ELEC273 Lecture Notes Set 11 AC Circuit Theoremstrueman/elec273files/ELEC273_11_… · • Find the...
Transcript of ELEC273 Lecture Notes Set 11 AC Circuit Theoremstrueman/elec273files/ELEC273_11_… · • Find the...
ELEC273 Lecture Notes Set 11 AC Circuit Theorems
Homework on phasors and impedance.
The course web site is: http://users.encs.concordia.ca/~trueman/web_page_273.htmFinal Exam (confirmed): Friday December 15, 2017 from 9:00 to 12:00 (confirmed)
The final exam in ELEC 273 is December 14, 2018 from 2:00 to 5:00.
Network Theorems:
Chapter 10 Sinusoidal Steady State Analysis• Superposition• Thevenin’s Theorem• Norton’s Theorem
Chapter 11 AC Power Analysis• The Maximum Power Transfer Theorem
Thevenin’ s Theorem for AC Circuits
Thevenin’s TheoremAny two-terminal network consisting of voltage sources, current sources, dependent sources, inductors, capacitors and resistors is equivalent to a voltage source 𝑉𝑉𝑇𝑇 in series with an impedance 𝑍𝑍𝑇𝑇.
Procedure to find the Thevenin Equivalent Circuit:1.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, 𝑉𝑉𝑜𝑜𝑜𝑜.2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, 𝐼𝐼𝑠𝑠𝑜𝑜. 3.The Thevenin equivalent voltage source is 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜.4.The Thevenin equivalent impedance is 𝑍𝑍𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜
𝐼𝐼𝑠𝑠𝑜𝑜.
Note that 𝑍𝑍𝑇𝑇 can also be found as the impedance of the “dead circuit” with the independent sources set equal to zero.
TwoTerminal Circuit
𝑉𝑉𝑇𝑇
𝑍𝑍𝑇𝑇
Example: Thevenin’ s Theorem
1)Find the Thevenin Equivalent Circuit at a frequency of 𝜔𝜔=3 rad/sec. Use phasors relative to “sine”.
2)Find the voltage across a load consisting of a 1 ohm resistor in series with a capacitor of value C=0.16667 F, using: • The Thevenin equivalent circuit• The original circuitThe answer must be the same in both cases!
2 Ω
1 Ω
12
H10 sin 3𝑡𝑡
A
B
Find the Thevenin Equivalent Circuit.
1) Find the open-circuit voltage.2) Find the short-circuit current.
Convert the circuit to phasors and impedance at frequency 𝜔𝜔 = 3 rad/sec.The impedance of the inductance is 𝑗𝑗𝜔𝜔𝜔𝜔 = j3x 1
2= 𝑗𝑗𝑗.5 ohms
In this example I have written my phasors relative to “sine”.
Hence 10 sin 3t becomes phasor 10 angle zero.
However, we must remember to convert phasors back to “sine”.
Thus phasor 𝑉𝑉 = 𝐴𝐴𝑒𝑒𝑗𝑗𝜃𝜃 converts to time function𝑣𝑣 𝑡𝑡 = 𝐴𝐴 sin(𝜔𝜔𝑡𝑡 + 𝜃𝜃).
2 Ω
1 Ω
12
H10 sin 3𝑡𝑡
A
B
2 Ω
1 Ω𝑗𝑗𝑗.5 Ω10
A
B
+𝑉𝑉𝑜𝑜𝑜𝑜−
Find the open-circuit voltage
015.12
10=−−
− ocococ VjVV
01
65.1
62
106 =−−− ocococ Vj
jVjVj
064)10(3 =−−− ocococ jVVVj
064330 =−−− ocococ jVVjVj
jjVV ococ 3094 =+
°∠=∠=∠
∠=
+= 24046.34182.0046.3
153.1849.9571.130
9430
jjVoc
°∠== 24046.3ocT VV
2 Ω
1 Ω𝑗𝑗𝑗.5 Ω10
A
B
+𝑉𝑉𝑜𝑜𝑜𝑜−
Solve for 𝑉𝑉𝑜𝑜𝑜𝑜:
2)Find the short-circuit current
°∠== 052
10scI
2 Ω
1 Ω𝑗𝑗𝑗.5 Ω10
A
B
𝐼𝐼𝑠𝑠𝑜𝑜
3)Find the impedance
°∠= 05scI°∠= 24046.3ocV
°∠== 24046.3ocT VV
2477.05565.00.246092.005
0.24046.3 jIVZ
sc
ocT +=∠=
∠∠
==
The reactance of 0.2477 ohms is equivalent to an inductor of value 𝜔𝜔𝜔𝜔 = 0.2477 so 𝜔𝜔 = 0.2477𝜔𝜔
= 0.24773
= 0.08257 H
This phasor is written relative to sine so the time function is 3.046 sin(𝜔𝜔𝑡𝑡 + 24°)
𝑉𝑉𝑇𝑇 = 3.046∠24°
𝑍𝑍𝑇𝑇 = 0.5565 + 𝑗𝑗𝑗.2477 Ω
𝑉𝑉𝑇𝑇 = 3.046∠24°
0.5565 Ω 0.08257 H
The impedance of the capacitor is 𝑍𝑍𝐶𝐶 = 1𝐽𝐽𝜔𝜔𝐶𝐶
= 1𝑗𝑗3𝑗𝑗0.16667
= −𝑗𝑗2 ohms
Draw the circuit in the frequency domain using phasors and impedances.
