero which means open
circuit the independent current source in figure P9.51. This leaves a series RLC with Rth= 10 Ω. Hence
s2 +Rth
Ls +
1
LC= s2 +100s + 2.5 ×104 = 0
Using MATLAB, we find»Rth = 10;C = 0.4e-3; L = 0.1;»s12=roots([1 Rth/L 1/(L*C)])s12 = -50 -50
Since for t > 0, the source is off, we use table 9.1, case 3 to obtain
iL (t) = K1 + K2t( )e−50 t A
It follows that 1 = iL (0+ ) = K1 and
diL
dt(0+ ) = −50K1 + K2 = −50 + K2 =
1
LvL (0+ ) =10vL (0+ )
To find vL (0+) we consider the equivalent circuit valid at 0+:
It follows that
vL (0+) = 5 −10 ×1 = −5 VHence
−50 + K2 = −50or K2 = 0. Finally
iL (t) = e−50tu(t) A
SOLUTION TO PROBLEM 9.52. To find the initial conditions, use the following equivalent
circuit at t = 0–.
By inspection iL (0+ ) = iL( 0− ) = 1 A and vC (0+ ) = vC( 0− ) = 5 V.To find the characteristic roots, set independent source to zero which means open
circuit the independent current source in figure P9.51. This leaves a series RLC with Rth= 10 Ω. Hence
s2 +RthL
s +1
LC= s2 + 20s + 5 ×103 = 0
Using MATLAB, we find»Rth = 10;C = 0.4e-3; L = 0.5;»s12=roots([1 Rth/L 1/(L*C)])s12 = -1.0000e+01 + 7.0000e+01i -1.0000e+01 - 7.0000e+01i
Since for t > 0, the source is off, we use table 9.1, case 2 to obtain
iL ( t) = e−10 t Acos(70t) + Bsin(70t)[ ] A
It follows that 1 = iL (0+) = A and
diLdt
(0+ ) = −10A + 70B = −10 + 70B =1L
vL (0+ ) = 2vL (0+ )
To find vL (0+) we consider the equivalent circuit valid at 0+:
It follows that
vL (0+) = 5 −10 ×1 = −5 VHence −10 + 70B = −10 or B = 0. Finally
iL ( t) = e−10 t cos(70t)u(t) A
0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
SOLUTION TO PROBLEM 9.53. (a) vC (0−) = vC (0+) = 150 ×10−3 × 80 =12 V.
(b) iL(0-) = iL(0+) = 150 mA
(c) For these values at 0+, the independent current source becomes an open circuit.Replace the inductor and capacitor with current and voltage sources to represent theinitial conditions. Solve the resulting simple circuit to obtain:
vL(0+) = –6 = 12 – 120*0.15 V, iC(0+) = iL(0+) = –150 mA
(d)»Rth = 120;C = 2/9 * 1e-3; L = 0.6;»s12=roots([1 Rth/L 1/(L*C)])s12 = -1.5000e+02 -5.0000e+01
(e)
vC (t) = K1e−150 t + K2e−50t V
From, the initial conditionsvC(0+) = 12 = K1 + K2
and
vC '(0) = s1K1 + s2K2 = −150K1 − 50K2 =iC (0+)
C= −675
A = [1 1;-150 -50];b= [12; -675];K = A\bK = 7.5000e-01 1.1250e+01
in which case
vC (t) = 0.75e−150 t +11.25e−50t V(f)
iL ( t) = −iC (t) = −CdvC (t)
dt=
29
×10−3 −150 × 0.75( )e−150 t + −50 ×11.25( )e−50 t[ ]= 0.025e−150t + 0.125e−50t A
SOLUTION TO PROBLEM 9.54. (a) At t = 0-, the independent current source is off, theinductor is a short circuit, and the capacitor is an open circuit. By voltage division,
vC(0-) = vC(0+) = (80/100)50 = 40 V(b)
iL(0-) = iL(0+) = 50/100 = 0.5 A
(c) At t = 0+, we have an independent current source. Also, we replace the inductor witha current source and the capacitor with a voltage source to represent the initial conditions.
1.5 – iL(0+) = iC(0+)⇒ iC(0+) = 1 A
Further,vC(0+) + 40 iC(0+) = 80 iL(0+) + vL(0+)
⇒ vL(0+) = 40 V
(d)Rth = 120;C = 2/9 * 1e-3; L = 0.6;s12=roots([1 Rth/L 1/(L*C)])s12 = -1.5000e+02 -5.0000e+01
(e) In steady state, the capacitor is open and the inductor is a short in which case, Xf = 80* 1.5 = 120 V.
vC (t) = K1e−150 t + K2e−50t +120 V
From, the initial conditionsvC(0+) = 40 = K1 + K2 + 120
and
vC '(0+) = −150K1 − 50K2 =iC (0+)
C= 4.5 ×103
A = [1 1;-150 -50];b= [-80; 4.5e3];K = A\bK = -5 -75
in which case
vC (t) = −5e−150t − 75e−50t +120 V
(f)
iL ( t) = 1.5 − iC (t) = 1.5 − CdvC (t)
dt= −0.16667e−150t − 0.83333e−50 t +1.5 A
SOLUTION TO PROBLEM 9.55.(a) At t = 0-, we replace the inductor by a short circuit and the capacitor by an opencircuit; hence
iL(0-) = iL(0+) = 1 Aand
vC(0-) = vC(0+) = (40)(1) – 20 = 20 V
(b) At t = 0+, we replace the capacitor by a voltage source of value 20 V and theinductor by a current source of value 1 A. Since the inductor current is 1 A and theindependent current source outputs 1 A, no current flows through the branch containingthe capacitor. Therefore,
iC(0+) = 0
Also, because of the zero current in the branch containing the capacitor, no voltage dropoccurs across the resistance in series with the capacitor. Therefore, the voltage across theindependent current source is vC(0+). Therefore,
vL(0+) = vC(0+) – 40*iL(0+) = – 20 V
(c) At steady state (large t), the capacitor becomes an open circuit and the inductorbecomes a short circuit. By inspection,
vC(∞) = 40 V
(d)Rth = 80;C = 1/15 * 1e-3; L = 0.1;s12=roots([1 Rth/L 1/(L*C)])s12 = -5.0000e+02 -3.0000e+02
vC (t) = K1e−500 t + K2e−300t + 40 V
(e) From, the initial conditionsvC(0+) = 20 = K1 + K2 + 40
and
vC '(0+) = −500K1 − 300K2 =iC (0+)
C= 0
A = [1 1;-500 -300];b= [-20; 0];K = A\bK = 3.0000e+01 -5.0000e+01
in which case
vC (t) = 30e−500t − 50e−300t + 40 V
SOLUTION TO PROBLEM 9.56.(a) At t = 0-, we replace the inductor by a short circuit and the capacitor by an opencircuit; hence
iL(0-) = iL(0+) = 1 Aand
vC(0-) = vC(0+) = (40)(1) – 20 = 20 V
(b) At t = 0+, we replace the capacitor by a voltage source of value 20 V and theinductor by a current source of value 1 A. Since the inductor current is 1 A and theindependent current source outputs 1 A, no current flows through the branch containingthe capacitor. Therefore,
iC(0+) = 0
Also, because of the zero current in the branch containing the capacitor, no voltage dropoccurs across the resistance in series with the capacitor. Therefore, the voltage across theindependent current source is vC(0+). Therefore,
vL(0+) = vC(0+) – 40*iL(0+) = – 20 V
(c) At steady state (large t), the capacitor becomes an open circuit and the inductorbecomes a short circuit. By inspection,
vC(∞) = 40 V
(d)Rth = 80;C = 62.5e-6; L = 0.1;s12=roots([1 Rth/L 1/(L*C)])s12 = -400 -400
vC (t) = K1 + K2t( )e−400t + 40 V
(e) From, the initial conditionsvC(0+) = 20 = K1 + 40 ⇒ K1 = – 20
and
vC '(0+) = −400K1 + K2 = 8000 + K2 =iC (0+)
C= 0
in which case
vC (t) = − 20 + 8000t( )e−400 t + 40 V
SOLUTION TO PROBLEM 9.57.Step 1:
dvC/dt = iC/C = 2iC
diL/dt = vL/L = 0.5vL
Step 2: iR = vC/2. From KCL, iL – iR – ic = 0. Therefore, ic = iL – vc/2.
Step 3: Similarly, vin – vL – vc = 0, which implies vL = vin – vC. Hence,
dvC/dt = 2 iL – vC
diL/dt = 0.5 vin – 0.5 vC
Step 4. Eliminate terms in vC (see equation 9.47 in text) to obtain:
d2iL/dt2 + diL/dt + iL = 0.5 dvin/dt + 0.5 vin
SOLUTION TO PROBLEM 9.58.(a) At t = 0-,
vC(0-) = vC(0+) = 0iL(0-) = iL(0+) = 0
For t between 0 and 1, we have a parallel RLC circuit, with Req being the parallelcombination of the two 21.1333 Ω resistors.
R = 21.1333/2; C = 15.7729e-3; L = 0.1;si = roots([1 1/(R*C) 1/(L*C)])si = -3.0000e+00 + 2.5000e+01i -3.0000e+00 - 2.5000e+01i
If there were no further switchings, then Xf = 1 A. Hence,
iL ( t) = e−3t A cos(25t) + Bsin(25t)[ ] +1 A
Applying the initial conditions,
iL(0+) = A + 1 = 0 ⇒ A = –1
and
iL' (0+) = −3A + 25B = 3 + 25B =
vL (0+)L
= 0
Hence, for 0 ≤ t < 1,
iL ( t) = −e−3t cos(25 t) + 0.12sin(25 t)[ ] +1 A
Now, for the next interval, we need initial conditions. These are obtained from the aboveequation for iL(t) at t = 1.
iL(1) = e-3[-cos(25) –3/25 sin(25)] + 1 = 0.9514and
iL' (1) = 3e−3 cos(25) + 0.12sin(25)[ ] − e−3 −25sin(25) + 0.12 × 25cos(25)[ ] = −0.1671
The circuit is still a parallel RLC circuit, but now there is no source and R = 1.268:R = 1.268; C = 15.7729e-3; L = 0.1;si = roots([1 1/(R*C) 1/(L*C)])si = -2.5000e+01 + 3.0002e+00i -2.5000e+01 - 3.0002e+00i
Hence,
iL ( t) = e−25(t−1) A cos(3(t −1)) + B sin(3(t −1))[ ]
Using initial conditionsA = 0.9514
–0.1671 = –25A+ 3B ⇒ B = 7.8726
Thus, for 1 ≤ t < 2
iL ( t) = e−25(t−1) 0.9514cos(3(t −1)) + 7.8726sin(3( t −1))[ ] A
(b)In period between 1 and 2 seconds, the response has a time constant of 1/25. So, when t =2, 25 time constants would have passed from the time the switch is turned (t = 1). Thismeans that the L and C currents and voltages would have settled almost identically totheir values at 0–. A similar argument can be made for the other cycle. Thus the overallresponse effectively becomes a periodic response equal to the response over 0 ≤ t < 2that reflects the periodicity of the switching.
(c)
0 1 2 3 4 5 60
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
Time in s
Indu
ctor
cur
rent
in A
TextEnd
SOLUTION TO PROBLEM 9.59.(a)
dvc/dt = ic/C = 0.707icdiL/dt = vL/L = 0.707vL
To find expressions for iC and vL we use the following figure.
From this resistive circuit, iC = – vC/1 + iL and vL = vin – 1×iL – vC
˙ v C = −0.707vC + 0.707iL˙ i L = −0.707vC − 0.707iL + 0.707vin
Using equation 9.47,
d2vC
dt2 +1.414dvCdt
+ vC = 0.5vin
»si = roots([1 sqrt(2) 1])si = -7.0711e-01 + 7.0711e-01i -7.0711e-01 - 7.0711e-01i
(b) At steady state (large t), vC = 0.5
vC(t) = e-0.707t[A cos(0.707t) + B sin(0.707t)] + 0.5 V
At t = 0–,vC(0-) = vC(0+) = A + 0.5 = 0 ⇒ A = –0.5
vC' (0) =
iC (0)C
=iL (0) − vC (0)
C= 0 = −0.707A + 0.707B
⇒ B = –0.5
Thusvc(t) = e-0.707t[–0.5 cos(0.707t) – 0.5 sin(0.707t)] + 0.5 V
SOLUTION TO PROBLEM 9.60.(a)
dvC
dt= 3iC
diLdt
=vL
3
To eliminate iC and vL, consider
HenceiC = iin – vC/2 – iL
vL = vC – 12iL
˙ v C = −1.5vC − 3iL + 3iin
˙ i L =1
3vC − 4iL
Using equation 9.47,
d2vC
dt2 + 5.5dvCdt
+ 7vC = 3iin' +12iin
»si = roots([1 5.5 7])si = -3.5000e+00 -2.0000e+00
Note: iin’(t) = 0 for t>0.
At t = 0-, current source is off, inductor is a short circuit, and capacitor is an open circuit.
vc(0-) = vc(0+) = 0iL(0-) = iL(0+) = 0
At t = 0+, the current source is on. Replace the inductor and capacitor by current andvoltage sources to represent the initial conditions. Hence, iC(0+) = 1 A. At t = ∞,vC (∞) =1× 12 / / 2( ) =1.7143. Thus,
vC (t) = K1e−3.5t + K2e−2t +1.7143 VTo find the constants,
vC (0) = K1 + K2 +1.7143 = 0and
vC' (0) =
iC (0)C
= 3 = −3.5K1 − 2K2
»A = [1 1;-3.5 -2];»b = [-1.7143;3];»K = A\bK = 2.8573e-01 -2.0000e+00
Therefore
vC (t) = 0.28573e−3.5t − 2e−2t +1.7143
(b) From equation 9.47b,
d2iLdt2 + 5.5
diLdt
+ 7vC = iin
Since iL (∞) = 0.14286 A,
iL ( t) = K1e−3.5t + K2e−2t + 0.14286 A
Using the initial conditions,
iL (0) = K1 + K2 + 0.14286 = 0and
iL' (0) =
vL (0)L
=vC (0) −12iL (0)
L= 0 = −3.5K1 − 2K2
»A = [1 1;-3.5 -2];»b = [-1/7; 0];»K = A\bK = 1.9048e-01 -3.3333e-01
Therefore,
iL ( t) = 0.19048e−3.5t − 0.3333e−2 t + 0.14286 A
SOLUTION TO PROBLEM 9.61.(a)
dvC1/dt = 0.5iC1
dvC2/dt = 0.5iC2
Writing a node equation,iC1 + vC1/0.5 + (vC1 – vC2)/0.5 = 0
⇒ iC1 = –4vC1 + 2vC2
By symmetry,iC2 = –4vC2 + 2vC1
Hence
˙ v C1 = −2vC1 + vC 2
˙ v C 2 = vC1 − 2vC2
From equation 9.47a,
d2vC
dt2 + 4dvCdt
+ 3vC = 0
(b)»si = roots([1 4 3])si = -3 -1Hence
vC (t) = K1e−3t + K2e−t
(c)
vC1(0) = 2 ⇒ K1 + K2 = 2
and
vC1' (0) =
iC1(0)C
=2(vC 2(0) − vC1(0)) − 2vC1(0)
C=
8 − 8C
= 0 = −3K1 − K2
Hence
vC (t) = −e−3t + 3e−t V
*SOLUTION P9.62. (a) For this problem we need to define a voltage at the output of thefirst op amp as shown below.
Also, let us relate the input and output voltages for an arbitrary leaky integrator as shownbelow.
We now write a node equation at the inverting terminal of the op amp. Here
va
Ra+
vb
Rb+ C
dvb
dt= 0
Equivalentlydvb
dt= −
vb
RbC−
va
RaC(1)
Now we apply the formula of (*) to the second stage of our given op amp circuit toobtain:
dvout
dt= −
vout
RC2−
v1
RC2(2)
where R = 1 MΩ and C2 is to be determined.Applying the formula of (*) to the first stage we obtain:
dv1
dt= −
v1
RC1−
vs
R1C1(3)
The equations (2) and (3) form a coupled set of state equations which we can write as
˙ v out
˙ v 1
=
− 1RC2
− 1RC2
0 − 1RC1
vout
v1
+
0−1
R1C1
vs
Using equation 9.47a of text we can write down the characteristic equation as
s2 +1
RC2+
1
RC1
s +
1
RC1RC2= s +
1
RC2
s +
1
RC1
= 0
We require that the natural frequencies be –4 and –12 in which case
1
RC2= 4,
1
RC1=12
From MATLAB»R = 1e6;»C2 = 1/(R*4)C2 = 2.5000e-07»C1 = 1/(R*12)C1 = 8.3333e-08(b) For this part, the overall dc gain must be 10. The dc gain of the second stage is –1.
The dc gain of the first stage must be −10 = −106
R1; hence R1 =100 kΩ.
(c) Since the roots are distinct and real, for t > 0
vout (t) = K1e−4t + K2e−12t + X f = K1e−4t + K2e−12 t + 10 V
where X f =10 V by part (b). The problem states that the capacitor voltages are initially
zero. Hence
0 = vC2(0)= vout (0) = K1 + K2 +10Equivalently
K1 + K2 = −10 (1)Also,
dvout
dt(0) =
1
C2iC2(0)= −4K1 −12K2 = 0 (2)
because iC2(0)= 0 . This is so because vC1(0)= 0 = v1(0) means no current flowsthrough the 1 MΩ input resistor to stage 2. This fact and the fact that vC2 (0) = 0, means
that no current flows through C2.
From equation 2, K1 = –3 K2. Substituting into equation (1) yields K2 = 5 and
hence K1 = –15. Finally,
vout (t) = −15e−4t + 5e−12 t +10 V
*SOLUTION 9.63. This problem requires the characteristic equation in terms of A. For
this we may set vin = 0 and the circuit becomes the one given below. Note the new label
v1.
The first step is to write a node equation at v1:
v1
100+10−4 d
dtv1 − vout( ) +
1
100v1 −
vout
A
= 0
Equivalentlydv1
dt−
dvout
dt=
100
Avout − 200v1 (*)
Now we write a node equation at vg = vout/A. Here,
10−4
A
dvout
dt+
1
100
vout
A− v1
= 0
Equivalently,dvout
dt= −100vout +100 Av1 (**)
Let us put (*) and (**) in matrix form to obtain:
1 −1
0 1
˙ v 1˙ v out
=
−200 100 / A
100A −100
v1
vout
We can now solve this to obtain the state equations
˙ v 1˙ v out
=
1 1
0 1
−200 100 / A
100A −100
v1
vout
=
−200 +100A −100 +100 / A
100 A −100
v1
vout
Compare these equations with equation 9.37 and use the formula of 9.47b to obtain the
following second order differential equation in vout:
d2vout
dt2 + 300 −100A( ) dvout
dt+104 vout = 0
The discriminant of this characteristic equation is plotted below for 0 < A < 3. For valuesof A > 3, the circuit is unstable. A negative value of the discriminant indicatesunderdamped (1 < A < 3) and a positive value overdamped (0 < A < 1). For A = 1, wehave critical damping.
Solution 9.64 We can write two state equations as follows:(i) From the definition of a capacitor,
dvC1
dt=107iC1
dvC 2
dt= 109iC 2
(ii) From KVL and Ohm's law
dvC1
dt=106 vi − vC1 − vC 2( ) = −106vC1 −106vC 2 +106vi
dvC 2
dt= 108 vi − vC1 − vC 2 − 0.01vC 2( ) = −108vC1 −1.01×108vC2 +108vi
Casting these two equations into a second order differential equation, as described in thetext:
d2vC 2
dt2 +1.02 ×108 dvC 2dt
+1012vC 2 =108 dvidt
The characteristic equation for this differential equation has real roots:»si = roots([1 1.02e8 1e12])si = -1.0199e+08 -9.8049e+03
Since the capacitors become open circuits, vC2(∞) = 0 and vC1(∞) = vi.
vC 2 t( ) = K1e−1.02×108 t + K2e−9.8×103 t V
Applying IC's:
vC 2 0( ) = K1 + K2 = 0Also,
vC 2' (0) =
iC 2(0+)C2
=0.1
10−9 = −1.02 ×108K1 − 9.8 ×103K2
Thus
vC 2' (0) =1 = −1.02K1 − 9.8 ×10−5K2
»b = [0; 1];»A = [1 1; -1.02 -9.8e-5];»K = A\bK = -9.8049e-01 9.8049e-01
vC 2 t( ) = −0.9805e−1.02×108 t + 0.9805e−9.805×103 t V
»t = 0:1/(abs(s1)*100):100/abs(s1);vc = -0.9805*exp(s1*t) + 0.9805*exp(s2*t);»plot(t,vc)»grid»xlabel('Time in s')»ylabel('vout in V')
0 0.2 0.4 0.6 0.8 1x 10-6
0
0.2
0.4
0.6
0.8
1
Time in s
vout
in V
TextEnd
Solution 9.65. First, derive the differential equation by writing state equations:
dvC
dt= 3iC
diLdt
=vL
3
Now, assume that the capacitor is a voltage source and the inductor is a current source,and write by KCL
iC = −vC2
+vCRN
− iL
And by KVL:
vL = vC −12iL
Substitute into the differential equations:
dvCdt
= −32
+3
RN
vC − 3iL
diLdt
=1
3vC − 4iL
Using equation 9.47 we obtain
d2vC
dt2 − −32
+3
RN
− 4
dvCdt
+ −32
+3
RN
−4( ) +1
vC = 0
or equivalently
d2vC
dt2 + 5.5 −3
RN
dvCdt
+ 7 −3
RN
vC = 0
For constant amplitude oscillations, the middle term should be zero, which means that RN
= 3/5.5 = 0.54545 Ω. Thus the negative resistance is – RN = –0.54545 Ω.
SOLUTION 9.66. The problem data is
i1(t) = Im sin(ωt + θ) A
R1 = 500 + 100(Im – 0.01), R2 = 500
Suppose it starts out with exponentially growing amplitude. R1 will increase withincreasing amplitude. This changes the location of the roots until equilibrium is reachedwhere the roots and the amplitude are stable. This is achieved when the roots of thecharacteristic equation describing the output voltage are purely imaginary, i.e.,
(R1 – R2)/(R1R2C) = 0
⇒ R1 = R2 = 500 = 500 + 100(Im – 0.01)
⇒ Im = 0.01
(a)ω0 = 1/[(500)(1µ)] = 2 k-rad/s
(b) Amplitude of i1 is 0.01. Thus i1 = 0.01 sin(ω0t) A. Let v1 = Vm cos(ω0t + φ) V.
Then
dv1/dt = –ω0Vmsin(ω0t + φ) = (0.01/C) sin(ω0t)
⇒ Vm = 0.01/(ω0C) = 5 V and φ = π rad
Now, v2 + v1 + i1R1 – 3v2 = 0 ⇒ v2 = v1/2 + i1R1/2. Finally,
vout = 3v2 = 1.5v1 +1.5R1i1 = 7.5cos( 0t + π) + 7.5sin( 0t) V
Hence the amplitude of vout(t) is: 7.5 2 V.
SOLUTION 9.67. (a)
n =1
RC=104 rad/s
(b) From equation 9.47a,d2v1
dt2 +1
R2C2 v1 = 0
Hence,
v1 t( ) = A cos1
RCt
+ Bsin
1RC
t
V
Using the value of v1(0) we have,
v1 0( ) = A = 5
To compute the second initial condition,
v1' 0 +( ) =
iC1(0+)C
=iR1(0+)
0.1×10−6 = 104 B
But,
iR1(0+) =3v2(0+) − v2(0+) − v1(0+)
R1=
2v2(0+) − v1(0+)
103 = −0.005 Amps
Hence
v1' 0 +( ) =
−5 ×10−3
0.1×10−6 = 104 B
Hence B = –5.
v1 t( ) = 5cos 10,000t( ) − 5sin 10,000 t( ) V
0 0.2 0.4 0.6 0.8 1x 10-3
-8
-6
-4
-2
0
2
4
6
8
Time in s
Vol
tage
v1
in V
TextEnd
*SOLUTION 9.68. For this problem, R2 = 10 kΩ should be Rf = 10 kΩ and R1 = 1 kΩshould be R2 = 1 kΩ.(a) For sustained sinusoidal oscillation, R1 = R2 = 1 kΩ. From equation 9.59,
ω0 =1
R1C=104 rad/s or 1.5915 kHz
(b) From figure P9.68, to obtain an R1 = 1 kΩ, IR1,peak = 0.2 mA. Therefore,
iR1(t) = 0.2sin(ω0t + θ) mA for appropriate θ. Since Cdv1
dt= iC1 = iR1 = 0.2sin(ω0t + θ)
mA, we know that v1 has the following form:v1(t) = V1m cos(ω0t + φ)
In which case
Cdv1
dt= C
d
dtV1m cos(ω0t + φ)( ) = CV1mω0 sin(ω0t + φ+ π) = 0.2sin(ω0t + θ) mA
Therefore CV1mω0 = 0.2 ×10−3. It follows that V1m = 0.2 volts. Here V1m is the peakvalue of the sinusoid. However, the op amp peak output voltage with respect to ground,as shown in problem 66, is 1.5 2V1m . Also, for such a small amplitude, we expect the
output waveform to be quite close to sinusoidal. By choosing a different lamp (R1) witha different characteristic, we can obtain larger peak output voltages.
SOLUTION 9.69. (a) Note that the capacitor is like an open circuit and the inductor islike a short circuit at t=0-. Thus, we can obtain the capacitor voltage by voltage division:
vC 0 −( ) = 10 ×45
= 8 = vC 0 +( )
Similarly, the inductor current is obtained by applying Ohm’s Law:
iL 0 −( ) =105
= 2 = iL 0 +( )
(b) Here, we note that the new initial conditions are just 2.5 times the values that we justobtained in part (a). This can be achieved by simply changing the input voltage source,from 10 to 25 V.
1
PROBLEM SOLUTIONS CHAPTER 10
SOLUTION 10.1 Using KCL, we can write
CdvCdt
+vCR
= iin t( )
Dividing by C:
dvCdt
+1
RCvC =
iin t( )C
We know that iin t( ) = 20sin 400t( ) mA, which can be represented by a complex exponential,
iin t( ) = Re 20e j400te− jπ/2[ ] mA. For convenience we will simply let iin t( ) = 20e j400 te− jπ/2 mA, knowing
that we must take the real part to complete our solution. The output voltage will also be reparesented as a
complex exponential:
vC t( ) = Vme j 400t+( ) = Vme j400te j
Substituting this expression into the differential equation and canceling e j400 t :
j400Vme j +VmRC
e j =20 ×10−3e− jπ/2
C
Thus
Vme j 1RC
+ j 400
=
20 ×10−3
C ⇒ Vme j =
− j40001000 + j400
= 3.714∠− 111.8o
where the values for R = 100 Ω and C = 5 mF were substituted in. Thus,
Vm = 3.714
= 0 − tan−1 4001000
= −111.8°
Taking into account a 90o phase shift we obtain
vC t( ) = 3.714cos 400t −111.8°( ) = 3.714sin 400t − 21.8°( ) V
and
iout t( ) = 18.57sin 400t − 21.8°( ) mA
SOLUTION 10.2 From KCL and component definitions:
2
iin t( ) −vL25
− iL = 0 ⇒ 0.125
diLdt
+ iL = iin t( ) ⇒ diLdt
+ 250iL = 250iin t( )
We represent the input signal by the complex exponential: iin t( ) = 0.2e j250 t A and the unknown current
can be represented as iL ( t) = IL e j 250t+( ) .
Substituting this into the differential equation and canceling e j250 t :
j250IL e j + 250IL e j = 50
Thus
ILe j j250 + 250( ) = 50 ⇒ IL e j =50
250 + j250= 0.14142∠ − 45o
and
IL = 0.141, = −45° ⇒ iL t( ) = 0.141cos 250t − 45°( ) A
SOLUTION 10.3. Construct differential equation by KVL and device definitions:
vin t( ) − 0.5diLdt
− 200iL = 0 ⇒ diLdt
+ 400iL = 2vin t( )
We represent vin t( ) as the complex exponential function, vin t( ) = 20e j400t V. The current in the inductor
has the form: iL = ILe j 400t+( ) . Substituting into the differential equation and canceling e j400 t :
j400ILe j + 400IL e j = 40
Thus
ILe j j + 400( ) = 40 ⇒ IL e j =40
400 + j 400= 0.070711∠− 45o
and
IL = 0.0707, = −45°, ⇒ iL t( ) = 70.7cos 400t − 45°( ) mA
Hence,
vout t( ) = 14.14cos 400t − 45°( ) V
SOLUTION 10.4. Construct differential equation using KVL and device definitions:
3
vin t( ) − vC − CdvCdt
R = 0 ⇒ RCdvCdt
+ vC = vin t( )
The output voltage is defined as:
vout t( ) = vin t( ) − vC t( )
This means that finding vC is enough to be able to obtain the output voltage. The input voltage is
represented by the complex exponential:
vin t( ) = 20e j250te− jπ/2 V
and vC t( ) = Vme j 250t+( ) . Substituting into the differential equation, dividing by e j250 t , and rearranging:
j250RCVCe j + VCe j = − j20 ⇒ VCe j j250RC +1( ) = − j20
⇒ VCe j =− j201 + j
= 14.142∠− 135o
Now
Voutej = − j20 − VCe j ⇒ Voute
j =10 − j10 =14.142∠ − 45o
Thus, in the time-domain,
vout t( ) = 14.142cos 250t − 45°( ) V
SOLUTION 10.5. The circuit is identical to that of problem 10.1. Thus,
RCdvsdt
+ vs = Ris t( ) ⇒ Ris t( ) − RCdvsdt
= vs
Moreover, the complex exponential solution is given by
vs(t) = Vme j = 223.6e− j63.43o V
Hence
R − jRC223.6e− j63.43o= 223.6e− j63.43o
= 100 − j200
i.e.,
R − jRC(100− j200) = R − 200RC − j100RC =100 − j200
4
Thus RC = 2 s, R =100 + 200RC = 500 Ω, and C = 4 mF.
SOLUTION 10.6. First use KVL
vin − vL − vc − vout = 0 ⇒ vin − LdiLdt
−1C
iLdt−∞
t
∫ − RiL = 0
Differentiating this equation, rearranging, and dividing by L,
d2iLdt2 +
RL
diLdt
+1
LCiL =
1L
dvindt
We represent the input signal with = 104 as vin t( ) =100e j te− jπ/2 V and iL ( t) = IL e j te j A.
Substituting these two expressions into the differential equation and dividing out e j t :
− 2IL e j + jRL
IL e j +1
LCIL e j =
− j100× jL
⇒ ILe j 1LC
− 2 + jRL
=108
⇒ ILe j =108
−75 ×106 + j108= −0.48 − j0.64 = 0.8∠− 126.87o
Solving for the magnitude and angle (by hand or using MATLAB):
iL t( ) = 0.8cos 10,000t −126.87°( ) = 0.8sin 10,000t − 36.87°( ) A
and
vout t( ) = 80sin 10,000t − 36.87°( ) V
SOLUTION 10.7. Using standard reference directions, from KCL and component definitions,
iin = iR + iL + iC =voutR
+ iL (0) +1L
vout ( ) d0
t
∫ + Cdvout
dt
Taking a second derivative and dividing by C yields
d2vout
dt2 +1
RCdvout
dt+
1LC
vout =1C
diindt
5
We now let ω = 2500 rad/s and represent iin(t) by the real part of the complex exponential 0.02e j t .
Further we represent vout(t) as the real part of the complex exponential Vme j( t+ ) = Vme j te j .
Substituting these expressions into the differential equation and taking the indicated derivatives yields
Vm( j )2e j te j +VmRC
j e j te j +1
LCVme j te j =
0.02C
j e j t
Observe that e j t cancels out on both sides of this equation producing
Vme j 1LC
− 2 +jRC
=
j0.02C
Hence
Vme j =j0.02
C1
LC− 2 +
j
RC
=1.28 + j0.96
This was obtained using MATLAB as follows:
»w = 2500;
»L = 40e-3;
»C = 1e-6;
»R = 100;
»Vout = j*0.02*w/(1/L - C*w^2 +j*w/R)
Vout = 1.2800e+00 + 9.6000e-01i
»magVout = abs(Vout)
magVout = 1.6000e+00
»angVout = angle(Vout)*180/pi
angVout = 3.6870e+01
Therefore
vout (t) = 1.6cos(2500t + 36.87o) V
SOLUTION 10.8. (a) From KCL: 3∠15° − 5∠45°− I = 0 ⇒ I = 3∠15°− 5∠45° . In MATLAB,
»Ibar = 3*exp(j*15*pi/180) - 5*exp(j*45*pi/180)
6
Ibar =-6.3776e-01 - 2.7591e+00i
»abs(Ibar)
ans =2.8318e+00
»angle(Ibar)*180/pi
ans =-1.0302e+02
Therefore,
i t( ) = Re Ibar = 2.83cos t −103°( ) A
(b) From KCL,
»Ibar = (1 + 2*j) - (-2 + j*6)
Ibar =
3.0000e+00 - 4.0000e+00i
»abs(Ibar)
ans = 5
»angle(Ibar)*180/pi
ans =-5.3130e+01
Therefore i t( ) = 5cos 50πt − 53°( ) A.
SOLUTION 10.9. We define a Gaussian surface encompassing the three bottom nodes. Thus, KCL
dictates that the sum of 4 currents be zero:
»Ibar = -(-2-j*8) + (3 +j*12) + 10
Ibar =
1.5000e+01 + 2.0000e+01i
»abs(Ibar)
ans =
25
»angle(Ibar)*180/pi
ans =
5.3130e+01
Therefore i t( ) = 25cos 1000t + 53.13°( ) A.
SOLUTION 10.10. First represent the time-domain functions as phasors:
7
V1 = 2∠0°, V2 = 2 2∠− 45°
Then, by KVL
»V1 = 2; V2 = 2*sqrt(2)*exp(-j*pi/4);
»VL = V1 - V2
VL =
0 + 2.0000e+00i
Therefore, VL = 2∠90° V and vL t( ) = 2cos t + 90°( ) = −2sin( t) V
SOLUTION 10.11. Apply KVL by simply following the loop defined by the independent voltage sources:
Vx = 4 j − 2 j −1−1+ (1− j) − (1+ j) = −2
SOLUTION 10.12. First note that Iout =VRR
. VR = − j10 + (5 − j5) −10 ⇒ VR = −5 − j15 V.
Thus, Iout =−5 − j15
5= −1− j 3 = 3.16∠ −108.4° ⇒ iout (t) = 3.16cos 500 t −108.4°( ) A.
SOLUTION 10.13. Using a Gaussian surface,
Iy = −(2 + j3) − (1+ j2) + (1− j5) = −2 − j10 = 10.2∠− 101.3o A.
Therefore, i t( ) =10.2cos 2000 t −101.3°( ) A. Now applying KVL,
Vx = (2 + j2) + (2 + j 3)− (1− 4 j) = 3 + 9 j
»Vx=2+2*j+2+j*3-1+4*j
Vx =
3.0000e+00 + 9.0000e+00i
»abs(Vx)
ans =
9.4868e+00
»angle(Vx)*180/pi
8
ans =
7.1565e+01
Thus, vx t( ) = 9.487cos 2000 t + 71.6°( ) V.
SOLUTION 10.14. (a) At = 1000 rad/s,
YC ( j ) = j C = j4.7 ×10−3 ⇒ C =1.496 F
So, at = 50 , YC ( j ) = j C = j2.3499 ×10−4 S ⇒ ZC ( j ) = − j4255 Ω .
(b) At = 1000 rad/s, ZL = j L = j18.85 ⇒ L = 6 mH. Since impedance is proportional to
frequency, multiplying the frequency by 20 means the impedance is multiplied by 20. Thus, at
= 20 × (103π) rad/s,
ZL = j18.85 × 20 = j 377 ⇒ YL = − j2.652 mS
SOLUTION 10.15. The input voltagephasor as 2∠0° . By inspection:
I1 = j C VC = j10 × 0.1× 2 = j2
From the definition of a dependent V-source and an inductor:
I2 =5I1j L
=j10j2
= 5
Finally,
Vout = 5 × 2 I1 + I2( ) = 10 5 + j2( ) = 50 + j20 = 53.85∠21.8°
vout t( ) = 53.85cos 10t + 21.5°( ) V
SOLUTION 10.16. Using KCL and the definition of a resistor:
V10Ω = 10 6∠0° − 3∠90°( ) = 10(6− 3j) = 60 − j 30 V
Thus,
Z j 0( ) =60 − j 30
j3= −10 − j20 Ω
9
And, the combination of this Z( j 0) with the 10 Ω resistance, at this frequency, is 10 – j5.
»Z=-10-j*20; R = 10;
»ZZ = R*Z/(R+Z)
ZZ =1.0000e+01 - 5.0000e+00i
»C = 1/(-5*j*j*2000*pi)
C =3.1831e-05
This is equivalent to a 10 Ω resistor in series with a 31.83 µF capacitor.
SOLUTION 10.17.
»w = 2*pi*60;
»VL = 3 +12*exp(-j*30*pi/180) + 6 -12*exp(j*30*pi/180)
VL =9.0000e+00 - 1.2000e+01i
»ZL = j*w/60
ZL =0 + 6.2832e+00i
»IL = VL/ZL
IL =-1.9099e+00 - 1.4324e+00i
»abs(IL)
ans =2.3873e+00
»angle(IL)*180/pi
ans =-1.4313e+02
Therefore, iL ( t) = 2.387cos(120πt −143.1o) A.
Solution 10.18 (a)
First represent the inputs with their phasors: Is1 =10∠30° = 8.66 + j5 and Vs2 =100∠0° .
The admittance of the RC combination is: Y1 =1R
+ j C = 0.1+ j0.1. Using superposition and noting that
a 0-volt V-source is a short circuit and a 0-amp current source is an open circuit,
Ix = Y1Vs2 − Is1 =1.3397 + j5 = 5.1764∠75o A
Therefore,
ix t( ) = 5.176cos 100t + 75°( )(b) Let ix1 be the contribution to ix generated by the current source and ix2 the contribution generated by
the voltage source. Then ix1(t) = −10cos(50t + 30o) A, and ix2(t) = −14.142cos(100 t + 45o) A since
10
»Vs2 = 100;
»w = 100;
»R = 10; C = 1e-3;
»Y1 = 1/R + j*w*C
Y1 =1.0000e-01 + 1.0000e-01i
»Ix2 = Y1*Vs2
Ix2 =1.0000e+01 + 1.0000e+01i
»abs(Ix2)
ans =1.4142e+01
»angle(Ix2)*180/pi
ans =45
Therefore ix (t) = −10cos(50t + 30o) −14.142cos(100t + 45o) A.
Solution 10.19 First of all, write out the given values: = 200 rad/s, I1 = 0.5∠90° = 0.5 j A, and
Vs2 = 4∠0° . From KVL Vs1 = 3I1 + j LI1 +I1
j C+ Vs2 which leads to:
»w = 200; I1 = 0.5j; Vs2 = 4;
»Vs1 = 3*I1 + j*w*0.04*I1 + I1/(j*w*1e-3) +Vs2
Vs1 =2.5000e+00 + 1.5000e+00i
»abs(Vs1)
ans =
2.9155e+00
»angle(Vs1)*180/pi
ans =
3.0964e+01
Therefore vs1(t) = 2.9155cos(200t + 30.964o) V.
Solution 10.20 The phasor for the input can be written as = 1000 rad/s and Vin = 10∠60° = 5 + j8.66
V. The currents can be obtained easily by applying Ohm’s law for phasors:
11
IR =10∠60
500= 0.02∠60° A, IL =
10∠60j1000 × 0.25
= 0.04∠− 30° A, and
IC = 10∠60 × j1000 × 2 ×10−6( ) = 0.02∠150° A. Thus
Iin = 0.02∠60° + 0.04∠ − 30° + 0.02∠150° = 0.0283∠15° A
and
iin t( ) = 0.0283cos 1000t +15°( ) A
using the following MATLAB code:
»Vin = 10*exp(j*60*pi/180)
Vin =5.0000e+00 + 8.6603e+00i
»R = 500; L = 0.25; C = 2e-6;
»w = 1e3;
»IR = Vin/500
IR =1.0000e-02 + 1.7321e-02i
»IL = Vin/(j*w*L)
IL =3.4641e-02 - 2.0000e-02i
»IC = j*w*C*Vin
IC =-1.7321e-02 + 1.0000e-02i
»Iin = IR + IL + IC
Iin =2.7321e-02 + 7.3205e-03i
»abs(Iin)
ans =2.8284e-02
»angle(Iin)*180/pi
ans =1.5000e+01
Solution 10.21 In MATLAB
»Iin = -100*j*1e-3; R = 100;
»L = 0.04; C = 2e-6; w = 2500;
»VR = R*Iin
VR =0 - 1.0000e+01i
»VL = j*w*L*Iin
VL =10
»VC = Iin/(j*w*C)
VC =-20
»Vin = VR + VL + VC
12
Vin =-1.0000e+01 - 1.0000e+01i
»abs(Vin)
ans =1.4142e+01
»angle(Vin)*180/pi
ans =-135
Therefore, vin t( ) =14.14cos 2500t −135°( ) V.
Solution 10.22 (a) Here, i1 t( ) = 0.6cos 200t( ) A and vout t( ) = 20sin 200t( ) = 20cos 200t − 90°( ) V.
For = 200 rad/s, the phasors are by inspection: I1 = 0.6∠0° A, Vout = 20∠ − 90° V .
(b) Write down the resistor, inductor, and capacitor current phasors, given Vout:
IR =20∠− 90
1/0.03= 0.6∠ − 90° = − j0.6
IL =20∠− 90j200 × 0.1
=1∠ −180° = −1
IC = 20∠ − 90 × j200 × 0.4 ×10−3( ) = 1.6∠0°
Now, by KCL
I2 = IR + IL + IC − I1 = − j0.6
where we have substituted the above values of branch currents. The time-domain function is:
i2 t( ) = 0.6cos 200t − 90°( ) = 0.6sin 200t( ) A
Solution 10.23 (a) = 400 rad/s and Vin = 20∠0° V. We can easily use the voltage divider formula for
phasors and substitute values to obtain:
Vout =200
200 + j LVin =
201 + j
= 14.14∠ − 45°
in which casevout t( ) = 14.14cos 400t − 45°( ) V
(b) = 250 rad/s and Vin = 20∠− 90° = − j20 V. Again, we can easily use voltage division:
Vout =400
400 +1
j C
Vin =20∠ − 90°
1− j=14.14∠− 45° V
Thus, in the time-domain,
vout t( ) = 14.142cos 250t − 45°( ) V
13
Solution 10.24 (a) = 10,000 rad/s and Vin = 100∠− 90° = − j100 V. Apply the voltage divider
formula:
Vout =100
100 + j L + 1j C
Vin =100
100 + j100 − j0.25Vin = 80∠ −126.87°
The steady-state response is thus,
vout t( ) = 80sin 10,000t − 36.87°( ) V
The phasor method provides for a much easier way of obtaining the steady-state response.
(b) Here, = 2500 rad/s and Iin = 0.02∠0° A. Now, apply current division:
IR =0.01
0.01 + j C + 1j L
Iin =0.01
0.01 + j0.0025 − j0.01Iin = 0.016∠36.86°
By Ohm’s law:
Vout = 100IR =1.6∠36.87° V
Therefore
vout (t) = 1.6cos(2500t + 36.87o) V
Solution 10.25 (a) It is easier to find the admittance first:
Yin ( j100) =1
j L+ j C = − j0.2 + j12.5 = j12.3
⇒ Zin ( j100) = − j0.0813
(b) Yin ( j100) =1
25 j=
1j100 × 0.05
+ j100C . Hence, in MATLAB,
»C = (1/(25*j) - 1/(j*w*L) )/(j*100)
C = 1.6000e-03
Solution 10.26 (a)
Zin ( j ) =1
j C+
R × j LR + j L
=− jC
+j RL
R + j L= 9.975∠− 0.0072°
»R = 10; L = 0.1; C = 1e-3; w = 2e3;
»Zin = -j/(w*C) + j*w*R*L/(R + j*w*L)
Zin = 9.9751e+00 - 1.2469e-03i
14
»abs(Zin)
ans = 9.9751e+00
»angle(Zin)*180/pi
ans = -7.1620e-03
(b) As the frequency increases, the capacitor becomes a short circuit and the inductor becomes an open
circuit. Thus, the impedance approaches R. Analytically,
lim →∞ Zin ( j ) = lim →∞1
j C+ lim →∞
RR
j L+1
= R
(c) Zin ( j ) =− jC
+j RL(R − j L)
R2 + 2L2 = jLR2
R2 + 2L2 −1C
+R 2L2
R2 + 2L2 . It follows that w must satisfy
2LR2C = R2 + 2L2 or equivalently 2 =R2
LR2C − L2 =1
LC 1−L
CR2
. For the given component
values 1−L
CR2
= 0; hence there is no finite value of frequency for which the impedance is real.
SOLUTION 10.27. We note that the input admittance is given by:
Yin ( j ) =1
100 + j 0.1+ j ×10−6 =
100 − j 0.1
104 + 0.01 2 + j ×10−6
=100
104 + 0.01 2+ j ×10−6 − j
0.1
104 + 0.01 2
Thus, ω must satisfy 105 =104 + 0.01 2 or equivalently, = 107 −106 = 3000 rad/s.
SOLUTION 10.28. (a) As usual we will use MATLAB.
»R1 = 20; R2 = 10;
»L = 0.04; C = 0.6e-3;
»w = 250;
15
»Yin = 1/R1 + 1/(R2 + j*w*L) + j*w*C
Yin = 1.0000e-01 + 1.0000e-01i
Hence
Yin ( j250) = 0.1+ j0.1 S
(b) For this part we observe that
Yin ( j250) =1
R1+
1
R2 + jωL+ jωC = 0.1 − 0.05 j + j250C
For this to be real, the imaginary part must be zero, i.e.,
C =0.05
250=
1
5000= 0.2 mF
Solution 10.29 (a) We can derive an expression for the input impedance by noting that it is the series
combination of the resistance and the inductor/capacitor pair connected in parallel. Thus,
Zin ( j ) = R +
1j C
× j L
1
j C+ j L
= R − j
L
C
L −1
C
Equating the real and imaginary parts of the given impedance, R = 4 Ω and
−LLC −1/
=−L
2L − 0.25= 2 ⇒ L = 0.1 H
(b) At zero inductance, the above reactance is zero. Also, at L = 0.125, the denominator of the above
reactance is zero, which means that the reactance is infinite.
Solution 10.30 (a) First derive an expression for the input impedance as a function of frequency:
Zin j( ) = 5 + j L +1
j C= 5 + j L −
1C
We want the imaginary part to be equal to zero. Thus,
L −1C
= 0 ⇒ 2 =1
LC ⇒ = 2500 rad/s
16
where we have substituted the values of L and C.
The magnitude of the impedance is minimum when the imaginary part is zero, and Zin(j2500) = 5.
(b) Derive an expression for the admittance:
Yin ( ) =15
+ j C +1
j L=
15
+ j C −1L
Again, the imaginary part is equal to zero when =1
LC or = 2500, at which point the admittance is
0.2.
Solution 10.31 The input admittance is Yin ( ) =1R
+ j C = 0.008+0.004. Equating this to the given
admittance at = 500yields R = 125 Ω, ωC = 0.004 or C = 8 µF.
Solution 10.32 Derive an expression for the admittance at = 1000:
Yin ( j ) =1
j L+
0.25 × j C0.25 + j C
= − j0.5 +j0.125
0.25 + j0.5= 0.2 − j0.4 S
Note that this is equivalent to a 0.2 S conductance (i.e. 5 Ω resistance) in parallel with a 2.5 mH inductance
(at the given frequency!). Now, the impedance is:
Zin =1
0.2 − j0.4= 1+ j2 Ω
This is equivalent to a resistance of 1 Ω in series with a 2 mH inductance (at the given frequency).
Solution 10.33 The current is zero when the input impedance of the parallel combination of inductor and
capacitor is infinite. The latter is given by:
ZLC( j ) =
1j C
× j L
1
j C+ j L
= − j
L
C
L −1
C
The magnitude of this is infinite when rL −1C
= 0 ⇒ r2 =
1LC
⇒ r =15,811 rad/s. Observe that
Is =Vs
R + ZLC ( j r )= 0
17
Hence is(t) = 0 at r = 15,811 rad/s. At this frequency, the voltage across the LC tank is equal to the input
voltage (since there is no drop across the resistor).
Solution 10.34 (a) By inspection:
Zin =2 × j 0.4
2 + j 0.4=
j 0.8
2 + j 0.4
Yin =1
R+
1
j L= 0.5 − j
2.5
(b) Use the current divider formula, and substitute the given frequency, to obtain:
IL =R
R + j L2 =
2 22 + j2
=1∠ − 45°
and iL ( t) = cos(5t − 45o) A.
Solution 10.35 (a) Using voltage division, Vout =R
R − jC
Vin =j RC
1+ j RCVin which
⇒ ∠Vout = 90o − in − tan−1 RC( ) and Vout =RC Vin
1+ 2R2C2
(b) For this part, we need to make sure that tan−1 RC( ) = 45° ⇒ RC = 1 ⇒ =1/ RC .
(c) At this frequency,
VoutVin
=
1
RCRC
1+1
R2C2 R2C2=
1
2
Solution 10.36 Here, = 1/ RC and VCVin
=
1j C
1
j C+ R
=1
1 + j RC=
11 + j
. Therefore
VC =1
1 + jVin = 0.707Vm∠ − 45°
The time-domain function is
vC t( ) = 0.707Vm cos1
RCt − 45°
V
18
Solution 10.37 (a) The magnitude of the capacitor voltage is 10/14.14 = 0.707 times the magnitude of the
input signal. We just showed in the above problem that
VCVin
=1
1+ j10RC ⇒
VCVin
=1
1 +100R2C2
And we also showed that the ratio is 0.707 when the frequency is 1/RC. So, C = 1/(10R) = 0.01 F.
(b) Again, from the results of the previous problem, the angle is –45 degrees.
Solution 10.38 = 1000 rad/s and Iin = 2∠45° A. The equivalent admittance is
Yeq j( ) =1R
+ j C −1L
= 0.25 + j 0.25 − 0.25( ) = 0.25 S
In MATLAB
»Yeq = 0.25;
»Iin = 2*exp(j*pi/4)
Iin =
1.4142e+00 + 1.4142e+00i
»Vout = Iin/Yeq
Vout =
5.6569e+00 + 5.6569e+00i
»abs(Vout)
ans = 8
»angle(Vout)*180/pi
ans = 4.5000e+01
»% Using Current Division
»IL = (1/(j*1000*4e-3)/Yeq)*Iin
IL =
1.4142e+00 - 1.4142e+00i
»abs(IL)
ans = 2
»angle(IL)*180/pi
ans = -45
19
Therefore, vout (t) = 8cos(1000t + 45o) V and iL ( t) = 2cos(1000t − 45o) A.
(b) If = 618 rad/s, then
Yeq( ) =14
− j0.405 + j0.1545 = 0.25 − j0.25 = 0.3536∠ − 45°
and
»Vout = Iin/Yeq
Vout =
-3.4779e-04 + 5.6565e+00i
»abs(Vout)
ans =
5.6565e+00
»angle(Vout)*180/pi
ans =
9.0004e+01
»IL = Iin*(1/(j*618*4e-3))/Yeq
IL =
2.2882e+00 + 1.4069e-04i
»abs(IL)
ans =
2.2882e+00
»angle(IL)*180/pi
ans =
3.5228e-03
Therefore, vout (t) = 5.657cos(1000t + 90o) = −5.657sin(1000t) V and iL ( t) = 2.288cos(1000t) A.
Solution 10.39 Write the input phasor: = 1000 rad/s and Iin = 0.01 2∠60°A.
»w =1000;
»Iin = 0.01*sqrt(2)*exp(j*60*pi/180)
Iin =
7.0711e-03 + 1.2247e-02i
»Yeq = 1/500 +1/(j*w*0.25) +j*w*2e-6
Yeq =
2.0000e-03 - 2.0000e-03i
»Vin = Iin/Yeq
20
Vin =
-1.2941e+00 + 4.8296e+00i
»abs(Vin)
ans =
5
»angle(Vin)*180/pi
ans =
1.0500e+02
»IR = Iin*(1/500)/Yeq
IR =
-2.5882e-03 + 9.6593e-03i
»abs(IR)
ans =
1.0000e-02
»angle(IR)*180/pi
ans =
1.0500e+02
»IL = Iin*(1/(j*w*0.25))/Yeq
IL =
1.9319e-02 + 5.1764e-03i
»abs(IL)
ans =
2.0000e-02
»angle(IL)*180/pi
ans =
1.5000e+01
»IC = Iin*j*w*2e-6/Yeq
IC =
-9.6593e-03 - 2.5882e-03i
»abs(IC)
ans =
1.0000e-02
»angle(IC)*180/pi
ans =
-1.6500e+02
21
Therefore, vin t( ) = 5cos 1000t −105°( ) V, iC t( ) = 0.01cos 1000t −165°( ) A, iL t( ) = 0.02cos 1000t +15°( )A, and iR t( ) = 0.01cos 1000t +105°( ) A.
SOLUTION 10.40. = 4 . Using voltage division
V1 =R
R +1
j C+ j L
Vin =2
2 − 4 j + 2 j−8 j( ) = 5.657∠− 45o
Converting back to time:
vout t( ) = 5.657cos 4t − 45°( ) V
SOLUTION 10.41. = 25 rad/s, Vs = 10∠0° V. By voltage division,
VC =− j
0.02 × 25
10 + j0.08 × 25 − j0.02 × 25
10∠0° = −2 j = 2∠ − 90°
Thus, vC t( ) = 2cos 25t − 90°( ) V.
SOLUTION 10.42. VoutVin
=1
3 + j8 + 1j 4C
= 0.2 ⇒ 25 = 3 + j 8 −1
4C
2
.
Thus 8 −1
4C
2
= 25 − 9 ⇒ C =1
16= 0.0625 F.
SOLUTION 10.43. Here, = 3.33 ×103 rad/s and Vin = 50∠0° V. Using phasors,
VR =400
400 −j
3.33×103 ×10−6
× 50 = 40∠36.897° V
and
Vout =100VR j C
100 +1
j C
=100VR
j100 C +1= 3985∠32.1° V
22
Hence, vout (t) = 3985cos(3.33×103t + 32.1°) V.
SOLUTION 10.44. Here, = 104 rad/s, Iin = 0.01∠0° A.
Z1 =100 × j 0.1100 + j 0.1
= 1+ j1000 ⇒ VL = ZLIin = 0.01× ZL = 0.01 + j10
Now, in MATLAB
»w=1e4; R = 100;L = 0.1; C = 0.1e-6;
»Z1 = R*j*w*L/(R+j*w*L)
Z1 = 9.9010e+01 + 9.9010e+00i
»Iin = 0.01;
»VL = Z1*Iin
VL =9.9010e-01 + 9.9010e-02i
»Z2 = 1/(1/R + j*w*C)
Z2 = 9.9010e+01 - 9.9010e+00i
»VC = Z2*VL
VC = 9.9010e+01 + 1.7764e-15i
»abs(VC)
ans = 9.9010e+01
»angle(VC)*180/pi
ans = 1.0280e-15
Thus, vC t( ) = 99cos 10000t( ) V.
SOLUTION 10.45. Here = 40 rad/s and Vin = 120∠0° . This problem is best done in MATLAB usingparallel impedance computation, voltage division, and Ohm's law for phasors:
»R1 = 500; R2 = 80;»C = 0.1e-3; L = 2;»w = 40; Vin = 120;»Z1 = R1/(j*w*C)/(R1 + 1/(j*w*C))Z1 = 1.0000e+02 - 2.0000e+02i»Z2 = R2*j*w*L/(R2 + j*w*L)Z2 = 4.0000e+01 + 4.0000e+01i» Use voltage division»VC = Z1*Vin/(Z1+Z2)VC = 1.2212e+02 - 3.1858e+01i»abs(VC)ans = 1.2621e+02»angle(VC)*180/pians = -1.4621e+01
23
» Use voltage division and Ohm's law for inductors»VL = Z2*Vin/(Z1+Z2)VL =-2.1239e+00 + 3.1858e+01i»IL = VL/(j*w*L)IL =3.9823e-01 + 2.6549e-02i»abs(IL)ans = 3.9911e-01»angle(IL)*180/pians = 3.8141e+00
Therefore, vC (t) = 126.21cos(40 t −14.621o) V and iL ( t) = 0.399cos(40 t + 3.814) A.
SOLUTION 10.46. Here = 40 rad/s and Iin = 0.120∠0° A. This problem is best solved usingMATLAB.
»w = 40; Iin = 0.12;R = 5;»C = 0.004; L = 0.1;»Y1 = 1/R + j*w*CY1 = 2.0000e-01 + 1.6000e-01i
»Y2 = 1/R + 1/(j*w*L)Y2 = 2.0000e-01 - 2.5000e-01i
»% USING CURRENT DIVISION»IL = Iin*Y2/(Y1 + Y2)IL = 7.3171e-02 - 5.8537e-02i»abs(IL)ans = 9.3704e-02»angle(IL)*180/pians = -3.8660e+01
»% AGAIN USING CURRENT DIVISION»IC = Iin*Y1/(Y1 + Y2)IC = 4.6829e-02 + 5.8537e-02i
»% USING OHM'S LAW FOR CAPACITORS»VC = IC/(j*w*C)VC = 3.6585e-01 - 2.9268e-01i»abs(VC)ans = 4.6852e-01»angle(VC)*180/pians = -3.8660e+01
SOLUTION 10.47 Here, = 25 rad/s and IS = 2∠0° A. Now perform a source transformation. The
combination of current source in parallel with resistor is changed into a voltage source in series with the
same resistor. The voltage source value is: VS = 1× IS = 2∠0° V. Apply Ohm’s law to obtain:
I1 =2∠0°
2 + 1j25 × 0.02
= 0.5 + j0.5 = 0.707∠45°
24
Hence, i1(t) = 0.707cos(25 t + 45o) A.
SOLUTION 10.48 Apply a source transformation to obtain:
The impedance of the inductor branch is ZL( j 4) = 2 + j4 × 0.5 = 2 + j2 Ω and
YL( j 4) =1
ZL ( j4)= 0.25 − j0.25 S. Now, the total admittance seen by the source:
Yeq( j 4) =12
+ j0.25 + YL ( j4) = 0.75 S
Thus, by inspection, VC =6∠0°0.75
= 8∠0° V and IC = 0.25∠90°× 8∠0° = 2∠90° A. Therefore
iC ( t) = 2cos(4 t + 90o) = −2sin(4 t) A.
SOLUTION 10.49. Apply a source transformation to obtain:
Then, by inspection, Yeq( j500) =12
+11
+1j2
−1j2
=1.5 S. Thus,
25
VC =2.5∠0°
1.5= 1.667∠0° V = VL ⇒ IL = 1.667/ j2 = 0.833∠− 90° A
In the time-domain:
vC t( ) = 1.667cos 500t( ) V and iL t( ) = 0.833cos 500t − 90°( ) = 0.833sin 500t( ) A.
1
PROBLEM SOLUTIONS CHAPTER 10
SOLUTION 10.50. The input voltage phasor is = 2000 rad/s and VS = 20∠0° V. Now, do a source
transformation on the phasor circuit:
where
IS =VSZL
=20∠0°
j2000 ×10 ×10−3 = 1∠− 90° = − j A
Now,
Yeq =120
+1
j20+ j2000 × 50 ×10−6 = 0.0707∠45° S
and
VC =ISYeq
=1∠− 90°
0.0707∠45°= 14.14∠ −135° V
SOLUTION 10.51. Use superposition. First, find response to current source using circuit below:
Vx_1 = Is1ZRC = Is13× (− j 3)
3 − j3= 2∠0°× 2.121∠− 45° = 4.242∠− 45°
Now, find the response due to the voltage source using the following circuit:
2
The voltage across the inductor is the same as the input source, and this voltage divides between the series
combination of capacitor and resistor:
Vx_2 =j3
3 − j3Vs2 =
j 33 − j3
3∠90o = 2.121∠ −135o V
Combining the two contributions implies that:
Vx = 4.242∠ − 45o + 2.121∠− 135o = 4.74∠ − 71.6o V
SOLUTION 10.52. (a) As stated, VL = aVs1 + bIs2 . To find a, set Is2 = 0 and use voltage division:
VL =−j30
j30 + j30Vs1 = −0.5Vs1 = −aVs1
To find b, set Vs1 = 0 and use parallel impedance and Ohm's law:
VL = j30/ / j30( )Is2 = j15Is2 = bIs2
Hence
VL = –0.5Vs1 +j15Is2
(b) For this part, Vs1 =10 and Is2 = 0.5∠ − 90o . Hence from the formula,
VL =−0.5Vs1 + j15Is2 = −0.5 ×10 + j15 × 0.5(− j) =−5 + 7.5 = 2.5 V
3
Therefore vL(t) = 2.5 cos(100πt) V.
SOLUTION 10.53. In this problem, we can make use of the linearity property for phasors. Specifically,
from the given information, we can write
V1Iin
=20∠45°10∠0°
= 2∠45° = a and V1Vin
=5∠90°10∠45°
= 0.5∠45° = b
Hence,
V1 = aIin + bVin
Substituting the new values of input current and voltage, we obtain:
V1 = 10∠0° +10∠45° =18.48∠22.5° V
SOLUTION 10.54. (a) For Vout to be zero, we want
RC
RC −j
C
=j L
RL + j L ⇒ RCRL =
LC
(b) Substituting RCC = 2 s and RL = 3 Ω into the above expression gives: L = RCCRL = 6 H.
(c) The bridge circuit can be represented by an impedance Zbridge(jω). The voltage that appears across the
bridge, say Vbridge, is obtained by voltage division. Hence, by the voltage substitution theorem, the problem
may be solved as in part (a) with this new source voltage Vbridge appearing across Zbridge(jω).
SOLUTION 10.55 The input phasor is: = 1000 rad/s and Iin = 2∠45° A, assuming peak value. First
compute
Yth ( j103) = 0.25 −j
1000 × 4 ×10−3 + j1000 × 0.25 ×10−3 = 0.25S ⇒ Zth ( j103) = 4 Ω
Then,
Voc = Iin × 4 = 8∠45° V
4
in which case voc ( t) = 8cos(103t + 45o) V. The final equivalent is a voltage source (having value Voc) in
series with a resistance of 4 Ω.
SOLUTION 10.56. (a) First note that the frequency is given in Hz, so, = 1281.77rad/s and Iin =10∠0°A. Then, in MATLAB,
»R = 0.25; L = 1.17e-3; C = 520e-6;
»w = 2*pi*204;
»Yin = j*w*C +1/(R + j*w*L)
Yin =
1.0815e-01 + 1.7737e-02i
»Zin = 1/Yin
Zin =
9.0039e+00 - 1.4766e+00i
»abs(Zin)
ans =
9.1242e+00
»angle(Zin)*180/pi
ans =
-9.3134e+00
Therefore Zth = 9.1242∠− 9.313° Ω. Finally, Voc = ZthIin = 91.2∠− 9.313° V.
(b) Now, the circuit looks like the following:
Simple voltage division can yield:
VL =ZL
ZL + ZTHVoc = 46.219∠0o ⇒ vL t( ) = 46.2cos 1281.77t( ) V
5
SOLUTION 10.57. For this problem we short the V-source and compute Zth and use voltage division to
find Vo c . Specifically
Zth =1
jωC + 1
0.1+ jωL
= 103 − j10 Ω
and
Voc =
1
jωC1
jωC+ 0.1 + jωL
× 2 = −200 j V
where
»w=1000;
»L = 0.01;
»C = 0.1e-3;
»Zth = 1/(j*w*C + 1/(0.1 + j*w*L))
Zth = 1.0000e+03 - 1.0000e+01i
»Voc = 2*(1/(j*w*C))/(0.1 + j*w*L + 1/(j*w*C))
Voc = 0 - 2.0000e+02i
To compute the load voltage, define
Zload =103 + j10−2 ω = 103 + j10 Ω
Again using voltage division and MATLAB we have
»Zload = 1e3 + j*10;
»Vload = Voc*Zload/(Zload + Zth)
Vload =
1.0000e+00 - 1.0000e+02i
»magVload = abs(Vload)
magVload =
1.0000e+02 (volts)
6
SOLUTION 10.58. Here, = 3.33 ×103 rad/s and Vin = 50∠0° V. Here we note that we already found
Voc in Problem 10.43. Thus, Voc = 3985∠32.1° V. In order to find Zth, we introduce a fictitious 1 A
current source at the A and B terminals:
Now, we note that the VR voltage phasor is zero. Thus, the dependent source has zero volts across it. This
way, the temporary current source sees the parallel combination of a resistor and a capacitor:
VAB =100 × − j
3.33×103× 0.25×10−6
100 − j1200= 99.65∠ − 4.77° V
Thus, Zth = 99.65∠− 4.77° Ω. The Thevenin equivalent consists of Voc in series with
Zth = 99.65∠− 4.77° = 99.3 − j8.2865 Ω. Thus the Norton equivalent is the parallel combination of Zth
and the current source with value
Isc =VocZth
=3985∠32.1°
99.65∠ − 4.77°= 39.99∠36.91o A.
SOLUTION 10.59. = 10,000 rad/s and Vin = 10∠0° V. Again, we have already found Voc in Problem10.44: Voc = 99∠0° V. Now, to compute the impedance, we introduce the temporary current source of 1A:
7
Again, the inductor has no voltage across it. So, the dependent source generates no current. Hence, the
independent source sees the parallel combination of a resistor and a capacitor:
VAB =100 × − j1000( )
100 − j1000= 99.5∠− 5.71° V
Thus Zth ( j104 ) = 99.5∠ − 5.71° Ω. So, the Thevenin equivalent is the series combination of the Voc
source and the above Zth ( j104 ) = 99.5∠ − 5.71° Ω.
SOLUTION 10.60. Inject a current source at terminals A and B. Then, write a KCL equation at node A:
IR = IS − I1 ⇒ VR = IS − I1 ⇒ VC = VAB = VR + I1 = IS − I1 + I1 = IS
Since the voltage across the current source is equal to its current, the equivalent impedance across this
current source is 1 Ω.
SOLUTION 10.61. Inject a current source as usual. Then, write Ohm’s law for phasors for the equivalent
series RLC circuit. Note that the controlling current for the dependent source is the input current:
VAB = j0.01× 200 × IS −j
200 × 0.005IS − 2IS = 2 + j( )IS
Therefore, Zth =VABIs
= 2 + j Ω.
SOLUTION 10.62. Inject a current source Is :
8
Apply KCL: j2I− 6I
6= Is − I ⇒ I =
6j2
Is . Now, VAB is the voltage across the inductor:
VAB =6 × j2
j2Is = 6Is ⇒ Zth = 6 Ω.
Solution 10.63. = 2000 rad/s and VS = 25∠0° V. Now, inject a current source, and express VAB as a
function of this current source and Voc:
Now, write two nodal equations at A and the top of the dependent current source:
V1 − VS
103 + gmVAB + j C V1 − VAB( ) = 0
j C VAB − V1( ) +VABj L
= IS
Substituting values, these can be cast into the following matrix equation:
9
10−3 1 + j 3 − j
− j − j
V1
VAB
=
10-3VS
IS
This can be solved in MATLAB to obtain:
V1
VAB
=100 ×
−2.5 − j2.5 −5 + j10
2.5 + j2.5 5
10-3VS
IS
Thus
VAB = (0.25 + j0.25)Vs + 500Is = 500Is + 6.25 + j6.25
Therefore Zth = 500 Ω and Voc = 6.25 + j6.25 = 8.839∠45o V. For the Norton equivalent we need
Isc =VocZth
=8.839500
∠45o = 0.01768∠45o A.
SOLUTION 10.64. Inject an upward current source, IS2, at VAB. Then, write the following two nodal
equations at Vx and VA: let R1 = R2 = R, then
VxR
+ j C1Vx + j C2 Vx − VAB( ) = IS
j C2 VAB − Vx( ) + gmVx +VAB
R= IS2
which after grouping terms becomes
1
R+ j C1 + j C2
Vx − j C2VAB = IS
− j C2 + gm( )Vx + j C2 +1
R
VAB = IS2
In MATLAB
»R = 100e3; C1 = 1e-9; C2 = 1e-10;
»gm = 0.1e-3; w = 1e3;
»Nodal = [1/R+j*w*C1+j*w*C2 -j*w*C2;-j*w*C2+gm j*w*C2+1/R]
Nodal =
1.0000e-05 + 1.1000e-06i 0 - 1.0000e-07i
10
1.0000e-04 - 1.0000e-07i 1.0000e-05 + 1.0000e-07i
»Nodalinv = inv(Nodal)
Nodalinv =
9.5680e+04 - 2.0070e+04i 2.1024e+02 + 9.5470e+02i
-9.5449e+05 + 2.1120e+05i 9.7783e+04 - 1.0523e+04i
Thus
Vx
VAB
=103 ×
95.68 − j20 0.21+ j0.955
−954.5 + j211.2 97.78 − j10.52
40 ×10-6
IS2
Therefore
VAB = −38.18 + j8.448 + 97.8 − j10.5( ) ×103IS2
from which we identify
Voc = −38.18 + j8.448 = 39.1∠167.5o V
and Zth = 97.8 − j10.5( ) ×103 = 98.35 ×103∠− 6.14o Ω.
SOLUTION 10.65. We solve this problem by the method illustrated in example 6.3 where a fictitious
source is applied and the repsonse is calculated. One can either apply a voltage source or a current source
(see figure 6.10 a and b). Generally speaking, neither choice can be claimed as superior to the other. But
for a specific circuit, one choice can lead to a much simpler solution than the other. To illustrate this point,
we solve the problem with both choices below. Note that although the first method is much simpler than
the second, it lacks the generality. If one more resistor were inserted into the circuit, the simplicity of
solution of solution may disappear totally, whereas the second method will proceed with very few changes.
Method 1. For this solution we apply an arbitrary voltage source, labeled in phasor form as Vs as
indicated in the circuit below.
For this circuit we will compute an equation of the form ofequation 6.11:
11
IA =1
ZthVs − Isc (1)
By inspection of the circuit,
IA = IC + IL −1 = jωCVs +VLjωL
−1 = j2Vs − jVL −1 (2)
ButVL = 0.25IC + Vs = 0.25 jωCVs + Vs = 0.5 j +1( )Vs (3)
Substituting (3) into (2) producesIA = 0.5 + j( )Vs −1 (4)
By comparing (4) with (1), we obtain the answers for the Norton equivlent circuit: Is c = 1 A and Zth =1/(0.5 + j) =
0.4 – j0.8 Ω.
Method 2: For this method we apply an arbitrary current source, labeled in phasor form as Is as indicated
in the circuit below.
Notice that we have added a current label Ix as we plan to use modified nodal analysis method to obtain the
desired answer. For this we will compute an equation of the form
VAB = ZthIs + Voc
Because of the addition of Ix , we can write the modified nodal equations more or less by inspection:
1/ jωL 0 1
0 jωC −1
1 −(1+ j0.25ωC) 0
VL
VAB
Ix
=− j 0 1
0 2 j −1
1 −1 − 0.5 j 0
VL
VAB
Ix
=1
Is
0
where from MATLAB,
12
»w=2000;
»L = 0.5e-3;
»C = 1e-3;
»Y11 = 1/(j*w*L)
Y11 = 0 - 1.0000e+00i
»Y22 = j*w*C
Y22 = 0 + 2.0000e+00i
»Y32 = -(1+j*0.25*w*C)
Y32 = -1.0000e+00 - 5.0000e-01i
Solving the equations in MATLAB produces,
»A = [-j 0 1;0 2*j -1; 1 -1-0.5*j 0];
»Ainv = inv(A)
Ainv =
8.0000e-01 - 6.0000e-01i 8.0000e-01 - 6.0000e-01i 1.6000e+00 + 8.0000e-01i
4.0000e-01 - 8.0000e-01i 4.0000e-01 - 8.0000e-01i 8.0000e-01 + 4.0000e-01i
1.6000e+00 + 8.0000e-01i 6.0000e-01 + 8.0000e-01i -8.0000e-01 + 1.6000e+00i
Multiplying the second row of Ainv times the right-most vector of our equations produces
VAB = (0.4 − j 0.8)Is + (0.4 − 0.8j )
This implies that Zth = 0.4 – j0.8 Ω and Vo c = 0.4–j0.8 V. For the Norton equivalent we need
Isc =VocZth
=1 A
SOLUTION 10.66. (a) = 1000 rad/s and VS = 20∠45° V. By voltage division:
Vout =1
1+ j20∠45° =14.14∠0° V ⇒ IL =14.14∠0° A
(b) At dc, Vout = Vin. We want the frequency at which Vout = 0.1Vin. Thus, we want:
13
VoutVin
=1
1+ j × 0.001= 0.1 ⇒
1
1+ 2 ×10−6 =1
100 ⇒ = 9950 rad/s
SOLUTION 10.67. Z = 25 − j20 Ω. By KCL:
VZ −14∠0°j15
+VZ
25 − j20+
VZ + 8∠90°− j20
= 0
This equation simply needs to be manipulated in order to obtain:
VZ = 41.35∠ − 73.46°
SOLUTION 10.68. In this problem, we note that the impedance, jX, is in series with the parallel RLC
circuit to the right. Thus, all we need to do is to find an expression for the equivalent impedance of the
parallel RLC circuit:
YRLC =130
−j
20+ j0.025 ⇒ ZRLC =19.2 + j14.4
Now, the total impedance seen by the source is 19.2 + j14.4 +jX. Therefore, for this to be real, the
unknown reactance has to be –14.4 Ω. Also, the input current now is I = 96/19.2 = 5 A. Hence
i( t) = 5cos(10 t) A.
SOLUTION 10.69. First, compute the current through the series RC section: IC =1
1− j A.
Now, by KCL, we can write
0 = IC − Ix +1− 2Ix
j=
11− j
− Ix +1− 2Ix
j ⇒ Ix = 0.3 + j0.1= 0.316∠18.44o A
SOLUTION 10.70. = 20 rad/s and Yin = 0.05 + j0.0866 S.
(a) Since Yin =1R
−jL
+ j C , equating the real parts of the above two expressions implies that R =
1/0.05 = 20 Ω.
(b) Similarly, equating the imaginary parts and substituting, we obtain:
14
j C −1L
= j0.0866 ⇒ L = 3.73 H
(c)
VC =IinYin
=20∠30°
0.05 + j0.0866= 200∠− 30° V ⇒ vC t( ) = 200cos 20t − 30o( ) V
(d)
IL =200∠30°j20 × 3.73
= 2.68∠ −120°A ⇒ iL t( ) = 2.68cos 20t −120o( ) A
(e) Voc is just the voltage VC, which was obtained in (c), and Zth =1
Yin=10∠− 60° Ω.
(f) Zth = 5 – j8.66 Ω. This is equivalent to a series combination of a 5 Ω resistance and a 5.77 mF
capacitance at the given frequency ω = 20 rad/s.
SOLUTION 10.71. (a) Vs = 2 + j0V, =1000 rad/s. By KCL
VA− j1.33
+VA
2 + j2+
VA − Vs4
= 0 ⇒ VA = 0.5 − j0.5 = 0.707∠ − 45° V
The time-domain expression is:
vC t( ) = vA t( ) = 0.707cos 1000t − 45o( ) V
(b) IL =0.707∠ − 45°
2 + j2= 0.25∠− 90° A implies that iL t( ) = 0.25cos 1000t − 90o( ) = 0.25sin(1000t) A.
(c) We already determined Voc in part (a). Now, turn off the source to compute the equivalent impedance:
Yth = 0.75 j +1
2 + j2+ 0.25 = 0.5 + j0.5 ⇒ Zth =1− j Ω. This is the series connection of a 1 Ω resistor
with a capacitor of value C =1
=10−3 F. This completes the definition of the Thevenin equivalent.
SOLUTION 10.72. In MATLAB,
»R = 5; L = 1e-3; C = 20e-6;Vs1 = 5; Is2 = 0.5*j;
»G = 1/R; w = 10e3;
»YL = 1/(j*w*L)
YL =
0 - 1.0000e-01i
15
»YC = j*w*C
YC =
0 + 2.0000e-01i
Hence
G(VC − Vs1) + YCVC + YL (VC − Vs1 − 2VR) = Is2
Substituting for VR,
G(VC − Vs1) + YCVC + YL (VC − Vs1 − 2(Vs1 − VC)) = Is2
Therefore
G + YC + 3YL( )VC = G + 3YL( )Vs1 + Is2
Again, using MATLAB,
»a = G + YC + 3*YL
a =
2.0000e-01 - 1.0000e-01i
»b = (G+3*YL)*Vs1 + Is2
b =
1.0000e+00 - 1.0000e+00i
»VC = b/a
VC =
6.0000e+00 - 2.0000e+00i
»abs(VC)
ans =
6.3246e+00
»angle(VC)*180/pi
ans =
-1.8435e+01
Hence, vC (t) = 6.325cos(104 t −18.44o) V.
SOLUTION 10.73. Denote by vC1, the node voltage of the 2.5 mF capacitor . Note that at ω = 800 rad/s,
Vs = 20∠0o V. From this we write a set of nodal equations by inspection after observing the following
from MATLAB:
16
»w = 800;
»C1 = 2.5e-3;
»L = 1.25e-3;
»Y1 = 0.5+j*w*C1-j/(L*w)
Y1 = 5.0000e-01 + 1.0000e+00i
»Yoff=j/(L*w)
Yoff = 0 + 1.0000e+00i
»Y2 = 0.25+j*w*3.75e-3 - j*w*1.25e-3
Y2 = 2.5000e-01 + 2.0000e+00i
This information leads to the following matrix nodal equation:
0.5 + j j
j 0.25 + j2
VC1
Vout
=
0.5Vs
0.25Vs
To solve these equations we again use MATLAB:
»A = [Y1, Yoff;Yoff, Y2]
A =
5.0000e-01 + 1.0000e+00i 0 + 1.0000e+00i
0 + 1.0000e+00i 2.5000e-01 + 2.0000e+00i
»b = [0.5*20;0.25*20]
b =
10
5
»Vnodes = inv(A)*b
Vnodes =
7.1141e+00 - 6.9799e+00i
-3.6242e+00 + 5.3691e-01i
»magVnodes = abs(Vnodes)
magVnodes =
9.9664e+00
17
3.6637e+00
»angVnodes = angle(Vnodes)*180/pi
angVnodes =
-4.4454e+01
1.7157e+02
Therefore
vout (t) = 3.664cos(800t +171.57o) V
SOLUTION 10.74. For this problem we use loop analysis with loops indicated in the figure below.
Since there are no controlled sources, we can write down the loop equations by inspection:
10 + 9 j 6 + 5j
6 + 5 j 12 + 9 j
IA
IC
=
120 −120∠ −120o
120∠120o −120∠− 120o
=180 +103.92
j207.85
The solution of this equation is done in MATLAB as follows:
»b1=120-120*exp(-j*2*pi/3)
b1 = 1.8000e+02 + 1.0392e+02i
»b2=120*exp(j*2*pi/3)-120*exp(-j*2*pi/3)
b2 = 0 + 2.0785e+02i
18
»A = [10+j*9, 6+j*5;6+j*5,12+j*9]
A =
1.0000e+01 + 9.0000e+00i 6.0000e+00 + 5.0000e+00i
6.0000e+00 + 5.0000e+00i 1.2000e+01 + 9.0000e+00i
»I=inv(A)*[b1;b2]
I =
1.4472e+01 - 1.3469e+01i
4.2926e-01 + 1.7703e+01i
»% Please note that using the commands I=inv(A)*[b1,b2]'
»% will lead to the wrong answer because a conjugate is
»% inserted along with the transpose.
»magI = abs(I)
magI =
1.9770e+01
1.7708e+01
»angleI = angle(I)*180/pi
angleI =
-4.2944e+01
8.8611e+01
»IB = -I(1)-I(2)
IB = 1.8288e+00 + 1.7234e+01i
»magIB = abs(IB)
magIB = 1.7331e+01
»angleIB = angle(IB)*180/pi
angleIB = 8.3943e+01
Changing the sign on each source amounts to multiplying its value by "–1". This means that all
magnitudes remain the same, but there is a 180o phase shift for each current, i.e., add 180
o to each current
angle.
19
SOLUTION P10.75. For this problem we have both a transient component to the response and a steady
state component. The steady state component is computed in the usual way because the circuit is stable,
i.e., the time constant is positive. Once the steady state part is computed, we use initial conditions to obtain
the coefficient B in the response.
Part 1: Compute steady state response. For this we use MATLAB,
»R = 0.5; L = 0.866;
»Vs = 10;
»w = 1;
»Zin = R + j*w*L
Zin = 5.0000e-01 + 8.6600e-01i
»IL = Vs/Zin
IL = 5.0002e+00 - 8.6604e+00i
»magIL = abs(IL)
magIL = 1.0000e+01
»angIL = angle(IL)*180/pi
angIL = -5.9999e+01
Hence
iL (t) = 10cos(t − 60o) + Be−0.577t A
Part 2: From the initial conditions we have
iL (0) = 1 = 10cos(t − 60o) + Be−0.577t[ ]t=0= 10cos(−60o ) + B = 5 + B
Hence B = –4. It follows that
iL (t) = 10cos(t − 60o) − 4e−0.577t A
SOLUTION 10.76. KCL dictates that:
IL + IR + IC = IS ⇒ IL = IS – IR – IC
We can perform this sum graphically as follows:
20
j2
-IC
-IR
IL=-j
SOLUTION 10.77. (a) Note that VS = VR + VL, but that VL leads VR by 90 degrees. Similarly, IS = IR +
IC, but IC leads VS by 90 degrees. Also note that the inductor current is also IR, and the capacitor voltage is
VS.
VRIR
VL
VS
ICIS
(b) Using graph paper to construct the phasor diagram to scale, we find the difference between the phase
angles of Is and Vs is zero.
Solution 10.78. First note that VC, the capacitor voltage, will lag IS by 90 degrees. Now, VC plus the
unknown element voltage should result in a vector that runs diagonally between the two vectors. From the
following illustration, it follows that the unknown voltage should have the same phase as the input current:
VC
IS
Vu
VS
VC
21
This means that the unknown element is a resistor. The 45o phase difference implies that VC = Vu or
IsC
= IsR . Therefore R =1C
=1
103 ×10−6 =1000 Ω.
SOLUTION 10.79. The student can construct the phasor diagram using graph paper. The diagram is going
to look like that in the problem statement, except that the proper lengths and angles will be used.
SOLUTION 10.80. As the frequency approaches infinity, the capacitor shorts and the inductor opens. So,
the output voltage is zero. As the frequency approaches zero, the capacitor opens, but the inductor shorts,
so the output is also zero. A plot of the complete response is shown below. (Note that the magnitude
response at 10rad/s is infinite):
»L = 0.04; C = 0.25;
»w = 0: 30/300:30;
»% Vout = Zin * Iin
»Zin = j*w*L ./(j*w*L*j.*w*C + 1);
»plot(w, abs(Zin))
»grid
»ylabel('Magnitude Zin')
»xlabel('Frequency in rad/s')
»plot(w,angle(Zin)*180/pi)
»grid
»xlabel('Phase in degrees')
»ylabel('Phase in degrees')
»xlabel('Frequency in rad/s')
22
0 5 10 15 20 25 300
5
10
15
20
25M
agni
tude
Zin
TextEnd
Frequency in rad/s0 5 10 15 20 25 30
-100
-80
-60
-40
-20
0
20
40
60
80
100
Frequency in rad/s
Pha
se in
deg
rees
TextEnd
SOLUTION 10.81. At infinite frequency, the resistor current is zero (because the inductor opens). So, theoutput voltage is zero. At DC, the inductor is short, and the output voltage is equal to the input voltage. Theplot of the frequency response is shown below (a logarithmic x-axis is used):
SOLUTION 10.82. The circuit inside the black box is
23
At DC, the capacitor is an open circuit. Thus, the voltage across the resistor is 1mA×R. But we know that
this voltage is 1 from the graph. This means that R = 1 kΩ. Now, in general for the above diagram:
VI
=R
1+ j RC
The magnitude of this function is R/sqrt(2) when = 1/ RC . Substituting the frequency from the graph
(1000 rad/s), we get C = 1 µF.
