APPLICATION OF THE LAPLACE TRANSFORMTO CIRCUIT ANALYSIS

LEARNING GOALSLaplace circuit solutions

Showing the usefulness of the Laplace transform

Circuit Element ModelsTransforming circuits into the Laplace domain

Analysis TechniquesAll standard analysis techniques, KVL, KCL, node,loop analysis, Thevenin’s theorem are applicable

Transfer FunctionThe concept is revisited and given a formal meaning

Pole-Zero Plots/Bode PlotsEstablishing the connection between them

Steady State ResponseAC analysis revisited

LAPLACE CIRCUIT SOLUTIONSWe compare a conventional approach to solve differential equations with a technique using the Laplace transform

)()()( tdtdiLtRitvS :KVL

tCC

CC eKtit

dtdiLtRi )(0)()(

equationary Complement

pC iii

LReLKeRK t

Ct

C 0)(

pp Kti )(case thisfor solution ParticularpS RKv 1

tLR

CeKR

ti

1)( 0000 ) i( for t(t)vS

conditionsboundary Use

0;11)(

te

Rti

tLR

“Take Laplace transform” of the equation

dtdiLsRIsVS L)()(

)()0()( ssIissIdtdi

L

)()(1 sLsIsRIs

)(

1)(LsRs

sI

LRsK

sK

sLRsLsI

/)/(/1)( 21

RsILRsK

RssIK

LRs

s

1|)()/(

1|)(

/2

01

0;11)(

te

Rti

tLR

)()()( tdtdiLtRitvS

LInitial conditionsare automaticallyincluded

No need tosearch forparticularor comple-mentarysolutions

Only algebrais needed

Particular

Complementary

LEARNING BY DOING 0),( ttv Find

KCL using Model

dtdvC

Rvv S

Sv

0

Rvv

dtdvC S

SvvdtdvRC

L)()( sVsV

dtdvRC S

L

)()0()( ssVvssVdtdv

L

ssVtuv SS

1)()(

0)0(0,0)( vttvSInitial conditiongiven in implicitform

In the Laplace domain the differentialequation is now an algebraic equation

ssVsRCsV 1)()(

)/1(/1

)1(1)(

RCssRC

RCsssV

Use partial fractions to determine inverse

RCsK

sK

RCssRCsV

/1)/1(/1)( 21

1|)()/1(1|)(

/12

01

RCs

s

sVRCsKssVK

0,1)(

tetv RCt

CIRCUIT ELEMENT MODELSThe method used so far follows the steps:1. Write the differential equation model2. Use Laplace transform to convert the model to an algebraic form

For a more efficient approach:1. Develop s-domain models for circuit elements2. Draw the “Laplace equivalent circuit” keeping the interconnections and replacing the elements by their s-domain models3. Analyze the Laplace equivalent circuit. All usual circuit tools are applicable and all equations are algebraic.

)()()()( sRIsVtRitv

)()()()(

sItisVtv

SS

SS

sourcest Independen

...)()()()()()()()(

sBVsItBvtisAIsVtAitv

CDCD

CDCD

sources Dependent

Resistor

Capacitor: Model 1

)0()(1)(0

vdxxiC

tvt

)0()()( CvsCsVsI

Source transformation

)0(1

)0(

Cv

Cs

sv

Ieq

svsI

CssV )0()(1)(

Impedance in serieswith voltage sourceCapacitor: Model 2

Impedance in parallelwith current source

ssIdxxi

t )()(0

L

Inductor Models

))0()(()()()( issILsVtdtdiLtv

)0()( issIdtdi

L

si

LssVsI )0()()(

LEARNING BY DOING

Ai 1)0(

Determine the model in the s-domain and the expression forthe voltage across the inductor

Inductor withinitial current

Equivalent circuit in s-domain

11)()(1)(

s

sVsIsV

Law sOhm')()1(1 sIs :KVL

Steady state for t<0

ANALYSIS TECHNIQUESAll the analysis techniques are applicable in the s-domain

LEARNING EXAMPLEDraw the s-domain equivalent and find the voltage in boths-domain and time domain