2)Find the voltage across a load consisting of a 1 ohm resistor in series with a capacitor of value C=0.16667 F, using: • The original circuit• The Thevenin equivalent circuit
Using the original circuit:
𝑗𝑗𝑗.5 Ω
2 Ω
1 Ω
12
H10 sin 3𝑡𝑡
A
B
1 Ω
0.16667 F
2 Ω
1 Ω10
B
1 Ω
−𝑗𝑗2 Ω
Using the original circuit, find 𝑉𝑉𝐿𝐿:
02115.12
10=
−−−−
−j
VVjVV LLLL
Write a node equation:
210
2115.12=
−+++
jVV
jVV LLLL
( ) ( ) ( )( ) 5
21)5.1(221)5.1(2212215.1=
−+−+−+−
jjjjjjjVL
536
5.311=
++
jjVL
°∠=++
= 9.8906.25.311
365jjVL
Do we get the same answer using the Thevenin Equivalent Circuit?
𝑗𝑗𝑗.5 Ω2 Ω
1 Ω10
B
1 Ω
−𝑗𝑗2 Ω
𝑉𝑉𝐿𝐿
2477.05565.00.246092.0 jZT +=∠=
Find the load voltage using the Thevenin Equivalent Circuit. Define 𝑍𝑍𝐿𝐿 = 1 − 𝑗𝑗2 Ω
°∠== 24046.3ocT VV
0=−−
L
L
T
LT
ZV
ZVV
T
T
LTL Z
VZZ
V =
+
11
T
T
LT
LTL Z
VZZ
ZZV =
+
LT
TLL ZZ
VZV+
=
21 jZL −=
)21(2477.05565.0)0.24046.3)(21(jj
jVL −++∠−
=
7523.15565.1)0.24046.3)(4.63236.2(
jVL −
∠−∠=
8.48343.2)0.24046.3)(4.63236.2(
−∠∠−∠
=LV
°∠= 915.8906.2LV
°∠= 9.8906.2LVThis agrees with the answer obtained above of
Note that this equation is simply a voltage divider between 𝑍𝑍𝑇𝑇 and 𝑍𝑍𝐿𝐿.
𝑉𝑉𝑇𝑇𝑍𝑍𝑇𝑇
𝑉𝑉𝐿𝐿
−𝑗𝑗2 Ω
1 Ω
Norton’s Theorem for AC Circuits.
Norton’s TheoremAny two-terminal network consisting of voltage sources, current sources, dependent sources, capacitors, inductors and resistors is equivalent to a current source 𝐼𝐼𝑁𝑁 in parallel with an admittance 𝑌𝑌𝑁𝑁.
Procedure to find the Norton Equivalent Circuit:1.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, 𝑉𝑉𝑜𝑜𝑜𝑜.2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, 𝐼𝐼𝑠𝑠𝑜𝑜. 3.The Norton equivalent current source is 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜.4.The Norton equivalent susceptance is 𝑌𝑌𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜
𝑉𝑉𝑜𝑜𝑜𝑜.
Note that 𝑌𝑌𝑁𝑁 can also be found as the admittance of the circuit with the independent sources set equal to zero.
Two TerminalCircuit
A
B
A
B
𝐼𝐼𝑁𝑁 𝑌𝑌𝑁𝑁
Norton Equivalent from the Thevenin EquivalentFind the open-circuit voltage 𝑉𝑉𝑜𝑜𝑜𝑜 and the short-circuit current 𝐼𝐼𝑠𝑠𝑜𝑜.
Thevenin: 𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜𝑍𝑍𝑇𝑇 =
𝑉𝑉𝑜𝑜𝑜𝑜𝐼𝐼𝑠𝑠𝑜𝑜
Norton:𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝐶𝐶𝑌𝑌𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜
𝑉𝑉𝑜𝑜𝑜𝑜
We can change The Thevenin Equivalent Circuit to the Norton Equivalent Circuits using:𝑌𝑌𝑁𝑁 = 1
𝑍𝑍𝑇𝑇and𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝐶𝐶 = 𝑉𝑉𝑇𝑇
𝑍𝑍𝑇𝑇
𝑉𝑉𝑇𝑇
𝑍𝑍𝑇𝑇
𝐼𝐼𝑁𝑁 𝑌𝑌𝑁𝑁
𝑉𝑉𝑇𝑇
𝑍𝑍𝑇𝑇
𝐼𝐼𝑆𝑆𝐶𝐶 =𝑉𝑉𝑇𝑇𝑍𝑍𝑇𝑇
Example: Find the Norton Equivalent Circuit
Find the Norton Equivalent Circuit at a frequency of 𝜔𝜔=3 rad/sec.
Method:1.Find the open-circuit voltage.
2.Find the short-circuit current.
3.Find the Norton Equivalent Circuit:𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑆𝑆𝐶𝐶𝑌𝑌𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜
𝑉𝑉𝑜𝑜𝑜𝑜
We found the Thevenin Equivalent Circuit earlier in this set of notes.
