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3Electrical Circuits
3.1 Basic Concepts
Electric charge
1 coulomb of negative change contains 6=241 1018 electrons. Current
1 ampere is a steady flow of 1 coulomb of change pass a given point in a con-
ductor in 1 second.
L(amperes) =T(coulombs)
w(seconds)= (3.1)
Time varying current
l(w) = lim{w
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Sec 3.2. Circuit Elements
Resistor Capacitor Inductor
iR vR vC vLiC iL
Current Source Voltage Sourse
i v 5V
DC Voltage Sourse
GND
Figure 3.1: Basic circuit elements.
vI
vi (pure AC)
VI (pure DC)
t
v
vI = VI + vi
Figure 3.2: DC signals versus AC signals.
Voltage source provides a specified voltage across the two terminals and does not
depend on the current flowing through the source.
The output impedance of ideal voltage source is zero.
The current flowing through an ideal voltage source is completely deter-mined by the circuit connected to the source.
Ideal current source Current source provides a specified current and does not depend on the voltage
across the source.
The output impedance of ideal current source is infinite.
The voltage across an ideal current source is completely determined by thecircuit connected to the source.
DC signals v.s. AC signals
Figure 3.2 shows the definitions of DC signals and AC signals.
Resistor U
Resistance is the property of materials that resists the movement of electrons.
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Lecture 3. Electrical Circuits
Ohms Law
U(ohms) =Y(volts)
L(amperes)(3.5)
Conductance is the inverse of resistance.
For parallel resistors,
UW =1
1@U1 + 1@U2 + 1@U3==== + 1@UQ(3.6)
When two resistors are connected in parallel, the equivalent resistance issmaller than any of the two resistors.
UW =U1U2
U1 + U2? U1
UW = U1U2U1 + U2? U2 (3.7)
For series resistors,
UW = U1 + U2 + U3==== + UQ (3.8)
Capacitor F
Capacitance is the ability of a capacitor to store charges on its two conductors.
F(farad) =T(coulombs)
Y(volts)
(3.9)
For time-varying voltage yF(w) across the capacitor
The capacitor is an open circuit, i.e., lF(w) = 0, when yF(w) is a constant. The voltage across the capacitor can not jump. However, the current flow-
ing through the capacitor does not have such constraint.
lF(w) = lim{w
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Sec 3.3. Circuit Laws
For time-varying current lO(w) flowing through the inductor
The inductor is a short circuit, i.e., yO(w) = 0, when lO(w) is a constant. The current flowing through the inductor can not jump. However, the
voltage across the inductor does not have such constraint.
yO(w) = lim{w
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Lecture 3. Electrical Circuits
iS vORCvS
iOR
L
Current Source Voltage Source
Parallel Serial
v i
C L
R R
Figure 3.3: Similarity between capacitor and inductor.
3.4 Network TheoremsDefinition 3.1 Linear circuit is formed by interconnecting the terminals of independent sources,
controlled sources, and linear passive elements to form one or more closed paths.
Linear passive elements include Resistor, Capacitor, and Inductor.
The l y characteristics of these elements satisfy the conditions of linearity. Resistor
y1(w) = l1(w)U
y2(w) = l2(w)U (3.20)
d y1(w) + e y2(w) = (d l1(w) + e l2(w))U
Capacitor
l1(w) = Fg
gw(y1(w))
l2(w) = Fg
gw(y2(w)) (3.21)
d l1(w) + e l2(w) = F ggw
(d y1(w) + e y2(w))
Inductor
l1(w) = O
Zw0
y1()g
l2(w) = O
Zw0
y2()g (3.22)
d l1(w) + e l2(w) = OZw
0
(d y1() + e y2())g
Theorem 3.2 In a linear network containing multiple sources, the voltage across or current
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Sec 3.4. Network Theorems
AV
1R 2R
AI 3R V
AV
1R 2R
3R
1V
1R 2R
AI 3R
2V
+
=(a)
(b)
(c)
Figure 3.4: Example of linear circuit with multiple independent sources.
through any passive element may be found as the algebraic sum of the individual voltages
or currents due to each of the independent sources action along, with all other independentsources deactivated.
Voltage source is deactivated by replacing it with a short circuit.
Current source is deactivated by replacing it with an open circuit.
Controlled sources remain active when the superposition theorem is applied.
Example 3.3 Given the circuit in Figure 3.4, find the voltage across the resistorU3 using
the superposition theorem of linear network.
1. The voltage across the resistor U3 is the superposition of the voltage when each
independent source actions alone, as shown in Figure 3.4 (b) and (c).
Y = Y1 + Y2 (3.23)
2. The contribution of the voltage source YD=
The current source LD is replaced with an open circuit.
