Lecture 3 Electrical Circuits

7/29/2019 Lecture 3 Electrical Circuits

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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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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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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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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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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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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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\ 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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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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]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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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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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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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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)(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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)(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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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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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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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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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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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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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) = UF v1

1 + vUF(3.121)

The factor UF

Magnitude is a constant and equal to 20 log10 1033 = 60dB=


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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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) = 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=


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) =

Lecture 3 Electrical Circuits

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Transcript of Lecture 3 Electrical Circuits