Open-circuit “load”.
Short-circuit “load”.
2 Ω
1 Ω12
H10 sin 3𝑡𝑡
A
B
2 Ω
1 Ω12
H10 sin 3𝑡𝑡
A
B
+𝑉𝑉𝑜𝑜𝑜𝑜−
𝐼𝐼𝑠𝑠𝑜𝑜10 sin 3𝑡𝑡
2 Ω12
H1 Ω
A
B
Find the open-circuit voltage:
1) Find the open-circuit voltage
015.12
10=−−
− ocococ VjVV
Convert the circuit to phasors and impedance.The impedance of the inductance is 𝑗𝑗𝜔𝜔𝜔𝜔 = j3x 1
2= 𝑗𝑗𝑗.5 ohms
°∠== 24046.3ocT VV
We solved for the open-circuit voltage and the short-circuit current when we found the Thevenin equivalent earlier in these notes.
2 Ω
1 Ω
12
H10 sin 3𝑡𝑡
A
B
2 Ω
1 Ω𝑗𝑗𝑗.5 Ω10
A
B
+𝑉𝑉𝑜𝑜𝑜𝑜−
Find the short-circuit current:
°∠== 052
10scI
2 Ω
1 Ω𝑗𝑗𝑗.5 Ω10
A
B
𝐼𝐼𝑠𝑠𝑜𝑜
3)Find the admittance:
°∠= 05scI°∠= 24046.3ocV
°∠== 05scN II
°−∠=∠∠
== 0.24641.10.24046.3
05oc
scN V
IY
2 Ω
1 Ω𝑗𝑗𝑗.5 Ω10
A
B
A
B
𝐼𝐼𝑁𝑁 𝑌𝑌𝑁𝑁
Norton Equivalent, example #2:
• Find the Norton Equivalent Circuit at 60 Hz.• Convert the Norton Equivalent to the Thevenin Equivalent Circuit.
Procedure to find the Norton Equivalent Circuit:1.Open-circuit test: terminate the circuit with an open circuit and find the open-circuit voltage, 𝑉𝑉𝑜𝑜𝑜𝑜.2.Short-circuit test: terminate the circuit with a short circuit and find the short-circuit current, 𝐼𝐼𝑠𝑠𝑜𝑜. 3.The Norton equivalent current source is 𝐼𝐼𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜.4.The Norton equivalent susceptance is 𝑌𝑌𝑁𝑁 = 𝐼𝐼𝑠𝑠𝑜𝑜
𝑉𝑉𝑜𝑜𝑜𝑜.
110 cos𝜔𝜔𝑡𝑡1 Ω
10.62 mH
5.31 mH
1330 𝜇𝜇F
1 Ω A
B
Convert to phasors and impedances:
Voltage source:𝑣𝑣𝑠𝑠 𝑡𝑡 = 110 cos𝜔𝜔𝑡𝑡 becomes phasor 𝑉𝑉𝑠𝑠 = 110 volts
At 𝑓𝑓 =60 Hz the radian frequency is 𝜔𝜔 = 2𝜋𝜋𝑓𝑓 = 377 r/s
𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝑗𝑗3𝑗𝑗𝑗𝑗𝑗.3𝑗𝑗𝑗10−3 = 𝑗𝑗2 ohms
𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝑗𝑗3𝑗𝑗𝑗𝑗10.62𝑗𝑗10−3 = 𝑗𝑗4 ohms
1𝑗𝑗𝜔𝜔𝐶𝐶
= −𝑗𝑗𝜔𝜔𝐶𝐶
= −𝑗𝑗377𝑗𝑗1330𝑗𝑗10−6
= −𝑗𝑗2 ohms
110 cos𝜔𝜔𝑡𝑡1 Ω
10.62 mH
5.31 mH
1330 𝜇𝜇F
1 Ω A
B
1.Find the open-circuit voltage:
𝑉𝑉𝑜𝑜𝑜𝑜5 + 𝑗𝑗28 − 𝑗𝑗2 = 110
𝑉𝑉𝑜𝑜𝑜𝑜 = 1108 − 𝑗𝑗25 + 2𝑗𝑗