Solution 10.83. The admittance of the parallel RLC circuit is:
Yin =1R
+1
j L+ j C
Zin =1
Yin=
VsIs
The function we want to find the frequency response for is nothing but the input admittance of the circuit.
Using MATLAB, the following plot can be obtained:
»R = 100; L = 0.1; C = 1e-3;
»w = 0:0.5:300;
»w = 0.01:0.5:300;
»Yin = 1/R + 1. ./(j*w*L) + j*w*C;
»Zin = 1 ./Yin;
»plot(w,abs(Zin))
»grid
»xlabel('Frequency in rad/s')
»ylabel('Magnitude Zin')
24
0 50 100 150 200 250 3000
10
20
30
40
50
60
70
80
90
100
Frequency in rad/s
Mag
nitu
de Z
in
TextEnd
»plot(w,angle(Zin)*180/pi)
»grid
»xlabel('Frequency in rad/s')
»ylabel('Phase Zin in degrees')
0 50 100 150 200 250 300-100
-80
-60
-40
-20
0
20
40
60
80
100
Frequency in rad/s
Pha
se Z
in in
deg
rees
TextEnd
25
SOLUTION 10.84. The circuit inside the box is a series RLC circuit. It cannot be a parallel RLC, because
as per problem 83, the admittance of a parallel RLC does not approach zero as w approaches infinity. Thus,
IV
= Yin =1
R +1
j C+ j L
The resonance frequency is 50 rad/s and is determined by 1/sqrt(LC). Given L = 0.4 H, C = 1 mF.
To obtain R, we make use of the fact that, from the given graph at ω = 57 rad/s, the current
magnitude is approximately 0.2 times the peak magnitude. Therefore
1
R2 + 57L −1
57C
2≅
0.2
R2 + 50L −1
50C
2=
0.2R
Hence
R2 + 57 × 0.4 −1
57 ×10−3
2
=R2
0.04= 25R2
R2 + 57x0.4 - 157x0.001
2 ≅ R2
0.04 = 25R2
From which R =5.2561
24=1.0728 Ω.
SOLUTION 10.85. Create three mesh currents in the three planar loops. All currents are clockwise: I1 in
the voltage source loop, I2 in the top bridge loop, and I3 in the bottom one. The three mesh equations are:
V − I1R1 − R2 I1 − I2( ) − R3 I1 − I3( ) = 0
R2 I2 − I1( ) +1
j C1I2 + Rmeter I2 − I3( ) = 0
R3 I3 − I1( ) + Rmeter I3 − I2( ) +1
j C2I3 = 0
The plots of the magnitude and phase of VB − VC = Rmeter I3 − I2( ) are shown in the text.
26
SOLUTION 10.86. We can see all ranges by plotting on a logarithmic scale:
Note that the output will decay when we start to reach the bandwidth of the op-amp. In other words, the
inverting amplifier says that the output is –1 times the input (provided the op amp works properly). Once
the op amp’s gain starts dropping, the output voltage also decays with it.
SOLUTION 10.87. Correction: Change the 0.01 µF capacitor to 1 µF. (a) For this part consider the
diagram below,
From the problem statement, ω = 320π rad/s, and Iin = 0.01∠0o A. Observe that the 50 kΩ resistor input
to the inverting op amp terminal is in parallel with the 100 Ω resistor because of the virtual ground at the op
amp terminals. However, for all practical purposes, this has no effect on the 100 Ω resistor, hence from
Ohm's law
27
VL =100 × j 0.1ω100 + j0.1ω
Iin = 50.265 + j50( ) × 0.01 = 0.50265 + j0.5
From the inverting op amp configuration,
ˆ V =-10VL = −5.0265 − j5
From voltage division,
VC =10−3
10−3 + jω10−9ˆ V = 0.49735 − j0.5( ) × −5.0265 − j5( ) = 5∠179.7o V
Therefore
vC (t) = 5cos(320πt +179.7o )V
Parts (b) and (c). For the SPICE simulation we have the following circuit in B2-SPICE:
which leads to the response below
28
MAG(V(IVM))
Frequency (Hz)Prb10-87-Small Signal AC-2
(V)
+0.000e+000
+100.000m
+200.000m
+300.000m
+400.000m
+500.000m
+1.000 +10.000 +100.000 +1.000k +10.000k
The magnitude at 160 Hz is 0.499 for a 1 mA current input. Thus a 10 mA input current should lead to
4.99 V by linearity which approximates the 5 V computed analytically in part (a). Hence with a 15 V
saturation limit, the input magnitude may increase by a factor of 3 to 30 mA.
SOLUTION 10.88. (a) = 400 rad/s and Vin = 10−3∠ − 90° V. By the virtual short property:
Iin =Vin
1/ j C= j CVin
All this current flows through the 1 MΩ resistor: Vin = − j C106 Vin = 0.4∠ −180° . Thus,
vout(t) = – 0.4cos(400t) V
(b) = 200 rad/s, Vin = 10−3∠0° V, and Vout = 10−3∠− 90° V. Again,
Vout = − j10−3 = − j C × 4 ×105Vin = − j80 ×103C ⇒ C =12.5 nF
29
SOLUTION 10.89. (a) = 800rad/s and Vin = 1∠− 90°V ⇒ Iin =Vin
106 . No current flows into Op-
Amp terminals:
Vout = −Iin1
j C=
j × (− j)800
= 1.25 ×10−3 V
Thus, vout(t) = 1.25cos(800t) mV.
(b) Again, Iin =Vin
200 ×103 A, and Vout = −Iin1
j C ⇒ 10 =
j × (− j)
C2 ×105 ⇒ C = 2.5 nF .
Solution 10.90 (a) = 2 700rad /s and Vin = 1∠0° V ⇒ Iin =1
150 ×103 A. Further,
Vout = −150 ×103 − j
C
150 ×103 − jC
Iin =−1
150 ×103 ×150 ×103 − j
C
150 ×103 − jC
= 0.0015∠90o V
Thus, vout(t) = 1.5cos(2π700t + 90o) mV.
(b)
(c) The output lags the DC response by 45 degrees (note that at DC, the amplifier is inverting, or has a
phase of –180 degrees). Now, the frequency response is really determined by the RC circuit in the
feedback path of the op-amp. The first resistance at the input simply converts the input into a current that
30
drives this RC circuit. It can be shown that a 45 degree phase shift occurs in an RC circuit when the
frequency is 1/RfC (directly from the results of an analysis on an RC circuit). So,
RfC = 1/2000π ⇒ C = 0.016 nF (We know Rf = 10 MΩ.)
Also, at this frequency, the response is 0.707×DC response. The DC response is –107/R. So, the DC gain
is 14.14. Thus, R = 7.07 kΩ.
Solution 10.91 (a) Use the virtual short property: Iin =Vin
103 + 1j C
. All of current flows through the
feedback path: Vout = −104Iin =104 Vin
103 − jC
⇒ VoutVin
= 7.07∠ −135o .
(b)
(c) The spice result looks the same at low frequencies. However, at high frequencies, the response falls
back to zero as the op-amp non-ideal frequency response starts to affect the behavior of the circuit.
SOLUTION 10.92. (a) The negative terminal of the op-amp is at Vs. This implies IR =VSR
. By KVL,
31
VS + ZRCIR = Vo ⇒ VoVS
=1 +1R
− j3RC
3R − jC
=1 +3
j3 RC +1
The MATLAB plot for the given values is:
(c) The spice result looks pretty much the same, especially since the cut-off frequency of this circuit is
much lower than the frequency at which the op-amp ceases to operate as an ideal op-amp.
SOLUTION 10.93. To compute the gain as a function of ω we observe that by the properties of an ideal opamp,
Gain =VoutVin
= −2 × 10−5 + j0.5 ×10−6ω
10−4 + j0.1 ×10−6ω=
20 + j0.5ω100 + j 0.1ω
In MATLAB
»G1 = 1/50e3;»G2 = 1/10e3;»C1 = 0.5e-6;»C2 = 0.1e-6;»w = logspace(-1,5,1500);»Y1 = G1 +j*w*C1;»Y2 = G2 + j*w*C2;»H = Y1 ./ Y2;»semilogx(w,abs(H))»grid
32
SOLUTION 10.94. (a) The equations are:
V1 − Vin1000
+V1 − V2
1000+
V1 − V21/ jwC f
= 0
V2 − V11000
+V2
1/ j C2= 0
V2 = Vout
Substituting values and solving for V1 and V2 ,
V1 = 0.8535∠ − 82oVin
V2 = 0.723∠− 114.7o Vin
The second one is the relation that we are looking for.
(b) At 100 Hz,
V1 = 1.00∠ − 3.68o Vin
V2 = 1.00∠ − 7.3oVin
At 3000 Hz,
V1 = 0.15∠ −102oVin
V2 = 0.0728∠ −164o Vin
33
(c) The response is that of a low-pass filter, as predicted from the results of part (b) above.
SOLUTION 10.95. (a) The two nodal equations:
Vx − Vin200
+ j C1Vx + j C2 Vx − Vout( ) = 0
j C2 0 − Vx( ) −Vout
28 ×103 = 0
In MATLAB,
»w = 2*pi*1.34e3; C = 0.05e-6; R1 = 200; R2 = 28e3;
»A = [1/R1+j*w*C+j*w*C -j*w*C;-j*w*C -1/R2]
A =
5.0000e-03 + 8.4195e-04i 0 - 4.2097e-04i
0 - 4.2097e-04i -3.5714e-05
»b = [1/R1; 0];
»V = A\b
V =
2.6664e-01 - 5.9266e+00i
-6.9859e+01 - 3.1429e+00i
»Vout = V(2)
Vout =
-6.9859e+01 - 3.1429e+00i
»abs(Vout)
ans =
6.9929e+01
»angle(Vout)*180/pi
ans =
-1.7742e+02
Hence, VoutVin
= 69.93∠− 177.4o .
When the capacitors are shorts, the output is shorted to the virtual ground input, at 0 V. Similarly, when
they are opens, the virtual ground makes sure that vout is zero, since there is no drop across the feedback
resistor.
34
(b) The band-pass response can be computed using any SPICE program.
SOLUTION 10.96. First, note the input-output relationship:
VoutVin
= −Z f
R1
where Zf is the impedance of the parallel RLC circuit. We have studied this circuit extensively earlier, and
we have shown that at “resonance”, the impedance of this circuit is going to be real and equal to the value
of resistance, in this case R2. That’s exactly the requirement of this problem, since we want VoutVin
= −Z f
R1
to simply be equal to R2/R1. It remains to note that this resonance occurs at a frequency ω = 1/sqrt(LC).
SOLUTION 10.97. (a) First, analyze the feedback amplifier circuit. The output of this op-amp circuit is:
Vop2 = −1/ j C
2 ×106 Vout
Also, by voltage division, the voltage at the resistive voltage divider (+ terminal of first op-amp):
VRR =2 ×103
2 ×105 × −1
j C22 ×106 Vout =−0.01
j C22 ×106 Vout
Now, the first op-amp circuit is an inverting amplifier, but it’s + terminal is at VRR now. Thus,
Vout = −10 Vin − VRR( ) + VRR
Substituting the above VRR means that:
Vout =−10
1+1.1
j
Vin
Since the input voltage has unity magnitude and zero phase, the above expression gives the required
magnitude and phase of the output voltage.
(b)
35
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
1
2
3
4
5
6
7
8
9
10
Frequency Hz
Mag
nitu
de
TextEnd
(c) As can be seen, the response to zero frequency (i.e. DC) is zero. Also, the circuit goes back very quickly
(less than 2 Hz) to provide the required operation, which is to achieve a gain of 10.
Complex Power Prbs 2/5/00 P11-1 @ DeCarlo & P. M. Lin
CHAPTER 11. PROBLEM SOLUTIONS
SOLUTION 11.1. Using equation 11.3, Pav =12
(et −1)2Rdt =e2 t
2+ t − 2et
0
1
= 0.7580
1
∫ W.
SOLUTION 11.2.
(a) From 11.6, Pav =Vm
2
2R= 50 mW for a sinusoidal input.
(b) From 11.3,
Pav =R2
(10cos(10t))2dt0
/20
∫ + (−10cos(10t))2dt/20
3 /20
∫ + (10cos(10t))2 dt3 /20
2 /10
∫
=Im2 R2
= 50 mW
just as the previous case since the square of the absolute cos(10t) is the same as the square of cos(10t).
(c)
Pav =10R2
0.01cos2(10t)[ ]2dt
0
2 /10
∫ =10−3
2R 0.5cos(20 t) + 0.125cos(40 t) + 0.375[ ]dt
0
2 /10
∫
=12
0.520
sin(20t) +0.125
40sin(40t) + 0.375t
0
2 /10
= 37.5 mW
(d)t=0:1/1000:1;R=1e3;
pta= (0.01*cos(10.*t)).^2.*R;ptb= (0.01*abs(cos(10.*t))).^2.*R;ptc= (0.01*cos(10.*t)).^4.*R;
subplot(3,1,1);plot(t,pta);gridylabel('W');subplot(3,1,2);plot(t,ptb);gridylabel('W');subplot(3,1,3);plot(t,ptc);gridylabel('W');xlabel('time in seconds');
Complex Power Prbs 2/5/00 P11-2 @ DeCarlo & P. M. Lin
0 0.2 0.4 0.6 0.8 10
0.05
0.1
W TextEnd
0 0.2 0.4 0.6 0.8 10
0.05
0.1
W TextEnd
0 0.2 0.4 0.6 0.8 10
0.5
1x 10-5
W TextEnd
time in seconds
SOLUTION 11.3. (a) For figure a, the period is 2, and Pav =12
v2(t)R0
2∫ dt =
12R
400 +100( ) = 25 W.
In figure b, the period is 1, and Pav =20t( )2
R0
1∫ dt =
400R
t3
3
0
1
=13.3 W.
(b)
Complex Power Prbs 2/5/00 P11-3 @ DeCarlo & P. M. Lin
0 0.5 1 1.5 2 2.5 3 3.5 410
15
20
25
30
35
40
Time in s
Inst
anta
neou
s P
ower
in W
TextEnd
0 0.5 1 1.5 2 2.5 3 3.5 40
5
10
15
20
25
30
35
40
Time in s
Inst
anta
neou
s P
ower
in W
TextEnd
SOLUTION 11.4. (a) For (a), looking the definition for the effective voltage, one sees graphically thatthe integral over one period, 2, of the squared waveform, is 500. Dividing by the period, and taking the
square root, Veff =15.81 V. For fig. b, Veff = 20t( )2 dt0
1
∫ = 40013
= 11.55 V
(b) Pav = Ieff2 R =
Veff
10
2
8 = 20 W.
(c) Pav =Veff
10
2
8 =10.67 W
Complex Power Prbs 2/5/00 P11-4 @ DeCarlo & P. M. Lin
SOLUTION 11.5. (a) This can be done graphically quite easily. The period of fig a, is 9s. The total
area of one period of the squared waveform is 75. This yields Ieff =759
= 2.89 A. In fig. b, the area
over one period is 25 which yields Ieff =253
= 2.89 A.
(b) Using current division, Pav = Ieff6090
2
30 =111.36 W.
(c) The same result as (b) is obtained since the effective current is the same.
SOLUTION 11.6. (a) Ieff =12
et −1( )2
0
1∫ dt = 0.615 A.
(b) Pav = Ieff2 R = 0.758 W.
SOLUTION 11.7. (a)
Veff2 =
202
10 + 2cos(20t)( )20
2 /20∫ dt =
202
100t + 2t +240
sin(40t) + 2sin(20t)
0
2 /20
= 102.01
Hence Veff =10.1 V.
(b)
Veff2 =
110cos(2t) + 5cos(4 t)( )2
0∫ dt =1
62.5t[ ]0 = 62.568
Hence Veff = 7.91 V.
(c)
Without going into detailed calculation, note the following fact about v32( t) . Only the product terms that
have the same frequency will produce a non-zero result when integrated. Thus the integral reducesto the following:
Veff2 =
1100cos2(2t) + 25cos 2(4 t) + 25cos2(4 t − π /4) + 50cos(4 t)cos(4t − π /4)( )0∫ dt
Hence
Veff = 50 +12.5 +12.5 + 25cos(−π / 4)[ ] = 9.627 V
SOLUTION 11.8. The voltage is V = 50∠0 V and the impedance Zeq =100 −100 j =141.42∠ − 45o
Ω. Thus I =V
Zeq= 353.6∠45o mA. Hence, Pav = R Ieff
2 =100 ×0.3536
2
2
= 6.2516 W.
Complex Power Prbs 2/5/00 P11-5 @ DeCarlo & P. M. Lin
SOLUTION 11.9. (a) The equivalent load seen by the source, Zeq =1
j C +1/ R= 2∠ − 36.87o . Thus
VL = IZeq = 10∠− 36.87o V and vL (t) = 10cos(30 t − 36.87o) V.
(b) Pav =VmIm
2cos( v − i ) =
10 × 52
cos(−36.87o) = 20 W, and from 11.4
pL ( t) = 25cos(−36.87o) + 25cos(60t − 36.87o) W
SOLUTION 11.10. (a) First find the impedance Zeq = 3 + j4 = 5∠53.13o Ω, then
Is =VsZeq
=10∠ −143.13o A. The magnitude is 10 A rms or 14.14 A peak-value.
(b) Pav = 10 ⋅ 50cos(53.13o) = 300 W.
(c) 3Is2 = 300 W
SOLUTION 11.11. (a) By KVL, Vs = 10IL + j100IL + 9 ⋅10IL and
IL =Vs
100 + j100=
100∠0
141.42∠45o =1
2∠− 45o A.
(b) PavVs = Vs IL cos(45o) = 50 W and Pav9V1 = 9 ⋅10IL IL cos(180o) = −45 W.
SOLUTION 11.12. (a) By KCL the current through the resistor is 5I. So by KVL,
Vs = j1000I − j500I + 200 ⋅ 5I and hence
I =Vs
1000 + j500= 107.33∠ − 26.57o A
The power absorbed by the resistor is, Pav = 200 ⋅ 5I 5I = 57.6 W.
(b) PavVs = Vs I cos(26.57o) = 11.52 W and Pav4 I = 4I 200 ⋅5I cos(0) = 46.08 W.
SOLUTION 11.13. First find the input impedance, Zeq = 2 − j8 + 6 + j2 = 8 − j6 Ω. Then calculate the
current IL =100
8 − j6= 8 + j6 A. The complex power is then computed:
»ZL = 6 + j*2;»IL = 8 + j*6;»VL = ZL*ILVL = 3.6000e+01 + 5.2000e+01i»SL = VL * conj(IL)SL = 6.0000e+02 + 2.0000e+02i
Complex Power Prbs 2/5/00 P11-6 @ DeCarlo & P. M. Lin
»abs(SL)ans = 6.3246e+02
Thus the apparent power is 790 VA, the average power 600 W, the reactive power 200 var, and theapparent power is 632.46 VA.
SOLUTION 11.14. (a) Using MATLAB
»Vseff = 100*exp(j*pi/6)Vseff = 8.6603e+01 + 5.0000e+01i
»ZL = 350 +j*1*300;»Zin = 50 + ZLZin = 4.0000e+02 + 3.0000e+02i
»ILeff = Vseff/ZinILeff = 1.9856e-01 - 2.3923e-02i
»ILpk = sqrt(2)*abs(ILeff)ILpk = 2.8284e-01
»ILang = angle(ILeff)*180/piILang = -6.8699e+00
»SL = ZL*ILeff*conj(ILeff)SL = 1.4000e+01 + 1.2000e+01i
Therefore, iL ( t) = 2 0.2( )cos(300 t − 6.87o) = 0.2828cos(300t − 6.87o) A. SL = 14 + j12 VA, and theaverage power is 14 W.
SOLUTION 11.15. First use voltage division and observe that VL =12
Vs = 60∠60° V. Now the
complex power is SL = VLVL
*
(6 + j8)* = 360∠53.13° = 216 + j288 VA. Thus the average power is 216
W.
*SOLUTION 11.16. (a) To find V2 we write a node equation. First we note that Yi = 1/Zi is thecorresponding admittance. Hence
Y1 V2 − Va( ) +Y2V2 + Y3 V2 − Vb( ) = 0Hence
V2 =Y1Va + Y3VbY1 +Y2 +Y3
=100 + j50 =111.8∠26.57o V
obtained using MATLAB as follows
»Z1 = 0.1+j*0.1; Z2 = 0.4+j*2.2;»Z3 = 0.2 + j*0.2; Va = 104 + j*50; Vb = 106 + j*48;»Y1 = 1/Z1
Complex Power Prbs 2/5/00 P11-7 @ DeCarlo & P. M. Lin
Y1 = 5.0000e+00 - 5.0000e+00i
»Y2 = 1/Z2Y2 = 8.0000e-02 - 4.4000e-01i
»Y3 = 1/Z3Y3 = 2.5000e+00 - 2.5000e+00i
»V2 = (Y1*Va + Y3*Vb)/(Y1 + Y2 + Y3)V2 = 1.0000e+02 + 5.0000e+01i
»magV2 = abs(V2)magV2 = 1.1180e+02»angleV2 = angle(V2)*180/piangleV2 = 2.6565e+01
(b) Again working strictly in MATLAB we have the following complex powers of the loads and the twosources:
»Sz1 = (V2 - Va)*conj((V2-Va)*Y1)Sz1 = 8.0000e+01 + 8.0000e+01i
»Sz2 = V2*conj(V2*Y2)Sz2 = 1.0000e+03 + 5.5000e+03i
»Sz3 = (V2 - Vb)*conj((V2-Vb)*Y3)Sz3 = 1.0000e+02 + 1.0000e+02i
»Sva = Va*conj((Va - V2)*Y1)Sva = 1.0800e+03 + 3.0800e+03i
»Svb = Vb*conj((Vb - V2)*Y3)Svb = 1.0000e+02 + 2.6000e+03i
(c) To verify conservation of power observe that:
»TotSrsPwr = Sva + SvbTotSrsPwr = 1.1800e+03 + 5.6800e+03i
»TotLdPwer = Sz1 + Sz2 + Sz3TotLdPwer = 1.1800e+03 + 5.6800e+03i
which provides the desired verification.
SOLUTION 11.17. Use MATLAB and refer to the following figure:
Complex Power Prbs 2/5/00 P11-8 @ DeCarlo & P. M. Lin
Z1
Z2 Z3
Z4
%(a)%Bundle the impedances as per the following figure and%obtain the following.Z1=2+2*j;Y1=1/Z1;Y2=2+0.5*j;Z2=1/Y2;ZL3=4*j;YL3=1/ZL3;Y3=4+0.25*j;Z3=1/Y3;Z4=4+4*j;Y4=1/Z4;V1=10+2*j;V2=12+2*j;
%Write out KCL for node 1 and 2%(Va-V1)*Y1=V1*Y2+(V1-V2)*YL3%(Vb-V2)*Y4=V2*Y3+(V2-V1)*YL3Va= (V1*Y2+(V1-V2)*YL3)/Y1+V1Vb= (V2*Y3+(V2-V1)*YL3)/Y4+V2Va = 29.0000 +59.0000iVb = 1.6000e+02+ 2.3400e+02i
(b)Sr3=( (Va-V1)*Y1*2)*conj((Va-V1)*Y1)Sl1= ((Va-V1)*Y1*(2*j))*conj((Va-V1)*Y1)Sc1= V1*conj(V1*0.5*j)Sr1= V1*conj(V1/0.5)Sl3= (V1-V2)*conj((V1-V2)*YL3)Sc2= V2*conj(V2*0.25*j)Sr2= V2*conj(V2/0.25)Sl2= (V2-Vb)*Y4*4*j*conj((V2-Vb)*Y4)Sr4= (V2-Vb)*Y4*4*conj((V2-Vb)*Y4)SVA = Va * conj((Va-V1)*Y1)SVB = Vb * conj((Vb-V2)*Y4)
Sr3 = 9.0250e+02
Sl1 = 0 + 9.0250e+02i
Sc1 = 0 - 5.2000e+01i
Complex Power Prbs 2/5/00 P11-9 @ DeCarlo & P. M. Lin
Sr1 = 208
Sl3 = 0 + 1.0000e+00i
Sc2 = 0 - 3.7000e+01i
Sr2 = 592
Sl2 = 0 + 9.4660e+03i
Sr4 = 9466
SVA = 1.1115e+03 + 8.4550e+02i
SVB = 1.0057e+04 + 9.4350e+03i
(c) Take the real part of each of the complex power found in (b). The only components with non-zeroaverage power will be the resistors which have 208 W, 592 W, 902.5 W, and 9466 W average powerrespectively.
%(d)Total_passive=Sr1+Sr2+Sr3+Sr4+Sl1+Sl2+Sl3+Sc1+Sc2Total_active= Va*conj((Va-V1)*Y1)+Vb*conj((Vb-V2)*Y4)Total_passive = 1.1168e+04+ 1.0280e+04i
Total_active = 1.1168e+04+ 1.0280e+04i
which verifies the conservation of power.
SOLUTION 11.18. (a) From conservation of energy, the complex power is the sum of the complexpower absorbed by every circuit elements. Thus Ss =1240 + j145 VA, and the apparent power is 1248.4VA. The average power is 1240 W.
(b) From Ss = VsIs* , Is =
SsVs
= 5.428 A.
SOLUTION 11.19. (a) The complex power delivered by the source is the sum of the complex powerconsumed by the circuit elements. Thus Ss = 44 + j28 kVA.
(b) Is =Ss
Vs= 22.675 A
(c) The total power delivered to the three groups of impedance following V1 is S1 = 41.5 + j22. From
the current obtained in (b) V1 =S1 + S2 + S4
Is= 2071.5 V.
(d) From V1, and the total power delivered to Z4 and Z2, I2 =S4 + S2
V1= 12.385 A. Finally
V2 =S2
I2= 1805.5 V.
Complex Power Prbs 2/5/00 P11-10 @ DeCarlo & P. M. Lin
SOLUTION 11.20. (a) Is =44 + j28
2.3
*
= 22.675∠− (32.47 − 0) = 22.675∠− 32.47° A.
(b) V1 =S1 + S2 + S4
Is*
= 2071.5∠ − 4.54° V
(c) Similarly as before I2 =S1 + S2
V1
*
=12.385∠ − 37.61° A, and V2 = Z2I2 =1805.5∠− 11.05° V.
SOLUTION 11.21. From equation 11.30, we find Q = P1
pf 2 −1 = 455.61 var; thus
S = 1000 + j455.61 =1099∠24.5° VA.
SOLUTION 11.22. (a) S1 =1000 W, and S2 = 800 + j600 VA. Thus the total power delivered by the
source is Stot = 1800 + j600 =1897.37∠18.43° VA, and Is =Stot
*
Vs* = 15.81∠− 18.43° = 15 − j5 A rms.
(b) V1 =S1
Is* = 63.25∠ −18.43° V.
(c) V2 =S2
Is* = 63.25∠18.43° V.
SOLUTION 11.23. Using 11.25, find SL =Pavepf
=30000.75
= 4000 VA, and the load current
IL =SL
VL
=4000120
= 33.33 A. The power absorbed by the transmission line is then from
Pline = R IL2 = 0.2 × 33.33( )2 = 222.2 W.
SOLUTION 11.24. The capacitor must absorb a reactive power of Qnew − Qold = −17.9 kvar. Thus
jQC = − j17.9 = − j CVs2, and C =
−QC
Vs2 = 0.897 mF.
SOLUTION 11.25. From equation 11.30, Qnew = 86.61
(0.94)2 −1 = 31.43 VA. Thus the reactive
power absorbed by the capacitor is –18.57 var. Hence C =−QC
Vs2 = 3.714 µF.
Complex Power Prbs 2/5/00 P11-11 @ DeCarlo & P. M. Lin
SOLUTION 11.26. Device 1 has a complex power of S1 = P1 + jQ1 = 360 + j 0 VA. Recall equation
11.29, pf =Pave
S, and equation 11.30 Qnew = Pave
1
(pf )2 −1
where with a lagging pf, Q > 0, and with a leading pf for Q < 0. Using equation 11.30, we have for
device 2:
Q2 =14401
(0.8)2 −1 =1080 var
S2 = 1440 + j 1080 VA.S1,2 = S1 + S2 = 1800 + j1080 VA
As an aside we compute the magnitude of the current without the capacitor attached.
Is =S1,2
120= 17.493 A
The capacitor is used to achieve a lower source current with the same average power. The first step is tofind the desired QC. Here
Stot = S1 + S2 + SC =1800 + j1080 + jQCHence
Stot120
=1800 + j1080 + jQC
120= 15
In MATLAB we have:
»QC = sqrt((15*120)^2 - 1800^2) - 1080QC = -1080
From the formula on page 451 of the text,
QC = − C Vsource2
Hence C = 0.2 mF. Finally
pf =Pave
S1 + S2 + SC=
18001800
= 1
SOLUTION 11.27. (a) From equation 11.30, Qold = 71
(0.65)2 −1 = 8.1839 kvar. Therefore, the
power absorbed without the capacitor bank is Sold = 7 + j8.1839 kVA. When the bank is added we
want Qnew = 71
(0.8)2 −1 = 5.25 kvar, and henceSnew = 7 + j5.25 kVA. Thus the reactance that must
be absorbed by the bank is –2.934 kvar, and Ceq =−QC
Vs2 = 0.13511 mF.
(b) As was just determined 5.25 kVA.
Complex Power Prbs 2/5/00 P11-12 @ DeCarlo & P. M. Lin
(c) Sold =7
0.65=10.77 kVA. Snew =
70.8
= 8.75 kVA. The kVA saving is: 10.77 – 8.75 = 2.02
kVA.»Savings = 20*2.02*12Savings = 4.8480e+02
i.e., $484.80.
SOLUTION 11.28. (a) The apparent power is simply 94kW/0.78=120.51 kVA.
(b) Sm =120.51∠38.74o = 94 + j 75.41 kVA.(c) 75.41 kvar.
(d) Ieff =Sm
*
Veff* = 523.96∠− 38.74 A.
(e) By KVL, Vs = RlineIeff + Veff = 516.08 − j229.52 V.
(f) Ss = VsIeff* = 295.94∠14.76° kVA.
(g) The efficiency is =120.51cos(38.74)295.94cos(14.76)
⋅100 = 32.96% . Note that the line resistance of 0.7 Ω is much
to large for practical usage. This value is chosen for pedagogical reasons.
(h) With a power factor of 0.94, Smnew = 94 + j 34.12 kVA. The average power of the motor must be kept
the same. The reactance that must be provided by the capacitor is, Qnew − Qold = QC = −41.29 kvar. The
proper capacitor current will be IC =j42.29k
Veff* , and ZC =
1j C
=VeffIC
=Veff
2
j 42.29 ×103 . Solving for
C =42.29 ×103
Veff2 = 2.12 mF.
(i) Ieffnew =
Smnew
Veff
*
= 434.79∠ −19.95° A
(j) By KVL, Vs = RlineIeffnew + Veff = 516.1− j103.84 = 526.43∠− 11.38o V.
(k) Ssnew = Vs
new Ieffnew( )*
= 228.89∠8.57° = 226.33 + j 34.12 kVA, and the efficiency is 41.5%.
SOLUTION 11.29. (a) The Thevenin equivalent seen at the output is, Zth = 5Ω− j /(0.1 ). For
maximum power transfer, ZL = Zth* = 5 + j . Note that VOC = 50 V rms, and that Ieff =
VOCZth + ZL
= 5 A
rms. Thus SL = ZLIeffIeff* = 125 + j25 VA, and the average power is 125 W.
SOLUTION 11.30. First find the following Thevenin equivalent,
Zth = 2 + j 4 Ω, Voc =10 2
3 V.
By the maximum transfer property, RL = 2Ω , and C = 1/(4 ) = 0.25 mF.
Complex Power Prbs 2/5/00 P11-13 @ DeCarlo & P. M. Lin
SOLUTION 11.31. (a) The thevenin equivalent left of the load is by KCL,Itest + Is = VR( j0.001+ 0.001) and Vtest = 3VR + j500Itest V. Substituting for VR,Vtest = Itest 1500 − j1000[ ] + Is 1500 − j1500[ ] V, and Zth = 1500 − j1000 Ω withVoc = Is × 2121.3∠ − 45° V. The value of the load for maximum power transfer is then,ZL = 1500 + j1000 Ω.
(b) The complex power absorbed by the load is, SL = ZLVoc
Zth + ZL
2
= 750 + j500 VA, and the average
power 750 W.
SOLUTION 11.32. Consider
V1 V2
(a) Using nodal analysis get the following equations:
V1 − V2 = 4(V1 /8)
Is + Itest = V1(1/8 + j0.25) + V2(1/16 + j0.5)
Vtest = V2 + 8Itest
Using MATLAB we get the following expression,
Vtest = (8.2847- j0.9110)Itest + (0.2847- j0.9110)Is
From this expression Voc = 28.47 − j91.1 V, and Zth = 8.2847- j0.9110 Ω. The phasor equivalent ofthis circuit is a Vocsource in series with a 8.28 Ω resistor and a 7.32 mF capacitor.(b) ZL = 8.2847+ j0.9110 , which is a 8.28 Ohm resistor in series with a 6.07mH inductor. Sameconfiguration as before.
(c) SL = ZLVoc
Zth + ZL
2
= 275.2 + j30.26 VA and the average power is 275.2 W.
SOLUTION 11.33. (a) The Norton equivalent may be found by inspection as Zth = 10 + j20 Ω andIsc =10 A. Thus for maximum power transfer, ZL = 20 − j20 Ω. This is a 10 Ω resistor and a 0.005 F
capacitor in series. The maximum power is SL = ZLIscZth
Zth + ZL
2
=1250 − j2500 VA. The maximum
average power is 1250 W.(b) If R is set to 20 Ω, the closest that can be achieved to maximum power transfer is ZL = 20 − j20 , orC equal to 0.005 F. With ZL as above, by current division
Complex Power Prbs 2/5/00 P11-14 @ DeCarlo & P. M. Lin
Iload =10 ×10 + j20
(10 + j20) + (20 − j20)= 3.333 − j6.6667 = 7.45∠63.434o A
The maximum average power then 20 × Iload2 = 1111 W.
(c) Using 11.38, RL = 31.62 Ω. Pav = RLIscZth
Zth + ZL
2
= 600.63 W.
SOLUTION 11.34. (a) From Thevenin Zth = 19.2 − j14.4 Ω, and from 11.38, set RL = 24 Ω.
Voc =Vs(− j 40)30 − j40
= 80∠− 36.87° . The maximum power is Pav =Voc
Zth + RL
2
RL = 74.07 W.
(b) The voltage is V = VocRL
Zth + RL, from this relationship, one sees that as the load resistance increase
to infinity the output voltage goes to Voc, which is the maximum output voltage.
SOLUTION 11.35. Correction: the inductor symbol in the load should be a resistor. Since the sourceresistance is variable, example 6.21 serves as a reference suggesting that R = 0 is the answer. To see thisconsider that
P = 2Iload2 =
2 ×100
(R + 2)2 + L −1C
2 =2 ×100
(R + 2)2 + (2 − 2)2
Hence, decreasing R produces increasing power and the maximum power is transferred when R = 0 withPmax = 50 W assuming that the source voltage is given in rms V.
SOLUTION 11.36. As per problem 35,
P = 10Iload2 =
10 × 50 2( )2
(R +10)2 + L −1C
2 =12.5 ×103
(R +10)2 + L −1C
2
Here, again R = 0 with C chosen to eliminate the reactive term maximizes power transfer. Hence
C =12L
= 0.01 F
with Pmax =125 W.
SOLUTION 11.37. (a) By the maximum power transfer theorem, P1 is maximized when ZL is chosenas the conjugate of Zsource, i.e.
ZL = 10 + j1000 Ω
(b) To find the appropriate values of L and C observe that
Complex Power Prbs 2/5/00 P11-15 @ DeCarlo & P. M. Lin
ZL( j ) = j L +1
10−4 + j C= j107 L +
10−4 − j C
10−8 +1014C2
=10−4
10−8 +1014 C2 + j 107 L −C
10−8 +1014C2
= 10 + j1000
Equating real parts leads to:
»%10^-4 = 10^-7 + 10^15*C^2»»C = sqrt(1e-4 - 1e-7)/sqrt(1e15)C = 3.1607e-10
Thus C = 0.31607 nF.
Equating imaginary parts using the above value of C leads to:
»w = 1e7;»L = (1e3 + w*C/(1e-8 + 1e14*C^2))/1e7L = 1.3161e-04
Thus L = 0.1316 mH.
(c) In part (b), L and C are chosen to maximize P1, the power delivered to ZL. Since L and C consumeno average power, this maximum power is transferred to the 10 kΩ fixed resistor with thecomputed values of L and C. Thus ZL is the same as in part (b) or ZL = 10 + j1000 Ω.
Since we know ZL,
Pmax =(0.1)2
4 ×10= 0.25 mW
This is the average power consumed by the 10 kΩ resistor. Therefore
V22 = 0.25 ×10−3 ×104 = 2.5
It follows that V2 =1.5811 V. Power to the 10 kΩ fixed resistor is
P10kΩ =V2
2
104
Thus if P10kΩ is maximized, then V2 is maximized.
SOLUTION 11.38. (a) From equation 11.4, if we substitute 2T
⇒ , then the resulting instantaneous
power will be p( t) =VmIm
2cos( v − i ) +
VmIm2
cos(4T
t + v + i ) W where it is clearly seen that the
fundamental period will now be halved. Note that by the same argument the fundamental frequency ofthe instantaneous power is double that of the voltage and current.
Complex Power Prbs 2/5/00 P11-16 @ DeCarlo & P. M. Lin
(b) As a sinusoid, the fundamental period is 2π/10, any integer multiple of this period will also beperiodic.(c) This is the same as (b) with an offset of 1 V added.
SOLUTION 11.39. First, Feff =1T
f 2(t)dtt0
to +T∫ . Without going into detailed calculation, f 2(t) will
give a summation of two types of products, a product of each element with themselves, and products ofeach element with the other element. In the later case, we know that two cosines multiplied with oneanother and integrated over one period will yield zero if their angular frequency are different. As for theformer case the integral will yields the result we are looking for. For example look at the first two terms,
Feff =1T
f 2(t)dtt0
to +T∫ =
1T
Fo2 + 2F1
2 cos2( 1t + 1) + ...( )dtt0
t0 +T∫
=1T
Fo2 + F1
2 + F12 cos(2 1t + 2 1) + ...( ) dt
t0
t0 +T∫
=1
TFo
2t + F12t + F1
2 sin(2 1t + 2 1)+ ...[ ]0T
= Fo2 + F1
2 ...
SOLUTION 11.40. (a) We are given that vC (t) = Vm sin( t) V. Hence
iC ( t) = CdvCdt
= CVm cos( t) A
It follows that
p( t) = vC ( t)iC (t) = CVm2 sin( t)cos( t) = 0.5 CVm
2 sin(2 t) Watts
Clearly, p(t) has a peak value of 0.5 CVm2 and the integral of the sign over one period is zero implying
that the average value of p(t) is zero.
(b) Here
WC (t) = 0.5CvC2 (t) = 0.5CVm
2 sin2( t) = 0.25CVm2 1− cos(2 t)( ) J
Here the peak value occurs when cos(2ωt) = –1 in which case the peak value is 0.5CVm2 . Further, the
average value of cos(2ωt) over one period, T = π/ω, is zero whereas the average of a constant over the
same period is simply the constant. Hence, WC,ave = 0.25CVm2 J.
(c) From example 11.6,
QC = −IC ,eff VC ,eff = − CVC,eff2 = −0.5 CVm
2 = −2 0.25CVm2( ) = −2 WC,ave .
Therefore, WC,ave = −QC2
.
Complex Power Prbs 2/5/00 P11-17 @ DeCarlo & P. M. Lin
SOLUTION 11.41. (a) We are given that iL ( t) = Im sin( t) A. Hence
vL (t) = LdiLdt
= LIm cos( t) V
It follows that
p( t) = vL ( t)iL (t) = LIm2 sin( t)cos( t) = 0.5 LIm
2 sin(2 t) watts
Clearly, p(t) has a peak value of 0.5 LIm2 and the integral of the sign over one period is zero implying
that the average value of p(t) is zero.
(b) Here
WL (t) = 0.5LiL2(t) = 0.5LIm
2 sin2( t) = 0.25LIm2 1− cos(2 t)( ) J
Here the peak value occurs when cos(2ωt) = –1 in which case the peak value is 0.5LIm2 . Further, the
average value of cos(2ωt) over one period, T = π/ω, is zero whereas the average of a constant over the
same period is simply the constant. Hence, WL,ave = 0.25LIm2 J.
(c) As an extension to example 11.5,
QL = IL,eff VL,eff = LIL,eff2 = 0.5 LIm
2 = 2 0.25CIm2( ) = 2 WL,ave .
Therefore, WL,ave =QL2
.
SOLUTION 11.42. (a) The complex power absorbed by the load is, SL = VI* = ZII* = Z I2 . Now notethat the average power is the real part of the complex power. Also note that a complex number
multiplied by its complex conjugate will yield a real value. Therefore the real part of ZI 2is just the
real part of Z, R, multiplied by I 2 , Pav = R I2 . With the same reasoning, the reactance is the
imaginary part of Z, X, multiplied by I 2 , Q = X I 2 .
(b) The complex power absorbed by the load is SL = VI* = V(YV)* = Y V 2 . The same reasoning as in
(a) holds thus the real part of the admittance times V 2 , yields Pav = GV 2 . Using the imaginary
part, Q = B V 2 .
SOLUTION 11.43. (a) The equivalent resistance seen by the source is Req = 6 – j9 Ω. So the currentdelivered by the source is:
Is =Vs
6 − j9=
− j1106 − j9
= 10.17∠− 33.69° A
or iL ( t) = 10.17 2 cos(120 t − 33.69o) A. Similarly
VC = − j15 × Is =−15 ×110
6 − j9=152.54∠− 123.69° V
Complex Power Prbs 2/5/00 P11-18 @ DeCarlo & P. M. Lin
or vC (t) = 152.54 2 cos(120 t −123.69o) V. Further, L =6
=15.9 mH and C =1
15= 176.8 µF.
The instantaneous energy stored in the inductor is
WL (t) = 0.5LiL2(t) = 1.646cos2(120 t − 33.69o) J
and the instantaneous energy stored in the capacitor is
WC (t) = 0.5CvC2 (t) = 4.115cos2(120 t −123.69o) J
The source voltage is zero when t = 0. Therefore
WL (0) = 1.646cos2(−33.69o) = 1.1395 Jand
WC (0) = 4.115cos 2(−123.69o) =1.266 J
(b) and (c) Observe that iL(t) and vC(t) are 90o out of phase. When one is zero, the other has a peakvalue. Therefore whenWC = 0 implies vC(t0) = 0 for appropriate t0; hence WL (t0) = 1.646 J.Similarly, when WL = 0, say at t0, then WC (t0) = 4.115 J.
SOLUTION 11.44. In order to solve this problem, we want to express the power in terms or R's and L'sin both circuits. First, looking at the circuit with just the coil and the 110 V source: I = V / Zcoil ,Zcoil = R + j L , and
Pcoil = I 2R =1102
R2 + 2L2 R = 300 watts (*)
Next, looking at the circuit when a resistance is added in series with the coil, I = V /(8 + Zcoil ),Zcoil = R + j L ,
Pcoil = I 2R =2202
(8 + R)2 + 2L2 R = 300 watts (**)
To find R, solve equation (*) for R2 + 2L2 and substitute into equation (**) to obtain
»R = 300*64/(220^2 - 300*16 -110^2)R = 6.0952e-01
Substituting R into equation (*) yields L = 13.05 mH.
SOLUTION 11.45. The average power consumed by the 2.7 Ω resistor is 250 watts. This allows us tocompute the magnitude of | Icoil |. We know that | Vcoil | is 150 Vrms. Thus we can computethe magnitude of the coil impedance and hence L as follows:
»magIcoil = sqrt(250/2.7)magIcoil = 9.6225e+00
»magZcoil = 150/magIcoilmagZcoil = 1.5588e+01
Complex Power Prbs 2/5/00 P11-19 @ DeCarlo & P. M. Lin
»% magZcoil^2 = 2.7^2 + (w*L)^2»»w = 2*pi*60;»L = sqrt( magZcoil^2 - 2.7^2)/wL = 4.0725e-02
»% Magnitude of impedance seen by 220 V source is:»»magZin = 220/magIcoilmagZin = 2.2863e+01
»% magZin^2 = (R + 2.7)^2 + (w*L)^2»»Realpart = sqrt(magZin^2 - (w*L)^2)Realpart = 1.6941e+01
»R = Realpart - 2.7R = 1.4241e+01
9/26/01 P12-1 @ DeCarlo & P. M. Lin
CHAPTER 12. PROBLEM SOLUTIONSSOLUTION P12.1. By conservation of energy, the instantaneous power consumed by each load whensummed together is equivalent to the total power consumed by the three phase load. Thus writing out
ptot (t) = pAB (t) + pBC ( t) + pCA ( t) = v AB( t)iAB( t) + vBC (t)iBC ( t) + vCA ( t)iCA (t)
= VL 2 cos( t + v ) ⋅VL 2
Zcos( t + i ) +VL 2 cos( t + v −120o) ⋅
VL 2Z
cos( t + i −120o)
+ VL 2 cos( t + v +120o) ⋅VL 2
Zcos( t + i +120o)
=VL
2
Zcos( v − i ) + cos(2 t + v + i )( ) +
VL2
Zcos( v − i ) + cos(2 t + v + i +120o)( )
+VL
2
Zcos( v − i ) + cos(2 t + v + i −120o)( )
=3VL
2
Zcos( v − i ) =
3VL2
Z× pf
SOLUTION P12.2. To justify the point of this problem we equate the following two equations:(i) For the 3 phase system:
Ploss' = 3× IL
2R ' = 3 ×
39
×PL
2
VL2 R ' =
PL2
VL2 R '
(ii) For the single phase system:
Ploss = 2 × IL2R( ) =
PL2
VL2 2R
It follows that R' = 2R. Since both systems have the same distance of transmission and the resistance ofa wire is inversely proportional to the cross sectional area, the condition R' = 2R implies that the crosssection A' of each wire in the three-phase system need only be half of the area A of the wire in the singlephase system. But there are two wires in the single phase system and three wires in the three-phasesystem. Therefore the ratio of the materials used is:
material in 3− phase systemmaterial in 1− phase system
=3A '
2A=
32
×12
= 75%
SOLUTION P12.3. For row 1 of table 12.1, the impedance in (a) seen between each pair of terminal is
Zik = Z∆ || 2Z∆ = 2Z∆3
. In (b) the impedance seen between any two terminals is
Z jk =Z∆3
+Z∆3
= 2Z∆3
.
9/26/01 P12-2 @ DeCarlo & P. M. Lin
In row 2, the impedance between any two terminals is Z jk = ZY + ZY = 2ZY for (c), andZik = 3ZY || 6ZY = 2ZY for (d).
SOLUTION P12.4. Consider the ∆-Y relationship of the figures below (row 3 of table 12.1):
Let us consider the terminal pair (1,2):
(i) For the Y-connected case, Zth = 2 ×Z∆3
and
Voc =Vp
3∠ − 30o( ) −
Vp
3∠ −150o( ) =
Vp
3∠ 1∠− 30o −1∠ −150o( ) = Vp∠
(ii) For the D-connected case,
Zth = Z∆ / / 2Z∆( ) =2(Z∆)2
3Z∆=
23
Z∆
Now note that there is no load connected to the ∆-configuration. Applying KVL to the indicated loopimplies that:
0 = 3Z∆Iloop +Vp∠ +Vp∠ −120o + Vp∠ +120o = 3Z∆Iloop
Hence Iloop = 0 . Finally, Voc = V12 = Z∆Iloop +Vp∠ = Vp∠ . Therefore, looking into terminals 1-2,
both the ∆-configuration and the Y-configuration have the same Thevenin equivalent. For terminal pairs(1-3) and (2-3), the proof is virtually the same. Hence this establishes the equivalence in row 3 of table12.1.
To establish the equivalence in row 4 of table 12.1, we do all the same computations with theslightly different labeling to obtain the same result, i.e., the circuits are equivalent.
SOLUTION 12.5. For (a) first note the following relationship VN =V1Z1
+V2Z2
Z1 || Z2 || Z3( ) , which is
obtained by KCL at the center node, N, with node 3 as the reference node.. Write out KCL at terminal 1,
I1 =V1Z1
−V1
Z1Z1+
V2Z2Z1
Z1 || Z2 || Z3( ) = V1
Z3 + Z2Z2Z3 + Z1Z3 + Z1Z2
− V2
Z3
Z1Z2 + Z2Z3 + Z3Z1
. Do the
9/26/01 P12-3 @ DeCarlo & P. M. Lin
same for terminal 1 in (b), I1 = V11
Z31+
1Z12
−
V2Z12
. Now substitute the formulas in the problem
statement into the later equation, I1 = V1Z2 + Z3
Z1Z2 + Z2Z3 + Z3Z1
− V2
Z3
Z1Z2 + Z2Z3 + Z3Z1
, which is
the same as the equation for (a).Using the same method, the same result is obtained for node 2. So the fact that the substitution
of the equivalence in (b) yields the same equation as in (a) proves that the equivalences are accurate.
SOLUTION 12.6. By Ohm’s law IA =Vp
Z=
200∠0°10 + j5
= 17.89∠− 26.57° A. By the same relationship
IB = 17.89∠ −146.57° A, and IC = 17.89∠93.43° A. The neutral line current is the sum of the otherthree and is zero. The power of each phase is the same, as they have the same load, and current
magnitude. Using the current at terminal A, the total power is, P = 3R IA2 = 9602 W.
SOLUTION 12.7. For this balanced Y-Y connection,we follow the method of example 12.6. UsingMATLAB:
»Z = 20 + j*10'Z = 2.0000e+01 + 1.0000e+01i
»VAN = 200;»IA = VAN/ZIA = 8.0000e+00 - 4.0000e+00i
»IAmag = abs(IA)IAmag = 8.9443e+00
»IAangle = angle(IA)*180/piIAangle = -2.6565e+01
»% Use phase inference to obtain:»IBmag = IAmagIBmag = 8.9443e+00
»IBangle = IAangle - 120IBangle = -1.4657e+02
»ICmag = IAmagICmag = 8.9443e+00
»ICangle = IAangle + 120ICangle = 9.3435e+01
»Ptotave = 3*real(Z)*IAmag^2Ptotave = 4.8000e+03
9/26/01 P12-4 @ DeCarlo & P. M. Lin
SOLUTION 12.8. Using the same approach as in the example 12.6 we compute
VAN = VsA ×Z
Z + Zg= 120 ×
4 + j34 + j3 + 0.1 + j0.2
= 116.94 V
Thus the drop is: 120 −116.94
120×100 = 2.55 % .
SOLUTION 12.9. A 5.5 % drop corresponds to VAN = 113.4 V. This means
Z
Z + Z1 + Zg=
VAN120
=0.945 ×120
120= 0.945
From this equation,
Z + Z1 + Zg =Z
0.945= 5.291.
So 4.05 + 0.1( )2 + 3.15 + 0.2( )2 = 27.99 . Solving the quadratic equation results in = 0.7893.
SOLUTION 12.10. Solving the single phase equivalent circuit for phase A,
IA =Vp −Vp∠120°
10 + j5= 58.87∠− 56.57°
Using phase inference,
IB = IA × ∠− 120° = 58.87∠ −176.57°IC = IA × ∠120° = 58.87∠64.43°
Alternately, using much more work,
IB =Vp∠ −120°− Vp
10 + j5= 58.87∠ −176.57°
IC =Vp∠120° −Vp∠ −120°
10 + j5= 58.87∠64.43°
The total power, using the previously derived relationship, is
P = 3VL IL pf = 3 × 380 × 58.87 × pf = 34.655 kWwhere
pf = cos tan−1 510
= 0.8944
9/26/01 P12-5 @ DeCarlo & P. M. Lin
SOLUTION 12.11. (a) By KVL and Ohm's law,
IAB =Vp
30 + j15= 6.57∠− 26.57°
By phase inference,
IBC = IAB × ∠ −120° = 6.57∠ −146.57°ICA = IAB × ∠120° = 6.57∠93.43°
Alternately,
IBC =Vp∠ −120°
30 + j15= 6.57∠ −146.57°
ICA =Vp∠120°30 + j15
= 6.57∠93.43°
To compute the line currents we have in Amps:
IA = IAB − ICA =11.36∠ − 56.57°IB = IBC − IAB = 11.36∠− 176.57°IC = ICA − IBC = 11.36∠63.43°
(b) The average power is P = 3 × 30 IAB2 = 3.88 kW, and the total reactive power is
Q = 3 ×15 IAB2 = 1.94 kvar.
SOLUTION 12.12. From the efficiency we know that 25 ⋅ 746Pdeliverd
= 0.85 , so the total power being
delivered to the load is 21941 W. Using 12.4, IL =21941
3 ⋅VL ⋅ pf= 63.31 A.
SOLUTION 12.13. (a) Using the power efficiency relationship, Ptot =300 ⋅ 746
0.935= 239.36 kW.
(b) Using 12.4, IL =239.36 ×103
3 ⋅VL ⋅ pf= 68.28 A.
SOLUTION 12.14. The magnitude of the power delivered to each delta connected load is 79.787 kW.We can now perform the analysis on a single phase. Using equation 11.30, we obtain
9/26/01 P12-6 @ DeCarlo & P. M. Lin
Qold = 797871
0.88
2
−1 = 43064 vars
and
Qnew = 797871
0.95
2
−1 = 26225 vars
Hence
Sold = 79787 + j 43064
Snew = 79787 + j26225
So the reactive power to be supplied by the capacitor is QC = 16839 = 43064 – 26225. Thecapacitance
C =QC
Vl2 = 8.44 uF.
SOLUTION 12.15. Replacing the delta configuration of the source by its Y-equivalent, the followingvoltages (in V) are obtained,
VAN =Vp
3∠ − 30°, VBN =
Vp
3∠− 150°, VCN =
Vp
3∠90°
From the single phase equivalent,
IA =VAN
Z=19.63∠− 56.57°
From phase inferenceIB = IA × ∠− 120° = 19.63∠− 176.57°IC = IA × ∠120° =19.63∠63.43°
The power factor of the load is
pf = cos tan−1 510
= 0.8944
Hence, the total power in the balanced system is
P = 3VL IL ⋅ pf = 3 × 380 ×19.63 × 0.8944 =11.556 kW.
SOLUTION 12.16. First replace the delta connected load by it Y-equivalent. Then analyzing a singlephase and using phase inference, we obtain in amps,
9/26/01 P12-7 @ DeCarlo & P. M. Lin
IA =Vp
(Z / 3)= 34.08∠− 26.56°
IB = IA × ∠− 120° = 34.08∠− 146.56°IC = IA × ∠120° = 34.08∠93.43°
The power factor of the load is
pf = cos tan−1 510
= 0.8944
The total power is P = 3VL IL ⋅ pf = 3 ⋅127 ⋅ 34.08 ⋅ pf = 11.61 kW. The voltage across each load is
3 ⋅Vp = VL = 219.97 V.
SOLUTION 12.17. (a) Referring to figure 12.2, the following relationships may be pointed out:
VAB = VAN + VNB = VAN − VBN
VBC = VBN + VNC = VBN − VCN
VCA = VCN + VNA = VCN − VAN
(b) Perform a delta to Y-transformation and use Ohm’s law and phase inference to obtain in amps:
IA =220∠− 30°
3 ⋅ Z=11.36∠ − 56.57°
IB = 11.36∠ −176.57°IC = 11.36∠63.43°
(c) First,
pf = cos tan−1 510
= 0.8944
The average power, Pav = 3VL IL pf = 3 × 220 × IA × 0.8944 = 3.87 kW.
The reactive power may be found as follows, Q = 3 × 5IL2 =1.936 kvar.
SOLUTION 12.18. Part (a) Apply ∆−Y transformation to the source.Apply the transformation of row 3, Table 12.1 to the source. The result is the circuit shown in figure12.19 of example 12.8. Therefore we may use many of the calculated values in example 12.9. Inparticular, the line currents are
IA = 19.49/-68.9o A, IB = 19.49/-188.9o A, and IC= 19.49/51.09o A
Part (b). Compute line-to-neutral voltage. From figure 12.21 and by Ohm's law
9/26/01 P12-8 @ DeCarlo & P. M. Lin
VAN = IA ZY = 19.49/-68.9o × 5∠36.87o = 97.45∠−32.03o VBy inference on phase angles
VBN = 97.45∠−152.03o V and VCN = 97.45∠87.97o V
Part (c). Compute line voltages VAB, VBC, and VCA. From the voltage phasor diagram of figure 12.2.
VAB = 3 VANe j30o= 168.8e− j2.03o
V
By inference on phase angles,
VBC = VABe− j120o=168.8e− j122.03o
V
VCA = VABe j120o=168.8e− j117.97o
V
Part (d). Compute the total power. From equation 12.4
Ptotal = 3 ×VL × IL × pf = 3 ×168.8 ×19.49 × 0.8 = 4558 W
SOLUTION 12.19. Perform a delta to Y transformation on the source. This yields
V1N =Vp
3∠− 30o =
380
3∠ − 30o V. Perform a delta to Y-transformation on the load. This yields a
new impedance Znew =Z3
= 4 + j 3 Ω. Now analyzing a single phase and finding the others by
inference yields, in amps,
IA =V1N
Z1 + Znew= 42.18∠− 67.97°
IB = 42.18∠− 187.97°IC = 42.18∠52.03°
The average power delivered to the load is Pav = 3 × IA2 × 4 = 21.35 kW.
SOLUTION 12.20. Performing a delta-to-Y-transformation on the source and the load, the new sourcemagnitudes are multiplied by 1 3 , and the loads are divided by 3. Observe that Zg = 0.15 + j0.45 Ω,Z1 = 0.1 + j0.2 Ω, and Z =12 + j9 Ω. Using Ohm’s law on phase one, and then inference on thephase angles for the other currents, yields in amps,
IA =Vp∠− 30°
3Zg
3+ Z1 +
Z
3
= 41.14∠− 68.91°
IB = 41.14∠ −188.91°IC = 41.14∠51.09°
9/26/01 P12-9 @ DeCarlo & P. M. Lin
The total power may be calculated as follows, Pav = 3123
IL
2 = 12 × (41.14)2 = 20.310 kW.
SOLUTION 12.21. Perform delta-to-Y-transformation on the delta load. Because of the property of theneutral to be ground in balance circuit, the two loads in the same phase combine in parallel. Thus the newimpedance seen in one branch is
ZYnew = ZY ||Z∆3
=(5 + j5) × (5 − j 3)(5 + j5) + (5 − j3)
= 4.044 + j0.1923 Ω
Now looking at a single phase and inferring for the others,
IA =Vp∠ − 30°
3 ⋅ ZYnew=
230∠ − 30°3 4.044 + j0.1923( ) = 32.84∠ − 32.73° A
IB = 32.84∠ −152.73° AIC = 32.84∠87.27° A
The complex power is S = 3 ⋅ ZYnew IL2 =13.07 + j0.615 kVA.
SOLUTION 12.22. For this problem we do not need to use any 3-phase knowledge if we are clever.We will provide a clever solution here. VDN = VDB – VNB. Using B as the reference we write a nodeequation at N to obtain:
0 =VNBZ1
+VNAZ1
+VNCZ1
=VNBZ1
+VNB − VAB
Z1+
VNB − VCBZ1
Therefore
3VNB = VAB + VCB = 115 −115∠− 120o =172.5 + j99.59
From voltage division
VDB =Z3
Z3 + Z2VAB =
2 + j42 + j4 + 4 − j2
115 = 57.5 + j57.5 V
Now using MATLAB we have
»115-115*exp(-j*2*pi/3)ans = 1.7250e+02 + 9.9593e+01i»VNB =ans/3VNB = 5.7500e+01 + 3.3198e+01i»VDB = 115*(2+j*4)/(6+j*2)VDB = 5.7500e+01 + 5.7500e+01i
»VDN = VDB - VNBVDN = 7.1054e-15 + 2.4302e+01i
»magVDN = abs(VDN)magVDN = 2.4302e+01
»angVDN = angle(VDN)*180/piangVDN = 9.0000e+01
9/26/01 P12-10 @ DeCarlo & P. M. Lin
SOLUTION P12.23. First we write the following two loop equations,
VsA − VsA∠− 120 = IA Zg + Z1 + ZA( ) + IC + IA( ) Z + Z1 + Zg( )VsA∠120 − VsA∠ −120 = IA + 2IC( ) Zg + Z1 + Z( )
and solve in MATLAB:(a)%Problem 12.23aZg=0.05+0.15i;Z=4+3i;Z1=0.1+0.2i;ZA=5+2i;
VA= 120;VB= 120*exp(-120*i*pi/180);VC= 120*exp(120*i*pi/180);
%Write mesh equations in matrix form and solve:% i.e. [IA;IC] = A - 1*CA=[ 2*(Z1+Zg)+ZA+Z Z+Z1+Zg; Zg+Z1+Z 2*(Zg+Z1+Z)];Ainv= A^(-1);C=[VA-VB;VC-VB];B=Ainv*C;IA= B(1);IC= B(2);IB= -IA-IC;IAabs(IA)180*angle(IA)/piIBabs(IB)180*angle(IB)/piICabs(IC)180*angle(IC)/pi
%And By Ohm's law,VAN= IA*ZAabs(VAN)180*angle(VAN)/piVBN= IB*Zabs(VBN)180*angle(VBN)/piVCN= IC*Zabs(VCN)180*angle(VCN)/pi
IA = 21.76∠–29.12° A
IB = 23.88∠–155.60° A
IC = 20.64∠ 82.39° A
9/26/01 P12-11 @ DeCarlo & P. M. Lin
VAN = 117.20∠-7.32° V
VBN = 119.4∠–118.73° V
VCN = 103.2∠119.26° V(b)% From phasor diagramVAB = VAN - VBNabs(VAB)180*angle(VAB)/piVBC = VBN - VCNabs(VBC)180*angle(VBC)/piVCA = VCN-VANabs(VCA)180*angle(VCA)/pi
VAB = 195.5 ∠27.34°VBC = 194.9 ∠–92.04°VCA = 196.99 ∠147.8°(c) Calculating the average power delivered to each load using the general formula
Ptot = R I2∑ = 5 × 21.76( )2 + 4 × 23.88( )2 + 4 × 20.64( )2 = 6.353 kW
SOLUTION P12.24. Doing this problem in MATLAB,%Problem 12.24ZDg=0.15+0.45i;ZD=12+9i;Z1=0.1+0.2i;ZDA=15+6i;%Apply the delta to Y transformation to the load as follows,ZA=(ZD*ZDA)/(2*ZD+ZDA);ZB=(ZDA*ZD)/(2*ZD+ZDA);ZC=(ZD*ZD)/(2*ZD+ZDA);
%Apply delta to Y trans. to the source,V1= 180/sqrt(3)*exp(-30*i*pi/180);V2= 180/sqrt(3)*exp(-150*i*pi/180);V3= 180/sqrt(3)*exp(90*i*pi/180);Zg= ZDg/3;
%Write nodal equations in matrix form and solve%i.e. [IA;IC]=A-1*CA=[ 2*(Z1+Zg)+ZA+ZB ZB+Z1+Zg; Zg+Z1+ZB 2*(Zg+Z1)+ZC+ZB];Ainv= A^(-1);C=[V1-V2;V3-V2];B=Ainv*C;IA= B(1);IC= B(2);IB= -IA-IC;IAabs(IA)
9/26/01 P12-12 @ DeCarlo & P. M. Lin
180*angle(IA)/piIBabs(IB)180*angle(IB)/piICabs(IC)180*angle(IC)/pi
IA = 17.4 ∠–63.1° A
IB = 20.1 ∠178.9° A
IC = 19.5 ∠51° A(b)% By Ohm's law,VAN= IA*ZA;VBN= IB*ZB;VCN= IC*ZC;%From phasor diagramVAB=VAN - VBNabs(VAB)180*angle(VAB)/piVBC = VBN-VCNabs(VBC)180*angle(VBC)/piVCA =VCN - VANabs(VCA)180*angle(VCA)/pi
VAB = 170.40 ∠–2.5° V
VBC = 168.58 ∠–122.4° V
VCA = 169.75 ∠118.1° V(c)% by Ohm's lawIAB=VAB/ZDAabs(IAB)180*angle(IAB)/piIBC=VBC/ZDabs(IBC)180*angle(IBC)/piICA=VCA/ZDabs(ICA)180*angle(ICA)/pi
IAB = 10.55 ∠–24.3o A
IBC = 11.24 ∠–159.3° A
ICA = 11.32 ∠81.2° A(d)
Ptot = R I2∑ =15 × 10.55( )2 +12 × 11.24( )2 +12 × 11.32( )2 = 4.721 kW
9/26/01 P12-13 @ DeCarlo & P. M. Lin
SOLUTION P12.25. The average power that the load draws is P = 440 ⋅ 40 ⋅ pf = 8.8 kW. From this weobtain the old and new complex powers:
Sold = 8.8 + j15.242 kVA and Snew = 8.8 + j15.242 + jQC kVA
Here the reactance supplied by the capacitor is QC = − CVl2 = −7299 var. So the new power factor is
pf = cos tan−1 Qnew8800
= 0.742 . From S = Vl Il = 11.855 kVA, we solve for
Il = IA = IB = 26.94 A.
SOLUTION P12.26. Using Ohm’s law and the power factor of the load,
IC = −440 ⋅ C∠90° = 16.59∠ − 90° A, IA = 40∠60° A, and IB = −IC − IA = 26.94∠ −137.9° A.
SOLUTION P12.27. By simply applying Ohm’s law to each load and then KCL at node N, we get thefollowing:
IA =VA50
= 4.4 A, IB =VB50
= 4.4∠ −120° A, IC =VC250
= 0.88∠120° A, and
IN = IA + IB + IC = 3.52∠− 60° A.
Finally,
PA = 50 × 4.4( )2 = 968 W, PB = 50 × 4.4( )2 = 968 W, PC = 250 × 0.88( )2 = 193.6 W.
SOLUTION P12.28. Using MATLAB,
VA = 220VB = 220*exp(-120*pi*i/180);VC = 220*exp(120*pi*i/180);
%Use KVL to solve for the different currentsIA = (VA - VB)/50 + (VA - VC)/50abs(IA)180*angle(IA)/pi
IB = (VB - VC)/100 + (VB - VA)/50abs(IB)180*angle(IB)/pi
IC = (VC - VB)/100 + (VC - VA)/50abs(IC)180*angle(IC)/pi
IA = 13.2 AIB = 10.1 ∠–130.9° A
9/26/01 P12-14 @ DeCarlo & P. M. Lin
IC = 10.1 ∠130.9° A
The total power delivered to each resistor may be calculated using the following relationship P =Vl
2
R,
where Vl2 = 3Vp
2 using Y to Delta source transformation. Thus PAB = 2904 W, PBC =1452 W, and
PCA = 2904 W.
SOLUTION P12.29. Consider the circuit of figure 12.29 with two additional node labeling:
Choose N as the reference node and apply KCL to node M. This yields the following node equation:
0.02 VM − 220( ) + 0.02 VM − 220e− j120o
+ 0.004 VM − 220e j120o
= 0
Equivalently2.2VM = 88 − j152.42 or VM = 40 − j69.282 V
The remaining calculations proceed in MATLAB as follows:»% Compute node voltage VM»X = +220 +220*exp(-j*2*pi/3) + 0.2*220*exp(j*2*pi/3)X = 8.8000e+01 - 1.5242e+02i»VM = X/2.2VM = 4.0000e+01 - 6.9282e+01i
»% Compute line currents»IA = 0.02*(220 - VM)IA = 3.6000e+00 + 1.3856e+00i»magIA = abs(IA)magIA = 3.8575e+00»angIA = angle(IA)*180/piangIA = 2.1052e+01»IB = 0.02*(220*exp(-j*2*pi/3) - VM)IB = -3.0000e+00 - 2.4249e+00i»magIB = abs(IB)magIB = 3.8575e+00»angIB = angle(IB)*180/piangIB = -1.4105e+02»IC = 0.004*(220*exp(j*2*pi/3) - VM)IC = -6.0000e-01 + 1.0392e+00i»magIC = abs(IC)
9/26/01 P12-15 @ DeCarlo & P. M. Lin
magIC = 1.2000e+00»angIC = angle(IC)*180/piangIC = 1.2000e+02
»% Compute total power»Ptot = 50*magIA^2 + 50*magIB^2 + 250*magIC^2Ptot = 1.8480e+03
SOLUTION P12.30. (a) The equivalent resistance of a 40 ft size #14 wire is
Req1 = 40 ⋅2.5751000
= 0.103Ω . The voltage at the light and appliance is then
Vapp = 115 − 2 ⋅ Req110 = 112.94 V, which is equivalent to a 1.8% drop, thus the wire gauze isappropriate.
(b) Repeating the previous calculations, Req2 = 50 ⋅1.6191000
= 0.08095 Ω.
Vapp = 115 − 2 ⋅ Req214 = 112.7334 V, which corresponds to 1.97 %. So the wire gauze is appropriate.
(c) Under normal operating condition the current in the live wire should equal the current in the neutralwire; thus no current should be in the ground wire.
(d) Note the following relationship, VAB = VAN + VNB = VAN − VBN . This means thatVAN =115 V and VBN = 115∠180° V.
SOLUTION P12.31. Assuming a very large resistance for the person who touches the prong. For prongA, the voltage is approximately V = 115 V. For prong N, the voltage is approximately V = 0. For theground prong G, no current flows through the wire, so the voltage should be ground or zero.
SOLUTION P12.32. For an approximate analysis, we use the circuit models below given the followingassumptions:
(i) The resistance from body to ground is very large (possibly due to rubber shoes);(ii) The resistances of all connecting wires are negligible.(iii) The motor winding is represented by R1 in series with R2. Here we can further estimate
that. R1 + R2 = 115/3 = 38.3 Ω, and R1 = R2 = 19.15 Ω.(iv) The trigger switch of the drill has been depressed.
(a) The point P is connected to the metal case as shown below. It is obvious that Vcase = Vp = 0.
9/26/01 P12-16 @ DeCarlo & P. M. Lin
(b) The point M is connected to the metal case as shown below. The current through the hot wire is115/19.15 = 6 A. If the fuse capacity is smaller than 6 A, then it will blow and Vcase = VM = 0. On the
other hand , if the fuse capacity is greater then 6 A, then R2 is nearly shorted and Vcase = VM = 0.
(c) The point Q is connected to the metal case as shown below.
9/26/01 P12-17 @ DeCarlo & P. M. Lin
Both R1 and R2 are nearly shorted. The line current is so large that it will blow the fuse. Hence Vcase= VQ = 0.
CONCLUSION: In all three cases, the person touching the metal case of the defective appliance willexperience zero or a very low voltage. The circuit is safe.
SOLUTION P12.33. For an approximate analysis, we use the circuit models below given the followingassumptions:
(i) The resistance from body to ground is very large (possibly due to rubber shoes);(ii) The resistances of all connecting wires are negligible.(iii) The motor winding is represented by R1 in series with R2. Here we can further estimate
that. R1 + R2 = 115/3 = 38.3 Ω, and R1 = R2 = 19.15 Ω.(iv) The trigger switch of the drill has been depressed.
(a) The point P is connected to the metal case as shown below. It is obvious Vcase = Vp = 0.
(b) The point M is connected to the metal case as shown below. Simple voltage divider action leads to
9/26/01 P12-18 @ DeCarlo & P. M. Lin
Vcase = VM = 0.5×115 = 57.5 V
(c) The point Q is connected to the metal case as shown below.
It is obvious that Vcase = VQ = 115 V.
CONCLUSION: In all three cases, the fuse will not blow. For cases 2 and 3, the voltage appearing onthe metal case may cause serious injury to the person.
Prbs Ch 13 March 18, 2002 P13-1 ©R. A. DeCarlo, P. M. Lin
1
PROBLEM SOLUTIONS CHAPTER 13
SOLUTION 13 .1 . Given
′ ′ i ( t) +16 ′ i (t) + 4Bi( t) = ′ v (t) + 8v(t)
(a) with v(t) = ′ v ( t) = 0 and i1(0) + i2(0) = 0
(i) the characteristic equation is
s2 +16s + 48 = 0(ii) the characteristic equation has factors
(s + 4) (s + 12) = 0
and hence
s1,s2 = −4,−12
(iii) Equivalent circuit at t = 0
(iv) Here by KCL
i1(0) = i2(0) = 6A
At point (1)
i1(0) + i2Ω(0) = i2(0)
6A + i2Ω(0) = 6A
i2Ω(0) = 0
and
vL1 = vR2(0) = 2i2Ω (0) = 0V
Then by KVL at t = 0
vL1(0) − v6Ω (0) + vL 2(0) = 0 ⇒ −vL 2 = −36V
L2di2dt
= −36V
′ i (0) = ′ i 2(0) = −36A / s
(v) from part ii
i( t) = Ae−4 t + Be−12 t
and then
′ i (t) = −4 Ae−4 t −12Be−12 t
Then using the initial conditions
A + B = 6
−4A −12B = −36
Prbs Ch 13 March 18, 2002 P13-2 ©R. A. DeCarlo, P. M. Lin
2
solving yields B = 1.5 and A = 4.5. Then
i( t) = 4.5e−4 t +1.5e−12t A
(b) If v(t) = 12V and i(0) = ′ i (0) = 0 , then ′ v ( t) = 0 and
′ ′ i ( t) +16 ′ i (t) + 48i(t) = 8v(t)
with
i( t) = Ae−4 t + Be−12 t
′ i ( t) = 4Ae−45 −12Be−12t
′ ′ i ( t) =16Ae−4 t +144Be−12t
Then at t = ∞ when v(∞) = 12V
i(∞) = C, ′ i (∞) = 0 and ′ ′ i (∞) = 0
Thus
48C = 96 ⇒ C = 2Then
i( t) = Ae−4 t + Be−12 t + 2
and
i(0) = 0 = A + B + 2
′ i (0) = 0 = −4 A −12B
Multiply the first of these by 4 yields
4 A + 4B = −8
−4A −12B = 0
Solving yields B = 1 and A = -3. Thus for t > 0
i( t) = −3e−4 t + e−12 t + 2 A
(c) If v(t) is as in fig. 13.1(b) then
v(t) =2t 0 < t ≤ 2
0 t ≥ 2
The value of C in part (b) will change and at t = 2s, a new set of initial conditions will be required
(obtainable from the solution at t = 2s) and these would be used in the decay portion described by
′ ′ i ( t) +16 ′ i (t) + 48i(t) = 0 t ≥ 2s
Prbs Ch 13 March 18, 2002 P13-3 ©R. A. DeCarlo, P. M. Lin
3
SOLUTION 13 .2 .
(a) Use the figure with the currents i1through i5 designated in the circuit below.
Then work from v0 to vin using repeated applications of KVL, KCL and the elemental equations:
i1 = 2vo
i2 = 2 ′ ′ v oi3 = i1 + i2 = 2 ′ v o + vo( )
with
v2 = vo
v1 = 0.5i3 + v2 = 0.5 2 ′ v o + vo( ) + vo[ ] = ′ v o + 2vo
then
i4 = 2 ′ v 1 = 2d
dtvo1 + 2vo( ) = 2 ′ ′ v o + 2 ′ v o( )
i5 = i3 + i4 = 2 ′ v o + vo( ) + 2 ′ ′ v o + 2 ′ v o( ) = 2 ′ ′ v o + 6 ′ v o + 2vo
Finally
vin = 0.5i5 + v1 = 0.5 2 ′ ′ v o + 6 ′ v o + 2vo( ) + ′ v o + 2vo = ′ ′ v o + 4 ′ v o + 3vo
Hence
˙ v out (t) + 4 ˙ v out (t) + 3vout ( t) = vin (t)
(b) Note from part (a) that
vout (t) = v2(t)
and
v1( t) = ˙ v out (t) + vout ( t)
Hence
vout (o) = v2(o) =1V
˙ v out (o) = v1(o) − vout (o) = 7 −1 = 6V
(c) The characteristic equation
s2 + 4s + 3 = 0has factors
s +1( ) s + 3( ) = 0
and roots
s1,v2 = −1,−3
Prbs Ch 13 March 18, 2002 P13-4 ©R. A. DeCarlo, P. M. Lin
4
Thus, because of the input, vin (t) = 6V
vout (t) = Ae−t + Be−3t + C
˙ v out (t) = −Ae−t − 3Be−3t
˙ v out (t) = Ae−t + 9Be−3t
and at t = ∞
˙ v out (∞) + 4vout (∞) + 3vout (∞) = 6V
0 + 0 + 3C = 6V
C = 2V
and
vout (t) = Ae−t + Be−3t + 2V
SOLUTION 13 .3 . (a) From the given differential equation, the characteristic equation is
s3 +14s2 + 52s + 24 = (s + 6)(s2 + 8s + 4) = 0
Therefore the roots a = −6, b = −4 + 2 3 = −0.5359,c = −4 − 2 3 = −7.4641.
(b) (i) v(0) = vC3(0) = 6 V, as given.
(ii) To compute v'(0) we write a nodal equation at node 3. In particular
(v − vC1) + (v − vC2) + 0.5v '+v = 0 which implies that
v '(t) = 2vC1(t) + 2vC 2( t) − 6v(t) (*)
Hence v '(0) = 2vC1(0) + 2vC 2(0) − 6v(0) = 24 +18 − 36 = 6 V/s.
(iii) To compute v' '(0) we first differentiate equation (*). This yields
v ''(t) = 2v 'C1( t) + 2v 'C 2 ( t) − 6 2vC1(t) + 2vC 2( t) − 6v( t)( ) (**)
To express v 'C1( t) and v 'C 2 (t) in terms of the node voltages we write node equations at nodes 1and
2 respectively. At node 1
(vC1(t) − vC 2(t)) + (vC1(t) − v(t)) + 0.5vC1' (t) = i(t)
and at node 2
(vC 2( t) − vC1(t)) + (vC 2( t) − v( t)) + 0.5vC 2' (t) = 0
Hence
v 'C1( t) = −4vC1(t) + 2vC 2(t)) + 2v(t) + 2i(t)
Prbs Ch 13 March 18, 2002 P13-5 ©R. A. DeCarlo, P. M. Lin
5
and
vC 2' ( t) = 2vC1(t)) − 4vC 2(t) + 2v( t)
Substituting these two equations into (**) yields the desired result when t is set to 0. However, this
quantity has no direct physical meaning.
v ''(t) = 2 −4vC1( t) + 2vC 2(t)) + 2v( t) + 2i(t)( ) + 2 2vC1(t)) − 4vC 2(t) + 2v( t)( ) − 6 2vC1(t) + 2vC 2( t) − 6v(t)( )
Hence, v' '(0) = 2(–48 + 18 + 12 + 2i(0)) + 2(24 – 36 + 12) – 6(24 + 18 – 36) = –72 + 4i(0) V/s2.
Finally, using the characteristic roots found in part (a) and assuming a constant input, the form
of the solution is
v(t) = Ae−6t + Be−0.5359t + Ce−7.4641t + D
Following the methods of example 13.2,
A = 0, B = (7.3301 – 1.0774×i(0)), C = (–1.3301 + 0.0774×i(0)), and D = i(0).
(c) (i) Not proportional to a single voltage but it is proportional to iC3(0).
(ii) Much more complex.
(iii) No.
(iv) No. This is why we use the Laplace transform approach.
SOLUTION 13 .4 .
(a) f (q + T0)]q=t = f ( t + T0)
(b) e−5q cos 0.5πq + 0.25π( )]q=2 t= e−10t cos πt + 0.25π( )
SOLUTION 13 .5 .
(a) Let T = 1.
Prbs Ch 13 March 18, 2002 P13-6 ©R. A. DeCarlo, P. M. Lin
6
-1 -0.5 0 0.5 1 1.5 2 2.5 30
0.5
1
1.5
2
(b) Again let T = 1.
-1 -0.5 0 0.5 1 1.5 2 2.5 3-2
-1.5
-1
-0.5
0
(c)
-1 -0.5 0 0.5 1 1.5 2 2.5 30
0.05
0.1
0.15
0.2
0.25
(d)
Prbs Ch 13 March 18, 2002 P13-7 ©R. A. DeCarlo, P. M. Lin
7
-1 -0.5 0 0.5 1 1.5 2 2.5 30
0.2
0.4
0.6
0.8
1
(e)
-1 -0.5 0 0.5 1 1.5 2 2.5 30
1
2
3
4
5
6
7
(f) i ( t − i)i=0
r
∑(g)
-1 -0.5 0 0.5 1 1.5 2 2.5 30
1
2
3
4
5
(h) Pulses of height 1 and width T.
Prbs Ch 13 March 18, 2002 P13-8 ©R. A. DeCarlo, P. M. Lin
8
SOLUTION 13 .6 .
(a) L[ f1( t)] = F1(s) = f1(t)e−stdt0−∞
∫ == e−stdtT1
T2∫ =e−st
−s
T1
T2
=1s
e−sT1 − e−sT2( )(b) f2( t) = f1( t) . Hence, the answer is the same as in (a).
(c) L[ f3( t)] = F3(s) = f3(t)e−stdt0−∞
∫ = −2 ( t)cos(4πt − 0.25π)e−stdt0−∞
∫ = −222
= − 2
(d) F4(s) = f4(t)e−stdt0−∞
∫ = −2 (t − T )cos(4πt − 0.25π)e−stdt0−∞
∫ = −2cos(4πT − 0.25π)e−sT
(e) F5(s) = f5(t)e−stdt0−∞
∫ = (t) − ( t − T )[ ]e−stdt0−∞
∫ = 1− e−sT
SOLUTION 13 .7 .
