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Applications of LaplaceTransforms

•Introduction•Circuit Element Models•Circuit Analysis•Transfer Functions•State Variables•Network Stability•Summary

Introduction•To learn how easy it is to work with circuits in the s

domain.•To learn the concept of modeling circuits in the s

domain.•To learn the concept of transfer function in the s

domain.•To learn how to apply the state variable method for

analyzing linear systems with multiple inputs andmultiple outputs.•To learn how the Laplace transform can be used in

stability analysis.

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Circuit Element Models•Steps in applying the Laplace transform:

–Transform the circuit from the time domain to the sdomain (a new step to be discussed later).

–Solve the circuit using circuit analysis technique(nodal/mesh analysis, source transformation, etc.).

–Take the inverse Laplace transform of the solutionand thus obtain the solution in the time domain.

s-Domain Models for R and L

)()()()(

)()(resistor,aFor

sRIsVtRiLtvL

tRitv

(b))0(

)(1

)(or

(a))0()(

)0()()(

.)(

)(inductor,anFor

si

sVsL

sI

LissLI

issILsVdt

tdiLtv

Time domain (a) (b)

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s-Domain Model for C

(b))0(

)(1

)(or

(a))0()(

)0()()(

.)(

)(capacitor,aFor

sv

sIsC

sV

CvssCV

vssVCsIdt

tdvCti

Time domain (a) (b)

For inductor: For capacitor:

si

sVsL

sI)0(

)(1

)(

)0()()( LissLIsVs

vsI

sCsV

)0()(

1)(

)0()()( CvssCVsI

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Summary

Element Z(s)

Resistor R

Inductor sL

Capacitor 1/sC

*Assuming zero initial conditions

•Impedance in the s domain.–Z(s)=V(s)/I(s)

•Admittance in the s domain.–Y(s)=1/Z(s)=I(s)/V(s)

Example 1

)(2sin2

3)(

)2()4(2

23

1883

)(

1883

4

22

22

232

tutetv

s

sssIsV

sssI

to

o

03

53

033

11

:analysismeshApplying(2)

31)F

31

(

)H1(

1)(

:domainthetion toTransforma(1)

21

21

Is

sIs

Is

Iss

ssCZ

ssLZs

tu

s

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Example 2

V5)0( ov

)0(oCv

Example 2 (Cont’d)

)()1510()(

15)()2(

10)()1(

method,residuetheApplying

2

2

1

||

tueetv

sVsB

sVsA

tto

so

so

21)2)(1(3525

10105.02

10)1(10

:nodeatanalysisnodalApplying

sB

sA

sss

V

sVVVsa

o

ooo

a

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0)0( Ii

Example 3

)()()0(

,)(

)()()(

0))((

/

0/00

000

0000

00

ieii

RL

RV

eRV

Iti

LRsRVI

sRV

LRssLVsI

sLRsV

sLRLI

sI

sV

LIsLRsI

t

t

Circuit Analysis•Operators (derivatives and integrals) into

simple multipliers of s and 1/s.

•Use algebra to solve the circuit equations.

•All of the circuit theorems and relationshipsdeveloped for dc circuits are perfectly valid inthe s domain.

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Example 1

V5)0(A1)0(

10

C

L

s

vi

sV

)(10 tu

)(3035)(

230

135

)2)(1(540

0)1.0(1

)0()0(5

0310

haveweanalysisnodalusingBy

21

1

111

tueetv

sssss

V

ssvV

si

sVVV

tt

s

Example 2

)(3035

)()101030()51030(

)()()()(

)(105)(

210

15

)2)(1(5

)(1010)(

210

110

)2)(1(10

)(3030)(

230

130

)2)(1(30

ion.superpositusingby1exampleSolve

2

2

321

23

3

22

2

21

1

tuee

tuee

tvtvtvtv

tueetv

sssss

V

tueetv

ssssV

tueetv

ssssV

tt

tt

tt

tt

tt

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Example 3Assume that no initial energy is stored.(a) Find Vo(s) using Thevenin’s theorem.(b) Find vo(0+) and vo() by applying the

initial- and final-value theorems.(c) Determine vo(t).