𝑉𝑉𝑜𝑜𝑜𝑜 = 1108.25∠ − 14.0°5.39∠ − 21.8°
𝑉𝑉𝑜𝑜𝑜𝑜 = 168.4∠ − 35.8°
110 − 𝑉𝑉𝑜𝑜𝑜𝑜1 + 𝑗𝑗2 −
𝑉𝑉𝑜𝑜𝑜𝑜1 + 𝑗𝑗4 −
𝑉𝑉𝑜𝑜𝑜𝑜−𝑗𝑗2 = 0
𝑉𝑉𝑜𝑜𝑜𝑜1
1 + 𝑗𝑗2 +1
1 + 𝑗𝑗4 +1−𝑗𝑗2 =
1101 + 𝑗𝑗2
𝑉𝑉𝑜𝑜𝑜𝑜1 + 𝑗𝑗4 −𝑗𝑗2 + 1 + 𝑗𝑗2 −𝑗𝑗2 + (1 + 𝑗𝑗2)(1 + 𝑗𝑗4)
(1 + 𝑗𝑗2)(1 + 𝑗𝑗4)(−𝑗𝑗2) =110
1 + 𝑗𝑗2
𝑉𝑉𝑜𝑜𝑜𝑜1 + 𝑗𝑗4 −𝑗𝑗2 + 1 + 𝑗𝑗2 −𝑗𝑗2 + (1 + 𝑗𝑗2)(1 + 𝑗𝑗4)
(1 + 𝑗𝑗4)(−𝑗𝑗2) = 110
𝑉𝑉𝑜𝑜𝑜𝑜−2𝑗𝑗 + 8 − 2𝑗𝑗 + 4 + 1 + 𝑗𝑗2 + 𝑗𝑗4 − 8
−𝑗𝑗2 + 8 = 110
𝑉𝑉𝑜𝑜𝑜𝑜5 + 𝑗𝑗28 − 𝑗𝑗2 = 110
𝑉𝑉𝑜𝑜𝑜𝑜
+
-
𝑉𝑉𝑜𝑜𝑜𝑜
2.Find the short-circuit current:
𝐼𝐼𝑠𝑠𝑜𝑜 =110
1 + 𝑗𝑗2 = 22 − j44 = 49.19∠ − 63.4°
𝐼𝐼𝑠𝑠𝑜𝑜
3 and 4.Find the Norton Equivalent Circuit:
°−∠== 4.6319.49scN II
6.272921.08.354.1684.6319.49
−∠=°−∠°−∠
==oc
scN V
IY
𝐼𝐼𝑠𝑠𝑜𝑜 =110
1 + 𝑗𝑗2= 22 − j44 = 49.19∠ − 63.4°
𝑉𝑉𝑜𝑜𝑜𝑜 = 168.4∠ − 35.8°
Siemens
Amps
A
B
𝐼𝐼𝑁𝑁 𝑌𝑌𝑁𝑁
Check by finding the dead-circuit impedance:
𝑍𝑍𝑇𝑇
𝑍𝑍𝑇𝑇
1 + 𝑗𝑗4 ∥ −𝑗𝑗2 = 1+𝑗𝑗4 −𝑗𝑗21+𝑗𝑗4−𝑗𝑗2
= 0.8 − 𝑗𝑗3.6 ohms
𝑍𝑍𝑇𝑇 = 1 + 𝑗𝑗2 ∥ 0.8 − 𝑗𝑗3.6 =1 + 𝑗𝑗2 0.8 − 𝑗𝑗3.61 + 𝑗𝑗2 + 0.8 − 𝑗𝑗3.6
𝑍𝑍𝑇𝑇 = 3.034 + 𝑗𝑗1.586 = 3.423∠27.6° ohms
𝑌𝑌𝑁𝑁 = 1𝑍𝑍𝑇𝑇
= 13.423∠27.6°
= 0.2920∠ − 27.6° Siemens
This agrees with the value found on the previous slide.
Convert the Norton Equivalent to the Thevenin Equivalent Circuit:
°−∠= 4.6319.49NI°−∠= 6.272921.0NY
𝑉𝑉𝑜𝑜𝑜𝑜 = 𝐼𝐼𝑁𝑁𝑌𝑌𝑁𝑁
= 49.19∠−63.4°0.2921∠−27.6°
= 136.6 − 𝑗𝑗𝑗𝑗.51 = 168.4∠ − 35.8° volts
𝑉𝑉𝑇𝑇 = 𝑉𝑉𝑜𝑜𝑜𝑜 = 98.51 = 168.4∠ − 35.8°𝑍𝑍𝑇𝑇 = 1
𝑌𝑌𝑁𝑁= 1
0.2921∠−27.6°= 3.423∠27.6° ohms
+𝑉𝑉𝑜𝑜𝑜𝑜-
The Superposition Theorem for AC Circuits
The Superposition Theorem is exactly the same for AC circuits as for DC circuits.
The response of a circuit 𝑉𝑉0 with two sources 𝑉𝑉𝑠𝑠1 and 𝑉𝑉𝑠𝑠2 acting together is equal to the sum of:• the response 𝑉𝑉01 with source 𝑉𝑉𝑠𝑠1 acting alone and 𝑉𝑉𝑠𝑠2=0, • plus the response 𝑉𝑉02 with source 𝑉𝑉𝑠𝑠2 acting alone and 𝑉𝑉𝑠𝑠1=0,𝑉𝑉0 = 𝑉𝑉01 + 𝑉𝑉02.
Two sources acting together. Source #1 acting alone. Source #2 acting alone. Two sources acting together. Source #1 acting alone. Source #2 acting alone.