Y1
=U3
U1 + U2 + U3YD
(3.24)
3. The contribution of the current source LD=
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Lecture 3. Electrical Circuits
The voltage source YD is replaced with an short circuit.
Y2 =U1U3
U1 + U2 + U3LD (3.25)
3.4.1 Equivalent Circuits of One-Port Networks
Linear
Network
A
Linear
Network
B
1
2
LinearNetwork
B
1
2
T HZ
T HV
Linear
Network
B
1
2N
I NY
(a)
(b)
(c)
Figure 3.5: Equivalent circuits. (a) The original circuit. (b) Thevenins equivalent. (c)Nortons equivalent.
Equivalent circuit
A reduction of a complex linear circuit into a simpler form.
A model of a complex linear circuit contained in a black box.
Theorem 3.4 Thevenins theorem states that an arbitrary linear, one port network such
as network A in Figure 3.5 (a) can be replaced at terminals 1> 2 with an equivalent series-
connected voltage source YWK and impedance ]WK as in Figure 3.5 (b).
YWK is the open-circuit voltage of network A at terminals 1> 2.
]WK is the ratio of the open-circuit voltage over short circuit current determined at
terminals 1> 2.
The equivalent impedance looking into network A through terminals 1> 2 with
all independent sources deactivated.
Voltage sources are replaced by short circuits. Current sources are replaced by open circuits.
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Sec 3.4. Network Theorems
Theorem 3.5 Nortons theorem states that an arbitrary linear, one port network such as
network A in Figure 3.5 (a) can be replaced at terminals 1> 2 with an equivalent parallel-
connected current source LQ and admittance \Q as in Figure 3.5 (c).
LQ is the short-circuit current flowing through terminals 1> 2 due to network A. \Q is the ratio of short-circuit current over open-circuit voltage at terminals 1> 2.
Conversion of equivalent circuits.
Any method for determining ]WK is equally valid for finding \Q=
LQ = LVF =YWK]WK
(3.26)
YWK = YRF = LQ1
\Q(3.27)
]WK =1
\Q(3.28)
Example 3.6 In Figure 3.6, YD = 4Y, LD = 2D, U1 = 2l, U2 = 3l, find the Thevenins
equivalent circuit and Nortons equivalent circuit for the network to the left of terminals
1> 2=
1
2
AV
1R 2R
AI
Figure 3.6: Examples of Thevenins and Nortons equivalent circuits.
1. Thevenins equivalent YWK is the open-circuit voltage at terminals 1> 2=
YWK = YD + LD U1 = 4 + 4 = 8Y= (3.29)
]WK is the ratio of the open-circuit voltage over short circuit current determined
at terminals 1> 2 with network B disconnected.
By the superposition of the short-circuit current caused by YD and LD> the
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Lecture 3. Electrical Circuits
short circuit current can be found.
L1>2 = L(YD) + L(LD)
=1
U1 + U2YD +
U1
U1 + U2LD (3.30)
=8
5D=
By definition, ]WK can be derived as YWK@L1>2 = 5l.
Alternatively, ]WK can be found as the equivalent impedance for the circuit to
the left of terminals 1> 2=
YD is replaced with short circuit.
LD is replaced with open circuit.
]WK = U1 + U2 = 5l= (3.31)
2. Nortons equivalent
LQ is the short-circuit current at terminals 1> 2> which can be derived as in Eq.
(3.30).
\Q is the ratio of the short-circuit current over the open-circuit voltage with
network B disconnected. From Eq. (3.29), the open circuit voltage is 8Y.
Thus, \Q =85
D@8Y = 1@5V=
Alternatively, \Q = 1@]WK = 1@5 = 0=2V=
3.4.2 Equivalent Circuits of Two-Port Networks
A two-port network is an electrical circuit or device with two pairs of terminals.
Figure 3.7 depicts a two-port linear network.
Only two of the four variables Y1, Y2, L1, L2 can be independent.
Linear
Network 2V
1V
+
-
+
-
1I
2I
1I 2I
Figure 3.7: Two-port linear network.
Characterization of Two-Port Networks
] parameters (Y1, Y2 depend on L1, L2)
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Sec 3.4. Network Theorems
The four }lm parameters represent impedance.
}12 and }21 are transfer impedances.
Each of the }lm parameters can be evaluated by open-circuiting an appropriate
port of the network.
Y1 = }11L1 + }12L2
Y2 = }21L1 + }22L2 (3.32)
}11 =Y1L1|L2=0
}12 =Y1L2|L1=0
}21 = Y2L1|L2=0 (3.33)
}21 =Y2L2|L1=0
] parameters in matrix form.
"Y1
Y2
#=
"}11 }12
}21 }22
#"L1
L2
#(3.34)
\ parameters (L1, L2 depend on Y1, Y2) The four |lm parameters represent admittance.
|12 and |21 are transfer admittances.