(a) F(s) = f (t)e−stdt0−∞
∫ = 5e−4 te−stdt0−∞
∫ = −5e−(s+ 4)t
s + 4
0−
∞
=5
s + 4
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
t
f(t)
TextEnd
Prbs Ch 13 March 18, 2002 P13-9 ©R. A. DeCarlo, P. M. Lin
9
(b) F(s) = f (t)e−stdt0−∞
∫ = 5e−4 te−stdt1−∞
∫ = −5e−(s+4) t
s + 4
1−
∞
=5e−(s+4)
s + 4= e−s 5e−4
s + 4
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.01
0.02
0.03
0.04
0.05
0.06
0.07
0.08
0.09
0.1
t
f(t)
TextEnd
(c) F(s) = f (t)e−stdt0−∞
∫ = 5e−4(t−1)e−stdt1
∞∫ = −5e4 e−(s+ 4)t
s + 4
1−
∞
= e−s 5s + 4
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.5
1
1.5
2
2.5
3
3.5
4
4.5
5
t
f(t)
TextEnd
(d) F(s) = f (t)e−stdt0−∞
∫ = 5e−4( t−1) (t)e−stdt0−∞
∫ = 5e4
(e) F(s) = f (t)e−stdt0−∞
∫ = 5e−4( t−1) (t −1)e−stdt0−∞
∫ = 5e−s
Prbs Ch 13 March 18, 2002 P13-10 ©R. A. DeCarlo, P. M. Lin
10
(f)
F(s) = f (t)e−stdt0−∞
∫ = 2 u(t)u(1− t)[ ]e−0.25te−stdt0−∞
∫ = 2 e−0.25te−stdt0−1
∫ =2
s + 0.251− e−(s+0.25)( )
= 2 e−0.25te−stdt0−∞
∫ − 2 e−0.25te−stdt1−∞
∫ =2
s + 0.251− e−s( )
0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.2
0.4
0.6
0.8
1
1.2
1.4
1.6
1.8
2
t
f(t)
TextEnd
F(s) = f (t)e−stdt0−∞
∫ =
SOLUTION 13 .8 .
(a) F(s) = f (t)e−stdt0−∞
∫ = Ae t e−stdt0−∞
∫ = Ae−(s− )tdt0−∞
∫ =A
s−
(b) F(s) = f (t)e−stdt0−∞
∫ = Ae (t−1) u( t −1)e−stdt0−∞
∫ = Ae− e−(s− )tdt1−∞
∫ = e−s As−
(c) F(s) = f (t)e−stdt0−∞
∫ = Ae (t−1) ( t)e−stdt0−∞
∫ = Ae−
(d) F(s) = f (t)e−stdt0−∞
∫ = Ae te−stdt0−1
∫ = −Ae−(s− )t
(s − )
0−
1
=A
s −1− e−(s− )( )
Prbs Ch 13 March 18, 2002 P13-11 ©R. A. DeCarlo, P. M. Lin
11
SOLUTION 13 .9 . Consider the following in which q = at and t = q/a.
L[ f (at)] = f (at)e−st dt
0−
∞
∫ =1a
f (q)e−(s/a )q dq
0−
∞
∫ =1a
Fsa
SOLUTION 13 .10 .
(a)
e j t = cos t + j sin t
e− j t = cos t − j sin t
Add these equations and divide by 2 to obtain
cos( t) =e j t + e− j t
2
Similarly, subtract the equations and divide by 2j to obtain
sin( t) =e j t − e− j t
2 j
Note that
L e j t[ ] =1
s − j, L e− j t[ ] =
1s + j
(b)
L cos t[ ] = Le j t + e− j t
2
=0.5
s− j+
0.5s + j
=s
s2 + 2
and
L sin t[ ] =−0.5 js − j
+0.5 j
s + j=
s2 + 2
SOLUTION 13.11 . From the time frequency scaling property,
Prbs Ch 13 March 18, 2002 P13-12 ©R. A. DeCarlo, P. M. Lin
12
L sin( t) u(t)[ ] =1 1
s
2
+1
=s2 + 2
Using the time differentiation property,
L[ cos( t)u( t)] = Lddt
sin( t)u(t)( )
= sL[sin( t) u(t)]− sin(0) =
s
s2 + 2
Finally using the multiplication by t property
L[− t cos( t)u(t)] =dds
s
s2 + 2
=
s2 + 2 −2 s2
s2 + 2( )2
Hence
L[sin( t)u(t) − t cos( t)u( t)] =2
s2 + 2 −2 s
s2 + 2( )2 =2 s2 + 2( ) − 2 s2
s2 + 2( )2 =2 3
s2 + 2( )2
SOLUTION 13.12. (a)
g1(t) = Atsin( t), f ( t) = sin(t)u(t), F (s) =1
s2 +1
By the frequency scaling property
L f ( t)[ ] =1 1
s
2
+1
=1 2
s2 + 2
=s2 + 2
By the multiplication by t property
L t sin( t)u( t)[ ] = −dds s2 + 2
= −
−2 s
s2 + 2( )2
=2 s
s2 + 2( )2
By the linearity property
L Atsin( t)[ ] = A2 s
s2 + 2( )2
Prbs Ch 13 March 18, 2002 P13-13 ©R. A. DeCarlo, P. M. Lin
13
(b)
g2( t) = Aeat sin( t)u( t), f ( t) = sin(t)u(t), F(s) =1
s2 +1
By the frequency scaling property,
L sin( t)[ ] =s2 + 2
By the damping property
L eat sin( t)[ ] =s − a( )2 + 2
and by the linearity property
L g2( t)[ ] = As − a( )2 + 2
(c) Before beginning, note that
sin t +( ) = cos( )sin( t) + sin( )cos( t)
Here
g3(t) = Aeat sin t +( )u(t) = cos( )Aeat sin( t)u( t) + sin( )Aeat cos( t)u( t)
f (t) = sin(t)u(t), F(s) =1
s2 +1
From the linearity property and part (b),
G3(s) = Acos( )s − a( )2 + 2 + Lsin( )Aeat cos( t)u(t)
By the differentiation in the time domain property cos(t) =ddt
sin(t) which implies
L cos(t)[ ] = s L sin(t)[ ] − f (0) =s
s2 +1− 0 =
s
s2 +1
By the frequency scaling property
L cos t[ ] =1 s
s( )2+1
=12
s 2
s2 + 2
=s
s2 + 2
Then by the linearity property
Prbs Ch 13 March 18, 2002 P13-14 ©R. A. DeCarlo, P. M. Lin
14
L Asin( )cos( t)[ ] = A sin( )s
s2 + 2
Then by the frequency-shift property
L Asin( )eat cos( t)[ ] = Asin( )s − a
(s− a)2 + 2
G3(s) = L Aeat sin t +( )[ ] = Acos( )s − a( )2 + 2
+ Asin( )s − a
(s− a)2 + 2
=A
s − a( )2 + 2cos( ) + sin( )(s − a)( )
SOLUTION 13.13. (a) We are given that
f t( ) = sin t( ) ⇒ F(s) =1
s2 +1
And must find the transform of
g1(t) = Atcos( )tu( t)
By the time differentation property
cos t( ) =ddt
sin t( ) ⇒ L cos t( )[ ] =s
s2 +1− sin 0−( ) =
s
s2 +1
By the frequency scaling property with
L cos t( )[ ] =1
Fs
=
1 s
s( )2+1
=s 2
2 s2 + 2( ) =s
s2 + 2
Using the multiplication-by-t property
L t cos t( )[ ] = −dds
s
s2 + 2
= −
s2 + 2 − s(2s)
s2 + 2( )2 =s2 − 2
s2 + 2( )2
Finally, by the linearity property
L Atcos t( )u(t)[ ] =A 2 − s2( )s2 + 2( )2
Prbs Ch 13 March 18, 2002 P13-15 ©R. A. DeCarlo, P. M. Lin
15
(b) Let g2( t) = Aeat cos t( )u( t) . Recall from part (a) that
L Acos t( )[ ] = As
s2 + 2
By the Frequency-shift property
L Aeat cos( t)[ ] = As− a
s − a( )2 + 2
(c) From Trig identities,
cos t +( ) = cos( t)cos( ) − sin( t)sin( )
Recall from part (b) that
L Aeat cos( t)[ ] = As− a
s − a( )2 + 2
Hence
L Aeat cos( )cos( t)[ ] = Acos( )s − a
s− a( )2 + 2
It follows that
L Aeat cos t +( )[ ] = Acos( )s− a
s − a( )2 + 2− A sin( )L eat sin t( )[ ]
= A cos( )s− a
s − a( )2 + 2− A sin( )
(s− a)2 + 2 = Acos( )(s− a) − sin( )
s − a( )2 + 2
SOLUTION 13 .14. (a) We are given that
L cos t( )[ ] =s
s2 +1
which implies by the frequency scaling property that
L cos t( )[ ] =s
s2 + 2
Using the multiplication-by-t property
L t cos t( )[ ] = −dds
s
s2 + 2
= −
s2 + 2 − 2s2
s2 + 2( )2 =s2 − 2
s2 + 2( )2
Prbs Ch 13 March 18, 2002 P13-16 ©R. A. DeCarlo, P. M. Lin
16
(b) If g( t) = teat cos( t), then using part (a) and frequency shift property,
L teat cos( t)[ ] =s − a( )2 − 2
s− a( )2 + 2[ ]2
SOLUTION 13 .15 . (a) With
sinh(at) =eat − e−at
2
L sinh at( )[ ] =12
1s− a
−1
s + a
=
12
s + a − s− a( )s2 − a2
=
12
2a
s2 − a2
=
a
s2 − a2
(b) With
cosh(at) =eat + e−at
2
L cosh(at)[ ] =12
1s + a
+1
s − a
=
12
s − a + s + a
s2 − a2
=
12
2s
s2 − a2
=
s
s2 − a2
SOLUTION 13 .16 . (a) From Problem 15
L sinh at( )[ ] =a
s2 − a2
So that by the multiplication by t property
L t sinh at( )[ ] = −dds
a
s2 − a2
= −
−a(2s)
s2 − a2( )2 =2as
s2 − a2( )2
(b) From Problem 15
L cosh(at)[ ] =s
s2 − a2
So that by the multiplication-by-t property
Prbs Ch 13 March 18, 2002 P13-17 ©R. A. DeCarlo, P. M. Lin
17
L t cosh(at)[ ] = −dds
s
s2 − a2
= −
s2 − a2( ) − s(2s)
s2 − a2( )2 =−s2 + a2 + 2s2
s2 − a2( )2 =s2 + a2
s2 − a2( )2
SOLUTION 13.17 . Here
F(s) =s + 2s +1
(a) Since g1(t) = 5 f ( t − 2), use the time shift and linearity properties to obtain
L g1(t)[ ] = 5e−2s s + 2s +1
(b) Since g2( t) = 5e−2t f (t), use the frequency shift and linearity properties,
L g2( t)[ ] = 5s + 2( ) + 2s + 2( ) +1
= 5
s + 4s + 3
(c) From part (a),
L 5 f ( t − 2)[ ] = 5e−2s s + 2s +1
Therefore, since g3(t) = 5e−2 t f ( t − 2) , by the frequency shift property
L g3(t)[ ] = G1(s + 2) = 5e−2(s+ 2) (s + 2) + 2(s + 2) +1
= 5e−2(s+2) s + 4
s + 3
(d) Since g4 (t) = 5tf ( t) , use the multiplication-by-t and linearity principle to obtain
L g4 (t)[ ] = −dds
5 s + 2( )s +1
= −
5 (s +1)− (s + 2)[ ](s +1)2 =
5
(s +1)2
SOLUTION 13 .18 . In all parts
L f ( t)u(t)[ ] = F (s) =s
s2 + 4
Prbs Ch 13 March 18, 2002 P13-18 ©R. A. DeCarlo, P. M. Lin
18
(a) L Af ( t − T )u( t − T )[ ] = Ae−Tss
s2 + 4
(b) L Atf ( t)u(t)[ ] = −dds
Af (s) = −Adds
s
s2 + 4
= −A
s2 + 4( ) − s(2s)
s2 + 4( )2
= As2 − 4( )
s2 + 4( )2
(c) Note that the answer is simply a time shift of the function given in (b).
L A( t − T ) f (t − T)u(t − T )[ ] = Ae−sTs2 − 4( )
s2 + 4( )2
(d) This function is that of part (a) multiplied by t. Hence, by the multiplication by t property,
L Atf ( t − T )u(t − T )[ ] = −Adds
e−Tss
s2 + 4
= −Ae−Ts − Tse−Ts( )(s2 + 4) − 2s × se−Ts
s2 + 4( )2
= Ae−Ts 2s2 − 1− Ts( )(s2 + 4)
s2 + 4( )2
= Ae−Ts Ts3 + s2 + 4Ts − 4
s2 + 4( )2
SOLUTION 13 .19 . In all parts
F(s) = L f ( t)u( t)[ ] =s2 + 2
(a) By the time shift property, L Af ( t − T )u( t − T )[ ] = Ae−Ts
s2 + 2( )
Using the multiplication-by-t property,
L Atf ( t)u(t)[ ] = −Adds
F (s)[ ] = −Adds s2 + 2
= A
(2s)
s2 + 2( )2 = A2 s
s2 + 2( )2
Prbs Ch 13 March 18, 2002 P13-19 ©R. A. DeCarlo, P. M. Lin
19
© The answer here is an application of the time shift property to the answer of part (b).
L A( t − T ) f (t − T)u(t − T )[ ] = Ae−sT 2 s
s2 + 2( )2
(d) The answer here uses the multiplication-by-t property applied to the answer of part (a).
L Atf ( t − T )u(t − T )[ ] = −Adds
e−Ts
s2 + 2( )
= ATe−Ts(s2 + 2) + 2se−Ts
s2 + 2( )2
= A e−Ts Ts2 + 2s + T 2
s2 + 2( )2
Solution 13.20. (a)
f (2 t) = (2t) + (2 t −1)
For the first term on the right, the peak occurs at t = 0 and
(2 t)dt
0−
0+
∫ = 0.5 ( )d = 0.5
0−
0+
∫
under the transformation τ = 2t. For the second term, the function peaks at t = 0.5 and
(2 t −1)dt
0.5−
0.5+
∫ = 0.5 ( )d = 0.5
0−
0+
∫
under the transformation τ = 2t – 1. Hence
f (2 t) = (2t) + (2 t −1) = 0.5 ( t) + (t − 0.5)[ ]Therefore, a = 0.5 = b.
Prbs Ch 13 March 18, 2002 P13-20 ©R. A. DeCarlo, P. M. Lin
20
(b) (i) Here F(s) = L f ( t)[ ] =1 + e−s . By the time scaling property
L f (2 t)[ ] =12
F (s /2) = 0.5 1+ e−0.5s( )(b)-(ii) For this part,
L f (2 t)[ ] = 0.5L (t) + (t − 0.5)[ ] = 0.5 1+ e−0.5s( )
SOLUTION 13.21. (a)
L v(t)[ ] = 2L g"( t)[ ] − L g'(t)[ ] = 2s2F (s) − 2sg(0−) − 2g'(0− ) − sF(s) + g(0−)
= 2s2 − s( )F (s) − 2sg(0−) − 2g'(0− ) + g(0− ) =(2s−1)(s +1)
s2
(b)
L v(t)[ ] = 2L f "( t)[ ] − L f '(t)[ ] = 2s2F (s) − sF(s) − 2sf (0−) − 2 f '(0 −) + f (0−)
= 2s2 − s( )F (s) − 2s − 2 +1 =2s−1( )(s +1)
s2 − 2s − 2 +1
(c)
L v(t)[ ] = L g'( t)[ ] − L g(q)dq−∞
t
∫
= sG(s) − g(0− ) −G(s)
s−
1s
g(q)dq−∞
0−
∫ = s−1s
G(s)
= s −1s
s +1
s3 = s +1( ) 1
s2 −1
s4
(d)
L v(t)[ ] = L f '( t)[ ] − L f (q)dq−∞
t
∫
= sF(s) − f (0−) −F(s)
s−
1s
f (q)dq−∞
0−
∫
= s +1( ) 1
s2 −1
s4
−1−
−1
s
assuming λ > 0. The expression is ill-defined if λ ≤ 0.
Prbs Ch 13 March 18, 2002 P13-21 ©R. A. DeCarlo, P. M. Lin
21
SOLUTION 13.22 .
(a) (i) If
F(s) = L f ( t)u( t)[ ] = ln
s2 + 4
s2
= ln s2 + 4( ) − ln s2( )Then by the multiplication-by-t property
L −tf (t)u(t)[ ] = +dds
ln s2 + 4( ) − ln s2( )[ ]=
2s
s2 + 4−
2s
s2 =2s
s2 + 4−
2
s=
2s2 − 2s2 − 8
s s2 + 4( ) =−8
s s2 + 4( ) (ii) Using the solution to (a)-(i), by the frequency shift property
L −te−2 t f ( t)u(t)[ ] =−8
s + 2( ) s + 2( )2 + 4( ) =−8
s + 2( ) s2 + 4s + 8( )=
−8
s3 + 6s +16s +16
(b) If
G(s) =−8
s s2 + 4( )
a partial fraction expansion may be employed
G(s) =−8
s s2 + 4( ) =K1s
+As + B
s2 + 4=
−2s
+2
s2 + 4
Hence,
g( t) = 2cos(2t) − 2[ ]u( t)
and
f ( t) = −g(t)
t=
2t
−2t
cos(2t)
SOLUTION 13.23: Part (a)-(i): From table 13.2, the multiplication by t property implies that
Prbs Ch 13 March 18, 2002 P13-22 ©R. A. DeCarlo, P. M. Lin
22
L[−tf (t)u(t)] =
dds
F(s) =dds
lns + as − a
=
dds
ln s + a[ ] − ln s − a[ ]( ) =1
s + a−
1s − a
=−2a
s2 − a2
Part (a)-(ii): Let us make use of the answer to part (a)-1. Let G(s) = L[−tf ( t)u(t)]=−2a
s2 − a2 .
Then by the frequency shift property in table 13.2,
L[−te−at f (t)u(t)] = L[e−at − tf (t)u( t)( )] = G(s + a) =−2a
(s + a)2 − a2 =−2a
s(s + 2a)
Part (b): g( t) = L–1[G(s)]= L–1 −2a
s2 − a2
= L–1 1
s + a
− L–1 1
s − a
= e−at − eat( )u(t)
More specifically,
g( t) = e−at − eat( )u(t) = −2eat − e−at
2
u( t) = −2sinh(at) u(t)
Hence f ( t) =g(t)−t
=2sinh(at)
t.
SOLUTION 13 .24 .
(a)-(i) If
F(s) = L f ( t)u( t)[ ] = ln
s + as + b
= ln(s + a) − ln(s + b)
Then by the multiplication-by-t property
L −tf (t)u(t)[ ] = +
dds
ln s + a( ) − ln s + b( )[ ] =1
s + a−
1s + b
=s + b − (s + a)(s + a)(s + b)
=b − a
(s + a)(s + b)
(ii) By the frequency-shift property
L −te−at f (t)u(t)[ ] =b − a
(s + 2a)(s + a + b)=
b − a
s2 + s(3a + b) + 2a(a + b)
(b) If
Prbs Ch 13 March 18, 2002 P13-23 ©R. A. DeCarlo, P. M. Lin
23
G(s) = L −tf ( t)u(t)[ ] =b − a
(s + a)(s + b)
a partial fraction expansion may be employed
G(s) =b − a
(s + a)(s + b)=
K1s + a
+K2
s + b=
1s + a
+−1
s + b
Hence, g( t) = e−at − e−bt( )u(t)
and
f ( t) =g(t)−t
=e−bt
t−
e−at
t
SOLUTION 13.25. The relationship is f ( t) =ddt
g(t) or equivalently, g( t) = f (q)dq−∞
t
∫ .
Now we have that f ( t) = 6 (t) −12 (t − 2) + 6 (t − 4) . Therefore,
F(s) = 6 −12e−2s + 6e−4s
From the time integration property,
G(s) =F (s)
s=
6s
−12e−2s
s+
6e−4s
s
SOLUTION 13 .26 . For 0 ≤ t < T1, we see that g( t) = f (q)dq−∞
t
∫ . Thus one presupposes here that
the relationship is f ( t) =ddt
g(t) or equivalently, g( t) = f (q)dq−∞
t
∫ . As such E = A – B and D = A –
B + C.
Further, f ( t) = A (t) − B ( t − T1) + C (t − T2) which implies that
F(s) = A − Be−T1s + Ce−T2s
Thus
Prbs Ch 13 March 18, 2002 P13-24 ©R. A. DeCarlo, P. M. Lin
24
G(s) =F (s)
s=
As
−Be−T1s
s+
Ce−T2s
s
SOLUTION 13 .27 .
(a) f ( t) = u(t) + u(t −1) ⇒ f (s) =1s
+e−s
s=
1s
1− e−s( )(b)
f ( t) = u(t) + u(t −1)− u(t − 3) ⇒ f (s) =1s
+e−s
s−
e−3s
s=
1s
1+ e−s − e−3s( )(c)
f ( t) = u(t) + u(t −1)− 2u(t − 3) ⇒ F (s) =1s
+e−s
s−
2e−3s
s=
1s
1 + e−s − 2e−3s( )(d)
f ( t) = 2u(t) − u(t − 2) − u( t − 3) ⇒ F(s) =2s
−e−2s
s−
e−3s
s=
1s
2 − e−2s − e−3s( )
SOLUTION 13 .28 .
(a) f ( t) = 2r(t) − 2r(t −1) ⇒ F (s) =2
s2 −2e−s
s2 =2
s2 1− e−s( )
(b) f ( t) = 2r(t) − 2r(t −1)+ r(t − 2) ⇒ F (t) =2
s2 −2e−s
s2 +e−2s
s2 =1
s2 2 − 2e−s + e−2s( )
(c) f ( t) = 2r(t) − 2r(t −1)− 2r( t − 2) + 2r(t − 3). It follows that
F(s) =2
s2 −2e−s
s2 −2e−2s
s2 +2e−3s
s2 =2
s2 1− e−s − e−2s + e−3s( )
Prbs Ch 13 March 18, 2002 P13-25 ©R. A. DeCarlo, P. M. Lin
25
SOLUTION 13 .29 . (a) Here f ( t) =32
r(t) − 3r(t − 2) +32
r(t − 4) . Thus,
F(s) =3
2s2 −3e−2s
s2 +3e−4s
2s2 =3
2s2 1− 2e−2s + e−4s( )
(b) Here f ( t) =VoT
r(t) −2VoT
r(t − T ) +VoT
r(t − 2T ). Thus
F(s) =VoT
1
s2 −2e−T
s2 +e−2T
s2
(c) Here f ( t) = 2r(t −1)− 4r(t − 2) + 4r(t − 4) − 2r(t − 5)
F(s) =1
s2 2e−s − 4e−2s + 4e−4s − 2e−5s( )
SOLUTION 13 .30 .
(a) Here f ( t) = 2r(t) − 2r(t −1)− 2u( t − 4) implies
F(s) =2
s2 −2e−s
s2 −2e−4s
s=
2
s2 1− e−s − se−4s( )
(b) Here f ( t) = 2u(t) − r(t − 2) + r(t − 4) implies
F(s) =2s
−e−2s
s2 +e−4s
s2 =2
s2 s− e−2s − e−4s( )
SOLUTION 13.31. (a) Here f ( t) = 2u(t) − r(t) + 2r(t − 2) − 2r(t − 4) − 2u( t − 4)
Thus F(s) =2s
−1
s2 +2e−2s
s2 −2e−4s
s2 −2e−4s
s.
(b) f ( t) = −u(t) + r(t) − r(t − 2) − u( t − 2). Hence F(s) =−1s
+1
s2 −e−2s
s2 −e−2s
s.
Prbs Ch 13 March 18, 2002 P13-26 ©R. A. DeCarlo, P. M. Lin
26
(c) f ( t) = 2r(t) − 2r(t −1)− 2u( t −1) + 2u(t − 2) − 2r( t − 2) + 2r(t − 3).
Hence F(s) =2
s2 −2e−s
s2 −2e−s
s+
2e−2s
s−
2e−2s
s2 +2e−3s
s2 .
2/23/02 page P14.1 © R. A. DeCarlo, P. M. Lin
PROBLEM SOLUTIONS CHAPTER 14
SOLUTION 14 .1 .(a)
Z(s) =R(Ls +
1
Cs)
R + Ls +1
Cs
=Cs(RLCs2 + R)
Cs(RCs + LCs2 +1)=
RLC(s2 +1 LC)
LC(s2 +R
Ls +
1
LC)
=R(s2 +1 LC)
s2 +R
Ls +
1
LC(b)
Z(s) = R +(Ls)(
1
Cs)
Ls +1
Cs
= R +LCs
C(LCs2 +1)= R +
Ls
LCs2 +1
= RLCs2 + Ls + R
LCs2 +1=
RLC(s2 +1
RCs +
1
LC)
LC(s2 +1
LC)
Hence,
Z(s) =R s2 +
1
RCs +
1
LC
s2 +1
LC
SOLUTION 14.2.(a)
Zin (s) =Vs(s)Is(s)
=(10 + 0.2s)(
80
s)
10 + 0.2s +80
s
=s(800 +16s)
s(0.2s2 +10s + 80)=
800s + 4000
s2 + 50s + 400
(b) If is( t) = 3e−20 tu(t) A then
Is(s) =3
s + 20and
Vs(s) = Zin (s)Is(s) =800s + 4000
s2 + 50s + 400−
3
s + 20=
2400s +12,000
(s +10)(s + 20)(s + 40)
=K1
s +10+
K2
s + 20+
K3
s + 40Here
K1 =2400s +12,000(s + 20)(s + 40) s=−10
=+12,000 − 24,000
(10)(30)= −40
K2 =2400s +12,000(s +10)(s + 40) s=−20
=12,000 − 48,000
(−10)(20)=180
2/23/02 page P14.2 © R. A. DeCarlo, P. M. Lin
K3 =2400s +12,000(s +10)(s + 20) s=−40
=12,000 − 96,000
(−30)(−20)= −140
Vs(s) =180
s + 20−
40s +10
−140
s + 40and for t > 0,
vs(t) = 180e−20 t − 40e−10t −140e−40 t V
SOLUTION 14.3.(a)
Yp(s) = Cs +1R
= 2 ×10−3s +1
0.10= 2 ×10−3(s + 50)
Then
Zp(s) =500
s + 50and
Zin (s) =1.25s +500
s + 50=
1.25s2 + 62.5s + 500s + 50
and
Yin (s) =Is(s)Vs(s)
=1
Zin (s)=
s + 50
1.25s2 + 62.5s + 500=
0.80s + 40
s2 + 50s + 400
With vs(t) = 90e−40 tu( t) , then
Vs(s) =90
s + 40and
Is(s) = Vs(s)Yin (s) =90
s + 40⋅
0.80s + 40(s +10)(s + 40)
=72s + 3600
(s +10)(s + 40)2 =K1
s +10+
C1s + 40
+C2
(s + 40)2
Here
K1 =72s + 3600
(s + 40)2s=−10
=3600 − 720
(30)2 =2880900
= 3.20
and with
p(s) =72s + 3600
s +10→ p(−40) =
720−30
= −24
′ p (s) =(s +10)(72) − (72s + 3600)
(s +10)2
′ p (−40) =−30(72) − (−2880 + 3600)
(−30)2 =−2160 − 720
900= −3.20
Then
C2 =p(−40)
0!= −24 , C1 =
′ p (−40)1!
= −3.20 ,
2/23/02 page P14.3 © R. A. DeCarlo, P. M. Lin
Is(s) =3.20s +10
−3.20
s + 40+
24
(s + 40)2
and for t > 0
is( t) = 3.20e−10t − 3.20e−40t + 24te−40t A
SOLUTION 14.4.(a) Find Zin (s) vis Yin (s)
Yin (s) = Cs +1
Ls + 20+
1
10=
200Cs +10LCs2 +10 + 20 + Ls
10(Ls + 20)
=10LCs2 + (200C + L)s + 30
10Ls + 200and
Zin (s) =1
Yin (s)=
10Ls + 200
10LCs2 + (200C + L)s + 30Ω
With C = 10−3F and L = 0.05 H
Zin (s) =0.50s + 200
0.0005s2 + 0.25s + 30=
1000s + 4 ×105
s2 + 500s + 60,000Ω
(b) If is( t) = 0.3u(t) A, then
Is(s) =0.30
sand
Vin (s) = Zin (s)Is(s) = 300s + 400
s(s2 + 500s + 60,000)
= 300s + 400
s(s + 2000)(s + 300)
= 300K1s
+K2200
+K3300
It follows that
K1 =s + 400
(s + 200)(s + 300) s=0=
400200(300)
=1
150
K2 =s + 400
s(s + 300) s=−200=
200(−200)(100)
= −1
100and
K3 =s + 400
s(s + 200) s=−300=
100(−300)(−100)
=1
300Thus
2/23/02 page P14.4 © R. A. DeCarlo, P. M. Lin
Vin (s) = 300
1
150s
−
1
100s + 200
+
1
300s + 300
=2
s−
3
s + 200+
1
s + 300and for t > 0
vin (t) = 2 − 3e−200t + e−300tV
SOLUTION 14.5.
Z(s) =s + 20s + 40
and the network is “at rest”(a) If
vin (t) = 20u(t) → Vin (s) =20s
then
Iin (s) =Vin (s)Z(s)
=20s
s + 40s + 20
= 20
s + 40s(s + 20)
Using a partial function expansion
Iin (s) = 20s + 40
s(s + 20)
= 20
K1s
+K2
s + 20
in which case
K1 =s + 40s + 20 s=0
=4020
= 2, K2 =s + 40
s s=−20=
20−20
= −1
Thus
Iin (s) = 202s
−1
s + 20
and
iin (t) = 20(2 − e−20t )u( t) A(b) Note that
vin (t) = 20e−40t → Vin (s) =20
s + 40Then
Iin (s) =Vin (s)Z(s)
=20
s + 40s + 40s + 20
=
20s + 20
in which case
iin (t) = 20e−20 tu( t) A(c) Note that
vin (t) = 20e−20t → Vin (s) =20
s + 20Then
2/23/02 page P14.5 © R. A. DeCarlo, P. M. Lin
Iin (s) =Vin(s)Z (s)
=20
s + 20s + 40s + 20
= 20
s + 40
(s + 20)2
Using a partial fraction expansion
Iin (s) = 20s + 40
(s + 20)2
= 20C1
s + 20+
C2
(s + 20)2
Herep(s) = s + 40 → p(−20) = 20
p'( s) =1 → p'(−20) =1and
C1 =p'(−20)
1!=1, C2 =
p(−20)0!
=201
= 20
in which case
Iin (s) = 201
s + 20+
20
(s + 20)2
so that
iin (t) = 20e−20t + 400te−20 t( )u( t) A
SOLUTION 14.6.(a) Apply an arbitrary Iin (s) to the upper terminal of Fig. P14.6a. Assuming branch currents Ia(s) andIb(s), it follows by KCL that
Iin (s) = Ia(s) + Ib(s) = 0.020sVa(s) + 0.005s Va (s) − sVa (s)[ ]= (0.020 + 0.005 − 0.015)sVa(s) = 0.010sVa (s)
Hence,
Zin (s) =Va(s)Iin (s)
=1
0.010s=
100s
Ω
(b) Similarly apply an arbitrary Iin (s) to Fig P14.6b to obtain, in the s-domain, by KCL
Iin (s) = 10sVin (s) +1
50Vin (s) + 0.10s Vin (s) −
Vin (s)
4
= 10s +
1
50+
30
4s
Vin (s)
=2000s2 + 4s +1500
200s
Vin (s)
Hence
Yin (s) =Iin (s)Vin (s)
=10s2 + 0.02s + 7.50
s
(c) Here we apply Vin (s) to the input terminals of figure P14.6c. By KCL
Vin (s) =10Iin (s) + 0.2sIin (s) +400
sIin (s) −
12
Iin (s)
= 10 + 0.2s +
200s
Iin
in which case
2/23/02 page P14.6 © R. A. DeCarlo, P. M. Lin
Zin (s) =Vin (s)Iin (s)
=0.2s2 +10s + 200
sΩ
SOLUTION 14 .7 . Writing two loop equations we obtain:
Vin (s) =100I1(s) + 200I2(s)and
100I1(s) + 100 +100
s
I2(s) = 0
In matrix form (dropping the s-dependence)
100 200
100 100 +100
s
I1I2
=
Vin
0
Using Cramer's rule,
I1 =
detVin 200
0 100 +100
s
det100 200
100 100 +100
s
=100s +100
s×
1
100100s +100
s
− 200
Vin
Hence
Zin (s) =VinI1
= −100s −1s +1
SOLUTION 14.8.Working in the s-domain, apply KVL to the left side of the circuit to obtain
Vin (s) =100
sIin (s) +
s100
Iin (s) + V2(s)
Now apply KCL to the right side to obtain
Iin (s) =s
100V2(s) +
100s
V2(s)
Thus
Vin (s) =s2 +104
100s
Iin (s) +V2(s)
To find V2(s) note that
Iin (s) =s2 +104
100sV2(s)
implying that
2/23/02 page P14.7 © R. A. DeCarlo, P. M. Lin
V2(s) =100s
s2 +104
Iin (s)
Thus
Vin (s) =s2 +104
100s+
100s
s2 +104
Iin (s) =
(s2 +104 )2 +104 s2
100s(s2 +104)Iin (s)
implying that
Zin (s) =Vin (s)Iin (s)
=s4 + 3 ×104 s2 +108
100s(s2 +104 ) Ω
and
Yin (s) =1
Zin (s)=
100s(s2 +104 )
s4 + 3 ×104 s2 +108 S
SOLUTION 14.9.Three mesh equations for the circuit
R + Z2(s) −Z2(s) −R
−Z2(s) 2R + Z2(s) −R
−R −R 2R + Z1(s)
I1(s)
I2(s)
I3(s)
=Vin (s)
0
0
Solve for I1(s) via Cramer’s rule
I1 =
det
Vin −Z2(s) −R
0 2R + Z2(s) −R
0 −R 2R + Z1(s)
det
R + Z2(s) −Z2(s) −R
−Z2(s) 2R + Z2(s) −R
−R −R 2R + Z1(s)
=Vin 3R2 + 2R Z1 + Z2( ) + Z1Z2( )
d(s)
where
d(s) = R + Z2( ) 3R2 + 2R Z1 + Z2( ) + Z1Z2( ) + Z2 −Z2(2R + Z1) − R2( ) − 2R2Z2 + 2R3( )= R3 + 2R2Z1 + 2R2Z2 + 3RZ1Z2
Under the condition that Z1(s)Z2(s) = R2, we have
Zin (s) =VinI1
=R3 + 2R2Z1 + 2R2Z2 + 3RZ1Z2
3R2 + 2R Z1 + Z2( ) + Z1Z2=
4R3 + 2R2 Z1 + Z2( )4R2 + 2R Z1 + Z2( ) = R
SOLUTION 14.10.
(a) Yin (s) = Cs +1R
implies a parallel RC circuit with values R and C respectively.
2/23/02 page P14.8 © R. A. DeCarlo, P. M. Lin
(b) Yin (s) =1
Zin (s)= 2s +
14
which is a parallel RC circuit of values 4 Ω and 2 F respectively.
(c) Zin (s) =1 +1
2s + 0.25=1 +
1Yb(s)
. Using the result of part (b), this circuit is a 1 Ω resistor in series
with the parallel RC of part (b).
(d) Zin (s) =2s + 8s + 2
= 2 +4
s + 2= 2 +
1
0.25s +1
2
. Using the results of parts (b) and (c), this circuit is a 2 Ω
resistor in series with a parallel combination of a 0.25 F capacitor and a 2 Ω resistor.
(e) Zin (s) =s + 3s +1
+s + 6s + 4
= 2 +2
s +1+
2s + 4
= 2 +1
0.5s +1 2+
10.5s +1 0.5
. Using the above results, this
circuit is a 2 Ω resistor in series with a parallel combination of a 0.5 F capacitor and a 2 Ω resistor which isin series with another parallel combination of a 0.5 F capacitor and a 0.5 Ω resistor.
SOLUTION 14.11.(a) Clearly this is an inductor of value L in series with a resistor of value R.(b) Inverting the admittance we have Zin(s) of the form of part (a). Hence the circuit is a 0.5 H inductorin series with a 10 Ω resistor.
(c) Yin (s) = 0.2 +1
0.5s +10= 0.2 +
1Zb(s)
. Using the result of part (b), the circuit is 0.2 S resistor in
parallel with a series connection of a 0.5 H inductor and a 10 Ω resistor.
(d) Yin (s) =10s + 50
s +1= 10 +
40s +1
= 10 +1
0.025s + 0.025. This is similar to part (c). Hence the circuit is a
10 S resistor in parallel with a series connection of a 25 mH inductor and a 0.025 Ω resistor.
(e) Yin (s) =s + 3s +1
+s + 6s + 4
= 2 +2
s +1+
2s + 4
= 2 +1
0.5s + 0.5+
10.5s + 2
. Hence, the circuit is a 2 S
resistor in parallel with the series connection of a 0.5 H inductor and a 0.5 Ω resistor which in turn is inparallel with a 0.5 H inductor and 2 Ω resistor.
SOLTUION 14.12.
(a) Zin (s) = Ls +1Cs
represents a series connection of an inductance L and a capacitance C.
(b) Yin (s) = Cs +1Ls
represents a parallel connection of an inductance L and a capacitance C.
(c) Zin (s) =0.125s2 +1
0.25s= 0.5s +
10.25s
which is a series connection of a 0.5 H inductor and a 0.25 F
capacitor.
(d) Yin (s) =0.125s2 +1
0.25s= 0.5s +
10.25s
which is a parallel connection of a 0.5 F capacitor and 0.25 H
inductor.
2/23/02 page P14.9 © R. A. DeCarlo, P. M. Lin
(e) Zin (s) =s2 +1
s+
0.25s2 +10.25s
= 2s +1s
+4s
= 2s +1
0.2s which is a 2 H inductor in series with a 0.2 F
capacitor.
(f) Zin (s) =s2 +1
s+
0.25s
0.25s2 +1= s +
1s
+1
s +1
0.25s
. This circuit is a 1 H inductor in series with a 1 F
capacitor which is in series with a parallel connection of a 1 F capacitor and a 0.25 H inductor.
(g) Yin (s) =s2 +1
s+
0.25s
0.25s2 +1= s +
1s
+1
s +1
0.25s
. This circuit is a 1 F capacitor in parallel with a 1 H
inductor which is in parallel with a series connection of a 1 H inductor and a 0.25 F capacitor.
SOLUTION 14.13.With L[vout ( t)] = Vo(s) and L[vin (t)] = Vi (s) and vout (0− ) = 0,
sV0(s) + 25V0(s) +100
sV0(s) = 5Vi (s) −
10s
Vi (s)
which implies that
s2 + 25s +100s
V0(s) =
5s −10s
Vi (s)
The transfer function is
H(s) =V0(s)Vi (s)
=5s−10
s2 + 25s +100=
5s −10(s + 5)(s + 20)
(a) If vin (t) = te−5tu(t) V, then
Vi (s) =1
(s + 5)2
and
Vout (s) =5s−10
(s + 20)(s + 5)3 =K1
s + 20+
C1s + 5
+C2
(s + 5)2 +C3
(s + 5)3
K1 =5s −10
(s + 5)3s=−20
=−110
(−15)3 =−110−3375
=22675
p(s) =5s−10s + 20
′ p (s) =(s + 20)(5) − (5s = 10)
(s + 20)2 =110
(s + 20)2 = 110(s + 20)−2
′ ′ p (s) =−220
(s + 20)3
p(−5) =−25 −10
15=
−3515
= −73
= −1575675
′ p (−5) =110
(−15)2 =110225
=330675
2/23/02 page P14.10 © R. A. DeCarlo, P. M. Lin
′ ′ p (−5) = −220
(−15)3 = −2203375
= −44675
Then
C3 =p(−5)
01= −
1575675
, C2 =′ p (−5)1!
=330675
, C1 =′ ′ p (−5)2!
=12
−44675
= −
22675
and
Vout (s) =1
67522
5 + 20−
22s + 5
+330
(s + 5)2 −1575
(s + 5)3
This yields
vout (t) =1
67522e−20t − 22e−5t + 330te−5t −
15752
t2e−5t
u( t) V
(b) If vin (t) = u(t) V,
Vin (s) =1s
and
Vout (s) =5s −10
s(s + 5)(s + 20)=
K1s
+K2s + 5
+K3
s + 20
K1 =5s =10
(s + 5)(s + 20) s=0=
−10100
= −110
K2 =5s −10
s(s + 20) s=−5=
−25 −10(−5)(+15)
=−35−75
=7
15and
K3 =5s−10s(s + 5) s=−20
=−100 −10(−20)(−15)
=−110300
=−1130
Thus
vout (t) =715
e−5t −1130
e−20t −1
10
u(t) V
By virtue of linearity and time invariance, if vin (t) = [u( t) − u(t − 0.5)] V,
vout (t) =7
15e−5t −
11
30e−20t −
1
10
u(t)
−7
15e−5(t−0.5) −
11
30e−20(t−0.5) −
1
10
u(t - 0.5) V
SOLUTION 14.14.Here vin (t) = cos( t) u(t) V and iout ( t) = 2sin(t)u( t) A, in which case
2/23/02 page P14.11 © R. A. DeCarlo, P. M. Lin
H(s) =Iout (s)Vin (s)
=
2
s2 +1s
s2 +1
=2s
SOLUTION 14.15.
Here vin (t) = te−tu( t) V which implies that Vin (s) =1
(s +1)2 . Further,
vout (t) = (1+ t − 0.5t2)e−tu( t) + sin(t)u(t) − cos(t)u(t) V in which case
Vout (s) =1
s +1+
1
(s +1)2+
1
(s +1)3 +1
s2 +1−
s
s2 +1(a) Hence
H(s) =Vout (s)Vin (s)
=1
s +1+
1
(s +1)2+
1
(s +1)3 −s −1
s2 +1
(s +1)2
= (s +1)+1 +1
(s +1)−
(s−1)(s +1)2
s2 +1Simplifying
H(s) =s3 + 2s2 + 5s + 2
(s +1)(s2 +1)
(b) If vin (t) = (1+ t)u( t) V, then Vin (s) =1s
+1
s2 =s +1
s2 . Hence
Vout (s) = H (s)Vin (s) =s3 + 2s2 + 5s + 2
s2(s2 +1)=
5s
+2
s2 −4s
s2 +1implying that
vout (t) = 5 + 2t − 4cos(t)[ ]u( t) V
SOLTUION 14.16.(a) By a voltage divider (Fig. P14.16a)
Vout (s) =Z4 (s)
Z3(s) + Z4 (s)Vin (s)
and
H(s) =Z4 (s)
Z3(s) + Z4 (s)(b) In Fig. P14.16b,
Yin (s) = Y1(s) + Y2(s)and
Vout (s) =1
Yin (s)Iin (s) =
1Y1(s) + Y2(s)
Iin (s)
2/23/02 page P14.12 © R. A. DeCarlo, P. M. Lin
Hence
H(s) =Vout (s)Iin (s)
=1
Y1(s) +Y2(s)(c) By current division,
Iout (s) =
1
Z3(s) + Z4(s)
Y1(s) + Y2(s) +1
Z3(s) + Z4(s)
Iin (s) =1
Y1(s) + Y2(s)[ ] Z3(s) + Z4 (s)[ ] +1Iin (s)
Hence
Vout (s) = Z4(s)Iout (s) =Z4(s)
Y1(s) +Y2(s)[ ] Z3(s) + Z4(s)[ ] +1Iin (s) .
and
H(s) =VoutIin
=Z4 (s)
[Y1(s) +Y2(s)] [Z3(s) + Z4 (s)] +1
SOLUTION 14.17. With Vin (s) = Vi and Vout (s) = V0 , H(s) =V0Vi
. By voltage division,
H(s) =V0Vi
=
1
10−4 s +10−3
103 +1
10−4 s +10−3
=1
0.1s + 2=
10s + 20
(a) Vout (s) =400
s(s + 20)=
20s
−20
s + 20 ⇒ vout (t) = 20 − 20e−20 t( )u( t) V. Plot omitted.
(b) If vin (t) = 40 u(t) − u( t − 0.2)[ ]V , then by linearity and time invariance
vout (t) = 20(1− e−20t )u( t) − 20 1− e−20(t−0.2)[ ]u(t − 0.2) V
(c) If vin (t) = 40 u(t) + u(t − 0.2)[ ]V, then by linearity and time invariance
vout (t) = 20(1− e−20t )u( t) + 201 − e−20(t−0.2)[ ]u( t − 0.2) V
(d) If vin (t) = 40e−20 tu(t) V, Vi (s) =40
(s + 20). Hence,
V0(s) = H (s)Vi (s) =400
(s + 20)2 ⇒ vout (t) = 400te−20 tu(t) V
(e) If vin = 40te−20 tu(t) V , then Vi (s) =40
(s + 20)2 . Hence,
V0(s) = H (s)Vi (s) =400
(s + 20)3 ⇒ vout ( t) = 200t2e−20tu( t) V
2/23/02 page P14.13 © R. A. DeCarlo, P. M. Lin
SOLUTION 14.18.(a) By voltage division
Vout (s) =
2s +1
s
2 +2
s+
2s +1
s
Vin (s) =
2s +1
s2s + 2 + 2s +1
s
Vin (s) =2s +14s + 3
Vin (s)
Hence
H(s) =Vout (s)Vin (s)
=2s +14s + 3
(b) With vin (t) = 8u(t) then
Vout (s) = H (s)Vin (s) =2s +14s + 3
8s
=
16s + 8s(4s + 3)
Using MATLAB
»n = [16 8]; d = [4 3 0];»[r,p,k] = residue(n,d)r = 1.3333e+00 2.6667e+00p = -7.5000e-01 0k = []Then
vout (t) =83
+43
e−0.75t
u( t) V
(c) If vin (t) = 8sin(2 t)u(t) , then
Vout (s) = H (s)Vin (s) =2s +14s + 3
16
s2 + 4
Using MATLAB,
»ilaplace( (32*s+16)/((4*s+3)*(s^2+4)) )ans =-32/73*exp(-3/4*t)+32/73*cos(2*t)+280/73*sin(2*t)
Hence,
vout (t) = −0.43836e−0.75t + 0.43836cos(2t) + 3.8356sin(2t)( )u(t) V
(d) With vin (t) = 8sin(8 t)u(t)
Vout (s) = H (s)Vin (s) =2s +1
4s + 3
64
s2 + 64
=
128s + 64
(4 s + 3)(s2 + 64)
=0.12391s + 31.907
s2 + 64−
0.12391
s + 0.75Using MATLAB
»ilaplace((128*s + 64)/((4*s+3)*(s^2+64)))ans =-128/1033*exp(-3/4*t)+128/1033*cos(8*t)+4120/1033*sin(8*t)
2/23/02 page P14.14 © R. A. DeCarlo, P. M. Lin
»»128/1033ans = 1.2391e-01»4120/1033ans = 3.9884e+00
vout (t) = 0.12391cos(8t) + 3.9884sin(8t) − 0.12391e−0.75t( )u( t) V
SOLUTION 14.19.
With a source transformation Iin (s) =Vin (s)
R.
(a) By current division,
IC (s) =Cs
1
R+ Cs +
1
Ls
Vin (s)R
=LCs2
LCs2 +L
Rs +1
Vin (s)R
=s2
s2 +1
RCs +
1
LC
Vin (s)R
Here
H(s) =IC (s)Vin (s)
=1R
s2
s2 +1
RCs +
1
LC
(b) With R =23
Ω, C = 0.5F and L = 1H ,
H(s) =32
s2
s2 + 3s + 2
If vin (t) = e−tu( t) V, then Vin (s) =1
s +1. Hence
IC (s) = H (s)Iin (s) =32
s2
s2 + 3s + 2
1
s +1
=
3s2
2(s +1)2(s + 2)=
K1s + 2
+C1s +1
+C2
(s +1)2
Using MATLAB,»n = [3 0 0]; d = conv([2 4],[1 2 1]);»[r,p,k] = residue(n,d)r = 6.0000e+00 -4.5000e+00 1.5000e+00p = -2.0000e+00 -1.0000e+00 -1.0000e+00k = []»
2/23/02 page P14.15 © R. A. DeCarlo, P. M. Lin
IC (s) =6
s + 2−
4.5s +1
+1.5
(s +1)2
and
iC ( t) = 6e−2t −92
e−t +32
te−t
u(t) A
SOLUTION 14.20.(a) Make a source transformation:
Vin (s) =1Cs
Iin (s) =250
sIin (s)
By voltage division
Vout (s) =10
250s
+250
s+
120
s +10
250
sIin (s)
=
2500
s120
s2 +10s + 500
Iin (s)
=50,000
s s2 + 200s +10,000( ) Iin (s)
and
H(s) =Vout (s)Iin (s)
=50,000
s(s2 + 200s +10,000)(b) If iin (t) = ( t) implies Iin (s) = 1. Using MATLAB»n = 50e3; d = [1 200 10e3 0];»[r,p,k] = residue(n,d)r = -5 -500 5p = -100 -100 0k = []Hence
Vout (s) =50,000
s(s +100)2 =5s
−5
s +100−
500
(s +100)2
and
vout (t) = 5 − 5e−100t − 500te−100 t( )u( t) V
This is the impulse response
(c) If Lin (t) = 100u(t) mA so that Iin (s) =0.1s
. Therefore
Vout (s) =5000
s2(s +100)2
In MATLAB,
2/23/02 page P14.16 © R. A. DeCarlo, P. M. Lin
»n = 5000; d = conv([1 0 0],[1 200 1e4])d = 1 200 10000 0 0»[r,p,k] = residue(n,d)r = 1.0000e-02 5.0000e-01 -1.0000e-02 5.0000e-01p = -100 -100 0 0k = []Hence
Vout (s) =−0.01
s+
0.5
s2 +0.01
s +100+
0.5
(s +100)2
and
vout (t) = 0.01e−100t + 0.5te−100 t − 0.01 + 0.5t[ ]u( t)
(d) By superposition and time invariance, ifiin (t) = 100 u(t) + u( t −1)[ ] mA
then the result of part (c) can be adjusted to
vout (t) = 0.01e−100t + 0.5te−100 t − 0.01 + 0.5t[ ]u( t)
− 0.01e−100(t−1) + 0.5te−100(t−1) − 0.01 + 0.5(t −1)[ ]u( t −1)
V
SOLUTION 14.21. For this problem change the 20 mH inductor to one of 0.3 H. (a)
Yin = 115
+ 10.3s + 90
+ 10.1s + 10
= ( s + 200) (s + 400)
15( s+ 100) (s + 300)
and
H(s) = IoutIin
= 1/15Yin
= ( s + 100) (s + 300)( s+ 200) (s + 400)
(b) If iin(t) = δ(t), then Iin(s) = 1 and
Iout(s) = H(s) = ( s + 100) (s + 300)( s+ 200) (s + 400)
= 1 - 50s + 200
- 150 s + 400
Henceiout(t) = δ(t) + (- 50 e-200t - 150 e-400 t) u(t) A
(c) We first find the response to iin(t) = 16u(t) mA. Here Iin(s) = 0.016/s and
2/23/02 page P14.17 © R. A. DeCarlo, P. M. Lin
Iout(s) = H(s)Iin(s) = 0.016( s + 100) (s + 300)
s( s+ 200) (s + 400) = 0.006
s + 0.004s + 200
+ 0.006 s + 400
Henceiout(t) = (6 + 4 e-200t + 6 e-400 t)u(t) mA
By linearity and time invariance, the response to iin(t) = 16[ut) – u(t – 0.01)] mA is
iout(t) = (6 + 4 e-200t + 6 e-400 t)u(t) - (6 + 4 e-200(t - 0.01) + 6 e-400 (t - 0.01))u(t - 0.01) mA
A plot of iout(t) using MATLAB is given below.
t= 0: 0.0005: 0.05;f1= (6 + 4*exp(-200*t) + 6*exp(-400*t)).*u(t);f2= (6 + 4*exp(-200*(t-0.01)) + 6*exp(-400*(t-0.01))).*u(t-0.01);iout= f1 - f2;plot(t, iout)gridylabel('iout in mA')xlabel(' time in second')
-10
-5
0
5
10
15
20
0 0.01 0.02 0.03 0.04 0.05
iout
in m
A
time in second
REMARK: Notice that the resistor current is not continuous.
SOLUTION 14.22 . (a) First observe that the admittance of a parallel LC is
YLC(s) = Cs +1Ls
Using voltage division,
2/23/02 page P14.18 © R. A. DeCarlo, P. M. Lin
Vout (s) =C1s +
1
Ls
C1s +1
Ls+ C2s +
1
Ls
Vin (s) =C1s +
1
Ls
C1 + C2( )s +2
Ls
Vin (s) =LC1s2 +1
L C1 + C2( )s2 + 2Vin (s)
Finally
H(s) =Vout (s)Vin (s)
=C1
C1 + C2( )s2 +
1
LC1
s2 +2
L C1 + C2( )= 0.2
s2 + 4 ×106
s2 +1.6 ×106
(b) Using MATLAB,»syms s t»ilaplace(0.2*(s^2+4e6)/(s^2+1.6e6))ans =1/5*Dirac(t)+120*10^(1/2)*sin(400*10^(1/2)*t)»120*10^(1/2)ans = 3.7947e+02
h(t) = 0.2L−1 s2 + 4 ×106
s2 +1.6 ×106
= 0.2 (t) + 379.47sin(1264.9t)u(t) V
SOLUTION 14.23.
Y1(s) = C1s +1R1
=R1C1s +1
R1
Y2(s) = C2s +1
R2=
R2C2s +1R2
Then Z1(s) =R1
R1C1s +1 and Z2(s) =
R2R2C2s +1
. By voltage division,
Vout (s) =
R2
R2C2S +1R1
R1C1s +1+
R2
R2C1s +1
Vin (s) =
R2
R2C2s +1
Vin (s)
R1R2C2s + R1 + R1R2C1s + R2
(R1C1s +1)(R2C2s +1)
=R2(R1C1s +1)
(C1 + C2)R1R2s + R1 + R2Vin (s)
Thus the transfer function is:
H(s) =Vout (s)Vin (s)
=R2(R1C1s +1)
(C1 + C2)R1R2s + R1 + R2
(b) If C1 = 0.5 F , C2 =1.0 F and vin (t) =10 u(t) V, then
2/23/02 page P14.19 © R. A. DeCarlo, P. M. Lin
Vout (s) = H (s)Vin (s) =0.5R1R2s + R2
1.5R1R2s + R1 + R2
10s
Moreover, with R1R2
= 4 so that R1 = 4R2
Vout (s) =2R2
2s + R2
6R22s + 5R2
10s
=
20R22(s +
1
2R2)
6R22 s(s +
5
6R2)
=103
s +1
2R2
s(s +5
6R2)
The partial fraction expansion is
Vout (s) =103
s +1
2R2
s s +5
6R2
=103
K1s
+K2
s +5
6R2
Observe that
K2
s +5
6R2
→L−1
K2e−
5
6R2t
and that it is required that
−5
6R2= −
53
ThusR2 = 0.5 Ω, R1 = 4R1 = 2 Ω
and
Vout (s) =103
s +1
s s +5
3
=103
0.6s
+0.4
s +5
3
=2s
+4 / 3
s +5
3Thus,
vout (t) = 2 +43
e− 5
3t
u(t)
(c) If R1C1 = R2C2 = , then the transfer function is
H(s) =R2(R1C1s +1)
R1R2C1s + R1R2C2s + R1 + R2=
R2 s + R2
R2(R1C1) + R1(R2C2)[ ]s + R1 + R2
= R2( s +1)(R1R2)( s +1)
= R2R2 + R1
The zero-state response is
2/23/02 page P14.20 © R. A. DeCarlo, P. M. Lin
Vout (s) =R2
R2 + R110u(t)
(d) Using H(s) from part (c) with the requirement that R1C1 = R2C2 , then
H(s) =R2
R1 + R2=
110
With R2 = 106 Ω, then 10R2 = R1 + R2 ⇒ R1 = 9R2 = 9 MΩ. Since C2 = 5 ×10−12 F, then
C1 =R2C2
R1= 0.556 ×10−12 F
SOLUTION 14.24.
(a) H(s) =Vout (s)Vin (s)
. Here Ib(s) =Vin (s)2000
. The parallel admittance at the right is
YR(s) = Cs +1R
=RCs +1
Rso that
ZR(s) =R
RCs +1Then
Vout (s) = −ZR(s) Ib(s) = −R
2000(RCs +1)
(b) With V1(s) = Vin (s) −Vout (s) , then
Cs Vout (s) −Vin (s)( ) +1sVout (s) +
12
Vout (s) − 3Vin (s) + 3Vout (s)[ ] = 0
Hence
Cs +32
Vin (s) = Cs +
1s
+ 2
Vout (s)
and
H(s) =Vout (s)Vin (s)
=s Cs +1.5( )Cs2 + 2s +1
=s s + 0.75( )s2 + s + 0.5
(c) Transform the current source iin (t) into a voltage source. In the s-domain with Iin (s) = 2Vin (s)
Here ZC (s) =1
Cs=
12s
which implies YC (s) = 2s . A single node equation yields
23
Vout (s) −Vin (s)[ ]−23
Vout (s) + 2sVout (s) = −23
Vin (s) +23
−23
+ 2s
Vout (s) = 0
2/23/02 page P14.21 © R. A. DeCarlo, P. M. Lin
23
Vin (s) = 2sVout (s) implies Vout (s) =Vin (s)
3s
But Vin (s) =Iin (s)
2 in which case
Vout (s) =Iin (s)
6sand
H(s) =Vout (s)Iin (s)
=16s
SOLUTION 14.25. Here in the s-domain
Iin (s) =Vin (s) −VC (s)
0.5s +10and with a node at VC (s)
−Iin (s) +sVC (s)500
+12
Iin (s) = 0 implies sVC (s)
50=
12
Iin (s) =Vin (s) −VC (s)
s + 20Hence,
s2 + 20s + 500500
VC (s) = Vin (s)
and the transfer function is
H(s) =VC (s)Vin (s)
=500
s2 + 20s + 500
With vin (s) = 4 5 u(t) V which implies Vin (s) =4 5
s, and
VC (s) =2000 5
s(s +10 − j20)(s +10 + j20)=
K1s
+K
s +10 − j20+
K*
s +10 + j20
where K* designates the complex conjugate of K
K1 =2000 5
s2 + 20s + 500 s=0
=2000 5
500= 8.944
K =2000 5
s(s +10 + j20)s=− (10− j20)
=2000 5
(−10 + j20)( j40)=
2000 5
−(800 + j400)= −4.472 + j2236 = Jeij
where = 153.44o . Then with
K* = −4.472 − j2.236 = 5e− j
VC (s) =8.944
s+
A + jBs +10 + j20
+A − jB
s +10 − j20Here
2/23/02 page P14.22 © R. A. DeCarlo, P. M. Lin
A = −4.472, B = 2.236, A2 + B2 = 5and
arc tanBA
= arc tan2.236−4.472
= arc tan1
−2= −153.44o
With the help of Table 13.1.
vC (t) = 8.944 +10e−10t cos(20 t +153.44o)[ ]u(t) V
SOLUTION 14.26. In the s-domain we first find Vx (s) in terms of Vin (s) via voltage division:
Vx (s) =Zp(s)
40 + Zp(s)Vin (s)
where
Zp(s) =(0.40s)(40)0.40s + 40
=40s
s +100Hence
Vx (s) =
40s
s +100
40 +40s
s +100
Vin (s) =40s
80s + 4000=
0.5ss + 50
Vin (s)
and
IL (s) =Vx (s)0.4s
=2.5s
Vx (s)
Then from the right hand side by another voltage division
Vout (s) =10
1000
s+10
0.25Vx (s) =10s
10s +10000.25Vx (s) =
0.25ss +100
Vx (s)
(a) If vin (s) = 20(1− e−40t )u(t), then Vin (s) =20s
−20
s + 40. Hence
Vx (s) =0.5s
s + 50
20s
−20
s + 40
=
ss + 50
10s
−10
s + 40
and
IL (s) =2.5s
Vx (s) =2.5
s + 50
1s
−1
s + 40
=
0.05s
+−0.25s + 40
+0.2
s + 50Hence
iL ( t) = 0.05 + 0.2e−50 t − 0.25e−40t[ ]u( t) A
(b) Vin (s) =20s
−20
s + 40. From, part (a), it was found that
Vout (s) =0.25ss +100
Vx (s) =0.25ss +100
×0.5ss + 50
Vin (s) =s
s +100×
ss + 50
×2.5s
−2.5
s + 40
2/23/02 page P14.23 © R. A. DeCarlo, P. M. Lin
=100s
s + 40( ) s + 50( ) s +100( ) =−20/ 3s + 40( ) +
10s + 50( ) +
−10 /3s +100( )
Thus,
Vout ( t) = 10e−50t −103
e−100 t −203
e−40 t
u(t) V
SOLUTION 14.27. In both parts (a) and (b), the op-ampis ideal. It will not draw current and thevirtual groundprincipal requires that
v + = v− = 0
(a) For a note at the inverting terminal with mode voltage v1( t) = 0 , KCL gives in the s-domain
Vin (s)R1
= −CsVout (s) −Vout (s)
R2 implies
Vin (s)R1
= − Cs +1R2
Vout (s)
in which caseVin (s)
R1= −
R2Cs +1R2
Vout (s)
Then,
H(s) =Vout (s)Vin (s)
= −R2R1
1R2Cs +1
= −
1R1C
1
s +1
R2C
To make
H(s) = −20
s + 4make
1R1C
= 20 and 1
R2C=
14
If
C = 1 F = 10−6F
1R2C
= 4 or R2 =1
4C=
106
4= 250 kΩ
and1
R1C= 20 or R1 =
220C
=10620
= 50 kΩ
(b) With Fig. 14.27b in the s-domain and v1( t) at the inverting terminal, KCL gives
C1sVin (s) +Vin (s)
R1= −C2sVout (s) −
Vout (s)R2
R1C1s +1R1
Vin (s) = −
R2C2s +1R2
Vout (s)
2/23/02 page P14.24 © R. A. DeCarlo, P. M. Lin
Then
H(s) =Vout (s)Vin (s)
= −R2R1
R1C1s +1R2C2s +1
= −
R1R2C1R1RC2
s +1
R1C1
s +1
R2C2
= −C1C2
s +1
R1C1
s +1
R2C2
(c) IfH(s) = −5, C2 =1 F and R2 =1MΩ
thenC1C2
= 5 and C1 = 5 F
The bracketed term must cancel and with
R2C2 = 106(10−6) = 1Then with C1 = 5 F
R1C1 = 1
R1 =1C1
=1
5 ×10−6 = 200kΩ
(d) Using H(s) in part (b)
H(s) = −C1C2
s +1
R1C1
s +1
R2C2
to obtain
H(s) = −5s +1s + 2
with C2 =1 F
C1 = 5C2 = 5 ×106F (5 F)
1R1C1
= 1 or R1 =1C1
=106
5= 200kΩ
1R2C2
= 2 or R2 =1
2C2= −
106
2= 500kΩ
SOLUTION 14.28. Here, the op-amp will not draw current at the non-inverting terminal and theprincipal of the virtual ground demand that
v1 = v2 = Vin (s)For Fig. 14.28 in the s-domain with a node V1(s)taken at the inverting terminal
Vin (s)R1
+Vin (s) −Vout (s)
Zp(s)= 0
Here
2/23/02 page P14.25 © R. A. DeCarlo, P. M. Lin
Zp(s) =
R2
Cs1
Cs+ R2
=R2
R2Cs +1=
1
C(s +1
R2C)
and1R1
+ C(s +1
R2C)
Vin (s) = C(s +
1R2C
)Vout (s)
R1C s +1
R2C
+1
Vin (s) = R1C s +
1R2C
Vout (s)
and
H(s) =Vout (s)Vin (s)
=R1C(s +
1
R2C) +1
R1C(s +1
R2C)
=s +
1
R2C+
1
R1C
s +1
R1C
Here 1
R2C+
1R1C
= 4
and1
R1C= 2
If C = 1 F then R1 =1
2C=
106
2= 500 kΩ
and 1
R2C= 4 − 2 = 2 implies R2 = 500 kΩ .
SOLUTION 14.29. For the non-inverting configuration in the s-domain, each of the two op-amps incascade have a transfer function
H(s) = −Z f (s)
Zin (s)Then for the two op-amps
H0(s) = −Z f 1(s)
Zin1(s)
−
Z f 2(s)
Zin,2
=Z f ,1(s)Z f ,2(s)
Zin,1(s)Zin,2(s)
For Fig. P14.29a in the s-domain
Zin,1 = 25kΩ, Zin,2 = 50kΩ , Z f ,1 =1
C s +1
RC
=250,000
s + 5, and Z f ,2 =
1
C s +1
RC
=250,000s + 2.5
Hence for Fig. P14.29a
2/23/02 page P14.26 © R. A. DeCarlo, P. M. Lin
Ha(s) =
250,000
s + 2.525,000
250,000
s + 5.050,000
=50
s + 2.5( )(s + 5)
If vin (t) = u(t), then Vin (s) =1s
and
Vout (s) =50
s s + 2.5( )(s + 5)=
4s
−8
s + 2.5+
4s + 5
Hence,
vout (t) = 4 − 8e−2.5t + 4e−5t[ ]u(t) V
(b) For Fig. P14.29b in the s-domain, Zin,1, Z f1 , and Z f 2 are in part (a). However,
Zin,2 =1Cs
=250,000
sThus
Hb(s) =
250,000
s + 2.525,000
250,000
s + 5250,000
s
=10
s + 2.5
ss + 5
=
10s
s +5
2
(s + 5)
With Vin (s) =1s
,
Vout (s) =10s
(s + 2.5)(s + 5)
1s
=
20(s + 2.5)(s + 5)
=4
s + 2.5−
4s + 5
and
vout (t) = 4e−2.5t − 4e−5t( )u( t) V
SOLUTION 14.30.(a) -(b). The subcircuit is an integrator, with
Vout(t)V1 (s)
= - 1s
(c) This subcircuit is again an integrator, with
V1(t)V2 (s)
= - 1s
(d) Applying KCL to the inverting input terminal of the top left op amp, we have
G2Vout(s) + V3(s) + V2(s) = 0 or
V2(s) = - G2Vout(s) - V3(s)
2/23/02 page P14.27 © R. A. DeCarlo, P. M. Lin
(e) Applying KCL to the inverting input terminal of the bottom op amp, we have
G3V1(s) + G1 Vin(s) + V3(s) = 0 or
V3(s) = - G3V1(s) - G1 Vin(s)
(f) The results of parts (b), (c) and (d) do not involve Vin. Therefore, , we can solve for V1, V2 and V3in terms of Vout from these threes equations:
V1(s) = - sV out(s)
V2(s) = - sV 1(s) = s2Vout(s) and
V3(s) = - V 2(s) - G2Vout(s) = - s2 + G2 Vout(s)
Substituting these relationships into the result of part (e), we obtain
- sG3Vout(s) + G1 Vin(s) - s2 + G2 Vout(s) = 0
Therefore
H(s) = Vout(s)Vin (s)
= G1
s2 + G3s + G2
SOLUTION 14.31. Use the parallel equivalent circuit model for the capacitor with the standard
directions for voltage and current as given in figure 14.16. For the single node with vC (0− ) = 20 V,
VC (s)10
s
+VC (s)40 +10
−110
vC (0− ) = 0 implies s
10+
150
VC (s) =
110
(20) = 2
Equivalently, (50s +10)VC (s) = (s + 0.2)VC (s) = 20 or VC (s) =20
s + 0.2. Therefore,
vC (t) = 20e−0.2tu(t) V
SOLUTION 14.32. Using the equivalent model for the inductor in figure 14.19, we can compute thetotal admittance as
Y(s) =52s
+140
+110
=52s
+18
=2s + 40
16s=
s + 208s
Using current division,
2/23/02 page P14.28 © R. A. DeCarlo, P. M. Lin
Iout (s) =0.1
s + 20
8s
×−iL (0)
s= −
0.8iL (0−)s + 20
= −1.6
s + 20
Thus
iout ( t) = −1.6e−20t u( t) A
SOLUTION 14.33. Using the equivalent model for the inductor in figure 14.19 and for the capacitorusing figure 14.16, we may combine the current sources to form an equivalent source (with C = 0.1 F) toobtain
Ieq(s) =110
vC (0−) −iL (0−)
s= 0.2 −
1s
Note that
Y(s) = Cs +1LS
=LCs2 +1
Ls= C
s2 +1
LCs
With C = 0.1 F and L = 0.4 H, 1
LC= 25 and
Z(s) =1C
s
s2 + 25
=
10s
s2 + 25Thus
VC (s) = Z(s)Ieq(s) =10s
s2 + 25
s− 55s
=
2(s− 5)
s2 + 25=
2s
s2 + 25−
10
s2 + 25and
vC (t) = 2cos(5t) − 2sin(5 t)( ) u(t) V
SOLUTION 14.34. Consider the equivalent circuit below:
Writing a single node equation we have,
2/23/02 page P14.29 © R. A. DeCarlo, P. M. Lin
0 = 0.5Vout − 0.2VR1 +1
5 + 0.4sVout + 0.8( )
= 0.5Vout + 5 × 0.2 × 15 + 0.4s
Vout + 0.8( ) + 15 + 0.4s
Vout + 0.8( )
= 0.5Vout +2
5 + 0.4sVout + 0.8( )
Therefore
Vout =−8
s + 22.5and
vout (t) = −8e−22.5tu(t) V
SOLUTION 14.35. Redraw the circuit in the s-domain and use an equivalent circuit for the capacitor(figure 14.16) that accounts for the initial condition. By KCL
VC (s)R
+ CsVC (s) = Cv(0−) + Iin (s)
With R = 50 Ω and C = 0.02 F,VC (s)
50+ 0.02sVC (s) = 0.02vC (0−) + Iin (s)
or
VC (s) + sVC (s) = vC (0−) + 50Iin (s)which is equivalent to
(s +1)VC (s) = vC (0−) + 50Iin (s)
(a) With vC (0− ) = 8 V and iin (t) = 40 (t) mA so that Iin (s) = 0.04 ,
(s +1)VC (s) = 8 + 2 implies VC (s) =10s +1
and
vC (t) = 10e−10 tu( t) V
(b) With vC (0− ) = 1 V and iin (t) = 200e−tu(t) mA we have that Iin (s) =0.2s +1
. Thus
VC (s) =vC (0−)(s +1)
+50
(s +1)Iin (s) =
1(s +1)
+10
(s +1)2
and
vC (t) = e−t +10te−t( )u( t) V
SOLUTION 14.36.
2/23/02 page P14.30 © R. A. DeCarlo, P. M. Lin
(a) By current division
H(s) = IL(s) Iin (s)
= 1Ls + 1
(b) The given data in Laplace transforms are:
Iin(s) = I0
s2 and IL (s) = 15
s2 - 3
s + 3
s+ 5 = 75
s2 s+ 5
Under the assumption of zero initial inductor current,
IL (s) = H(s) Iin (s) = H(s) = IL(s) Iin (s)
= I0
Ls + 1 s2 = I0/L
s + 1/L s2 = 75
s2 s+ 5
Equating coefficients, we obtain the answers
L = 1/5 = 0.2 H and I0 = 75L =15 A
(c) The s-domain equivalent is shown below.
Applying KVL to the right mesh, we have
LsIL(s) - L iL(0-) + 1 ×[IL(s) - 1] = 0
Solving for IL(s) from this equation, and equating it to 10/(s+ 5), we have
IL(s) = L iL(0-) + 1L s + 1
= iL(0-) + 1/L s + 1/L
= 10 s + 5
from which L= 0.2 H and iL (0-) = 5 A.
SOLUTION 14.37.
(a) By inspection,
2/23/02 page P14.31 © R. A. DeCarlo, P. M. Lin
H(s) = IL(s) Vin (s)
= 12s + 200
= 0.5s + 100
Given vin(t) = 2u(t) V, then Vin(s) = 2/s , and
IL(s) = 1s s + 100
= 0.01s
- 0.01s + 100
in which case iL(t) = 0.01 ( 1 - e-100t) A
Plots are omitted.
(b) By linearity and time invariance,
iL(t) = 0.01 ( 1 - e-100t)u(t) - 0.01( 1 - e-100(t- 0.05))u(t - 0.05) A
(c) Correction: (a) should be (c). With nonzero initial inducto current, the s-domain equivalentbecomes:
Given iL(0-) = 0.01 A and vin(t) = 2e-200tu(t) V, then Vin(s) = 2/(s + 200) and
IL(s) = 2
s + 200 + 0.02
2s + 200 = 0.01s + 3
( s + 100)( s + 200)
= 0.02 s + 100
- 0.01 s + 200
in which caseiL( t) = 0.02e-100t - 0.01e-200t u(t) A
(d) Correction . (b) should be (d).Given iL(0-) = 0.01 A and vin(t)= 2cos(200t)u(t) V, then
Vin(s) = 2ss2 + 40000
and
2/23/02 page P14.32 © R. A. DeCarlo, P. M. Lin
IL(s) =
2ss2 + 40000
+ 0.02
2s + 200 = 0.01s2 + s + 400
( s + 100)( s2 + 40000)
We use MATLAB to do the partial fraction expansion.
n= [ 0.01 1 400];d= conv([ 1 100], [ 1 0 40000]);[ r p k ] = residue (n,d)r = 0.0010 - 0.0020i 0.0010 + 0.0020i 0.0080p = 1.0e+02 * 0.0000 + 2.0000i 0.0000 - 2.0000i -1.0000
From the above MATLAB output,
IL(s) = 0.008 s + 100
+ 0.001-j0.002 s - j200
+ 0.001+j0.002 s + j200
= 0.008 s + 100
+ 0.002s + 0.8 s2 + 2002
From table 13.1, item 18,
iL( t) = 0.008e-100t + 0.002 cos(200t) + 0.004 sin(200t) u(t) A
2/23/02 page P14.1 © R. A. DeCarlo, P. M. Lin
SOLUTION 14.38. In the s-domain, we break the response up into the part due to the initial conditionand the part due to the source with the initial condition set to zero. The transfer function with the initialcondition set to zero is
H(s) =VC (s)Vin (s)
=1 Cs
R +1 Cs=
1 RCs +1 RC
=0.25
s + 0.25
Using the parallel equivalent circuit for the charged capacitor while setting the source voltage to zero,the capacitor voltage due only the initial condition is:
VC,IC (s) =1
1
R+ Cs
CvC (0− )[ ] =vC (0−)s + 0.25
Hence,
VC (s) =0.25
s + 0.25Vin (s) +
vC (0−)s + 0.25
and
IC (s) =Vin (s) −VC (s)
20= 0.05 1−
0.25s + 0.25
Vin (s) −
0.05vC (0−)s + 0.25
=0.05s
s + 0.25Vin (s) −
0.05vC (0−)s + 0.25
for all inputs and initial conditions.
(a) If vin (t) = 20u(t) and vC (0− ) = 10 V, then Vin (s) =20s
and
VC (s) =5
s(s + 0.25)+
10s + 0.25
=20s
−10
s + 0.25 ⇒ vC (t) = 20 −10e−0.25t( )u( t) V
and
IC (s) =1
s + 0.25−
0.5s + 0.25
=0.5
s + 0.25 ⇒ iC ( t) = 0.5e−0.25tu(t) A
(b) If vin (t) = 5e−0.25tu( t) V, then Vin (s) =5
s + 0.25. Hence,
VC (s) =1.25
(s + 0.25)2 +10
s + 0.25 ⇒ vC (t) = 10 +1.25t( )e−0.25tu(t) V
and
IC (s) =0.25s
(s + 0.25)2 −0.5
s + 0.25=
−0.25s + 0.25
−0.0625
(s + 0.25)2
Hence
iC ( t) = − 0.25 + 0.0625t( )e−0.25tu(t) A
2/23/02 page P14.2 © R. A. DeCarlo, P. M. Lin
0 5 10 15 200
1
2
3
4
5
6
7
8
9
10
Time in s
Inpu
t and
Cap
acito
r vo
ltage
s, V
TextEnd
SOLUTION 14.39. The figure which accounts for the initial conditions is given below.
(a) For the zero-input response, the above circuit reduces to a parallel RLC driven by two currentsources.
Hence VC(s) equals the total current divided by the total admittance, i.e.,
2/23/02 page P14.3 © R. A. DeCarlo, P. M. Lin
VC (s) =CvC (0−) +
iL (0−)
s
Cs +1
R+
1
Ls
=svC (0−) +
iL (0− )
C
s2 +1
RCs +
1
LC
=20s +10
s2 + 250s +104 =26.6
s + 200−
6.6s + 50
Hence
vC(t) = [26.6e–200t
– 6.6e–50t
]u(t) V
(b) For the zero-state response, the current sources disappear. Executing a source transformation on
the remaining voltage source, we obtain a current, I(s) = Vin(s)/(Ls), driving a parallel RLC circuit.Hence, the zero input response is
VC (s) =
Vin (s)
Ls
Cs +1
R+
1
Ls
=1
LC
Vin (s)
s2 +1
RCs +
1
LC
=20000
s3 + 250s2 +104 s=
2s
+0.6667s + 200
−2.6667s + 50
Hence
vC(t) = [2 + 0.6667e–200t
– 2.6667e–50t
]u(t) V(c) By superposition, the complete response is the sum of the answers to (a) and (b). Hence
vC(t) = [2 + 27.267e–200t
– 9.2667e–50t
]u(t) V(d) By linearity and time-invariance,
vC(t) = [2 + 0.6667e–200t
– 2.6667e–50t
]u(t)
+ [4 + 1.3334e–200(t–0.01)
– 5.3334e–50(t–0.01)
]u(t – 0.01) V
2/23/02 page P14.4 © R. A. DeCarlo, P. M. Lin
SOLUTION 14.40. f ( t) = Lin (t) = e−2 tu(t)A [part (c)]
for the zero-input response, f ( t) = Lin (t) =L2(0− )
s [part (d)]
for the zero-input response, f ( t) = Lin (t) = CvC (0)A [part (e)]
(a) Yin (s) =1 +1s
+ s +1=s2 + 2s +1
s=
(s +1)2
s
(b) Iout (s) =Y1(s)Yin (s)
Iin (s) =s +1
s2 + 2s +1
s
=s(s +1)
(s +1)2 =s
s +1Iin (s)
and
H(s) =Iout (s)Iin (s)
=s
s +1
(c) If iin (t) = e−2 tu(t) A, then Iin (s) =1
s + 2, then
Iout (s) = H (s)Iin (s) =s
(s +1)(s + 2)=
−1s +1
+2
s + 2
which implies that the zero-state response is
iout ( t) = 2e−2 t − e−t( )u(t) A
(d) If iL (0−) = 2 A, vC (0− ) = 0, and iin (t) = 0. Using the parallel equivalent circuit for the inductor,figure 14.19, we have
Iout (s) = H (s)i(0−)
s=
ss +1
−
2s
= −
2s +1
⇒ iout ( t) = −2e−tu( t) A
(e) Use the parallel equivalent circuit for the capacitor, figure 14.16, to obtain by current division,
Iout (s) = −
1
s+1
1
s+1 + s +1
[CvC (0−)] = 4s +1
s2 + 2s +1=
4s +1
⇒ iout (t) = 4e−tu(t) A
(f) By superposition, the complete response is the sum of the answers to parts (c), (d) and (e).
SOLUTION 14.41. With vin (t) = 4u(t)V and vC (0− ) = 1 V, a single node equation at the front half of
the circuit yields with CvC (0− ) = 1×1= 1:
2/23/02 page P14.5 © R. A. DeCarlo, P. M. Lin
−245
+ 2 +
4s
+ s
VC1(s) −1= 0 ⇒
s2 + 2s + 4s
VC1(s) =8s
+1 =s + 8
s
or
VC1(s) =s + 8
s2 + 2s + 4
For the rear-half, represent the capacitor by a series equivalent circuit. Thus we can obtain an equivalentvoltage source with value:
Veq(s) = 2VC1(s) −VC (0− )
s=
2(s + 8)
s2 + 2s + 4−
1s
=2s(s + 8)− (s2 + 2s + 4)
s s2 + 2s + 4( )or equivalently
Veq(s) =s2 +14s − 4
s s2 + 2s + 4( )By a voltage division,
Vout (s) =
1
s1
s+ 0.5
Veq(s) +vC (0−)
s=
2s + 2
×s2 +14s− 4
s s2 + 2s + 4( ) +1s
=2s2 + 28s− 8 + s + 2( ) s2 + 2s + 4( )
s s + 2( ) s2 + 2s + 4( ) =s2 + 6s + 36( )
s + 2( ) s2 + 2s + 4( )Using MATLAB»num = [1 6 36];»den = conv([1 2],[1 2 4])den = 1 4 8 8»[r,p,k] = residue(num,den)r = 7.0000e+00 -3.0000e+00 - 2.8868e+00i -3.0000e+00 + 2.8868e+00ip = -2.0000e+00 -1.0000e+00 + 1.7321e+00i -1.0000e+00 - 1.7321e+00ik =
However, it would appear easier here to use ilaplace:»syms t s»ilaplace((s^2+6*s+36)/((s+2)*(s^2+2*s+4)))ans =7*exp(-2*t)-6*exp(-t)*cos(3^(1/2)*t)+10/3*exp(-t)*3^(1/2)*sin(3^(1/2)*t)
2/23/02 page P14.6 © R. A. DeCarlo, P. M. Lin
»Hence
vout (t) = 7e−2t + e−t 3.334 3 sin( 3t( ) − 6cos( 3t)[ ]u(t) V
SOLUTION 14.42. Using the series equivalent circuit (figure 14.17) for C1, we have
I1K(s) = vC1(0-)/s
R + 1sC1
= - 0.25/s1000 + 50/s
= -0.251000s + 50
Next, since vC 2(0−) = 0 , we have
VC 2(s) =I1k (s)C2s
=−0.25 × 500s 1000s + 50( ) =
−0 / 1 2 5s(s + 0.05)
Finally,
Vout(s) = -VC2(s) = 0.125s(s + 0.05)
= 2.5( 1s - 1s + 0.05
)
andvout(t) = 2.5( 1 - e-0.05t )u(t) V
SOLUTION 14.43. (a) It is preferable to use the series equivalent circuit (figure 14.17) for C1, and theparallel equivalent circuit (figure 14.16) for C2.(b) The current through the 2.5 kΩ resistor is given by
I2.5K (s) = vC1(0-)/s
R1 + 1sC1
= - 2/s2500 + 5000/s
= - 22500s + 5000
=-8×10-4
s + 2
Next,
Vout(s) =VC2(s) = I2.5K (s) 1sC2 + 1
R2
= -8×10-4
s + 2 × 1
0.0002s + 0.0002 = - 4
(s + 2) (s + 1)
(c) Hence
Vout(s) = - 4(s + 2) (s + 1)
= - 4 1s + 1
- 1s +2
and
vout (t) = 4 e−2t − e−t( )u(t) V
(d) SPICE plot omitted.
SOLUTION 14.44. (a) From voltage division,
2/23/02 page P14.7 © R. A. DeCarlo, P. M. Lin
H(s) = VC2Vin
= Z2Z1+ Z2
=
R2R2C2s + 1
1C1s
+ R1 + R2R2C2s + 1
= R2C1sR1C1R2C2s2 + ( R1C1 +R2C2 + R2C1)s + 1
(b) If vin(t) = 15u(t) V, then Vin(s) = 15/s and
VC2 = H(s)Vin = 1.75ss2 + 4.25 s + 1
×15s = 7
s+ 0.25 - 7
s+ 4Hence
vC 2( t) = 7 e−0.25t − e−4 t( )u( t) V
0 1 2 3 4 5 6 7 80
5
10
15
Time in s
vin
and
vc2
TextEnd
(c) Using the series equivalent for C1, we have
VC2 = H(s)vC1(0-)
s = 1.75ss2 + 4.25 s + 1
×15s
which is the same as result in part (b). Therefore
vC 2( t) = 7 e−0.25t − e−4 t( )u( t) V
(d) Using the parallel equivalent for C2, we have
2/23/02 page P14.8 © R. A. DeCarlo, P. M. Lin
VC2(s) = C2vC2(0-) 127
s + 47
+ 12 + 1/s
=27
×15× 127
s + 47
+ s2s + 1
= 1s + 0.25
+ 14s + 4
Hence
vC 2( t) = e−0.25t +14e−4 t( )u(t) V
(e) By linearity, the answer is the sum of parts (b), (c) and (d).