𝑉𝑉𝑠𝑠1 𝑉𝑉𝑠𝑠2
+𝑉𝑉𝑜𝑜 −
Linear Circuit 𝑉𝑉𝑠𝑠2
+𝑉𝑉𝑜𝑜2 −
Linear Circuit𝑉𝑉𝑠𝑠1
+𝑉𝑉𝑜𝑜1 −
Linear Circuit
Superposition: DC Sources and AC Source
Use Superposition to separate this problem into:1)The DC response 𝒗𝒗𝑫𝑫𝑫𝑫 to the DC sources 𝑉𝑉2 = 3 volts and 𝐼𝐼1 = 1 amp2)The AC response 𝒗𝒗𝑨𝑨𝑫𝑫(t) to the AC source 𝑣𝑣1 𝑡𝑡 = 0.1 cos(𝑗𝑗𝑗𝑗𝑡𝑡) volts at 𝜔𝜔=1000 r/sThen the response of the circuit is𝒗𝒗 𝒕𝒕 = 𝒗𝒗𝑫𝑫𝑫𝑫 + 𝒗𝒗𝑨𝑨𝑫𝑫(t)
In the circuit below there are two DC sources, 𝑉𝑉2 = 3 volts and 𝐼𝐼1 = 1 amp. There is also an AC source, 𝑣𝑣1 𝑡𝑡 = 0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡 volts.
Use Superposition to find the voltage across the current generator 𝑣𝑣 𝑡𝑡 .
The component values are 𝑅𝑅1 = 1 Ω, 𝑅𝑅2 = 2 Ω, 𝐶𝐶1 = 500 microFarads, 𝐶𝐶2 = 40 microFarads, 𝜔𝜔 = 1 mH.
0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡
𝐶𝐶1 = 500 𝜇𝜇F
𝑅𝑅1 = 1 Ω
𝐶𝐶2 = 40 𝜇𝜇F 𝑅𝑅2 = 2 Ω 𝜔𝜔 = 1 mH
5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−
𝑉𝑉2 = 3 volts𝐼𝐼1 = 1 amp+𝑣𝑣−
𝑣𝑣(𝑡𝑡)
Question from an old final exam!
Solve with the DC sources:
• The AC source is set to zero.• The capacitors become open circuits.• The inductor becomes a short circuit. • Since 𝐶𝐶1 and 𝐶𝐶2 are open-circuits at DC, there is zero DC current in 𝑅𝑅1 and
so 𝑣𝑣𝑗𝑗 = 0• Then the dependent source is zero: 5𝑣𝑣𝑗𝑗 = 0
Node equation:
−𝑣𝑣𝐷𝐷𝐶𝐶 − 3
2 − 1 = 03 − 𝑣𝑣𝐷𝐷𝐶𝐶 − 2 = 0
𝑣𝑣𝐷𝐷𝐶𝐶 = 1 volt at DC
Draw the DC circuit:• AC source becomes zero volts = short circuit• Capacitor: 𝑖𝑖 = 𝐶𝐶 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑= 0, C’s become open
circuits at DC. • Inductor: 𝑣𝑣 = 𝜔𝜔 𝑑𝑑𝑑𝑑
𝑑𝑑𝑑𝑑= 0, L’s become short
circuits at DC.• Note that 𝑣𝑣𝑗𝑗 = 0 at DC.
+𝑣𝑣𝑗𝑗−
5𝑣𝑣𝑗𝑗 12 Ω
3
3 volts0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡
𝐶𝐶1
𝑅𝑅1
𝐶𝐶2 𝑅𝑅2 𝜔𝜔
5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−
1 amp+𝑣𝑣−
𝑣𝑣(𝑡𝑡)
𝑣𝑣𝐷𝐷𝐶𝐶
Solve with the AC source: use phasors and impedance.
Draw the AC circuit.
0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡
𝐶𝐶1 = 500 𝜇𝜇F
1 Ω
𝐶𝐶2 = 40 𝜇𝜇F 2 Ω 1 mH
5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−
Use Superposition to separate this problem into:1)The DC response 𝑣𝑣𝐷𝐷𝐶𝐶 to the DC sources 𝑉𝑉2 = 3 volts and 𝐼𝐼1 = 1 amp2)The AC response 𝑣𝑣𝐴𝐴𝐶𝐶(𝑡𝑡) to the AC source 𝑣𝑣1 𝑡𝑡 = 0.1 cos(𝑗𝑗𝑗𝑗𝑡𝑡) volts at 𝜔𝜔=1000 r/s