Each of the |lm parameters can be evaluated by short-circuiting an appropriate
port of the network.
L1 = |11Y1 + |12Y2
L2 = |21Y1 + |22Y2 (3.35)
|11 =L1Y1|Y2=0
|12 =L1Y2|Y1=0
|21 =L2Y1|Y2=0 (3.36)
|21 =L2Y2|Y1=0
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Lecture 3. Electrical Circuits
\ parameters in matrix form.
"L1
L2
#=
"|11 |12
|21 |22
#"Y1
Y2
#(3.37)
Relation between |lm parameters and }lm parameters.
"|11 |12
|21 |22
#=
"}11 }12
}21 }22
#31=
1
}11}22 }12}21
"}22 }21}12 }11
#(3.38)
K parameters (Y1, L2 depend on L1, Y2)
The k11 represents impedance.
The k22 represents admittance.
Parameters k11 and k21 are obtained by short-circuiting port 2. Parameters k12 and k22 are obtained by open-circuiting port 1.
Y1 = k11L1 + k12Y2
L2 = k21L1 + k22Y2 (3.39)
k11 =Y1L1|Y2=0
k12 = Y1Y2|L1=0
k21 =L2L1|Y2=0 (3.40)
k22 =L2Y2|L1=0
K parameters in matrix form.
" Y1L2# = " k11 k12
k21 k22#" L1
Y2# (3.41)
Example 3.7 In Figure 3.8, U1 = 10l, U2 = 6l, find the } parameters and the k
parameters for the network=
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Sec 3.4. Network Theorems
1R
2RaIaI3.01V
+
-
1I
1I
2V
+
-
2I
2I
Figure 3.8: Example of two-port linear network.
From Eq. (3.33), the }lm parameters can be obtained as follows.
}11 =Y1
L1 |L2=0 =
(10 + 6)Ld
Ld + 0=3Ld =16
1=3 = 12=31l
}12 =Y1L2|L1=0 =
6Ld 0=3Ld 10Ld + 0=3Ld
=3
1=3= 2=31l
}21 =Y2L1|L2=0 =
6LdLd + 0=3Ld
=6
1=3= 4=62l
}22 =Y2L2|L1=0 =
6LdLd + 0=3Ld
=6
1=3= 4=62l
From Eq. (3.40), the klm parameters can be calculated as follows.
k11 =Y1L1|Y2=0 =
L1 U1L1
= U1 = 10l
k12 =Y1Y2|L1=0 =
Ld U2 0=3Ld U1Ld U2
=3
6= 0=5
k21 =L2L1|Y2=0 = 1
k22 =L2Y2|L1=0 =
Ld + 0=3LdLd U2
=1=3
6= 0=217V=
Equivalent Circuits of Special Two-Port Networks
T-Model network. (Figure 3.9 (a))
]1> ]2>and ]3 can be derived from the ] parameters of a two-port networks.
"}11 }12
}21 }22
#=
"]1 + ]3 ]3
]3 ]2 + ]3
#(3.42)
-Model network. (Figure 3.9 (b))
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Lecture 3. Electrical Circuits
]d>]e>and ]f can be derived from the \ parameters of a two-port networks.
"|11 |12
|21 |22
#=
"1]d
+ 1]f
1]f
1]f
1]e
+ 1]f
#(3.43)
Transformation from -Model to T-Model.
]1 =]d]f
]d + ]e + ]f
]2 =]e]f
]d + ]e + ]f(3.44)
]3 =]d]e
]d + ]e + ]f
Transformation from T-Model to -Model.
]d =]1]3 + ]1]2 + ]2]3
]2
]e =]1]3 + ]1]2 + ]2]3
]1(3.45)
]f =]1]3 + ]1]2 + ]2]3
]3
Z1 Z2
Z31V 2V
1I
2
I
(a) T-Model Network
Zb1
V2
V
1I2
I
Za
Zc
(b) Model Network
Figure 3.9: Equivalent circuits of two port networks. (a) T-Model Network. (b) Pi-Model Network.
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Sec 3.5. Laplace Transforms
3.5 Laplace Transforms
Laplace Transform
A mathematical tool for representing system transfer functions of causal LTI
systems. A generalization of Fourier transform.
The result of Fourier transform can be obtained by evaluating the Laplacetransform along the imaginary axis.
Definition 3.8 Given i(w) with i(w) = 0 for w ? 0, its Laplace Transform, which is
defined as follows, is a complex function over a complex-number domain.
I(v)
L[i(w)] , Z
"
03
i(w)h3vwgw = Z"
03i(w)h3w h3m$wgw= (3.46)
Definition 3.9 Correspondingly, the Inverse Laplace Transform is as follows:
i(w) L31[I(v)] = 12m
If
I(v)hvwgv= (3.47)
Example 3.10 Given i(w) & j(w) (i(w) = j(w) = 0 for w ? 0) and their Laplace Trans-
forms I(v) & J(v),the following shows the properties of Laplace Transform.