0)0(0,0)( oS vtti

One needs to determine the initial voltageacross the capacitor

13)(

s

sIS

)(1||)( sICs

RsV So

1103

/1/1)(1)(

3

sRCsCsI

CsR

CsR

sV So

25.0)1025)(1010( 63 RC

14)1)(4(120)( 21

sK

sK

sssVo

40|)()1(40|)()4(

12

41

so

so

sVsKsVsK

)(40)( 4 tueetv tto

LEARNING EXAMPLEtheorem. sNorton' and sThevenin' tion,transforma source

ion,superposit analysis, loop analysis, node using Find )(tvo

Assume all initial conditions are zero

Node Analysis

01)()(

12)(4 11

s

sVsVs

ssV

so

1 V@KCL

01)()(

2)( 1

s

sVsVsV oo

oKCL@V

s

2

0)()21()(2

124)()()1(

1

21

2

sVsssVs

ssVssVs

o

o

)1( 2s

s2

2)1()3(8)(

sssVo

Could haveused voltagedivider here

s

sV12

)(1

Loop Analysis

ssI 4)(1

1 Loop

ssIsI

ssIsIs 12)(2)(1))()(( 2212

2 Loop

22 )1()3(4)(

sssI

22 )1()3(8)(2)(

sssIsVo

Source Superposition Applying current source

Current divider

'2I

' 4( ) 2 12o

sV sss

s

Voltage divider

sss

sVo12

12

2)("

2

"'

)1()3(8)()()(

sssVsVsV

ooo

Applying voltage source

Source Transformation

The resistance is redundant

Combine the sources and use currentdivider

2

124

212)(ss

ss

ssVo

2)1()3(8)(

sssVo

Using Thevenin’s TheoremReduce this part

ss

ss

ssVOC

124412)(

Only independent sources

sss

sZTh

11 2

ss

ss

sVo124

12

2)( 2

Voltagedivider

2)1()3(8)(

sssVo

Using Norton’s Theorem

2124/124)(

ss

ss

ssISC

Reduce this part

sZTh

2124

212)(s

s

ss

ssVo

2)1()3(8)(

sssVo

Currentdivision

LEARNING EXAMPLE zero be to conditions initial all Assume voltagethe Determine ).(tvo

. Three loops, three non-reference nodes

. One voltage source between non-reference nodes - supernode. One current source. One loop current known or supermesh. If v_2 is known, v_o can be obtained with a voltage divider

Selecting the analysis technique:

Transforming the circuit to s-domain

ssVsV 12)()( 12 :constraint Supernode

01)()(2

/2)(

2)( 211

ssVsI

ssVsV :supernode KCL@

2)()( 1 sVsI : variablegControllin

)(1

1)( 20 sVs

sV

:divider Voltage

ssVsV /12)()( 21 :algebra the DoingssVsI /62/)()( 2

0)1/()(

)/62/)((2/12)()1)(2/1(

2

22

ssVssVssVs

)54()3)(1(12)( 22

ssssssV

)54()3(12)( 2

sssssVo

Continued ... theorem sThevenin' using Compute )(sVo

-keep dependent source and controlling variable in the same sub-circuit-Make sub-circuit to be reduced as simple as possible-Try to leave a simple voltage divider after reduction to Thevenin equivalent

sVOC /12

2/12'

0'2/2

/122

/12

sVI

Is

sVsV

OC

OCOC

ssVOC

12)(

' 0I

0)/2/("2""2 sIIIISC

s/12

sI /6"

ssISC

)3(6 32

)()(

ssIsVZ

SC

OCTH

ss

ssVo

12

321

1)(

Continued … Computing the inverse Laplace transform

Analysis in the s-domain has established that the Laplace transform of the output voltage is

)54()3(12)( 2

sssssVo

)12)(12(542 jsjsss

)12()12()12)(12()3(12)(

*11

jsK

jsK

sK

jsjssssV o

o

)()cos(||2)()( 11

*11 tuKteK

jsK

jsK t

536|)( 0 soo ssVK

57.16179.343.19879.3)902(43.1535

45212)2)(12(

)11(1212|)()12(1 jj

jjssoVjsK

)(57.161cos(59.75

36)( 2 tutetv to

1)2( 2 s

1)2(1)2()2(

12)3(12)( 2

22

12

s

Cs

sCs

Css

ssV oo

)(]sincos[)()(

)(2122

222

1 tutCtCes

Cs

sC t

5/36|)( 0 soo ssVC

])2([)1)2(()3(12 212 CsCssCs o 5/12610/362122 22 CCCs o

5/360: 11 CCCo2s of tscoefficien Equating

)(sin5

12)cos1(5

36)( 22 tutetetv tto

One can also usequadratic factors...