+𝑣𝑣𝐴𝐴𝐶𝐶(t)−
0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡
𝐶𝐶1 = 500 𝜇𝜇F
𝑅𝑅1 = 1 Ω
𝐶𝐶2 = 40 𝜇𝜇F 𝑅𝑅2 = 2 Ω 𝜔𝜔 = 1 mH
5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−
𝑉𝑉2 = 3 volts𝐼𝐼1 = 1 amp+𝑣𝑣−
𝑣𝑣𝐴𝐴𝐶𝐶(𝑡𝑡)
Use phasors and impedance:Draw the AC circuit: • The DC sources are set to zero.• 𝑣𝑣1 𝑡𝑡 = 0.1 cos(𝑗𝑗𝑗𝑗𝑡𝑡)
becomes phasor 𝑉𝑉1 = 0.1• The frequency is 𝜔𝜔 = 1000 r/s.• 1
𝑗𝑗𝜔𝜔𝐶𝐶1= 1
𝑗𝑗𝑗𝑗1000𝑗𝑗𝑗00𝑗𝑗10−6= −𝑗𝑗2
• 1𝑗𝑗𝜔𝜔𝐶𝐶2
= 1𝑗𝑗𝑗𝑗1000𝑗𝑗40𝑗𝑗10−6
= −𝑗𝑗2𝑗• 𝑗𝑗𝜔𝜔𝜔𝜔 = 𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗𝑗10−3 = 𝑗𝑗𝑗
Node equations:
0.1 − 𝑉𝑉𝑗𝑗−𝑗𝑗2 −
𝑉𝑉𝑗𝑗1 −
𝑉𝑉𝑗𝑗 − 𝑉𝑉𝐴𝐴𝐶𝐶−𝑗𝑗2𝑗 = 0
𝑉𝑉𝑗𝑗 − 𝑉𝑉𝐴𝐴𝐶𝐶−𝑗𝑗2𝑗 − 5𝑉𝑉𝑗𝑗 −
𝑉𝑉𝐴𝐴𝐶𝐶2 + 𝑗𝑗 = 0
0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡
𝐶𝐶1 = 500 𝜇𝜇F
1 Ω
𝐶𝐶2 = 40 𝜇𝜇F 2 Ω 1 mH
5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−
0.1
−𝑗𝑗2
1 Ω
−𝑗𝑗2𝑗 2 Ω 𝑗𝑗𝑗
5𝑉𝑉𝑗𝑗+𝑉𝑉𝑗𝑗−
𝑉𝑉𝑗𝑗 𝑉𝑉𝐴𝐴𝐶𝐶
+𝑉𝑉𝐴𝐴𝐶𝐶−
Solve the equations to find 𝑉𝑉𝐴𝐴𝐶𝐶 = 0.4𝑗36∠ − 118.1°
Superposition: add the DC solution to the AC solution.
DC Solution:
𝑣𝑣𝐷𝐷𝐶𝐶 = 1 volt
AC Solution:𝑉𝑉𝐴𝐴𝐶𝐶 = 0.4𝑗36∠ − 118.1°
𝑣𝑣𝐴𝐴𝐶𝐶 𝑡𝑡 = 0.4536 cos 𝑗𝑗𝑗𝑗𝑡𝑡 − 118.1° volts
With both the DC sources and the AC source active: 𝑣𝑣 𝑡𝑡 = 𝑣𝑣𝐷𝐷𝐶𝐶 + 𝑣𝑣𝐴𝐴𝐶𝐶 𝑡𝑡So𝑣𝑣(𝑡𝑡) = 1 + 0.4536 cos 𝑗𝑗𝑗𝑗𝑡𝑡 − 118.1° volts
+𝑣𝑣𝑗𝑗−
5𝑣𝑣𝑗𝑗 12 Ω
3+𝑣𝑣−
0.1−𝑗𝑗2
1 Ω
−𝑗𝑗2𝑗 2 Ω 𝑗𝑗𝑗5𝑉𝑉𝑗𝑗
+𝑉𝑉𝑗𝑗−
𝑉𝑉𝑗𝑗 𝑉𝑉
+𝑉𝑉−
3 volts0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡
𝐶𝐶1
𝑅𝑅1
𝐶𝐶2 𝑅𝑅2 𝜔𝜔
5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−
1 amp+𝑣𝑣−
𝑣𝑣(𝑡𝑡)
Superposition with two frequencies:
An antenna receives two signals at the same time, one at frequency , 𝑓𝑓1 = 1 GHz and the other at frequency 𝑓𝑓2 = 2 GHz. The antenna is modelled with two voltage sources in series. The circuit uses two ideal op-amps. The component values are 𝑅𝑅 = 1 ohm, 𝑅𝑅1 = 1ohm, 𝐶𝐶 = 1 pF and 𝜔𝜔 = 6.333 nH.
Use Superposition to find the output voltage 𝑣𝑣4(𝑡𝑡).
𝑣𝑣𝑠𝑠1 𝑡𝑡 = 0.1 cos𝜔𝜔1𝑡𝑡
𝑣𝑣𝑠𝑠2 𝑡𝑡 = 0.1 cos𝜔𝜔2𝑡𝑡
50 Ω50 Ω
𝑅𝑅
𝑅𝑅1
𝑣𝑣4𝑣𝑣3
+𝑣𝑣4(𝑡𝑡)−
𝜔𝜔𝐶𝐶𝑣𝑣2𝑣𝑣1
Analysis of the first stage:𝑣𝑣𝑠𝑠1 + 𝑣𝑣𝑠𝑠2 − 𝑣𝑣1
50−𝑣𝑣1 − 𝑣𝑣2
50= 0
The op-amps are ideal with infinite gain and very high input impedance. Consider 𝑣𝑣1 to be a virtual ground, so 𝑣𝑣1 = 0. Then
𝑣𝑣𝑠𝑠1 + 𝑣𝑣𝑠𝑠250
+𝑣𝑣250
= 0
𝑣𝑣2 = − 𝑣𝑣𝑠𝑠1 + 𝑣𝑣𝑠𝑠2
The output of the first stage is the sum of the two voltages.