SOLUTION 14.45. (a) Z1 = R1 + L1s and Z2 = R2L2sR2 + L2s
. From Ohm's law
IL1 = VinZ1+ Z2
= Vin
R1 + L1s + R2L2sR2 + L2s
= (R2 + L2s)Vin
(R1 + L1s) (R2 + L2s) + + R2L2s
Using current division, we have
IL2 = R2R2 + L2s
IL1= R2Vin(R1 + L1s) (R2 + L2s) + + R2L2s
Therefore
H(s) = IL2Vin
= R2(R1 + L1s) (R2 + L2s) + + R2L2s
= G1
G1L1G2L2s2 + (G1L1+ G2L2 + G1L2)s + 1
With the given element values,
H(s) = 22×1×4
7×7
8s2 + (2×1+ 4
7×7
8 + 2×7
8)s + 1
= 2s2 + 4.25s + 1
(b) If vin(t) = 15u(t) V, then Vin(s) = 15/s and
IL2 = H(s)Vin = 2s2 + 4.25 s + 1
×15s = 30
s - 32s+ 0.25
- 2s+ 4
Hence
iL 2(t) = 30 − 32e−0.25t + e−4 t( )u( t) A
Plot omitted.
(c) Using the series equivalent for L1, we have
IL2 = H(s) L1iL1(0-) = 2s2 + 4.25 s + 1
×1×15 = 8s + 0.25
- 8s +4
Therefore
iL 2(t) = 8 e−0.25t − e−4 t( )u(t) A
2/23/02 page P14.9 © R. A. DeCarlo, P. M. Lin
(d) Using the parallel equivalent for L2, we have
IL2(s) = L2iL2(0-)
L2s + R2(R1 + L1s)
R2 + (R1 + L1s)
= 78
×15
78
s + 1.75(0.5 + s)
1.75 + (0.5 + s)
= 15( s + 2.25)
s2 +4.25s + 1 = 8
s + 0.25 + 7
s + 4
HenceiL2(t) = (8e-0.25t + 7 e-4t) u(t) A
(e) By linearity, the answer is the sum of parts (b), (c) and (d).
SOLUTION 14.46. (a) Using the result of problem 14.44(a)
VC22Iin
= R2C1sR1C1R2C2s2 + ( R1C1 +R2C2 + R2C1)s + 1
= 1.75ss2 + 4.25 s + 1
Therefore
H1(s) = VC2Iin
= = 3.5ss2 + 4.25 s + 1
(b) Using the result of problem 14.45(a),
H2(s) = IL2VC2
= R2(R1 + L1s) (R2 + L2s) + + R2L2s
= 2s2 + 4.25s + 1
(c)
H(s) = IL2Iin
=H1(s)H2(s) = 3.5ss2 + 4.25s + 1
× 2s2 + 4.25s + 1
= 7s(s2 + 4.25s + 1)2
(d) We first represent the initialized capacitor by the series equivalent (figure 14.17), and then apply asource transformation. From this circuit, by utilizing the expression derived in part (c), we have
IL2vC1(0)
2 s
=sIL27.5
H(s) =H 1(s)H2(s) = 7s(s2 + 4.25s + 1)2
ThereforeIL2(s) = 52.5
(s2 + 4.25s + 1)2 = 1.9911
s + 4 + 3.7333
( s + 4)2 - 1.9911
s + 0.25 + 3.7333
( s + 0.25)2
and
iL2(t) = [(1.9911 + 3.7333t) e-4t +(- 1.9911 + 3.7333t) e-0.25t ] u(t) A
Note: the book answer for part (d) should be divided by 2.
(e) Since Iin(s) = 15/s, we have
2/23/02 page P14.10 © R. A. DeCarlo, P. M. Lin
IL2(s) = H(s)Iin(s) = 105
(s2 + 4.25s + 1)2 = 3.9822
s + 4 + 7.4666
( s + 4)2 - 3.9822
s + 0.25 + 7.4666
( s + 0.25)2
and
iL2(t) = [(3.9822 + 7.4666t) e-4t +(- 3.9822 + 7.4666t) e-0.25t ] u(t) A
SOLUTION 14.47. (a) For this passive circuit, we may write the nodal equations by inspection.
0.8s + 2 + 10s - 10
s
- 10s 1 + 10
s
VCVR
= 2Vs1-Is2
(b) Vs1= 3/s and Is2 = 3/s. We solve for VR by Cramer's rule to obtain
VR =
0.8s + 2 + 10s
6s
- 10s - 3s
0.8s + 2 + 10s - 10
s
- 10s 1 + 10
s
= - 2.4s2 - 6s + 30s(0.8s2 + 10s + 30)
= - 4s + 7.5
+ 1s
and
vR( t) = 1− 4e−7.5t( )u(t) V
(c) We represent the initialized capacitor by the parallel equivalent circuit ( figures 14.16) In this casethe nodal equations becomes
0.8s + 2 + 10s - 10
s
- 10s 1 + 10
s
VCVR
=
6s + 2.4
- 3s
Solve for VR by Cramer's rule to obtain
VR =
0.8s + 2 + 10s
6s + 2.4
- 10s - 3s
0.8s + 2 + 10s - 10
s
- 10s 1 + 10
s
= - 2.4s2 +18s + 30s(0.8s2 + 10s + 30)
= 1s - 16
s + 7.5 + 12
s + 5
2/23/02 page P14.11 © R. A. DeCarlo, P. M. Lin
vR(t) = (1 - 16e-7.5t + 12e- 5t) u(t) V
SOLUTION 14.48. (a) After performing the suggested source transformation, and representing theinitialized capacitor and inductor by their series equivalent circuits, we can write two mesh equations byinspection:
0.5 + 1.25s - 1.25
s
- 1.25 s 1 + 0.1s + 1.25
s
Is1IL
= Vs1 -
vC(0)s
vC(0)s + LiL(0) + Is2
(b) With Vs1= 3/s, Is2 = 3/s, vC(0) = 0, and iL(0) = 3 A, the above mesh equation becomes
0.5 + 1.25s - 1.25
s
- 1.25 s 1 + 0.1s + 1.25
s
Is1IL
= Vs1 -
vC(0)s
vC(0)s + LiL(0) + Is2
=
3s
0.3 + 3s
Solve for IL(s) by Cramer's rule to obtain
IL(s) = 0.15s2 +1.875s + 7.50.05s 3 + 0.625s2 + 1.875s
= 2s + 7.5
+ -3s + 5
+ 4 s
Therefore
iL(t) = (4 + 2e-7.5t -3 e-5t) u(t) A
SOLUTION 14.49. (a) Represent the initialized capacitors by their parallel equivalent circuits.(b) Write two nodals equation by inspection
0.001s + 0.4 - 0.2- 0.2 0.001s + 0.4
VC1VC2
= 0.2Vin + 0.001vC1(0-)0.001vC2(0-)
=
2.4s + 0.006
0.002
(c) Solve for VC2(s) by Cramer's rule to obtain
VC2 =
0.001s + 0.4 2.4/s + 0.006- 0.2 0.002
0.001s + 0.4 - 0.2- 0.2 0.001s + 0.4
= 2s2 + 2×103s + 48×104
s(s2 + 8×102s + 12×104) = 0
s + 600 + -2
s + 200 + + 4
s
(d)
2/23/02 page P14.12 © R. A. DeCarlo, P. M. Lin
vC2(t) = (4 - 2e-200t ) u(t) V
SOLUTION 14.50. (a) Let VC denote the node voltage across the capacitor. By inspection the nodalequations in matrix form are:
1+1 /R + 4s −1 /R
−1 /R 1 +1/ R +1 / ( 4s)
VC
Vout
=
Vin
Vin / ( 4s)
(b) By Cramer's rule,
H(s) =Vout (s)Vin (s)
=det
1 +1/ R + 4s 1
−1 /R 1 / ( 4s)
det1+1 /R + 4s −1 /R
−1 /R 1 +1/ R +1 / ( 4s)
=
( 4s +1)(1 +1/ R)
4s (1+1/ R + 4s)(1+ 4s / R + 4s) −1/ R2( )
=(4s +1)(1 +1/ R)
(1 + 8s +16s2)(1+1 /R)=
1(1+ 4s)
Clearly, R does not affect the transfer function. The question is why? Note that the circuit can beredrawn as a balanced Wheatstone bridge circuit in which there is no voltage across R and no currentthrough R. Hence R has no effect on the transfer function and on the impedance at the input. Hence Rcan be removed in the analysis of the circuit. In this case, the transfer function follows trivially byvoltage division.(c) In view of the answer to (b), the impedance can be calculated with R removed. Hence
Zin (s) =1+
1
4s
1+ 4s( )
1+1
4s
+ 1+ 4s( )
=1 + 4s( )2
1 + 4s( )2 =1
Hence, the input impedance is a constant resistance and the network is called a constant resistancenetwork.
(d) The input is vin(t) = 10e–atu(t) V and R = 5 Ω. Find vout(t) for t ≥ 0 for the three cases, a = 0,
0.5, 0.25.(d) From part (b), for s ≠ 0.25,
Vout (s) =0.25
(s + 0.25)×
10s + a
=2 . 5 / ( 0 . 2 5− a)
s + a−
2 . 5 / ( 0 . 2 5− a)(s + 0.25)
which leads to
vout (t) =2.5
0.25 − a
e−at − e−0.25t( )u( t) V
For a = 0.25,
2/23/02 page P14.13 © R. A. DeCarlo, P. M. Lin
Vout (s) =2.5
(s + 0.25)2 implying that vout (t) = 2.5te−0.25tu( t) V
SOLUTION 14.51.(a) This bridged-T circuit was analyzed in problem 14.9. Here R = 1 Ω, Z1(s) = 0.25s and Z2(s) = 4/s.
Since the condition Z1(s) Z2(s) = R2 is met, we have Zin(s)= 1Ω.
(b) The s-domain equivalent circuit accounting for initial conditions is given below.
(c) Two nodal equations at Vc and Vout are:
VC - Vin + VC - Vout + 0.25sVC = 0.25 vC(0-)and
Vout - VC + Vout - Vin
0.25s + Vout = iL(0-)
s
Writing these in matrix form, we have
0.25s + 2 -1
-1 4s
+ 2
VC
Vout
= Vin + 0.25vC(0-)
4Vins
+ iL(0-)s
Solving for Vout by Cramer's rule yields
2/23/02 page P14.14 © R. A. DeCarlo, P. M. Lin
Vout(s) = 4s+4
Vin(s) + 0.5(s+ 8)
(s+4)2 iL(0-) + 0.5s
(s+4)2 vC(0-)
(d) Given vin(t) = 4u(t) - 3e-t u(t) V, then
Vin(s) = 4s - 3s+1
= s + 4s( s + 1)
and
Vout(s) = 4s(s + 1)
+ 0.25(s+ 8)
(s+4)2 + 0.75s
(s+4)2
Taking the inverse Laplace transform, we obtain, for t ≥ 0,
vout(t) = (4 - 4 e-t) + (0.25 e-4t + te-4t) + (0.75 e-4t - 3 te-4t) = 4 - 4 e-t + e-4t - 2te-4t V
SOLUTION 14.52. A supernode is defined by drawing a curve to enclose the controlled voltage source.One node within the supernode has voltage Vout and the other has voltage V1Ω that is equal to
V1Ω = -2I1 - Vout = -2 Vin - VC 2
= - Vin + VC - Vout
Next, we write nodal equations at VC and the supernode: At node VC
0.5 (VC - Vin ) + 0.5sVC + 0.5s (VC- Vout ) = 0
At the supernode
12s
(Vout- VC ) + 12
Vout + - Vin + VC + Vout 1
= 0
In matrix form, the nodal equations are:
0.5(s + 1 + 1/s) -0.5/s1 - 0.5/s 1.5 +0.5/s
VC Vout
= 0.5 Vin Vin
Solving by Cramer's rule yields
H(s) = VoutVin
= 0.5s + 0.75/s0.75s + 1+ 1.5/s
= 2( s2 + 1.5)
3s2 + 4s + 6
SOLUTION 14.53. A supernode is defined by drawing a curve to enclose the controlled voltage source.One node within the supernode has voltage Vout and the other has voltage V1Ω which is equal to
2/23/02 page P14.15 © R. A. DeCarlo, P. M. Lin
V1Ω = -2I1 - Vout = -2 Vin - VC 2
= - Vin + VC - Vout
Next, write nodal equations at VC and the supernode: At node VC
0.5 (VC - Vin ) + (0.25s + 4s )VC
+ ss2 + 16
(VC- Vout ) = 0
At the supernode
ss2 + 16
(Vout- VC) + 12
Vout + - Vin + VC + Vout 1
= 0
In matrix form, the nodal equation are:
0.25s + 0.5 +s/( s2 + 16) -s/(s2 + 16)
1 - s/(s2 + 16) 1.5 + s/(s2 + 16) VC
Vout = 0.5 Vin
Vin
Solving by Cramer's rule yields
H(s) = VoutVin
= 0.25s 3 +5.5s0.375s3 + s2+ 9s + 12
SOLUTION 14.54. Write nodal equations at V1 and V2:
0.5 (V1 - Vin ) + 0.125( V1 + 0.2V2 ) + 0.1s(V1- V2 ) = 0 and
0.1s(V2- V1) + 14
V2 + V21
- 5V1 = 0
In matrix form, the nodal equations are:
(0.1s + 0.625) (-0.1s + 0.025)(-0.1s - 5) (0.1s + 1.25)
VC Vout
= 0.5 Vin 0
Solving by Cramer's rule yields
H(s) = IoutVin
= V2Vin
= 0.05s + 2.5- 0.31s + 0.9062
SOLUTION 14.55.(a) Simply replace each capacitor by the parallel form circuit model given in figure 14.16.(b) For this passive circuit, we can write the nodal equation by inspection.
2/23/02 page P14.16 © R. A. DeCarlo, P. M. Lin
0.5s +2 -1 0-1 0.5s +2 -10 -1 0.5s +2
VC1VC2VC3
= Vin + 0.5vC1(0)
0.5vC2(0)0.5vC3(0)
Solving for VC3 by Cramer's rule yields
VC3 (s) = Vin(s) + 0.5vC1(0) + 0.25s + 1 vC2(0) +( 0.125s2 + s +1.5 )vC3(0)
0.125s3 + 1.5s2 + 5s + 4
(c) Substituting Vin(s) = 12/s, vC1(0) = 0, vC2(0)= 6, and vC3(0) = 2 into the above expression, weobtain
VC3 (s) = 12/s + 6 0.25s + 1 + 2( 0.125s2 + s +1.5 )
0.125s 3 + 1.5s2 + 5s + 4 = 0.25s 3 + 3.5s2 + 9s +12
s 0.125s 3 + 1.5s2 + 5s + 4
Now use MATLAB to do the partial fraction expansion.n= [ 0.25 3.5 9 12];d= [ 0.125 1.5 5 4 0];[ r p k ] = residue (n,d)r = -2.5000 4.0000 -2.5000 3.0000p = -6.8284 -4.0000 -1.1716 0
From the MATLAB output, we have, for t ≥ 0,
vC3(t) = 3 - 2.5 e-6.828t + 4e-4t - 2.5e-1.1716t V
SOLUTION 14.56. For this problem we utilize loop analysis with loops as indicated below.
2/23/02 page P14.17 © R. A. DeCarlo, P. M. Lin
In doing the following loop analysis, note that we will use gmVout = I3 and that due to our judiciouschoice of loops
Vout =1Cs
+ Ls
I1
or equivalently,
0 =1
Cs+ Ls
I1 − Vout
For loop 1,
V1 = 2R +1Cs
+ Ls
I1 + 2R I2 + RgmVout
For loop 2,V1 − V2 = 2R I1 + 7R I2 + 5RgmVout
In Matrix form1000
s+ 0.016s 0 −1
2 +1000
s+ 0.016s 2 2
2 7 10
I1I2
Vout
=0
V1
V1 − V2
By Cramer's rule
2/23/02 page P14.18 © R. A. DeCarlo, P. M. Lin
Vout =
det
1000s
+ 0.016s 0 0
2 +1000
s+ 0.016s 2 V1
2 7 V1 − V2
det
1000
s+ 0.016s 0 −1
2 +1000
s+ 0.016s 2 2
2 7 10
=−
1000
s+ 0.016s
(5V1 + 2V2)
6000
s+ 0.096s − 10 +
7000
s+ 0.112s
Hence
Vout =
1000
s+ 0.016s
(5V1 + 2V2)
10 +1000
s+ 0.016s
=s2 + 62500
s2 + 625s + 62500(5V1 + 2V2)
The answers to (a) and (b) are clear at this point.(c) Using MATLAB»n = [21 0 21*62500];»d = [1 625 62500 0];»[r,p,k] = residue(n,d)r = 35 -35 21p = -500 -125 0k = []Hence
vout(t) = 21− 35e−125t + 35e−500t( )u t( ) V.
SOLUTION 14.57. (a) Replace the LC combination by a 1 V source after setting V1 and V2 to zero.
We need to compute the current leaving the 1 V source which will be 1/Rth. Let the left node be denoted
by Va and the right node by Vb. Also let G = 1/R. The nodal equations are by inspection
2G 0
0 1.25G
Va
Vb
=
gm + G
−gm + 0.25G
⇒
Va
Vb
=
gm + G( ) 2G
−gm + 0.25G( ) 1.25G
2/23/02 page P14.19 © R. A. DeCarlo, P. M. Lin
Thus, the current leaving the 1 V source isI1V = G 1−Va( ) + 0.25G 1− Vb( ) = G 1− gm + G( ) 2G( ) + 0.25G 1− −gm + 0.25G( ) 1.25G( ) = 0.5G − 0.5gm + 0.25G + 0.2gm − 0.05G = 0.7G − 0.3gm
Substituting G = 1/R we obtain
Rth =1
I1V=
10.7G − 0.3gm
=R
0.7 − 0.3gmR=
10.7 − 0.3 × 2
= 10 Ω
(b) Replace the LC combination by a short circuit and compute Isc. This makes the controlled sourcezero. By inspection
Isc =V12R
+V25R
Thus
Voc = RthIsc = RthV12R
+V25R
= 10 0.5V1 + 0.2V2( )
(c) By voltage division
Vout =ZLC
Rth + ZLCVoc =
1
Cs+ Ls
1
Cs+ Ls + Rth
Voc =s2 +1 LC
s2 +Rth
Ls +1 LC
5V1 + 2V2( )
=s2 + 62500
s2 + 625s + 625005V1 + 2V2( )
SOLUTION 14.58. (a) The last equation is the constraint equation for the controlled floating voltagesource. Hence, we have
V1 − V2 − z0(s)I0 = 0(b) By Cramer's rule,
V2 =
det
1
R+ Cs
Iin 1
0 0 −1
1 0 − z0(s)
det
1
R+ Cs
1 R 1
0 Cs −1
1 −1 −z0(s)
=− Iin
−21
R+ Cs
− z0(s)Cs
1
R+ Cs
=Iin
1
R+ Cs
2 + z0(s)Cs( )
(c) Here
2/23/02 page P14.20 © R. A. DeCarlo, P. M. Lin
V2 =Iin
1
R+ Cs
2 + LCs2( )
in which case =2
LC.
SOLUTION 14.59. (a) Since the switch has been at position A for a very long time, the inductor lookslike a short and iL(5-) = iL(5+) = 10/4 = 2.5 A. For t > 5, the switch moves to position B and theinductor current decays according to
iL ( t) = iL (5+ )e−(t−5)/ = 2.5e−(t−5)/0.1 = 2.5e−10(t−5) A
(b) Note that iL(0-) = iL(0+) = 0. Hence
IL (s) =1
10s + 4Vin (s) =
0.1s + 0.4
50s
−50
s + 0.5
=
12.5s
−62.5
s + 0.4+
50s + 0.5
Hence for 0 ≤ t ≤ 5s, iL ( t) = 12.5 − 62.5e−0.4 t + 50e−0.5t A. Here iL(5-) = iL(5+) = 8.1458 A.For t > 5, the inductor decays with a time constant of 0.1 s. Thus
iL ( t) = 8.1458e−10(t−5)
SOLUTION 14.60. (a) Since the switch has been at position A for a very long time, the capacitorlooks like an open and vC(5-) = vC(5+) = 40 V. For t > 5, the switch moves to position B and thecapacitor voltage decays according to
vC (t) = vC (5+ )e−(t−5)/ = 40e−(t−5)/2 = 40e−0.5(t−5) V
(b) Note that vC(0-) = vC(0+) = 0. Hence
VC (s) =1 /Cs
1 /Cs + 40Vin (s) =
12.5s +12.5
50s
−50
s +12.5
=
7812.5
s s +12.5( )2
In MATLAB,»syms t s»ilaplace(7.8125e3/(s*(s+12.5)^2))ans =50-625*t*exp(-25/2*t)-50*exp(-25/2*t)Hence for 0 ≤ t ≤ 5s, vC (t) = 50 − 625te−12.5t − 50e−12.5t V. Here vC(5-) = vC(5+) = 50 V.For t > 5, the capacitor voltage decays with a time constant of 0.08 s. Thus
vC (t) = 50e−12.5(t−5) V
2/23/02 page P14.21 © R. A. DeCarlo, P. M. Lin
SOLUTION 14.61. (a) Since the switch has been closed for a very long time, the capacitor looks likean open and vC(5-) = vC(5+) = 32 V. For t > 5, the switch opens and the capacitor voltage decaysaccording to
vC (t) = vC (5+ )e−(t−5)/ = 32e−(t−5)/0.4 = 32e−2.5(t−5) V
(b) Note that vC(0-) = vC(0+) = 0 and vout = vC. Hence for 0 ≤ t ≤ 5,
VC (s) =1
1
50+
1
200+ 0.002s
Vin (s)50
=10
s +12.5Vin (s) =
10s +12.5
50s
−50
s +12.5
=6250
s(s +12.5)2
In MATLAB,»syms t s»ilaplace(6250/(s*(s+12.5)^2))ans =40-500*t*exp(-25/2*t)-40*exp(-25/2*t)
Hence for 0 ≤ t ≤ 5s, vC (t) = 40 − 500te−12.5t − 40e−12.5t V. Here vC(5-) = vC(5+) = 40 V.For t > 5, the capacitor voltage decays with a time constant of 0.4 s. Thus
vC (t) = 40e−2.5(t−5) V
SOLUTION 14.62. (a) At 0-, vC( 0-) = 0 and iL(0
-) = 50/10 = 5 A.
(b) For this part, consider the equivalent circuit below.
By inspection,
VC =5
s 1+ 0.5s +1
0.184s
=5
0.5s2 + s +1
0.184
=10
s +1( )2 + π2
From table 13.1,
2/23/02 page P14.22 © R. A. DeCarlo, P. M. Lin
vC (t) = 3.1831e−t sin(πt)u( t) V(c) Using MATLAB,
SOLUTION 14.63. (a) Since the switch has been closed for a long time, iL(1-) = iL(1
+) = 30/0.8 = 37.5 A
and vC(1-) = vC(1
+) = 0. Represent the initialized inductor by its parallel equivalent circuit. Then
VC (s) =−37.5
s×
1
Cs +1
Ls
=−150
s2 + 4
Hence from table 13.1,
vC (t ') = −75sin(2 t') V ⇒ vC (t) = −75sin(2( t −1 ) ) V for t > 1s.
(b) All initial conditions at t = 0 are zero. For 0 ≤ t ≤ 1s,
VC (s) =1
1
0.8+ 0.25s +
1
s
Vin (s)0.8
=5s
s2 + 5s + 4Vin (s) =
150s
s2 + 5s + 4
1s
−1
s + 2
=300s
s2 + 5s + 4
1s(s + 2)
=
100s +1
−150s + 2
+50
s + 4Hence, for 0 ≤ t ≤ 1s,
vC (t) = 100e−t −150e−2 t + 50e−4 t V
Here, vC(1-) = vC(1+) = 17.403 V. Note, vin(1-) = 25.94 V. Next,
2/23/02 page P14.23 © R. A. DeCarlo, P. M. Lin
iC (1−) = CdvCdt
t=1
= 0.25 −100e−1 + 300e−2 − 200e−4[ ] = 0.037378
Thus in MATLAB»vin1 = 30*(1 - exp(-2))vin1 = 2.5940e+01»vc1 = 17.403vc1 = 1.7403e+01»ic1 = 0.25*(-100*exp(-1) + 300*exp(-2)-200*exp(-4))ic1 = 3.7378e-02»iL1 = (vin1 - vc1)/0.8 - ic1iL1 = 1.0634e+01
Therefore, iL (1−) = iL (1+ ) =10.634 A. For t ≥ 1, we use the parallel equivalent circuit for both theinductor and the capacitor:
VC (s) = e−s 4s
s2 + 4
−iL (1+ )s
+ CvC (1+ )
= e−s 4s
s2 + 4
−10.634s
+ 4.3507
Therefore from table 13.1, for t > 1,
vC (t) = −21.268sin 2(t −1)( ) +17.403cos 2(t −1)( ) V
Plots omitted.
SOLUTION 14.64. Here vC(0) = 0 for both capacitors.Part 1: 0 ≤ t ≤ 1s.
Vout (s) =
10
s
20 +10
s
20s
−20
s + 2
=
20s(s + 0.5)(s + 2)
From MATLAB»num = 20;»den = [1 2.5 1 0];»[r,p,k] = residue(num,den)r = 6.6667e+00 -2.6667e+01 2.0000e+01p = -2.0000e+00 -5.0000e-01 0k = []
2/23/02 page P14.24 © R. A. DeCarlo, P. M. Lin
Therefore for 0 ≤ t ≤ 1,
vout (t) = 20 − 26.667e−0.5t + 6.667e−2t( ) u( t) − u( t −1)( )
Part 2. 1 ≤ t. Here the initial condition for the right-most capacitor is vout (1− ) = vout (1 + ) = 4.7281 V.As above, the left-most capacitor has zero value at t = 1s. Let us use the series equivalent circuit for theright capacitor. Then,
IC (s) = e−s −4.7281s
×1
10 +20
s
= e−s −0.47281s + 2
Therefore,
Vout (s) =10s
IC (s) + e−s 4.7281s
= e−s −4.7281
s(s + 2)+
4.7281s
= e−s 2.3641
s+
2.3641s + 2
and for t ≥ 1,
vout (t) = 2.3641+ 2.3641e−2(t−1)( )u(t −1) V
SOLUTION 14.65. Assume the switch has been in position A for a long time. Both capacitors behave asopen circuits and both capacitors have initial voltages at t = 0 of 10 V. For t ≥ 0, use the parallelequivalent circuits for both capacitors and write nodal equations. Let the left capacitor have voltageVCa.
0.005s + 0.03 −0.01
−0.01 0.0025s + 0.01
VCa
Vout
=
0.005 ×10
0.0025 ×10
=
0.05
0.025
By Cramer's rule,
Vout (s) =det
0.005s + 0.03 0.05
−0.01 0.025
det0.005s + 0.03 −0.01
−0.01 0.0025s + 0.01
= 10s +10
s2 +10s +16=
4 0 /3s + 2
−10 / 3s + 8
Therefore for t ≥ 0,
vout (t) =403
e−2t −103
e−8 t V
SOLUTION 14.66. (a) Zin (s) = 2 +1
0.5s +2
s
= 2 +2s
s2 + 4.
(b) Here, the initial condition is zero and
2/23/02 page P14.25 © R. A. DeCarlo, P. M. Lin
VC (s) =
2s
s2 + 4
2 +2s
s2 + 4
×10s
=10
s2 + s + 4
»syms t s»ilaplace(10/(s^2+s+4))ans =4/3*exp(-1/2*t)*15^(1/2)*sin(1/2*15^(1/2)*t)Hence using MATLAB above or table 13.1 we have for 0 ≥ t ≤ 1.5s,
vC (t) = 5.164e−0.5t sin(1.9365t) Vand
vC (1.5− ) = vC (1.5 + ) = 0.57237 V(c) Use the parallel equivalent circuit for the left capacitor. The right capacitor has a zero initial voltageat t = 1.5. Hence, we do not use an equivalent circuit for the right capacitor.
(d) Therefore
e1.5sVC (s) =1
0.5s + 2 +s
s + 4
× 0.5 × 0.57237 = 0.57237s + 4
s2 +10s +16
(e) In MATLAB»[r,p,k] = residue(0.57237*[1 4],[1 10 16])r =V 3.8158e-01 1.9079e-01p = -8 -2k = []
Hence
VC (s) = e−1.5s0.57237s + 4
s2 +10s +16= e−1.5s 0.19079
s + 2+
0.38158s + 8
(f) Finally
vC (t) = 0.19079e−2(t−1.5) + 0.38158e−8(t−1.5)[ ]u(t −1.5) V
SOLUTION 14.67. (a) v1(0−) = v1(0+ ) = v2(0− ) = v2(0+ ) =R2R
16 = 8 V.
(b) For 0 ≤ t ≤ 1, v1( t) = 8 V and v2(t) = 8e−t /RC = 8e−0.6931t V.
(c) v1(1−) = 8 V and v2(1− ) = 8e−0.6931 = 4 V.
2/23/02 page P14.26 © R. A. DeCarlo, P. M. Lin
(d) From KVL, v1(1+ ) = v2(1 + ). From conservation of charge,
1× v1(0−) +1× v2(0−) =12 = 2 × v1(0+ ) = 2 × v2(0+ ) . Therefore v1(1+ ) = v2(1 + ) = 6 V.(e) This response represents a decay with time constant τ = 2R = 2.8854 s. Hence
v1( t) = v2(t) = 6e−0.34657( t−1)u(t −1) VIt follows that
v1(3) = v2(3) = 6e−0.34657×2 = 3 V
(f) Both capacitor voltages change abruptly at t = 1.
SOLUTION 14.68. Label the current down through the first inductor as i1(t).
(a) i1(0−) = i1(0+ ) =1 A a n d iout (0− ) = iout (0+ ) = 0 .(b) For 0 ≤ t we use a parallel equivalent for the first inductor. By current division
Iout (s) =
1
5 + 0.1s
0.057143 +1
0.35s+
1
5 + 0.1s
×−1s
=−175
s2 + 275s + 2500
Use MATLAB to do the partial fraction expansion
num = -175; den = [ 1 275 2500]; [ r, p, k] = residue (num, den)
r = 0.6831 -0.6831p = -265.5869 -9.4131
From the MATLAB output
Iout (s) = - 0.68313s+ 9.4131
+ 0.68313s+ 265.59
Therefore ,
iout ( t) = 0.68313 e−265.59t − e−9.413t( )u(t) A
SOLUTION 14.69. (a) Here we use voltage division:
2/23/02 page P14.27 © R. A. DeCarlo, P. M. Lin
V1(s) =
1
4 ×10−6s1
4 ×10−6s+
1
4 ×10−6 s+
1
1×10−6s
×30s
=306s
=5s
Therefore, v1(0+ ) = 5 V.(b) Again use voltage division:
V1(s) =
1
5 ×10−6 s1
5 ×10−6s+
1
1×10−6 s+
1
2 ×10−6s
×40s
=8017s
Therefore, v1(0+ ) = 4.7059 V.
SOLUTION 14.70. (a) Consider a mesh current I(s) in the usual direction and use the series equivalentcircuit for each capacitor. Thus
I(s) =1
1
4 ×10−6 s+
1
4 ×10−6s+
1
1×10−6 s
3s
−0.3s
−0.9s
−0.6s
=
1.21.5
×10−6 = 0.8 ×10−6
Therefore for t > 0,
V1(s) =1
4 ×10−6sI(s) +
0.3s
=0.8 ×10−6
4 ×10−6s+
0.3s
=0.5s
⇒ v1( t) = 0.5 V
Similarly for t > 0
V2(s) =0.8 ×10−6
4 ×10−6s+
0.9s
=1.1s
⇒ v2(t) =1.1 V
and
V3(s) =0.8 ×10−6
1×10−6s+
0.6s
=1.4
s ⇒ v3( t) = 1.4 V
(b) Consider a mesh current I(s) in the usual direction and use the series equivalent circuit for eachcapacitor. Thus
I(s) =1
1
5 ×10−6 s+
1
1×10−6s+
1
2 ×10−6 s
4s
−0.3s
−0.9s
−0.6s
=
2.21.7
×10−6 =1.2941×10−6
Therefore for t > 0,
V1(s) =1
5 ×10−6sI(s) +
0.3s
=1.2941×10−6
5 ×10−6s+
0.3s
=0.55882
s ⇒ v1( t) = 0.55882 V
Similarly for t > 0
2/23/02 page P14.28 © R. A. DeCarlo, P. M. Lin
V2(s) =1.2941×10−6
1×10−6 s+
0.9s
=2.1941
s ⇒ v2(t) = 2.1941 V
and
V3(s) =1.2941×10−6
2 ×10−6s+
0.6s
=1.2471
s ⇒ v3(t) =1.2471 V
SOLUTION 14.71. (a) For 0 < t < 2, the 150 mF capacitor is charged to 25 V. From conservation ofcharge,
0.15 × 25 = 0.15vC (2+ ) + 0.1vC (2+ ) = 0.25vC (2+ )Therefore
vC (2+ ) =0.15 × 25
0.25=15 V
This voltage remains constant for t > 2 s.(b) For 0 < t < 2, the 150 mF capacitor is charged to 25 V. From conservation of charge,
0.15 × 25 + 0.1×10 = 0.15vC (2+ ) + 0.1vC (2+ ) = 0.25vC (2+ )Therefore
vC (2+ ) =0.15 × 25 + 0.1×10
0.25= 19 V
This voltage remains constant for t > 2 s.
SOLUTION 14.72. (a) Let the middle node have voltage Va(s). Then writing node equations
8s −2s
−2s 6s
Va
Vout
=
1.144
0
⇒
Va
Vout
=
144
6 2
2 8
1.144 s
0
=
0.156 / s
0.052 / s
V
Thus for t > 0, vout (t) = 0.052 V.
(b) Again define Va(s) as the middle node. Then
Va(s) =
3
10s1
10s+
1
4s+
3
10s
×0.286
s=
613
×0.286
s=
0.132s
Hence for t > 0,
Vout (s) =
1
4s1
2s+
1
4s
×0.132
s=
0.1323s
=0.044
s ⇒ vout (t) = 0.044 V
2/23/02 page P14.29 © R. A. DeCarlo, P. M. Lin
SOLUTION 14.73. With switches in position A, the equivalent capacitance to the right of v2 is 4 mF.Therefore at t = 0+, by voltage division
V1(s) = V2(s) =10s
⇒ v1( t) = v2( t) =10 V for 0 < t < 1.
Hence with the switches in position B, let us write a single node equation using the parallel equivalentcircuit the initialized capacitors:
0.002sV2(s) − 0.002 ×10 + 0.004sV2(s) + 0.004 ×10 = 0.004s×11s
Equivalently0.006sV2(s) = 0.044 − 0.02 = 0.024 ⇒ V2(s) = 4 / s
Hence, for t > 1 s, v2(t) = 4 V and v1( t) = 7 V.
SOLUTION 14.74. (a) At t = 0+, the frequency domain equivalent circuit is given below.
To compute Ceq, we observe
Ceq = 1+1
1
3+
1
2+
1
6
= 2 mF
By voltage division
V1 =Ceqs
Ceqs + 0.002s×
20s
=10s
and V2 =0.002s
Ceqs + 0.002s×
20s
=10s
Hence, for 0 < t <1, v1(t) = v2(t) = 10 V.(b) When the switch moves to B, the pertinent part of the equivalent frequency domain circuit is givenbelow.
2/23/02 page P14.30 © R. A. DeCarlo, P. M. Lin
By superposition,
V1 =1
1 + 2×
8s
+1
0.003s× 0.02 −
10.003s
× 0.01=6s
Hence, for 1 < t < 2, v1(t) = 6 V and by KVL, v2(t) = 2 V.
SOLUTION 14.75. When the switch in position A, the 2 µF capacitor is charged to –2 V. Hence, the
charge on the top plate is C*vC = –4 µC. When the switch is moved to position B, due to the virtual
ground, the 2 µF capacitor voltage is zero meaning it cannot retain any charge. Hence, assuming an
ideal op amp, all charge moves to the 1 µF capacitor with –4 µC on the left plate. Hence, vout = −(–4
µC)/1 µF = 4 V.
SOLUTION 14.76. (a) vout(t) = 0 for 0 < t < 1 ms. Every time the switch moves to position A, thecapacitor, C, charges to 8 V. When the switch moves to position B, because of the virtual ground, allcharges moves to the kC capacitor. Hence with k = 1, vout(t) = – 8 V for 1 ms < t < 3 ms. For 3 ms < t <
5 ms, vout(t) = – 16 V. Repeating the pattern implies that for 5 ms < t < 7 ms, vout(t) = – 24 V, etc. Seefor example figure 14.51.(b) With k = 0.5, the voltages computed in part (a) double.(c) With k = 2, the voltages computed in part (a) are halved.
SOLUTION 14.77.(a) For this part consider the circuit below.
2/23/02 page P14.31 © R. A. DeCarlo, P. M. Lin
For the normalized design C = 1 F, G1 = 0.5 S, G2 = 2 S, and G3 = 1.5 S. After magnitude scaling with
Km = 106, then C = 1 F , R1 = 2 MΩ, R2 = 5 0 0 kΩ, R3 = 6 6 6 . 7 kΩ.
(b) For this part, consider the circuit below.
For the normalized design C = 1 F, G1 = 1 S, G3 = 2 S, G2 = 0.5 S, and G4 = 2.5 S. After magnitude
scaling with Km = 106, then C = 1 F , R1 =1 MΩ, R3 = 5 0 0 kΩ, R2 = 2 MΩ, R4 = 4 0 0 kΩ.
(c) Now consider the circuit below.
2/23/02 page P14.32 © R. A. DeCarlo, P. M. Lin
For the normalized design C = 1 F, G1 = 0.25 S, G3 = 0.5 S, G2 = 0.75 S, G4 = 1 S, and G5 = 1 S. After
magnitude scaling with Km = 106, then C = 1 F , R1 = 4 MΩ, R3 = 2 MΩ, R2 = 4 / 3 MΩ,R4 =1 MΩ, R5 = 1 MΩ.
(d) Finally, we consider the following circuit.
For the normalized design C = 1 F, G1 = 2 S, G3 = 1.5 S, G6 = 1 S, G2 = 0.5 S, G4 = 2 S, and G5 = 2 S.
After magnitude scaling with Km = 106, then C = 1 F , R1 = 0 . 5 MΩ, R3 = 2 /3 MΩ ,R6 = 1 MΩ, R2 = 2 MΩ , R4 = 0 . 5 MΩ, R5 = 0 . 5 MΩ.
2/23/02 page P14.33 © R. A. DeCarlo, P. M. Lin
SOLUTION 14.78. With V1 = Vout a prototype design is given by the topology below.
Using MATLAB»Km = 1e7;»Gin = 1; G1 = 1.405; G2 = 0.402; G3 = 0.942;»DG = 1.865;»Rinnew = Km/GinRinnew = 10000000»R1new = Km/G1R1new = 7.1174e+06»R2new = Km/G2R2new = 2.4876e+07»R3new = Km/G3R3new = 1.0616e+07»DRnew = Km/DGDRnew = 5.3619e+06
Later, when we study frequency scaling, Km will be smaller and the filter will have a cutoff frequency ina more reasonable range.
SOLUTION TO 14.79. Note corrections to problem statement. W(0+) should be W(∞) in part (b) and inpart (c) one should calculate W(0-) – W(∞). The frequency domain equivalent circuit is given by thefigure below.
2/23/02 page P14.34 © R. A. DeCarlo, P. M. Lin
(a) To find the required time functions, we first find their Laplace equivalents.
I(s) =
a
s−
b
s
R +1
C1s+
1
C2s
=
a − b
R
s +1
RC1+
1
RC2
≡K1
s + p1
in which case i( t) = K1e− p1tu(t) where
K1 =a − b
R and p1 =
1RC1
+1
RC2
.
Further,
VC1(s) =− I(s)C1s
+as
=as
−
K1
C1s s + p1( ) = a −
K1C1p1
×
1s
+K1
C1p1×
1s + p1( )
in which case
vC1(t) = a −K1
C1p1
+
K1C1p1
e− p1t
u( t)
Also, by symmetry,
VC 2(s) =I(s)C2s
+bs
=bs
+
K1
C2s s + p1( ) = b +
K1C2 p1
×
1s
−K1
C2p1×
1s + p1( )
in which case
vC 2( t) = b +K1
C2 p1
−
K1C2 p1
e− p1t
u( t)
(b) The total energy stored in the capacitors at time 0- is
W (0− ) =12
C1a2 +12
C2b2
Also at t = ∞,
vC1(∞) = a −K1
C1p1
and vC 2(∞) = b +
K1C2p1
.
Hence
2/23/02 page P14.35 © R. A. DeCarlo, P. M. Lin
W (∞) =12
C1 a −K1
C1p1
2
+12
C2 b +K1
C2 p1
2
(c)
Ri2( t)dt0
∞
∫ = RK12e−2 p1tdt
0
∞
∫ =RK1
2
2p1=
(a − b)2
2Rp1=
(a − b)2
21
C1+
1
C2
Observe that
W (0− ) −W (∞) =12
C1a2 +12
C2b2 −12
C1 a −K1
C1p1
2
−12
C2 b +K1
C2 p1
2
=aK1p1
−bK1p1
−12
K12
C1p12 −
12
K12
C2p12 =
(a − b)2
Rp1−
12
×(a − b)2
Rp1=
(a − b)2
2Rp1=
(a − b)2
21
C1+
1
C2
This indicates that the total energy lost between 0- and infinity is the energy dissipated in the resistorand the result is independent of the value of R.(d) When R → 0,
limR→0
I(s) =
a
s−
b
s1
C1s+
1
C2s
=a − b
1
C1+
1
C2
=C1C2
(C1 +C2)(a − b)
Therefore
i( t) =C1C2
(C1 +C2)(a − b) (t)
Further
VC1(s) =−C2
s(C1 +C2)(a − b) +
as
=C1
s(C1 +C2)a +
C2s(C1 +C2)
b
and
VC 2(s) =C1
s(C1 +C2)(a − b) +
bs
=C1
s(C1 +C2)a +
C2s(C1 +C2)
b
Therefore
vC1(t) = vC 2(t) =C1
(C1 +C2)a +
C2(C1 +C2)
b
u(t)
SOLUTION TO 14.80. (a) From conservation of charge
C1vC1(0−) + C2vC 2(0− ) = C1vC1(0+ ) + C2vC 2(0+ ) = C1 + C2( )vC1(0+) = C1 + C2( )vC 2(0+ )Therefore
2/23/02 page P14.36 © R. A. DeCarlo, P. M. Lin
vC1(0+ ) = vC2(0+ ) =C1vC1(0− ) + C2vC 2(0− )
C1 + C2( )(b) Inserting values into out answer for part (a) yields
vC1(0+ ) = vC2(0+ ) =C1
C1 + C2( ) = 0.5 V
and the voltage remains the same for t > 0.(c) Before the switch is closed, the energy in C2 is zero and the energy in C1 is the total stored energy:
Wtot (0−) = WC1(0− ) = 0.5C1vC1
2 (0− ) = 0.5 JAfter the switch is closed,
Wtot (0+ ) = WC1(0+ ) + WC 2(0+ ) = 0.5C1vC1
2 (0+ ) + 0.5C2vC 22 (0+ ) = 0.25 J
(d) (i) Using the series equivalent circuit for C1, we have
I(s) =1
R +2
s
×1s
=1 /R
s + 2 / R ⇒ i( t) =
1R
e−2 t Ru(t) A
Thus
VC 2(s) =I(s)
s=
1/ Rs(s + 2 / R)
=0.5s
−0.5
s + 2 / R ⇒ vC 2( t) = 0.5 1− e−2 t R( )u(t) V
and
VC1(s) = −I(s)s
+1s
=−1/ R
s(s + 2 / R)+
1s
=0.5s
+0.5
s + 2 / R ⇒ vC1(t) = 0 . 5 1+ e−2 t R( )u(t) V
(ii) The energy dissipated in the resistor is given by
WR(0,∞) = R i2( )d0
∞
∫ =1R
e−4 t Rd0
∞
∫ = −e−4 t R
4
0
∞
=14
J
(iii) For all R,
Area under i(t) = i( )d0
∞
∫ =1R
e−2 t Rd0
∞
∫ = −e−2 t R
2
0
∞
=12
Further, as R → 0, i( t) =1R
e−2 t Ru(t) has a decay that becomes infinitely fast and its magnitude (1/R)
approaches ∞. Thus we have infinite height, zero-width, but a finite area of 0.5. Thus as R → 0,i( t) → 0.5 (t) A. (We have avoided a more rigorous explanation as the above argument is moreplausible to sophomores.). As R → 0, the exponential terms in the expressions for vC1(t) and vC 2( t)have infinitely fast decays and hence disappear from the expressions yielding the stated result.
2/23/02 page P14.37 © R. A. DeCarlo, P. M. Lin
SOLUTION TO 14.81.(a)
Zin (s) = 2s + 4.5(s+ 0.5)(s + 4)
= 1s+ 0.5
+ 1 s + 4
= Za (s) + Zb (s)
Ya (s) = 1Za (s)
= s+ 0.5
Yb (s) = 1Zb (s)
= s + 4
From the above expressions, the RC circuit consists of a series connection of (a 1 farad capacitor inparallel with a 2 Ω resistor) and (a 1 F capacitor in parallel with a 0.25 Ω resistor).
(b)Yin (s) = 12s + 440
(s+ 120)(s + 20) = 10
s+ 120 + 2
s + 20 = Ya (s) + Yb (s)
Za (s) = 1Ya (s)
= 0.1s + 12
Zb (s) = 1Yb (s)
= 0.5s + 10
From the above expressions, the RL circuit consists of a parallel connection of (0.1 H inductor in serieswith a 12 Ω resistor) and (a 0.5 H inductor in series with a 10 Ω resistor).
(c) Following the hint, we have
Yin (s)s = 0.225s +0.075
(s+ 0.2)(s + 0.5) = 0.1
s+ 0.2 + 0.125
s + 0.5
Hence
Yin (s) = 0.1ss+ 0.2
+ 0.125s s + 0.5
= Ya (s) + Yb (s)
Za (s) = 1Ya (s)
= s + 0.20.1s
= 10 + 10.5s
Zb (s) = 1Yb (s)
= s + 0.50.125s
= 8 + 10.25s
From the above expressions, we see that each term in Yin (s) represents a series RC circuit. The RCcircuit for Yin(s) consists of a parallel connection of (a 0.5 F capacitor in series with a 10 Ω resistor)and (a 0.25 F capacitor in series with a 8 Ω resistor).
(d) CORRECTION: for part (d),change the second term to 2s/(s2 + 2).
2/23/02 page P14.38 © R. A. DeCarlo, P. M. Lin
Yin (s) = 0.5ss2+ 1
+ 2s s2+ 2
= Ya (s) + Yb (s)
Za (s) = 1Ya (s)
= s2 + 1 0.5s
= 2s + 1 0.5s
Zb (s) = 1Yb (s)
= s2 +22s
= 0.5s + 1s
From the above expressions, we see that each term in Yin (s) represents a series LC circuit. The LCcircuit for Yin(s) consists of a parallel connection of (a 0.5 F capacitor in series with a 2 H inductor)and (a 1 F capacitor in series with a 0.5 H inductor).
SOLUTION 14.82. CORRECTIONS TO PROBLEM STATEMENT: (i) v0(t), should read vout(t)and (ii) there should be a connection from the circuit inside the shaded box to the bottom line orreference node.(a) (i) 0 ≤ t < 1 ms. Since the capacitor voltage is initially zero and the switch is in position (a), asimple source transformation yields a Norton equivalent (seen by the capacitor) consisting of a 20 mAcurrent source in parallel with 9.8039 kΩ resistor. Hence
Vout (s) =1
1
98039+10−6s
×.02s
=20 ×103
s s +102( )
Using MATLABn = .02*1e6;p1 = 1e6/Rthp1 = 102»d = [1 p1 0];»[r,p,k] = residue(n,d)r = -1.9608e+02 1.9608e+02p = -102 0k = []
Hence, for 0 ≤ t < 1 ms, vout (t) = 196.08 1− e−102t( )u( t) V. It follows that vout (1 ms) = 19.014 V.
(ii) 1 ms ≤ t < 1.05 ms. The frequency domain equivalent circuit is given below.
2/23/02 page P14.39 © R. A. DeCarlo, P. M. Lin
Writing a single node equation yields
19.014 ×10−6 =Vout −
200
s10000
+Vout −
1
s10
+10−6 sVout
Equivalently 19.014 +1.2 ×105
s= s +100100( )Vout or
ˆ V out =19.014s +1.2 ×105
s s +100100( )Therefore, ˆ v out (t ') = 1.1988 +17.815e−100100t'( )u(t ') , and for 1 ms ≤ t < 1.05 ms,
vout (t) = ˆ v out (t − 0.001) in which case
vout (t) = 1.1988 +17.815e−100100(t−0.001)( )u( t − 0.001)
Again using MATLAB,
(b) Part 1: for t > 0 up to t1 which denotes the time when vout(t) reaches 80 V, i.e., the capacitor ischarging. The frequency domain equivalent circuit is
2/23/02 page P14.40 © R. A. DeCarlo, P. M. Lin
Using our knowledge of part (a), this circuit simplifies to
Hence
Vout (s) =5 ×10−6 +
0.02
s
10−6 s +1
9803.9
=5s + 20 ×103
s s +102( )
and from MATLAB»syms s t»ilaplace((5*s+20e3)/(s^2+102*s))ans =10000/51-9745/51*exp(-102*t)
in which case
vout (t) = 196.08 −191.08e−102t( )u( t) V
From this expression,
vout (t1) = 80 = 196.08 −191.08e−102 t1 Vandt1 = log((80-10000/51)/(-9745/51))/(-102)t1 = 4.8864e-03
This part of the problem considers t1 ≤ t < t2, i.e., the capacitor is discharging where vout(t2) = 5.The equivalent frequency domain circuit is given below which is a slight modification of the circuit of(a)-(ii):
2/23/02 page P14.41 © R. A. DeCarlo, P. M. Lin
Making use of our knowledge of part (a)-(ii), we have
ˆ V out =80s +1.2 ×105
s s +100100( )
in which case ˆ v out (t ') = 1.1988 + 78.801e−100100t '( )u( t') V, and
vout (t) = 1.1988 + 78.801e−100100(t−t1)( )u( t − t1)
Here t2' = 3.0286 ×10−5s and t2 = t1 + t2
' = 4.9167 ms where t2' is the duration of the discharge cycle.
As a final point, note that the frequency of the sawtooth is 1/t2 = 203.39 Hz. Finally a plot is givenbelow.
2/23/02 page P14.42 © R. A. DeCarlo, P. M. Lin
SOLUTION 14.83. CORRECTION: In example 14.10, page 560, delete the four minus signs in the
equation for VC(s) and one more for vC(t).
We use MATLAB instead of SPICE to solve this problem. Applying voltage division to the circuit offigure P14.83, we have
VC (s) = 1
Cs1
Cs + Ls + R
LiL(0-) = iL(0-)
C 1s2 + R
Ls + 1
LC
= 108
s2 + 125s + 1.25 ×109
From table 13.1, item 18
vC (t) = 2828e-62.5t sin(35,355t) u(t) V
A plot of vC(t) is given below with the vertical axis in V and the horizontal axis in seconds.
-3000
-2000
-1000
0
1000
2000
3000
0 0.2 0.4 0.6 0.8 1
x10-3
The waveform for the first few cycles is essentially the same as the example 14.10. Thus for the firs fewcycles, the lossless circuit of example 14.10 is a good approximation to the more accurate circuit modelof this problem. The effect of the presence of 100 Ω resistance is a slow decay (with respect to msintervals) of the peak values.
1/25/02 P16-1 © R. A. DeCarlo, P. M. Lin
PROBLEM SOLUTIONS
Solution 16.1.(a) By the definition of the convolution integral
f2( t) ∗ f2( t) = f2(t − ) f2( )d =−∞
∞
∫ 2u( t − )2u( )d = 4 u( t − )d0
∞
∫−∞
∞
∫The integrand, u( t − ) , is nonzero only when ≤ t. This suggests that there are two regions ofconsideration: t < 0 and t ≥ 0.Case 1: t < 0. Here u( t − ) = 0 since is restricted to the interval [0,∞). Hence
f2( t) ∗ f2( t) = 0 , for t < 0.Case 2: t ≥ 0.
f2( t) ∗ f2( t) = 4 u(t − )d = 4 d0
t
∫−∞
∞
∫ = 4t , for t ≥ 0.
In sum,
f2( t) ∗ f2( t) =0, t < 0
4t, t ≥ 0
(b) By the definition of the convolution integral
f2( t) ∗ f3(t) = f2(t − ) f3( )d =−∞
∞
∫ 2u( t − )4e−2 u( )d = 8 e−2 u( t − )d0
∞
∫−∞
∞
∫
The integrand, u( t − ) , is nonzero only when ≤ t. This suggests that there are two cases toconsider: t < 0 and t ≥ 0.Case 1: t < 0. Here u( t − ) = 0 since is restricted to the interval [0,∞). Hence
f2( t) ∗ f3(t) = 0, for t < 0.
Case 2: t ≥ 0.
f2( t) ∗ f3(t) = 8 e−2 d = −4e−2 ]0
t
0
t
∫ = 4(1− e−2 t )
In sum,
f2( t) ∗ f2( t) =0, t < 0
4(1− e−2t ), t ≥ 0
(c) By the definition of the convolution integral and the sifting property of the delta function
f1(t) ∗ f3( t) = f1(t − ) f3( )d =−∞
∞
∫ 5 ( t − )4e−2 u( )d =−∞
∞
∫= 20e−2 u( )] =t
= 20e−2tu( t)
(d) By the definition of the convolution integral
1/25/02 P16-2 © R. A. DeCarlo, P. M. Lin
f3(t) ∗ f3( t) = f3(t − ) f3( )d =−∞
∞
∫ 4e−2(t− )u( t − )4e−2 u( )d =16e−2 t u( t − )d0
∞
∫−∞
∞
∫
The integrand, u( t − ) , is nonzero only when ≤ t. This suggests that there are two cases toconsider: t < 0 and t ≥ 0.Case 1: t < 0. Here u( t − ) = 0 since is restricted to the interval [0,∞). Hence
f3(t) ∗ f3( t) = 0 , for t < 0.Case 2: t ≥ 0.
f3(t) ∗ f3( t) =16e−2t d = 16te−2 t
0
t
∫In sum,
f3(t) ∗ f3( t) =0, t < 0
16te−2t , t ≥ 0
(e) By the definition of the convolution integral and the sifting property of the delta function
f1(t + 2) ∗ f2( t + 4) = f1(t + 2 − ) f2( + 4)d = 5 (t + 2 − )2u( + 4)d =−∞
∞
∫−∞
∞
∫= 10u( + 4)] =t+2 =10u(t + 6)
(f) By the distributive property of convolution
f2( t) ∗ f2(t) + f3(t)[ ] = f2(t) ∗ f2(t) + f2( t) ∗ f3(t)
Using the results of parts (a) and (b) the result follows immediately
f2( t) ∗ f2(t) + f3(t)[ ] =0, t < 0
4(1+ t − e−2t ), t ≥ 0
Solution 16.2.(a) By definition
f3(t) = f1( ) f2( t − )d =−∞
∞
∫ K1u( − T1)K2u(t − − T2)d = K1K2 u( t − − T2)dT1
∞
∫−∞
∞
∫Here observe that u( t − − T2) = 0 for > t − T2 . Hence there are two cases to consider:t − T2 < T1 and t − T2 ≥ T1 .
Case 1: t − T2 < T1 . Here u( t − − T2) = 0 , since is restricted to the interval [T1,∞) .f3(t) = 0
Case 2: t − T2 ≥ T1 .
1/25/02 P16-3 © R. A. DeCarlo, P. M. Lin
f3(t) = K1K2 d = K1K2( t − T2 − T1)T1
t−T2
∫In sum,
f3(t) =0, t < T1 + T2
K1K2(t − T2 − T1), t ≥ T1 + T2
(b) By definition
f3(t) = f1( t − ) f2( )d =−∞
∞
∫ K1u(t − + T1)K2u( + T2)d = K1K2 u(t − + T1)d−T2
∞
∫−∞
∞
∫
Here observe that u( t − + T1) = 0 for > t + T1. Hence there are two cases to consider:t + T1 < −T2 and t + T1 ≥ −T2 .
Case 1: t + T1 < −T2 . Here u( t − + T1) = 0 , since is restricted to the interval [−T2 ,∞): f3(t) = 0.
Case 2: t + T1 ≥ −T2 .
f3(t) = K1K2 d = K1K2(t + T1 + T2)−T2
t+T1
∫In sum,
f3(t) =0, t < −T1 − T2
K1K2(t − T2 − T1), t ≥ −T1 − T2
(c) By definition
f3(t) = f1( t − ) f2( )d =−∞
∞
∫ K1u(t − )K2e−a u( )d = K1K2 e−a u( t − )d0
∞
∫−∞
∞
∫
The integrand is nonzero only when ≤ t. Hence, there are two cases to consider: t < 0 and t ≥ 0.
Case 1: t < 0. f3(t) = 0, for t < 0.Case 2: t ≥ 0.
f3(t) = K1K2 e−a
0
t
∫ d = K1K2e−a
−a
0
t
=K1K2
a(1− e−at ).
Therefore
f3(t) =K1K2
a(1− e−at )u(t). for t ≥ 0.
(d) By definition
f3(t) = f1( t − ) f2( )d =−∞
∞
∫ K1u(t − + T1)K2e−a u( )d = K1K2 e−a u( t − + T1)d0
∞
∫−∞
∞
∫
1/25/02 P16-4 © R. A. DeCarlo, P. M. Lin
The integrand is nonzero only when ≤ t + T1 . Hence, there are two cases to consider: t + T1 < 0and t + T1 ≥ 0.
Case 1: t + T1 < 0. Here u( t − + T1) = 0 , since is restricted to the interval [0,∞). Hencef3(t) = 0.
Case 2: t + T1 ≥ 0.
f3(t) = K1K2 e−a
0
t+T1
∫ d = K1K2e−a
−a
0
t+T1
=K1K2
a1− e−a (t+T1)[ ].
Therefore
f3(t) =K1K2
a1− e−a (t+T1 )[ ]u(t + T1).
(e) By definition
f3(t) = f1( t − ) f2( )d =−∞
∞
∫ K1u(−t + )K2e−a u( )d = K1K2 e−a u(−t + )d0
∞
∫−∞
∞
∫
The integrand is nonzero only when ≥ t. Hence, there are two cases to consider: t ≤ 0 and t > 0.Case 1: t ≤ 0. Here u(− t + ) = 1, since ≥ 0. Hence
f3(t) = K1K2 e−a
0
∞
∫ d = K1K2e−a
−a
0
∞
=K1K2
a, for t ≤ 0.
Case 2: t > 0.
f3(t) = K1K2 e−a
t
∞
∫ d = K1K2e−a
−a
t
∞
=K1K2
ae−at , for t > 0.
In sum,
f3(t) =
K1K2
a, t ≤ 0
K1K2
ae−at , t > 0
Solution 16.3.(a) By definition
f3(t) = f1( ) f2( t − )d =−∞
∞
∫ K1e−a u( )K2e−a (t− )u( t − )d
−∞
∞
∫ = K1K2 e−atu( t − )d0
∞
∫
The integrand, u( t − ) , is nonzero only when ≥ t. Hence, there are two cases to consider: t < 0and t ≥ 0.
Case 1: t < 0. Here u( t − ) = 0 , since ≥ 0. Hence f3(t) = 0, for t < 0.Case 2: t ≥ 0.
1/25/02 P16-5 © R. A. DeCarlo, P. M. Lin
f3(t) = K1K2e−at d0
t
∫ = K1K2e−at t , for t ≥ 0.
In sum,
Ic(s) =Cs
Cs +1
R
Iin (s)
(b) By definitiona = 1K =1
The integrand, u( t − ) , is nonzero only when ≥ t. Hence, there are two cases to consider: t < 0and t ≥ 0.
Case 1: t < 0. Here u( t − ) = 0 , since ≥ 0. Hence f3(t) = 0, for t < 0.Case 2: t ≥ 0.
f3(t) = K1K2e−bt e(b−a ) d0
t
∫ =t if a = b
1b − a
e(b−a )t −1[ ] if a ≠ b
Therefore, for t ≥ 0,
f3(t) =K1K2e−bt tu( t) if a = bK1K2
b − ae−at − e−bt[ ] if a ≠ b
(c) By replacing K1 = 50, K2 = 20 and a = 10 in the formula of f3(t) in part (a) the answer forpart (i) is easily obtained as
f3(t) =0, t < 0
1000e−10t t, t ≥ 0
For part (ii) the parameters have the following values: vc ( t) = e−(t− )
−∞
t
∫ d = e −t ]−∞
t=1,
K2 = 0.2, a = 10 and b = 0.2. Using these values in the formula developed in part (b) for f3(t) theanswer follows immediately
f3(t) = 0.102 e−0.2t − e−10 t( )u( t)
SOLUTION 16.4.(a) Using the impulse response theorem and the definition of the convolution integral the responseof the system, y(t), can be computed as follows
y(t) = h( t − )v( )d−∞
∞
∫ = 2e−2(t− )u( t − ) u( −1)− u( − 3)[ ]d−∞
∞
∫
Observing that u( −1)− u( − 3) is nonzero only when 1 ≤ < 3 yields
1/25/02 P16-6 © R. A. DeCarlo, P. M. Lin
y(t) = 2 e2( −t)u(t − )d1
3
∫
The integrand in the above equation is nonzero only when ≤ t. This suggests three regions ofconsideration: t <1, 1 ≤ t ≤ 3, and 3 < t.Case 1: t <1. Here u( t − ) = 0 , since is restricted to the interval [1,3]. Hence y(t) = 0, for t <1.Case 2: 1 ≤ t ≤ 3.
y(t) = 2 e2( −t)
1
t
∫ d = e2( −t) ]1t
=1− e2(1−t ) , for 1 ≤ t ≤ 3
Case 3: 3 < t.
y(t) = 2 e2( −t)
1
3
∫ d = e2( −t) ]13
= e2(1−t )(e4 −1), for 3 < t
In sum,
y(t) =0, t < 1
1− e2(1−t ), 1≤ t ≤ 3
e2(1−t )(e4 −1), 3 < t
A picture of y(t) is sketched in the next figure.
(b) Using the impulse response theorem and the definition of the convolution integral the responseof the system, y(t), can be computed as follows
y(t) = h( t − )v( )d−∞
∞
∫ = 2e−2(t− )u( t − )u(2 − t + ) u( −1)− u( − 3)[ ]d−∞
∞
∫
Observing that u( −1)− u( − 3) is nonzero only when 1 ≤ < 3 yields
y(t) = 2 e2( −t)u(t − )u(2 − t + )d1
3
∫The integrand in the above equation is nonzero only when t − 2 ≤ ≤ t. This suggests four regionsof consideration: t <1, 1 ≤ t ≤ 3, 3 < t ≤ 5and 5 < t.Case 1: t <1. Here u( t − ) = 0 , since is restricted to the interval [1,3]. Hence y(t) = 0, for t <1.Case 2: 1 ≤ t ≤ 3. Here u( t − )u(2 − t + ) is nonzero only when 1 ≤ ≤ t. Therefore,
1/25/02 P16-7 © R. A. DeCarlo, P. M. Lin
y(t) = 2 e2( −t)
1
t
∫ d = e2( −t) ]1t
=1− e2(1−t ) , for 1 ≤ t ≤ 3.
Case 3: 3 < t ≤ 5. Here u( t − )u(2 − t + ) is nonzero only when t − 2 ≤ ≤ 3. Therefore,
y(t) = 2 e2( −t)
t−2
3
∫ d = e2( −t) ]t−2
3= e2(3−t) − e−4 , for 3 < t ≤ 5.
Case 4: 5 < t. Here u( t − )u(2 − t + ) = 0 , since is restricted to the interval [1,3]. Therefore,y(t) = 0, for 5 < t.
A picture of y(t) is sketched in the next figure.
(c) By the impulse response theorem, the zero-state response of the circuit y(t) isy(t) = h(t) ∗v( t)
Using the definition of the convolution integral and the sifting property of delta function it followsthat
y(t) = 2h(t) − 2h( t −1) + h(t − 2) =
0, t < 0
2, 0 ≤ t < 1
−2, 1≤ t < 2
0, 2 ≤ t < 3
1, 3 ≤ t < 4
0, 4 ≤ t
Using the waveform of h(t) given in figure P16.4, y(t) is sketched in the next picture.
1/25/02 P16-8 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.5.(a) By definition
f4(t) = f1( t − ) f2( )d = ( t − − 2) ⋅2 ⋅ u( +1)d−∞
∞
∫−∞
∞
∫
By the sifting property of delta function it follows that f4(t) = 2u( +1)] =t−2 = 2u( t −1).
(b) By the definition of convolution and the sifting property of delta function
f5(t) = f1( t − ) f3( )d = (t − − 2)e−2 u( )d−∞
∞
∫−∞
∞
∫= e−2 u( )] =t−2
= e−2(t−2)u( t − 2)
(c) By definition
f6( t) = f2( t − ) f3( )d = 2u(t − +1)e−2 u( )d−∞
∞
∫−∞
∞
∫ = 2 e−2 u(t − +1)d0
∞
∫
The integrand in the above equation is nonzero only when ≤ t +1. This suggests two regions ofconsideration: t < −1and −1≤ t .Case 1: t < −1. Here u( t − +1)= 0, since 0 ≤ . Hence f6( t) = 0 , for t < −1.Case 2: −1≤ t . Here u( t − +1) is nonzero only when ≤ t +1. Therefore,
f6( t) = 2 e−2 d = e−2 ]0
t+1= 1− e−2(t+1)
0
t+1
∫It follows that
f6( t) = 1− e−2(t+1)[ ]u( t +1)
(d) By the definition of convolution and the sifting property of delta function we have
1/25/02 P16-9 © R. A. DeCarlo, P. M. Lin
f7( t) = f1(t − ) f3( − 2)d = ( t − − 2)e−2( −2)u( − 2)d−∞
∞
∫−∞
∞
∫ =
= e−2( −2)u( − 2)] =t−2= e−2(t−4)u( t − 4) .
SOLUTION 16.6.(a) By definition
f6( t) = 1− e−2(t+1)[ ]u( t +1)
Here observe that u(t – τ) = 0 for τ > t. Hence, there are two cases to consider: t ≤ 0 and t > 0.
Case 1: t ≤ 0 y ( t ) = K e a τ d τ
−∞
t
∫ = K e a τ
a
−∞
t = K
e at
a
Case 2: t > 0
y(t) = K eaτdτ−∞
t
∫ = K eaτdτ−∞
0
∫ = Keaτ
a
−∞
0
=K
a
(b) By definition
y(t) = K u(τ − t)−∞
∞
∫ e−aτ
u(τ)dτ = K u(τ)t
∞
∫ e− aτ
dτ
Here observe that u(τ – t) = 0 for τ < t; hence the lower limit of integration is t. Also, because of the
presence of u(τ) in the integrand, there are two cases to consider: t < 0 and t ≥ 0.
Case 1: t < 0
y(t) = K e−aτdτ0
∞
∫ = Ke−aτ
−a
0
∞
=K
a
Case 2: t ≥ 0
y(t) = K e−aτdτt
∞
∫ = Ke− aτ
−a
t
∞
=K
ae−at
SOLUTION 16.7.(a) Using the definition of the convolution integral and the sifting property of delta function,f5(t)can be computed as below
f5(t) = f2(t − ) f4( )d = e−a (t− )u(t − ) ( − 4)d−∞
∞
∫−∞
∞
∫ =
= e−a (t− )u(t − )] =4= e−a (t−4)u( t − 4)
A picture of f5(t), for a = 1, is sketched in the next figure.
1/25/02 P16-10 © R. A. DeCarlo, P. M. Lin
(b) By definition
f6( t) = f1(t − ) f1( )d = K 2 u(t − )u( )d−∞
∞
∫−∞
∞
∫ = K2 u( t − )d0
∞
∫
Since u( t − ) is nonzero only when ≤ t, there are two regions of consideration: t < 0and 0 ≤ t.Case 1: t < 0. Here u( t − ) = 0 , since ≥ 0. Hence
f6( t) = 0 , for t < 0.Case 2: 0 ≤ t.
f6( t) = K 2 d0
t
∫ = K 2t , for 0 ≤ t.
A picture of f6( t) , for K =1, is sketched in the next figure.
(c) By definition
f7( t) = f1(t − ) f2( )d = Ku( t − )e−a u( )d−∞
∞
∫−∞
∞
∫ = K e−a u(t − )d0
∞
∫
Since u( t − ) is nonzero only when ≤ t, there are two regions of consideration: t < 0and 0 ≤ t.Case 1: t < 0. Here u( t − ) = 0 , since ≥ 0. Therefore, f7( t) = 0 , for t < 0.Case 2: 0 ≤ t.
f7( t) = K e−a d0
t
∫ =−Ka
e−a ]0
t=
Ka
1− e−at( ) , for 0 ≤ t.
1/25/02 P16-11 © R. A. DeCarlo, P. M. Lin
A picture of f7( t) , for t < 0 and a = 1, is sketched in the next figure.
(d) By definition
f8(t) = f1( t − ) f3( )d = Ku( t − )ea u(− )d−∞
∞
∫−∞
∞
∫ = K ea u(t − )d−∞
0
∫
The integrand, )( −tu , is nonzero only when t≤ . This suggests two regions of consideration:0≤t and t<0 .
Case 1: 0≤t .
f8(t) = K ea u(t − )d−∞
t
∫ =Ka
ea ]−∞
t=
Ka
eat , for t ≤ 0.
Case 2: 0 < t.
f8(t) = K ea u(t − )d−∞
0
∫ =Ka
ea ]−∞
0=
Ka
, for 0 < t.
In sum,
f8(t) =
K
aeat , t ≤ 0
K
a, 0 < t
A picture of f8(t), for K =1 and a = 1, is sketched in the next figure.
1/25/02 P16-12 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.8.(a) Using the current division formula
Ic(s) =Cs
Cs +1
R
Iin (s)
By Ohm’s law the Laplace transform of capacitor’s voltage
Vc (s) =1Cs
Ic (s)
Therefore the transfer function of the circuit
H(s) =Vc (s)Iin (s)
=1
Cs +1
R
=1
s + 4
Taking the inverse Laplace transform of H(s) yields the impulse response h(t) = e−4 tu( t) .
(b) By the impulse response theorem
vc ( t) = iin (t) ∗ h( t) = iin ( t −−∞
∞
∫ )h( )d = 3e−(t− )
−∞
∞
∫ u(t − )e−4 u( )d =
= 3 e−(t+ 3 )
0
∞
∫ u(t − )d
The integrand is nonzero only when ≤ t. Therefore there are two regions of consideration:t < 0and 0 ≤ t.Case 1: t < 0. Here u( t − ) = 0 , since 0 ≤ . Hence vc ( t) = 0 , for t < 0.Case 2: 0 ≤ t.
1/25/02 P16-13 © R. A. DeCarlo, P. M. Lin
vc ( t) = 3 e−(t+ 3 )
0
t
∫ d = −e−(t+ 3 )]0
t= e−t − e−4 t , for 0 ≤ t.
In sum,
vc ( t) = e−t − e−4 t( )u(t)V.
SOLUTION 16.9.(a) By voltage division formula
Vout (s) =
1
Cs
R +1
Cs
Vin (s)
Therefore the transfer function
H(s) =Vout (s)Vin (s)
=
1
Cs
R +1
Cs
=1
s +1
Taking the inverse Laplace transform of H(s) yields yields the impulse response h(t) = e−tu(t).By the impulse response theorem and the convolution definition
vout (t) = h( t − )vin ( )−∞
∞
∫ d = e−(t− )
−∞
∞
∫ u( t − ) u(− ) + 2e−2 u( )[ ]d =
= e−(t− )
−∞
0
∫ u( t − )d + 2 e−(t+ )
0
∞
∫ u(t − )d
For both integrals the integrand is nonzero only when ≤ t. This suggests two regions ofconsideration: t < 0and 0 ≤ t. Case 1: t < 0. Here the second integral is zero since, for this integral, is restricted to [0,∞).
vc ( t) = e−(t− )
−∞
t
∫ d = e −t ]−∞
t=1, for t < 0.
Case 2: 0 ≤ t.
vc ( t) = e−(t− )
−∞
0
∫ d + 2 e−(t+ )
0
t
∫ d
= e −t ]−∞
0− 2e−(t− )]0
t= e−t − 2 e−2t − e−t( ) = 3e−t − 2e−2t , for 0 ≤ t
(c) By the impulse response theorem and the definition of convolution
1/25/02 P16-14 © R. A. DeCarlo, P. M. Lin
vout (t) = h( t − )vin ( )−∞
∞
∫ d = e−(t− )
−∞
∞
∫ u( t − )e−a| |d
= e−(t− )
−∞
0
∫ u( t − )ea d + e−(t− )
0
∞
∫ u( t − )e−a d
= e−t e (a+1)
−∞
0
∫ u( t − )d + e−t e (1−a )
0
∞
∫ u(t − )d
For both integrals the integrand is nonzero only when ≤ t. This suggests two regions ofconsideration: t < 0and 0 ≤ t. Case 1: t < 0. Here the second integral is zero since, for this integral, is restricted to [0,∞).
vout (t) = e−t e (a+1)
−∞
t
∫ d = e−t e (a+1)
a +1
−∞
t
=eat
a +1, for t < 0.
Case 2: 0 ≤ t.
vout (t) = e−t e (a+1)
−∞
0
∫ d + e−t e (1−a )
0
t
∫ d =
= e−t e (a+1)
a +1
−∞
0
+ e−t e (1−a)
0
t
∫ d =e−t
a +1+ e−t e (1−a)
0
t
∫ d
Here observe that a +1 is nonzero since a > 0.For computing the second integral, in case 2, we need to distinguish two subcases: a = 1
and a ≠ 1.
e−t e (1−a)
0
t
∫ d =e−t t if a =1
11− a
e−at − e−t( ) if a ≠1
Therefore, for 0 ≤ t,
vout (t) =
e−t
a +1+ e−t t if a = 1
e−t
a +1+
1
1− ae−at − e−t( ) if a ≠ 1
SOLUTION 16.10.(a) The impulse response is obtained by taking the inverse Laplace transform of the transferfunction
h(t) = −2e−0.2tu(t)
By the impulse response theorem the response y(t)equals
1/25/02 P16-15 © R. A. DeCarlo, P. M. Lin
y(t) = h(t) ∗v( t) = h( t − )v( )d−∞
∞
∫Substituting v(t) = u( t +1)− u( t −1) in the above integral yields
y(t) = h( t − ) u( +1)− u( −1)[ ]d−∞
∞
∫
Here observe that u( +1)− u( −1) is nonzero only when −1≤ ≤ 1. Hence
y(t) = h( t − )d = −2 e−0.2( t− )
−1
1
∫ u(t − )d−1
1
∫
The integrand is nonzero only when ≤ t. This suggests three regions of consideration: t < −1,−1≤ t < 1 and 1 ≤ t.Case 1: t < −1. y(t) = 0Case 2: −1≤ t < 1.
y(t) = −2 e−0.2( t− )
−1
t
∫ d =10 e−0.2(t+1) −1[ ] , for −1≤ t < 1.
Case 3: 1 ≤ t.
y(t) = −2 e−0.2( t− )
−1
1
∫ d =10e−0.2t e−0.2 − e0.2( ) , for 1 ≤ t.
(b) The transfer function of the leaky integrator (see equation 14.14 in the textbook) is given by
H(s) =−
1
R1
Cs +1
R2
where R2 is the leakage resistance of the capacitor C and R1 is the resistance connected at theinverting input of the op amp. Equating the two expressions of H(s) we obtain that
−1
R1
Cs +1
R2
=−2
s + 0.2
Matching the coefficients and taking into account that the smallest resistor is 10kΩ the following
values are obtained: R1 =10kΩ, R2 = 100kΩ and C = 5 ⋅10−5F .
(c) The impulse response is obtained by taking the inverse Laplace transform of the transferfunction
h(t) = Ke−atu(t)
By the impulse response theorem the response y(t)equals
1/25/02 P16-16 © R. A. DeCarlo, P. M. Lin
y(t) = h(t) ∗v( t) = h( t − )v( )d−∞
∞
∫
Substituting v(t) = u( t + T ) − u( t − T ) in the above integral yields
y(t) = h( t − ) u( + T ) − u( − T )[ ]d−∞
∞
∫
Here observe that u( + T ) − u( − T ) is nonzero only when −T ≤ ≤ T . Hence
y(t) = h(t − )d = K e−a (t− )
−T
T
∫ u(t − )d−T
T
∫
The integrand is nonzero only when ≤ t. This suggests three regions of consideration: t < −T ,−T ≤ t < T and T ≤ t.Case 1: t < −T . y(t) = 0Case 2: −T ≤ t < T .
y(t) = K e−a (t− )
−T
t
∫ d =Ka
1− e−a (t+T )[ ] , for −T ≤ t < T .
Case 3: T ≤ t.
y(t) = K e−a (t− )
−T
T
∫ d =Ka
e−at eaT − e−aT( ) , for T ≤ t.
In sum,
y(t) =
0, t < −TKa
1− e−a (t+T )[ ], −T ≤ t < T
K
ae−at eaT − e−aT( ), T ≤ t
SOLUTION 16.11.(a) First observe, from figure P16.11(a), that
f2( t) = (−2t + 4) u(t) − u( t − 2)[ ]By definition
f3(t) = f1( t − ) f2( )d = 4u( t − )(−2 + 4) u( ) − u( − 2)[ ]d = −8 ( − 2)u( t − )d0
2
∫−∞
∞
∫−∞
∞
∫
The integrand is nonzero only when ≤ t. This suggests three regions of consideration: t < 0,0 ≤ t < 2 and 2 ≤ t.Case 1: t < 0. Here u( t − ) = 0 due to the fact that is restricted to the interval [0,2]. Hence
f3(t) = 0, for t < 0.
1/25/02 P16-17 © R. A. DeCarlo, P. M. Lin
Case 2: 0 ≤ t < 2.
f3(t) = −8 ( − 2)d = −82
2− 2
0
t
∫0
t
= −4 t2 − 4t( ) , for 0 ≤ t < 2.
Case 3: 2 ≤ t.
f3(t) = −8 ( − 2)d = −82
2− 2
0
2
∫0
2
= 16, for 2 ≤ t.
In sum,
f3(t) =0, t < 0
−4 t2 − 4t( ), 0 ≤ t < 2
16, 2 ≤ t
A picture of f3(t) is sketched in the next figure.
(b) First observe, from figure P16.11(b), that
f2( t) = t u( t) − u( t − 2)[ ] + (4 − t) u(t − 2) − u( t − 4)[ ]By definition
f3(t) = f1( t − ) f2( )d−∞
∞
∫By replacing f1(t) and f2( t) with their expressions we have
f3(t) = 4u(t − ) u( ) − u( − 2)[ ] + (4 − ) u( − 2) − u( − 4)[ ] −∞
∞
∫ d =
= 4u( t − ) u( ) − u( − 2)[ ]d + 4u(t − )−∞
∞
∫ (4 − ) u( − 2) − u( − 4)[ ]−∞
∞
∫ d =
= 4 u(t − )d + 4 (4 − )u(t − )2
4
∫0
2
∫ d
1/25/02 P16-18 © R. A. DeCarlo, P. M. Lin
The integrands are nonzero only when ≤ t. This suggests four regions of consideration: t < 0,0 ≤ t < 2, 2 ≤ t < 4, and 4 ≤ t.Case 1: t < 0. Here u( t − ) = 0 due to the fact that is restricted to the interval [0,4]. Hence
f3(t) = 0, for t < 0.
Case 2: 0 ≤ t < 2. here observe that the second integral is zero since, for this integral, is restrictedto the interval [2,4]. Therefore
f3(t) = 4 d = 2t2
0
t
∫ , for 0 ≤ t < 2.
Case 3: 2 ≤ t < 4.
f3(t) = 4 d + 4 (4 − )2
t
∫0
2
∫ d = 8 + 4 4 −2
2
2
t
= −2t2 +16t −16, for 2 ≤ t < 4.
Case 4: 4 ≤ t.
f3(t) = 4 d + 4 (4 − )2
4
∫0
2
∫ d = 8 + 4 4 −2
2
2
4
=16, for 4 ≤ t.
In sum,
f3(t) =
0, t < 0
2t2, 0 ≤ t < 2
−2t2 +16t −16, 2 ≤ t < 4
16, 4 ≤ t
A picture of f3(t) is sketched in the next figure.
SOLUTION 16.12. (a) By voltage division,
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H(s) =Vout
Vin=
2
s
2s + 5 + 2
s
=1
s2 + 2.5s +1=
2 / 3
(s + 0.5)−
2 / 3
(s + 2)
Hence, the impulse response is
h(t) =2
3e−0.5tu(t) −
2
3e−2 tu(t)
(b) By definition
h(t) * vin (t) = h(t −τ )vin (τ)−∞
∞
∫ dτ =20
3e−0.5(t−τ)u(t −τ )
−∞
∞
∫ eτu(−τ) dτ
−20
3e−2(t−τ)u(t −τ )
−∞
∞
∫ eτu(−τ) dτ
Case 1: t ≤ 0.
vout (t) =20
3e−0.5(t−τ)
−∞
t
∫ eτ dτ −20
3e−2(t−τ)
−∞
t
∫ eτdτ =20
3e−0.5t
−∞
t
∫ e1.5τ dτ −20
3e−2t
−∞
t
∫ e3τdτ
=20e−0.5t
4.5e1.5τ[ ]−∞
t−
20e−2t
9e3τ[ ]−∞
t= 4.444et − 2.222et = 2.222et
Case 2: t > 0.
vout (t) =20e−0.5t
4.5e1.5τ[ ]−∞
0−
20e−2t
9e3τ[ ]−∞
0=
20e−0.5t
4.5−
20e−2t
9
SOLUTION 16.13.(a) The impulse response of the circuit has been computed in problem 16.12
h(t) =23
e−0.5t ( t) −23
e−2tu(t)
By the impulse response theorem and the convolution definition
vout (t) = h(t) ∗vin (t) = h(t − )vin ( )d =−∞
∞
∫
1/25/02 P16-20 © R. A. DeCarlo, P. M. Lin
=203
e−0.5(t− ) − e−2(t− )[ ]u(t − )e−| |d−∞
∞
∫
The integrand is nonzero only when ≤ t. Hence
vout (t) =203
e−0.5(t− ) − e−2(t− )[ ]e−| |d−∞
t
∫The existence of the function e−| | under the integral suggests two regions of consideration: t ≤ 0and 0 < t.Case 1: t ≤ 0.
vout (t) =203
e−0.5(t− ) − e−2(t− )[ ]e d−∞
t
∫ =
=203
e−0.5t e1.5
−∞
t
∫ d −203
e−2 t e3 d =−∞
t
∫
=204.5
e−0.5t e1.5[ ]−∞
t−
209
e−2t e3[ ]−∞
t=
= 4.444et − 2.222et = 2.222et , for t ≤ 0.Case 2 : 0 < t.
vout (t) =203
e−0.5(t− ) − e−2(t− )[ ]e d−∞
0
∫ +203
e−0.5(t− ) − e−2(t− )[ ]e− d0
t
∫ =
=203
e−0.5t+1.5 − e−2t+ 3[ ]d−∞
0
∫ +203
e−0.5t−0.5 − e−2 t+[ ]d0
t
∫ =
=203
e−0.5t+1.5
1.5−
e−2t+ 3
3
−∞
0
+203
e−0.5t−0.5
−0.5− e−2 t+
0
t
=
= 17.778e−0.5t − 20e−t + 4.444e−2 t , for 0 < t.