Linearity
|(w) = di(w) + ej(w) L#$ \(v) = dI(v) + eJ(v) (3.48)
Time Derivatives
|(w) =gi(w)
gwL#$ \(v) = vI(v) i(03) (3.49)
|(w) =gqi(w)
gwL#$ \(v) = vqI(v) vq31i(03) vq32gi(0
3)
gw======= gq31i(0
3)
gw
Time Integrals
|(w) =
Zw03
i()g + |(03)L#$ \(v) = I(v)
v+
|(03)
v(3.50)
Time Scaling
|(w) = i(w)L#$ 1
I(
v
)> where A 0= (3.51)
Time Delay
|(w) = i(w w0) L#$ \(v) = h3vw0I(v)> where w0 A 0= (3.52)
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Lecture 3. Electrical Circuits
w Multiplication
|(w) = wqi(w)L#$ \(v) = (1)q g
qI(v)
gvq(3.53)
v Shift
|(w) = hdwi(w)L#$ \(v) = I(v d) (3.54)
Convolution
|(w) = {(w) k(w) L#$ \(v) = [(v)K(v) (3.55)
Product
|(w) = i(w){(w)L#$ \(v) = 1
2m
If
I(v)[(v )g= (3.56)
Example 3.11 Laplace Transform Pairs1
i(w)L#$ I(v)
(w)L#$ 1
DL#$ D
v
DwqL#$ D( q!
vq+1)
DhdwL#$ D
v
d
cos $0w L#$ vv2 + $20
sin $0wL#$ $0
v2 + $20
2Dhdw cos($0w + )L#$ Dh
m
v (d +m$0)+
Dh3m
v (dm$0)=
2D cos (v d $0 tan )(v d)2 + $20
3.6 Equivalent Circuits in Laplace Domain
Capacitor
l y characteristic in Laplace domain. Admittance of capacitor in Laplace domain is vF=
lF(w) = FgyF(w)
gwL#$ LF(v) = F(vYF(v) yF(03)) (3.57)
Figure 3.10 (a) depicts the KCL equivalent circuit in Laplace domain. y l characteristic in Laplace domain.
1Footnote
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Sec 3.6. Equivalent Circuits in Laplace Domain
)(sIC
)(sVC s C )0(
CC v
)(sIC
)(sVC
s C/1
)0(1
Cv
s
(a) (b)
)(sIC
)(sVL s L/1
)0(1
Li
s
)(sIL
)(sVL
s L
)0( L
L i
(c) (d)Figure 3.10: (a) KCL equivalent circuit of a capacitor in Laplace domain. (b) KVLequivalent circuit of a capacitor in Laplace domain. (c) KCL equivalent circuit of aninductor in Laplace domain. (b) KVL equivalent circuit of an inductor in Laplace domain.
Impedance of capacitor in Laplace domain is 1@vF=
yF(w) =1
FZw
3"
lF()g (3.58)
=1
F
Zw0
lF()g + yF(03)
L#$ YF(v) =1
vFLF(v) +
1
vyF(0
3)
Figure 3.10 (b) depicts the KVL equivalent circuit in Laplace domain. Inductor
l y characteristic in Laplace domain. Admittance of inductor in Laplace domain is 1@vO=
lO(w) =1
OZw3" y
O()g (3.59)
=1
O
Zw0
yO()g + lO(03)
L#$ LO(v) =1
vOYO(v) +
1
vlO(0
3)
Figure 3.10 (c) depicts the KCL equivalent circuit in Laplace domain. y l characteristic in Laplace domain.
Impedance of inductor in Laplace domain is vO=
yO
(w) = OglO(w)
gw
L
#$YO
(v) = O(vLO
(v)
lO
(03)) (3.60)
Figure 3.10 (d) depicts the KVL equivalent circuit of an inductor in Laplace
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Lecture 3. Electrical Circuits
domain.
3.7 System Transfer Function in Time and Laplace
Domains
Two ways to deduce the system transfer function in time and Laplace domains.
Start in time domain and then take the Laplace transform.
Start in Laplace domain and use Inverse Laplace transform to return to time
domain.
Example 3.12 Consider the RC circuit in Figure 3.11, the input is the current source
lv(w) and the output yr(w) is the voltage across the RC components. Find out the system
transfer function and the outputs w.r.t. digerent inputs.
Si
ovR C
Figure 3.11: The RC circuit to be solved in time domain and Laplace domain.
By writing the node equation with KCL in time domain, we obtain the system
equation.
lV(w) =yr(w)
U+ F
gyr(w)
gw(3.61)
By taking Laplace transform to both sides, the system equation in Laplace domain
can be derived.