𝑣𝑣𝑠𝑠1 𝑡𝑡
𝑣𝑣𝑠𝑠2 𝑡𝑡
50 Ω50 Ω
𝑣𝑣2
𝑣𝑣1 ≈ 0
≈ 0
𝑣𝑣𝑠𝑠1 + 𝑣𝑣𝑠𝑠2
Use phasors to analyze the second stage at some frequency 𝜔𝜔:
Write a node equation at 𝑉𝑉3:
𝑉𝑉2 − 𝑉𝑉3
𝑅𝑅 + 𝑗𝑗 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶
−𝑉𝑉3 − 𝑉𝑉4𝑅𝑅1
= 0
Amplifier input 𝑉𝑉3 is a virtual ground, 𝑉𝑉3 = 0
𝑉𝑉2
𝑅𝑅 + 𝑗𝑗 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶
+𝑉𝑉4𝑅𝑅1
= 0
phasor
Drive the circuit with a generator at frequency 𝜔𝜔:𝑣𝑣𝑠𝑠 𝑡𝑡 = 𝐴𝐴 cos𝜔𝜔𝑡𝑡
The phasor is𝑉𝑉𝑠𝑠 = 𝐴𝐴𝑒𝑒𝑗𝑗0
𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠
𝑅𝑅 + 𝑗𝑗 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶
The output of the first stage is 𝑉𝑉2 = −𝑉𝑉𝑠𝑠
𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠
𝑅𝑅 + 𝑗𝑗 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶
𝑣𝑣𝑠𝑠 𝑡𝑡 = 𝐴𝐴 cos𝜔𝜔𝑡𝑡
50 Ω50 Ω
𝑅𝑅
𝑅𝑅1
𝑉𝑉4𝑉𝑉3 ≈ 0
+𝑉𝑉4−
𝑗𝑗𝜔𝜔𝜔𝜔1𝑗𝑗𝜔𝜔𝐶𝐶
𝑉𝑉2 = −𝑉𝑉𝑠𝑠𝑉𝑉1 ≈ 0
𝑉𝑉𝑠𝑠 = 𝐴𝐴
≈ 0 ≈ 0
At frequency 𝑓𝑓1 = 1 GHz, 𝜔𝜔1 = 2𝜋𝜋𝑓𝑓1 = 6.2𝑗3𝑗𝑗109:
𝑅𝑅 = 1 ohm, 𝑅𝑅1 = 1 ohm, 𝐶𝐶 = 1 pF and 𝜔𝜔 = 6.333 nH, 𝑉𝑉𝑠𝑠 = 0.1
𝜔𝜔1𝜔𝜔 −1
𝜔𝜔1𝐶𝐶= 6.2𝑗3𝑗𝑗109𝑗𝑗6.333𝑗𝑗10−9 −
16.2𝑗3𝑗𝑗109𝑗𝑗𝑗𝑗𝑗10−12 = 39.79 − 159.15 = −119.36
𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠
𝑅𝑅+𝑗𝑗 𝜔𝜔1𝐿𝐿−1
𝜔𝜔1𝐶𝐶
= 0.11−𝑗𝑗119.36
= 8.3𝑗𝑗𝑗𝑗10−4∠𝑗𝑗.5 degrees
𝑣𝑣4 𝑡𝑡 = 8.3𝑗𝑗𝑗𝑗10−4 cos 𝜔𝜔1𝑡𝑡 + 89.5°
𝑣𝑣𝑠𝑠 𝑡𝑡 = 0.1 cos𝜔𝜔1𝑡𝑡
50 Ω50 Ω
𝑅𝑅
𝑅𝑅1
𝑉𝑉4𝑉𝑉3 ≈ 0
+
𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠
𝑅𝑅 + 𝑗𝑗 𝜔𝜔1𝜔𝜔 −1
𝜔𝜔1𝐶𝐶−
𝑗𝑗𝜔𝜔1𝜔𝜔1𝑗𝑗𝜔𝜔1𝐶𝐶
𝑉𝑉2𝑉𝑉1 ≈ 0
𝑉𝑉𝑠𝑠
≈ 0 ≈ 0
The output voltage is small at 1 GHz.