In sum,
vout (t) =2.222et , 0 ≤ t
17.778e−0.5t − 20e−t + 4.444e−2t , 0 < t
SOLUTION 16.14.(a) The impulse response of the circuit is obtained by taking the inverse Laplace transform of H(s)
h(t) = 2e−t − 2e−2t + 4e−4 t( )u( t)
(b) The result follows from the following MATLAB code:
1/25/02 P16-21 © R. A. DeCarlo, P. M. Lin
>> p = [-1,-2,-4];>> r = [2,-2,4];>> k = 0;>> [n,d] = residue(r,p,k)n =
4 14 16
d =
1 7 14 8
Therefore,
H(s) =4s2 +14s +16
s3 + 7s2 +14s + 8(c) By the impulse response theorem
y(t) = u( t) ∗ h(t) = u(t)* 2e−tu(t) − 2e−2 tu(t) + 4e−4 tu( t)[ ]Using the distributive property of convolution we have
y(t) = u( t) ∗ 2e−tu(t)[ ] + u( t) ∗ −2e−2tu( t)[ ] + u(t) ∗ 4e−4 tu(t)[ ]In problem P16.2(c) the following equation has been obtained
K1u(t)[ ]∗ K2e−atu(t)[ ] =K1K2
a(1− e−at )u(t)
Using the above equation y(t)is immediately obtained
y(t) = 2 1− e−t( )u( t) − 1− e−2t( )u(t) + 1− e−4 t( )u(t) =
= 2 − 2e−t + e−2t − e−4 t( )u(t)
(d) By the impulse response theorem
y(t) = f ( t) ∗ h(t) = 8u(− t) − 8u(− t − T )[ ]* 2e−tu(t) − 2e−2 tu(t) + 4e−4 tu( t)[ ]Using the distributive property of convolution we have
y(t) = 8u(− t) ∗ 2e−tu(t) − 2e−2tu( t) + 4e−4 tu( t)[ ]−
−8u(−t − T )* 2e−tu(t) − 2e−2 tu(t) + 4e−4 tu( t)[ ]We denote
1/25/02 P16-22 © R. A. DeCarlo, P. M. Lin
y1( t) = 8u(−t) ∗ 2e−tu( t) − 2e−2 tu(t) + 4e−4 tu( t)[ ]By the time invariance property it follows that
y(t) = y1( t) − y1( t − T )
In order to compute y1( t) we will use the following equation which has been obtained in problem16.2(e)
K1u(−t)[ ]∗ K2e−atu( t)[ ] =
K1K2
a, t ≤ 0
K1K2
ae−at , t > 0
Therefore,
y1( t) =16, t ≤ 0
16e−t − 8e−2 t + 8e−4 t , 0 < t
The zero-state response to the input f ( t) can now be computed
y(t) =0, t ≤ 0
16e−t − 8e−2t + 8e−4 t −16, 0 < t ≤ T
16 e−t − e−(t−T )[ ] − 8 e−2t − e−2(t−T )[ ] + 8 e−4 t − e−4(t−T )[ ], T < t
SOLUTION 16.15.(a) Using the convolution theorem the transfer function of the cascade is
H(s) = L h(t)[ ] = L h1( t) ∗ h2( t) ∗ h3( t)[ ] = L h1( t)[ ]⋅ L h2( t)[ ]⋅ L h3( t)[ ] = H1(s) ⋅ H2(s) ⋅ H3(s)
From table 13.1
H1(s) =1s
H2(s) =10
s + 2
H3(s) =2
s2
Therefore,
H(s) =20
s3(s + 2)
A partial fraction expansion of H(s) can be obtained using the residue command in MATLAB:
>> num = [20];>> den = [1 2 0 0 0];>> [r,p,k] = residue(num,den)
1/25/02 P16-23 © R. A. DeCarlo, P. M. Lin
r =
-2.5000 2.5000 -5.0000 10.0000
p =
-2 0 0 0
k =
[]
Hence
H(s) =−2.5s + 2
+2.5s
+−5
s2 +10
s3
Taking the inverse Laplace transform yields the impulse response of the cascade
h(t) = −2.5e−2tu( t) + 2.5u(t) − 5tu( t) + 5t2u(t)
(b) By the impulse response theorem and the convolution theorem, the Laplace transform of thestep response of the cascade equals
Y(s) = H (s) ⋅U (s) =20
s3(s + 2)⋅1s
=20
s4 (s + 2)
A partial fraction expansion of H(s) can be obtained using the residue command in MATLAB:
>> num = [20];>> den = [1 2 0 0 0 0];>> [r,p,k] = residue(num,den)
r =
1.2500 -1.2500 2.5000 -5.0000 10.0000
p =
-2 0 0
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0 0
k =
[]
Hence
Y(s) =1.25s + 2
+−1.25
s+
2.5
s2 +−5
s3 +10
s4
Taking the inverse Laplace transform yields the step response of the cascade
y(t) = 1.25e−2 tu(t) −1.25u(t) + 2.5tu(t) − 2.5t2u(t) +1.667t3u( t) .
SOLUTION 16.16.(a) By the voltage division formula
Vout (s) =
1
Cs
R +1
Cs
Vi (s) =1
CRs +1Vin (s)
Therefore, the transfer function of the circuit is
H(s) =Vout (s)Vin (s)
=1
CRs +1=
1s +1
Taking the inverse Laplace transform yields the impulse response h(t) = e−tu(t).(b) From table 13.1 the Laplace transform of vin (t) is
Vin (s) =1s
+1
(s +1)2
By the impulse response theorem and the convolution theorem it follows that
Vout (s) = H (s) ⋅Vin (s) =1
s +11s
+1
(s +1)2
=1
s(s +1)+
1
(s +1)3
A partial fraction expansion of Vout (s) is
Vout (s) =1s
−1
s +1+
1
(s +1)3
Taking the inverse Laplace transform yields
1/25/02 P16-25 © R. A. DeCarlo, P. M. Lin
vout (t) = u( t) − e−tu(t) − 0.5t2e−tu( t).
Using the time domain convolution method vout (t) can be computed as follows
vout (t) = h( t − )vin ( )d−∞
∞
∫ =
= e−(t− )u( t − ) u( ) + e− u( )[ ]d−∞
∞
∫From the experience earned by computing convolution integrals we know that the computation of theabove integral requires more computational work than the Laplace transform method. Morecomputations imply, of course, more sources of errors.From the solution of this problem we have seen that, in the case of the Laplace transform method,the computational burden consists in computing Laplace and inverse Laplace transforms. For alarge class of functions these transforms can be found in tables(for example table 13.1). The onlycomputation that we did, in the solution of this problem, was the partial fraction expansion ofVout (s) .
(c) In this case the (unilateral) transform method cannot be used because vin (t) ≠ 0 for t < 0.
SOLUTION 16.17.(a) The impulse response can be obtained by taking the inverse Laplace transform of H(s) .Therefore
h(t) = 8e−10 tu(t)
(b) From table 13.1
Vin (s) =8
s2 +16
By the impulse response theorem and the convolution theorem it follows that
Vout (s) = H (s) ⋅Vin (s) =64
(s +10)(s2 +16)
Vout (s) can be further written as
Vout (s) =−0.5517s + 5.517
s2 +16+
0.5517s +10
= −0.5517s
s2 +16+1.379
4
s2 +16+
0.5517s +10
The above expansion of Vout (s) can be obtained by using the technique of example 13.14, page
514. Taking the inverse Laplace transform yields
vout (t) = −0.5517cos(4 t)u( t) +1.379sin(4 t)u(t) + 0.5517e−10tu( t).
1/25/02 P16-26 © R. A. DeCarlo, P. M. Lin
(c) In this case
Vin (s) = L 2e−2t sin(4t)u(t)[ ] =8
(s + 2)2 +16Therefore
Vout (s) = H (s) ⋅Vin (s) =64
(s + 2)2 +16[ ](s +10)
Using again the technique of example 13.14, page 514, Vout (s) can be written as
Vout (s) =−0.8s + 4.8
(s + 2)2 +16+
0.8s +10
= −0.8s + 2
(s + 2)2 +16+1.6
4
(s + 2)2 +16+
0.8s +10
Taking the inverse Laplace transform yields
vout (t) = −0.8e−2t cos(4 t)u(t) +1.6e−2t sin(4 t)u(t) + 0.8e−10 tu(t) V.
In this context the Laplace transform method is faster than the time domain convolution. This isdue to the fact that vin (t) has a relatively complicated expression and the convolution integral willrequire more computational work than the Laplace transform method.
(d) In this context the Laplace transform method cannot be used because vin (t) ≠ 0 for t < 0. Thetime domain convolution method will be used to compute the response vout (t). By the impulseresponse thorem
vout (t) = h(t) ∗vin (t) = 8e−10 tu(t)[ ] ∗ u(− t)[ ]
Using the result of problem 16.2, part (e), it follows that
vout (t) =0.8, t ≤ 0
0.8e−10t , 0 < t
SOLUTION 16.18.(a) Replacing R1, R2 , C1 and C2 with their values the transfer function can be obtained
H(s) =s
s2+5s + 2
The only zero of H(s) is 0 and the poles of H(s) are –0.5 and –2. A partial fraction expansion ofH(s) is:
H(s) =−0.167s + 0.5
+0.667s + 2
The impulse response can be obtained by taking the inverse Laplace transform of H(s)
h(t) = −0.167e−0.5t + 0.667e−2t
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(b) vout (t) will be computed using the Laplace transform method. This approach is valid becauseh(t) and vin (t) are zero for t < 0.From table 13.1 the Lapace transform of vin (t) is
Vin (s) =1
(s + 2)2
By the impulse response theorem and the convolution theorem it follows that
Vout (s) = H (s) ⋅V (s) =s
2s2 + 5s + 2
1
(s + 2)2 =s
2s4 +13s3 + 30s2 + 28s + 8
A partial fraction expansion of Vout (s) can be obtained using the residue command in MATLAB:>> a = [1 0];>> b = [2 13 30 28 8];>> [r,p,k] = residue(a,b)
r = 0.0741 0.1111 0.6667 -0.0741
p = -2.0000 -2.0000 -2.0000 -0.5000
k = []Therefore,
Vout (s) =−0.0741s + 0.5
+0.0741s + 2
+0.1111
(s + 2)2 +0.6667
(s + 2)3
Taking the inverse Laplace transform yields
vout (t) = −0.0741e−0.5t + 0.0741e−2t + 0.1111te−2t + 0.3333t2e−2 t[ ]u( t) V.
One would prefer the time domain convolution method to compute vout (t), but the computationsmay require more work relatively to the Laplace transform method.
(c) In this part vin (t) ≠ 0 for t < 0. Therefore the time domain convolution method will be used tocompute vout (t).By the impulse response theorem
vout (t) = h(t) ∗vin (t) =
= −0.1667e−0.5(t− ) + 0.6667e−2(t− )[ ]u−∞
∞
∫ (t − )e2 u(− )d =
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= −0.1667e−0.5(t− ) + 0.6667e−2(t− )[ ]−∞
0
∫ u(t − )e2 d
The integrand of the previous integral is nonzero only when ≤ t. This suggests two regions ofconsideration: t < 0 and 0 ≤ t.Case 1: t < 0.
vout (t) = −0.1667e−0.5t e2.5
−∞
t
∫ d + 0.6667e−2 t e4
−∞
t
∫ d =
= −0.1667e−0.5t e2.5
2.5
−∞
t
+ 0.6667e−2t e4
4
−∞
t
=
= 0.1e−2t , for t < 0.Case 2: 0 ≤ t.
vout (t) = −0.1667e−0.5t e2.5
−∞
0
∫ d + 0.6667e−2 t e4
−∞
0
∫ d =
= −0.1667e−0.5t e2.5
2.5
−∞
0
+ 0.6667e−2t e4
4
−∞
0
=
= −0.1667e−0.5t + 0.6667e−2 t , for 0 ≤ t.
SOLUTION 16.19.Replacing Rand C with their values
H(s) =s − 5s + 5
= 1−10
s + 5
The zero-state response vout (t) will be computed using the time domain convolution methodbecause vin (t) ≠ 0 for t < 0.The impulse response of the circuit is
h(t) = ( t) −10e−5tu( t)By the impulse response theorem
vout (t) = h(t) ∗vin (t) = h( )vin ( t − )d =−∞
∞
∫
= 10 ( ) −10e−5 u( )[ ]co−∞
∞
∫ s10( t − )[ ]d =
= 10 ( )co−∞
∞
∫ s( t − )d −10 10e−5 u( )co−∞
∞
∫ s 10(t − )[ ]d
Using the sifting property of the delta function and expanding cos(t − )it follows that
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vout (t) = 10cos( t − ) −100cos(10t) e−5 co0
∞
∫ s(10 )d −100sin(10t) e−5
0
∞
∫ sin(10 )d
Using the definition of the Laplace transform we observe that
e−5 cos0
∞
∫ (10 )d = L cos(10t)u(t)[ ]s=5 =s
s2 +100
s=5
= 0.04
and
e−5 sin0
∞
∫ (10 )d = L sin(10t)u( t)[ ]s=5 =10
s2 +100
s=5
= 0.08
Thereforevout (t) = 10cos(10t) − 4cos(10 t) − 8sin(10t)
= 6cos( t) − 8sin(10t) =10cos 10t + tan−1(43)[ ]
Notice that vout (t) and vin (t) have the same frequency and magnitude.
SOLUTION 16.20.(a) First notice that vin (t − T) = u(t). Therefore w(t) = u( t) and
W (s) =1s
(b) Using the properties of the Laplace transform it follows that
Voutw (s) = H (s) ⋅W (s) =
2s(s + 2)
A partial fraction expansion of Voutw (s) is
Voutw (s) =
1s
−1
s + 2
Taking the inverse Laplace transform yields
voutw (t) = u( t) − e−2 tu(t)
(b) Sincevin (t) = w( t + T )
it follows, by the time invariance property, that
voutv (t) = vout
w (t + T ).Therefore,
voutv (t) = u( t + T ) − e−2(t+T )u(t + T ) V.
SOLUTION 16.21.(a) First observe from figure P16.21 that
vin (t) = u(t + T ) − u(t − T )From the definition of w(t) it follows that
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w(t) = vin ( t − T ) = u(t) − u(t − 2T )Therefore
W (s) =1s
−1s
e−2sT
(a) By the impulse response theorem and the convolution theorem it follows that
Voutw (s) = H (s) ⋅W (s) =
2s(s + 2)
1− e−2sT( ) =1s
−1
s + 2
1− e−2sT( ) =
=1s
−1
s + 2−
1s
−1
s + 2
e
−2sT
Taking the inverse Laplace transform and using the time shift property of the Laplace transformyields
voutw (t) = 1− e−2 t( )u( t) − 1− e−2(t−2T )[ ]u( t − 2T )
Becausevin (t) = w( t + T )
it follows, by the time invariance property, that
voutv (t) = vout
w (t + T ) == 1− e−2(t+T )( )u(t + T ) − 1− e−2(t−T )[ ]u( t − T ) V.
SOLUTION 16.22. (a) The use of t = t + T1 in the problem statement means replace t by t + T1 .
However, strictly speaking we should have used a statement of the form t = t ' +T1 which is done in
the proof below. By definition of the convolution and the property of commutivity,
f (t − T1) * g(t) = f (t − T1 −τ )g(τ) dτ−∞
∞
∫ = f (t' −τ)g(τ) dτ−∞
∞
∫
t'=t−T1
= f (t' ) * g(t' )[ ]t'=t−T1
Observe that t = t ' +T1 . Hence
f (t − T1)* g( t)[ ]t=t'+T1= f (t' ) * g(t' )
Realizing that t and t' are simply dummy variables, we immediately obtain the result. From a
systems perspective, this corresponds to the property of time-invariance where a shift of an input
function by T1 yields a corresponding shift of the output function by T1.
(b) The steps in this part are similar to those of part (a).
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f (t − T1) * g(t − T2) = f (t − T1 −τ )g(τ − T2) dτ−∞
∞
∫
Let λ = τ − T2 in which case τ =λ + T2 and dλ = dτ . Hence
f (t − T1) * g(t − T2) = f (t − T1 − T2 −λ )g(λ)dλ−∞
∞
∫ = f (t' −λ)g(λ )dλ−∞
∞
∫
t'=t−T1−T2
= f (t' )* g( t')[ ]t'=t−T1−T2
Since t = t ' +T1 + T2 , and t and t' are dummy variables, we have
f (t − T1)* g( t − T2 )[ ]t=t'+T1+T2= f (t' )* g(t')
and the result follows.
SOLUTION 16.23. From table 13.1, G(s) =1
s + 2, H(s) =
1
(s + 2)2 .
(a) Define p(t) = f(t – 2) = u(t).
Then, P(s) =1
s. Consider
P(s)G(s) =1
s(s + 2)=
0.5
s−
0.5
s + 2
Hence
p(t) * g(t) = 0.5u(t) − 0.5e−2 tu(t)
From problem 16.22 part (a),
f (t)* g(t) = p(t) * g(t)[ ]t=t+2 = 0.5u(t) − 0.5e−2tu(t)[ ]t=t+2= 0.5 − 0.5e−2(t+2)[ ]u(t + 2)
(b) Define p(t) = f(t – 2) = u(t).
Then, P(s) =1
s. Consider
P(s)H(s) =1
s(s + 2)2 =0.25
s−
0.25
s + 2−
0.5
(s + 2)2
Hence
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p(t) * h(t) = 0.25 − 0.25e−2t − 0.5te−2t( )u(t)
From problem 16.22 part (a),
f (t)* h( t) = p(t)* h( t)[ ]t=t+2 = 0.25 − 0.25e−2 t − 0.5te−2t( )u(t)[ ]t=t+2
= 0.25 − 0.25e−2(t+2) − 0.5(t + 2)e−2(t+2)[ ]u(t + 2)
SOLUTION 16.24.(a) The pictures of f ( t) and g( t) are sketched in the next figures
(b) Using the convolution theorem it follows thatL f (t − 2 ) ∗ g( t − )[ ] = L f (t − 2 )[ ]⋅L g( t − )[ ]
From table 13.1
L f (t − 2 )[ ] = L u( t)[ ] =1s
L g(t − )[ ] = L sin(t)u( t)[ ] =1
s2 +1Therefore
L f (t − 2 ) ∗ g( t − )[ ] =1
s(s2 +1)=
1s
−s
s2 +1Taking the inverse Laplace transform yields
f ( t − 2 ) ∗ g(t − ) = u(t) − cos( t)u(t) = 1− cos(t)[ ]u(t)
Using the property give in problem 16.22(b) it follows that
f ( t) ∗ g( t) = f (t − 2 ) ∗ g( t − )[ ]t=t+2 + = 1− cos(t + 3 )[ ]u(t + 3 )
1/25/02 P16-33 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.25.Define
w(t) = vin ( t − 2)Hence
w(t) = u( t)and, from table 13.1,
W (s) =1s
From table 13.1 we also have that
H1(s) =1
s +1 and H2(s) =
1
(s +1)2
The impulse response of the cascade ish(t) = h1(t) ∗ h2(t)
Hence the transfer function of the cascade is
H(s) = H1(s)H2(s) =1
(s +1)3
We denote by voutw (t)the zero state response due to the input w(t). Hence,
Voutw (s) = H (s) ⋅W (s) =
1
s(s +1)3
A partial fraction expansion of Voutw (s) is obtained using the residue command in MATLAB:
>> a = [1];>> b = [1 3 3 1 0];>> [r,p,k] = residue(a,b)
r = -1.0000 -1.0000 -1.0000 1.0000
p = -1.0000 -1.0000 -1.0000 0
k = []
Therefore
Voutw (s) =
1s
+−1
s +1+
−1
(s +1)2 +−1
(s +1)3
Taking the inverse Laplace transform yields
1/25/02 P16-34 © R. A. DeCarlo, P. M. Lin
voutw (t) = 1− e−t − te−t − 0.5t2e−t[ ]u(t) V
Due to the fact thatvin (t) = w( t + 2)
the time invariance property implies that
vout (t) = voutw (t + 2) =
= 1− e−(t+ 2) − ( t + 2)e−(t+2) − 0.5(t + 2)2e−(t+ 2)[ ]u(t + 2) V.
SOLUTION 16.26.(a) Using the sifting property of the delta function it follows that
f4(t) = (t) + (t − 4)[ ] ∗ f2(t) = f2(t) + f2(t − 4)
The right-hand side of the above equation interprets as a graphical sum of shifted pictures of f2( t) .A picture of f4(t) is sketched in the next figure.
0 1 2 3 4 5 6 70
2
4
(b) In order to compute the area beneath f2( t − ) ⋅ f2( ) four regions will be considered: t < 0,0 ≤ t <1, 1 ≤ t < 2 and 2 ≤ t.Step 1: t < 0. In this case f2( t − ) ⋅ f2( ) = 0 for all . Therefore
f2( t) ∗ f2( t) = 0 for t < 0.
Step 2: 0 ≤ t <1. In this case f2( t − ) ⋅ f2( ) = 16 for 0 ≤ ≤ t and is zero elsewhere. The areabeneath f2( t − ) ⋅ f2( ) equals 16t . Therefore
f2( t) ∗ f2( t) =16t for 0 ≤ t <1.
Step 3: 1 ≤ t < 2. In this case f2( t − ) ⋅ f2( ) = 16 for t −1 < ≤1 and is zero elsewhere. Hencethe area beneath f2( t − ) ⋅ f2( )equals
f2( t) ∗ f2( t) =16(2 − t) for 1 ≤ t < 2.Step 4: 2 ≤ t. In this case f2( t − ) ⋅ f2( ) = 0 for all . Therefore
f2( t) ∗ f2( t) = 0 for2 ≤ t.
A picture of f5(t)is sketched in the next figure.
1/25/02 P16-35 © R. A. DeCarlo, P. M. Lin
(c) In order to compute the area beneath f2( t − ) ⋅ f3( ) five regions will be considered: t < 0,0 ≤ t <1, 1 ≤ t < 2, 2 ≤ t < 3 and 3 ≤ t.Step 1: t < 0. In this case f2( t − ) ⋅ f3( ) = 0 for all . Therefore f2( t) ∗ f3(t) = 0 for t < 0.Step 2 : 0 ≤ t <1. In this case f2( t − ) ⋅ f3( ) = 8 for 0 ≤ ≤ t and is zero elsewhere. Thereforethe area beneath f2( t − ) ⋅ f3( ) equals
f2( t) ∗ f3( ) = 8t for 0 ≤ t <1.Step 3: 1 ≤ t < 2. Here
f2( t − ) ⋅ f3( ) =8, t −1 < <1
24, 1 ≤ < t
0, otherwise
Hence, the area beneath f2( t − ) ⋅ f3( ) equals
f2( t) ∗ f3( ) = 81 − (t −1)[ ] + 24( t −1) = 8(2t −1) for 1 ≤ t < 2.
Step 4: 2 ≤ t < 3. In this case f2( t − ) ⋅ f3( ) = 24 for t −1 < < 2 and is zero otherwise. Hence,the area beneath f2( t − ) ⋅ f3( ) equals
f2( t) ∗ f3(t) = 24 2 − (t −1)[ ] = 24(3− t) for 2 ≤ t < 3.
Step 5: 3 ≤ t. Here f2( t − ) ⋅ f3( ) = 0 for all . Therefore f2( t) ∗ f3(t) = 0 for t < 0.A picture of f6( t) is sketched in the next figure.
1/25/02 P16-36 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.27.By the impulse response theorem, it follows that the response is
y(t) = h(t) ∗ f (t) == h( t) ∗ (t) − (t −1)[ ]
Using the distributive property of convolution and the sifting property of delta function y(t) can bewritten as
y(t) = h(t) − h(t −1)The right-hand side of the above equation interprets as a graphical sum of (shifted) pictures of h(t). The pictures of h(t), h(t −1) and y(t) are sketched in the next figures.
SOLUTION 16.28.(a) From the picture of f ( t) and h(t) in figure P16.28 we observe that, in order to compute thearea beneath h(t − ) f ( ) , we need to consider four cases: t < 0, 0 ≤ t < 4, 4 ≤ t < 8 and 8 ≤ t.Step 1: t < 0. Here h(t − ) f ( ) = 0 for all . Therefore the area beneath h(t − ) f ( ) equals zeroand
h(t) ∗ f (t) = 0 for t < 0.Step 2: 0 ≤ t < 4. In this case h(t − ) f ( ) =1 for 0 ≤ ≤ t and is zero otherwise. Hence the areabeneath h(t − ) f ( ) equals
h(t) ∗ f (t) = t for 0 ≤ t < 4.Step 3: 4 ≤ t < 8. In this case
h(t − ) f ( ) =1, t − 4 < ≤ 4
2, 4 ≤ ≤ t
0, otherwise
Therefore the area beneath h(t − ) f ( ) equalsh(t) ∗ f (t) = 4 − ( t − 4)[ ] + 2(t − 4) = t for 4 ≤ t < 8.
1/25/02 P16-37 © R. A. DeCarlo, P. M. Lin
Step 4: 8 ≤ t. Here h(t − ) f ( ) = 2 for t − 4 < ≤ t and is zero otherwise. Henceh(t) ∗ f (t) = 2 t − (t − 4) = 8[ ] for 8 ≤ t.
A picture of y(t) is sketched in the next figure.
(b) The impulse response ish(t) = u( t) − u(t − 4)
By the impulse response theorem
y(t) = x(t) ∗ h( t) = x( )h( t − )d−∞
∞
∫ = x( ) u(t − ) − u(t − − 4)[ ]−∞
∞
∫ d
Here observe that u( t − ) − u(t − − 4) is nonzero only when t ≤ < t − 4.Therefore
y(t) = x( )dt
t−4
∫which interprets as the running area under x(t) over the interval [t − 4,t].
SOLUTION 16.29.The response, y(t), is obtained as indicated in the statement of the problem, by using the followingMATLAB code:
>> tstep = 1;>> vin = [1];>> h = [0, 2, 3, 1, 1];>> y = tstep*conv(vin, h);>> y = [0 y 0];>> t = 0:tstep:tstep*(length(vin)+length(h));>> plot(t,y)>> grid
The response is plotted in the next figure.
1/25/02 P16-38 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.30.A picture of vin (t) sketched in the next figure.
In order to plot the response, y(t), the MATLAB code of problem 16.29 will be used with only onemodification. Namely
vin = [1, 1, 2, 2]
as it can be observed from the picture of vin (t) with the time step tstep = 1.Therefore the MATLAB code is:
>> tstep = 1;>> vin = [1, 1, 2, 2];>> h = [0, 2, 3, 1, 1];>> y = tstep*conv(vin,h);>> y = [0 y 0];>> t = 0:tstep:tstep*(length(vin)+length(h));>> plot(t,y)>> grid
The response is plotted in the next figure.
1/25/02 P16-39 © R. A. DeCarlo, P. M. Lin
1/25/02 P16-1 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.31.In order to compute the area beneath v(t − )h( ) seven regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, 2 ≤ t < 3, 3 ≤ t < 4, 4 ≤ t < 5 and 5 ≤ t.
Step 1: t < 0. For t in this region v(t − )h( ) = 0 for all . Hence
y(t) = v(t) ∗ h( t) = 0 for t < 0.
Step 2: 0 ≤ t <1. In this case v(t − )h( ) = v0 × h0 for 0 ≤ ≤ t and is zero otherwise. Therefore thearea beneath v(t − )h( ) equals
y(t) = v(t) ∗ h( t) = v0 × h0 × t for 0 ≤ t <1.
Step 3: 1 ≤ t < 2. For t in this region we have
v(t − )h( ) =
v1× h0, 0 ≤ ≤ t −1
v0 × h0, t −1< < 1
v0 × h1, 1≤ < t
0, otherwise
Therefore the area beneath v(t − )h( ) equals
y(t) = v(t) ∗ h( t) = v1× h0 × ( t −1)− 0[ ] + v0 × h0 × 1− ( t −1)[ ] + v0 × h1× ( t −1)== t × (v1× h0 − v0 × h0 + v0 × h1)− v1× h0 + 2 × v0 × h0 − v0 × h1, for 1 ≤ t < 2
Step 4: 2 ≤ t < 3. In this case
v(t − )h( ) =
v1× h0, t − 2 < < 1
v1× h1, 1≤ ≤ t −1
v0 × h1, t −1< < 2
v0 × h2, 2 ≤ < t
0, otherwise
Hence, for 2 ≤ t < 3,y(t) = v(t) ∗ h( t) =
= v1× h0 × 1− (t − 2)[ ] + v1× h1× (t −1)−1[ ] + v0 × h1× 2 − ( t −1)[ ] + v0 × h2 × (t − 2) == t × (−v1× h0 + v1× h1− v0 × h1+ v0 × h2) + 3× v1× h0 − 2 × v1× h1 + 3 × v0 × h1− 2 × v0 × h2
Step 5: 3 ≤ t < 4. In this case
v(t − )h( ) =
v1× h1, t − 2 < < 2
v1× h2, 2 ≤ ≤ t −1
v0 × h2, t −1< < 3
0, otherwise
Hence, for 3 ≤ t < 4,y(t) = v(t) ∗ h( t) =
= v1× h1× 2 − (t − 2)[ ] + v1× h2 × (t −1)− 2[ ] + v0 × h2 × 3− ( t −1)[ ] == t × (−v1× h1 + v1× h2 − v0 × h2) + 4 × v1× h1− 3 × v1× h2 + 4 × v0 × h2
Step 6: 4 ≤ t < 5. In this case v(t − )h( ) = v1× h2 for t − 2 < < 3 and is zero otherwise. Thereforey(t) = v(t) ∗ h( t) = v1× h2 × 3− ( t − 2)[ ] = v1× h2 × (5 − t) for 4 ≤ t < 5.
1/25/02 P16-2 © R. A. DeCarlo, P. M. Lin
Step 7: 5 ≤ t. For t in this region v(t − )h( ) = 0 for all . Hence
y(t) = v(t) ∗ h( t) = 0 for 5 ≤ t.In sum,
y( t) =
v0 × h0 × t , 0 ≤ t < 1
t × (v1 × h0 − v0 × h0 + v0 × h1) − v1 × h0 + 2 × v0 × h0 − v0 × h1, 1 ≤ t < 2
t × (−v1 × h0 + v1 × h1 − v0 × h1 + v0 × h2) + 3 × v1 × h0 − 2 × v1 × h1 + 3 × v0 × h1 − 2 × v0 × h2, 2 ≤ t < 3
t × (−v1 × h1 + v1 × h2 − v0 × h2) + 4 × v1 × h1 − 3 × v1 × h2 + 4 × v0 × h2, 3 ≤ t < 4
v1 × h2 × (5 − t), 4 ≤ t < 5
0, otherwise
Hence,
y1 = y(1) = v0 × h0 = 6y2 = y(2) = v0 × h1+ v1× h0 = 8y3 = y(3) = v0 × h2 + v1× h1 = −6y4 = y(4) = v1× h2 = 4
(b) Using the expressions of p(x) and q(x) it follows that
p(x) × q(x) = x 3 × (v0 × h0) + x 2 × (v0 × h1+ v1× h0) + x × (v0 × h0 + v1× h1)+ v1× h2We observe that the coefficients of p(x) × q(x) are exactly y1 , y2 , y3 and y4 , respectively. Therefore
r(x) = p(x) × q(x) .
SOLUTION 16.32.(a) This part will be solved using the techniques of convolution algebra. Therefore we can write f3(t) as
f3(t) = f1(−1)(t) ∗ f2
(1)( t)Where the superscript (-1) means integration and the superscript (1) means differentiation. From figureP16.32 we observe that
f1(t) = 4 u(t) − u(t − 4)[ ]Hence
f1(−1)(t) = 4 tu(t) − (t − 4)u(t − 4)[ ] =
= 4 r( t) − r(t − 4)[ ]
By inspection, from the same figure, we have
f2(1)( t) = 4 (t) − 2 ( t − 2) + 2 (t − 4) − 2 ( t − 6) + ( t − 8)[ ]
Using the sifting property of the delta function f3(t) can be computed as follows
f3(t) = 4 r(t) − r(t − 4)[ ] ∗ 4 ( t) − 2 (t − 2) + 2 ( t − 4) − 2 (t − 6) + ( t − 8)[ ] == 16 r( t) − 2r(t − 2) + r( t − 4) − r(t − 8) + 2r( t −10) − r(t −12)[ ]
A picture of f3(t) is sketched in the next figure.
1/25/02 P16-3 © R. A. DeCarlo, P. M. Lin
(b) Using the techniques of problem 16.31 and considering the time step tstep = 2, the polynomials p(x) ,q(x) and r(x) can be associated with the functions f1(t), f2( t) andf3(t), respectively, as below:
p(x) = 4x + 4
q(x) = 4 x3 − 4 x2 + 4 x − 4
r(x) = 32x4 − 32
We need to verify that the equalityp(x) ⋅q(x) ⋅tstep = r(x)
holds. The equality indeed holds because
p(x) ⋅q(x) ⋅tstep = 32(x +1)(x 3 − x 2 + x −1) = 32(x4 −1)= r(x).
The results obtained in part (a) and part (b) coincide.
SOLUTION 16.33.(a) Using the techniques of convolution algebra f3(t) can be written as
f3(t) = f1(−1)(t) ∗ f2
(1)( t)
Where the superscript (-1) means integration and the superscript (1) means differentiation. From figureP16.33 we observe that
f1(t) = 2 u(t +1)− u( t − 4)[ ]Therefore
f1(−1)(t) = 4 (t +1)u(t +1)− ( t − 4)u( t − 4)[ ] =
= 4 r( t +1)− r( t − 4)[ ] = g(t)
By inspection, from the same figure, we have
f2(1)( t) = 4 (t) − 8 (t − 2) + 6 (t − 4) − 2 (t − 6)
Using the sifting property of the delta function f3(t) can be computed as follows
f3(t) = 4g( t) − 8g( t − 2) + 6g( t − 4) − 2g( t − 6)
1/25/02 P16-4 © R. A. DeCarlo, P. M. Lin
This is plotted in MATLAB as follows:
>> t = -2:0.01:14;>> g = 2*(t+1).*u(t+1)-2*(t-4).*u(t-4);>> g1 = 4*g;>> g2 = -8*( 2*(t-1).*u(t-1)-2*(t-6).*u(t-6) );>> g3 = 6*( 2*(t-3).*u(t-3)-2*(t-8).*u(t-8) );>> g4 = -2*( 2*(t-5).*u(t-5)-2*(t-10).*u(t-10) );>> f3 = g1+g2+g3+g4;>> plot(t,f3);>> grid;
A picture of f3(t) is sketched in the next figure.
(b) To account for the fact that f1(t)is nonzero for negative t the following formula (see problem 16.22)f1(t) ∗ f2(t) = f1(t −1)∗ f2(t)[ ]t=t+1
will be used to compute f3(t). Using a slightly modified version of the code of problem 16.31, we have
>> f1 = [2, 2, 2, 2, 2];>> f2 = [4, 4, -4, -4, 2, 2];>> T = 1;>> tstep = T;>> f3 = tstep*conv(f1,f2);>> f3 = [0 f3 0];>> t = -1:tstep:tstep*(length(f1)+length(f2))-1;>> plot(t,f3)>> grid
1/25/02 P16-5 © R. A. DeCarlo, P. M. Lin
The results of parts (a) and (b) coincide.
SOLUTION 16.34. This problem is solved using the techniques of the convolution algebra with the
graphical method left to the student.
f3(t) = f1(t) * f2(t) = f1(t)[ ](−1)* f2(t)[ ](1)
where the superscript (-1) means integration and the superscript (1) means differentiation. By inspection,
f1(t )[ ](−1) = 4tu(t) − 4(t − 6)u(t − 6) = g(t)
and
f2(t)[ ](1) = 4δ(t) − 8δ( t − 2) + 8δ(t − 6) − 4δ(t − 8)
Hence the response say y(t) satisfies
f3(t) = 4g(t) −8g( t − 2) +8g(t − 6) − 4g(t −8)
This is plotted in MATLAB as follows:
»t=0:.05:20;
»g = 4*t .* u(t) - 4*(t-6) .*u(t-6);
»g1=4*g;
»g2 = -8*(4*(t-2) .* u(t-2) - 4*(t-8) .*u(t-8));
»g3 = 8*(4*(t-6) .* u(t-6) - 4*(t-12) .*u(t-12));
»g4 = -4*(4*(t-8) .* u(t-8) - 4*(t-14) .*u(t-14));
»f3 = g1+g2+g3+g4;
»plot(t,f3)
»grid
1/25/02 P16-6 © R. A. DeCarlo, P. M. Lin
0 2 4 6 8 10 12 14 16 18 20-40
-30
-20
-10
0
10
20
30
40
(b) Using the code of problem 31, we have
»f1 = [4 4 4];
»f2 = [4 -4 -4 4];
»T = 2;
»tstep = T;
»f3 = [0 conv(f1,f2)*tstep 0];
»t = 0: tstep : tstep* (length(f1) + length(f2));
»plot(t,f3)
»grid
1/25/02 P16-7 © R. A. DeCarlo, P. M. Lin
0 2 4 6 8 10 12 14-40
-30
-20
-10
0
10
20
30
40
The results of parts (a) and (b) coincide.
SOLUTION 16.35.(a) In order to compute the area beneath f1(t − ) ⋅ f1( ) four regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, and 2 ≤ t.Step 1: t < 0. Here f1(t − ) ⋅ f1( ) = 0 for all . Hence
f3(t) = f1(t) ∗ f1( t) = 0 for t < 0.
Step 2: 0 ≤ t <1. In this case f1(t − ) ⋅ f1( ) = 1 for 0 ≤ ≤ t and is zero otherwise. Therefore the areabeneath f1(t − ) ⋅ f1( ) equals
f3(t) = t for 0 ≤ t <1.
Step 3: 1 ≤ t < 2. In this case f1(t − ) ⋅ f1( ) = 1 for t −1 < <1 and is zero otherwise. Therefore thearea beneath f1(t − ) ⋅ f1( ) equals
f3(t) = 1− (t −1)= 2 − t for 1 ≤ t < 2.
Step 4: 2 ≤ t. Here f1(t − ) ⋅ f1( ) = 0 for all . Hence
f3(t) = f1(t) ∗ f1( t) = 0 for2 ≤ t.In sum,
f3(t) =t, 0 ≤ t <1
2 − t, 1 ≤ t < 2
0, otherwise
1/25/02 P16-8 © R. A. DeCarlo, P. M. Lin
(b) In order to compute the area beneath f1(t − ) ⋅ f2( ) four regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, and 2 ≤ t.Step 1: t < 0. Here f1(t − ) ⋅ f2( ) = 0 for all . Hence
f4(t) = f1(t) ∗ f2(t) = 0 for t < 0.
Step 2: 0 ≤ t <1. In this case f1(t − ) ⋅ f2( ) = for 0 ≤ ≤ t and is zero otherwise. Therefore the areabeneath f1(t − ) ⋅ f2( ) equals
f4(t) = 0.5t2 for 0 ≤ t <1.
Step 3: 1 ≤ t < 2. For t in this region f1(t − ) ⋅ f2( ) = for t −1 < <1 and is zero otherwise.Therefore the area beneath f1(t − ) ⋅ f2( ) equals
f4(t) = 0.5 − 0.5t2 for 1 ≤ t < 2.
Step 4: 2 ≤ t. Here f1(t − ) ⋅ f2( ) = 0 for all . Hence
f4(t) = f1(t) ∗ f1( t) = 0 for2 ≤ t.
In sum,
f4(t) =0.5t2, 0 ≤ t < 1
0.5 − 0.5t2, 1≤ t < 2
0, otherwise
(c) In order to compute the area beneath f2( t − ) ⋅ f2( ) four regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, and 2 ≤ t.
Step 1: t < 0. Here f2( t − ) ⋅ f2( ) = 0 for all . Hence
f5(t) = f2( t) ∗ f2( t) = 0 for t < 0.
Step 2: 0 ≤ t <1. In this case f2( t − ) ⋅ f2( ) = (t − ) for 0 ≤ ≤ t and is zero otherwise. Thereforethe area beneath f2( t − ) ⋅ f2( ) equals
f5(t) = ( t − ) d = −0.3333 3 + 0.5t 2[ ]0
t=
0
t
∫ 0.1667t3 for 0 ≤ t <1.
Step 3: 1 ≤ t < 2. For t in this region f2( t − ) ⋅ f2( ) = (t − ) for t −1 < <1 and is zero otherwise.Therefore the area beneath f2( t − ) ⋅ f2( ) equals
f5(t) = (t − ) d =t−1
1
∫ −0.3333 3 + 0.5t 2[ ]t−1
1
= −0.1667t3 + t − 0.6667 for 1 ≤ t < 2.
Step 4: 2 ≤ t. Here f2( t − ) ⋅ f2( ) = 0 for all . Hence
1/25/02 P16-9 © R. A. DeCarlo, P. M. Lin
f5(t) = f2( t) ∗ f2( t) = 0 for2 ≤ t.In sum,
f5(t) =0.1667t3, 0 ≤ t <1
−0.1667t3 + t − 0.6667, 1 ≤ t < 2
0, otherwise
SOLUTION 16.36.In order to compute the area beneath f1(t − ) ⋅ f2( ) five regions will be considered: t < 0, 0 ≤ t <1,1 ≤ t < 2, 2 ≤ t < 3and 3 ≤ t.
Step 1: t < 0. Here f1(t − ) ⋅ f2( ) = 0 for all . Hencef3(t) = f1(t) ∗ f2(t) = 0 for t < 0.
Step 2: 0 ≤ t <1. In this case f1(t − ) ⋅ f2( ) = for 0 ≤ ≤ t and is zero otherwise. Therefore the areabeneath f1(t − ) ⋅ f2( ) equals
f3(t) = 0.5t2 for 0 ≤ t <1.
Step 3: 1 ≤ t < 2. For t in this region
f1(t − ) ⋅ f2( ) =, t −1< < 1
2 − , 1≤ ≤ t
0, otherwise
Therefore the area beneath f1(t − ) ⋅ f2( ) equals
f3(t) = 0.5 − 0.5( t −1)2[ ] + 0.5 − 0.5(2 − t)2[ ]= −t2 + 3t −1.5 for 1 ≤ t < 2.
Step 4: 2 ≤ t < 3. For t in this region f1(t − ) ⋅ f2( ) = 2 − for all t −1 < < 2. Hence
f3(t) = f1(t) ∗ f2(t) = 0.5 − 0.5(2 − t)2 = −0.5t2 + 2t −1.5 for 2 ≤ t < 3.
Step 5: 3 ≤ t. Here f1(t − ) ⋅ f2( ) = 0 for all . Hencef3(t) = f1(t) ∗ f2(t) = 0 for 3 ≤ t.
In sum,
f3(t) =
0.5t2, 0 ≤ t < 1
− t2 + 3t −1.5, 1≤ t < 2
−0.5t2 + 2t −1.5, 2 ≤ t < 3
0, otherwise
SOLUTION 16.37.
1/25/02 P16-10 © R. A. DeCarlo, P. M. Lin
(a) In order to compute the area beneath f1( ) ⋅ f2( t − ) three regions will be considered: t < 0,0 ≤ t < 2, and 2 ≤ t.Step 1: t < 0. Here f1( ) ⋅ f2( t − ) = 0 for all . Hence
f3(t) = f1(t) ∗ f2(t) = 0 for t < 0.
Step 2: 0 ≤ t < 2. In this case f1( ) ⋅ f2( t − ) = 8 for 0 ≤ ≤ t and is zero otherwise. Therefore thearea beneath f1( ) ⋅ f2( t − ) equals
f3(t) = 0.5(t ⋅ 8t) = 4t2 for 0 ≤ t < 2.
Step 3: 2 ≤ t. For t in this region
f1( ) ⋅ f2( t − ) =8 , 0 ≤ < 2
16, 2 ≤ < t
0, otherwise
Therefore the area beneath f1( ) ⋅ f2( t − ) equals
f3(t) = 16 +16( t − 2) =16( t −1) for 2 ≤ t.In sum,
f3(t) =4t2, 0 ≤ t < 2
16( t −1), 2 ≤ t
0, otherwise
A picture of f3(t) is sketched in the next figure.
(b) From figure P16.37 we observe that
f1(t) = 2tu(t) − 2( t − 2)u(t − 2)
From table 13.1 and the time shift property of the Laplace transform it follows that
F1(s) =2
s2 1− e−2s( )
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F2(s) =4s
By the convolution theoremF3(s) = F1(s)F2(s)
Therefore
F3(s) =8
s3 1− e−2s( )Taking the inverse Laplace transform yields
f3(t) = 4t2u(t) − 4( t − 2)2u( t − 2)
The results of parts (a) and (b) coincide.
SOLUTION 16.38.(a) In order to compute the area beneath f1(t − ) ⋅ f2( ) six regions will be considered: t < 0, 0 ≤ t < 2,2 ≤ t < 6, 6 ≤ t < 8, 8 ≤ t <10, and 10 ≤ t.Step 1: t < 0. Here f1(t − ) ⋅ f2( ) = 0 for all . Hence
f3(t) = f1(t) ∗ f2(t) = 0 for t < 0.
Step 2: 0 ≤ t < 2. In this case f1(t − ) ⋅ f2( ) = 8( t − ) for 0 ≤ ≤ t and is zero otherwise. Thereforethe area beneath f1( ) ⋅ f2( t − ) equals
f3(t) = 0.5(t ⋅ 8t) = 4t2 for 0 ≤ t < 2.
Step 3:2 ≤ t < 6. For t in this region
f1(t − ) ⋅ f2( ) =16, 0 ≤ < t − 2
8(t − ), t − 2 ≤ < t
0, otherwise
Therefore the area beneath f1(t − ) ⋅ f2( ) equals
f3(t) = 16 +16( t − 2) =16( t −1) for 2 ≤ t < 6.
Step 4: 6 ≤ t < 8. For t in this region
f1(t − ) ⋅ f2( ) =
16, 0 ≤ < t − 2
8(t − ), t − 2 ≤ < 6
−8( t − ), 6 ≤ < t
0, otherwise
Therefore, for 6 ≤ t < 8, the area beneath f1(t − ) ⋅ f2( ) equals
f3(t) = 16(t − 2) + 16 −8(t − 6)2
2
−
8( t − 6)2
2
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= −8(t − 6)2 +16t −16 = −8t2 +112t − 304
Step 5: 8 ≤ t <10. For t in this region
f1(t − ) ⋅ f2( ) =
16, 0 ≤ < 6
−16, 6 ≤ < t − 2
−8( t − ), t − 2 ≤ < 8
0, otherwise
Therefore, for 8 ≤ t <10, the area beneath f1(t − ) ⋅ f2( ) equals
f3(t) = 96 −16(t − 8) − 16 −8(t − 8)2
2
= 4t2 − 80t + 464
Step 6: 10 ≤ t. For t in this region
f1(t − ) f2( ) =16, 0 ≤ < 6
−16, 6 ≤ < 8
0, otherwise
Therefore,f3(t) = 96 − 32 = 64 for 10 ≤ t.
In sum,
f3(t) =
0, t < 0
4t2, 0 ≤ t < 2
16t −16, 2 ≤ t < 6
−8t2 +112t − 304, 6 ≤ t < 8
4t2 − 80t + 464, 8 ≤ t < 10
64, 10 ≤ t
A picture of f3(t) is sketched in the next figure.
(b) From figure P16.38 we observe that
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f2( t) = 4 u( t) − u( t − 6)[ ] − 4 u(t − 6)u − (t − 8)[ ]= 4u(t) − 8u( t − 6) + 4u(t − 8)
From table 13.1 and the time shift property of the Laplace transform it follows that
F2(s) =4s
−8s
e−6s +4s
e−8s
From problem 16.37 we have that
F1(s) =2
s2 1− e−2s( )By the convolution theorem
F3(s) = F1(s)F2(s)Therefore
F3(s) =8
s3 −8
s3 e−2s −16
s3 e−6s +24
s3 e−8s −8
s3 e−10s
Taking the inverse Laplace transform yields
f3(t) = 4t2u(t) − 4( t − 2)2u( t − 2) − 8( t − 6)2 u(t − 6)
+12(t − 8)2 u(t − 8) − 4( t −10)2u( t −10)
The results of parts (a) and (b) coincide.
SOLUTION 16.39.p( t) ∗ q(t) will be computed using the techniques of convolution algebra. Therefore we can write
p( t) ∗ q(t) = p (−1)( t) ∗ q(1) (t)
where the superscript (−1)means integration and the superscript (1)means differentiation. By inspection
p( t) = ( t + 4) u(t + 4) − u( t)[ ] + (− t + 4) u( t) − u(t − 4)[ ] + 4 u( t − 4) − u(t − 8)[ ] == (t + 4)u( t + 4) + (−2t)u(t) + tu(t − 4) − 4u( t − 8)
Therefore,
p(−1)(t) = 0.5(t + 4)2 u(t + 4) − t2u( t) + (0.5t2 − 8)u( t − 4) − (4t − 32)u(t − 8)
By inspection we also have
q (1)(t) = 4 ( t)
By the sifting property of the delta function it follows that
p( t) ∗ q(t) = 4 p (−1)( t) == 2(t + 4)2 u(t + 4) − 4t2u(t) + 2( t2 −16)u(t − 4) −16(t − 8)u( t − 8).
SOLUTION 16.40.(a) First observe that
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h(t) = 0.1u( t − 0.1) + 0.2u(t − 0.2) + 0.2u(t − 0.3) + 0.2u(t − 0.4) + 0.2u( t − 0.5) ++0.1u( t − 0.6) − 0.1u(t −1)− 0.2u(t −1.3)− 0.2u(t −1.5) −
−0.2u(t −1.7) − 0.2u( t − 2) − 0.1u( t − 2.2)Due to the fact that h(t) is a linear combination of terms of the type Ku(t − T ), the convolution of h(t)and vin (t) reduces to a linear combination of terms of the following type: K1u(t)[ ]∗ K2u( t − T )[ ] . Usingthe definition of the convolution, the previous convolution product is computed below
K1u(t)[ ]∗ K2u( t − T )[ ] = K1K2 u( )u(t − − T )d−∞
∞
∫ =
= K1K2 u(t − − T )d0
∞
∫ =
0, t < T
K1K2 d0
t−T
∫ , T ≤ t
=0, t < T
K1K2( t − T ), T ≤ t
= K1K2( t − T )u( t − T )
Therefore vout (t) is a linear combination of functions of type K1K2( t − T )u(t − T ),
vout (t) = h(t) ∗vin (t) == 10(t − 0.1)u(t − 0.1)+ 20(t − 0.2)u( t − 0.2) + 20(t − 0.3)u( t − 0.3) + 20( t − 0.4)u(t − 0.4) +
+20(t − 0.5)u( t − 0.5) +10(t − 0.6)u(t − 0.6) −10(t −1)u(t −1)− 20(t −1.3)u( t −1.3) − −20(t −1.5)u(t −1.5) − 20(t −1.7)u(t −1.7) − 20(t − 2)u(t − 2) −10(t − 2.2)u(t − 2.2)
At t = 0svout (0) = 0 V.
At t = 0.5svout (0.5) = 16 V.
At t =1svout (1)= 65 V.
At t =1.5svout (1.5)= 106 V.
(b) In this case vout (t) will be computed using the techniques of the convolution algebra. Hence we have
vout (t) = vin ( t) ∗ h(t) = vin(−1)(t) ∗ h (1)(t)
= 50t2u(t)[ ]∗ 0.1 ( t − 0.1) + 0.2 ( t − 0.2) + 0.2 (t − 0.3) + 0.2 ( t − 0.4) +[+0.2 ( t − 0.5) + 0.1 ( t − 0.6) − 0.1 ( t −1)− 0.2 (t −1.3)−
−0.2 ( t −1.5) − 0.2 (t −1.7)− 0.2 (t − 2) − 0.1 ( t − 2.2)]
Using the sifting property of delta function it follows that
+−−+−−+−−= )3.0()3.0(10)2.0()2.0(10)1.0()1.0(5)( 222 tuttuttuttvout
+10(t − 0.4)2u( t − 0.4) +10( t − 0.5)2 u(t − 0.5) + 5( t − 0.6)2u(t − 0.6) −−5(t −1)2u( t −1)−10( t −1.3)2u( t −1.3) −10(t −1.5)2u( t −1.5) −
−10(t −1.7)2u( t −1.7) −10( t − 2)2 u(t − 2) − 5(t − 2.2)2u( t − 2.2) V.
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At t =1svout (1)= 22.25 V.
(c) From the expression of vout (t)obtained in part (a) we observe that vout (t)does not change aftert = 2.2s . Therefore it is sufficient to compute vout (t)for t ≤ 2.2s . Hence vin (t) can be considered to beequal to
vin (t) =100 u( t) − u(t − 2.2)[ ] V.Using the code of problem 16.31 we have
>> vin = 100*ones(1,22);>> h = [0, 0.1, 0.3, 0.5, 0.7, 0.9, 1, 1, 1, 1, 0.9, 0.9, 0.9, 0.7, 0.7, 0.5, 0.5, 0.3, 0.3, … …0.3, 0.1, 0.1];>> T = 0.1;>> tstep = T;>> y = tstep*conv(vin,h);>> y = [0 y 0];>> t = 0:tstep:tstep*(length(h)+length(vin));% After t = 2.2s vout(t) does not change>> t = t(1:length(h)+1);>> y = y(1:length(h)+1);>> plot(t,y)>> grid
A picture of y(t) is sketched in the next figure.
Using the previous MATLAB code we the values of y(t) at the specified instants of time are:At t = 0s
y(0) = 0 V
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At t = 0.5sy(0.5) = 16 V
At t =1sy(1)= 65 V
At t =1.5sy(1.5)= 106 V
The results of parts (a) and (c) coincide.
SOLUTION 16.41.Using the MATLAB code of problem 16.31 we have:
>> vin = [1];>> h = [9, -6, 3, -2];>> T = 1;>> tstep = T;>> y = tstep*conv(vin,h);>> y = [0 y 0];>> t = 0:tstep:tstep*(length(h)+length(vin));>> plot(t,y)>> grid
The breakpoints in y(t) of the above figure are [9, -6, 3 –2] as expected because the polynomialassociated with vin (t) is the constant 1 and the polynomial associated with h(t) is the polynomial
9x3 − 6x2 + 3x − 2, as it can be observed from figure P16.41.
SOLUTION 16.42.(a) Let vout ,40( t) denote the response that has been obtained in problem 16.40, part (a), to the input100u( t) .The expression of vout ,40( t) is (see problem 16.40, part (a)):
vout ,40( t) = h( t) ∗ 100u( t)[ ] == 10(t − 0.1)u(t − 0.1)+ 20(t − 0.2)u( t − 0.2) + 20(t − 0.3)u( t − 0.3) + 20( t − 0.4)u(t − 0.4) +
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+20(t − 0.5)u( t − 0.5) +10(t − 0.6)u(t − 0.6) −10(t −1)u(t −1)− 20(t −1.3)u( t −1.3) −−20(t −1.5)u(t −1.5) − 20(t −1.7)u(t −1.7) − 20(t − 2)u(t − 2) −10(t − 2.2)u(t − 2.2)
Using the distributive property of the convolution product and the time invariance property it follows thatvout (t) = h(t) ∗vin (t) = h( t) ∗ 100u(t) −100u(t − 0.2)[ ] =
= h( t) ∗ 100u(t)[ ] − h(t) ∗ 100u( t − 0.2)[ ] == vout,40 (t) − vout,40 (t − 0.2)
Using the above expression of vout ,40( t) we have:vout ,40(0) = 0 V and vout ,40(−0.2) = 0V,
vout ,40(0.5) =16 V and vout ,40(0.3) = 4 V,vout ,40(1) = 65 V and vout ,40(0.8) = 45V,
vout ,40(1.5) =106 V and vout ,40(1.3)= 92V.
At t = 0svout (0) = vout ,40(0) − vout ,40(−0.2) = 0 V.
At t = 0.5svout (0.5) = vout ,40(0.5) − vout ,40(0.3) =12 V.
At t =1svout (1)= vout ,40(1)− vout ,40(0.8) = 20 V.
At t =1.5svout (1.5)= vout ,40(1.5)− vout ,40(1.3) =14 V.
(b) In this case vout (t) will be computed using the techniques of convolution algebra.We have
vout (t) = vin ( t) ∗ h(t) = vin(−1)(t) ∗ h (1)(t)
where the superscript (-1) means integration and the superscript (1) means differentiation.From figure P16.42 observe that
vin (t) =100t u( t) − u(t − 0.5)[ ] +100(1− t) u(t − 0.5) − u( t −1)[ ] == 100tu( t) +100(1− 2t)u(t − 0.5) +100(t −1)u( t −1)
Therefore
vin(−1)( t) = 50t2u(t) + (−100t2 +100t − 25)u( t − 0.5) + (50 t2 −100t + 50)u(t −1) = g(t)
By the sifting property of the delta function we have
vout (t) = vin(−1)(t) ∗ h (1)( t) = g( t) ∗ h (1) (t)
= g(t) ∗ 0.1 ( t − 0.1) + 0.2 (t − 0.2) + 0.2 (t − 0.3) + 0.2 (t − 0.4) +[+0.2 ( t − 0.5) + 0.1 ( t − 0.6) − 0.1 ( t −1)− 0.2 (t −1.3)−
−0.2 ( t −1.5) − 0.2 (t −1.7)− 0.2 (t − 2) − 0.1 ( t − 2.2)]=
= 0.1g(t − 0.1)+ 0.2g(t − 0.2) + 0.2g( t − 0.3) + 0.2g(t − 0.4) +
+0.2g( t − 0.5) + 0.1g( t − 0.6) − 0.1g( t −1)− 0.2g(t −1.3)−
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−0.2g(t −1.5) − 0.2g( t −1.7) − 0.2g( t − 2) − 0.1g(t − 2.2)
The values ofvout (t) at the specified instants of time can be computed using MATLAB. The results are:vout (0) = 0V
vout (0.5) = 2.2Vvout (1)= 17.85V
vout (1.5)= 23.3V.
SOLUTION 16.43.
Using the techniques of convolution algebra we have
vout (t) = h(t) ∗vin (t) = h (−1)(t) ∗vin(1)(t)
where the superscript (-1) means integration and the superscript (1) means differentiation.We have
h (−1)( t) = 2(1− e−2t )u(t)and
vin(1)(t) = ( t) − ( t −1)
Using the sifting property of the delta function it follows that
vout (t) = 2(1− e−2 t )u(t)[ ]∗ ( t) − (t −1)[ ] =
= 2(1− e−2t )u( t) − 2(1− e−2(t−1))u( t −1)A picture of vout (t)is sketched in the next figure.
SOLUTION 16.44.(a) From table 13.1 it follows that the Laplace transform of vin (t) is
Vin (s) =1s
−1
s +1−
1s + 2
=−s2 + 2
s(s +1)(s + 2)
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And the Laplace transform of vout (t) is
Vout (s) =1s
−2
s +1−
1
(s +1)2 +1
s + 2=
−s2 + 2
s(s +1)2(s + 2)
Therefore the transfer function of the circuit is
H(s) =Vout (s)Vin (s)
=1
s +1
A simple RC circuit that represents this transfer function is a series RC circuit with R =1Ω and C = 1F .vout (t)is represented by the capacitor voltage and vin (t) is the source voltage.
(b) The impulse response of the circuit is
h(t) = L−1 H (s)[ ] = L−1 1s +1
= e−tu( t).
(c) Assuming zero initial conditions we have
vout (t) = vin ( t) ∗ h(t) = vin(1)( t) ∗ h (−1)(t) = (t) ∗ (1− e−t )u( t)[ ] = (1− e−t )u( t) V.
(d) Using the techniques of convolution algebra the zero-state response can be computed as
vout (t) = vin ( t) ∗ h(t) = vin(2) (t) ∗ h (−2)(t)
where the superscript (2) means double differentiation and the superscript (-2) means double integration.First, from figure P16.44, observe that
vin(1)(t) = u(t −1)− u(t − 2)[ ] + u(t − 3)− u(t − 4)[ ]
Therefore
vin(2)( t) = (t −1)− (t − 2) + (t − 3) − ( t − 4)
h (−2) (t) is computed as the integral of h (−1)( t) .
h (−2) (t) = h (−1)( )d−∞
t
∫ = (1− e− )u( )d−∞
t
∫ = (1− e− )d
0−
t
∫ = (t + e−t −1)u(t)
The zero-state response can now be computed
vout (t) = vin(2) (t) ∗ h (−2)( t) =
= ( t −1)− (t − 2) + (t − 3)− (t − 4)[ ]∗ ( t + e−t −1)u( t)[ ]By the sifting property of the delta function it follows that
vout (t) = t − 2 + e−(t−1)( )u(t −1)− t − 3 + e−(t−2)( )u( t − 2)
+ t − 4 + e−(t−3)( )u(t − 3)− t − 5 + e−(t−4)( )u( t − 4) V.
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SOLUTION 16.45.Using the techniques of the convolution algebra we have
y(t) = f ( t) ∗ g( t) = f (2) (t) ∗ g(−2)( t)where
g(−1)(t) = 2 cos( )d = sin( t)u( t)
0−
t
∫and
g(−2)( t) = 2 sin( )d = 1− cos( t)[ ]u( t)
0−
t
∫Differentiating f ( t) twice leads to
f (2) (t) = − ( t) + 2 (t −1)− 2 ( t − 3)+ (t − 4)Therefore
y(t) = f (2) (t) ∗ g(−2)( t) == − ( t) + 2 ( t −1)− 2 ( t − 3) + ( t − 4)[ ]∗ 1− cos( t)[ ]u(t)
Using the sifting property of the delta function it follows that
y(t) = − 1− cos( t)[ ]u( t) + 2 1− cos[ (t −1)] u(t −1)−2 1− cos[ (t − 3)] u( t − 3) + 1− cos[ ( t − 4)] u(t − 4)
Simplifying the expression of y(t) yields
y(t) = − 1− cos( t)[ ]⋅ u(t) − u( t − 4)[ ] + 2 ⋅ 1 + cos( t)[ ]⋅ u(t −1)− u(t − 3)[ ] .
A picture of y(t) is sketched in the next figure.
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SOLUTION 16.46.(a) By the current division formula
IC (s) =Cs
Cs +1
Ls
⋅ Iin (s) =s2
s2 +1⋅ Iin (s)
The transfer function of the circuit can be computed as below
H(s) =VC (s)Iin (s)
=IC (s) ⋅ ZC (s)
Iin (s)=
s2
s2 +1⋅1s
=s
s2 +1
(b) The impulse response is computed as the inverse Laplace transform of the transfer function
h(t) = L−1 H (s)[ ] = cos(t)u( t).
(c) Assuming zero initial conditions it follows, by the impulse response theorem, that
vout (t) = iin (t) ∗ h(t)
Using the techniques of the convolution algebra we have
vout (t) = iin(2) (t) ∗ h (−2) (t)
By inspection
iin(1)( t) = u( t) − u( t − 2 )[ ] + u(t − 4 ) − u( t − 6 )[ ]
Therefore
iin(2) (t) = ( t) − ( t − 2 ) + (t − 4 ) − ( t − 6 )
And
h (−1)( t) = cos( )d = sin(t)u( t)
0−
t
∫Hence
h (−2) (t) = sin( )d = 1− cos(t)[ ]u(t)
0−
t
∫
Using the sifting property of the delta function we have
vout (t) = (t) − (t − 2 ) + ( t − 4 ) − (t − 6 )[ ] ∗ 1− cos(t)[ ]u( t) = 1− cos(t)[ ]u(t) − 1− cos(t − 2 )[ ]u(t − 2 )
+ 1− cos(t − 4 )[ ]u(t − 4 ) − 1− cos(t − 6 )[ ]u(t − 6 )= 1− cos(t)[ ]⋅ u( t) − u( t − 2 ) + u(t − 4 ) − u( t − 6 )[ ]
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V.(d) A picture of vout (t) is sketched in the next figure.
SOLUTION 16.47.(a) The step response, vout (t), is computed using the convolution algebra techniques.We have
vout (t) = h(t) ∗vin (t) = h (1)( t) ∗ vin(−1)(t)
From figure P16.47 observe that
h(t) = 2u( t) − u(t −1)− 2u(t − 2) − u( t − 3)+ u(t − 5) + 2u(t − 6) + u( t − 7) − 2u( t − 8)
Differentiating we have
h (1)(t) = 2 ( t) − ( t −1)− 2 (t − 2) − (t − 3) + (t − 5) + 2 (t − 6) + ( t − 7) − 2 ( t − 8)Since
vin (t) = u(t),by integration it follows that
vin(−1)( t) = tu(t).
Using the sifting property of the delta function it follows that
vout (t) = h (1) (t) ∗ vin(−1)( t)
= 2 ( t) − (t −1)− 2 (t − 2) − ( t − 3) + (t − 5) + 2 (t − 6) + (t − 7) − 2 ( t − 8)[ ] ∗ tu( t)[ ]= 2tu( t) − ( t −1)u( t −1)− 2( t − 2)u( t − 2) − ( t − 3)u(t − 3) +
+( t − 5)u(t − 5) + 2(t − 6)u( t − 6) + (t − 7)u( t − 7) − 2(t − 8)u(t − 8)V.
A picture of the step response is sketched in the next figure.
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(b) Using the convolution algebra techniques we have
vout (t) = h(t) ∗vin (t) = h (1)( t) ∗ vin(−1)(t)
For computing vin(−1)( t) we have
vin(−1)( t) = e u(− )d =
e d−∞
t
∫ , t < 0
e d−∞
0
∫ , 0 ≤ t
−∞
t
∫ =et , t < 0
1, 0 ≤ t
Using the sifting property of the delta function it follows that
vout (t) = h (1) (t) ∗ vin(−1)( t) =
= 2 ( t) − (t −1)− 2 (t − 2) − ( t − 3) + (t − 5) + 2 (t − 6) + (t − 7) − 2 ( t − 8)[ ] ∗ vin(−1)(t)[ ] =
= 2vin(−1)(t) − vin
(−1)(t −1)− 2vin(−1)(t − 2) − vin
(−1)( t − 3) + +vin
(−1)(t − 5) + 2vin(−1)( t − 6) + vin
(−1)(t − 7) − 2vin(−1)(t − 8) V.
Using the expression of vin(−1)( t) computed above it follows that
vout (7.5) = 0.7869 V,vout (6.5) = 1.1603V,vout (5.5) = 0.2720V,vout (0.5) = 0.8848V.
SOLUTION 16.48.First observe, from figure P16.48(a), that
h(t) = (1− t) ⋅ u(t) − u(t −1)[ ] =
1/25/02 P16-24 © R. A. DeCarlo, P. M. Lin
= (1− t)u(t) − (1− t)u(t −1).Using the convolution algebra techniques it follows that the step response y(t) is
y(t) = v(t) ∗ h( t) = v(1) (t) ∗ h (−1)( t) == (t) ∗ h (−1)(t) = h (−1)(t)
where
h (−1)( t) = (1− )u( )d − (1− )u( −1)d =−∞
t
∫−∞
t
∫
= − 0.5 2[ ]0
t⋅ u(t) − − 0.5 2[ ]1
t⋅ u( t −1) =
= (t − 0.5t2)u( t) − ( t − 0.5t2 − 0.5)u(t −1).Hence the step response is
y(t) = (t − 0.5t2)u(t) − ( t − 0.5t2 − 0.5)u( t −1).
Let y v( t) denote the zero-state response to the input v(t).From figure P16.48(b) we observe that
v(t) = u( t) + u( t −1)− 2u( t − 2) .Using the distributive property of the convolution it follows that
y v( t) = h( t) ∗ v(t) == h( t) ∗ u(t) + u(t −1)− 2u(t − 2)[ ] =
= h( t) ∗ u( t) + h(t) ∗ u( t −1)− 2 ⋅ h( t) ∗ u(t − 2).Due to the fact that
y(t) = h(t) ∗ u(t)by the linearity and time invariance properties it follows that
y v( t) = y( t) + y(t −1)− 2y( t − 2) = = (t − 0.5t2)u( t) + (t −1)u(t −1)+
+(1.5t2 − 8t +12)u(t − 2) + (0.5t2 − 3t + 4)u( t − 3).
SOLUTION 16.49.(a) By the voltage division formula it follows that
Vout (s) =
1
Cs1
Cs+ Ls
⋅Vin (s) =1
LCs2 +1⋅Vin (s)
Therefore the transfer function is
H(s) =Vout (s)Vin (s)
=1
LCs2 +1Taking the inverse Laplace transform yields
h(t) = L−1 1
LCs2 +1
=
1
LCL−1
1
LC
s2 +1
LC
=1
LCsin
1
LCt
u( t) .
(b) The step response is computed as the convolution of the impulse response and the step function.vout (t) = h(t) ∗ u( t)
Using the techniques of the convolution algebra it follows that
1/25/02 P16-25 © R. A. DeCarlo, P. M. Lin
vout (t) = h (−1)(t) ∗ u(1) (t)where the superscript (-1) means integration and the superscript (1) means differentiation.Taking the integral of h(t) we have
h (−1)( t) =1
LCsin
1
LC
u( )d
−∞
t
∫ =
= u(t) ⋅ −cos1
LC
0
t
= 1− cos1
LCt
⋅ u( t) .
Therefore
vout (t) = h (−1)(t) ∗ u(1) (t) =
= 1− cos1
LCt
⋅ u(t)
∗ (t) =
= 1− cos1
LCt
⋅ u( t) V.
(c) We denote by voutT (t) the output to the rectangular pulse in figure P16.49(b).
Observe, from figure P16.49(b), that
vin (t) =1T
⋅ u( t) − u(t − T )[ ] V.
By linearity and time invariance it follows that
voutT (t) =
1T
⋅ vout (t) − vout (t − T )[ ]where vout (t)is the step response obtained in part (b). Therefore
voutT (t) =
1
2 LC1− cos
1
LCt
⋅ u(t) −
1
2 LC1− cos
1
LCt − 2
⋅ u( t − 2 LC )
=1
2 LC1− cos
1
LCt
⋅ u(t) − u( t − 2 LC[ ]
A picture of voutT (t), for L = 1H and C = 1F , is sketched in the next figure.
1/25/02 P16-26 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.50.The impulse response of the configuration in the figure P16.50 is
h(t) = h1(t) ∗ h2(t) + h3( t)[ ] ∗ h4( t)Due to the fact that h4(t) = 2 ( t) , the sifting property of the delta function can be applied and it followsthat
h(t) = 2 ⋅ h1(t) ∗ h2(t) + h3( t)[ ]Using the techniques of the convolution algebra we can further write
h(t) = 2 ⋅ h1(1) (t) ∗ h2(t) + h3( t)[ ](−1) =
h(t) = 2 ⋅ h1(1) (t) ∗ h2
(−1)(t) + h3(−1)( t)[ ]
We have
h1(1)(t) = u (1)(t) = ( t)
h2(−1)( t) = 2e−2 u( )d = u( t) ⋅ −e−2[ ]0
t= 1− e−2t( )u(t)
−∞
t
∫
h3(−1)( t) = 8e−4 u( )d = u(t) ⋅ −2e−4[ ]0
t= 2 1− e−4 t( )u(t)
−∞
t
∫
Substituting the expressions of h1(−1)( t) , h2
(−1)( t) and h3(−1)( t) in the expression of h(t), and using the
sifting property of the delta function we have
h(t) = 2 1− e−2 t( )u( t) + 4 1− e−4 t( )u(t).
A picture of h(t) is sketched in the next figure.
1/25/02 P16-27 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.51.Observe first that h2( t) , h3(t) and h4(t) have the same expressions as in the problem 16.50. In problem16.50 the following convolution has been computed
u( t) ∗ h2(t) + h3(t)[ ]∗ h4 (t) = 2 1− e−2t( )u( t) + 4 1− e−4 t( )u(t)
Using the time shift property of the convolution (see problem 16.22, part (a)) it follows that
u( t − 2) ∗ h2(t) + h3(t)[ ]∗ h4 (t) = 2 1− e−2(t−2)( )u(t − 2) + 4 1− e−4( t−2)( )u( t − 2)
Using the above expressions and the distributive property of the convolution it follows that the overallimpulse response can be computed as below
h(t) = h1(t) ∗ h2(t) + h3( t)[ ] ∗ h4( t) = u( t) − u(t − 2)[ ]∗ h2( t) + h3(t)[ ]∗ h4( t)
= u(t) ∗ h2( t) + h3(t)[ ] ∗ h4( t) − u(t − 2)∗ h2(t) + h3( t)[ ]∗ h4 (t) =
= 21 − e−2t( )u( t) + 4 1− e−4 t( )u(t) − 2 1− e−2(t−2)( )u(t − 2) + 4 1− e−4(t−2)( )u(t − 2).
A picture of h(t) is sketched in the next figure.
1/25/02 P16-28 © R. A. DeCarlo, P. M. Lin
SOLUTION 16.52.The overall impulse response of the configuration is
h(t) = h1(t) ∗ h2(t) + h3( t)[ ] ∗ h4( t)
Using the distributive property of the convolution we have
h(t) = h1(t) ∗ h2(t) ∗ h4 (t) + h1( t) ∗ h3(t) ∗ h4 (t)
Replacing the expressions for h2( t) and h3(t), and using the sifting property of the delta function itfollows that
h(t) = 2 ⋅ h1(t) ∗ (t) ∗ h4 (t) − 2 ⋅ h1(t) ∗ (t − 2) ∗ h4 (t)= 2 ⋅ h1( t) ∗ h4 (t) − 2 ⋅ h1(t) ∗ h4( t)[ ]t=t−2
Using the techniques of the convolution algebra it follows that
h1(t) ∗ h4( t) = h1(1)(t) ∗ h4
(−1)(t)
where the superscript (1) means differentiation and the superscript (-1) means integration.We have
h1(1)(t) = u (1)(t) = ( t)
and
h4(−1)( t) = 2e− u( )d
−∞
t
∫ = 2u( t) e− d0
t
∫ = 2u(t) −e−[ ]0
t= 2(1− e−t )u(t)
Therefore
h1(t) ∗ h4( t) = (t) ∗ 2(1− e−t )u(t)[ ] = 2(1− e−t )u( t)
1/25/02 P16-29 © R. A. DeCarlo, P. M. Lin
Replacing the above expression into the expression of h(t) it follows that
h(t) = 4(1− e−t )u( t) − 4 (1− e−t )u( t)[ ]t=t−2= 4(1− e−t )u( t) − 41 − e−(t−2)[ ]u(t − 2).
A picture of h(t) is sketched in the next picture.
SOLUTION 16.53.(a) The overall impulse response of the configuration in figure P16.53 is
h(t) = h1( t) + h2(t)[ ]∗ h3(t)
By the distributive property of convolution it follows that
h(t) = h1(t) ∗ h3( t) + h2(t) ∗ h3( t) = (t) ∗cos( t)u( t) + (t −1)∗cos( t)u( t)
By the sifting property of the delta function it follows that
h(t) = cos( t)u(t) + cos (t −1)[ ]u( t −1).
A picture of h(t) is sketched in the next figure.
1/25/02 P16-30 © R. A. DeCarlo, P. M. Lin
(b) The response y(t) is computed as
y(t) = h(t) ∗ u(t) = cos( t)u( t) + cos ( t −1)[ ]u(t −1) ∗ u(t)
Using the distributive property and the time shift property of convolution(see problem 16.22, part (a) ) wehave
y(t) = cos( t)u(t)[ ]∗ u( t) + cos ( t −1)[ ]u(t −1) ∗ u(t) == cos( t)u(t)[ ]∗ u(t) + cos( t)u(t)[ ] ∗ u(t) t=t−1
Using the techniques of convolution algebra, the convolution cos( t)u(t) ∗ u(t) is computed as
cos( t)u( t)[ ] ∗ u( t) = cos( t)u(t)[ ](−1) ∗ u (1)(t)
where the superscript (-1) means integration and the superscript (1) means differentiation.We have
cos( t)u( t)[ ](−1) = cos( )u( )d−∞
t
∫ = u(t) cos( )d =0
t
∫ u(t)1
sin( t)[ ]0t =
sin( t)u( t) .
Therefore
cos( t)u( t)[ ] ∗ u( t) =sin( t)
u( t)
∗ (t) =
sin( t)u( t)
Hence the step response is
y(t) =sin( t)
u(t) +sin (t −1)[ ]
u(t −1) =sin( t)
u(t) − u( t −1)[ ] .
SOLUTION 16.54.(a) The overall impulse response of the configuration in figure P16.53 is
1/25/02 P16-31 © R. A. DeCarlo, P. M. Lin
h(t) = h1( t) + h2(t)[ ]∗ h3(t)
By the distributive property of convolution it follows that
h(t) = h1(t) ∗ h3( t) + h2(t) ∗ h3( t) = (t) ∗cos( t)u( t) + (t − 3) ∗cos( t)u( t)
By the sifting property of the delta function it follows that
h(t) = cos( t)u(t) + cos (t − 3)[ ]u( t − 3).
A picture of h(t) is sketched in the next figure.
(b) The response y(t) is computed as
y(t) = h(t) ∗ u(t) = cos( t)u( t) + cos ( t − 3)[ ]u(t − 3) ∗ u(t)
Using the distributive property and the time shift property of convolution(see problem 16.22, part (a) wehave
y(t) = cos( t)u(t)[ ]∗ u( t) + cos ( t − 3)[ ]u(t − 3) ∗ u(t)
= cos( t)u(t)[ ]∗ u(t) + cos( t)u(t)[ ] ∗ u(t) t=t−3
Using the techniques of convolution algebra, the convolution cos( t)u(t) ∗ u(t) is computed as
cos( t)u( t)[ ] ∗ u( t) = cos( t)u(t)[ ](−1) ∗ u (1)(t)
where the superscript (-1) means integration and the superscript (1) means differentiation. We have
cos( t)u( t)[ ](−1) = cos( )u( )d−∞
t
∫ = u(t) cos( )d =0
t
∫ u(t)1
sin( t)[ ]0t =
sin( t)u( t) .
Therefore
1/25/02 P16-32 © R. A. DeCarlo, P. M. Lin
cos( t)u( t)[ ] ∗ u( t) =sin( t)
u( t)
∗ (t) =
sin( t)u( t)
Hence the step response is
y(t) =sin( t)
u(t) +sin (t − 3)[ ]
u(t − 3) =sin( t)
u(t) − u( t − 3)[ ] .
SOLUTION 16.55.(a) The overall impulse response of the configuration in figure P16.55 is
h(t) = h1( t) + h2(t) + h3( t) + h4 (t)[ ] ∗ h5( t)
By the distributive property of convolution it follows that
h(t) = h1(t) ∗ h5( t) + h2(t) ∗ h5( t) + h3(t) ∗ h5(t) + h4( t) ∗ h5( t) = = (t) ∗cos( t)u( t) + (t −1)∗cos( t)u( t) −
− ( t − 3)∗ cos( t)u(t) − (t − 4) ∗cos( t)u(t)
By the sifting property of the delta function it follows thath(t) = cos( t)u(t) + cos (t −1)[ ]u( t −1)−
−cos (t − 3)[ ]u( t − 3)− cos (t − 4)[ ]u(t − 4) .
A picture of h(t) is sketched in the next figure.
(b) The response y(t) is computed as
y(t) = h(t) ∗ u(t) = cos( t)u( t) + cos ( t −1)[ ]u(t −1)−cos (t − 3)[ ]u( t − 3)− cos (t − 4)[ ]u(t − 4)∗ u( t)
1/25/02 P16-33 © R. A. DeCarlo, P. M. Lin
Using the distributive property and the time shift property of convolution (see problem 16.22, part (a)) wehave
y(t) = cos( t)u(t)[ ]∗ u( t) + cos ( t −1)[ ]u(t −1) ∗ u(t) − − cos (t − 3)[ ]u( t − 3) ∗ u(t) − cos ( t − 4)[ ]u( t − 4) ∗ u(t) =
= cos( t)u(t)[ ]∗ u(t) + cos( t)u(t)[ ] ∗ u(t) t=t−1 − − cos( t)u( t)[ ]∗ u( t) t=t−3 − cos( t)u( t)[ ] ∗ u(t) t=t−4
Using the techniques of convolution algebra, the convolution cos( t)u(t) ∗ u(t) is computed as
cos( t)u( t)[ ] ∗ u( t) = cos( t)u(t)[ ](−1) ∗ u (1)(t)
where the superscript (-1) means integration and the superscript (1) means differentiation. We have
cos( t)u( t)[ ](−1) = cos( )u( )d−∞
t
∫ = u(t) cos( )d =0
t
∫ u(t)1
sin( t)[ ]0t =
sin( t)u( t) .
Therefore
cos( t)u( t)[ ] ∗ u( t) =sin( t)
u( t)
∗ (t) =
sin( t)u( t)
Hence the step response is
y(t) =sin( t)
u(t) +sin (t −1)[ ]
u(t −1) −sin (t − 3)[ ]
u(t − 3)−sin ( t − 4)[ ]
u(t − 4)
=sin( t)
u(t) − u( t −1)+ u(t − 3) − u( t − 4)[ ] .
SOLUTION 16.56.(a) By definition
h(t) ∗ f (t) = h(t − ) f ( )d−∞
∞
∫By making a change of variable
1 = t −we have
h(t) ∗ f (t) = h( 1) f (t − 1)d 1−∞
∞
∫ = f ( t − 1)h( 1)d 1−∞
∞
∫By definition
f ( t) ∗ h(t) = f (t − )h( )d−∞
∞
∫From the above expressions we observe that
h(t) ∗ f (t) = f ( t) ∗ h( t)because and 1 are only variables of integration.
1/25/02 P16-34 © R. A. DeCarlo, P. M. Lin
(b) Using the definition of convolution we have
h( t) ∗ f ( t)[ ] ∗ g( t) = h( ) f (t − )d−∞
∞
∫
∗ g(t) = h( ) f (t − 1 − )d−∞
∞
∫
g( 1)
d 1
−∞
∞
∫ =
= h( ) f (t − 1 − )g( 1)[ ]d−∞
∞
∫
d 1
−∞
∞
∫
Changing the order of integration we have
h( t) ∗ f ( t)[ ] ∗ g( t) = h( ) f (t − 1 − )g( 1)d 1−∞
∞
∫
d
−∞
∞
∫
By the definition of the convolution we have
f ( t) ∗ g( t) = f (t − )g( )d−∞
∞
∫Therefore
f (t) ∗ g(t)[ ]t=t− = f ( t − − )g( )d−∞
∞
∫Hence
h( t) ∗ f ( t)[ ] ∗ g( t) = h( ) ⋅ f ( t) ∗ g( t)[ ]t=t− d−∞
∞
∫
By the definition of the convolution product we have
h(t) ∗ f ( t) ∗ g(t)[ ] = h( ) ⋅ f ( t) ∗ g( t)[ ]t=t− d−∞
∞
∫Therefore
h( t) ∗ f ( t)[ ] ∗ g( t) = h( t) ∗ f (t) ∗ g( t)[ ] .
Thus the associative property of convolution is proved.