=10u(t)

Example 3: (a)

)4(12550

3255

55

32)32(50

50)32(

502

,32

1002

and

02

05

0)2(10:analysisnodalApplying

use,findTo

50105

11

1

11

ssssV

ZV

ssss

IV

Z

sssV

IIs

V

sV

I

sVIV

s

IVZZss

VV

ThTh

o

sc

ocTh

scx

x

x

scocThTh

Thoc

V1

=0

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Example 3: (b), (c)

4125

4125

lim)(lim)(

givestheoremvalue-finalThe

04

125lim)(lim)0(

givestheoremvalue-initialThe)4(

125)(

:(b)Solution

00

sssVv

sssVv

sssV

soso

soso

o

)()1(25.31)(

25314

125)()4(

25314

125)(

method,residuetheApplying4)4(

125)(

:(c)Solution

4

4

0

tuetv

.sVsB

.ssVA

sB

sA

sssV

to

so

so

o

|

|

Transfer Functions•The transfer function H(s) is the ratio of the output

response Y(s) to the input excitation X(s), assumingall initial conditions are zero.

)(oftransformLaplacethe:)(networktheofresponseimpulseunit:)(

output:)(inputorexcitation:)(

)()(

)(or)()()(

)(*)()(:integralnconvolutioFrom

thsHthtytx

sXsY

sHsHsXsY

thtxty

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Transfer Functions (Cont’d)•Two ways to find H(s).–Assume an input and find the output.

–Assume an output and find the input (the laddermethod: Ohm’s law + KCL).

•Four kinds of transfer functions.

)()(

Admittance)(,)()(

Impedance)(

)()(

gainCurrent)(,)()(

gainVoltage)(

sVsI

sHsIsV

sH

sIsI

sHsVsV

sHi

o

i

o

Example 1

)(4sin40)(10)(

4)1(4

40104)1(

)1(10)()(

)(

4)1()1(10

)(,1

1)(

:Solution

response.impulseitsandfunctiontransfertheFind).()(when)(4cos10)(If:Q

2222

2

22

tutetth

sss

sXsY

sH

ss

sYs

sX

tuetxtutety

t

tt

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Example 2

)212()4()4(

division,currentBy:Solution

02 ss

IsI

1122)4(4

)()(

)(

216)4(2

2

20

0

020

ssss

sIsV

sH

ssIs

IV

Q: FindH(s)=V0(s)/I0(s).

Example 2 (The Ladder Method)

1122)4(41

)(

)4(41122

givesKCLApplying)4(4

144

414

41

121

2

212law,sOhm'By

V,1Let

200

0

2

210

11

21

02

0

ssss

IIV

sH

ssss

III

sss

sV

I

ss

ssIV

VI

V

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Example 3Find(a) H(s) = Vo/Vi,(b) the impulse response,(c) the response when vi(t) = u(t) V,(d) the response when vi(t) = 8cos2t V.

Example 3: (a), (b)

i

i

iab

abo

Vss

Vss

ss

Vs

sV

Vs

V

321

)2()1(1)2()1(

)1(||11)1(||1

11

division,By voltage:Solution

)(5.0)(

231

21

321

)(

32321

11

23 tueth

ssVV

sH

sV

Vss

sV

t

i

o

iio

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Example 3: (c), (d)

V)(131

)(

31

,31

23

23

2

1

)()()(

1)()()(

(c):Sol

23 tuetv

BA

s

BsA

ss

sVsHsVs

sVtutv

to

io

ii

V)(2sin34

2cos2524

)(

42

34

4231

2524

)(

2564

,2524

,2524

423

423

4

)()()(4

8)(2cos8)(

(d):Sol

23

22

22

2

tuttetv

sss

ssV

CBA

sCBs

s

A

ss

s

sVsHsVs

ssVttv

to

o

io

ii

State Variables•The state variables are those variables which, if

known, allow all other system parameters to bedetermined by using only algebraic equations.

•In an electric circuit, the state variables are theinductor current and the capacitor voltage sincethey collectively describe the energy state of thesystem.