At frequency 𝑓𝑓2 = 2 GHz, 𝜔𝜔2 = 2𝜋𝜋𝑓𝑓2 = 1.2𝑗𝑗𝑗𝑗1010:
𝑅𝑅 = 1 ohm, 𝑅𝑅1 = 1 ohm, 𝐶𝐶 = 1 pF and 𝜔𝜔 = 6.333 nH, 𝑉𝑉𝑠𝑠 = 0.1
𝜔𝜔2𝜔𝜔 −1
𝜔𝜔2𝐶𝐶= 1.2𝑗𝑗𝑗𝑗1010𝑗𝑗6.333𝑗𝑗10−9 −
11.2𝑗𝑗𝑗𝑗1010𝑗𝑗𝑗𝑗𝑗10−12 = 79.54 − 79.55 = −0.01 ≈ 0
𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠
𝑅𝑅 + 𝑗𝑗 𝜔𝜔2𝜔𝜔 −1
𝜔𝜔2𝐶𝐶=
0.11 + 𝑗𝑗𝑗 = 0.𝑗∠𝑗 degrees
𝑣𝑣4 𝑡𝑡 = 0.1 cos 𝜔𝜔2𝑡𝑡
𝑣𝑣𝑠𝑠 𝑡𝑡 = 0.1 cos𝜔𝜔2𝑡𝑡
50 Ω50 Ω
𝑅𝑅
𝑅𝑅1
𝑉𝑉4𝑉𝑉3 ≈ 0
+
𝑉𝑉4 = 𝑅𝑅1𝑉𝑉𝑠𝑠
𝑅𝑅 + 𝑗𝑗 𝜔𝜔2𝜔𝜔 −1
𝜔𝜔2𝐶𝐶−
𝑗𝑗𝜔𝜔2𝜔𝜔1𝑗𝑗𝜔𝜔2𝐶𝐶
𝑉𝑉2𝑉𝑉1 ≈ 0
𝑉𝑉𝑠𝑠
≈ 0 ≈ 0
Superposition with two frequencies:
By superposition, with both sources active:𝑣𝑣4 𝑡𝑡 = 8.3𝑗𝑗𝑗𝑗10−4 cos 𝜔𝜔1𝑡𝑡 + 89.5° + 0.1 cos 𝜔𝜔2𝑡𝑡
The design of the circuit makes 𝜔𝜔𝜔𝜔 − 1𝜔𝜔𝐶𝐶
=0 at 2 GHz.Then the circuit “passes” 2 GHz but “rejects” 1 GHz.The circuit behaves as a ‘filter’ to recover the 2 GHz signal from the two signals received by the antenna.
𝑣𝑣𝑠𝑠1 𝑡𝑡 = 0.1 cos𝜔𝜔1𝑡𝑡
𝑣𝑣𝑠𝑠2 𝑡𝑡 = 0.1 cos𝜔𝜔2𝑡𝑡
50 Ω50 Ω
𝑅𝑅
𝑅𝑅1
𝑣𝑣4𝑣𝑣3
+𝑣𝑣4(𝑡𝑡)−
𝜔𝜔𝐶𝐶𝑣𝑣2𝑣𝑣1
To master this set of lecture notes, solve the problems yourself:
1)Find the Thevenin Equivalent Circuit at a frequency of 𝜔𝜔=3 rad/sec. Use phasors relative to “sine”.
2)Find the voltage across a load consisting of a 1 ohm resistor in series with a capacitor of value C=0.16667 F, using: • The Thevenin equivalent circuit• The original circuitThe answer must be the same in both cases!
2 Ω
1 Ω
12
H10 sin 3𝑡𝑡
A
B
Find the Norton Equivalent Circuit
Find the Norton Equivalent Circuit at a frequency of 𝜔𝜔=3 rad/sec.
2 Ω
1 Ω12
H10 sin 3𝑡𝑡
A
B
1) Find the Norton Equivalent Circuit at 60 Hz.2)Convert the Norton Equivalent to the Thevenin Equivalent Circuit.
110 cos𝜔𝜔𝑡𝑡1 Ω
10.62 mH
5.31 mH
1330 𝜇𝜇F
1 Ω A
B
SuperpositionIn the circuit below there are two DC sources, 𝑉𝑉2 = 3 volts and 𝐼𝐼1 = 1 amp. There is also an AC source, 𝑣𝑣1 𝑡𝑡 = 0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡 volts.
Use Superposition to find the voltage across the current generator 𝑣𝑣 𝑡𝑡 .
The component values are 𝑅𝑅1 = 1 Ω, 𝑅𝑅2 = 2 Ω, 𝐶𝐶1 = 500 microFarads, 𝐶𝐶2 = 40 microFarads, 𝜔𝜔 = 1 mH.
0.1 cos 𝑗𝑗𝑗𝑗𝑡𝑡
𝐶𝐶1 = 500 𝜇𝜇F
𝑅𝑅1 = 1 Ω
𝐶𝐶2 = 40 𝜇𝜇F 𝑅𝑅2 = 2 Ω 𝜔𝜔 = 1 mH
5𝑣𝑣𝑗𝑗+𝑣𝑣𝑗𝑗−
𝑉𝑉2 = 3 volts𝐼𝐼1 = 1 amp+𝑣𝑣−
𝑣𝑣(𝑡𝑡)
Superposition:
An antenna receives two signals at the same time, one at frequency , 𝑓𝑓1 = 1 GHz and the other at frequency 𝑓𝑓2 = 2 GHz. The antenna is modelled with two voltage sources in series. The circuit uses two ideal op-amps. The component values are 𝑅𝑅 = 1 ohm, 𝑅𝑅1 = 1ohm, 𝐶𝐶 = 1 pF and 𝜔𝜔 = 6.333 nH.
Use Superposition to find the output voltage 𝑣𝑣4(𝑡𝑡).
𝑣𝑣𝑠𝑠1 𝑡𝑡 = 0.1 cos𝜔𝜔1𝑡𝑡
𝑣𝑣𝑠𝑠2 𝑡𝑡 = 0.1 cos𝜔𝜔2𝑡𝑡
50 Ω50 Ω
𝑅𝑅
𝑅𝑅1
𝑣𝑣4𝑣𝑣3
+𝑣𝑣4(𝑡𝑡)−
𝜔𝜔𝐶𝐶𝑣𝑣2𝑣𝑣1