SOLUTION 16.57.We have
f ( t) ⋅ h( t) = f (t) (t − kT )k=0
∞∑
For some nonnegative k , the Laplace transform of f ( t) (t − kT) is
L f (t) ( t − kT)[ ] = f (kT)e−skT
1/25/02 P16-35 © R. A. DeCarlo, P. M. Lin
by the sifting property of the delta function. Therefore we have
L f (t)h(t)[ ] = L f ( t) ( t − kT)k =0
∞∑
= = f (kT )e−skT
k =0
∞∑
Using the notation z = esT we have
L f (t)h(t)[ ] = f (kT)z−k
k =0
∞∑ .
SOLUTION 16.58.(a) By the voltage division formula we have
Vout (s) =
1
Cs
R +1
Cs
⋅Vin (s)
Therefore the transfer function of the circuit is
H(s) =Vout (s)Vin (s)
=
1
Cs
R +1
Cs
=1
CRs +1=
12s +1
.
Taking the inverse Laplace transform yields
h(t) = L-1 12s +1
= = L-1
1
2
s +1
2
= 0.5e−0.5tu(t).
(b) By the impulse response theorem it follows thatvout (t) = h(t) ∗vin (t) =
= 0.5e−0.5tu(t)[ ]∗ ( t) + ( t −1) + (t − 2) +K[ ]
Using the sifting property of the delta function it follows that
vout (t) = 0.5e−0.5tu( t) + 0.5e−0.5(t−1)u(t −1) + 0.5e−0.5(t−2)u(t − 2) +K V.
Therefore for 0 < t <1
vout (t) = 0.5e−0.5t V
because only u( t) is nonzero for 0 < t <1.
(c) From the expression of vout (t)obtained in the part (b) it follows that
1/25/02 P16-36 © R. A. DeCarlo, P. M. Lin
vout (t) = 0.5e−0.5t + 0.5e−0.5(t−1) V, for 1 < t < 2
because only u( t) and u( t −1) are nonzero for 1 < t < 2.
(d) For t in the interval (4,5) , only u( t) , u( t −1), u( t − 2) , u( t − 3) and u( t − 4) are nonzero. Therefore
vout (t) = 0.5e−0.5t + 0.5e−0.5(t−1) + 0.5e−0.5(t−2) + 0.5e−0.5(t−3) + 0.5e−0.5(t−4) V.
The above expression can be written as
vout (t) = 0.5e−0.5(t−4) 1+ e−0.5( ) + e−0.5( )2+ e−0.5( )3
+ e−0.5( )4
V.
(e) Using the sum formula for geometric series with = e−0.5 we have
1 + e−0.5( ) + e−0.5( )2+ e−0.5( )3
+ e−0.5( )4=
1− e−0.5( )5
1− e−0.5
vout (t) = 0.5e−0.5(t−4) ⋅1− e−0.5( )5
1− e−0.5 V.
Therefore
vout (t) = 8.6189 ⋅ e−0.5t V.
(f) Using the expression of vout (t)obtained in part (b) it follows that, for n < t < n +1,
vout (t) = 0.5e−0.5t + 0.5e−0.5(t−1) + 0.5e−0.5(t−2) +K + 0.5e−0.5(t−n) =
= 0.5e−0.5(t−n) 1+ e−0.5( ) + e−0.5( )2
+K + e−0.5( )n
V.
Using the sum formula for geometric series we have
vout (t) = 0.5e−0.5(t−n) ⋅1− e−0.5( )n+1
1− e−0.5 V.
(g) For large n , e−0.5( )n+1≅ 0. Therefore, for large n we have
vout (t) ≅ e−0.5(t−n) ⋅0.5
1− e−0.5 V for n < t < n +1.
A picture of vout (t), for large n , is sketched in the next figure.
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SOLUTION 16.59.By the impulse response theorem we have
vout (t) = h(t) ∗vin (t) = 0.5e−0.5tu(t)[ ]∗ ( t) + ( t +1)+ (t + 2) +K[ ] V.
Using the sifting property of the delta function it follows that
vout (t) = 0.5e−0.5tu( t) + 0.5e−0.5(t+1)u(t +1) + 0.5e−0.5(t+ 2)u(t + 2) +K V.
For 0 < t we have
vout (t) = 0.5e−0.5t + 0.5e−0.5(t+1) + 0.5e−0.5(t+ 2) +K =
= 0.5e−0.5t 1+ e−0.5( ) + e−0.5( )2
+K
V.
Using the sum formula for geometric series( for n = ∞ ) we have
vout (t) = 0.5e−0.5t ⋅1
1− e−0.5 V for 0 < t.
Simplifying the expression of vout (t) it follows that
vout (t) = 1.2707 ⋅ e−0.5t V for 0 < t.
SOLUTION 16.60. First we plot for reader convenience vin(t) and its staircase approximation.
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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 11
1.2
1.4
1.6
1.8
2
2.2
2.4
2.6
2.8vin and its Staircase Approximation
The transfer function of the circuit of figure 16.58 is H(s) =1
2s + 1 in which case
h(t) = 0.5e−0.5tu(t). Because we only want the output for 0 ≤ t ≤ 2, we only need h(t) for 0 ≤ t ≤ 2 s.
Hence we need to generate staircase approximations to both vin(t) and h(t) as follows:
t = 0:0.05:2;
vin = exp(t .^2) .* (u(t) - u(t - 1));
h = 0.5*exp( - 0.5*t) .* (u(t) - u(t - 2));
T = 0.05;
tstep = T;
y = [0 conv(vin,h)*tstep 0];
t = 0:tstep:tstep*(length(vin)+length(h));
% For plotting through time 2 s we set
t=t(1:41);
y = y(1:41);
plot(t,y)
grid
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0 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 20
0.1
0.2
0.3
0.4
0.5
0.6
0.7Convolution of vin and h
5/15/01 P17-1 R.A. DeCarlo & P.M. Lin
CHAPTER 17 PROBLEM SOLUTIONS
SOLUTION PROBLEM 17.11. Case 1: suppose R1 > R2. From example 17.3, page 696, if L
and C are connected as indicated in part (a), then Z1 can be made real and larger than R2. This
means we can solve the problem at least for Z1. Specifically, consider the figure below
From example 17.3, at a specified frequency, ωr, for which Z1 is real, then L, C, and R2 must
satisfy,
ωr =1
LC−
R22
L2 (1)
Further from example 17,3, at ωr,
Z1( jωr ) =L
R2C
We require that Z1( jωr ) = R1 in which case
R1R2 =L
C(2)
It is necessary to solve equations (1) and (2) simultaneously for L and C. From (2), L = R1R2C .
Substituting into the square of (1) yields
ωr2 =
1
R1R2C2 −R2
2
R12R2
2C2
Hence
C =1
ωrR1
R1 − R2
R2
It follows that
5/15/01 P17-2 R.A. DeCarlo & P.M. Lin
L = R1R21
ωr R1
R1 − R2
R2
=1
ωrR2 R1 − R2( )
Observe that since R1 > R2, both C and L are real, i.e., exist. Please note that this connection would
not result in real values of C and L had R1 < R2. If we can now show that Z2 = R2, then parts (a)
and (b) are valid for this case.
By direct computation
Z2 ( jωr) = jωr L +1
jωrC + 1
R1
= jωrL +R1
jωrCR1 +1= j R2 R1 − R2( ) +
R1
jR1 − R2
R2+1
= j R2 R1 − R2( ) +R1 − jR1
R1 − R2
R2
1+ R1 − R2
R2
= j R2 R1 − R2( ) + R2 − jR2R1 − R2
R2= R2
Thus, (a) and (b) are true for the case R1 > R2.
We can also arrive at the conclusion that Z2 = R2 using maximum power transfer concepts.
Since Z1 is constructed so that Z1 = R1, we have set up the conditions for maximum power transfer
of a V-source in series with R1 to the "load" Z1. Since the LC coupling network is lossless,
whatever average power is received by the network to the right of R1, will be dissipated by R2.
Therefore maximum power is transferred to the load R2. Looking back from R2, it must be that R2
sees a Thevenin resistance Z2 = R2 since it is known that there is a non-zero R1.
Case 2, R1 < R2 . Now consider the configuration
5/15/01 P17-3 R.A. DeCarlo & P.M. Lin
Interchanging the subscripts of 1 and 2 in case 1 produces the derivation for this case.
Combining cases 1 and 2 using the text notation produces
C =1
ωRlarge
Rlarge − Rsmall
RsmallL =
1
ωRsmall Rlarge − Rsmall( )
SOLUTION PROBLEM 17.16.
(a)
»C = 0.1e-6;
»R = 510;
»wr = 2*pi*1.56e3;
»%From equation 17.4 in example 17.3 which analyzes
»%this particular circuit,
»%L must satisfy
»% wr^2*L^2 - L/C + R^2 = 0.
»% Therefore
»
»L = roots([wr^2 -1/C R^2])
L =
5.3133e-02
5.0952e-02
(b) From HW problem 17.9,
ωr( )2 =1
LC−
1
R2C2
Hence, in MATLAB,
»Linv = C*(wr^2 + 1/(R^2*C^2))
Linv =
4.8054e+01
»L = 1/Linv
L = 2.0810e-02
5/15/01 P17-4 R.A. DeCarlo & P.M. Lin
SOLUTION PROBLEM 17.22. (Correction: change RL in figure P17.22 to R.) We solve this
problem using phasors rather than the Laplace transform approach as it is simpler. Consider
As such, since by definition ω0 = 1 LC , the resonant frequency,
VC = VL = Zin ( jω0 )Iin = RIin = RIm
Hence
vC (t) = vL (t) = RIm cos(ω0t) and iL (t) =RImω0 L
sin(ω0t).
(a) By direct computation
wC (t) =1
2CvC
2 (t) =1
2CR2Im
2 cos2(ω0t)
(b) By direct computation
wL( t) =1
2LiL
2(t) =R2Im
2
2ω02 L
sin2(ω0t)
(c) Also by direct computation, since ω02 = 1 LC ,
wC (t) + wL (t) =1
2CR2Im
2 cos2(ω0t) +R2Im
2
2ω02L
sin2(ω0t)
=1
2R2Im
2C cos2(ω0t) +1
ω02 LC
sin2(ω0t)
=1
2R2Im
2 C
(d) The energy dissipated in the resistance in one period is
wR(0,T ) = R iR2(t) dt
0
T
∫ = R iin2 (t) dt
0
T
∫ = RIm2 cos2(ω0t)dt
0
T
∫
5/15/01 P17-5 R.A. DeCarlo & P.M. Lin
=RIm
2
2dt
0
T
∫ +RIm
2
2cos(2ω0t) dt
0
T
∫ =RIm
2
2T =
RIm2 π
ω0
(e) Finally
2πmaximum energy stored
total energy lost per period= 2π
0.5R2Im2 C
RIm2 π
ω0
= ω0RC = Q
by equation 17.13.
SOLUTION PROBLEM 17.27.
(a) H1(s) =IL
Vs=
1
Zin (s)=
1
R + Ls + 1
Cs
=
1
Ls
s2 + R
Ls + 1
LC
H2(s) = H1(s) ×1
Cs=
1
VsIL ×
1
Cs
=
VC
Vs=
s
L×
1
Cs
s2 + R
Ls + 1
LC
=
1
LC
s2 + R
Ls + 1
LC
(b) H1(s) is precisely of the form of equation 17.18 with a single zero at the origin. It follows
that ω p2 =
1
LC and 2σ p =
R
L. Hence, from equation 17.19, ωm = ω p =
1
LC, H1( jωm ) =
1
R,
Bω = 2σ p =R
L, and Qcir = Qp =
ω p
2σ p=
1
LC×
L
R=
1
R
L
C.
(c) With s = jω,
H2( jω)2 =
1
LC
2
1
LC−ω 2
2+ R
Lω
2 =1
1− LCω2( )2+ RCω( )2
Instead of maximizing H2( jω)2
we minimize its reciprocal, through differentiation. Let
5/15/01 P17-6 R.A. DeCarlo & P.M. Lin
f (C) =1
H2( jω)2 = 1 − LCω2( )2
+ RCω( )2
Then
f '(C) = 2C(ωR)2 + 2 ω2LC −1( )ω2L = 0
implies that
C =L
R2(ωL)2 =1
Lω21
R2
L2ω2 +1
=1
Lω21
1
Qcoil2 +1
If the coil has high Q, then
C ≅1
Lω2
in which case
H2( jω) =VC
Vs≅
LωR
= Qcoil
Therefore
VC max ≅ Qcoil Vs
SOLUTION PROBLEM 17.31. Here
H( jω) 2 = Ka + jω
(ωp2 −ω 2) + j2σ pω
2
= K2 a2 +ω 2
(ω p2 −ω 2)2 + 4σ p
2ω2 ≡ K2 f ω2( )
(a) Differentiating with respect to ω2 and setting the derivative equal to zero yields
0 =d f ω2( )
dω2 =d
dω2a2 +ω2
(ω p2 −ω 2 )2 + 4σ p
2ω2
5/15/01 P17-7 R.A. DeCarlo & P.M. Lin
= 1
(ω p2 −ω 2)2 + 4σ p
2ω2 −a2 +ω 2( ) −2(ω p
2 −ω 2) + 4σ p2( )
(ω p2 −ω 2)2 + 4σp
2ω2[ ]2
Given that the denominator of the first term is non-zero, this is equivalent to
0 = (ω p2 −ω 2)2 + 4σp
2 ω2 − a2 +ω 2( ) −2(ω p2 −ω2 ) + 4σp
2( )
= ω2( )2+ω p
4 − 2ω p2ω2 + 4σ p
2ω2 + 2 a2 +ω 2( )(ω p2 −ω 2) − 4 a2 +ω 2( )σp
2
= ω2( )2+ω p
4 − 2ω p2ω2 + 4σ p
2ω2 + 2a2ω p2 − 2a2ω2 + 2ωp
2 ω2 − 2 ω2( )2− 4a2σp
2 − 4σ p2ω2
= − ω2( )2− 2a2ω2 +ω p
4 + 2a2ω p2 − 4a2σ p
2
Hence
ω2( )2+ 2a2ω2 + a4 =ω p
4 + 2a2ω p2 − 4a2σp
2 + a4
where we have added a4 to produce perfect squares, i.e.,
ω2 + a2( )2= ω p
2 + a2( )2− 2aσp( )2
This implies that
ω2 = −a2 ± ω p2 + a2( )2
− 2aσp( )2
Thus, to achieve a real positive solution we obtain,
ωm = −a2 + ω p2 + a2( )2
− 2aσ p( )2
(b) Here, as in part (a), the arithmetic is simpler if we deal with 1/H(s) rather than H(s).
Specifically
1
H(s)=
s2 + 2σ ps +ω p2
K(s + a)=
s2 + 2σ p(s + a) − 2σ pa +ω p2
K(s + a)=
2σp
K+
s2 − 2σ pa +ω p2
K(s + a)
5/15/01 P17-8 R.A. DeCarlo & P.M. Lin
The problem asks for a non-zero value of ω. For zero phase shift of H(jω), Im[H(jω)] = 0
Im[H(jω)] > 0. This necessarily requires that the imaginary part of the above expression at s = jω
must be zero. Thus
Im1
H( jω)
= Im
−ω2 − 2σ pa +ω p2
K( jω + a)
= Im−ω2 − 2σ pa +ω p
2( )(a − jω)
K(ω2 + a2)
=−ω2 − 2σ pa +ω p
2( )(−ω)
K(ω2 + a2)= 0
The solutions to this are dc, i.e., ω = 0, and
ω (for zero phase shift) = ω p2 − 2σ pa
Note that such a frequency may not exist if the quantity under the radical is negative.
SOLUTION PROBLEM 17.40. (a) Compute the transfer function:
H(s) =Vout
Iin=
1
Yin(s)=
11
Ls + Rs+ Cs + 1
Rp
=1
C×
s +Rs
L
s2 + RsL
+ 1RpC
s + RsRp
+1
1
LC
= 8 ×105 s +1.333 ×103
s2 + 2000s +1.0006ω02
where ω02 = 1.6 ×109 and ω p
2 = 1.0006ω02 ≅1.60089 ×109 and ω p = 4.001×104 rad/sec.
(b) Qp =ω p
2σ p= 20 which is high Qp. Further a =
Rs
L= 1.333 ×103 <<ω p = 40.011109 ×103 .
(c) Equation 17.28,
5/15/01 P17-9 R.A. DeCarlo & P.M. Lin
ωm = −a2 + ω p2 + a2( )2
− 2aσ p( )2
gives the exact value of ωm. In MATLAB,
»wmsqrd = -a^2 + sqrt((wp^2+a^2)^2 - (twosig*a)^2)
wmsqrd = 1.600886670362673e+09
»wm = sqrt(wmsqrd)
wm = 4.001108184444246e+04
Clearly, ωm approximates ωp. Finally,
H( jωm) ≅ H( jω p) = 8 ×105 jω p +1.333 ×103
−ω p2 + j2000ω p +ω p
2 =8 ×105
20001− j
1.333 ×103
ω p
≅ 400 Ω
(d) From equation 17.30,
Bω ≅ 2σ p = 2000 rad/sec, ω1 ≅ ωm − 0.5Bω = 39.011 ×103 rad/sec, and
ω2 ≅ω m + 0.5Bω = 41.011 ×103 rad/sec.
(e) From equation 17.30,
Qcir ≅ Qp =ω p
2σ p=
Rs
Rp+ 1
1
LC
RsL
+ 1RpC
≅ω0
RsL
+ 1RpC
1
Qcir≅
Rs
L+
1
RpC
ω0=
Rs
Lω0+
1
RpCω0=
1
Qcoil+
1
Qcap=
Qcoil + Qcap
QcoilQcap
Hence,
5/15/01 P17-10 R.A. DeCarlo & P.M. Lin
Qcir ≅QcoilQcap
Qcoil + Qcap
SOLUTION PROBLEM 17.65. Here we consider the equivalent circuit valid for t > 0. Note that
iL(0-) = iL(0
+) = E/Rs and vC(0
-) = vC(0
+) = 0. Hence
It follows that
VC (s) =1
1
R+ 1
Ls+ Cs
×E
Rss=
1
s2 + 1
RCs + 1
LC
×E
CRs
Complex roots occur when
1
RC
2−
4
LC=
1
RC
2− 4ω0
2 < 0
Equivalently,
ω0RC >1
2
as was to be shown. Further, since vC(0+) = 0, the general form of the capacitor voltage for constant
excitation is (as per chapter 10)
vC (t) = e−σt Acos(ωdt) + Bsin(ωdt)( ) = Be−σt sin(ωdt)
From the characteristic equation, the complex roots are
−σ ± jωd =−1
2RC± j
1
LC−
1
2RC
2=
−ω0
2Q± jω0 1−
1
4Q2
Here
5/15/01 P17-11 R.A. DeCarlo & P.M. Lin
a = σ =ω0
2Q ωd =ω 0 1−
1
4Q2
Letting B = Vm, for the appropriate value of Vm, we obtain the desired result.
(b) If Q is large, ωd ≅ω 0 . The Vme−at will drop to 1/e of its peak value in t =1
a=
2Q
ω0≅
2Q
ωd
seconds. The period of oscillation of the damped sinusoid is 2πωd
. Therefore the number of cycles
contained in this interval is
2Q
ωd2πωd
=Q
π
5/31/01 Mag Crt Probs P18-1 © R. A. DeCarlo & P.M. Lin
CHAPTER 18 PROBLEM SOLUTIONS
SOLUTION PROBLEM 18.34. There is a correction to this problem: set M = 3 H.
(a) The stored energy at t = 0 is:
W(0) = 0.5L1i12(0) + 0.5L2i2
2(0) + Mi1 (0)i2(0) = 8 J
»L1 = 10; L2 = 2; M = 3; i10 = 1; i20 = -3;»W0 = 0.5*L1*i10^2 + 0.5*L2*i20^2 + M*i10*i20W0 = 5
(b) Writing two differential mesh equations we obtain
L1di1dt
+ Mdi2dt
+ R1i1 = 10di1dt
+ 3di2dt
+ i1 = 0
and
L2di2dt
+ Mdi1dt
+ R2i2 = 2di2dt
+ 3di1dt
+ i2 = 0
Taking the Laplace transform of these equations yields
10sI1 −10i1(0) + 3sI2 − 3i2(0) + I1 = (10s +1)I1 + 3sI2 −10i1(0) − 3i2 (0) = 0and
2sI2 − 2i2(0) + 3sI1 − 3i1(0) + I2 = (2s +1)I2 + 3sI1 − 3i1(0) − 2i2(0) = 0
Putting these equations in matrix form yields
(10s +1) 3s
3s (2s +1)
I1I2
=
10i1(0) + 3i2(0)
3i1(0) + 2i2(0)
=
1
−3
Solving yields
I1I2
=
(10s +1) 3s
3s (2s + 1)
−1 1
−3
=
1
11s2 +12s +1
2s +1 −3s
−3s 10s +1
1
−3
=1
(11s +1)(s +1)
11s +1
−33s − 3
=
1
s +1
1
−3
Therefore, by inspection,
i1(t) = e−tu(t) A and i2(t) = −3e−tu(t) A
Remark: normally, i1(t) and i2(t) would have two exponential terms present. Because of the specialchoice of initial conditions, a pole cancelled out.
(c) From equation 18.24 with the lower limit changed to zero and the upper limit changed to ∞, we have
5/31/01 Mag Crt Probs P18-2 © R. A. DeCarlo & P.M. Lin
W(0,∞) = v1i1 + v2i2( )0
∞∫ dt = 0.5L1i1
2(∞) + 0.5L2i22(∞) + Mi1(∞)i2(∞)
−0.5L1i12(0) + 0.5L2i2
2(0) + Mi1(0)i2(0)
From part (b) all currents at t = ∞ are zero, hence
W(0,∞) = −0.5L1i12 (0) + 0.5L2i2
2(0) + Mi1(0)i2(0) = −5 J
The result of part (a) indicates that the initial store energy is 5 J. The result of part (c) indicates that theenergy returned to the circuit is also 5 J, i.e., the total energy accumulated in the inductors over [0,∞) is–5 J. Hence 5 J is dissipated in the resistors.
Remark: the interested student might computer the integral R1i12( t) + R2i2
2( t)( )0
∞
∫ dt , the actual energy
dissipated in the resistors over [0,∞), and show that this is 5 J.
SOLUTION PROBLEM 18.35.
(a)»L1 = 4; L2 = 9; M = 3;»I1 = 2; I2 = -3;»W = 0.5*L1*I1^2 + 0.5*L2*I2^2 - M*I1*I2W = 6.6500e+01
(b)»K = 0.5*L1*I1^2K = 8»% Minimize (over I2) K + 0.5*9*I2^2 - 3*2*I2»% Take Derivative and set to zero; then solve for I2.»% Derivative is: 9*I2 – 6 = 0»% The result is I2 = 2/3 A.»I2 = 2/3;»Wmin = 0.5*L1*I1^2 + 0.5*L2*I2^2 - M*I1*I2Wmin = 6
(c)
5/31/01 Mag Crt Probs P18-3 © R. A. DeCarlo & P.M. Lin
(d)»L1 = 4; L2 = 9; M = 3;»k = M/sqrt(L1*L2)k = 5.0e-01
SOLUTION PROBLEM 18.37.
»k = 0.5;»L1 = 9; L2 = 4;L3 = 1;»M = 0.5*sqrt(L1*L2)M = 3»Lcpld = L1 + L2 + 2*MLcpld = 19»Leq = Lcpld + L3Leq = 20»Imax = 2;»Wmax = 0.5*Leq*Imax^2Wmax = 40 J
SOLUTION PROBLEM 18.41.
»RL = 100; Rs = 300e3; R = 10e3;»m = 20; n = 5;
(a)»Z2 = RL*m^2Z2 = 40000»Req1 = Z2*R/(Z2+R)Req1 = 8000»Z1 = Req1*n^2Z1 = 200000
5/31/01 Mag Crt Probs P18-4 © R. A. DeCarlo & P.M. Lin
(b)»%Gv1 = v1/vin»%v1 = [Z1/(Rs + Z1)]vin»Gv1 = Z1/(Rs+Z1)Gv1 = 4.0000e-01
»% Gv2 = v2/vin»% Gv2 = v2/v1 * v1/vin = (1/n) * G1»Gv2 = G1/nGv2 = 8.0000e-02
»% Gv3 = v3/vin»Gv3 = -Gv2/mGv3 = -4.0000e-03
(c)»% Gi2 = i2/iin»Gi2 = n*R/(R+Z2)Gi2 = 1
»% Gi3 = i3/iin = i3/i2 * i2/iin = –m*Gi2»Gi3 = -m*Gi2Gi3 = -20
SOLUTION PROBLEM 18.55. (a) The parameters in the circuit of figure P18.55b are given byequations in figure 18.22b. Specifically, since k = M L1L2 = 0.16 3.5 × 0.008 = 0.95618
»M = 0.16; L1 = 3.5; L2 = 0.008;»k=M/sqrt(L1*L2)k = 9.5618e-01»La = (1 - k^2)*L1La = 3.0000e-01»Lb = k^2 * L1Lb = 3.2000e+00»N =M/L2N = 20
(b)»R = 500;»w = 2*pi*60;»Vseff = 110;»Zin = R + j*La*w + j*Lb*wZin = 5.0000e+02 + 1.3195e+03i»Iseff = Vseff/ZinIseff = 2.7624e-02 - 7.2899e-02i»Pave = R*abs(Iseff)^2
5/31/01 Mag Crt Probs P18-5 © R. A. DeCarlo & P.M. Lin
Pave = 3.0387e+00
(c)»Zin2 = R + j*La*wZin2 = 5.0000e+02 + 1.1310e+02i»Iseff2 = Vseff/Zin2Iseff2 = 2.0929e-01 - 4.7341e-02i»Is2mag = abs(Iseff2)Is2mag = 2.1458e-01
» % The current in the secondary is (in A):
»Isecmag = Ismag*NIsecmag = 4.2916e+00
(d)»% Our first step is to compute the reflected impedance:»Zrefl = 100*N^2Zrefl = 40000
»% We now compute the impedance of the parallel combination»% of Lb and Zrefl denoted Zpar»Zpar = 1/(1/Zrefl + 1/(j*w*Lb))Zpar = 3.6350e+01 + 1.2053e+03i
»% We now compute the input impedance:
»Zin = R+j*w*La + ZparZin = 5.3635e+02 + 1.3184e+03i
»% Now we compute the voltage across the primary of the»% ideal transformer, by voltage division:
»Vpar = Vseff*Zpar/ZinVpar = 8.7342e+01 + 3.2500e+01i
»% Now we compute the voltage across the load:»Vload = Vpar/NVload = 4.3671e+00 + 1.6250e+00i»Vloadmag = abs(Vload)Vloadmag = 4.6596e+00»Iloadmag = Vloadmag/100Iloadmag = 4.6596e-02
5/31/01 Mag Crt Probs P18-6 © R. A. DeCarlo & P.M. Lin
SOLUTION PROBLEM 18.65. (a) The equivalent circuit accounting for initial conditions is givenbelow:
(b) From the definition of coupled inductors
V1
V2
=
0.6s 0.1472s
0.1472s 0.1472s
I1'
I2'
(c) Hence
I1'
I2'
=0.6s 0.1472s
0.1472s 0.1472s
−1 V1
V2
=
15
s
0.1472 −0.1472
−0.1472 0.6
V1
V2
(d) Writing nodal equations we obtain,
Vs
0
=
2 −1
−1 2
V1
V2
+
I1
I2
=
2 −1
−1 2
V1
V2
+
1
s
i1(0− )
i2(0− )
+I1'
I2'
(e) Now we substitute our result of part (c):
Vs
0
=
2 −1
−1 2
V1
V2
+
1
s
−2
−2
+
1
s
2.208 −2.208
−2.208 9
V1
V2
which simplifies toVs
0
−
1
s
−2
−2
=
1
s
2s + 2.208 −(s + 2.208)
−(s + 2.208) 2s + 9
V1
V2
or equivalently
2s + 2.208 −(s + 2.208)
−(s + 2.208) 2s + 9
V1
V2
=
sVs + 2
2
(f) Solving these equations we obtain:
5/31/01 Mag Crt Probs P18-7 © R. A. DeCarlo & P.M. Lin
V1
V2
=
2s + 2.208 −(s + 2.208)
−(s + 2.208) 2s + 9
−1 sVs + 2
2
=1
3s2 +18s + 15
2s + 9 (s + 2.208)
(s + 2.208) 2s + 2.208
sVs + 2
2
(g) If vs(t) = 10u(t) V, then
V1
V2
=
1
3s2 +18s +15
2s + 9 (s + 2.208)
(s + 2.208) 2s + 2.208
12
2
=
1
3s2 +18s + 15
26s +112.42
16s + 30.912
From MATLAB»[r,p,k]=residue([16 30.912], [3 18 15])r = 4.0907e+00 1.2427e+00p = -5 -1k = []
Therefore,
v2( t) = 4.0907e−5t +1.2427e−t( )u(t) V
SOLUTION PROBLEM 18.67. The solution to this problem is based upon the following: (i) Leq = L1 +L2 + 2M for series aiding connection (see example 18.4) and (ii) k = M L1L2 (a definition), and (iii) k= 1 (an assumption).
(a) Given L1 = L2 = L and k = 1, Leq = L1 + L2 + 2M = L + L + 2k L2 = 4L . Hence, when thenumber of turns is doubled, the inductance is quadrupled.
(b) For this part, let us first consider L2 which has 2N turns. We can view L2 as two coils of N turnseach connected in series aiding with coupling coefficient k = 1. Hence, according to part (a), theinductance of L2 is four times that of L1 which only has N turns. Hence,
Leq = L1 + L2 + 2M = L + 4L + 2k 4L2 = 9L
Observe that the coil has 3N turns yielding an inductance of 9L = 32L.
(c) Suppose coil 1 and coil 2 consist of one turn each. Here the total number of turns is 2N where N =
1 turn. Suppose further that L1 = L. From part (a), Leq = 4L = (2)2 L . Now suppose coil one consistsof one turn and coil 2 consists of M turns. We assume here as an induction hypothesis that
5/31/01 Mag Crt Probs P18-8 © R. A. DeCarlo & P.M. Lin
Leq = M +1( )2 L
We must show that if coil 2 has (M+1)N turns then,
Leq = M + 2( )2 L
Our first step is to compute the equivalent inductance of coil 2. However, coil 2 consists of a single turncoupled to an M-turn coil, which by the induction hypothesis means that
L2 = M +1( )2 L
Thus coil 1 in a series aiding connection with L2 leads to
Leq = L1 + L2 + 2M = L + M + 1( )2 L + 2 M + 1( )2 L2 = L + M +1( )2 L + 2 M +1( )L
= L M +1( )2 + 2 M +1( ) +1[ ] = L M +1( ) +1[ ]2 = M + 2( )2 L
Given this relationship, if coil 1 consists of N1 turns, and one turn has an inductance L, then
L1 = N1( )2L . Similarly, L2 = N2( )2
L , and M = k L1L2 = L1L2 = N1( )2N2( )2
L2 = N1N2L . Itimmediately follows that
L1 : L2 : M = N12 : N2
2 : (N1N2)
SOLUTION PROBLEM 18.70. (a) k = M / L1L2»M = 1.5; L1 = 1.5; L2 = 6;»k = M/sqrt(L1*L2)k = 5.0000e-01
(b) For this part and the remaining parts consider the following equivalent circuit where the coupledcoils have been replaced by the model of figure 18.22(b).
»La = (1 - k^2)*L1La = 1.1250e+00»Lb = k^2 * L1Lb = 3.7500e-01
5/31/01 Mag Crt Probs P18-9 © R. A. DeCarlo & P.M. Lin
»% The turns ratio is M:L2, i.e.,»1.5/6ans = 2.5000e-01»% Therefore the turns ratio is 1:4. It follows that
»Rb = 200*(1/4)^2Rb = 1.2500e+01
To compute Z(s) we have,
Z(s) = 20 +1
Cs+ Las +
RbLbs
Lbs + Rb= 1.125s + 20 +
12.5s
s + 100
3
+1
Cs
Hence,
Z( jω) = j1.125ω + 20 +j12.5ω
jω + 100
3
−j
Cω= 20 +
12.5ω2
ω2 + 104
9
+ j 1.125ω −1
Cω+
416.67ω
ω2 + 104
9
(c) For this part we need to make the imaginary part of Z(jw) real. To this end:
»K1 = 12.5*100/3K1 = 4.1667e+02»w = 1333;»K2 = 1.125*w + K1*w/(w^2 + 1e4/9)K2 = 1.4999e+03»C = 1/(K2*w)C = 5.0015e-07
Hence, we take C = 5 µF.
(d) At resonance, we have
Z( jωr ) = Z( j1333) = 20 +12.5ωr
2
ωr2 + 104
9
= 32.5 Ω
and
Zb ( jωr) =j12.5ωr
jωr + 100
3
=j16,662
33.33 + j1333= 12.492 + j0.31238
By voltage division
Vout ( jωr)
Vs( jωr)=
Vout
Vb×
Vb
Vs=
4
1×
Zb( jωr )
Z( jωr )= 1.5379 + j0.038456
»Zb = j*12.5*w/(j*w + 100/3)Zb = 1.2492e+01 + 3.1238e-01i
»Zwr=20 + 12.5*w^2/(w^2 + 1e4/9)Zwr =
5/31/01 Mag Crt Probs P18-10 © R. A. DeCarlo & P.M. Lin
3.2492e+01
»Gv = 4*Zb/ZwrGv = 1.5379e+00 + 3.8456e-02i
»MagGv = abs(Gv)MagGv = 1.5384e+00
»AngGv = angle(Gv)*180/piAngGv = 1.4325e+00
SOLUTION PROBLEM 18.71.
(a) Following the hint we apply a source transformation to obtain
Writing two mesh equations we obtain the following matrix form of the mesh equations:
s +β+ 1 s ks
ks s +β+ 1 s
I1
I2
=
Iin s
0
Solving for I2 yields
I2 =det
s + β + 1 s Iin s
ks 0
dets + β + 1 s ks
ks s + β + 1 s
=−kIin
s + β + 1 s( )2 − ks( )2
To find H(s) we have
H(s) =Vout
Iin=
− I2 s
Iin=
k s
s + β + 1 s( )2 − ks( )2 =k s
s + ks + β + 1 s( ) s − ks + β + 1 s( )
=ks
(1+ k)s2 +β s +1[ ] (1− k)s2 +βs +1[ ]
5/31/01 Mag Crt Probs P18-11 © R. A. DeCarlo & P.M. Lin
(b) R = 0.02 Ω, b = R = 0.02 Ω, Q = 50, and k = 0.01, 0.02, and 0.04.
»beta = 0.02; k1 = 0.01; k2 = 0.02; k3 = 0.04;»p11 = roots([(1+k1) beta 1]);»p12 = roots([(1-k1) beta 1]);»p1 = [p11;p12]p1 = -9.9010e-03 + 9.9499e-01i -9.9010e-03 - 9.9499e-01i -1.0101e-02 + 1.0050e+00i -1.0101e-02 - 1.0050e+00i
»p21 = roots([(1+k2) beta 1]);»p22 = roots([(1-k2) beta 1]);»p2 = [p21;p22]p2 = -9.8039e-03 + 9.9010e-01i -9.8039e-03 - 9.9010e-01i -1.0204e-02 + 1.0101e+00i -1.0204e-02 - 1.0101e+00i
»p31 = roots([(1+k3) beta 1]);»p32 = roots([(1-k3) beta 1]);»p3 = [p31;p32]p3 = -9.6154e-03 + 9.8053e-01i -9.6154e-03 - 9.8053e-01i -1.0417e-02 + 1.0206e+00i -1.0417e-02 - 1.0206e+00i
(c)
»f = 0.14:.0001:.18;»n1 = [k1/(1-k1)^2 0];n2 = [k2/(1-k2)^2 0];n3 = [k3/(1-k3)^2 0];»h1 = freqs(n1, poly(p1), 2*pi*f);»h2 = freqs(n2, poly(p2), 2*pi*f);»h3 = freqs(n3, poly(p3), 2*pi*f);»plot(f, abs(h1), f, abs(h2), f, abs(h3))»grid
5/31/01 Mag Crt Probs P18-12 © R. A. DeCarlo & P.M. Lin
(d) An inspection of [f, abs(h2)] (i.e., a tabulation of the values) in part (c) indicates that fpeak = 0.15865Hz and Hpeak = 26.148 Ω. The frequency scale factor Kf is defined according to:»Kf = 455e3/fpeakKf = 2.8679e+06
Further,
»Km = Kf*2.35e-3Km = 6.7397e+03
»Lnew = Km*1/KfLnew = 2.3500e-03»Cnew = 1/(Km*Kf)Cnew = 5.1736e-11»Rnew = Km*0.02Rnew = 1.3479e+02
The 3 dB down value of h2 is Hpeak/sqrt(2). Hence
5/31/01 Mag Crt Probs P18-13 © R. A. DeCarlo & P.M. Lin
»Hmax = max(abs(h2))Hmax = 2.6148e+01»H3db = Hmax/sqrt(2)H3db = 1.8490e+01
Again, inspecting the tabulated values indicates that the 3 dB frequencies are: f1 = 0.1569 Hz andf2 = 0.1614 Hz. Finally»Bf = f2 - f1Bf = 4.5000e-03 Hz»Bfnew = Kf*BfBfnew = 1.2906e+04 Hz
(e) For this part we redo part (a) with R, L, and C as literals.
Writing two mesh equations we obtain the following matrix form of the mesh equations:
Ls + R +1 Cs kLs
kLs Ls + R +1 Cs
I1I2
=
Iin Cs
0
Solving for I2 yields
I2 =det
Ls + R + 1 Cs Iin Cs
kLs 0
detLs + R +1 Cs kLs
kLs Ls + R +1 Cs
=−kL C
Ls + R +1 Cs( )2 − kLs( )2 Iin
To find H(s) we have
H(s) =Vout
Iin=
− I2 Cs
Iin=
kL C2s
Ls + R + 1 Cs( )2 − kLs( )2 =kL C2s
Ls + kLs + R +1 Cs( ) Ls − kLs + R +1 Cs( )
=kLs
LC(1+ k)s2 + CRs +1( ) LC(1− k)s2 + CRs +1( ) =ks
LC2
(1+ k)s2 + R
Ls + 1
LC
(1− k)s2 + R
Ls + 1
LC
=kω0
2sC
(1+ k)s2 + ω0Q
s +ω 02
(1− k)s2 + ω0
Qs +ω 0
2
Evaluating this expression at s = jω0, yields
5/31/01 Mag Crt Probs P18-14 © R. A. DeCarlo & P.M. Lin
H( jω0 ) =jkω0
3
C
−(1+ k)ω02 + jω0
2
Q+ω 0
2
−(1− k)ω02 + jω0
2
Q+ω 0
2
=− jk
Cω0
k − jQ
k + j1
Q
=− jk
Cω0
k2 + 1
Q2
Therefore
H( jω0) =1
Cω0×
k
k 2 + 1
Q2
(f) To solve this part we differentiate and set equal to zero as follows:
d H ( jω0)
dk=
1
Cω0×
d
dk
k
k2 + 1
Q2
=1
Cω0×
1
k 2 + 1
Q2
−2k2
k2 + 1
Q2
2
= 0
It follows that2k2
k 2 + 1
Q2
= 1
Hence
k 2 =1
Q2
or k = 1/Q. With this value of k,
H( jω0) max =1
Cω0×
1/ Q2
Q2
=Q
2Cω0
At ω0, the magnitude of the transfer function increases with increasing k, reaching a peak at k = 1/Q andthen decreases with a further increase in k as born out in the plots of part (c).
Prbs Chap 19, 1/7/02 P19-1 © R. A. DeCarlo, P. M. Lin
CHAPTER 19 PROBLEM SOLUTIONS
SOLUTION PROBLEM 19.1. Refer to figure 19.3.
»Vs = 100; ZL = 20; Rs = 1e3;
»beta = 149;
»Zinbox = (beta + 1)*ZL
Zinbox =
3000
»% By voltage division
»V1 = Vs*Zinbox/(Zinbox + Rs)
V1 = 75
»% To obtain the power delivered by the source
»I1 = Vs/(Zinbox +Rs)
I1 = 2.5000e-02
»Psource = I1*Vs
Psource = 2.5000e+00
SOLUTION PROBLEM 19.2. Refer to figure 19.4.
»Vs = 100; Z1 = 30e3; Rs = 50; beta = 149;
»Zboxin = Z1/(beta+1)
Zboxin = 200
»V1 = Vs*Zboxin/(Zboxin + Rs)
V1 = 80
»Psource = Vs^2/(Rs + Zboxin)
Psource = 40
SOLUTION PROBLEM 19.3. Refer to figure 19.5.
(a) »C = 0.1e-3; vc0 = 10; Z1 = 300; Z2 = 1e3;
»Z3 = 1e3; gm = 9e-3;
»
»Zin = Z1 + (1 + gm*Z1)*Z2
Zin = 4.0000e+03
(b)
»tau = Zin*C
Prbs Chap 19, 1/7/02 P19-2 © R. A. DeCarlo, P. M. Lin
tau = 4.0000e-01
Hence, vC (t) = vC (0)e−t / =10e−2.5t V.
SOLUTION PROBLEM 19.4. (a) First observe that since no current can flow into the secondary
we have
Voc = aVpri = aRIin = 800 Vrms
Now
Zth =1
j C+ a2R = 640 − j 360 Ω
(b) ZL = Zth( )* = 640 + j360 Ω.
»Voc = 800; Rth = 640;
»Pmax = Voc^2/(4*Rth)
Pmax = 250
(c) By inspection the circuit is a 640 Ω resistor in series with a 3.6 H inductor.
SOLUTION PROBLEM 19.5. Because the output is open circuited, no current flows into the
secondary of the transformer, hence
voc = vsec + sin(3t)u( t) = 2v pri + sin(3t)u(t) = cos(3t) + sin(3t)[ ]u( t).
Additionally
Zth (s) =10s
+ s + 4 + 42.5s
+ 0.25s + 9
=
20s
+ 2s + 40
SOLUTION PROBLEM 19.6. Using Cramer's rule,
Prbs Chap 19, 1/7/02 P19-3 © R. A. DeCarlo, P. M. Lin
I1 =
det
V1 1 −a
0 0.5 0
0 0 0.5
1.5a=
0.25V11.5a
=16a
V1
Therefore Rin = 6a Ω.
To compute the average power, V1,eff =10 V. Hence Pave =V1,eff
2
Rin=
1006a
watts.
SOLUTION PROBLEM 19.7. As per the hint, we write loop equations as follows:
−Vout
10 s
−40 s
=s +1 −1 0
−1 6 −2
0 −2 4
Iout
I2
I3
Using equation 19.6,
−Vout = s +1− −1 0[ ] 0.2 0.1
0.1 0.3
−1
0
Iout + −1 0[ ] 0.2 0.1
0.1 0.3
10 s
−40 s
= (s + 0.8)Iout +2
s
Therefore
Vout = (s + 0.8) −Iout( ) −2s
= Zth − Iout( ) +Voc
i.e., Zth = s + 0.8, and Voc = −2s
.
SOLUTION PROBLEM 19.8. (a) Let the node voltages from left to right be V1, V2, and Vout.
Also inject a current I3 into node 3. Writing nodal equations by inspection we have:
Prbs Chap 19, 1/7/02 P19-4 © R. A. DeCarlo, P. M. Lin
Iin
0
I3
=1.5 −1 −0.25
−1 2 −0.5
−0.25 −0.5 0.0625s + 0.75
V1
V2
Vout
Using equation 19.11, we have
I3 = W22 − W21W11−1W12( )Vout + W21W11
−1 Iin
0
= 0.0625s + 0.75 − 0.25 0.5[ ] 1 0.5
0.5 0.75
0.25
0.5
Vout − 0.25 0.5[ ] 1 0.5
0.5 0.75
Iin
0
Thus
I3 = 0.0625s + 0.375( )Vout − 0.5Iin
Therefore Isc = −I3]Vout =0 = 0.5Iin . Further Zth (s) =1
0.0625s + 0.375=
16s + 6
Ω.
(b) Vout (s) = Zth (s)Isc =16Iscs + 6
=8Iins + 6
. By inspection, the impulse response is
vout ,imp(t) = 8e−6tu( t) V. Further, from MATLAB
»n = 8; d = [1 6 0];
»[r,p,k] = residue(n,d)
r =
-1.3334e+00
1.3334e+00
p =
-6
0
k =
[]
Hence the step response is:
vout ,step( t) =43
1− e−6t( )u(t) V
SOLUTION PROBLEM 19.9. (a) Consider the following figure:
Prbs Chap 19, 1/7/02 P19-5 © R. A. DeCarlo, P. M. Lin
Let Iout enter the output terminal and I1 and I2 be the currents entering the primary and secondary
of the transformer respectively. It follows that
Iin =bVout
R+ I1 +
(b −1)VoutR
=(2b −1)Vout
R+ I1
which implies that
I1 = Iin −(2b −1)Vout
R
Further,
Iout =(1− b)Vout
R+ I2 =
(1− b)VoutR
− bI1
Therefore
Iout =(1− b)Vout
R+ I2 =
(1− b)VoutR
− bI1 =(1− b)Vout
R− b Iin −
(2b −1)VoutR
=2b2 − 2b +1
RVout − bIin
Equivalently
Vout =R
2b2 − 2b +1Iout +
bR
2b2 − 2b +1Iin
Therefore
Zth =R
2b2 − 2b +1, Voc =
bR
2b2 − 2b +1Iin , Isc = bIin
(b) Isc = b∠45o , Zth =R
2b2 − 2b +1.
Prbs Chap 19, 1/7/02 P19-6 © R. A. DeCarlo, P. M. Lin
(c) B =10 =1
ZthC=
1R
2b2 − 2b +1C
=2b2 − 2b +1
RC=
525C
. Hence C = 0.02 F. Further,
02 = 25 =
1LC
=1
0.02L. Therefore L = 2 H.
(d) H(s) =VoutIin
=Vout
0.5Isc= 2
1
2b2 − 2b +1
R+
1
Ls+ Cs
=2
0.2 +1
2s+ 0.02s
=100s
s2 +10s + 25
and Iin (s) =2
2×
s−10
s2 +100. Hence,
Vout (s) =50 2 s(s−10)
(s2 +10s + 25)(s2 +100)
To compute vout(t) in MATLAB,
%partial fraction expansion of Vout/√2 = = n(s)/d(s)
n =50*[ 1 -10 0];d = conv([ 1 0 100], [ 1 10 25]);[r ,p, k] = residue(n,d)
%numerator polynomial of combined complex pole terms
num = [1 -p(1)]*r(2) +[1 -p(2)]*r(1)
Output from MATLAB
r = 2.8000 + 0.4000i 2.8000 - 0.4000i -5.6000 30.0000p = -0.0000 +10.0000i -0.0000 -10.0000i -5.0000 -5.0000num = 5.6000 -8.0000
Laplace transform of ouput
Vout (s) =−5.6 2
s + 5+
30 2
(s + 5)2 +(5.6s − 8) 2
s2 +100
Prbs Chap 19, 1/7/02 P19-7 © R. A. DeCarlo, P. M. Lin
Taking the inverse Laplace transform using table 13.1 on page 515:
vout (t) = 30 2te−5t − 5.6 2e−5t + 5.6 2 cos(10t) − 0.8 2 sin(10t)[ ]u( t) V
The steady state part consists of the cosine and sine terms only. Since the parallel RLC acts like
a band pass circuit and the peak value occurs at 5 rad/s, one expects the magnitude of the steady
state output to be much smaller at 100 rad/s.
SOLUTION PROBLEM 19.10. (a) Write two nodal equations by inspection:
I1
I2
=
Y1 + Y3 −Y3 + gm
−Y3 Y2 + Y3
V1
V2
=
y11 y12
y21 y22
V1
V2
(b) When port-2 is shorted, y11 is the input admittance. Therefore
Zin =1
y11=
1Y1 + Y3
and since V2 = 0,
I2 = y21V1 =−KY3
s
SOLUTION PROBLEM 19.11. (a) By inspection
I1
I2
=
Y1 + Y3 −Y3
−Y3 + gm Y2 + Y3
V1
V2
=
y11 y12
y21 y22
V1
V2
Clearly, Y3 = −y12 . Then, Y1 = y11 − Y3 = y11 + y12 and Y2 = y22 − Y3 = y22 + y12 . Finally,
gm = y21 +Y3 = y21 − y12 .
(b) Recall that
Prbs Chap 19, 1/7/02 P19-8 © R. A. DeCarlo, P. M. Lin
I1
I2
=
Y1 + Y3 −Y3 + gm
−Y3 Y2 + Y3
V1
V2
=
y11 y12
y21 y22
V1
V2
Clearly, Y3 = −y21 . Then, Y1 = y11 − Y3 = y11 + y21 and Y2 = y22 − Y3 = y22 + y21 . Finally,
gm = y12 + Y3 = y12 − y21 .
SOLUTION PROBLEM 19.12. (a) By definition of coupled inductors
V1
V2
=
L1s Ms
Ms L2s
I1I2
⇒
I1I2
=
1
L1L2 − M2( )s
L2 −M
−M L1
V1
V2
Hence, the y-parameters are:
1
L1L2 − M 2( )sL2 −M
−M L1
(b) By definition of coupled inductors
V1
V2
=
L1s −Ms
−Ms L2s
I1
I2
⇒
I1
I2
=
1
L1L2 − M 2( )sL2 M
M L1
V1
V2
Hence, the y-parameters are:
1
L1L2 − M 2( )sL2 M
M L1
If the coupling coefficient is 1, L1L2 = M2 and the y-parameters do not exist since the
determinant of the z-parameter matrix is zero.
SOLUTION PROBLEM 19.13. Let I2' denote the current entering the dotted terminal of the
secondary of the coupled inductors. Then using the result of problem 12a,
Prbs Chap 19, 1/7/02 P19-9 © R. A. DeCarlo, P. M. Lin
I1
I2'
=
1
L1L2 − M 2( )sL2 −M
−M L1
V1
V2
=
16s
4 −1
−1 1
V1
V2
From the given circuit I2 = I2' + 2I1 =
−16s
V1 +16s
V2 +86s
V1 −26s
V2 =76s
V1 −16s
V2 . Therefore
the y-parameter matrix is:
16s
4 −1
7 −1
SOLUTION PROBLEM 19.14.
(a) By definition and the properties of the ideal transformer y11 =I1V1
V2 =0
= G1 and
y22 =I2
V2
V1=0
=G1 + G2
a2 . Additionally, since the circuit is obviously reciprocal,
y12(= y21) =I1V2
V1=0
=G1
a.
(b) V2 reflected to the primary side, denoted by ˆ V 1, is
ˆ V 1 =−2K a
s2 + 4
Hence
I1 =2G1K a
s2 + 4
To compute I2, we reflect the parallel of G1 and G2 to the secondary of the ideal transformer.
Hence the impedance in parallel with V2, denoted Zsec, is
Zsec =a2
G1 + G2
Therefore,
Prbs Chap 19, 1/7/02 P19-1 0 © R. A. DeCarlo, P. M. Lin
I2 =V2
Zsec=
2K G1 + G2( ) a2
s2 + 4
SOLUTION PROBLEM 19.15. (a) By definition
y11 =I1V1
V2 =0
=1
Zin
V2 =0
=19
S
where
Zin = 6 +12/ /1
16× (320/ /80)
= 9 Ω
Similarly, by definition
y22 =I2V2
V1=0
=1
Zout
V1=0
=3
400 S
where
Zout = 80 + 320/ /16 × (6 //12)( ) =4003
Ω
SOLUTION PROBLEM 19.16. Write nodal equations:
I1
I2
0
=2s + 2 −1 −2s
−1 s + 2 −s
−2s −s 5s
V1
V2
V3
Using the matrix partitioning method, we obtain the 2-port y parameters
I1
I2
=
2s + 2 −1
−1 s + 2
−
15s
−2s
−s
−2s − s[ ]
V1
V2
=1.2s + 2 −0.4s−1
−0.4s−1 0.8s + 2
V1
V2
=
y11 y12
y21 y22
V1
V2
Prbs Chap 19, 1/7/02 P19-1 1 © R. A. DeCarlo, P. M. Lin
SOLUTION PROBLEM 19.17. (a) The ideal transformer yields the constraints
I1 = −aˆ I 2 and ˆ V 2 = aV1
The three resistors have nodal equations
− ˆ I 2I2
=2 −1
−1 2
ˆ V 2V2
Substituting the first two equations into the last one, we obtain
I1
I2
=
2a2 −a
−a 2
V1
V2
=
y11 y12
y21 y22
V1
V2
(b)
Yin = y11 −y12y21
y22 + YL= 2a2 −
a2
4=1.75a2 S
Zin =1
Yin=
4
7a2 Ω
and
Gv = V2
V1 = -y21
y22 + yL1 = a
4
SOLUTION PROBLEM 19.18. This problem is solved in MATLAB.
Part (a)
»% Parameter specification
»Ys = 1e-3;YL = 1e-3;»y11=4e-3; y12 = -0.1e-3;»y21 = 50e-3; y22 = 1e-3;
»% Calculation of input admittance and impedance
»Yin = y11 - y12*y21/(y22 + YL)Yin = 6.5000e-03»Zin = 1/YinZin = 1.5385e+02
»% Calculation of output admittance
Prbs Chap 19, 1/7/02 P19-1 2 © R. A. DeCarlo, P. M. Lin
»Yout = y22 - y12*y21/(y11 + Ys)Yout = 2.0000e-03
Part (b)»% Calculation of voltage gain
»Gv = (Ys/(Ys + Yin))*(-y21/(y22 + YL))Gv = -3.3333e+00
Part (c)»V2 = Gv*10V2 = -3.3333e+01
Therefore, v2(t) = −33.333u(t) V. Finally,
»PL = V2^2/1e3PL = 1.1111e+00
SOLUTION PROBLEM 19.19.
(a) V1
Vs = Z in
Z in + Zs = Z in
Z in + 10 = 0.5 ⇒ Z in = 10 Ω or 0.1 S
Now
Yin = y11 - y12y21
y22 +YL = y11 -
0.02×20.2 + 0.1
= 0.1
Solving for y11 yields y11 = 0.2333 S.
(b) v2v1
= - y21
y22 +YL = - 2
0.2 +0.1 = - 6.667
Hence v2(t) = - 6.667 v1(t) = - 3.333 vs(t) = - 33.33 u(t) V
and
PL = V22
RL = 33.332
10 = 111.11 W
SOLUTION PROBLEM 19.20.(a) Writing a node equation at port 1 and mesh equation at port 2, we obtain by inspection
I1 = 2V1 + 3I 2
V2 = 2V1 + 2I2
Rearranging in matrix form , we have
Prbs Chap 19, 1/7/02 P19-1 3 © R. A. DeCarlo, P. M. Lin
1 -3
0 2
I1
I2
= 2 0
-2 1
V1
V2Therefore
I1
I2
= -1 1.5
-1 0.5
V1
V2
= y11 y12
y21 y22
V1
V2
(b) Yin = y11 - y12y21
y22 +YL = -1 + 1.5×1
0.5 + 0.25 = 1 S
(c) Here we compute Yout seen looking into port-2, i.e.,
Yout = y22 - y12y21
y11 +Ys = 0.5 + 1.5×1
-1 + 0 = -1 S
From current division,
i2(t) = Yout
Yout+YLis(t) = -1
-1 + 0.25×5u(t) = 6.667u(t) A
Finally from Ohm's law
v1(t) = - 0.5 ×3i2(t) = - 0.5 ×3×6.667u(t) = - 10u(t) V
REMARK: Because the current source sees a negative resistance, the circuit is unstable as itstands.
SOLUTION PROBLEM 19.21. (a) With port 2 shorted, the Laplace transform of the given data
are:
I1(s) =1s, V1(s) =
1s
−1
s + 4=
4s(s + 4)
, I2(s) =−1
s + 3Hence
y11 =I1V1
=1 s4
s(s + 4)
=s + 4
4
and
y21 =I2V1
=−1 (s + 3)
4
s(s + 4)
=− s(s + 4)4(s + 3)
Next, with port-2 terminated in a 1-Ω reistor, the Laplace transform of the given data are:
Prbs Chap 19, 1/7/02 P19-1 4 © R. A. DeCarlo, P. M. Lin
ZL=YL = 1
I1(s) = 1/s,
V1(s) = 1s
- 1s+4
+ 1(s+4)2
= 5s + 16s(s + 4)2
I2 = - 1s +7
Now
I2 = - V2
ZL = - V 2 = y21V1
y22 +YL
Solving for y22, and using y21 expression found earlier, we obtain
y22 = y21V1/I2 - YL= - s(s + 4)4 (s + 3)
× 5s + 16s(s + 4)2
×s + 7 -1
- - 1
= s2 + 23s + 644(s + 3)(s+ 4)
Finally, to compute y12, we use the defining equation for y-parameters, i.e.,
I1 = y11V1 + y12V2 from which we obtain
y12 = I1 - y11V1
V2 =
1s - s + 4
4× 5s + 16
s(s + 4 )2
1s + 7
= - (s + 7)4 ( s + 4)
(b) Given YL =1 S, the input admittance is
Yin = y11 - y12y21
y22 +YL = s + 4
4 -
- (s + 7)4 ( s + 4)
×- s(s + 4)4(s + 3)
s2 + 23s + 644(s + 3)(s+ 4)
+1 = (s + 4)2
5(s+3.2)
and
Zin = 1Yin
= 5(s+3.2)
(s + 4)2
Prbs Chap 19, 1/7/02 P19-1 5 © R. A. DeCarlo, P. M. Lin
(c) For this part, we use phasors to do the sinusoidal steady state analysis: ω = 10 rad/s and I1 =1. Also,
V1
I1 = Zin = 5(j10+3.2)
(j10 + 4)2 = 0.1973 - j0.4072
and
V2
V1 = -y21
y22 +YL =
- j10(j10 + 4)4(j10 + 3)
-100 + 230j + 644(j + 3)(j+ 4)
+1 = -0.2873+ j0.1787
Thus
V2
I1 = V2
V1 ×V1
I1 = (-0.2873+ j0.1787)(0.1973 - j0.4072)
from which we obtain in Ohms,
V2
I1 = -0.2873+ j0.1787 ×0.1973 - j0.4072 = 0.8192
SOLUTION PROBLEM 19.22. Looking into port-1, the admittance is:
Yport 1 (s) = y11 - y12y21
y22 +YL = 0 + 2×2
0 + 0.1s = 40
s
andZport 1(s) = s
40
Therefore
Zin(s) = 1000s
+ s40
= s2 + 2002
40sHence
Zin ( j ) =4 ×104 − 2
j40= − j
4 ×104 − 2
40
The imaginary part is zero when = 200 rad/s. Hence, the resonant frequency is 200 rad/s.
SOLUTION PROBLEM 19.23. (a) By writing nodal equations for the boxed 2-port, we have by
inspection (note passive circuit in which y21 = y12):
Prbs Chap 19, 1/7/02 P19-1 6 © R. A. DeCarlo, P. M. Lin
I1
I2
= 3s + 2 -2s -2
-2s -2 3s + 3
V1
V2
= y11 y12
y21 y22
V1
V2
(b) In Ohms, ZL = s+1; Zs = 2. Therefore,
Yin = y11 - y12y21
y22 +YL = 3s+2 - (2s +2)2
3s + 3 + s + 1 = 2s2 + 3s + 1
s + 1 = 2s + 1
V1
Vs = Z in
Z in + Zs = 1
1 + Z s Yin = 1
1 + 2×(2s + 1) = 1
4s + 3and
V2
V1 = -y21
y22 + yL = 2s +2
3s + 3 + s + 1 = 0.5
Thus,
V2
Vs = V1
Vs ×V2
V1 = 1
4s + 3× 0.5 = 1
8s2 + 14s + 6 = 1
8s+ 6
(c) The impulse reponse is
h(t) = L−1 H (s) = L−1 18s + 6
= L−1 0.125s + 0.75
= 0.125e−0.75tu(t)
For the step response,
v2(t) = L−1 H (s)s
= L−1 0.125s(s + 0.75)
=16
1− e−0.75t( )u(t) V
(d) We must compute the complete Laplace transform and invert. Here
Vs(s) = 12.75 × 2s2 + 4
and
V2(s) = H(s)V2(s) = 0.125s+ 0.75
×12.75× 2s2 + 4
= 3.1875(s + 0.75) (s2 + 4)
We use MATLAB to compute the partial fraction expansion
n=3.1875;d= conv([1 0.75], [ 1 0 4]);[r p k ] = residue (n,d)r = -0.3493 - 0.1310i -0.3493 + 0.1310i 0.6986p = -0.0000 + 2.0000i
Prbs Chap 19, 1/7/02 P19-1 7 © R. A. DeCarlo, P. M. Lin
-0.0000 - 2.0000i -0.7500
Hence, after combining the two complex terms, we obtain
V2(s) = 0.6986s + 0.75
+ - 0.6986s + 0.524s2 + 4
From Table 13.1, the steady state response is
v2,ss(t) = [-0.6986cos(2t) + 0.262sin(2t)]u(t) V
and the transient response is
v2,tran(t) = 0.6986e-0.75tu(t) V
SOLUTION PROBLEM 19.24. (a) Using the y-parameters of stage 2,
Yin2 = y11 - y12y21
y22 +YL = 500 + 0.1×75000
185 +4000 = 501.8 µS
The load for stage 1 is the parallel combination of Zin2 and the 2 kΩ resistor. Hence, using the y-parameters of stage 1, we obtain
Yin1 = y11 - y12y21
y22 +YL = 2000 + 0.5×24×104
100 +501.8 + 500 = 2108.9 µS
(b) We compute the following voltage gains:
V1
Vs = Z in1
Z in1 + Zs = 1
1 + Z s Yin1 = 1
1 + 25×2108.9×10-6 = 0.9499
V2V1
stage1
=−y21
y22 + YL,stage1=
−0.24
(100 + 500 + 501.8) ×10−6 = −217.8
andV2V1
stage2
=−y21
y22 +YL,stage2=
−0.075
(185 + 4000) ×10−6 = −17.92
Finally Gv is the product of the three gains calculated above
Gv = 0.9499×(- 217.8) ×(-17.92) = 3708.2
Prbs Chap 19, 1/7/02 P19-1 8 © R. A. DeCarlo, P. M. Lin
SOLUTION PROBLEM 19.25. (a) With the switch in position A, the load to the 2-port is YL =Cs = 0.25s. Hence,
V2V1
=−y21
y22 +YL=
−y21y22 + 0.25s
(b) From the given data, V1(s) = –2/s and hence,
V2(s) =−y21
y22 + 0.25sV1(s) =
−1 s−1 s + 0.25s
×−2s
=8
s(s2 − 4)=
−2s
+1
s + 2+
1s − 2
Hence, for t ≥ 0,
v2(t) = ( -2 + e-2t + e2t) u(t) V
(c) The circuit is not stable in the time interval 0 to 1 s, because the transfer function has a polein the right half plane.
(d) v2(1-) = -2 + e-2 + e2 = 5.524 V
(e) Replace the charged capacitor by the parallel combination of an admittance of 0.25s and acurrent source of value 0.25×8.524 (in accordance with figure 14.16
(f) Z1(s) = b2 × 1 = 4 Ω.
(g) For t ≥ 1s, the capacitor is discharging through a 4-Ω equivalent resistance, with a timeconstant 0.25×4 = 1 s, and an initial voltage v2(1-) = 5.542 V.Hence
v2(t) = 5.524e−(t−1)u(t −1)Vand by the ideal transformer voltage ratio property,
v3( t) = 2.762e−(t−1)u(t −1)V
SOLUTION PROBLEM 19.26. (a) By definition of coupled inductors
V1
V2
=
L1s Ms
Ms L2s
I1I2
Hence, the z-parameters are:
Prbs Chap 19, 1/7/02 P19-1 9 © R. A. DeCarlo, P. M. Lin
L1s Ms
Ms L2s
(b) By definition of coupled inductors
V1
V2
=
L1s −Ms
−Ms L2s
I1
I2
Hence, the y-parameters are:
L1s −Ms
−Ms L2s
The z-parameters exist independent of the values of M, L1, and L2.
SOLUTION PROBLEM 19.27. (a) By definition and the properties of the ideal transformer
z11 =V1I1
I 2=0
= R1 + R2 and z22 =V2I2
I1=0
= a2R2 . Additionally, since the circuit is obviously
reciprocal, z21(= z12) =V2I1
I2=0
= aR2 .
(b) The input impedance is given by the formula
Zin = z11 −z12z21
z22 + ZL= R1 + R2 −
a2R22
a2R2 + R2= R1 +
1
a2 +1R2
(c) If port-1 is open circuited, I1 = 0. Hence,
V1 = z12I2 =2aKR2
s2 + 4 and V2 = z22I2 =
2a2KR2
s2 + 4
SOLUTION PROBLEM 19.28. (a) For this part consider the figure below:
Prbs Chap 19, 1/7/02 P19-2 0 © R. A. DeCarlo, P. M. Lin
With port 2 open, it follows that Zpr = 320/16 = 20 Ω and
z11 =V1
I1
I2 =0
= 6 +20 ×12
20 + 12=13.5 Ω
With port 1 open and I2 injected into port 2, we have Zsec = 12*16 = 192 Ω and
z12 =V1
I2
I1= 0
=Vpr
I2
I1=0
=−Vsec
4I2
I1=0
=−I2 320 / /192( )
4I2
I1=0
= −30 Ω
With port 2 open and I1 injected into port 1, we have Zpr = 20 Ω and
z21 =V2
I1
I2 =0
=Vsec
I1
I2 =0
=−4Vpr
I1
I2 =0
=−4I1 12 //20( )
I1
I2 =0
= −30 Ω
With port 1 open, it follows that Zsec = 12*16 = 192 Ω and
z22 =V2
I2
I1=0
= 80 +320 ×192
320 + 192= 200 Ω
(b) V1(s) = z12I2 (s) =−300
s2 + 4 and V2(s) = z22I2(s) =
2000
s2 + 4.
Prbs Chap 19, 1/7/02 P19-2 1 © R. A. DeCarlo, P. M. Lin
SOLUTION PROBLEM 19.29. (a) Writing two mesh equations we have by inspection,
V1
V2
=
Z1 + Z3 Z3
Z3 + rm Z2 + Z3
I1
I2
=
z11 z12
z21 z22
I1
I2
(b) V1 = z11I1 =K Z1 + Z3( )
s and V2 = z21I1 =
K Z3 + rm( )s
SOLUTION PROBLEM 19.30. From the result of problem 29,
Z1 + Z3 Z3
Z3 + rm Z2 + Z3
=
z11 z12
z21 z22
Therefore, Z3 = z12 , Z1 = z11 − Z3 = z11 − z12 , Z2 = z22 − Z3 = z22 − z12 , and
rm = z21 − Z3 = z21 − z12 .
SOLUTION PROBLEM 19.31. (a) The z-parameters can be computed by inspection (first writethe z-parameters of the passive part of the network, i.e., with the dependent source ignored; thenadd the effect of the dependent source to the resulting equations.) As such, using loop equations,
Z =5 +
10
s10 +
10
s10
s10 +
10
s
=10s
0.5s +1 s +1
1 s +1
Ω
(b) Zin =5s +10
s−
10
s×
10
s(s +1)
10(s +1)
s+10
=5s +10
s−
10(s +1)s(2s +1)
=(5s +10)(2s +1)−10(s +1)
2s(s + 0.5)
=5s2 + 7.5ss(s + 0.5)
=5s + 7.5s + 0.5
(c) I1(s) =V1(s)Zin (s)
=10s
×s + 0.5
5s + 7.5=
0.6667s
+1.3334s +1.5
. Hence
Prbs Chap 19, 1/7/02 P19-2 2 © R. A. DeCarlo, P. M. Lin
i1(t) =23
+43
e−1.5t
u(t) A
SOLUTION PROBLEM 19.32. (a) To find the resonant frequency we first find
Zin (s) = z11 −z12z21
z22 + ZL= z11 −
z11z11
z11 + 10=
10z11
z11 +10=
10
1 + 10
z11
=10
1 + s2 + 25
s
=10s
s2 + s + 25
This is of the form of equation 17.18 with K = 10. Here according to equation 17.19f,
ωr = ω p = 25 = 5 rad/s
(b) To find Q we use equation 17.19e, i.e.,
Q = Qp =ω p
2σ p=
5
1= 5
(c) Using MATLAB we obtain the frequency response plots below:
Prbs Chap 19, 1/7/02 P19-2 3 © R. A. DeCarlo, P. M. Lin
0 2 4 6 8 10 12 14 16 18 200
1
2
3
4
5
6
7
8
9
10
Frequency in rad/s
Mag
nitu
de o
f Zin
Ω
TextEnd
Prbs Chap 19, 1/7/02 P19-2 4 © R. A. DeCarlo, P. M. Lin
0 2 4 6 8 10 12 14 16 18 20-100
-80
-60
-40
-20
0
20
40
60
80
100
Frequency in rad/s
Pha
se o
f Zin
deg
rees
TextEnd
SOLUTION PROBLEM 19.33. (a) Since Z =z11 z12
z21 z22
=
2 −3
30 4
Ω. Assuming that Zin does
not include Zs, it follows that
Zin = z11 −z12z21
z22 + ZL= 2 +
9024
= 5.75 Ω
Assuming that Zout does not include the parallel connection of ZL, then
Zout = z22 −z12z21
z11 + Zs= 4 +
903
= 34 Ω
(b) From equation 19.27
GV = Gv2Gv1 =ZL
z22 + ZL
z21
Zin + ZS
.
Prbs Chap 19, 1/7/02 P19-2 5 © R. A. DeCarlo, P. M. Lin
Thus in MATLAB»Gv = (20/(4 + 20))*(30/(5.75+1))Gv = 3.7037e+00»% Therefore»v2 = Gv*30v2 = 1.1111e+02
Hence, v2(t) = 111.11u( t) V. The power absorbed by ZL is therefore (in watts):
»PZL = v2^2/20PZL = 6.1728e+02
SOLUTION PROBLEM 19.34. (a) Before determining b, it is necessary to compute the
impedance seen at the secondary of the 2-port. Here,
Zout = z22 −z12z21
z11 + Zs= 33 2R0 −
2 3R02
R0 + R0= 33 2R0 − 2R0 = 32 2R0
Thus b must be chosen so that the impedance reflected to the secondary of the transformer is 2
Ω, i.e.,
2 =32 2R0
b2 ⇒ b = 4 R0
(b) First do a source transformation on the front end of the two port to obtain
vs(t) = R0 2 cos(2t) = 32 2 cos(2 t) V
Therefore, Vs,eff = 32 V. Also note that the impedance looking into the primary of the two port
is
Zin = z11 −z12z21
z22 + ZL= R0 −
2 3R02
33 2R0 + 32 2R0= 31.015 Ω
Hence
GV = Gv2Gv1 =ZL
z22 + ZL
z21Zin + ZS
= 0.5
Consider the following MATLAB calculations:
»a=2; R0 = 16;
»Zin = a*R0-2*a^3 * R0^2/(65*a^2 *R0)
Prbs Chap 19, 1/7/02 P19-2 6 © R. A. DeCarlo, P. M. Lin
Zin = 3.1015e+01
»ZL = 32*a^2 * R0
ZL = 2048
»Zs = a*R0
Zs = 32
»z22 = 33*a^2*R0
z22 = 2112
»z21 = 2*a*R0
z21 = 64
»Gv = (ZL/(z22+ZL))*(z21/(Zin+Zs))
Gv = 5.0000e-01
»V1eff = 32;
»V2eff = Gv*V1eff
V2eff = 16
»b = 4*a*sqrt(R0)
b = 32
»VLoadeff=V2eff/b
VLoadeff = 5.0000e-01
»Pmax = VLoadeff^2/2
Pmax = 1.2500e-01
Hence, max power transferred to the load is 125 mW.
SOLUTION PROBLEM 19.35. (a) Using the usual formula
Z(s) = z11 −z12z21
z22 + ZL= 0 −
1000 × (−1000)
0 +108 s= 0.01s
Yes, the input to the 2-port looks like a 0.01 H inductor.
(b) Under the given conditions, the circuit reduces to parallel RLC with Req = 50 kΩ, L = 0.01
H, and C = 100 pF. Therefore, m =1
LC= 106 rad/s and B =
1ReqC
= 2 ×105 rad/s.
(c) For this circuit
Prbs Chap 19, 1/7/02 P19-2 7 © R. A. DeCarlo, P. M. Lin
V1(s) =1
1
Req+
1
Ls+ Cs
×VinR1
=s C
s2 +1
ReqCs +
1
LC
×VinR1
=106s
s2 + 2 ×106s +1012 Vin
For the impulse response, Vin = 1 and from MATLAB
»n = [1e6 0];d = [1 2e6 1e12];
»[r,p,k] = residue(n,d)
r =
1.0000e+06
-1.0000e+12
p =
-1000000
-1000000
k =
[]
Therefore, the impulse response is:
h(t) = 106 −1012t( )e−106 tu(t) V
(d) y =0 1000
−1000 0
−1
=0 −0.001
0.001 0
S
(e) V2(s) =−y21
y22 +YLV1(s) =
−0.001
0 +10−8 s×
106 s
s2 + 2 ×106s +1012 =−1011
(s +106)2
Hence
v2(t) = −1011te−106 tu( t) V.
SOLUTION PROBLEM 19.36.
Prbs Chap 19, 1/7/02 P19-2 8 © R. A. DeCarlo, P. M. Lin
(a) Zin2 = z11 −z12z21
z22 + ZL= 62.582 −
1.2075 × 63.751.25 + 0.016
= 1.7778 kΩ
Because z12 for stage 1 is zero,
Zin = z11 −z12z21
z22 + Zin2 / /2= z11 = 2 kΩ
(b) Gv1 =V1Vs
=Zin
Zin + 75= 0.96386 . Let ZL1 = Zin2 //2 = 0.94118 kΩ. Then
Gv2 =V2V1
=ZL1
ZL1 + z22×
z21Zin
= −22.472. Finally Gv3 =VoutV2
=ZL
ZL + z22×
z21Zin2
= 0.45319.
Thus
Gv = Gv1Gv 2Gv 3 = −9.816
SOLUTION PROBLEM 19.37. (a) By inspection via mesh standard equations
z =L1s Ms
Ms L2s
(b) Utilizing the properties of an ideal transformer,
z11 =V1I1
I 2=0
= L1s z22 =V2I2
I1=0
=k2L2L1
L1s + (1− k2)L2s = L2s
z12 =V1I2
I1=0
=k L2
L1L1s = k L1L2s = Ms
Finally,
z21 =V2I1
I2 =0
=k L2
L1L1s = k L1L2s = Ms
(c) Utilizing the properties of an ideal transformer,
Prbs Chap 19, 1/7/02 P19-2 9 © R. A. DeCarlo, P. M. Lin
z22 =V2I2
I1=0
= L2s z11 =V1I1
I 2=0
=k2L1L2
L2s + (1− k2)L1s = L1s
z12 =V1I2
I1=0
=k L1
L2L2s = k L1L2s = Ms
Finally,
z21 =V2I1
I2 =0
=k L1
L2L2s = k L1L2s = Ms
(d) For this circuit k = 1 and the turns ratio a =10−2
1= 0.1. Under this condition the given
circuit reduces to a current source of value Is = Vin 5000 driving a parallel RLC with R =
5000 Ω, L = L1 = 0.01 H, the capacitance reflected to the primary of value C = 10−8 F.
Therefore
m =1
LC= 105 rad/s and B =
1ReqC
= 2 ×104 rad/s.
Finally, at w = wm, the circuit is resonant and Vin appears across the primary of the transformer.
This voltage is then stepped up by a factor of 10. Therefore VoutVin max
= 10.
SOLUTION PROBLEM 19.38. (a) By inspection
V1
I2
=
Z1 1
−1 Y2
I1
V2
=
h11 h12
h21 h22
I1V2
(b) By inspection
V1 = V2 − Z1I2 ⇒ V1 + Z1I2 = V2
and
I1 = Y2V1 − I2 ⇒ Y2V1 − I2 = I1
Prbs Chap 19, 1/7/02 P19-3 0 © R. A. DeCarlo, P. M. Lin
In matrix form,
V1
I2
=
0 1
1 0
I1V2
⇒
V1
I2
=
1 Z1
Y2 −1
−1 0 1
1 0
I1
V2
Thus
V1
I2
=
1Z1Y2 +1
Z1 1
−1 Y2
I1
V2
SOLUTION PROBLEM 19.39. (a) From problem 19.38b we have
V1
I2
=
1Z1Y2 +1
Z1 1
−1 Y2
I1
V2
=
11
RCs+1
1
Cs1
−11
R
I1V2
=
1RCs +1
R RCs
−RCs Cs
I1
V2
(b) This part is a cascade of part (a) and an ideal transformer. Label the voltage and current at
the port 2 of N1 as ˆ V 2 and ˆ I 2 . From the properties of the ideal transformer, V2 = n ˆ V 2 and
I2 = ˆ I 2 n . Hence
V1ˆ I 2
=
1RCs +1
R RCs
−RCs Cs
I1ˆ V 2
⇒
V1
nI2
=
1RCs +1
R RCs
−RCs Cs
I1
V2 n
Therefore
V1
I2
=
1RCs +1
R RCs n
−RCs n Cs n2
I1V2
From table 19.1, if h22 = Cs
n2(RCs +1) ≠ 0, then the z-parameters exist and if h11 =
RRCs +1
≠ 0,
the y-parameters exist, i.e., if C ≠ 0 and R ≠ 0 respectively.
Prbs Chap 19, 1/13/02 P19-1 © R. A. DeCarlo, P. M. Lin
CHAPTER 19 PROBLEM SOLUTIONS
SOLUTION PROBLEM 19.40. (a) Let Z1 = R and Z2 = 1/Cs or Y2 = Cs. From problem 38,
h =Z1 1
−1 Y2
=
R 1
−1 Cs
(b) This part is a cascade of an ideal transformer and part (a). Label the voltage and current at
the port 1 of N1 as ˆ V 1 and ˆ I 1 . From the properties of the ideal transformer, V1 = −b ˆ V 1 and
I1 = − ˆ I 1 b . Hence
ˆ V 1I2
=R 1
−1 Cs
ˆ I 1V2
⇒
−V1 b
I2
=
R 1
−1 Cs
−bI1V2
Therefore
V1
I2
=
b2R −b
b Cs
I1
V2
From table 19.1, if h22 = Cs ≠ 0, then the z-parameters exist and if h11 = b2R ≠ 0, the y-
parameters exist, i.e., if C ≠ 0 and R ≠ 0 (assuming reasonably that b ≠ 0) respectively.
SOLUTION PROBLEM 19.41. For this solution we apply the definition of h-parameters: by
inspection
h11 =V1I1 V2 =0
=1
2 + 2s
h21 =I2I1 V2 =0
=2V1 − 2sV1
I1 V2=0
=(2 − 2s)V1(2 + 2s)V1 V2 =0
=1− ss +1
When I1 = 0, then I2 = 2V1 + 2V1 = 4V1 and V2 = 0.5(2V1) +12s
(2V1) =s +1
sV1.
Prbs Chap 19, 1/13/02 P19-2 © R. A. DeCarlo, P. M. Lin
Therefore, h12 =V1V2 I1=0
=s
s +1 and h22 =
I2V2 I1=0
=4V1V2 I1=0
=4s
s +1.
SOLUTION PROBLEM 19.42. (a) In MATLAB
»h11 = 250; h12 = 0.025; h21 = 12.5; h22 = 2.25e-3;
»Zs = 1e3; ZL = 500;
»YL = 1/ZL
YL = 2.0000e-03
»Zin = h11 - h12*h21/(h22 + (1/ZL))
Zin = 1.7647e+02
»Yout = h22 - h12*h21/(h11 + Zs)
Yout = 2.0000e-03
»Zout = 1/Yout
Zout = 500
(b)
»% Gv1 = V1/Vs
»Gv1 = Zin/(Zin + Zs)
Gv1 = 1.5000e-01
»% Gv2 = V2/V1
»Gv2 = -h21/(Zin*(h22 + YL))
Gv2 = -1.6667e+01
»Gv = Gv1*Gv2
Gv = -2.5000e+00
(c) Given the above, the Thevenin equivalent seen by the capacitor is Voc = −2.5Vin and
Rth = 500 Ω.
In MATLAB
»Zth = ZL*Zout/(ZL + Zout)
Zth = 250
»Vin = 10;
»Voc = -2.5*Vin;
»w = 400;
Prbs Chap 19, 1/13/02 P19-3 © R. A. DeCarlo, P. M. Lin
»Zc = 1/(j*w*10e-6)
Zc = 0 - 2.5000e+02i
»Vc = Voc*Zc/(Zth + Zc)
Vc = -1.2500e+01 + 1.2500e+01i»V2mag = abs(Vc)
»V2mag = abs(Vc)
V2mag =
1.7678e+01
»V2ang = angle(Vc)*180/pi
V2ang =
135
From above,
v2(t) = 17.678 2 cos(400 t +135o) V
Therefore
»Pave = V2mag^2/500
Pave =
6.2500e-01
SOLUTION PROBLEM 19.43.
(a) Using the h-parameters of stage 2
Zin2 = h11 - h12h21
h22 +YL = 1000+ 0.966×51
0.0008 + 1/64 = 4000 Ω
The load for stage 1 is the parallel combination of Zin2 and the 3 kΩ resistance. However,because h12 = 0, the input impedance is unaffected by the load, and hence for stage 1,
Zin1 = h11 = 2000 Ω
(b) For stage 1, because h12 = 0, the output impedance is unaffected by the source impedance.Thus,
Yout1 = h22 = 0.05 ×10-3 S,
Zout1 =1/Yout1 = 20 ×103 Ω
Prbs Chap 19, 1/13/02 P19-4 © R. A. DeCarlo, P. M. Lin
For stage 2, the source impedance is the parallel combination of Zout1 and the 3 kΩ resistance.Thus
Zs2 = 20000×300020000 + 3000
=2608.7 Ω
and
Yout2 = h22 - h12h21
h11 +Zs2 = 0.0008+ 0.966×51
1000 + 2608.7 = 0.0145 S
Zout2 =1/Yout2 = 69.19 Ω
(c)
V1
Vs = Z in1
Z in1 + Zs = 2000
2000 + 2000 = 0.5
The load of stage 1 is the parallel combination of Zin2 and Zm. Thus
YL1 = Y in2 + 1/3000 = 5.834 ×10-4 S
HenceZL1 = 1/ YL1 = 1714 Ω
and
V2
V1 stage 1 = 1
Z in1× -h21
h22 + YL1 = 1
2000× -50
(0.05 + 0.5834) ×10-3 = - 39.46
For stage 2, the load is 64 Ω. Hence
V2
V1 stage 2 = 1
Z in2× -h21
h22 + YL2 = 1
4000× 51
(0.8 + 1000/64) ×10-3 = 0.7762
Finally, the overall voltage gain is the product of the three gains calculated above
Vout
Vs = 0.5 ×(- 39.46) ×(0.7762) = - 15.32
(d) The input circuit consists of a series connection of Vs, Zs, C and Zin1. The remainder of thecircuit is resistive and has no effect on the frequency response. The magnitude response is of thehigh pass type with
f3dB =1
2π(Rs + Rin1)C=
1
2π(2000 + 2000)10−6 = 39.789 Hz
SOLUTION PROBLEM 19.44. To meet the required matching, we must have
Prbs Chap 19, 1/13/02 P19-5 © R. A. DeCarlo, P. M. Lin
Zout = Zout2 = ZL = 64 = 1Yout2
This requires that
Yout2 = h22 - h12h21
h11 +Zs2 = 0.0008+ 0.966×51
1000 +Zs2 = 1
64
Solving for Zs2, we obtain Zs2 = 2323.2 Ω. Now for stage 2, Zs2 is the parallel combination ofZout1 = 20 kΩ and Zm:
Zs2 = 20000×Zm
20000 +Zm = 2323.2 Ω
from which Zm = 2628.5 Ω. With this new value of Zm, we repeat the calculations of problem19.43 to obtain Zin = 2000 Ω and Vout/Vs = -14 .26. Details follow.