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State Variable Method

sintegratorofoutputs:)(

svariablestatengrepresenti

vectorstatethe

)(

)()(

)(x

whereBzAxx

asarrangedbecanequationstateThe

2

1

tx

n

tx

txtx

t

i

n

DzCxyBzAxx

)(

)()(

)(

)()(

x 2

1

2

1

tx

txtx

dttdx

dttdxdttdx

nn

)(

)()(

)(z 2

1

tz

tztz

t

m

)(

)()(

)(y 2

1

ty

tyty

t

p

State Variable Method (Cont’d)

DZ(s))BZ(A)IC(

DZ(s))CX()Y(

matrixidentitythe:I)BZ(A)I()X(

)BZ()A)X(I()BZ()AX()X(

,transformLaplacetheapplyingandconditionsinitialzeroAssuming

DzCxyBzAxx

1

1

ss

ss

sss

sssssss

B)AI(C)(H

.)(Hofrdenominatotheofdegreethan thelessis)(HofnumeratortheofdegreetheSo

.0Dcases,mostIn

matrixfeedforwadDmatrixoutputC

matrixcouplinginputBmatrixsystemA

where

DB)AI(C)(Z)(Y

)(H

1

1

ss

ss

sss

s

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Transfer Function/Matrix

SystemofPolesAofsEigenvalue

AIdetAI)(where

)()(N

AI)(N

)(H

AI1

AI

BAIC)(H

1

1

sssD

sDs

ss

s

ss

ss

How to Apply State Variable Method•Steps to apply the state variable method to circuit

analysis:

–Select the inductor current and capacitor voltage as thestate variables (define vector x, z).

–Apply KCL and KVL to obtain a set of first-orderdifferential equations (find matrix A, B).

–Obtain the output equation and put the final result in state-space representation (find matrix C).

–H(s)=C(sI-A)-1B

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Example: LRC Circuit

(4):KVL

(3))1(

)2(,byiablesoutput varandinputtheand

(1),byvariablesstatetheDefining*

122

21

1

21

uxRxxLCx

x

xeyeu

eCIxeVx

oi

oLoC

2

1

2

1

2

1

01and

10

1

10

)4()3(

x

xy

u

Lxx

LR

L

Cxx

Network Stability•A circuit is stable if its impulse response satisfies

•Two requirements for stability: H(s)=N(s)/D(s)(1) Degree of N(s) Degree of D(s).

(2) All the poles (pi) must lie in the left half of s plane.

ble)(unaccepta)(lim)()(

)1()()(

)()(

)(

satisfied,notis(1)If

01

1

jHsHjH

nsDsR

ksksksDsN

sH

js

nn

nn

0]Re[ifonly0)(

)()())((

)()()(

121

11

i

tpN

ii

N

ph

tueApspsps

sNLsHLth i

0)(lim

tht

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Network Stability (Cont’d)•A circuit is stable when all the poles of its transfer

function H(s) lie in the left half of the s plane.

•Circuits composed of passive elements (R, L, and C)and independent sources either are stable or havepoles with zero real parts.

•Active circuits or passive circuits with controlledsources can supply energy, and can be unstable.

Example 1

LC

LR

p

LCLRssL

sCsLRsC

VV

sHs

o

12where

11

11

)(

0

20

22,1

2

(unstable)0]Re[:0For

(stable)0]Re[:0,,For

2,1

2,1

pR

pCLR

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Example 2

sCkRCsR

sCk

sCsCR

II

sCR

sCk

sCsCRV

kIsCI

IsC

R

sCI

IsC

RV

i

i

2

111

istdeterminanThe

11

11

0

01

01

givesanalysismeshApplying

2

2

2

1

11

2

21

RkCR

Rkp

CRRk

p

2

02

operation,stableFor

2asgivenispolesingleThe

0:equationsticcharacteriThe

2

2

Find k for a stable circuit.

Summary•The methodology of circuit analysis using Laplace

transform–Convert each element to its s-domain model.–Obtain the s-domin solution.–Apply the inverse Laplace transform to obtain the t-domain

solution.

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Summary•The transfer function H(s) of a network is the Laplace

transform of the impulse response h(t).

•A circuit is stable when all the poles of its transferfunction H(s) lie in the left half of the s plane.

)()()(,)()(

)( sHsXsYsXsY

sH