LV(v) =Yr(v)
U + F(vYr(v) yr(03)) (3.62) The system transfer function in Laplace domain.
K(v) =Yr(v)
LV(v)|yr(03)=0 =
1
vF+ (1@U)=
U
1 + vUF(3.63)
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Sec 3.7. System Transfer Function in Time and Laplace Domains
Initial Value (Driving Free) Response
Given lV(w) = 0> LV(v) = 0, the output can be obtained from Eq. (3.62).
0 =
Yr(v)
U + F(vYr(v) yr(03)), Yr(v) =
yr(03))
v + (1@UF)(3.64)
The time domain response can be calculated by taking Inverse Laplace transform.
yr(w) = yr(03)h3w@UF (3.65)
Impulse Response
Given lV(w) = (w); LV(v) = 1, the output can be obtained from Eq. (3.63).
For simplicity, we do not consider initial condition, i.e., yr(03) = limw
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Lecture 3. Electrical Circuits
The term D@ (1 + vUF) stands for transient response, which can be trans-
formed into exponentially decaying time function.
D = U2F@
1 + ($0UF)
2
=
The term (Ev + G)@ (v2 + $20) represent the steady state response, which shall
yield the time-domain signal.
Ev + G
v2 + $20=
Ev
v2 + $20+
G
v2 + $20
, yr(w) = E cos($0w) +G
$0sin($0w)
= N0 cos($0w !) (3.71)
N0 = rE2 + G$0
2
=
! = tan31( G$0E
)=
E = U@
1 +
$01@UF
2=
G = 1F
$0
1@UF
2@
1 +
$0
1@UF
2=
$0 1
UF
E ' U, G ' 0> yr(w) = U cos($0w) = At low frequency, the capacitor is similar to an open circuit.
$0 1
UF
E ' 0, G ' 1F> yr(w) = 1$0F sin($0w) = 1$0F cos
$0w 2
=
At high frequency, the capacitor is similar to a short circuit. The circuit acts as a low-pass filter.
Example 3.13 Consider the RLC circuit in Figure 3.12, the input is the current source
lv(w) and the outputyr(w) is the voltage across the RLC components. Find out the system
transfer function and the outputs w.r.t. digerent inputs.
Si ovR C L
Figure 3.12: The RLC circuit to be solved in time domain and Laplace domain.
Start in Time Domain
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Sec 3.7. System Transfer Function in Time and Laplace Domains
)(tvoR C L
)(tiR
)(tiC
)(tiL
)(tiS
Figure 3.13: Time domain analysis.
By writing the node equation with KCL in time domain, we obtain
lV(w) = lU(w) + lF(w) + lO(w)> (3.72)
where
lU(w) =yr(w)
U
lF(w) = Fgyr(w)
gw(3.73)
lO(w) =1
O
Zw3"
yr()g =1
O
Zw0
yr()g + lO(03)
The system equation in time domain.
lV(w) =yr(w)
U+ F
gyr(w)
gw+
1
O
Zw0
yr()g + lO(03) (3.74)
Apply the Laplace transform to both sides of Eq.(3.74).
LV(v) =Yr(v)
U+ F(vYr(v) yr(03)) +
1
vOYr(v) +
1
vlO(0
3)> (3.75)
where LV(v) , L[lV(w)] and Yr(v) = L[yr(w)].
The system transfer function in Laplace domain. Assume no initial condition, i.e., yr(0
3) = limw
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Lecture 3. Electrical Circuits
)(sIS
)(sVo
R
1s C
s L
1
)0( o
C v
)0(1
Li
s
Figure 3.14: Laplace domain analysis.
Replace all the circuit elements by their KCL equivalent circuits in Laplace domain.
U $ 1U
F $ vFYr(v) Fyr(03) (3.77)O $ 1
vOYr(v) +
1
vlO(0
3)
LV(v) = LU(v) + LF(v) + LO(v)
=Yr(v)
U+
vFYr(v) Fyr(03)
+
1
vOYr(v) +
1
vlO(0
3)
(3.78)
The system transfer function.
LV(v) = LU(v) + LF(v) + LO(v)|yr(03)=0>lO(03)=0
=Yr(v)
U+ vFYr(v) +
1
vOYr(v) (3.79)
, Yr(v)LV(v)
=1
( 1U
+ vF+ 1vO
)
Impulse Response
Impulse response is the system output w.r.t an input signal of impulse function. It
is also the transfer function of the system which can be used to find system poles
and zeros.
lV(w) = (w); (w)L$ 1=
Assume no initial condition.