Using the h parameters of stage 2
Zin2 = h11 - h12h21
h22 +YL = 1000+ 0.966×51
0.0008 + 1/64 = 4000 Ω
For stage 1, because h12 = 0, the input impedance is not affected by the load and
Zin1 = h11 = 2000 Ω. The voltage gains of the various stages are:
V1
Vs = Z in1
Z in1 + Zs = 2000
2000 + 2000 = 0.5
The load of stage 1 is the parallel combination of Zin2 and Zm. Thus
YL1 = Y in2 + 1/2628.5 = 6.3044 ×10-4 SHence
V2
V1 stage 1 = 1
Z in1× -h21
h22 + YL1 = 1
2000× -50
(0.05 + 0.63044) ×10-3 = - 36.74
For stage 2, the load is 64 Ω. Hence
V2
V1 stage 2 = 1
Z in2× -h21
h22 + YL2 = 1
4000× 51
(0.8 + 1000/64) ×10-3 = 0.7762
Finally, the overall voltage gain is the product of the three gains calculated above
Vout
Vs = 0.5 ×(- 36.74) ×(0.7762) = - 14.26
Prbs Chap 19, 1/13/02 P19-6 © R. A. DeCarlo, P. M. Lin
SOLUTION PROBLEM 19.45.
(a) Using the h-parameters of stage 2
Zin2 = h11 - h12h21
h22 +YL = 500 + 0.966×51
0.0016 + 1/32 = 2000 Ω
The load for stage 1 is the parallel combination of Zin2 and the 1.5 kΩ resistance. However,
because h12 = 0, the input impedance is unaffected by the load, and Zin1 = h11 = 1000 Ω.
(b) For stage 1, because h12 =0, the output impedance is unaffected by the sourceimpedance, and
Yout1 = h22 = 0.1 ×10-3 S,
Zout1 =1/Yout1 =104 Ω
For stage 2, the source impedance is the parallel combination of Zout1 and the 1.5 kΩ resistor
Zs2 =104 ×1.5 ×103
104 +1.5 ×103 =1304.3 Ω
and
Yout2 = h22 - h12h21
h11 +Zs2 = 0.0016+ 0.966×51
500 + 1304.3 = 0.0289 S
Zout2 =1/Yout2 = 34.6 Ω
(c)
V1
Vs = Z in1
Z in1 + Zs = 1000
1000 + 1000 = 0.5
The load of stage 1 is the parallel combination of Zin2 and Zm. Thus
YL1 = Y in2 + 1/1500 = 1.1667 ×10-3 SHence
ZL1 = 1/ YL1 = 857.1 Ω
V2
V1 stage 1 = 1
Z in1× -h21
h22 + YL1 = 1
1000× -50
(0.1 + 1.1667) ×10-3 = - 39.47
For stage 2, the load is 32 Ω. Hence
Prbs Chap 19, 1/13/02 P19-7 © R. A. DeCarlo, P. M. Lin
V2
V1 stage 2 = 1
Z in2× -h21
h22 + YL2 = 1
2000× 51
0.0016 + 1/32 = 0.7762
Finally, the overall voltage gain is the product of the three gains calculated above
Vout
Vs = 0.5 ×(- 39.47) ×(0.7762) = - 15.32
(d) The input circuit consists of a series connection of Vs, Zs, C and Zin1. The remainder of thecircuit is resistive and has no effect on the frequency response. The magnitude response is of thehighpass type with
f3dB =1
2π(Rs + Rin1)C=
1
2π(1000 +1000)10−6 = 79.58 Hz
SOLUTION PROBLEM 19.46. (a) Since the currents through YL and h22 are the same, h22 = YL.
(b) From current division, I2 =YL
YL + h22h21I1 ⇒
I2I1
=YL h21
YL + h22.
(c) 150 =I2I1
=YL h21
YL + h22= 0.5h21 ⇒ h21 = 300.
(d) h12 =V1V2
I1=0
=−12
= −0.5 .
(e) I1Is
=Zs
Zs + Zin=
9 ×103
9 ×103 + Zin= 0.9 ⇒ Zin =1000 Ω. Given this quantity,
h11 = Zin +h12h21
h22 +YL=1000 −
1500.25
= 400 Ω.
SOLUTION PROBLEM 19.47. Recall that
V1
I2
=
h11 h12
h21 h22
I1
V2
Prbs Chap 19, 1/13/02 P19-8 © R. A. DeCarlo, P. M. Lin
(a) From this expression and specification 1, h12 =V1
V2
I1=0
= 0 .
(b) From the formula for Yout (equation 19.50), specification 2, and the result of part (a), we
have
Yout =1
Zout=
1
800= h22 −
h12h21
h11 + Zs= h22
Hence, h22 =1
800= 1.25 mS.
For maximum power transfer from amplifier to the load,
Zout = 800 = b2ZL = 8b2
Therefore, b = 10.
(c) and (d) Observe that
Zin = h11 −h12h21
h22 + YL b2 = h11
From specification 3 and voltage division,
V1
Vs=
24
25=
Z in
Zin + 40
Equivalently,
Vs
V1=
25
24= 1+
40
Zin
Hence h11 = Z in = 40 × 24 = 960 Ω.
(e) From equation 19.51,
Gv2 =V2
V1= −100 =
− h21
Zin h22 + YL b2( ) =−h21
960 1.25 ×10−3 +1.25 × 10−3( ) =−h21
2.4
Hence, h21 = 240 .
Prbs Chap 19, 1/13/02 P19-9 © R. A. DeCarlo, P. M. Lin
(f) The power delivered to the load is PL =V2
2
800 and the power delivered to the amplifier is
Pamp =V1
2
Zin=
V12
960. Therefore the power gain is
PL
Pamp=
96
80
V2
V1
2
= 1.2 ×104
SOLUTION PROBLEM 19.48. Recall that
V1
I2
=
h11 h12
h21 h22
I1
V2
(a) From this expression and specification 1, h12 =V1V2
I1=0
= 0.01.
(b) For maximum power transfer from amplifier to the load,
Zout = 800 = b2ZL = 8b2
Therefore, b = 10.
Now we find Zin. From specification 3 and voltage division,
V1
Vs=
24
25=
Z in
Zin + 40 (1)
Equivalently,
VsV1
=2524
=1 +40Zin
⇒ Zin = 40 × 24 = 960 Ω
Using the formula for Zin we have the following equation
Zin = 960 = h11 −0.01h21
h22 +1.25 ×10−3 ⇒ h11 = 960 + 0.01h21
h22 +1.25 ×10−3 (2)
Prbs Chap 19, 1/13/02 P19-10 © R. A. DeCarlo, P. M. Lin
But from the given specs,
V2V1
= −100 =−h21
Zin h22 + YL b2( ) =−h21
960 h22 +1.25 ×10−3( )which implies that
h21
h22 +1.25 ×10−3( ) = 960 ×102 (3)
Substituting (3) into (2) allows us to solve for h11:
h11 = 960 + 0.01× 960 ×102 =1920 Ω
Now let us rewrite equation (3) as:
h21 − 960 ×102h22 = 960 ×102 ×1.25 ×10−3 = 120 (4)
Also
Yout = 1.25 ×10−3 = h22 −h12h21
h11 + Zs= h22 −
0.01h211960
Equivalently,
1960 ×1.25 ×10−3 = 2.45 =1960h22 − 0.01h21 (5)
Solving equations (4) and (5) simultaneously in MATLAB yields
»A = [1 -960e2;-0.01 1960]
A =
1.0000e+00 -9.6000e+04
-1.0000e-02 1.9600e+03
»b = [120; 2.45]
b =
1.2000e+02
2.4500e+00
»x = A\b
x =
Prbs Chap 19, 1/13/02 P19-11 © R. A. DeCarlo, P. M. Lin
4.7040e+02
3.6500e-03
»h21 = x(1); h22 = x(2);
h = [1920 0.01;h21 h22]
h =
1.9200e+03 1.0000e-02
4.7040e+02 3.6500e-03
We can verify these results as well as compute the overall amplifier gain using the following m-
file:
% two-port analysis in terms of h-parameters
function [zin, zout] =twoport(h, zL, zs)
['twoport analysis using h-parameters']
h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2);
zin = h11 - h12*h21/(h22+ 1/zL)
yout= h22 - h12*h21/(h11+zs);
zout= 1/yout
v1tovs= zin/(zin+zs)
v2tov1= -h21/(zin*(h22+1/zL))
v2tovs= v1tovs*v2tov1
»twoporth(h,ZL,Zs)
ans =
twoport analysis using h-parameters
zin =
960
zout =
8.0000e+02
v1tovs =
9.6000e-01
v2tov1 =
-100
v2tovs =
-96
Prbs Chap 19, 1/13/02 P19-12 © R. A. DeCarlo, P. M. Lin
Hence the overall voltage gain is VL/Vs = –96/10 = –96 because of the transformer.
Finally to compute power gains,
»Vs = 1; Vin = 24/25;
»VL = -9.6;
»Pin = Vin^2/960
Pin =
9.6000e-04
»Pload = VL^2/8
Pload =
1.1520e+01
»Pgain = Pload/Pin
Pgain =
12000
SOLUTION PROBLEM 19.49.
(a) h21 =I2I1 V2 =0
=−C sVπ + gmVπ
1
Rπ+ Cπ + C( )s
Vπ
=−C s + gm
Cπ + C( )s +1
Rπ
(b) h11 =V1I1 V2 =0
=
Rx +1
1
Rπ+ Cπ + C( )s
I1
I1= Rx +
11
Rπ+ Cπ + C( )s
(c) Under the condition that I1 = 0, V1 = Vπ. Using voltage division from V2 to Vπ:
Prbs Chap 19, 1/13/02 P19-13 © R. A. DeCarlo, P. M. Lin
h12 =V1V2 I1=0
=VπV2
=
11
Rπ+ Cπs
1
C s+
11
Rπ+ Cπs
=C s
1
Rπ+ Cπs + C s
=C s
Cπ + C( )s +1
Rπ
(d) h22 =I2V2 I1=0
=gmVπ +
1
Rπ+ Cπs
Vπ
Cπ + C( )s +1
RπC s
Vπ
=C s Cπs + gm +
1
Rπ
Cπ + C( )s +1
Rπ
SOLUTION PROBLEM 19.50: (a) Recall t-parameter relationship:
V1
I1
=
t11 t12
t21 t22
V2
−I2
For the given network,V2 = ZL (−I2) and
V1 = t11V2 + t12(−I2) = t11ZL (− I2) + t12(− I2) = (t11ZL + t12)(−I2)
Further,
I1 = t21V2 + t22(− I2) = ( t21ZL + t22)(−I2)
Hence,
Zin =V1I1
=( t11ZL + t12)(−I2)( t21ZL + t22)(−I2)
=t11ZL + t12t21ZL + t22
(b) For the output impedance relationship, from the t-parameter relationships
V1 = Zs(− I1) = t11V2 + t12(−I2) = −Zs(t21V2 + t22(− I2))
Grouping V2 and I2 terms together on separate sides of the equation implies
(t21Zs + t11)V2 = (t22Zs + t12)I2Thus
Prbs Chap 19, 1/13/02 P19-14 © R. A. DeCarlo, P. M. Lin
Zout =V2I2
=t22Zs + t12t21Zs + t11
SOLUTION PROBLEM 19.51:
45. From the z-parameter relationships
V1 = z11I1 + z12I2 ⇒ V1 − z11I1 = z12I2 = −z12(− I2)and
V2 = z21I1 + z22I2 ⇒ z21I1 = V2 + z22(− I2)
These two equations in matrix form are:
1 −z11
0 z21
V1
I1
=
0 − z12
1 z22
V2
−I2
Solving for V1 I1[ ]T yields
V1
I1
=
1z21
z21 z11
0 1
0 − z12
1 z22
V2
− I2
=
1z21
z11 ∆z
1 z22
V2
− I2
SOLUTION PROBLEM 19.52: For figure 19.52a, by inspection
V1
I1
=
1 Z1
0 1
V2
−I2
Therefore T is as indicated in the problem.For figure 19.52b, by inspection
V1
I1
=
1 0
Y2 1
V2
− I2
Therefore T is as indicated in the problem.
SOLUTION PROBLEM 19.53: Here we use the results of problem 19.52:
(a) For figure (a)
Tnew =1 Z1
0 1
1 0
Y2 1
=
1+ Z1Y2 Z1
Y2 1
(b) For figure (b)
Prbs Chap 19, 1/13/02 P19-15 © R. A. DeCarlo, P. M. Lin
Tnew =1 0
Y2 1
1 Z1
0 1
=
1 Z1
Y2 1 + Z1Y2
SOLUTION PROBLEM 19.54: By the properties of an ideal transformer,
V1 = nV2 I1 =−1n
I2 =1n
(− I2)
ThereforeV1
I1
=
n 0
0 1 n
V2
−I2
with the t-parameters given by the 2x2 matrix.
SOLUTION PROBLEM 19.55: This problem uses the results of the previous two problems.(a)
T =1+ Z1Y2 Z1
Y2 1
n 0
0 1 n
=
n 1+ Z1Y2( ) Z1 n
nY2 1 n
(b)
T =n 0
0 1 n
1 + Z1Y2 Z1
Y2 1
=
n 1+ Z1Y2( ) nZ1
Y2 n 1 n
SOLUTION PROBLEM 19.56:(a)
T =0.25 0
0 4
8 4
2 5
=
2 1
8 20
(b)
T =1 0
1 R 1
8 4
2 5
=
8 4
2 + 8 R 5 + 4 R
(c)
T =1+ Z1Y2 Z1
Y2 1
8 4
2 5
=
1+ 0.25s 0.5s
0.5 1
8 4
2 5
=
8 + 3s 4 + 3.5s
6 7
SOLUTION PROBLEM 19.57: For each 2-port of the form of figure P19.53a, we have that the t-
parameters are given by
Prbs Chap 19, 1/13/02 P19-16 © R. A. DeCarlo, P. M. Lin
T =1+ Z1Y2 Z1
Y2 1
The given network consists of three such sections in cascade whose t-parameters are respectiely,
T1 =1 + s2 s
s 1
, T2 =
1+1× 0.5s 1
0.5s 1
=
1 + 0.5s 1
0.5s 1
, and
T3 =1 + 2s ×
1
4s 2s
1
4s 1
=1+
1
2s2 2s
1
4s 1
Observe that
T1T2 =
1
2s3 +
3
2s2 +
1
2s +1 s2 + s +1
1
2s2 +
3
2s s +1
and the overall t-parameters are
T = T1T2T3 =
1
4s5 +
3
4s4 + s3 +
9
4s2 +
3
4s +1 s4 + 3s3 + 2s2 + 3s +1
1
4s4 +
3
4s3 +
3
4s2 +
7
4s s3 + 3s2 + s +1
SOLUTION PROBLEM 19.58: This poblem is done primarily in MATLAB.
Part (a)% The following code solves part (a) of the problem.% Parameter Specificationt11= 0.895+j*0.022;t22= t11;t12= 40 + j*180;
% t21= (t11*t22 -1)/t12;% The above formula follows because it is a reciprocal network.% The actual value is specified.t21= -2.6175e-05+j*1.1023e-03;t=[t11 t12; t21 t22]
% Part (a) calculationsvr= 115200ir = 361
vs= t11*vr +t12*irmagvs=abs(vs)angvs= angle(vs)*180/pi
Prbs Chap 19, 1/13/02 P19-17 © R. A. DeCarlo, P. M. Lin
is=t21*vr+t22*irmagis= abs(is)angis= angle(is)*180/pipscomp=vs*conj(is)ps=real(pscomp)pr=real(vr*conj(ir))eff= pr/pspf= ps/abs(pscomp)ploss= ps- pr
The MATLAB output is as follows:
T = 8.9500e-01 + 2.2000e-02i 4.0000e+01 + 1.8000e+02i -2.6175e-05 + 1.1023e-03i 8.9500e-01 + 2.2000e-02ivr = 115200ir = 361
vs = 1.1754e+05 + 6.7514e+04imagvs = 1.3555e+05angvs = 2.9872e+01
is = 3.2008e+02 + 1.3493e+02imagis = 3.4736e+02angis = 2.2857e+01
pscomp = 4.6733e+07 + 5.7501e+06ips = 4.6733e+07pr = 41587200
eff = 8.8989e-01pf = 9.9252e-01ploss = 5.1458e+06
Part (b)% The following code solves part (b) of the problem.
zL=500;zin= (t11*zL + t12)/( t21*zL + t22)yin = 1/zinvsnew=134000;iin= yin*vsnewpsnew= vsnew^2*real(yin)m= inv(t)v2=m(1,1)*vsnew +m(1,2) *iinmagv2=abs(v2)iload= m(2,1) *vsnew + m(2,2)* iinmagild = abs(iload)% Check value of rloadrload=abs(v2)/abs(iload)
The MATLAB output for part (b) is:
Prbs Chap 19, 1/13/02 P19-18 © R. A. DeCarlo, P. M. Lin
zin = 4.8759e+02 - 1.0031e+02iyin = 1.9676e-03 + 4.0478e-04iiin = 2.6366e+02 + 5.4241e+01ipsnew = 3.5331e+07
% m is the inverse of T-matrix.
m = 8.9500e-01 + 2.2000e-02i -4.0000e+01 - 1.8000e+02i 2.6175e-05 - 1.1023e-03i 8.9500e-01 + 2.2000e-02i
v2 = 1.1915e+05 − 4.6681e+04imagv2 = 1.2796e+05iload = 2.3829e+02 - 9.3362e+01imagild = 2.5593e+02rload = 5.0000e+02
SOLUTION PROBLEM 19.59. From the given information, the circuit is linear and reciprocal.(a) Here i2( t) is the integral of i1(t). Therefore, the new v1( t) is the integral of the old v2(t).The result for t ≥ 0 is:
v1( t) = 3.005 − 3e−t + e−t[0.00865sin(500 t) − 0.005cos(500 t)]
= 3.005 − 3e−t + e−t −2 ⋅10−5 cos(500t − π /6) + 0.01sin(500t − π /6)[ ](b) From the problem statement
z21(s) =3
s +1+ 5
cosπ6
s +1( ) + 250
s +1( )2 + 500( )2
From reciprocity, z12(s) = z21(s). For steady state analysis, we use phasors to obtain
V1 = z12( j500)I2 =3
1+ j500+ 5
cosπ6
(1+ j500) + 250
(1+ j500)2 + (500)2
v1ss (t) = 2.505cos(500 t − 30.15o) V
SOLUTION PROBLEM 19.60. Writing loop equations we have:
(i) For the left loop,
V1 − aV2 − 3(I1 + I3) = 0 ⇒ V1 − aV2 = 3I1 + 3I3
(ii) For the right loop,
Prbs Chap 19, 1/13/02 P19-19 © R. A. DeCarlo, P. M. Lin
V2 + bI1 − (I2 − I3) = 0 ⇒ V2 = −bI1 + I2 − I3
(iii) For the middle loop,
bI1 + (I3 − I2) + I3 + 3(I1 + I3) = 0 ⇒ 0 = (b + 3)I1 − I2 + 5I3
Writing the first two equations in matrix form yields
1 −a
0 1
V1
V2
=
3 0 3
−b 1 −1
I1
I2
I3
whose solution is
V1
V2
=
1 a
0 1
3 0 3
−b 1 −1
I1
I2
I3
=3 − ab a 3 − a
−b 1 −1
I1
I2
I3
HenceV1
V2
=
3− ab a
−b 1
I1
I2
+
3 − a
−1
I3
From the third equation
I3 = −0.2b − 0.6 0.2[ ] I1
I2
ThusV1
V2
=
3 − ab a
−b 1
+
3− a
−1
−0.2b − 0.6 0.2[ ]
I1
I2
V1
V2
=
−0.6b − 0.8ab + 0.6a +1.2 0.6 + 0.8a
−0.8b + 0.6 0.8
I1I2
For reciprocity,z12 = z21 ⇒ a = −b
SOLUTION PROBLEM 19.61. (a) and (b) together: Observe that
V1 − 0.5V1 − I1 = V2 ⇒ I1 = 0.5V1 − V2 ⇒ V1 = 2I1 + 2V2and
I1 + I2 = 0.5(V2 − aI1) ⇒ I2 = −(0.5a +1)I1 + 0.5V2The h-parameters are
H =2 2
−(0.5a +1) 0.5
Prbs Chap 19, 1/13/02 P19-20 © R. A. DeCarlo, P. M. Lin
Reciprocity requires thath12 = −h21 ⇒ 2 = 0.5a +1 ⇒ a = 2
Thus
H =2 2
−2 0.5
SOLUTION PROBLEM 19.62. There are 2 corrections in the problem statement concerning thesecond set of expressions:
(1) v1(t) = 2e-t - 1.5e-1.5t V(2) i2(t) = 0.5 e-1.5t A
For both parts, recall, the y-parameters:
I1
I2
=
y11
y21
y12
y22
V1
V2
(a) Part-1: From the first set of given data (V2 = 0)
I1 =1s, I2 =
0.5s +1
, V1 =0.5s +1
+0.5
(s +1)2 =0.5(s + 2)
(s +1)2
Hence,
y11 =I1V1
V2 =0
=2(s +1)2
s(s + 2), y21 =
I2V1
V2=0
=s +1s + 2
(a) Part-2: From the second set of given data (ZL = 1 Ω, I1 = 1 s , etc.), we have
I2 =0.5
s +1.5, V1 =
2s +1
−1.5
s +1.5=
0.5(s + 3)(s +1)(s +1.5)
For a terminated 2-port,
I2 =−V2ZL
= −V2 =y21
y22 + YLV1 =
y21y22 +1
V1
Therefore
y22 = y21V1I2
−1 =s +1s + 2
×0.5(s + 3)
(s +1)(s +1.5)×
s +1.50.5
−1=(s + 3)(s + 2)
−1 =1
s + 2
Also, I1 = y11V1 + y12V2 ⇒ y12 =I1 − y11V1
V2=
I1 − y11V1−I2
. Hence
y12 = −s +1.5
0.5×
1s
−2(s +1)2
s(s + 2)0.5(s + 3)
(s +1)(s +1.5)
= −(2s + 3)(s + 2)
s(s + 2)+
2(s +1)(s + 3)s(s + 2)
Prbs Chap 19, 1/13/02 P19-21 © R. A. DeCarlo, P. M. Lin
=1
(s + 2)In conclusion
y11
y21
y12
y22
=
2(s +1)2
s(s + 2)
1
(s + 2)s +1
s + 2
1
(s + 2)
=1
(s + 2)
2(s +1)2
s1
s +1 1
(b) This is a straightforward application of the conversion table 19.1.
SOLUTION PROBLEM 19.63. (a) Consider figure (a). Write two mesh equations:
V1 =16I1 +2s
(I1 + I2) + 4I1 = 20 +2s
I1 +
2s
I2
V2 = sI2 +2s
(I1 + I2) + 4I1 = 4 +2s
I1 + s +
2s
I2
Therefore
Z =1s
20s + 2
4s + 2
2
s2 + 2
Taking the inverse yields the y-parameters
Y =1
∆Z
z22 −z12
−z21 z11
=
1
20s2 + 2s + 32
s2 + 2 −2
−(4s + 2) 20s + 2
where
∆Z =(20s + 2)(s2 + 2)− 2(4s + 2)
s=
20s3 + 2s2 + 32ss
= 20s2 + 2s + 32
Finally, the h-parameters are given as
H =1 y11 z12 z22
− z21 z22 1 z22
=
1
s2 + 2
20s2 + 2s + 32 2
−(4 s + 2) s
(b) Now consider figure (b). z11 is V1 when I2 = 0. But if I2 = 0, then because of the idealtransformer I1 = 0, meaning that the ratio is not defined. Hence the z-parameters do not exist.
To find the h-parameters, observe that because of the ideal transformer, I2 = −0.5I1,I1 = −2I2 , and Vpri = 0.5Vsec . Writing a mesh equation at the right mesh first we obtain
V2 = RI2 + Vsec + (I1 + I2)R = Vsec
Prbs Chap 19, 1/13/02 P19-22 © R. A. DeCarlo, P. M. Lin
Now writing a mesh equation on the left we have
V1 = RI1 + Vpri + (I1 + I2)R = 1.5RI1 +Vpri = 1.5RI1 + 0.5V2
Therefore
H =1.5R 0.5
−0.5 0
To obtain the y-parameters we use table 19.1:
Y =1
h11
1 −h12
h21 ∆h
=
16R
4 −2
−2 1
Note: the det[Y] = 0 implying again that the z-parameters do not exist.
(c) For this network we consider it as a cascade (left to right) of an ideal transformer, the middlenetwork of a transformer and an inductor across the top, and finally another ideal transformer.The t-parameters of these two ports are respectively:
T1 =a 0
0 1 a
, T2 =
A B
C D
, T3 =
1 b 0
0 b
To find T2, we replace the mutually coupled inductors by the pi-equivalent circuit of figure
18.25c where L1 = 4 H, L2 = 9 H, M = k L1L2 = 3 H, and ∆ = 27. Thus Lleft =∆
L2 − M= 4.5
H, Lright =∆
L1 − M= 27 H, and Ltop =
∆M
= 9 H. Notice that the external 9 H inductor is in
parallel with Ltop leading to Lpar = 4.5 H. The y-matrix of this new pi-network is by inspection:
Ymid =1s
2 4.5 −1 4.5
−1 4.5 1 27 +1 4.5
=
127s
12 −6
−6 7
To compute the t-parameters we have
T2 =
−y22
y21
−1
y21−∆y
y21
−y11
y21
=7 6 4.5s
8 27s 2
Therefore
T = T1T2T3 =a 0
0 1 a
7 6 4.5s
8 27s 2
1 b 0
0 b
=
7a 6b 4.5abs
8 27abs 2b a
From table 19.1, we obtain
Prbs Chap 19, 1/13/02 P19-23 © R. A. DeCarlo, P. M. Lin
H =a22.25s 0.5a / b
−0.5a / b 4 27b2
, Z = s
63a2
16
27ab
827ab
8
27b2
4
, Y =1
27a2b2s
12b2 −6ab
−6ab 7a2
SOLUTION PROBLEM 19.64.
(a) The defining equation for the g-parameters is:
I1V2
=
g11 g12
g21 g12
V1
I2
Because of the load impedance ZL, we have V2 = −ZLI2 . Hence substituting for V2 in thesecond equation yields
V2 = g21V1 + g22I2 = −ZLI2 ⇒ I2 =−g21V1
g22 + ZL
Substituting this equation into I1 = g11V1 + g12I2 we obtain
I1 = g11V1 − g12g21V1
g22 + ZL
= g11 −
g12g21g22 + ZL
V1
Therefore,
Yin = g11 −g12g21
g22 + ZL
(b) Because of the source impedance Zs, we have V1 = −ZsI1 or I1 = −YsV1 . Hence substitutingfor V1 in the first equation yields
I1 = g11V1 + g12I2 = −YsV1 ⇒ V1 =−g12I2g11 + Ys
Substituting this equation into V2 = g21V1 + g22I2 we obtain
V2 = g21−g12I2g11 + Ys
+ g22I2 = g22 −
g12g21g11 + Ys
I2 ⇒ Zout = g22 −
g12g21g11 + Ys
(c)
G1 =V1Vs
=Zin
Zs + Zin=
YsYs +Yin
=Ys
Ys + g11 −g12g21
g22 + ZL
=Ys g22 + ZL( )
g22 + ZL( ) g11 +Ys( ) − g12g21
(d) Refer to figure P19.64, where V2 =ZL
g22 + ZLg21V1 ⇒ G2 =
V2V1
=g21ZL
g22 + ZL.
Prbs Chap 19, 1/13/02 P19-24 © R. A. DeCarlo, P. M. Lin
(e) Gv = G1G2 =Ys g22 + ZL( )
g22 + ZL( ) g11 +Ys( ) − g12g21×
g21ZLg22 + ZL
=g21YsZL
g22 + ZL( ) g11 + Ys( ) − g12g21
SOLUTION PROBLEM 19.65. Recall
I1V2
=
g11 g12
g21 g22
V1
I2
For the given circuit
V2 =14
(I2 − 4V2) + V1 ⇒ V2 =12
V1 +18
I2
Also
V2 = V1 −14
I1 − sV1( )This implies that
I1 = (4 + s)V1 − 4V2 = (4 + s)V1 − 412
V1 +18
I2
= (2 + s)V1 −
12
I2
Thus the g-parameter matrix is:
G =2 + s −
1
21
2
1
8
SOLUTION PROBLEM 19.66. We first convert the y-parameters to t-parameters using table 19.1:
YN1 → TN 1 =
−y22
y21
−1
y21−∆y
y21
−y11
y21
=−0.1 −0.5
−3.82 −25.1
where ∆y = 50.2 × 0.2 − 2 ×1.2 = 7.64 . Now we convert the h-parameters to t-parameters usingtable 19.1:
HN 2 → TN 2 =
−∆h
h21
−h11
h21−h21
h21
−1
h21
=15
56 13
2 1
=
11.2 2.6
0.4 0.2
where ∆h =13 × 2 + 6 × 5 = 56 . To obtain the cascaded t-parameters we compute
Prbs Chap 19, 1/13/02 P19-25 © R. A. DeCarlo, P. M. Lin
Tcas = TN1TN 2 =15
−6.6 −1.8
−264.12 −74.76
=
−1.32 −0.36
−52.824 −14.952
Thus,
Yin =1
Zin=
t21ZL + t22t11ZL + t12
=
603
515
5
= 40.2 S
To obtain the voltage gain, we first convert the t-parameters back to y-parameters (table19.1) and then use the derived voltage gain formula:
Tcas =−1.32 −0.36
−52.824 −14.952
→ Ycas =
41.533 2
2.7778 3.6667
S
Hence,
Gv =VLV1
=−y21
y22 + YL=
−2.77783.6667 + 0.5
= −0.66667
Alternately, one could consider the load as a 2-port, compute its t-parameters, construct
the overall t-parameters as a cascade of three networks, and then use Gv =1t11
.
SOLUTION PROBLEM 19.67.
(a) Since only one C or L , weh have in general: H(s) =as + bcs + d
. Since H(∞) = 0 =ac
, we
have that a = 0. Therefore,
H(s) =
b
c
s +d
c
=K
s + c
(b) To prove that c =1
RthC or
RthL
where Rth is the Thevenin resistance seen by the energy
storage element, we refer the reader to problem 19.69 which provides a general derivation withH(∞) arbitrary; hence this problem is the special case of H(∞) = 0.
(c) HereRth = 2000/ /(50 +1000/ /200)) = 195.49 Ω. Hence
c =1
RthC= 25.58 ×106 rad/s
Prbs Chap 19, 1/13/02 P19-26 © R. A. DeCarlo, P. M. Lin
SOLUTION PROBLEM 19.68.
(a) Because there is only one energy storage element, it follows that the most general form of
the transfer function is: H(s) =as + bcs + d
. Since H(0) = 0 =bd
, we have b = 0. Further, since the
transfer function is high pass, c ≠ 0, and
H(s) =as
cs + d=
a
c
s
s +d
c
=Ks
s + c
(b) To prove that c =1
RthC or
RthL
where Rth is the Thevenin resistance seen by the energy
storage element, we refer the reader to problem 19.69 which provides a general derivation withH(0) arbitrary; hence this problem is the special case of H(0) = 0.
(c) Let us apply the result of example 19.1 to that part of the circuit to the right of the 1 kΩresistor and call the associated resistance Zin. Here Z1 = ∞, Z2 = 2 kΩ, ZL = 100 Ω, and beta =50. Hence
Zin = beta +1( )ZL = 5.1 kΩThus
Rth = 200 +1000 //(2000 + 5100) = 1076.5 ΩHence,
c =1
RthC=
1
1076.54 × 2 ×10−6 = 464.45 rad/s
SOLUTION PROBLEM 19.69. In this problem we assume (i) a single input single output system
and that linear circuit seen by the energy storage element has a Thevenin equivalent or a
Norton equivalent. For simplicity we will presume the existence of a Thevenin equivalent.
(a) The Thevenin equivalent seen by the dynamic element L or C consists of Zth(s) in series with
Voc(s). Since the remainder network seen by L or C is non-dynamic (resistive, resistive
with dependent sources and ideal op amps, etc), we have Zth(s) = Rth and Voc(s) =
Ko×Input(s), Rth and Ko being real constants.
Prbs Chap 19, 1/13/02 P19-27 © R. A. DeCarlo, P. M. Lin
By voltage division,
VL =Ls
Ls + RthVoc =
K0s
s + Rth
L
× Input(s) (1)
and
VC =
1
Cs1
Cs+ Rth
Voc =
K0
RthC
s + 1
RthC
× Input(s) (2)
After VL(s) or VC(s) has been determined, we can find the Laplace transform of any other output
(voltage or current) using the voltage source substitution theorem (chapter 6) and linearity
(chapter 5):
Ouput(s) = K1 × Input(s) + K2 VL (s) or VC(s)( ) (3)
For the case of VL(s),
Ouput(s) = K1 + K2K0s
s + Rth
L
× Input(s) (4)
and for the case of VC(s),
Ouput(s) = K1 + K2
K0
RthC
s + 1
RthC
× Input(s) (5)
For either case, the transfer function H(s) has the form
H(s) =Ouput(s)
Input(s)=
K3s + K4
s +ω c(6)
Prbs Chap 19, 1/13/02 P19-28 © R. A. DeCarlo, P. M. Lin
where ωc =Rth
L or
1
RthC with K3 and K4 real constants.
It remains to give K3 and K4 some physical interpretations. In (6), let s = ∞, we have
H(∞) =K3s + K4
s + ωc
s=∞= K3
On the other hand, letting s = 0 in (6) produces
H(0) =K3s + K4
s +ω c
s=0
=K4
ωc
Substituting K3 and K4 into (6), we obtain the desired result
H(s) =Ouput(s)
Input(s)=
H (∞)s +ω cH(0)
s +ω c(7)
(b) When s = ∞, the impedance of C is
ZC (∞) =1
Cs
s=∞= 0
and the impedance of L isZL (∞) = Ls]s=∞ = ∞
Therefore in calculating H(∞), we may replace C by a short circuit and L by an open circuit. Onthe other hand, when s = 0, the impedance of C is
ZC (0) =1
Cs
s=0= ∞
and the iimpedance of L is
ZL (0) = Ls]s=0 = 0
Therefore in calculating H(0), we may replace C by an open circuit and L by a short circuit.
Prbs Chap 19, 1/13/02 P19-29 © R. A. DeCarlo, P. M. Lin
(c) For figure P19.69a, by inspection
Rth = 3//(2 + 4) = 3//6 = 2 Ωωc = 1/(Rth C) = 1/(2 × 0.5) = 1H(∞) = 4/(2 + 4) = 2/3H(0) = 4/(2 + 3 + 4) = 4/9
Therefore
H(s) =H(∞)s + ωcH(0)
s +ω c=
2
3s +
4
9s + 1
=
2
3s +
2
3
s +1
For figure P19.69b
Rth = 3//(2 + 4) = 3//6 = 2 Ωωc = Rth/ L = 2/2 =1H(∞) = 4/(2 + 4) = 2/3H(0) = 4/(2 + 3 + 4) = 4/9
Therefore
H(s) =H(∞)s + ωcH(0)
s +ω c=
4
9s +
2
3s + 1
=
4
9s + 1.5( )
s +1
SOLUTION PROBLEM 19.70. According to problem 19.68, the transfer function of the circuit is
H(s) =Ks
s + c
where c =1
RthC and Rth is the Thevenin resistance seen by the storage element C. To find Rth
we make use of figure 19.4 and the associated formula. The details are in the MATLAB code
below:
»R1 = 200*1e3/1200
R1 =
1.6667e+02
»Z1 = R1+2e3
Z1 =
2.1667e+03
»Z3 = 100;
Prbs Chap 19, 1/13/02 P19-30 © R. A. DeCarlo, P. M. Lin
»beta = -50;
»Zout = Z1/(1+beta)
Zout =
-4.4218e+01
»Rth = 2000 + Zout
Rth =
1.9558e+03
»wc = 1/(Rth*2e-6)
wc =
2.5565e+02
Hence
c =1
1955.8 × 2 ×10−6 = 255.65 rad/s
SOLUTION PROBLEM 19.71. (a) Except for the terminating resistor, let the other element
branches of the circuit be given by
Ζ1 =1
3s1, Y2 =
1
2s +1
2s
=2s
4s2 +1, and Z3 =1s
Consider the circuit
Here from problem 19.53a and from 19.52 a we obtain
Prbs Chap 19, 1/13/02 P19-31 © R. A. DeCarlo, P. M. Lin
V1
I1
=
1+
23
4s2 +1
1
3s2s
4s2 +11
V3
−I3
and
V3
−I3
=
11
s0 1
V2
− I2
which implies that
V1
I1
=
1
4s2 +14s2 + 5/ 3
16s2 + 6( )3s
2s 4s2 + 3
V2
−I2
(b)
GV (s) =−y21
y22 + yL
(c)
y21 = −1t12
= −3s(4s2 +1)
16s2 + 6, y22 =
t11t12
=3s 4s2 +
5
3
16s2 + 6=
s(12s2 + 5)
16s2 + 6, and YL = 1
(d) Hence,
GV (s) =−y21
y22 + yL=
3s(4s2 +1)
16s2 + 6s(12s2 + s)
16s2 + 6+1
=3s(4 s2 +1)
12s3 +16s2 + 5s + 6
SOLUTION PROBLEM 19.72. For part (a), treating each capacitor as a short circuit yields the
equivalent circuit below.
Prbs Chap 19, 1/13/02 P19-32 © R. A. DeCarlo, P. M. Lin
»h11T = 4.2e3; h12T = 0; h21T = 150; h22T = 0.1e-3;
»hT = [h11T, h12T;h21T h22T];
»yT = htoy(hT)
yT =
2.3810e-04 0
3.5714e-02 1.0000e-04
By inspection, the y11 parameter of the overall two-port consists of the sum of y11T plus the
conductances of the two front end resistors. Also, the y22 parameter of the overall two port
is y22T plus the conductance of the 4.7k Ω resistor. Hence,
»y = yT + [1/1e4+1/1e5 0;0 1/4700]
y =
3.4810e-04 0
3.5714e-02 3.1277e-04
Hence, the overall h-parameters are:
»h = ytoh(y)
h =
2.8728e+03 0
1.0260e+02 3.1277e-04
(b)
»yout = h(2,2) - (h(1,2)*h(2,1)/(h(1,1)+50))
yout =
Prbs Chap 19, 1/13/02 P19-33 © R. A. DeCarlo, P. M. Lin
3.1277e-04
»Zout = 1/yout
Zout =
3.1973e+03
»a = sqrt(8/Zout)
a =
5.0021e-02
(c) To compute the gain we first need Zin . From equation 19.49, since h12 = 0, Zin = h11.
» Zin = h(1,1)
Zin =
2.8728e+03
From equation 19.51,
»Gv2 = -h(2,1)/(Zin*(h(2,2)+h(2,2)))
Gv2 =
-5.7094e+01
From equation 19.52,
»Gv1 = Zin/(Zin + 50)
Gv1 =
9.8289e-01
To compute VL/Vs we use:
»Gv = Gv1*Gv2*a
Gv =
-2.8071e+00
To compute VL/V1 we use:
»Gvv = Gv2*a
Gvv =
Prbs Chap 19, 1/13/02 P19-34 © R. A. DeCarlo, P. M. Lin
-2.8559e+00
(d) For these calculations, we assume Vs is normalized to 1 V. Since we are computing gains,
we may do this without loss of generality.
»Vs = 1;
»Is = 1/(50 + Zin)
Is =
3.4214e-04
Now we compute the normalized power delivered by the voltage source.
»Psnorm = Vs*Is
Psnorm =
3.4214e-04
Next we compute the normalized power absorbed by the load.
»VL = Gv*1
VL =
-2.8071e+00
»PLnorm = VL^2/8
PLnorm =
9.8496e-01
Next, the power gain from source to load is:
»GpLs = PLnorm/Psnorm
GpLs =
2.8788e+03
Further, we compute the power gain from input to the two port to the load as follows:
»V1 = Vs*Zin/(Zin + 50)
V1 =
Prbs Chap 19, 1/13/02 P19-35 © R. A. DeCarlo, P. M. Lin
9.8289e-01
»P1 = V1*Is
P1 =
3.3629e-04
»GpL1 = PLnorm/P1
GpL1 =
2.9289e+03
(e) SPICE Simulation Because the frequency response is flat for freqency above 800 Hz, we
only plotted up to 1.6k Hz. The circuit diagram reflects the load back to the primary of the
ideal transformer. In general, this is not possible. Hence for a SPICE simulation, it is
necessary to use one of the models given in Figure 18.15 consisting of two controlled
sources. For this example, this is not necessary. Note however that the actual output
voltage is 0.05 times the values on the graph given below. This simulation assumes a 1 V
source voltage and the parameter of GVCCS is 0.035714. Notice that in this problem
Prbs Chap 19, 1/13/02 P19-36 © R. A. DeCarlo, P. M. Lin
MAG(V(IVM2))
Frequency (Hz)Prb 19.72-Small Signal AC-3
+0.000e+000
+10.000
+20.000
+30.000
+40.000
+50.000
+60.000
+200.000 +400.000 +600.000 +800.000 +1.000k +1.200k +1.400k +1.600k
(f) For this part, we change 100 µF to 10 µF. The resulting plot shows degradation of the low
end frequency response.
MAG(V(IVM2))
Frequency (Hz)Prb 19.72-Small Signal AC-4
+0.000e+000
+10.000
+20.000
+30.000
+40.000
+50.000
+60.000
+200.000 +400.000 +600.000 +800.000 +1.000k +1.200k +1.400k +1.600k
SOLUTIONS PROBLEMS CHAPTER 20
USEFUL MATLAB M-FILES FOR USE IN THE SOLUTION TO PROBLEMS IN THIS CHAPTER.
Program 1: converts y-parameters to t-parameters% convert y parameters to t parametersfunction [t,t11,t12,t21,t22] = ytot(y)y11=y(1,1);y12=y(1,2);y21=y(2,1);y22=y(2,2);deltay= y11*y22-y12*y21;t11=-y22/y21;t12 = -1/y21;t21= -deltay/y21;t22= -y11/y21;t= [ t11 t12; t21 t22];
Program 2: Computes Zin, Zout, and gains using two port h-parameters.»% two-port analysis in terms of h-parameters»function [zin, zout] =twoport(h, zL, zs)»['twoport analysis using h-parameters']»h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2);»zin = h11 - h12*h21/(h22+ 1/zL)»yout= h22 - h12*h21/(h11+zs);»zout= 1/yout»v1tovs= zin/(zin+zs)»v2tov1= -h21/(zin*(h22+1/zL))»v2tovs= v1tovs*v2tov1
Program 3: Computes Zin, Zout, and gains using two port y-parameters.% two-port analysis in terms of y-parametersfunction [zin, zout] =twoporty(y, zL, zs)['twoport analysis using y-parameters']y11= y(1,1); y12=y(1,2); y21=y(2,1); y22=y(2,2);yin = y11 - y12*y21/(y22+ 1/zL)zin= 1/yinyout= y22 - y12*y21/(y11+1/zs)zout= 1/youtv1tovs= zin/(zin+zs)v2tov1= -y21/(y22+1/zL)v2tovs= v1tovs*v2tov1
Program 4: Computes Zin, Zout, and gains using two port t-parameters.% two-port analysis in terms of t-parametersfunction [zin, zout] =twoportt(t, zL, zs)
['analysis of terminated twoport using t-parameters']t11= t(1,1); t12=t(1,2); t21=t(2,1); t22=t(2,2);zin= (t11*zL + t12)/(t21*zL + t22)zout= (t22*zs + t12)/(t21*zs + t11)v2tov1= zL/(t11*zL + t12)v1tovs= zin/(zin+zs)v2tovs= v2tov1*v1tovs
Program 5: converts z-parameters to t-parameters%converting z to t paramters (same sormulas as%converting t to z parameters)
function [t,t11,t12,t21,t22] = ztot(z)z11=z(1,1); z12=z(1,2); z21=z(2,1); z22=z(2,2);deltaz= z11*z22 - z12*z21;t11= z11/z21;t12= deltaz/z21;t21= 1/z21;t22= z22/z21;t= [ t11 t12; t21 t22];
SOLUTION 20.1.
For network a, the z-parameters are by inspection
Za =R1 + R3 R3
R3 R2 + R3
Similarly, for network b, the z-parameters are the same:
Zb =R1 + R3 R3
R3 R2 + R3
The interconnection of networks a and b conforms to figure 20.2b, which is a seriesinterconnection. Hence, the new overall z-parameters are
Znew = Za + Zb = 2R1 + R3 R3
R3 R2 + R3
SOLUTION 20.2. For networks a and b, the y-parameters are by inspection
Ya = Yb =3 −2
−2 4
S
Hence, their z-parameters are the inverse of the y-parameter matrix:
Za = Zb =18
4 2
2 3
Ω
The interconnection of networks a and b conforms to figure 20.2b, which is a seriesinterconnection. Hence, the new overall z-parameters are
Znew = Za + Zb =14
4 2
2 3
Ω
SOLUTION 20.3. For network a consisting of the single inductor,
Za =s s
s s
Ω
For network b, we have
Zb = Yb[ ]−1 =12
1 1
2 4
=
0.5 0.5
1 2
Ω
The interconnection of networks a and b conforms to figure 20.2b, which is a seriesinterconnection. Hence, the new overall z-parameters are
Znew = Za + Zb =s + 0.5 s + 0.5
s +1 s + 2
Ω
SOLUTION 20.4. For network a consisting of the single capacitor,
Za =1s
1 1
1 1
Ω
The interconnection of networks a and b conforms to figure 20.2b, which is a seriesinterconnection. Hence, the new overall z-parameters are
Znew = Za + Zb =1s
1 + 0.5s 1+ 0.5s
1+ s 1 + 2s
Ω
SOLUTION 20.5. For network a, the y-parameters are by inspection
Ya =3 −2
−2 3
S
Hence, their z-parameters are the inverse of the y-parameter matrix:
Za =15
3 2
2 3
=
0.6 0.4
0.4 0.6
Ω
The interconnection of networks a and b conforms to figure 20.2b, which is a seriesinterconnection. Hence, the new overall z-parameters are
Znew = Za + Zb =1.6 0.9
0.4 −0.4
Ω
SOLUTION 20.6. For networks Na and Nb, the y-parameters are:
YNa = YNb =0.7 −0.2
−0.2 0.7
S
Hence, their z-parameters are the inverse of the y-parameter matrix:
ZNa = ZNb =209
0.7 0.2
0.2 0.7
=
19
14 4
4 14
Ω
The network Na* has the same z-parameters as Na and continues to act as a two when seriesinterconnected to another 2-port because of the transformer. Hence, the interconnection ofnetworks Na* and Nb forms a valid series interconnection in which cas
Znew = ZNa * + ZNb =29
14 4
4 14
Ω
SOLUTION 20.7. For network Nb, the y-parameters are the same as in problem 20.6, i.e.,
YNb =0.7 −0.2
−0.2 0.7
S and ZNb =
19
14 4
4 14
Ω
For network Na, consider the purely resistive part without the transformer. The y-parameters ofthis part are half the y-parameters of Nb, i.e.,
YR =0.35 −0.1
−0.1 0.35
S
In MATLAB we use the m-file which converts y-parameters to t-parameters:
»y = [0.35 -0.1;-0.1 0.35];»t = ytot(y)t = 3.5000e+00 1.0000e+01 1.1250e+00 3.5000e+00
From figure 19.34, the transformer has to parameters with a = 2 given by
ttrans =1 a 0
0 a
=
0.5 0
0 2
»ttrans = [0.5 0;0 2];»tNa = ttrans*ttNa = 1.7500e+00 5.0000e+00 2.2500e+00 7.0000e+00»»zNa = ttoz(tNa)zNa = 7.7778e-01 4.4444e-01 4.4444e-01 3.1111e+00»»zNb = [14 4;4 14]/9;»znew = zNa + zNbznew = 2.3333e+00 8.8889e-01 8.8889e-01 4.6667e+00»
SOLUTION 20.8. This problem is identical numerically to problem 20.6. Here however theisolation transformer is on the right hand side which makes no difference to the interconnection.Therefore,
Znew = ZNa * + ZNb =29
14 4
4 14
Ω
SOLUTION 20.9. Figure 20.4 is
Using the values in figure 20.3, we obtain the following mesh equation matrix
V1
V2
0
=4 + R1 2 −R1
2 4 + R2 R2
−R1 R2 R1 + R2
I1I2
I3
For I3 to be zero, the third equation implies that R1I1 = R2I2. Therefore,
V1 = 4 + R1( )I1 + 2R1R2
I1 = 4 + R1 +2R1R2
I1
Similarly
V2 = 2I1 + 4 + R2( ) R1R2
I1+ = 2 +(4 + R2)R1
R2
I1
Therefore
V1V2
=4 + R1 +
2R1
R2
2 +( 4 + R2)R1
R2
=4R2 + R1R2 + 2R1( )2R2 + R1R2 + 4R1( ) =
4248
=78
as was to be shown.
Now suppose, V1V2
=78
or equivalently V1 =78
V2. With specific values
V1
V2
0
= V2
7 8
1
0
=10 2 −6
2 7 3
−6 3 9
I1
I2
I3
In MATLAB»Z = [10 2 -6;2 7 3;-6 3 9];»b = [7/8 1 0]';»I = inv(Z)*bI = 6.2500e-02 1.2500e-01
2.7756e-17Thus
I1
I2
I3
= V2
0.0625
0.124
0
SOLUTION 20.10(a) Write two loop equations assuming I1 and I2 are the usual port currents. Here
V1 = Z1I1 + V13 + Z3 I1 + I2( )and
V2 = Z2I2 +V23 + Z3 I1 + I2( )which in matrix form are
V1
V2
=
Z1 + Z3 Z3
Z3 Z2 + Z3
I1I2
+
V13
V23
However,V13
V23
= ZA
I1I2
ThereforeV1
V2
=
Z1 + Z3 Z3
Z3 Z2 + Z3
I1I2
+ ZA
I1I2
=
Z1 + Z3 Z3
Z3 Z2 + Z3
+ ZA
I1I2
(b) The procedure of part (a) is repeated to produce the same result.
REMARK: this problem states that the two networks are equivalent two ports. Thus theconfigurations can be used interchangeably.
SOLUTION 20.11. Because of the isolation transformers, the overall Z-parameters are the sum of
the three component Z-parameters. For Na,
Za =2 1
1 2
Ω
For Nb,
Zb =8 1
1 5
Ω
For Nc,
Zc = Za =2 1
1 2
Ω
Therefore
Z = Za + Zb + Zc =12 3
3 9
Ω
SOLUTION 20.12. (a) Given the Z-parameters of N, then
YN = ZN−1 =
7 2
10 3
−1
=3 −2
−10 7
S
The 1-F capacitor considered as a two port has y-parameters
YC =s − s
− s s
S
Therefore, the parallel connection of the two ports has y-parameters
Y = YN +YC =s −s
−s s
+
3 −2
−10 7
=
s + 3 −(s + 2)
−(s +10) s + 7
S
(b) For this part, the same reasoning applies with s replaced by 1/s.
SOLUTION 20.13. The isolation transformer allows for the valid parallel connection of Na* and
Nb in the sense that the overall y-parameters are the sum of the individual y-parameters. Further,
because the ideal transformer is 1:1 with the standard dot locations, the y-parameters of Na* are
those of Na. Further, the y-parameters of Nb are the same as those of Na as the circuits aresimple vertical flips of each other. Therefore
Ya* = Ya =
0.7 −0.2
−0.2 0.7
= Yb
Hence, the overall y-parameters are:
Y = 2 ×0.7 −0.2
−0.2 0.7
=
1.4 −0.4
−0.4 1.4
S
True-False: FALSE because the connection does not conform to figure 20.2a.
SOLUTION 20.14. From the previous example, the y-parameters of Nb are
Yb =0.7 −0.2
−0.2 0.7
S
The resistance values of the resistive part of Na are twice those of Nb. Hence, the y-parameters
of the resistive part are half those of Nb, i.e.,
I1'
I2
= Ya ,RV1
'
V2
=0.35 −0.1
−0.1 0.35
V1'
V2
S
We obtain the y-parameters of Na by considering the effect of the transformer on these y-parameters. Observe that
I1 = 2I1' and V1 = 0.5V1
'
HenceI1
I2
=
0.7 −0.2
−0.1 0.35
V1'
V2
=0.7 −0.2
−0.1 0.35
2V1
V2
=
1.4 −0.2
−0.2 0.35
V1
V2
Thus
Ya =1.4 −0.2
−0.2 0.35
S
It follows that
Y = Ya + Yb =2.1 −0.4
−0.4 1.05
S
SOLUTION 20.15. (a)
MAG(V(IVM))
Frequency (Hz)Prob 20.15-Small Signal AC-10
+0.000e+000
+200.000m
+400.000m
+600.000m
+800.000m
+1.000
+100.000m +200.000m +300.000m +400.000m +500.000m
(b)
C31.8n
R1k
R01k
C015.9n
C115.9n
R1500
IVm
V0
MAG(V(IVM))
Frequency (Hz)Prob 20.15-Small Signal AC-11
+0.000e+000
+200.000m
+400.000m
+600.000m
+800.000m
+1.000
+10.000k +20.000k +30.000k
SOLUTION 20.16. (a) For network Na, the y-parameters by inspection are:
Ya =G1 + j C1 0
gm G0
=
2 + j10.21 0
95 0.07143
mS
For network Nb, the y-parameters by inspection are:
Yb = G f + j C2( ) 1 −1
−1 1
= 0.8333 + j1.021( ) 1 −1
−1 1
mS
Therefore
Y = Ya + Yb =2.8333 + j11.23 −0.8333− j1.021
94.167 − j1.021 0.9048 + j1.021
mS
(b), (c), and (d). Here we use the MATLAB m-file for two port analysis in terms of y-parameters:
»zs = 50; zL = 50;»twoporty(Y, zL, zs)ans =twoport analysis using y-parametersyin = 6.8501e-03 + 1.5594e-02izin = 2.3614e+01 - 5.3756e+01iyout = 5.3617e-03 + 3.0023e-03izout = 1.4199e+02 - 7.9506e+01iv1tovs = 5.5701e-01 - 3.2349e-01iv2tov1 = -4.4915e+00 + 2.6821e-01iv2tovs = -2.4150e+00 + 1.6023e+00i»
SOLUTION 20.17. The t-parameters of the LR circuit follow from problem 19.53 with Z1 = Ls =
s Ω and Z2 = 0.5 Ω:
TLR =1+ 2s s
2 1
Therefore
Tcascade = T * TLR =5 + 2s 2 + s
5 + 2s 2 + s
SOLUTION 20.18. This problem can be solved in many ways. Here we emphasize the cascadenature of the two ports.»% The y-parameters of Nb are:»Yb = [8 2;20 6];»% The z-parameters of Na are:»Za = [0.75 -0.25;-2.5 1];»% The t-parameters of Na are:»Ta = ztot(Za)Ta = -3.0000e-01 -5.0000e-02 -4.0000e-01 -4.0000e-01»% The t-parameters of Nb are:»Tb = ytot(Yb)Tb =
-3.0000e-01 -5.0000e-02 -4.0000e-01 -4.0000e-01»% The t-parameters of the cascaded two port are:»»Tab = Ta*TbTab = 1.1000e-01 3.5000e-02 2.8000e-01 1.8000e-01»% Doing a t-parameter analysis we obtain:»twoportt(Tab,0.25,0.5)ans =analysis of terminated twoport using t-parameterszin = 2.5000e-01zout = 5.0000e-01v2tov1 = 4v1tovs = 3.3333e-01v2tovs = 1.3333e+00ans = 2.5000e-01»
SOLUTION 20.19. Use the h-parameter analysis m-file.
»H1 = [1e3 0.001;50 6e-5]; H2 = [1e3 0.99;-5 8e-4];»zL = 100; zs = 2e3; zm = 10e3;»% For part (a), the common collector stage:»twoporth(H2,zL,0)ans =twoport analysis using h-parameterszin = 1.4583e+03zout = 1.7391e+02v1tovs = 1v2tov1 = 3.1746e-01v2tovs = 3.1746e-01
Remark: in the above, v2 is Vout and v1 is vm; zout has no signifigance.
»% We now compute the load on the first stage.»zin2 = 1.4583e+03;»zLm = zm*zin2/(zin2 + zm)zLm = 1.2727e+03»% zLm is the load impedance on stage 1.
»% For part (a), the common emitter stage:»twoporth(H1,zLm,zs)ans =twoport analysis using h-parameterszin = 9.4088e+02zout = 2.3077e+04v1tovs = 3.1993e-01v2tov1 = -6.2835e+01v2tovs = -2.0103e+01
Remark: in the above, v2 is vm and v1 is Vin; zout is the output impedance of stage 1.
Conclusion: the input impedance to stage 1 is 940.88 Ω and the input impedance to stage 2 is1.4583 kΩ.
(b) From the above output and remark, Vm/Vin = -62.835. Further Vout/Vm = 0.31746.
(c) From above Vm/Vs = -20.103. Therefore Vout/Vs = Vout/Vm * Vm/Vs = -6.38.
(d)»zout1 = 2.3077e+04;»zs2 = zm*zout1/(zm + zout1)zs2 = 6.9768e+03»twoporth(H2,zL,zs2)ans =twoport analysis using h-parameterszin = 1.4583e+03zout = 7.0395e+02v1tovs =
1.7289e-01v2tov1 = 3.1746e-01v2tovs = 5.4885e-02
Conclusion: Zout of amplifier is 704 Ω.
SOLUTION 20.20(a) Using MATLAB»na = 1.1514;»nb = 3.4012;»Zlprime = nb^2*75 + j*1042.94ZLprime = 8.6761e+02 + 1.0429e+03i
»Zsprime = 75/na^2 + j*30Zsprime = 5.6573e+01 + 3.0000e+01i
(b) Since the h-parameters of the transistors are given, we can again use MATLAB and the m-file twoporth defined earlier. Hence:
»h = [ 60-j*50 0.01; -j*2 0.0005+j*0.0004];»[Zin, Zout] = twoporth(h,ZLprime, Zsprime)Zin = 5.6569e+01 - 3.0000e+01iZout = 8.6763e+02 - 1.0429e+03i
(c) Observe that Zin = 5.6569e+01 - 3.0000e+01i and Zsprime = 5.6573e+01 + 3.0000e+01i,which are clearly conjugates of each other. Further, Zout = 8.6763e+02 - 1.0429e+03iand ZLprime = 8.6761e+02 + 1.0429e+03i, which are also conjugates of each other.Hence maximum power is transferred into and out of the transistor.
(d) For this part, we change all cascaded two ports to t-parameters. Specifically,
t0 = [1 75;0 1];t1 = [na 0; 0 1/na];t2 = [ 1 j*30; 0 1];t3 = htot(h);t4 = [ 1 j*1042.9; 0 1];t5 = [nb 0; 0 1/nb];t6 = [1 0;1/75 1];t = t0*t1*t2*t3*t4*t5*t6
t = 2.2176e-01 - 2.5959e-01i 8.3160e+00 - 9.7351e+00i 1.4785e-03 - 1.7307e-03i 6.6577e-02 - 7.4415e-02i»gain = 1/t(1,1)gain = 1.9024e+00 + 2.2270e+00i
»gainmag = abs(gain)gainmag = 2.9289e+00
»gainangle = angle(gain)*180/pigainangle = 4.9494e+01
In this case, Vout/Vs = gain = 1.9024e+00 + 2.2270e+00i = 2.9289∠49.494o
SOLUTION 20.21.(a)»Y2N = [1 0;20.1 0]*1e-3;»Y10k = [1 -1;-1 1]*1e-4;»Yshade = Y2N + Y10kYshade = 1.1000e-03 -1.0000e-04 2.0000e-02 1.0000e-04»(b) This is a series connection of two 2-ports. Hence we first convert the answer of part (a) to z-parameters.
»Y2N = [1 0;20.1 0]*1e-3;»Y10k = [1 -1;-1 1]*1e-4;»Yshade = Y2N + Y10kYshade = 1.1000e-03 -1.0000e-04 2.0000e-02 1.0000e-04»Zshade = inv(Yshade)Zshade = 4.7393e+01 4.7393e+01 -9.4787e+03 5.2133e+02»Z1k = [1 1;1 1]*1e3;»Zdashed = Zshade + Z1kZdashed = 1.0474e+03 1.0474e+03 -8.4787e+03 1.5213e+03»
(c)»twoportz(Zdashed,1e12,1e3)
ans =twoport analysis using z-parameterszin = 1.0474e+03zout = 5.8588e+03v1tovs = 5.1157e-01v2tov1 = -8.0950e+00v2tovs = -4.1412e+00
Conclusion: Vout/Vs = -4.1412.
SOLUTION 20.22.»Z = [3 1;5 2]*1e3;»Y = inv(Z)Y = 2.0000e-03 -1.0000e-03 -5.0000e-03 3.0000e-03»% Consider the parallel connection of Y with the 1 kΩ resistor»Y1 = Y + [1 -1;-1 1]*1e-3Y1 = 3.0000e-03 -2.0000e-03 -6.0000e-03 4.0000e-03»»% Now consider Y/Z in parallel with 1 kΩ connected between B and C»Y2 = Y + [0 0;0 1]*1e-3Y2 = 2.0000e-03 -1.0000e-03 -5.0000e-03 4.0000e-03»»% This combination is in series with another 1 kΩ resistor»% Hence we need to compute Z2 first»Z2 = inv(Y2)Z2 = 1.3333e+03 3.3333e+02 1.6667e+03 6.6667e+02»% Now we compute the series combo of Z2 and the 1 kΩ resistor»Z3 = Z2 + [1 1;1 1]*1e3Z3 = 2.3333e+03 1.3333e+03 2.6667e+03 1.6667e+03»
»»% We now convert Y1 and Z3 to t-parameters and then multiply»% together to obtain the overall t-parameters»»T1 = ytot(Y1)T1 = 6.6667e-01 1.6667e+02 0 5.0000e-01»T3 = ztot(Z3)T3 = 8.7500e-01 1.2500e+02 3.7500e-04 6.2500e-01»T = T1*T3T = 6.4583e-01 1.8750e+02 1.8750e-04 3.1250e-01»Y = ttoy(T)Y = 1.6667e-03 -8.8889e-04 -5.3333e-03 3.4444e-03»
SOLUTION 20.23. Refer to figure 20.13 for all problems.(a) By inspection,
I1
I2
I3
=6 −4 −2
−4 4 + 5s −5s
−2 −5s 2 + 5s
V1
V2
V3
The 3x3 coefficient matrix is the desired Yind(s).(b) Writing the nodal equation by inspection yields
I1
I2
I3
0
=
0.5 0 −0.25 −0.25
0 1 −0.5 −0.5
−0.25 −0.5 1.25 −0.5
−0.25 −0.5 −0.5 1.25
V1
V2
V3
V4
Using MATLAB,
»W11=[0.5, 0 -0.25;0 1 -0.5;-0.25 -0.5 1.25]W11 = 5.0000e-01 0 -2.5000e-01 0 1.0000e+00 -5.0000e-01 -2.5000e-01 -5.0000e-01 1.2500e+00»W12 = [-0.25 -0.5 -0.5]'
W12 = -2.5000e-01 -5.0000e-01 -5.0000e-01»W21 = [-0.25 -0.5 -0.5]W21 = -2.5000e-01 -5.0000e-01 -5.0000e-01»W22 = 1.25W22 = 1.2500e+00»Yind = W11 - W12*inv(W22)*W21Yind = 4.5000e-01 -1.0000e-01 -3.5000e-01 -1.0000e-01 8.0000e-01 -7.0000e-01 -3.5000e-01 -7.0000e-01 1.0500e+00»
(c) Again, writing the nodal equation by inspection yields
I1
I2
I3
0
=
5 0 0 −5
0 6 −7 1
0 −4 8 −4
−5 −2 −1 8
V1
V2
V3
V4
Again, using MATLAB,»W11 = [5 0 0;0 6 -7;0 -4 8]W11 = 5 0 0 0 6 -7 0 -4 8»W12 = [-5 1 -4]'W12 = -5 1 -4»W21 = [-5 -2 -1]W21 = -5 -2 -1»W22 = 8;»Yind = W11 - W12*inv(W22)*W21Yind = 1.8750e+00 -1.2500e+00 -6.2500e-01 6.2500e-01 6.2500e+00 -6.8750e+00 -2.5000e+00 -5.0000e+00 7.5000e+00»
(d) This part is similar to part (a) as it does not require the method of matrix partitioning. Byinspection,
IG
ID
IS
=(CGD + CGS )s −CGDs −CGS s
−CGDs + gm CGDs −gm
−CGSs − gm 0 CGSs + gm
VG
VD
VS
The 3x3 coefficient matrix is the desired Yind(s).
SOLUTION 20.24. (a) With regard to the given information, the associated indefinite admittancematrix is the coefficient matrix in the following nodal equation given reference to figure 20.13:
IG
ID
IS
=0 0 0
gm 0 −gm
−gm 0 gm
VG
VD
VS
We use property 5 to compute the remaining answers:
(a)-(a): YGD is as given.
(a)-(b): YSD =gm 0
−gm 0
S
(a)-(c): YGS =0 0
−gm gm
S
(a)-(d): YDG =0 gm
0 0
S
(a)-(e): YDS =0 −gm
0 gm
S
(a)-(e): YSG =gm −gm
0 0
S
(b) Transmission from port-1 to port-2 occurs when the 2-1 entry of the 2-port y-matrix isnonzero. Hence, the following all have the desired transmission: YGD , YSD, YGS .
SOLUTION 20.25. Using the zero-sum properties of the rows and columns, we have byinspection:
Yind =40 2 −42
? ? −50
? −22 92
=40 2 −42
30 20 −50
−70 −22 92
S
SOLUTION 20.26. (a) Yind =8 ? ?
−2 9 −7
−6 ? ?
=8 −2 −6
−2 9 −7
−6 −7 13
S, where the second equality arises
because a purely
resistive network has a symmetric indefinite admittance matrix.(b) Construct a common ground 2-port with input terminal C, common terminal A, and outputterminal B:
YCB =13 −7
−7 9
S
Therefore, since the output terminal B is shorted, YL = ∞, i.e.,
Yin = y11 −y12y21
y22 + YL= 13 S
Hence
Zin =1
13 Ω
(c) Construct a common ground 2-port with input terminal A, common terminal B, and outputterminal C:
YAC =8 −6
−6 13
S
Hence, the required voltage gain is
GV =V2V1
=−y21
y22 + YL=
−y21y22
=613
Therefore, V2 =613
V.
SOLUTION 20.27.(a) Using the zero-sum property of the rows and columns of an indefinite admittance matrix wecan write down by inspection (in mS)
IB
IE
IC
=1 −1 0
−100 100.1 −0.1
99 99.1 0.1
VB
VE
VC
(b) The y-parameters (also in mS) of the common emitter configuration are easily computed as
IB
IC
=
1 0
99 0.1
VBE
VCE
Hence in MATLAB»Y = [1 0;99 0.1];»Z = inv(Y)Z = 1.0000e+00 0 -9.9000e+02 1.0000e+01
where Z is in kΩ. It follows that VCE = −990IB + 10IC where IB and IC are in mA and VCE isin volts.
SOLUTION 20.28. Expanding the given y-parameter matrix into an indefinite admittance matrixyields
IG
ID
IS
= Yind
VG
VD
VS
=0.2 + j2.5 −0.01− j0.65 −0.19 − j1.85
3.1− j0.65 0.05 + j0.8 −3.15 − j0.15
−3.3 − j1.85 −0.04 − j0.15 3.34 + j2
VG
VD
VS
By inspection, with G as the common terminal, S as the input terminal, and D as the outputterminal, we obtain,
Ynew =3.34 + j2 −0.04 − j0.15
−3.15 − j0.15 0.05 + j0.8
mS
SOLUTION 20.29. Here we use the zero-sum properties of the columns and rows to complete theindefinite admittance matrix:
Yind =−s ? ?
s −1 2 ?
1 2s −1 −2s
=−s −2s −1 3s +1
s−1 2 −s−1
1 2s−1 −2s
S
In figure (b), the top 2-port has ytop =−s −2s−1
s −1 2
S and the bottom 2-port has y-parameters
ybot =−2s 2s −1
−s −1 2
S. Since these 2-ports are connected in parallel, the overall 2-port y-
parameters are
y = ytop + ybot =−s −2s−1
s −1 2
+
−2s 2s −1
−s −1 2
=
−3s −2
−2 4
S
SOLUTION 20.30. From problem 28,
IG
ID
IS
= Yind
VG
VD
VS
=0.2 + j2.5 −0.01− j0.65 −0.19 − j1.85
3.1− j0.65 0.05 + j0.8 −3.15 − j0.15
−3.3 − j1.85 −0.04 − j0.15 3.34 + j2
VG
VD
VS
Hence
yGS =0.2 + j2.5 −0.19 − j1.85
−3.3− j1.85 3.34 + j2
S
SOLUTION 20.31. (a) Here, by inspection we can compute the indefinite admittance matrix asthe coefficient matrix of the following nodal equations:
IA
IB
IC
= Yind
VA
VB
VC
=Y1 + Y2 + 2 −Y2 −Y1 − 2
Y2 − 2 −Y2 2
−Y1 − 2Y2 2Y2 Y1
VA
VB
VC
(b)
yAB =Y1 + Y2 + 2 −Y2
Y2 − 2 −Y2
(c)
yAC =Y1 + Y2 + 2 −Y1 − 2
−Y1 − 2Y2 Y1
SOLUTION 20.32. (a) Here, by inspection we can compute the indefinite admittance matrix asthe coefficient matrix of the following nodal equations:
IA
IB
IC
= Yind
VA
VB
VC
=Y1 + Y2 −Y2 −Y1
−Y2 + gm Y2 +Y3 −Y3 − gm
−Y1 − gm −Y3 Y1 + Y3 + gm
VA
VB
VC
(b)
yAB =Y1 + Y2 −Y2
−Y2 + gm Y2 +Y3
(c)
yAC =Y1 + Y2 −Y1
−Y1 − gm Y1 + Y3 + gm
SOLUTION 20.33. (a) Here the nodal equation matrix is:
IA
IB
IC
0
=W11 W12
W21 W22
VA
VB
VC
VD
=
8 0 0 −8
0 2 + 2s −2s −10 8
0 −2s 16 + 2s −16
−8 −2 −6 16
VA
VB
VC
VD
where VD is the internal node voltage. Using the method of matrix partitioning,
Yind = W11 − W12W22−1W21 =
8 0 0
0 2 + 2s −2s−10
0 −2s 16 + 2s
−
116
−8
8
−16
−8 −2 −6[ ]
=8 0 0
0 2 + 2s −2s−10
0 −2s 16 + 2s
−
−0.5
0.5
−1
−8 −2 −6[ ] =4 −1 −3
4 3 + 2s −2s − 7
−8 −2s− 2 10 + 2s
S
(b) When C is grounded,
yAB =4 −1
4 3 + 2s
S and zAB =
18s +16
3 + 2s 1
−4 4
Ω
(c) When B is grounded
yAC =4 −3
−8 10 + 2s
S
SOLUTION 20.34. (a) Writing the usual node equations we have,
IA
IB
IC
0
=W11 W12
W21 W22
VA
VB
VC
VD
=
s + 0.5 − s 0 −0.5
−s s + 0.5 0 −0.5
0 0 0.5s −0.5s
−0.5 −0.5 −0.5s 0.5s +1
VA
VB
VC
VD
Using the method of matrix partitioning,
Yind = W11 − W12W22−1W21 =
s + 0.5 − s 0
−s s + 0.5 0
0 0 0.5s
−
0.5s + 2
1
1
s
1 1 s[ ]
=s + 0.5 −s 0
− s s + 0.5 0
0 0 0.5s
−
0.5s + 2
1 1 s
1 1 s
s s s2
=1
s + 2
s2 + 2.5s + 0.5 −s2 − 2s− 0.5 −0.5s
− s2 − 2s − 0.5 s2 + 2.5s + 0.5 −0.5s
−0.5s −0.5s s
S
(b) Here
yAB =1
s + 2s2 + 2.5s + 0.5 − s2 − 2s − 0.5
−s2 − 2s− 0.5 s2 + 2.5s + 0.5
S
SOLUTION 20.35. (a) Rule 1: Consider the two networks NA (3 external nodes) and NB (4external nodes) given below:
The two indefinite admittance matrices are
YindN A=
G1 + G2 −G2 −G1
−G2 G2 + G3 −G3
−G1 −G3 G1 + G3
and YindN B
=
G1 + G2 −G2 −G1 0
−G2 G2 + G3 −G3 0
−G1 −G3 G1 + G3 0
0 0 0 0
Observe that YindN B can be obtained from YindN A
by adding a column of zeros and a row of
zeros to form a 4x4 matrix.
(b) Rule 2: Consider two networks NA and NB and a third network NC which combines
elements of NA and NB as given below:
The corresponding indefinite admittance matrices are:
YindN A=
6 −2 −4
−2 3 −1
−4 −1 5
, YindN B
=5 −3 −2
−3 5 −2
−2 −2 4
, and YindNC
=11 −5 −6
−5 8 −3
−6 −3 9
.
Clearly,YindNC
= YindN A+YindNB
(c) Rule 3: Consider the 3-terminal network
with indefinite admittance matrix
YindN A=
5 −3 −2
−3 5 −2
−2 −2 4
S
If we move node 3 inside to form a 2-terminal network and labeled as NB,
then from nodal analysis we have
I1
I2
0
=5 −3 −2
−3 5 −2
−2 −2 4
V1
V2
V3
Using the method of matrix partitioning,
YindN B=
5 −3
−3 5
−
14
−2
−2
−2 −2[ ] =
4 −4
−4 4
S
This computation is the one given by the formula in the problem. To see that this is correct, weobserve that the internal simplification of NB leads to the following:
SOLUTION 20.36.Part (a)»Yinda= [1/2 -1/4 0 -1/4 0; -1/4 1/2 0 0 -1/4; ...0 0 0 0 0; -1/4 0 0 1/2 -1/4; 0 -1/4 0 -1/4 1/2]
Yinda =
5.0000e-01 -2.5000e-01 0 -2.5000e-01 0 -2.5000e-01 5.0000e-01 0 0 -2.5000e-01 0 0 0 0 0 -2.5000e-01 0 0 5.0000e-01 -2.5000e-01 0 -2.5000e-01 0 -2.5000e-01 5.0000e-01
»Yindb= [0 0 0 0 0; 0 0 0 0 0; 0 0 3/4 -1/2 -1/4; ...0 0 -1/2 5/8 -1/8; 0 0 -1/4 -1/8 3/8]Yindb =
0 0 0 0 0 0 0 0 0 0 0 0 7.5000e-01 -5.0000e-01 -2.5000e-01 0 0 -5.0000e-01 6.2500e-01 -1.2500e-01 0 0 -2.5000e-01 -1.2500e-01 3.7500e-01
Part (b)»Yind = Yinda + YindbYind =
5.0000e-01 -2.5000e-01 0 -2.5000e-01 0 -2.5000e-01 5.0000e-01 0 0 -2.5000e-01 0 0 7.5000e-01 -5.0000e-01 -2.5000e-01 -2.5000e-01 0 -5.0000e-01 1.1250e+00 -3.7500e-01 0 -2.5000e-01 -2.5000e-01 -3.7500e-01 8.7500e-01
% To suppress nodes 4 and 5 we use the partitioned matrix formula as follows:
»W11= [Yind(1:3, 1:3)]W11 =
5.0000e-01 -2.5000e-01 0 -2.5000e-01 5.0000e-01 0 0 0 7.5000e-01
»W12=[Yind(1:3, 4:5)]W12 = -2.5000e-01 0 0 -2.5000e-01 -5.0000e-01 -2.5000e-01
»W21= [Yind(4:5, 1:3)]W21 = -2.5000e-01 0 -5.0000e-01 0 -2.5000e-01 -2.5000e-01
»W22= [Yind(4:5, 4:5)]W22 = 1.1250e+00 -3.7500e-01 -3.7500e-01 8.7500e-01
»Yind123 = W11 - W12*inv(W22)*W21Yind123 = 4.3519e-01 -2.7778e-01 -1.5741e-01 -2.7778e-01 4.1667e-01 -1.3889e-01 -1.5741e-01 -1.3889e-01 2.9630e-01
(d) For the required Y-matrix we delete row and column 3 to obtain»Ysc = Yind123(1:2,1:2)Ysc = 4.3519e-01 -2.7778e-01 -2.7778e-01 4.1667e-01
»Zoc = inv(Ysc)Zoc = 4.0000e+00 2.6667e+00 2.6667e+00 4.1778e+00
SOLUTION 20.37. Since complex roots must occur in conjugate pairs, we will only check jω0.
0 = p( j 0) = − ja3 03 − a2 0
2 + ja1 0 + a0 = a0 − a2 02 + j a1 0 − a3 0
3( )Both real and imaginary parts must be zero, i.e.,
a0 − a2 02 = 0 and a1 0 − a3 0
3 = 0 a1 − a3 02( ) = 0
From the first equation, a2 02 = a0. The second equation above must be true for arbitrary ω0
which implies that a1 = a3 02. Equivalently a1a2 = a3a2 0
2 = a3a0. Conclusion: this conditionleads to imaginary complex roots.
(b) Given the above condition, what are the resulting imaginary roots of the polynomial? Sincethe polynomial is cubic, we can assume a3 ≠ 0. In this case,
0 = p(s) = s3 +a2a3
s2 +a1a3
s +a0a3
= s s2 +a1a3
+
a2a3
s2 +a1a2
a32 = s s2 +
a1a3
+
a2a3
s2 +a1a3
= s2 +a1a3
s +
a2a3
Therefore, the roots are: s = ± ja1a3
, −a2a3
.
SOLUTION 20.38.
(a) The four 2-port equations arising from the interconnection are:
V1 = V1a −V1b I1 = I1a = −I1b V2 = V2a = V2b I2 = I2a + I2b
ThusV1 = V1a −V1b = h11aI1a + h12aV2a( ) − h11bI1b + h12bV2b( ) = h11a + h11b( )I1 + h12a − h12b( )V2
andI2 = I2a + I2b = h21aI1a + h22aV2a( ) + h21bI1b + h22bV2b( ) = h21a − h21b( )I1 + h22a + h22b( )V2
This proves that the series-parallel connection has the required h-parameters.
(b) The four 2-port equations arising from the interconnection are:
V2 = V2a −V2b I2 = I2a = −I2b V1 = V1a = V1b I1 = I1a + I1b
Thus,I1 = I1a + I1b = g11aV1a + g12aI2a( ) + g11bV1b + g12bI2b( ) = g11a + g11b( )V1 + g12a − g12b( )I2
andV2 = V2a −V2b = g21aV1a + g22aI2a( ) − g21bV1b + g22bI2b( ) = g21a − g21b( )V1 + g22a + g22b( )I2
This proves that the parallel-series connection has the required g-parameters.
SOLUTION 20.39.
(a) Refer here to Na in figure P20.39b. With reference to figure 19.28b, h11 = 11 kΩ, h12 = 0,
h21 = 95.9, and h22 = 1/105 = 10
-5 S. Similarly, by inspection with reference to equation 19.33,
h11 = 90||10 = 9 kΩ, h12 = 0.1 (reverse voltage division), h21 = –0.1, and h22 = 10-5
S.
(b) By problem 20.38, part (a),
h11 = h11a + h11b = 20 kΩ h12 = h12a – h12b = –0.1
h21 = h21a – h21b = 96 h22 = h22a + h22b = 2×10-5 S
(c) Recall from chapter 19 that Zin =V1
I1= h11 −
h12h21
h22 + YL and Yout =
I2
V2= h22 −
h12h21
h11 + Zs in
which case Zout is the reciprocal. Using our MATLAB script, we have
h = [20e3 -0.1; 96 0.02e-3 ];zL= 1e8;zs= 5e3;»twoporth(h,zL,zs)ans =twoport analysis using h-parameterszin = 4.9976e+05zout = 2.4752e+03v1tovs = 9.9009e-01v2tov1 = -9.5998e+00v2tovs = -9.5047e+00
REMARK: We have used the following m-file code for "twoporth":
»% two-port analysis in terms of h-parameters»function [zin, zout] =twoport(h, zL, zs)»['twoport analysis using h-parameters']»h11= h(1,1); h12=h(1,2); h21=h(2,1); h22=h(2,2);»zin = h11 - h12*h21/(h22+ 1/zL)»yout= h22 - h12*h21/(h11+zs);
»zout= 1/yout»v1tovs= zin/(zin+zs)»v2tov1= -h21/(zin*(h22+1/zL))»v2tovs= v1tovs*v2tov1
SOLUTION 20.40.(a) The y-parameters for Na are:
yAB =7 4
4 7
−1
=133
7 −4
−4 7
S
Hence the indefinite admittance matrix for Na is:
YindN a=
133
7 −4 −3
−4 7 −3
−3 −3 6
S
Let us consider the associate 2-port with port A grounded and B as the new port 1 input. Theproblem is then solved by computing the input impedance with port 2 open circuited. Hence, thenew y-parameters are:
yBC =133
7 −3
−3 6
S
Thus Yin = y11 −y12y21
y22=
133
7 −96
=
5.533
=16
S. Hence, Zin = 6 Ω is the unique reading.
(b) The answer is not unique as demonstrated in part (c).
(c) The following two networks have the given Z-parameters, but the meter reading for N1 is 4
Ω but for N2 it is 2 Ω.
SOLUTIONS CHAPTER 21 PROBLEMS
SOLUTION TO 21.1. (a) Low pass(b) High pass
SOLUTION TO 21.2.
SOLUTION TO 21.3.(a)»n = 0.65378;»d = [1 0.80381643 0.82306043];»w = 0:0.005:2;»h = freqs(n,d,w);»plot(w, 20*log10(abs(h)))»grid»xlabel('Frequency rads/s')»ylabel('dB Gain')»
0 0.5 1 1.5 2-15
-10
-5
0
Frequency rads/s
dB G
ain
TextEnd
(b) »poles = roots(d)poles = -4.0191e-01 + 8.1335e-01i -4.0191e-01 - 8.1335e-01i
(c)»% Poles of new transfer function»wp = 2*pi*750'wp = 4.7124e+03»wp = 2*pi*750;»polesnew = poles*wppolesnew = -1.8939e+03 + 3.8328e+03i -1.8939e+03 - 3.8328e+03i»% All zeros remain at infinity.