Yr(v) =1
( 1U
+ vF+ 1vO
)LV(v)
=1
( 1U
+ vF+ 1vO
) 1 (3.80)
= vUOv2UOF+ vO + U
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Sec 3.7. System Transfer Function in Time and Laplace Domains
System poles and zeros are the roots of denominator and numerator of the system
transfer function.
Zeros , the roots of Q(v).
Poles , the roots of G(v).
K(v) =Yr(v)
LV(v),
Q(v)
G(v)(3.81)
The simplest way to obtain K(v) is by analyzing the system impulse response
in Laplace domain.
K(v) =Yr(v)
1= Yr(v)
=
vUO
v2UOF+ vO + U (3.82)
=Q(v)
G(v)
The time-domain expression of yr(w) , L31[Yr(v)] depends on the nature of the
system poles, which are the roots of denominator.
Let the system characteristic equation be equal to 0.
v2UOF+ vO + U = 0 (3.83)
Rewrite it as
dv2 + ev + f = 0 with d = UOF, e = O, f = U (3.84)
The roots are
vS and v0S =
es
e2 4df2d
= e
2d r
(
e
2d)2 4df
4d2 (3.85)
= e2d
r(
e
2d)2 f
d
The nature of the roots (vS, v0S) shall depend on the value of the critical expression.
2 , (e
2d)2 f
d
= (O
2UOF)2 ( U
UOF) (3.86)
= ( 12UF
)2 1OF
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Lecture 3. Electrical Circuits
0 2 4 6 8 10 12 14
0.4
0.2
0.2
0.4
0.6
0.8
Underdamped
Criticall Damped
Overdamped
Impulse Response of RLC Circuit
t
vo(t)
Figure 3.15: Comparison of system impulse responses in digerent cases.
Traditionally, we define two more parameters.
Neper frequency = ,
e
2d=
1
2UF(3.87)
Resonant frequency $0=
$0 ,
rf
d=
1sOF
(3.88)
We can now rewrite the roots of the characteristics equation.
vS and v0S =
q2 $20 (3.89)
Three distinct cases are possible with respect to vS and v0S depending on the values
of and $0. Overdamped
,p
2 $20 is real, i.e., 2 A $20= vS and v
0S are two real, distinct and negative values.
Yr(v) can be factorized as follows.
D = UO@(1 v0SvS
) and E = UO@(1 vSv0S
)=
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Sec 3.7. System Transfer Function in Time and Laplace Domains
Both D and E are real numbers.
Yr(v) =vUO
UOF(v vS)(v v0S)
=D
v vS +E
v v0S (3.90)
=D
v + |vS|+
E
v + |v0S|
By taking the inverse Laplace transform, the response of yr(w) (without initial
condition) can be formulized as follows:
yr(w) = Dh3|vS|w + Eh3|v
0S|w (3.91)
Figure 3.15 depicts the yr(w) in the overdamped case. Underdamped
,p
2 $20 is imaginary, i.e., 2 ? $20= vS and v
0S are two complex conjugate poles.
The damped resonant frequency $g ,p
$20 2 is real.
vS and v0S = m$g (3.92)
Yr(v) can be factorized as follows.
D and E are complex conjugates.
Yr(v) =vUO
UOF(v ( +$g)) (v ($g))(3.93)
=D
v ( +$g)+
E
v ($g)(3.94)
Taking inverse Laplace transform, y0(w) can be obtained.
Note that (D + E) and m(D E) are real number since D and E arecomplex conjugates.
yr(w) = Dh(3+$g)w + Eh(33$g)w
= h3w
Dh$gw + Eh3$gw
= h3w [D (cos $gw +m sin $gw) + E (cos $gwm sin $gw)](3.95)= h3w [(D + E)cos $gw +m(DE)sin $gw]
Figure 3.15 depicts the yr(w) in the underdamped case.
Critically damped ,
p2 $20=0, i.e., 2 = $20=
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Lecture 3. Electrical Circuits
jZ
V
-DsP
+jZ
-jZ
sP' 0
Critically damped
Underdamped
Overdamped
Figure 3.16: Locations of poles and zeros.
vS and v
0
S = .Yr(v) =
vUO
UOF(v ())2 =D
v + +
E
(v + )2(3.96)
Taking inverse Laplace Transform, y0(w) can be obtained.
y0(w) = Dh3w + Ewh3w (3.97)
Figure 3.15 depicts the y0(w) in the case of critically damped.
Summary The positions of poles determine the nature of System Response.
vS and v0S =
s2 $2 with = 1
2UFand $0 =
!sOF
Response is overdamped
if 2 $20 A 0 and vS, v0S are two distinct real numbers. Response is underdamped
if 2
$2
0 ? 0 and vS, v0
S are complex conjugate numbers. Response is critically damped
if 2 = $20 and vS, v0S are real and the same. The positions of system poles move along the path (known as Root Locus)
shown in the following diagram.