Further
H(s) = HNLP (s p) =( p)2
s2 + 0.80381643 ps + 0.82306043( p)2 =2.2207 ×107
s2 + 3.7879 ×103s +1.8277 ×107
SOLUTION TO 21.4. (a) The 2nd order normalized LP transfer function is HNLP (s) =1
s2 + 2s +1. This
must be frequency scaled by Kf = 1000π. Hence,
H(s) = HNLP (s K f ) =(K f )2
s2 + K f 2s + (K f )2 =9.8696 ×106
s2 + 4.4429 ×103s + 9.8696 ×106
(b) Using MATLAB,»n = (1000*pi)^2;»d = [1 sqrt(2)*pi*1e3 (1000*pi)^2];»w = 0:1:2*pi*1500;»h = freqs(n,d,w);»plot(w/(2*pi),abs(h))»grid»xlabel('Frequency in Hz')»ylabel('Magnitude')»plot(w/(2*pi),20*log10(abs(h)))»grid»xlabel('Frequency in Hz')»ylabel('Magnitude in dB')
0 500 1000 15000.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency in Hz
Mag
nitu
de
TextEnd
0 500 1000 1500-20
-18
-16
-14
-12
-10
-8
-6
-4
-2
0
Frequency in Hz
Mag
nitu
de in
dB
TextEnd
(c)»nn = 9.8696e+06»dd = 1.0000e+00 4.4429e+03 9.8696e+06»w = j*2000*pi;»mag = abs(n/(w^2 + d(2)*w + d(3)))mag = 2.4254e-01
SOLUTION TO 21.5. (a) ∈max is that value of e that places the magnitude response curve through Amax
at ω = ωp. Therefore
Amax = 10log10 H ( j p)2
=10log10 1+ ∈max2 p
p
2n
= 10log10 1+ ∈max2( )
Therefore ∈max2 =100.1Amax −1 which upon a square root yields the final answer.
(b) Similarly, ∈min puts the magnitude response curve through the Amin spec. Hence
Amin = 10log10 H ( j s)2 =10log10 1+ ∈min
2 s
p
2n
Therefore
∈min2 =
100.1Amin −1
s
p
2n
which is equivalent to the required formula.
SOLUTION TO 21.6. The relationship of ε and ωc is given by the formula: c = p
(∈)1 n . Further, ∈max
in putting the magnitude response curve through the Amax spec produces c min , and ∈min in putting the
magnitude response curve through the Amin spec produces c max . Hence, from the solution to problem5,
c min = p
(∈max )1 n = p2n
100.1Amax −1 ≤ ω ≤ c max = p
(∈min )1 n = s2n
100.1Amin −1
SOLUTION TO 21.7. (a) From above material, the second order Butterworth NLP transfer function is
HNLP2(s) =1
s2 + 2s +1and from tables, the third order is
HNLP3(s) =1
s3 + 2s2 + 2s +1 (b)n1 = 1; d1 = [1 sqrt(2) 1];n2 = 1; d2 = [1 2 2 1];w = 0:.01:5;h1 = freqs(n1,d1,w);h2 = freqs(n2,d2,w);plot(w,abs(h1))gridxlabel('Normalized Frequency')ylabel('Magnitude')holdplot(w,abs(h2),'r')hold off
0 1 2 3 4 50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Normalized Frequency
Mag
nitu
de
TextEnd
Notice how the 3rd order filter has a sharper transition to zero.
(c) »% The simplest way to obtain the step response is as follows:»»syms s t»StepResp1 = ilaplace(1/(s^3 + sqrt(2)*s^2 + s))StepResp1 =1-exp(-1/2*2^(1/2)*t)*cos(1/2*2^(1/2)*t)-exp(-1/2*2^(1/2)*t)*sin(1/2*2^(1/2)*t)»StepResp2 = ilaplace(1/(s^4 + 2*s^3 + 2*s^2 + s))
StepResp2 =1-exp(-t)-2/3*exp(-1/2*t)*3^(1/2)*sin(1/2*3^(1/2)*t)»Thus the step response of the second order Butterworth normalized LP filter is:
v(t) = u( t) − e−0.70711t cos(0.70711t) − sin(0.70711t)[ ]u( t)
and that of the third order Butterworth normalized LP filter is:
v(t) = 1− e−t( )u(t) −1.1547e−0.5t sin(0.86603t)u( t)
SOLUTION TO 21.8.
fp = 100; fs = 1200;Amax = 0.3; Amin = 35;n = buttord(fp,fs,Amax,Amin,'s')emax = sqrt(10^(0.1*Amax) - 1)emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^nfcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n)))fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n)))wcmin = 2*pi*fcminwcmax = 2*pi*fcmaxwc = wcmin;fc = fcmin;[z,p,k] = buttap(n)% Numerators are each 1. Denominators are the polynomialsd1 = poly(p(1:2))d2 = poly(p(3))zplane(p)gridpauseznew = z*wcpnew = p*wcknew = k*wc^nf = 0:fc/50:1.2*fs;h = freqs(knew*poly(znew),poly(pnew),2*pi*f);plot(f,abs(h))gridxlabel('Frequency in Hz')ylabel('Gain magnitude')pauseplot(f,20*log10(abs(h)))xlabel('Frequency in Hz')ylabel('Gain in dB')
grid
n = 3emax = 2.6743e-01emin = 3.2538e-02fcmin = 1.5521e+02fcmax = 3.1324e+02wcmin = 9.7524e+02wcmax = 1.9681e+03z = []p = -5.0000e-01 + 8.6603e-01i -5.0000e-01 - 8.6603e-01i -1.0000e+00k = 1% Numerators are each 1. Denominators are the polynomialsd1 = 1.0000e+00 1.0000e+00 1.0000e+00d2 = 1 1
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real part
Imag
inar
y pa
rt
znew = []pnew = -4.8762e+02 + 8.4458e+02i -4.8762e+02 - 8.4458e+02i -9.7524e+02knew = 9.2753e+08
0 500 1000 15000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency in Hz
Gai
n m
agni
tude
0 500 1000 1500-60
-50
-40
-30
-20
-10
0
Frequency in Hz
Gai
n in
dB
SOLUTION TO 21.9.
fp = 100; fs = 1200;Amax = 0.3; Amin = 35;n = buttord(fp,fs,Amax,Amin,'s');emax = sqrt(10^(0.1*Amax) - 1);emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^n;fcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n)));fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n)));wcmin = 2*pi*fcmin;wcmax = 2*pi*fcmax;[z,p,k] = buttap(n);wc = wcmax;fc = fcmax;znew = z*wcpnew = p*wcknew = k*wc^nf = 0:fc/50:1.2*fs;h = freqs(knew*poly(znew),poly(pnew),2*pi*f);plot(f,abs(h))gridxlabel('Frequency in Hz')ylabel('Gain magnitude')pauseplot(f,20*log10(abs(h)))xlabel('Frequency in Hz')ylabel('Gain in dB')grid
znew = []pnew = -9.8406e+02 + 1.7044e+03i -9.8406e+02 - 1.7044e+03i -1.9681e+03knew = 7.6235e+09
0 500 1000 15000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency in Hz
Gai
n m
agni
tude
0 500 1000 1500-40
-35
-30
-25
-20
-15
-10
-5
0
5
Frequency in Hz
Gai
n in
dB
SOLUTION TO 21.10.
fp = 75; fs = 450;Amax = 1; Amin = 45;n = buttord(fp,fs,Amax,Amin,'s')emax = sqrt(10^(0.1*Amax) - 1)emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^nfcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n)))fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n)))wcmin = 2*pi*fcminwcmax = 2*pi*fcmax[z,p,k] = buttap(n)d1 = poly(p(1:2))d2 = poly(p(3:4))zplane(p)gridpausewc = wcmin;fc = fcmin;znew = z*wcpnew = p*wcknew = k*wc^nW = 0:0.01:fs/fp;h = freqs(k*poly(z),poly(p),W);plot(W*wc/(2*pi),abs(h))gridxlabel('Frequency in Hz')ylabel('Gain magnitude')pauseplot(W*wc/(2*pi),20*log10(abs(h)))xlabel('Frequency in Hz')ylabel('Gain in dB')grid
n = 4emax = 5.0885e-01emin = 1.3721e-01fcmin = 8.8800e+01fcmax = 1.2323e+02wcmin = 5.5795e+02wcmax = 7.7427e+02z = []p = -3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i -9.2388e-01 - 3.8268e-01ik = 1
d1 = 1.0000e+00 7.6537e-01 1.0000e+00d2 = 1.0000e+00 1.8478e+00 1.0000e+00znew = []pnew = -2.1352e+02 + 5.1548e+02i -2.1352e+02 - 5.1548e+02i -5.1548e+02 + 2.1352e+02i -5.1548e+02 - 2.1352e+02iknew = 9.6912e+10
-1 -0.5 0 0.5 1
-1
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
Real part
Imag
inar
y pa
rt
0 100 200 300 400 500 6000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency in Hz
Gai
n m
agni
tude
0 100 200 300 400 500 600-70
-60
-50
-40
-30
-20
-10
0
Frequency in Hz
Gai
n in
dB
SOLUTION TO 21.11.
fp = 75; fs = 450;Amax = 1; Amin = 45;
n = buttord(fp,fs,Amax,Amin,'s');% The order mfile may not be available in the student edition.emax = sqrt(10^(0.1*Amax) - 1);emin = sqrt(10^(0.1*Amin) - 1)/(fs/fp)^n;fcmin = fp/((10^(0.1*Amax)-1)^(1/(2*n)));fcmax = fs/((10^(0.1*Amin)-1)^(1/(2*n)));wcmin = 2*pi*fcmin;wcmax = 2*pi*fcmax;[z,p,k] = buttap(n);wc = wcmax;fc = fcmaxznew = z*wcpnew = p*wcknew = k*wc^nW = 0:0.01:fs/fp;h = freqs(k*poly(z),poly(p),W);plot(W*wc/(2*pi),abs(h))gridxlabel('Frequency in Hz')ylabel('Gain magnitude')pauseplot(W*wc/(2*pi),20*log10(abs(h)))xlabel('Frequency in Hz')ylabel('Gain in dB')grid
fc = 1.2323e+02znew = []pnew = -2.9630e+02 + 7.1533e+02i -2.9630e+02 - 7.1533e+02i -7.1533e+02 + 2.9630e+02i -7.1533e+02 - 2.9630e+02iknew = 3.5940e+11
0 100 200 300 400 500 600 700 8000
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Frequency in Hz
Gai
n m
agni
tude
0 100 200 300 400 500 600 700 800-70
-60
-50
-40
-30
-20
-10
0
Frequency in Hz
Gai
n in
dB
SOLUTION TO 21.12. Here we require that
s3 + 2s2 + 2s +1= s3 +C1 + C2
C1C2s2 +
C1 + C2 + LLC1C2
s +2
LC1C2Thus
LC1C2 = 2 ⇒ C1 + C2 + L = 4 and 2 =C1 + C2
C1C2×
LL
=LC1 + LC2
2
in which case, LC1 + LC2 = 4 . Therefore, L C1 + C2 + L( ) = 4L = 4 + L2. Equivalently
L2 − 4L + 4 = (L − 2)(L − 2) = 0
Hence L = 2 H is the only solution. Thus C1C2 = 1 ⇒ C1 =1
C2 ⇒
1C2
+ C2 + 2 = 4 or equivalently
C22 − 2C2 +1 = (C2 −1)(C2 −1) = 0 which implies that C1 = C2 = 1 F is the only solution, as was to be
shown.
SOLUTION TO 21.13. (a) By voltage division
H(s) =
1
Cs + G
Ls + 1
Cs + G
=
1
LC
s2 + 1
RCs + 1
LC (b) With R = 1 Ω,
H(s) =
1
LC
s2 + 1
RCs + 1
LC
=1
s2 + 2s +1
requires that C = 1/ 2 F and L = 2 H.
(c) »fc = 1000;»wc = 2*pi*fcwc = 6.2832e+03»Kf = wc;»Km = 1000;»C = 1/sqrt(2);»L = sqrt(2);»Cnew = C/(Kf*Km)Cnew = 1.1254e-07»Lnew = L*Km/KfLnew = 2.2508e-01
(d)
»Km = C/(wc*1e-6)Km = 1.1254e+02»Kf = wc;»Rnew = KmRnew = 1.1254e+02»Lnew = L*Km/wcLnew = 2.5330e-02
SOLUTION TO 21.14. (a) By voltage division
H(s) =
1
Cs
Ls + Rs + 1
Cs
=
1
LC
s2 + Rs
Ls + 1
LC (b) With R = 1 Ω,
H(s) =
1
LC
s2 + Rs
Ls + 1
LC
=1
s2 + 2s +1
requires that L = 1/ 2 F and C = 2 H.
(c)»L = 1/sqrt(2); C = 1/L;»Km = 10;»Kf = 2*pi*500;»Rnew = 10;»Cnew = C/(Km*Kf)Cnew = 4.5016e-05»Lnew = Km*L/KfLnew = 2.2508e-03
(d)»Km = C/(1e-6*Kf)Km = 4.5016e+02»Lnew = L*Km/KfLnew = 1.0132e-01»Cnew = C/(Km*Kf)Cnew = 1.0000e-06
SOLUTION TO 21.15.
(a) Since 1 = 2/(LC), L = 2/C. Since (1/L + 1/C) = (C/2 + 1/C) = sqrt(2), we have that C is a root of
the quadratic 0.5C2 – sqrt(2)C + 1 = 0. Hence»v = [0.5 -sqrt(2) 1];»r = roots(v)r = 1.4142e+00 1.4142e+00»C = r(1)C = 1.4142e+00»L = 2/CL = 1.4142e+00
(b)»Km = 1e3;»Kf = 2*pi*3500;»Cnew = C/(Km*Kf)Cnew = 6.4308e-08»Lnew = L*Km/KfLnew = 6.4308e-02
(c)»Km = C/(Kf*10e-9)Km = 6.4308e+03»Cnew = C/(Km*Kf)Cnew = 1.0000e-08»Lnew = L*Km/KfLnew = 4.1356e-01»Rs = KmRs = 6.4308e+03»RL = RsRL = 6.4308e+03
SOLUTION TO 21.16. (a) Let G = RL. Then by voltage division
H(s) =Vout
Vin=
1
Cs + G
Ls + Rs + 1
Cs + G
=
1
LC
s2 +1
CRL+
Rs
L
s +
1 + Rs RL
LC
(b) Since 1 = 1.25/(LC), L = 1.25/C. Since (Rs/L + 1/RLC) = (2C/1.25 + 1/8C) = sqrt(2), we have that
C is a root of the quadratic (16/1.25)C2 – 8sqrt(2)C + 1 = 0. Hence»C = roots([16/1.25 -8*sqrt(2) 1])C = 7.8427e-01 9.9615e-02»L = 1.25 ./CL = 1.5938e+00 1.2548e+01(c)»Km = 1e3;»Kf = 2*pi*5e3Kf = 3.1416e+04»Cnew = C/(Km*Kf)Cnew = 2.4964e-08 3.1708e-09»Lnew = L*Km/KfLnew = 5.0734e-02 3.9942e-01
SOLUTION TO 21.17. (a) Define G = 1/Rs, execute two source transformations, and apply voltage division to obtain
H(s) =Vout
Vin=
11
Cs + G+ Ls +1
×G
Cs + G=
1
RsLC
s2 + 1RsC
+ 1L
s + 1 +1 Rs
LC
(b) Since L = 1.5/C, the values of C are the roots of the quadratic, (2/1.5)C2 – 2sqrt(2)C + 1 = 0.Hence»C = roots([(2/1.5) -2*sqrt(2) 1])C = 1.6730e+00 4.4829e-01»L = 1.5 ./CL = 8.9658e-01 3.3461e+00»Km = 2e3;»Kf = 2*pi*5e3;»Cnew = C/(Km*Kf)
Cnew = 2.6627e-08 7.1347e-09»Lnew = Km*L/KfLnew = 5.7078e-02 2.1302e-01»Rsnew = 2*KmRsnew = 4000
SOLUTION TO 21.18. (a)
H( jω) =K
jω
ω p
+1
2 =K
ωω p
2
+1
The 3 dB down frequency, ωc, occurs when
1
ωcω p
2
+1
=1
2
Equivalently
ωc = ω p 2 −1 = 0.64359ω p = 6.4359 ×104 rad/sec.
(b) h(t) = Kω p2te
−ω ptu(t) since
H(s) =K
sω p
+1
2 =Kω p
2
s + ωp( )2 .
Further,
H(s)1
s=
Kω p2
s s + ωp( )2 =K
s−
K
s + ωp−
Kω p
s + ωp( )2
Hence, the step response is
K 1 − e−ω pt
− ωpte−ω pt
u(t)
SOLUTION TO 21.19.
(a) Using voltage division,
H(s) =VC
Vin×
Vout
VC=
1
Cs +1
L2s +1
1 + L1s +1
Cs +1
L2s +1
×1
L2s +1
=
1L1L2C
s3 + 1L1
+ 1L2
s
2 + L1 + L2 + CL1L2C
s + 2L1L2C
(b) Matching coefficients in
1
L1L2C
s3 +1
L1+
1
L2
s2 +
L1 + L2 + C
L1L2Cs +
2
L1L2C
=1
s3 + 2s2 + 2s +1
yields C =2
L1L2 and
L1 + L2 + CL1L2C
=L1 + L2 +
2
L1L22
= 2; equivalently, L1 + L2 +2
L1L2= 4. Further,
1L1
+1L2
=
L1 + L2L1L2
= 2 implies that L1 + L2 = 2L1L2 implies 2L1L2 +2
L1L2= 4 . This requires that
L1L2 = 1 and from earlier equations that L1 + L2 = 2L1L2 = 2 which forces L1 = L2 = 1 H and C = 2 F.The idea is to match the denominator coefficients and thus the dc gain is 0.5 instead of the desired 1. Atransformer or some amplifier device is needed to increase the gain to 1.
(c) Using MATLAB:
»Km = 1000;»Kf = 2*pi*20e3;»Lnew = Km/KfLnew = 7.9577e-03»Cnew = 2/(Km*Kf)Cnew = 1.5915e-08
»Hence, L1new = L2new = 7.96 mH and C = 15.9 nF.
(d) SPICE simulation
V0
R1K
L7.96m
L07.96m
R01K
C15.9n
IVm
MAG(V(IVM))
Frequency (Hz)Prob 21.19-Small Signal AC-1
+0.000e+000
+100.000m
+200.000m
+300.000m
+400.000m
+500.000m
+10.000k +20.000k +30.000k +40.000k
SOLUTION TO 21.20. (a) From figure P21.19a
Vin − V1Rs
= I1, V2 = Vout
in which caseV1 = z11I1 + z12 × 0 = Vin − RsI1
Also
V2 = z21I1 + z22 × 0 = Vout
This implies that Vin = (z11 + Rs)I1 and Vout = z21I1 . Finally we conclude that
VoutVin
=z21
z11 + Rs
(b) Now from figure P21.20b, we have V1 = Vin and V2 = Vout = −RL I2 . This implies that
I2 = −VoutRL
= y21V1 + y22V2
ThusVoutVin
=−y21
y22 +1
RL
=−y21
y22 + GL
Consider here that
VoutVin
=z21
z11 +1=
1
s2 + 2s +1=
1
s
s +1
s+ 2
=
1
2ss
2+
1
2s
Hence z21 =1
2s and z11 =
s
2+
1
2s
. This leads to the circuit
with L =1
2 H and C = 2 F.
( c) Similarly,
−y21y22 +1
=
1
2ss
2+
1
2s
+1
implies y21 = −1
2s and y22 =
s
2s+
1
2s.
This yields the same circuit as above with
L = 2 H and C =1
2 F.
(d) Here Km =1000 and K f = 5000rad / s
(d-i) For (b),
L =1
2×
103
5 ×103 =1
5 2= 0.1414 H
and
C = 2 ×1
5 ⋅106 = 0.2828 µF
(d-ii) For (c)
L = 0.2828 H and C = 0.1414 µF.
SOLUTION TO 21.21.
(a) From earlier developments
H(s) = −Yin
Yout= −
1
R1
Cs + 1
R2
(b) Let C = 1 F, and R1 = R2 = 1 Ω.(c) »Kf = 2*pi*3500
Kf = 2.1991e+04»Km = 1/(Kf*1e-9)Km = 4.5473e+04
In the final design, R1 = R2 = 45.5 kΩ.
SOLUTION TO 21.22.
(a) From problem 21 and voltage division,
H(s) = −
1
R1
Cs + 1
R2
×
1
C2s
R3 + 1
C2s
= −
1
C1R1
s + 1
C1R2
×
1
R3C2
s + 1
R3C2
(b) By inspection, let C1 = 0.1 F, R1 = 1 Ω, R2 = 10 Ω, R3 = 10 Ω, and C2 = 0.1 F, in which case
H(s) = −10
s + 1×
1
s +1=
−10
s +1( )2
(c)»Kf = 1e5;»Km = 0.1/(Kf*1e-9)Km = 1000Hence, in the final design
C1 = 1 nF, R1 = 1 kΩ, R2 = 10 kΩ, R3 = 10 kΩ, and C2 = 1 nF.
(d) Cascade the circuit of figure P21.22 with another op amp section. For the first part of the design,
again set ωp = 1 and use the same values as in part (c). The extra op amp section has the same values as
the first section. As such, final values are the same as in part (c).
SOLUTION TO 21.23. The 2nd order NLP Butterworth transfer function is: HNLP (s) =1
s2 + 2s +1. The
design parameters and steps are detailed in the excel spread sheet below. An additional design called
design C is also listed. For input attenuation, the resistor R1 is replaced by the voltage divider R3-R4combination.
w0^2 w0/Q Num w0 Q KNLP KMA Kf=wp KmR KMB KMC1 1.41 1.00 1.0000 0.7071 1.00 22507.86 6283.20 10000.00 22507.86 15915.46
KMS19492.37
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0 1 2Q 1/(2Q) 1 1 KNLP/K 1/alpha 1/(1-alpha)Design B 1 1 2 1 1/Q 1 Q KNLP/K 1/alpha 1/(1-alpha)Design C 1 1-1/Q 3-1/Q 1 1 1 1 KNLP/K 1/alpha 1/(1-alpha)Saraga RA RA/3 3-Apr rt(3)Q 1 1/Q 1/rt(3) KNLP/K 1/alpha 1/(1-alpha)
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 1.4142 0.7071 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.4142 1.0000 0.7071 0.5000 2.0000 2.0000Design C 1.00 -0.4142 1.5858 1.0000 1.0000 1.0000 1.0000 0.6306 1.5858 2.7071Saraga 3.00 1.0000 1.3333 1.2247 1.0000 1.4142 0.5774 0.7500 1.3333 5.6569
w0 scaleDesign A ∞ 0.0000 1.0000 1.4142 0.7071 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.4142 1.0000 0.7071 0.5000 2.0000 2.0000Design C 1.00 0.1140 2.1140 1.0000 1.0000 1.0000 1.0000 0.6306 1.5858 2.7071Saraga 3.00 1.0000 1.3333 1.2247 1.0000 1.4142 0.5774 0.7500 1.3333 5.6569
wp scaleDesign A ∞ 0.000E+00 1.000E+00 2.251E-04 1.125E-04 1.000E+00 1.000E+00 1.000E+00 1.000E+00 #DIV/0!Design B 1.00 1.000E+00 2.000E+00 1.592E-04 2.251E-04 1.000E+00 7.071E-01 5.000E-01 2.000E+00 2.000E+00Design C 1.00 1.140E-01 2.114E+00 1.592E-04 1.592E-04 1.000E+00 1.000E+00 6.306E-01 1.586E+00 2.707E+00Saraga 3.00 1.000E+00 1.333E+00 1.949E-04 1.592E-04 1.414E+00 5.774E-01 7.500E-01 1.333E+00 5.657E+00
Km scaleDesign A ∞ 0.000E+00 1.000E+00 1.000E-08 5.000E-09 2.251E+04 2.251E+04 1.000E+00 2.251E+04 #DIV/0!Design B 10000 1.000E+04 2.000E+00 7.0711E-09 1.000E-08 2.251E+04 1.592E+04 5.000E-01 4.502E+04 4.502E+04Design C 10000 1.140E+03 2.114E+00 1.000E-08 1.000E-08 1.592E+04 1.592E+04 6.306E-01 2.524E+04 4.308E+04Saraga 30000 1.000E+04 1.333E+00 1.000E-08 8.165E-09 2.757E+04 1.125E+04 7.500E-01 2.599E+04 1.103E+05
SOLUTION TO 21.24 AND 21.25. In problem 21.8, the transfer function information was computed inMATLAB as:
% Numerators are each 1. Denominators are the polynomialsd1 = 1.0000e+00 1.0000e+00 1.0000e+00d2 = 1 1
Further we know from MATLAB thatfcmin = 1.5521e+02
The Saraga design and Design A for d1, the second order section of each filter, are given by the excelspread sheet below, as well as two alternate designs labeled B and C.
w0^2 w0/Q Num w0 Q KNLP KMA Kf=wp KmR KMB KMC1 1.00 1.00 1.0000 1.0000 1.000 41016.58 975.22 10000.00 20508.29 20508.29
KMS35521.40
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0 1 2Q 1/(2Q) 1 1 KNLP/K 1/alpha 1/(1-alpha)Design B 1 1 2 1 1/Q 1 Q KNLP/K 1/alpha 1/(1-alpha)Design C 1 1-1/Q 3-1/Q 1 1 1 1 KNLP/K 1/alpha 1/(1-alpha)Saraga RA RA/3 3-Apr rt(3)Q 1 1/Q 1/rt(3) KNLP/K 1/alpha 1/(1-alpha)
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 2.0000 0.5000 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.0000 1.0000 1.0000 0.5000 2.0000 2.0000Design C 1.00 0.0000 2.0000 1.0000 1.0000 1.0000 1.0000 0.5000 2.0000 2.0000Saraga 3.00 1.0000 1.3333 1.7321 1.0000 1.0000 0.5774 0.7500 1.3333 4.0000
w0 scaleDesign A ∞ 0.0000 1.0000 2.0000 0.5000 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.0000 1.0000 1.0000 0.5000 2.0000 2.0000Design C 1.00 0.1140 2.1140 1.0000 1.0000 1.0000 1.0000 0.5000 2.0000 2.0000Saraga 3.00 1.0000 1.3333 1.7321 1.0000 1.0000 0.5774 0.7500 1.3333 4.0000
wp scaleDesign A ∞ 0.000E+00 1.000E+00 2.051E-03 5.127E-04 1.000E+00 1.000E+00 1.000E+00 1.000E+00 #DIV/0!Design B 1.00 1.000E+00 2.000E+00 1.025E-03 1.025E-03 1.000E+00 1.000E+00 5.000E-01 2.000E+00 2.000E+00Design C 1.00 1.140E-01 2.114E+00 1.025E-03 1.025E-03 1.000E+00 1.000E+00 5.000E-01 2.000E+00 2.000E+00Saraga 3.00 1.000E+00 1.333E+00 1.776E-03 1.025E-03 1.000E+00 5.774E-01 7.500E-01 1.333E+00 4.000E+00
Km scaleDesign A ∞ 0.000E+00 1.000E+00 5.000E-08 1.250E-08 4.102E+04 4.102E+04 1.000E+00 4.102E+04 #DIV/0!Design B 10000 1.000E+04 2.000E+00 5.000E-08 5.000E-08 2.051E+04 2.051E+04 5.000E-01 4.102E+04 4.102E+04Design C 10000 1.140E+03 2.114E+00 5.000E-08 5.000E-08 2.051E+04 2.051E+04 5.000E-01 4.102E+04 4.102E+04Saraga 30000 1.000E+04 1.333E+00 5.000E-08 2.887E-08 3.552E+04 2.051E+04 7.500E-01 4.736E+04 1.421E+05
The first order (leaky integrator) section is common to both problems. This section consists of an input
resistor (conductance) R1 (G1) connected to the inverting terminal with a parallel R2-C combination
feeding back from the output. The transfer function is: H(s) =G1
Cs + G2. For the normalized design we
set G1 = G2 = 1 S (R1 = R2 = 1 Ω) and C = 1 F. This design can be scaled independently of the S&K 2nd
order section. Hence we set Cnew = 50 nF. Thus Km = 20,508.29. Hence R1 = R2 = 20,508.29 Ω.
SOLUTION TO 21.26 AND 21.27. The relevent data from the solution of problem 21.10 is:
k = 1d1 = 1.0000e+00 7.6537e-01 1.0000e+00d2 = 1.0000e+00 1.8478e+00 1.0000e+00fcmin = 8.8800e+01wcmin = 5.5795e+02
In providing the designs, we set forth all the possible S&K designs using two excel spreadsheets, one foreach second order section.
The designs for denominator d1 with numerator equal to 1 are:
w0^2 w0/Q Num w0 Q KNLP KMA Kf=wp KmR KMB KMC1 0.7654 1.00000 1.0000 1.3066 1.0000 46834.37 557.95 10000.00 17922.81 17922.81
KMS40559.76
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0 1 2Q 1/(2Q) 1 1 KNLP/K 1/alpha 1/(1-alpha)Design B 1 1 2 1 1/Q 1 Q KNLP/K 1/alpha 1/(1-alpha)Design C 1 1-1/Q 3-1/Q 1 1 1 1 KNLP/K 1/alpha 1/(1-alpha)Saraga RA RA/3 3-Apr rt(3)Q 1 1/Q 1/rt(3) KNLP/K 1/alpha 1/(1-alpha)
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 2.6131 0.3827 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 0.7654 1.0000 1.3066 0.5000 2.0000 2.0000Design C 1.00 0.2346 2.2346 1.0000 1.0000 1.0000 1.0000 0.4475 2.2346 1.8100Saraga 3.00 1.0000 1.3333 2.2630 1.0000 0.7654 0.5774 0.7500 1.3333 3.0615
w0 scaleDesign A ∞ 0.0000 1.0000 2.6131 0.3827 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 0.7654 1.0000 1.3066 0.5000 2.0000 2.0000Design C 1.00 0.1140 2.1140 1.0000 1.0000 1.0000 1.0000 0.4475 2.2346 1.8100Saraga 3.00 1.0000 1.3333 2.2630 1.0000 0.7654 0.5774 0.7500 1.3333 3.0615
wp scaleDesign A ∞ 0.000E+00 1.000E+00 4.683E-03 6.859E-04 1.000E+00 1.000E+00 1.000E+00 1.000E+00 #DIV/0!Design B 1.00 1.000E+00 2.000E+00 1.792E-03 1.372E-03 1.000E+00 1.307E+00 5.000E-01 2.000E+00 2.000E+00Design C 1.00 1.140E-01 2.114E+00 1.792E-03 1.792E-03 1.000E+00 1.000E+00 4.475E-01 2.235E+00 1.810E+00Saraga 3.00 1.000E+00 1.333E+00 4.056E-03 1.792E-03 7.654E-01 5.774E-01 7.500E-01 1.333E+00 3.061E+00
Km scaleDesign A ∞ 0.000E+00 1.000E+00 1.000E-07 1.464E-08 4.683E+04 4.683E+04 1.000E+00 4.683E+04 #DIV/0!Design B 10000 1.000E+04 2.000E+00 1.000E-07 7.654E-08 1.792E+04 2.342E+04 5.000E-01 3.585E+04 3.585E+04Design C 10000 1.140E+03 2.114E+00 1.000E-07 1.000E-07 1.792E+04 1.792E+04 4.475E-01 4.005E+04 3.244E+04Saraga 30000 1.000E+04 1.333E+00 1.000E-07 4.419E-08 3.104E+04 2.342E+04 7.500E-01 5.408E+04 1.242E+05
The designs for denominator d2 with numerator equal to 1 are:
w0^2 w0/Q Num w0 Q KNLP KMA Kf=wp KmR KMB KMC1 1.8478 1.00 1.00 0.54 1.00 19399.08 557.95 10000.00 33117.77 17922.81
KMS17922.81
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0 1 2Q 1/(2Q) 1 1 KNLP/K 1/alpha 1/(1-alpha)Design B 1 1 2 1 1/Q 1 Q KNLP/K 1/alpha 1/(1-alpha)Design C 1 1-1/Q 3-1/Q 1 1 1 1 KNLP/K 1/alpha 1/(1-alpha)Saraga RA RA/3 3-Apr rt(3)Q 1 1/Q 1/rt(3) KNLP/K 1/alpha 1/(1-alpha)
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 1.0824 0.9239 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.8478 1.0000 0.5412 0.5000 2.0000 2.0000Design C 1.00 -0.8478 1.1522 1.0000 1.0000 1.0000 1.0000 0.8679 1.1522 7.5703Saraga 3.00 1.0000 1.3333 0.9374 1.0000 1.8478 0.5774 0.7500 1.3333 7.3912
w0 scaleDesign A ∞ 0.0000 1.0000 1.0824 0.9239 1.0000 1.0000 1.0000 1.0000 #DIV/0!Design B 1.00 1.0000 2.0000 1.0000 1.8478 1.0000 0.5412 0.5000 2.0000 2.0000Design C 1.00 0.1140 2.1140 1.0000 1.0000 1.0000 1.0000 0.8679 1.1522 7.5703Saraga 3.00 1.0000 1.3333 0.9374 1.0000 1.8478 0.5774 0.7500 1.3333 7.3912
wp scaleDesign A ∞ 0.000E+00 1.000E+00 1.940E-03 1.656E-03 1.000E+00 1.000E+00 1.000E+00 1.000E+00 #DIV/0!Design B 1.00 1.000E+00 2.000E+00 1.792E-03 3.312E-03 1.000E+00 5.412E-01 5.000E-01 2.000E+00 2.000E+00Design C 1.00 1.140E-01 2.114E+00 1.792E-03 1.792E-03 1.000E+00 1.000E+00 8.679E-01 1.152E+00 7.570E+00Saraga 3.00 1.000E+00 1.333E+00 1.680E-03 1.792E-03 1.848E+00 5.774E-01 7.500E-01 1.333E+00 7.391E+00
Km scaleDesign A ∞ 0.000E+00 1.000E+00 1.000E-07 8.536E-08 1.940E+04 1.940E+04 1.000E+00 1.940E+04 #DIV/0!Design B 10000 1.000E+04 2.000E+00 5.412E-08 1.000E-07 3.312E+04 1.792E+04 5.000E-01 6.624E+04 6.624E+04Design C 10000 1.140E+03 2.114E+00 1.000E-07 1.000E-07 1.792E+04 1.792E+04 8.679E-01 2.065E+04 1.357E+05Saraga 30000 1.000E+04 1.333E+00 9.374E-08 1.000E-07 3.312E+04 1.035E+04 7.500E-01 2.390E+04 1.325E+05
SOLUTION TO 21.28. For this problem we use the excel spread sheet given below. First we observe that
H(s) =ˆ K
s2 + 0
Qs + 0
2 → H ( 0s) =
ˆ K
0s( )2 + 0
Q 0s( ) + 02
=ˆ K 0
2 (= KNLP)
s2 +1
Qs +1
Thus after this type of frequency scaling, the new transfer function is:
Hnew(s) =0.7943
s2 +1
1.1286s +1
The dc gain is of course 0.7943 and the modification of the circuit to achieve the correct dc gain is givenin the spread sheet below via R3 and R4 which constitute a voltage divider that replaces R1.
w0^2 w0/Q Num w0 Q KNLP KMA Kf=wp KmR KMB KMC0.82306 0.8038 0.65378 0.9072 1.1286 0.7943 1131.42 43982.40 10000.00 501.23 501.23
KMS979.83
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0 1 2Q 1/(2Q) 1 1 KNLP/K 1/alpha 1/(1-alpha)Design B 1 1 2 1 1/Q 1 Q KNLP/K 1/alpha 1/(1-alpha)Design C 1 1-1/Q 3-1/Q 1 1 1 1 KNLP/K 1/alpha 1/(1-alpha)Saraga RA RA/3 3-Apr rt(3)Q 1 1/Q 1/rt(3) KNLP/K 1/alpha 1/(1-alpha)
RA RB K C1 C2 R1 R2 alpha R3 R4Design A ∞ 0.0000 1.0000 2.2573 0.4430 1.0000 1.0000 0.7943 1.2589 4.8621Design B 1.00 1.0000 2.0000 1.0000 0.8860 1.0000 1.1286 0.3972 2.5179 1.6588Design C 1.00 0.1140 2.1140 1.0000 1.0000 1.0000 1.0000 0.3758 2.6613 1.6019Saraga 3.00 1.0000 1.3333 1.9549 1.0000 0.8860 0.5774 0.5957 1.6786 2.1917
w0 scaleDesign A ∞ 0.0000 1.0000 2.4881 0.4883 1.0000 1.0000 0.7943 1.2589 4.8621Design B 1.00 1.0000 2.0000 1.1023 0.9766 1.0000 1.1286 0.3972 2.5179 1.6588Design C 1.00 0.1140 2.1140 1.1023 1.1023 1.0000 1.0000 0.3758 2.6613 1.6019Saraga 3.00 1.0000 1.3333 2.1548 1.1023 0.8860 0.5774 0.5957 1.6786 2.1917
wp scaleDesign A ∞ 0.000E+00 1.000E+00 5.657E-05 1.110E-05 1.000E+00 1.000E+00 7.943E-01 1.259E+00 4.862E+00Design B 1.00 1.000E+00 2.000E+00 2.506E-05 2.220E-05 1.000E+00 1.129E+00 3.972E-01 2.518E+00 1.659E+00Design C 1.00 1.140E-01 2.114E+00 2.506E-05 2.506E-05 1.000E+00 1.000E+00 3.758E-01 2.661E+00 1.602E+00Saraga 3.00 1.000E+00 1.333E+00 4.899E-05 2.506E-05 8.860E-01 5.774E-01 5.957E-01 1.679E+00 2.192E+00
Km scaleDesign A ∞ 0.000E+00 1.000E+00 5.000E-08 9.813E-09 1.131E+03 1.131E+03 7.943E-01 1.424E+03 5.501E+03Design B 10000 1.000E+04 2.000E+00 5.000E-08 4.430E-08 5.012E+02 5.657E+02 3.972E-01 1.262E+03 8.315E+02Design C 10000 1.140E+03 2.114E+00 5.000E-08 5.000E-08 5.012E+02 5.012E+02 3.758E-01 1.334E+03 8.029E+02Saraga 30000 1.000E+04 1.333E+00 5.000E-08 2.558E-08 8.682E+02 5.657E+02 5.957E-01 1.645E+03 2.148E+03
SOLUTION TO 21.29. (a) HHP (s) = HNLPc
s
=
1
cs
2
+ 2 cs
+1
. At s = j p ,
HHP ( j p) = HNLPc
j p
=1
c
j p
2
+ 2 c
j p
+1
=1
−5.57
2
− j 25.57
+1
Thus in MATLAB,»Magfp = 1/abs(1 - (5.5/7)^2 -j*sqrt(2)*(5.5/7))Magfp = 8.5091e-01»Attenfp = -20*log10(Magfp)
Attenfp = 1.4023e+00»Magfs = 1/abs(1 - (5.5/1)^2 -j*sqrt(2)*(5.5/1))Magfs = 3.3040e-02»Attenfs = -20*log10(Magfs)Attenfs = 2.9619e+01
Thus the attenuation at fp is 1.4023 dB and that at fs is 29.619 dB.(b) From problem 21.15, the transfer function is
Hcir(s) =1 LC
s2 +1
C+
1
L
s +
2
LCand the values of L and C realizing the 2nd order Butterworth NLP transfer function can be computedaccording to
1C
+1L
= 2, 2
LC=1 ⇒
1C
+C2
= 2 ⇒ C2 − 2 2C + 2 = 0
(C − 2)2 = 0 ⇒ C = 2 F ⇒ L = 2 H
(c) Here Km = 1000. Ls → L C
s →
11
LKm Cs
and 1
Cs →
sC C
→ Km
C Cs . Thus in
MATLAB,
»wc = 2*pi*5.5e3wc = 3.4558e+04»Km = 1000;»C = sqrt(2); L = sqrt(2);»Lhp = Km/(C*wc)Lhp = 2.0462e-02»Chp = 1/(Km*wc*L)Chp = 2.0462e-08
Therefore, the resistors take on values of 1 kΩ, the inductor is changed to a capacitor of value of Chp =
20.46 nF and the capacitor is changed to an inductor of value Lhp = 20.46 mH.
SOLUTION TO 21.30. (a)
Ω = p, Amax = 2 dB, = max = 100.1×2 −1 = 0.76478, Ωc =
16
100.1× 2 −1= 1.0935
»Wc=1/(10^0.2-1)^(1/6)Wc = 1.0935e+00
(b)»wchp = 2*pi*5e3/Wcwchp = 2.8730e+04»fchp = wchp/(2*pi)fchp = 4.5725e+03
Thus chp = p
Ωc= 28.73 krad/s.
( c ) (i) NLP → HP transformation: scale by wchp.
Chp =1
Llp chp=17.404 F, Lhp =
1Clp chp
= 34.807 H
(ii) Magnitude scale to obtain proper value of capacitors.»Clp = 1;»Llp = 2;»Chp = 1/(Llp*wchp)Chp = 1.7404e-05»Lhp = 1/(Clp*wchp)Lhp = 3.4807e-05»Cnewhp = 100e-9;»Km = Chp/CnewhpKm = 1.7404e+02»Lnewhp = Km*LhpLnewhp = 6.0578e-03
Km =174.02, Chp,new =Chp
Km= 100 nF , Lhp,new = KmLhp = 6.06 mH
SOLUTION TO 21.31.
The 2nd order NLP Butterworth transfer function is: HNLP (s) =1
s2 + 2s +1. Using the transformation
s to 1/s, we obtain the NHP Butterworth transfer function:
HNHP(s) = HNLP1s
=
s2
s2 + 2s +1= K∞
s2
s2 + d(1) s + d(2)
INPUT: d(1) d(2) K K Kf Km KmR1.414213562 1 1.33334 1 18849.6 12247.44871 30000
NHP Params w0 Q alpha1 0.70710678 0.74999625
NHP Crt Pars C1 = Q C2 = rt(3) R1 = 1 R2 =1/(rt(3)Q) R R/30.707106781 1.73205081 1 0.816496581 1 0.333333333
HP Crt Params C1new=C1/(Km*Kf) C2new R1new=R1*Km R2new=R2*Km Rnew=R*KmR Rnew/33.06293E-09 7.5026E-09 12247.44871 10000 30000 10000
A plot of the design without input attenuation is shown below. Notice that as predicted the gain is 4/3.
R130k
XOpAmp
Vpls15
Vminus15
Vin C07.503n
C3.063n
R012.247k
R10k
IVm
R210k
MAG(V(IVM))
Frequency (Hz)S&K HP-Small Signal AC-3
(V)
+0.000e+000
+500.000m
+1.000
+100.000 +316.228 +1.000k +3.162k +10.000k
Input attenuation requires that we replace C1 with a series combination of capacitors in which C1 = C3+ C4 and (1/C3)/(1/C4 + 1/C3) = alpha. Here then, C1 = C3 + C4 and alpha = C4/(C3 + C4) = C4/C1.Thus C4 = alpha*C1 and C3 = (1 – alpha)*C1. Thus
InputAttenuation
C3 = (1 - alpha)C1 C4 = alpha*C1
7.65744E-10 2.29719E-09
SOLUTION TO 21.32. The fourth order Butterworth NLP transfer function can be obtained from tables orfrom MATLAB as follows:
»[z,p,k] = buttap(4)z = []p = -3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i -9.2388e-01 - 3.8268e-01ik = 1»% Second Order Sections
»n1 = 1;»d1 = poly([p(1),conj(p(1))])d1 = 1.0000e+00 7.6537e-01 1.0000e+00»n2 = 1;»d2 = poly([p(3),conj(p(3))])d2 = 1.0000e+00 1.8478e+00 1.0000e+00»Thus,
HNLP (s) =1
s2 + 0.76537s +1×
1
s2 +1.8478s +1having frequency response
»w = 0:0.01:3.5;»h = freqs(k*poly(z),poly(p),w);»plot(w,abs(h))»grid»xlabel('Normalized Frequency')»ylabel('Magnitude 4th Order Butterworth')
0 0.5 1 1.5 2 2.5 3 3.50
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Normalized Frequency
Mag
nitu
de 4
th O
rder
But
terw
orth
TextEnd
The Saraga design parameters are given in the following Excel tables:
INPUT: d(1) d(2) KNLP/NHP w0LP/HP QNLP H(s) 0.76537 1 1 1 1.306557613NHP H(s) 0.76537 1 1 1 1.306557613
K Kf Km KmR alpha1.33334 18849.6 22630.24168 60000 0.74999625
NHP Crt Pars C1 = Q C2 = rt(3) R1 = 1 R2 =1/(rt(3)Q) R R/31.306557613 1.732050808 1 0.441886576 1 0.333333333
Scale by w0HP 1.306557613 1.732050808 1 0.441886576 1 0.333333333
HP Crt Params C1new=C1/(Km*Kf) C2new R1new=R1*Km R2new=R2*Km Rnew=R*KmR Rnew/33.06293E-09 4.0604E-09 22630.24168 10000 60000 20000
Input Attenuation C3 = (1 - alpha)C1 C4 = alpha*C17.65744E-10 2.29719E-09
INPUT: d(1) d(2) KNLP/NHP w0LP/HP QNLP H(s) 1.8478 1 1 1 0.541184111NHP H(s) 1.8478 1 1 1 0.541184111
K Kf Km KmR alpha1.33334 18849.6 10000 30000 0.74999625
NHP Crt Pars C1 = Q C2 = rt(3) R1 = 1 R2 =1/(rt(3)Q) R R/30.541184111 1.732050808 1 1.066827827 1 0.333333333
Scale by w0HP 0.541184111 1.732050808 1 1.066827827 1 0.333333333
HP Crt Params C1new=C1/(Km*Kf) C2new R1new=R1*Km R2new=R2*Km Rnew=R*KmR Rnew/32.87106E-09 9.18879E-09 10000 10668.27827 30000 10000
Input Attenuation C3 = (1 - alpha)C1 C4 = alpha*C17.17777E-10 2.15329E-09
SOLUTION TO 21.33. Using MATLAB,
»fp = 5e3; fs = 1.5e3;»wp = 2*pi*fp; ws = 2*pi*fs;»Amax = 3; Amin = 40;»n = buttord(wp,ws,Amax,Amin,'s')n =4
»[z,p,k] = buttap(n)z =[]p =-3.8268e-01 + 9.2388e-01i -3.8268e-01 - 9.2388e-01i -9.2388e-01 + 3.8268e-01i
-9.2388e-01 - 3.8268e-01ik = 1
»d1 = real(poly([p(1),p(2)]))d1 =1.0000e+00 7.6537e-01 1.0000e+00»d2 = real(poly([p(3),p(4)]))d2 =1.0000e+00 1.8478e+00 1.0000e+00
In general,
HNHP(s) = HNLP1s
=
1
1s
2
+ 0LPQ
1s
+ ( 0LP )2
=1
( 0LP )2 ×s2
s2 +1 0LP( )
Qs + 1 0LP( )2
=( 0HP )2 s2
s2 + 0HP
Qs + ( 0HP )2
The S&K Saraga design for d1 is given by the following excel spreadsheet:
INPUT: d(1) d(2) KNLP/NHP w0LP/HP QNLP H(s) 0.76537 1 1 1 1.306557613NHP H(s) 0.76537 1 1 1 1.306557613
K Kf Km KmR alpha1.33334 31416 45260.48336 60000 0.74999625
NHP Crt Pars C1 = Q C2 = rt(3) R1 = 1 R2 =1/(rt(3)Q) R R/31.306557613 1.732050808 1 0.441886576 1 0.333333333
Scale by w0HP 1.306557613 1.732050808 1 0.441886576 1 0.333333333
HP Crt Params C1new=C1/(Km*Kf) C2new R1new=R1*Km R2new=R2*Km Rnew=R*KmR Rnew/39.18879E-10 1.21812E-09 45260.48336 20000 60000 20000
Input Attenuation C3 = (1 - alpha)C1 C4 = alpha*C12.29723E-10 6.89156E-10
The S&K Saraga design for d2 is given by the following excel spreadsheet:
INPUT: d(1) d(2) KNLP/NHP w0LP/HP QNLP H(s) 1.8478 1 1 1 0.541184111NHP H(s) 1.8478 1 1 1 0.541184111
K Kf Km KmR alpha1.33334 31416 20000 60000 0.74999625
NHP Crt Pars C1 = Q C2 = rt(3) R1 = 1 R2 =1/(rt(3)Q) R R/30.541184111 1.732050808 1 1.066827827 1 0.333333333
Scale by w0HP 0.541184111 1.732050808 1 1.066827827 1 0.333333333
HP Crt Params C1new=C1/(Km*Kf) C2new R1new=R1*Km R2new=R2*Km Rnew=R*KmR Rnew/38.61319E-10 2.75664E-09 20000 21336.55655 60000 20000
Input Attenuation C3 = (1 - alpha)C1 C4 = alpha*C12.15333E-10 6.45986E-10
This completes the design.
SOLUTION TO 21.34. For the woofer,
H(s) =8
Ls + 8=
8
L
s +8
L
Thus, 8L
= 2000 × 2π ⇒ L = 636 µH.
For the tweeter,
H(s) =8
1
Cs+ 8
=8Cs
8Cs +1=
s
s +1
8C
Thus, 1
8C= 2000 × 2π ⇒ C = 9.95 µF.
SOLUTION TO 21.35. For the woofer,
HNLP (s) =1
s2 + 2s +1
The transfer function of the following circuit
is
H(s) =
1
LC
s2 +1
Cs +
1
LC
Thus 2 =1C
⇒ C = 0.70711 F and since 1
LC= 1, L = 2 = 1.4142 H. Frequency scaling the
element values by Kf = 4000π and magnitude scaling by Km = 8 yields C =0.70711KmK f
= 7.0337 µF and
L =1.4142Km
K f= 0.90032 mH:
For the tweeter we first realize the NLP Butterworth transfer function as above to obtain asabove
with Thus C = 0.70711 F and L = 2 = 1.4142 H. We now apply the frequency transformation
s → Cs
to each element (capacitors become inductors and inductors become capacitors according to
figure 21.24) and we obtain the HP circuit topology
where Cnew = 7.0337 µF and Lnew = 0.90032 mH.
SOLUTION TO 21.36. Consider figure (a). Let the current entering the RC network from Z1 be dentoted
by Ifa. Let the voltage from this point to ground be denoted Vfa. Then
Vout,a = Vfa + I faZ1 = Ha(s)VinFor figure (b) with a similar denotation of voltage and current, we have
Vout,b = Vfb + I fb +VfbZ1
k −1
Z1k
= V fb + I fbZ1k
+k −1
kVfb =
1k
(2k −1)Vfb +I fb
Z1
If(2k −1)Vfb ≅ Vfa and I fb ≅ I fa (**)
then
Vout,b ≅1k
Ha(s)Vin =Vout ,a
k
For gain enhancement, k < 1. However, for the (**) to be valid, we require that Z1
k −1 be large relative
to what it sees in the RC network. Hence, in general, k must be close to 1. Thus only small gainenhancements are possible. For such a potentially sensitive approach to gain enhancement, it might bebetter simply to add another op amp stage as op amps are comparatively inexpensive.
CHAPTER 22 PROBLEM SOLUTIONS
SOLUTION TO PROBLEM 22.1(a) For figure P22.1a, T0 = 2 and ω0 = π . Let t0 = -1 in equation 22.5b. Then f(t) = δ(t) and
cn = 0.5 ( t)e− jnπtdt−1
1
∫ = 0.5 for all n
From equation 22.6, an = 1 and bn = 0 for all n. Finally from equation 22.2
f ( t) = 0.5 + cos(nπt)n=1
∞∑
(b) For figure P22.1b, T0 = 2 and ω0 = π . Let t0 = 0 in equation 22.5b. Then f(t) = - δ(t -1) and
cn = −0.5 (t −1)e− jnπtdt−1
1
∫ = −0.5e− jnπ
From equation 22.6, bn = 0 for all n, and
an = -1 for n even
an = 1 for n odd
Finally from equation 22.2
f ( t) = −0.5 + cos(πt) − cos(2πt) + cos(3πt) − cos(4πt) +K
SOLUTION TO PROBLEM 22.2(a) T0 = 1 and ω0 = 2π . Let t0 = 0 in equation 22.5b. Then
f ( t) = e− ln (2)( )t
and from equation 22.5b
cn = e− ln (2)( )te− j2nπtdt0
1
∫ = e− ln (2)+ j2nπ( )tdt0
1
∫ =−1
ln(2) + j2nπ( ) e− ln (2)+ j2nπ( ) −1[ ] =
0.5ln(2) + j2nπ( )
(b) Using the above result for cnc0 =0.5
ln2 = 0.7213 ,
c1 = 0.5ln2+ j2π
= 0.7213e -j1.4609
c2 = 0.5ln2+ j4π
= 0.397e -j1.516
From equation 22.6
d0 = c0 = 0.7213
d1 =2c1 = 0.158,
θ1 =-1.461x180/π =-83.7o
d2 =2c2 = 0.0795
θ2 =-1.516x180/π =-86.84o
Thus, f(t) in the form of equation 22.3 is
f(t) = 0.7213 + 0.158cos(2πt - 83.7o) + 0.0795os(2πt - 86.84o)
SOLUTION TO PROBLEM 22.3.
(a) T0 = 1 and ω0 = 2π . Let t0 = 0 in equation 22.5b. Then
f ( t) = e− ln (2)( )t u(t) − u( t − 0.5)[ ]
and from equation 22.5b
cn = e− ln (2)( )te− j2nπtdt0
0.5
∫ = e− ln (2)+ j2nπ( )tdt0
0.5
∫ = −1ln(2) + j2nπ( ) e−0.5 ln (2)+ j2nπ( ) −1[ ]
=1−
(−1)n
2ln(2) + j2nπ( )
(b) Using the above result for cn, and MATLAB to evaluate the numerical result,»n= 0;»c0= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pic0 4.2256e-01»n=1;»c1= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi)c1 = 2.9612e-02- 2.6843e-01i»abs(c1)ans = 2.7006e-01»degreec1=angle(c1)*180/pidegreec1 = -8.3705e+01»n=2; »c2= (1- (-1)^n/sqrt(2))/(log(2) + j*2*n*pi)c2 = 1.2817e-03- 2.3237e-02i
»abs(c2)ans = 2.3272e-02degreec2= angle(c2)*180/pidegreec2 =-8.6843e+01
From equation 22.6 and equation 22.3
f(t) = 0.4226 + 0.54cos(2πt - 83.7o) + 0.04654os(2πt - 86.84o)
SOLUTION PROBLEM 22.4. (a) f(t) = cos(4t) sin(2t) = 0.5[ sin(6t) - sin (2t)] .The fundamental angular frequency of f(t) is ω0= 2 rad/s. The given f(t) can be expressed as
f(t) = -0.5 sin( ω0t) + 0.5sin(3ω0t) . Observe that b1= -0.5, b3= 0.5 and all other ai and bi are zero.
From equation 22.4 , d1= 0.5 /-90o and d3 = 0.5 /90o. From equations 22.6a and 22.6b.
c1 = 0.25j and c3 = -0.25j. All other cn are zero for n positive
(b) f(t) = sin2(4t) cos2(8t)= 0.5[1 - cos(8t)]x0.5 1+ cos(16t)
= 0.25 [ 1 - cos(8t) + cos(16t) - cos(8t) cos(16t)]
= 0.25 - 0.375 cos(8t) + 0.25cos(16t) - 0.125cos(24t) The fundamental angular frequency of f(t) is ω0= 8 rad/s. The given f(t) can be expressed as
f(t) = 0.25 -0.375 cos( ω0t) + 0.25cos(2ω0t) - 0.125sicos(3ω0t). Observe that a0= 0.5, a1=-0.375, a2=
0.25 , a3= -0.125 and all other ai and bi are zero.
Fom equation 22.4,d0= 0.25, d1= 0.375 /180o , d2 = 0.25 /0o, and d3 = 0.125 /180o,
From equations 22.6a and 22.6b.c0 = 0.25, c1 = -0.375 and c2 = 0.25, c3 = -0.125. All other cn are zero for n positive.
(c)f(t) = 2 + 1.5 sin(500t)- 2cos(2000t)]cos(106t)
=2cos(106t) + 0.75sin(1000500t) + 0.75sin(999500t)- cos(1002000t) - cos(998000t)
The fundamental angular frequency of f(t) is ω0= 500 rad/s. The given f(t) can be expressed as
f(t) =2cos( 2000ω0t) + 0.75sin(2001ω0t) + 0.75sin(1999ω0t) - cos(2004ω0t) - cos(1996ω0t)
Observe that: a1996 = -1, b1999= 0.75, a2000= 2, b2001= 0.75 , a2004= -1, and all other ai and bi are
zero. From equation 22.4 , d1996 = 1 /180o , d1999 = 0.75 /-90o, d2000 = 2 /0o ,
d2001 = 0.75/-90o, d2004 = 1 /180o , and all other di are zero.
From equations 22.6a and 22.6b.c1996 = -0.5 , c1999 = -j0.375, c2000 = 1 ,
c2001 = -j0.375o, dc2004 =-0.5 , and all other cn are zero for n positive..
SOLUTION PROBLEM 22.5
By inspesction, the derivative of f(t) is
f '( t) =AT
− A (t − nT )n=−∞
∞∑ =
AT
− f ( t)
where fδ(t) is shown in figure 22/7, with its Fourier series given by equation 22.20b, i.e.
f (t) =AT
+2AT
cos(n 0t)n=1
∞∑
Therefore
f '( t) = −2AT
cos(n 0t)n=1
∞∑
The dc component is the average value of f(t) and is given by 0.5A. Other terms in the Fourier series of
f(t) are obtained by integrating the cosine terms in the above expression. The result is
f ( t) = 0.5A −Anπ
sin(n 0t)n=1
∞∑
SOLUTION TO PROBLEM 22.6Denote by f5(t) the waveform of figure P22.5, with A= 0.5 and T =1. Then by inspection
f ( t) = f prob5(−t) + 0.5
Substituting the result of problem 22.5 into the above equation, we have
f ( t) = 0.75 +0.5nπ
sin(n 0t)n=1
∞∑
SOLUTION PROBLEM 22.7
Consider the square wave g(t) shown in figure 22.4 with its Fourier series given by equation 22.13. By
inspection, the derivative of f(t) is
f '( t) = −4T
g(t) f'(t) = - 4T
g(t)
Substituting equation 22.13 into the above expression, we have
f '( t) = −8AπT
sin(n 0t)n
n=1,odd
∞∑
The dc component is the average value of f(t) and is given by 0.5A. Other terms in the Fourier series of
f(t) are obtained by integrating the sine terms in the above expression. The result is
f ( t) = 0.5A +4A
π 2cos(n 0t)
n2n=1,odd
∞∑
SOLUTION PROBLEM 22.8
The method used below is simpler than that suggested in the hint.
We first sketch the waveform of f'(t) and observe that it may be expressed as the sum of two periodic rectangular
pulse trains:f'(t)= 1
αT[ fp(t + 0.5αT) +fp(t - 0.5αT)]
where fp(t) is sketched in figure 22.8. Using equation 22.23, we have
f'(t)= 1αT
[αA +2Asin(nαπ)
nπ∑n=1
∞
cos(nω0(t+0.5αT))]
-[αA +2Asin(nαπ)
nπ∑n=1
∞
cos(nω0(t-0.5αT))]
= 1αT
2Asin(nαπ)
nπ[∑
n=1
∞
cos(nω0(t+0.5αT))-cos(nω0(t-0.5αT)] \
= 1αT
[ - 4Asin2(nαπ)
nπ sin(nω0t)]∑
n=1
∞
The dc component is the average value of f(t) and is given by αA. Other terms in the Fourier series of
f(t) are obtained by integrating the sine terms in the above expression. The result is
f(t) = αA + 2Aαπ2
[sin(nαπ)
n ]2cos(nω0t)∑n=1
∞
SOLUTION TO PROBLEM 22.9
We first sketch the waveform of f'(t) and observe that it may be expressed as the sum of a periodic rectangular
pulse train and a periodic impulse train:
f'(t)= 1αT
fp(t + 0.5αT) - fδ(t )
where fp(t) is sketched in figure 22.8, and fδ(t) in figure 22.7.
Using equations 22.23 and 22.20b, we have
f'(t)= 1αT
[αA +2Asin(nαπ)
nπ∑n=1
∞
cos(nω0(t+0.5αT))
- [AT
+ 2AT ∑
n=1
∞cos(nω0t)]
= 2AT
sin(nαπ)
nαπ∑[n=1
∞
cos(nω0(t+0.5αT)) -cos(nω0t)]
The dc component is the average value of f(t) and is given by 0.5αA. Other terms in the Fourier series
of f(t) are obtained by integrating the sine terms in the above expression. The result is
f(t) =0.5αA + Anπ
[sin(nαπ)
nαπ∑n=1
∞
sin(nω0(t+0.5αT)) -sin(nω0t)]
It remains to rewrite the expression in the form of equation 22.2.To this end, let b = sin(nαπ)/(nαπ) and re-write the terms within [ ] as follows:
bsin(nαπ)cos(nω0t) + bcos(nαπ) - 1sin(nω0t)
Hence, f(t) in the form of equation 22.2 has the coefficients, for n=1,2...
an = Aαπ2n2
sin2(nαπ)
bn = Aαπ2n2
sin(nαπ)cos(nαπ) -nαπ
dn= an2 + bn2 = A
απ2n2sin4(nαπ) + sin(nαπ)cos(nαπ) -nαπ2
= Aαπ2n2
sin2(nαπ) +(nαπ)[nαπ - sin(2nαπ)]
θn= tan-1(-bnan
)
The result is item 6 of table 22.4
SOLUTION PROBLEM 22.10.
Observe that the present f(t) can be derived from that of problem 22.9 by (a) replacing t by -t; and (b)replacing α by β. Thus the Fourier series for f(t) is:
f(t) =0.5βA + ancos(nω0t) + bn∑[n=1
∞
sin(nω0t) ]
wherean = A
βπ2n2
sin2(nβπ)
bn = -Aβπ2n2
sin(nβπ)cos(nβπ) -nβπ
dn= an2 + bn2 = A
βπ2n2
sin4(nβπ) + sin(nβπ)cos(nβπ) -nβπ2
= A
βπ2n2
sin2(nβπ) +(nβπ)[nβπ - sin(2nβπ)]
θn= tan-1(-bnan
)
SOLUTION PROBLEM 22.11. Following the hint, we have the second derivative of f(t) given by
f ''( t) =1
(1− )Tf (t + T) −
1(1− )T
f ( t)
where fd(t) is given in example 22.5. Notice that we have focused on the part of the waveform over[–αT, (1–α)T]. By making use of equation 22.20b, we obtain
f ''( t) =2A
(1− )T 2 cos(n 0t + 2n π) − cos(n 0t)[ ]n=1
∞∑
=−4A
(1− )T 2 sin(n 0t + n π)sin(n π)[ ]n=1
∞∑
Therefore,
f ( t) = f ( t)[ ]ave +4A
(1− )T 2sin(n π)
n 0( )2 sin(n 0t + n π)
n=1
∞∑
=A2
+A sin(n π)
n2π 2 (1− )cos n 0t + (n − 0.5)π( )
n=1
∞∑
Letting T = 1 and α = 0.25 we obtain,
f ( t) =A2
+16A sin n
π4
3n2π 2 cos n2πt + (0.25n − 0.5)π( )
n=1
∞∑
Therefore, d0 = 0.5A , and
dn =16Asin n
π4
3n2π 2 .
It follows that d1 = 0.38211A and d2 = 0.13509A.
SOLUTION TO PROBLEM 22.12.
Denote by fp(t, α) the period rectangular waveform of figure 22.5, with A=1.
Then we can express the present f(t), with T= 4, as the sum of 3 terms:
f(t) = 3fp(t-0.125T, 0.25) + 4 fp(t - 0.5T, 0.5) -2
From equation 22.14b, and equation 22.12c, for n= 1,2,...
cn = 3πn
sin(0.25nπ)e-jnω0x0.125T+ 4πn
sin(0.5nπ)e-jnω0x0.5T
= 3πn
sin(0.25nπ)e-j0.25nπ+ 4πn
sin(0.5nπ)e-jnπ
The numerical values of the first few Fourier series coefficients are:
c0 = averge value of f(t) = 14
(1 + 4 - 2) = 0.75
c1= 3π
sin(0.25π)e-j0.25π+ 4π
sin(0.5π)e-jπ = -0.7958 -j 0.4775
c2= 32π
sin(0.5π)e-j0.5π+ 42π
sin(π)e-j2π =-j0.4775
Solution Problem 22.13
(a) For sinusoidal steady analysis, the transfer function is
H(jω) = YL YL +YC + YR
= 1 1 +ZLYC + ZLYR
= 1(1 - ω2LC) + jωL
R
= 1
(1 - 4×10 -5ω2) + j×10--3ω
The transfer function evaluated at various input frequencies are listed below.
H(0) = 1
H(j377) = 0.2128 /-175.4o
H(j3x377) = 0.0199/-178.7o
H(j5x377) = 0.0071/-179.2o
Using equation 15.7 and superposition, we obtain the steady state output voltage (in V):vout(t) = 200 + 200× 2×0.2128cos (377t -175.4o) +60× 2×0.0199cos (3×377t +30o -178.7o)
+80× 2×0.0071cos (5×377t +50o -17.2o)
= 200 + 2×42.55cos (377t -175.4o) + 2×1.196cos (3×377t -148.7o)
+ 2×0.5668cos (5×377t -129.2o)
(b) From equation derived in P11.39,
Vout,eff= 2002+42.552+ 1.1962 +0.56682 = 204.48 V
The average power absorbed by the 10 kΩ resistor is
Pav=204.482
104 = 4.1812 W
Solution Problem 22.14
One correction in the problem statement: in the angle expression, 5000 should be 10000.
(a) Using the identity cos(x)cos(y)= 0.5cos(x+y) + 0.5cos(x-y), we have
vin(t) = 0.1cos (998,000t) +0.2cos (999,000t) + 2cos (1,000,000t)
+ 0.2cos (1,001,000t) + 0.1cos (1,002,000t) V
(b) The transfer function has a constant magnitude of 10, and a phase shift proportional to the deviationfrom ωc. at ω = ωc + 2ωm, the phase shift is -9 degrees. From these facts, we can write directly
vout(t) = cos (998,000t +9o) +2cos (999,000t + 4.5o) + 20cos (1,000,000t)
+ 2cos (1,001,000 - 4. 5o) + cos (1,002,000t - 9o) V
(c) Using the identity
cos(x)cos(y) = 2 cos(x+y
2)cos(
x+y2
)
we can group the terms in vout(t) and re-write it as
vout(t) = 2 cos( 2ωmt - 9o)cos(ω ct) +4cos( ωmt - 4.5o)cos(ω ct) + 20cos(ω ct)= [20 + 4cos(ωmt - 4.5o) +2 cos(2ωm - 9o)]cos(ω ct) = g(t)cos( ω ct)
Thus g(t) = 20 + 4cos( ωmt - 4.5 o) +2cos(2ωmt - 90)
With td = 78.54 µs, then ωmtd = 1000 x 78.54x10-6 = 0.07854 rad, or 4.5 degrees. We have
10 f(t - td) = 20[1 + 0.2cos( ωm(t - td) +0.1cos( 2ωm(t - td)]
= 20[1 + 0.2cos(ωmt - 4.5o) + 0.1 cos( 2ωmt - 9o)] = g(t)
SOLUTION PROBLEM 22.15. (a) This proof is a special case of the general proof given in the
solution to Problem 22.16. See the solution to problem 22.16, below.
FOR THE REMAINING PARTS, WE USE THE FOLLOWING MATLAB CODE:
% chapter 22, problem 15.%part (b).Vmax= 30*pi;Vmin=0;T=4;R = 1; C = 1;voutmin= Vmin+(Vmax - Vmin)/(1+ exp(0.5*T/(R*C)))voutmax = Vmax - (Vmax -Vmin)/(1+ exp(0.5*T/(R*C)))t1= 0: 0.05: 2;vseg1= Vmax+(voutmin- Vmax)*exp((-t1/(R*C)));t2= 2:0.05: 4;vseg2= Vmin+(voutmax- Vmin)*exp(-(t2-T/2)/(R*C));t= [ t1 t2];v= [vseg1 vseg2];plot(t,v)gridxlabel('t in seconds')ylabel('Steady State Output Voltage in V')
%part (c)error1= 100*(12.235- voutmin)/voutminerror2= 100*(82.013 - voutmax)/voutmax
TO OBTAIN
(b)voutmin = 1.1235e+01voutmax = 8.3013e+01
(c)error1 = 8.9045e+00error2 = -1.2048e+00
SOLUTION PROBLEM 22.16. CORRECTION: In the problem statement, s
should be s .
For simplicity, let us consider the case when the transfer function is a voltage ratio, i.e., H(s) = Vout./Vin.
If a constant input vin(t) = Vcon is applied to the stable network, then the vout(∞) = KVcon ,
independent of the initial conditions. This is because the zero-input response for a stable network
approaches zero as t approaches infinite. To see this observe that the zero-state response is given by
vout (t) = L−1 Ks +1
×Vcon
s
= L−1 KVcons
−KVcons +1
= KVcon 1− e−t( )u( t)
from which vout(∞) = KVcon as asserted.
For the remainder of our proof we make use of the fact that in steady state, vout(t+T) = vout(t)
with t = 0 in our case. Specifically, after the first order network has reached steady state, the vout(t)
waveform will be periodic as shown in figure P22.15c, where the time reference has been chosen so thatvout,min occurs at t = 0.
Recall equation 8.19
x(t) = x(∞) + x(t0+ ) − x(∞)[ ]e−(t−t0 )
Applying this equation to the interval [0, T/2], we have
b0 = KVmax + a0 − KVmax( )e−0.5T / ≡ KVmax + a0 − KVmax( ) (1)
Note that = e−0.5T / . Similarly, applying equation 8.19 to the interval [ T/2, T] leads to
a0 = KVmin + b0 − KVmn( ) (2)
Equations (1) and (2) can be written as a single matrix equation
− 1
1 −
a0
b0
=
(1− )KVmax
(1− )KVmin
(3)
Solving equation (3) by matrix inverse (or Cramer's rule) results in
a0
b0
=
−12 −1
1
1
(1− )KVmax
(1− )KVmin
=
K+1
1
1
Vmax
Vmin
=
K+1
Vmax + Vmin
Vmax + Vmin
vout ( t)[ ]min = a0 = KVmax + Vmin
+1= K
Vmax + Vmin + Vmin − Vmin+1
= KVmin +Vmax −Vmin
+1= KVmin + K
Vmax −Vmin1+1/
= KVmin + KVmax − Vmin
1+ e0.5T /
and
vout ( t)[ ]max = b0 = KVmax + Vmin
+1= K
Vmax + Vmin + Vmax − Vmax+1
= KVmax − KVmax − Vmin
1 +1/= KVmax − K
Vmax − Vmin
1+ e0.5T /
This complete completes the derivation of the desired formulas.
SOLUTION TO PROBLEM 22.17
There is one correction in the problem statement. 1-kHz should be 5.1mHz.(a) The transfer function is, according to equation 4.3, and using the given element values R1 = 10 kΩ,
Rf = 50 kΩ and C = 20 mF:
H(s) = VoutVin
= - ZfZ1
= - Y1Yf
= -
1R1
Cs + 1Rf
= -
RfR1
1 + RfCs = -5
1 + 1000s
(b) Here we have We have K= -5, t = 1000 s, Vmax = 1 V, Vmin = -1 V and T = 1/f = 1/0.0051=
196.15 s. In using the equations derived in problem 22.16, we note that the subscripts min and max
should be switched in this case because K is negative. The following MATLAB codes perform the
needed numerical calculations.
f= 5.1e-3;Cf=20e-3;Rf=50e3;Rs= 10e3;T=1/f;K= -Rf/Rs;tau=Rf*Cf;['part (b)']Vmax=1;Vmin=-1;voutmax= K*( Vmin + (Vmax-Vmin)/(1+exp(0.5*T/tau)))voutmin= K*( Vmax - (Vmax-Vmin)/(1+exp(0.5*T/tau)))
answers from MATLAB are:
voutmax = 0.2449voutmin = -0.2449
(c)['part (c)']
t1= 0:0.005*T:0.5*T;v1= 5 + (-0.245 -5).*exp(-(t1/tau)) ;t2= 0.5*T:0.005*T:T;v2= -5 + (0.245 +5).*exp(-(t2-0.5*T)/tau) ;t=[t1,t2];v=[v1,v2];plot (t,v)xlabel('time in seconds')ylabel('vout in V')grid
-0.3
-0.2
-0.1
0
0.1
0.2
0.3
0 50 100 150 200
time in seconds
vout
in V
SOLUTION PROBLEM 22.18.
vi(t) = cos (ωt) = cos(6000t)
vo(t) = 10v i + vi2 + vi
3 = 10cos(ωt) + cos2(ωt) + cos3(ωt)
= 10cos(ωt) +0.5 +0.5cos(2ωt) + 0.75cos(ωt) + 0.25cos(3ωt)
= 0.5 + 10.75 cos(ωt) + 0.5cos(2ωt) 0.25cos(3ωt)
The effective values of various components of the output are:
dc fundamental 2nd harmonic 3rd harmonic total harmonic
0.5 10.75/√2 0.5/√2 0.25/√2 0.559/√2
The total harmonic distortion is0.55910.75
× 100% = 5.2%
and the average power at the fundamental frequency is
0.5× 10.752
100 = 0.5778 W
SOLUTION TO PROBLEM 22.19.
vo(t) = V0 + V1 cos(ωt) + V2cos(2ωt) + V3cos(3ωt)
We use equation 22.30 and the values of vo(t) read from the oscilloscope
to compute Vk. The total harmonic distortion is given by
H.D. = V2
2 + V32
V1×100 %
MATLAb codes:vo0= 10; vo60= 5.2; vo120= -4.6; vo180= -9.6;V0= (vo0+2*vo60+2*vo120 +vo180)/6V1= (vo0+ vo60 - vo120 -vo180)/3V2= (vo0 - vo60 - vo120 +vo180)/3V3= (vo0-2*vo60+2*vo120 -vo180)/6% total harmonic distortionHD= 100*sqrt( V2^2 +V3^2)/V1
The following answers are obtained frm MATLAB output.
V0 = 0.2667 V
V1 = 9.8000 V
V2 = -0.0667
V3 = 0 V
HD = 0.6803 (percent)
SOLUTION PROBLEM 22.20.
We shall follow the solution given in example 22.12, and only indicate the needed changes below. There are twocorrections in example 22.12: (1) 20cos(ωαT) should be 20cos(0.5ωαT) and (2) 1.842cos(ωt) should be 1.842
cos(3ωt).
The new input is
vi(t) = 0.9 cos(6000t) The positive peak of the output sine wave is clipped for αT.
0.9 ×20cos(0.5ωαT) = 15 → απ = cos-1 (15/18) → α = 0.1864
Equation 22.35 becomes
Equations 2.36 becomes
Using A=1, and α = 0.1864 in item #2 of table 22.4, equation 22.37 becomes
and equation 22.38 becomes
Equation 22.39 becomes
from which the harmonic distortions are:
third order 1.076/16.57 = 6.595%
fifth order 0.55/16.57 = 3.319%
SOLUTION PROBLEM 22.21. Corrections: (1) cos( 1)e
– 2− 1+πRC = cos( 2) should be
cos( 1)e− – 2 − 1+π
RC
= cos( 2) and (2) on page 943, equation 22.46 , θ2 should be –θ2, i.e., equation
22.46 should be cos( 1)e− – 2 − 1+π
RC
.
The proof is similar to that given on page 943 for the half-wave rectifier case, except for some
minor changes described below.
For the case of a full-wave rectifier, the output voltage waveform is a modification of figure 22.17
as shown below.
The exponential decay of vo(t) starts with the value Vmcos(θ1) at θ = θ1 – π (instead of θ1 – 2π, as in the
half-wave rectifier case). Therefore, in equation 22.45, change T to T/2 and 2π to π. In other words
v0 t( ) = Vm cos( 1)( )e−
t − 1 −T
2
RC
= Vm cos( 1)( )e−
t − 1−( )RC
In equation 22.46, make correction (2) above, and change 2π to π, i.e.,
cos( 1)e− – 2 − 1+π
RC
= cos( 2)
The desired proof is complete.
SOLUTION PROBLEM P22.22
Refer to figure 22.17. Assuming θ1 = 0, we have
v0(t) = Vmexp( -tRC
) = Vm 1 - ( tRC
) + ( tRC
)2 - ( tRC
)3 + ...
Assuming θ2 = 0, we compute the average value of vo(t) over the time interval [ 0, T]. For the case
RC >> T, we can approximate vo(t) over this interval by keeping only the first two terms of the infinite
series. Thus
v0(t) ≅ Vm 1 - tRC
which indicates that the plot of vo vs. t over the interval [ 0 T] is approximately a straight line.
Therefore the average of the vo(t) over [ 0 T] is equal to vo(T/2). Thus, for the case RC >> T, or
equivalently ωRC >> 2π,
Vdc ≅ voT2
= Vm 1 - T2RC
= Vm 1 - πωRC
SOLUTION PROBLEM 22.23Given values are: C = 20e-6 F, R =100kΩ, Vm = 20 V; f = 60 Hz. From equation 22.48b
Vdc ≅ (1 - πωRC
)Vm = (1 - 12fRC
)Vm = (1 - 12×60×105×20×10-6
)×20 = 19.916 V
To calculate the ripple factor, we first calculate θ2 from equation 22.47,
and then use the result in equation 22.50.
θ2 = cos-1( 2πRC
) = 0.1289 rad
ripple factor ≅ 1 - cos(θ2)
3 [ 1 +cos(θ2)] =0.2406%
For the diode average and peak currents, use equations 22.49a and 22.52
Idc = VdcR
= 19.917100,000
= 0.199 ×10-3 A
id,peak ≅ VmωCsin(θ2) = 19.4×10-3 A
SOLUTION PROBLEM 22.24
Equations 22.44 - 22.52 are derived for a half-wave rectifier. For a full-wave rectifier, some of these equations will
be modified slightly as given below. The difference arises from changing T to T/2.The new equation for θ2 is derived in problem 22.21, and repeated below.
The given values are: C = 20e-6 F, R =100kΩ, Vm = 20 V; f = 60 Hz.
A modification of equation 22.48b gives
Vdc = (1 - π2ωRC
)Vm = (1 - T4RC
)Vm = (1 - 14×60×105×20×10-6
)×20 = 19.96 V
To calculate the ripple factor, we first calculate θ2 from equation derived in problem 22.21, and then use the result
in equation 22.50.
θ2 ≅ cos-1(e- πωRC) =5.23 degrees
ripple factor ≅ 1 - cos(θ2)
3 [ 1 +cos(θ2)] =0.12%
For the diode average and peak currents, use equations 22.49a and 22.52
Idc = VdcR
= 19.96100,000
= 0.1996 ×10-3 A
id,peak ≅ VmωCsin(θ2) = 13.75×10-3 A
_
SOLUTION PROBLEM 22.25. CORRECTIONS: (1) 195 Ω should be 1950 Ω. (2) 100 µF should
be 10 µF.
Since this problem only requires an estimate of the answer, we can use reasonable approximations to simply
the solution. Let H(s) = Vo(s) I(s) be the transfer function for the linear circuit to the right of the diodes. Then,
H(s) =Vin (s)I(s)
×Vo(s)Vin (s)
= Zin (s)Zpar(s)
R + Zpar(s)
where Zpar(s) is the impedance of the parallel C-RL.
The first step is to find the magnitude H ( j ) . As long as 1/RL << ωC and 1/(ωC) << R, the
following approximations are valid:
Zpar( j ) =1
j C +1
RL
≅1C
which means that the parallel impedance is essentially that of the capacitor, and
Zpar( j )
R + Zpar ( j )≅
Zpar ( j )
R≅
1RC
and again since 1/(ωC) << R,
Zin ( j ) ≅1C
Using these approximations,
H ( j ) = Zin ( j ) ×Zpar ( j )
R + Zpar ( j )≅
1C
×1RC
=R
RC( )2
It is given that Idc = 0.01 A. From the short pulse property, the input current i(t) consists of very short pulses at
120 Hz and all ac components of i(t) have peak magnitudes approximately equal to twice the average dc value.
Hence the peak magnitudes are 0.02 A. Therefore the magnitude of 120-Hz component of the output voltage is
H ( j2π ×120) × 0.02 ≅R
RC( )2 × 0.02 =1950
2π ×120 ×1950 ×10 ×10−6( )2 × 0.02 = 0.18042 V
Hence, the effective value is 0.18042 2 = 0.12757 V and the ripple factor is
0.1275730
×100 = 0.42524 %
SOLUTION PROBLEM 22.26
Since this problem only requires an estimate of the answer, we can use reasonable approximations
to simply the solution. Let H(s) = Vo(s) I(s) be the transfer function for the linear circuit to the right of
the diodes. Then
H(s) =VoI
= VinI
× VoVin
Under the condition 1/(ωC) << R and 1/(ωC) << RL, we have
Zin(s) ≅ ZC(s) = 1Cs
and VoVin
≅ 1ωRC
2
Using these approximations,
H(jω) ≅Zin VoutVin
= 1ωC
( 1ωRC
)2 = R( 1ωRC
)3
For a full-wave rectifier, the fundamental frequency is 120 Hz, and ω = 240π rad/s. It is given that Idc
= 0.01 A. From the short pulse property, the input current i(t) consists of very short pulses at 120 Hz
and all ac components of i(t) have peak magnitudes approximately equal to twice the average dc value.
Hence the peak magnitudes are 0.02 A. Therefore the peak magnitude of 120-Hz component of the
output voltage is
H(j120π) x0.02 ≅ = R( 1ωRC
)3×0.02= 975×( 1120π×975×16×10-6)
3× 0.02 = 0.012 V
Hence, the effective value is 0.012/ 2 =0.0848 V and the ripple factor is
0.084830
×100% = 0.2827 %
SOLUTION PROBLEM 22.27Using H(jω) of figure P22.27 and the given formula for cn , we have
cn = 1ω s
H(jω) e-jnT0ωdω = 1ω s
e-jnT0ωdω = 1-jnT0ω s
[ e-jnT0ω]-0.25 ωs0.25ωs
-0.25 ωs
0.25ωs
-0.5 ωs
0.5ωs
= 1nπ
sin (nπ2
), n = 1,2,...
For n = 0, we have c0= d0 = a0 /2 = [Average value of H(jω) ]= 0.5 .
For n = 1,2,..., from equation 22.6
bn = 0, dn=an = 2
nπ sin (nπ2
),
The Fourier series representation of H(jω) is
H(jω) = 0.5 + 2π cos(T0ω) - - 1
3cos(3T0ω) + - 1
3cos(5T0ω) - - 1
7cos(7T0ω + ...
A plot of H(jω) vs. ω curve using the first 11 terms (n=0,1,...11) of the Fourier
series is given below together with the MATLAB codes.
-0.5
0
0.5
1
1.5
0 20 40 60 80 100
w in radians
H(j
w)
%chapter 2, problem 27.ws= 100;T0=2*pi/ws;w= 0: ws/200:ws;d0w = 0.5;d1w = (2/pi)*cos(T0*w);d3w= -(2/pi/3)*cos(3*T0*w);d5w= (2/pi/5)*cos(5*T0*w);d7w= -(2/pi/7)*cos(7*T0*w);d9w= +(2/pi/9)*cos(9*T0*w);d11w= -(2/pi/10)*cos(11*T0*w);H =d0w +d1w + d3w +d5w +d7w + d9w +d11w;plot(w,H)xlabel(' w in radians')ylabel('H(jw)')grid