Sinusoidal Response
Sinusoidal response is the system output w.r.t an input signal of sinusoidal function.
lV(w) = cos ($w) x(w) or LV(v) = L[lV(w)] = v@ (v2
+ $2
) =
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Sec 3.8. Bode Plots
For simplicity, we do not consider initial condition.
Yr(v) =v2UO
(v2 + $2)(v2UOF+ vO + U)(3.98)
= v2
UOUOF(v vS)(v v0S)(v2 + $2) = D(v vS) + E(v v0S) + Gv + H(v2 + $2)
Instead of solving for the coecients D>E>G>Hanalytically by hand (which is
a practically impossible task), we shall deduce the time and frequency domain
responses by reasoning.
Time-Domain Response
The terms D@(vvS) and E@(vv0S) stand for transient response. Both canbe transformed into damped response in time with exponentially decaying
envelop.h3w where =
1
2UF(3.99)
The term (Gv + H) @(v2 + $2) represent the steady state response, whichshall yield the time-domain signal N0 cos($w !). N0 =
sG2 + H2=
! = tan31(HG
)=
Gv + H
(v2 + $2)=
Gv
(v2 + $2)+
H
(v2 + $2)
, G cos($w)+Hsin($w)= N0 cos($w !) (3.100)
3.8 Bode Plots
A graph used to show the frequency response of an LTI system.
To plot the magnitude in decibels (dB) and use a log scale for $=
The log scale helps to compress a wide range of data.
Both the magnitude and the phase of the transfer function versus the angular
frequency $=
System transfer function K(v) can be written as a product of factors of the following
items.
1. Constant factor N=
2. Poles or zeros at the origin, vQ=
3. Real poles or zeros, (W v + 1)Q=
4. Complex-conjugate poles or zeros, (W2v2+2W v+1)Q, where is the damping
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Lecture 3. Electrical Circuits
ratio and 0 ? ? 1=
K(v) =Q(v)
G(v)(3.101)
The magnitude ofK(v) in dB allows us to plot the factors individually and sum the results
to obtain the complete plot.
|K(v)|gE = 20 log10 |K(v)| (3.102)
Given K(v) as follows,
K(v) = N(1 + W}1v)(1 + W}2v)
(1 + Ws1v)(1 + Ws2v)(3.103)
|K(v)|gE
can be obtained by plotting each factor individually
|K(v)|gE = 20 log10 |K(v)|
= 20 log10
N(1 + W}1v)(1 + W}2v)(1 + Ws1v)(1 + Ws2v)
(3.104)
= 20 log10 |N|+ 20 log10 |1 + W}1v|+ 20 log10 |1 + W}2v|
20log10 |1 + Ws1v| 20log10 |1 + Ws2v|
Bode plot for each of the factors.
1. Constant factor, N. Magnitude
The magnitude 20 log10 |N| is a constant. Phase
The phase is a constant and equal to 0 (hm0 = 1) or 180 (hm = 1)depending on whether N is positive or negative, respectively.
2. Poles or zeros at the origin, vQ=
Magnitude
The magnitude is a straight line that intersects the $ axis (0dB) at
$ = 1 and has a slope of20dB/decade.
Phase
The phase is a constant and equal to Q90 .
K(m$) = (m$)Q = ($hm@2)Q (3.105)
|K(m$)|gE = 20 log10 $Q
= 20Qlog10 $ (3.106)
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Sec 3.8. Bode Plots
]K(m$) = Q(m@2) = Q90 (3.107)
3. Real poles or zeros, (W v + 1)Q=
Poles or zeros locate at 1W
=
K(m$) = (m$W + 1)Q =
uhmQ
(3.108)
|K(m$)|gE = ugE
= 20 log10
q1 + ($W)2
Q(3.109)
= 10Qlog10
1 + ($W)2
Magnitude Low frequency response ($W 1; $ 1
W): A horizontal line of 0dB=
|K(m$)|gE = 10Qlog10 (1) = 0 dB (3.110)
High frequency response ($W 1; $ 1W
): A straight line of slope
20QdB/decade that intersects 0dB when $ = 1W
=
|K(m$)|gE = 10Qlog10 ($W)2
= 20Qlog10
$1@W
(3.111)
= 20Q
log10 $ log10
1
W
Corner frequency response $ = 1W
=
|K(m$)|gE = 10Qlog10 2 = 3QdB (3.112)
|K(m$)| = 10
3Q@20
= 0=7079
Q
(negative) or 1=4125
Q
( positive )(3.113)
Phase
]K(m$) = Q = Qtan31 $W (3.114)
Low frequency: ]K(m$) 0 = Corner frequency: ]K(m$) Q45 = High frequency: ]K(m$) Q90 =
4. Complex conjugate poles or zeros, (W2v2 + 2W v + 1)Q=
For simplicity, we consider only the case of a single pair of complex conju-gate poles.
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Lecture 3. Electrical Circuits
If the poles are repeated by Q times, all coordinates on the curves willbe multiplied by Q.
If we have zeros instead of poles, curves are mirror images through the$ axis.
K(v) = 1W2v2 + 2W v + 1
(3.115)
|K(m$)|gE = 20log10
1 W2$22
+ 42W2$21@2
= 10log10
1 W2$22
+ 42W2$2
(3.116)
Magnitude
Low frequency response ($W 1; $ 1
W): A horizontal line of 0dB=
High frequency response ($W 1; $ 1W): A straight line of slope40dB/decade that intersects 0dB when $ = 1
W=
|K(m$)|gE = 10log10
1 W2$22
+ 42W2$2
' 10log10
W2$22
+ 42W2$2' 40log10 (W $)
= 40
log10 $ log101
W
(3.117)
Phase 0 ? $ ? 1@W=]K(m$) = tan31 2W $
1 W2$2 (3.118)
$ A 1@W=]K(m$) = 180 + tan31 2W $
W2$2 1 (3.119)
Example 3.14 Consider the transfer function Yr(v) = U@(1 + vUF) where U = 1 kl
and F = F, express its frequency responses including both the magnitude and the phase
responses with Bode Plot.
Yr(v) can be firstly written as follows.
Yr(v) = U1
1 + vUF(3.120)
The factor U=
Magnitude is a constant and equal to 20 log10 103 = 60dB.
Phase is a constant and equal to 0 =
The factor 1@(1 + vUF)= $ ? 1@UF
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Sec 3.8. Bode Plots
Magnitude is a constant and equal to 0dB. Phase is around 0 =
$ 1@UF
Magnitude is a straight line of slope
20dB/decade.
Phase is around 45 when $ = 1@UF= Phase is around 90 when $ 1@UF=
20
30
40
50
60
Magnitude(dB)
101
102
103
104
105
-90
-45
0
Phase(deg)
Bode Diagram
Frequency (rad/sec)
Figure 3.17: Bode plot for the transfer function Yr(v) = U@(1 + vUF)=
Example 3.15 Consider the transfer function Yr(v) = vUF@(1 + vUF) where U = 1 kl
and F = F, express its frequency response including both the magnitude and the phase
responses with Bode Plot.
Yr(v) can be firstly written as follows.
Yr(v) = UF v1
1 + vUF(3.121)
The factor UF
Magnitude is a constant and equal to 20 log10 1033 = 60dB=
Phase is a constant and equal to 0 =
The factor v
Magnitude is a straight line of slope 20dB/decade and intersects the $-axis at
$ = 1=
Phase is 90 .
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Lecture 3. Electrical Circuits
The factor 1@(1 + vUF)=
$ ? 1@UF
Magnitude is a constant and equal to 0dB.
Phase is around 0 =
$ 1@UF Magnitude is a straight line of slope 20dB/decade. Phase is around 45 when $ = 1@UF= Phase is around 90 when $ 1@UF=
-40
-30
-20
-10
0
Magnitud
e(dB)
101
102
103
104
105
0
30
60
90
Phase(deg)
Bode Diagram
Frequency (rad/sec)
Figure 3.18: Bode plot for the system transfer function Yr(v) = vUF@(1 + vUF)=
Example 3.16 Consider the transfer function Yr(v) = vUO@ (v2UOF+ vO + U) where
U = 1 kl, F = 1 F and O = 1K, express its frequency response including both the
magnitude and the phase responses with Bode Plot.
Yr(v) can be firstly written as follows.
Yr(v) = UO v1
(v2UOF+ vO + U)= O v
1v2OF+ vO
U+ 1
(3.122) The factor O
Magnitude is a constant and equal to 20 log10 1 = 0dB=
Phase is a constant and equal to 0 =
The factor v
Magnitude is a straight line that intersects the $ axis (0dB) at $ = 1 and has
a slope of 20dB/decade.
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Sec 3.8. Bode Plots
Phase is a constant and equal to 90 .
The factor 1@(v2OF+ vOU
+ 1)
$ 1W
= 1IOF
Magnitude is a horizontal line of 0dB=
Phase is tan31 2W$13W2$2 .
$ 1W
= 1IOF
Magnitude is a straight line of slope 40dB/decade that intersects 0dBwhen $ = 1
W=
Phase is 180 + tan31 2W$W2$231 =
0
10
20
30
40
50
60
Magnitude(dB)
101
102
103
104
105
-90
-45
0
45
90
Phase(deg)
Bode Diagram
Frequency (rad/sec)
Figure 3.19: Bode plot for the system transfer function Yr(v) =vUO@ (v2UOF+ vO + U) =
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