LEARNING EXTENSION equations node using Find )(tio

supernode

oVSo VV

Assume zero initial conditionsImplicit circuit transformation to s-domainSV

KCL at supernode( ) ( )2( ( ) ( )) 0

2o o

o SV s V sCs V s V s

s s

2)()(,12)( sVsI

ssV o

oS

1615

41

6115.0

61)( 22

s

sssssIo

algebra the Doing

415

41

415

41

415

41

415

41

61)(*

1 1

js

K

js

K

jsjs

ssIo

)()cos(||2)()( 11

*11 tuKteK

jsK

jsK t

4152

415

4161

|)(415

41

415

411

j

jsIjsK

jso

72.15653.69097.0

72.6633.61K

72.156

415cos06.13)( 4 teti

t

o

LEARNING EXTENSION equations loop using Find )(tvo

supermesh

122 IIs

source to due constraint

0212212211 I

ssIII

s

supermesh onKVL

)73.3)(27.0(216

)14(216)( 22

sss

ssss

ssI

73.327.0)( 210

2

sK

sK

sKsI

2|)( 020 sssIK

48.2)73.327.0)(27.0(

2)27.0(16|)()27.0( 27.021

ssIsK

47.4)27.073.3)(73.3(

2)73.3(16|)()73.3( 73.322

ssIsK

)(2)(

)(47.448.22)(

2

73.327.02

titvtueeti

o

tt

Determine inverse transform

TRANSIENT CIRCUIT ANALYSIS USING LAPLACE TRANSFORM

For the study of transients, especially transients due to switching, it is importantto determine initial conditions. For this determination, one relies on the properties:

1. Voltage across capacitors cannot change discontinuously2. Current through inductors cannot change discontinuously

LEARNING EXAMPLE 0),( ttvo Determine

AiVv LC 1)0(,1)0( itshortcircu are inductors

circuit open are capacitors case DCFor

Assume steady state for t<0 and determinevoltage across capacitors and currents through inductors

)0( Li

)0(Cv

Circuit for t>0

Use mesh analysis

Laplace

14)1( 21 s

sIIs

11)21( 21 s

Is

ssI

23212)( 22

ssssI

ssI

ssVo

1)(2)( 2

23272)( 2

ss

ssVo

Now determine the inverse transform

roots conjugatecomplex 042 acb

47

43

47

43

)(*

1 1

js

K

js

KsVo

47

43

1 )(47

43

jso sVjsK

5.7614.2

)()cos(||2)()( 11

*11 tuKteK

jsK

jsK t

34 7( ) 4.28 cos( 76.5 )

4t

ov t e t

Circuit for t>0

LEARNING EXTENSION 0),(1 tti Determine

Initial current through inductorAii LL 1)0()0(

12

6 s2s1)(1 sI

ssssI 1182

2)(1

Currentdivider

91 1

1( ) ( ) ( )9

tI s i t e u ts

LEARNING EXTENSION 0),( ttvo Determine

Determine initial current through inductor)0(Li Use source

superpositionAi V 212

Ai V 32

4

AiL 34)0(

s2

V38

)(sVo

divider) (voltage

3812

242)(

sssVo

)2(3)368()(

ssssVo 2

21

sK

sK

6|)( 01 so ssVK

310|)()2( 22 so sVsK

)(386)( 2 tuetv t

o

TRANSFER FUNCTION

)(sX )(sYSystem with allinitial conditionsset to zero

)()()(

sXsYsH

xadtdxa

dtxda

dtxda

ybdtdyb

dtydb

dtydb

om

m

mm

m

m

on

n

nn

n

n

11

1

1

11

1

1

...

...

equationaldifferenti a is system thefor model the If

)(sYsdt

yd kk

k

L

zero are conditions initial all If

)()(...)(

)()(...)(

01

01

sXassXasXsa

sYbssYbsYsbm

m

nn

)(......)(

01

01 sXasasabsbsbsY m

m

nn

01

01

......)(

asasabsbsbsH m

m

nn

1)()()( sXttx function impulse theFor

The inverse transform of H(s) is alsocalled the impulse response of the system

If the impulse response is known then onecan determine the response of the systemto ANY other input

H(s) can also be interpreted as the Laplacetransform of the output when the input isan impulse and all initial conditions are zero

LEARNING EXAMPLE response impulse has networkA u(t)eth t)(

)(10)()( 2 tuetvtv tio

input thefor , response, the DetermineIn the Laplace domain, Y(s)=H(s)X(s)

)()()( sVsHsV io

11)()()(

ssHtueth t

210)()(10)( 2

ssVtuetv i

ti

)2)(1(10)(

sssVo 21

21

sK

sK

10|)()1( 11 so sVsK

10|)()2( 22 so sVsK

)(10)( 2 tueetv tto

Impulse response of first and second order systems

First order system

t

KethsKsH

)(

1)(

Normalized second order system

200

2

20

2)(

sssH

12002,1 s :poles

tt eKeKth )1(2

)1(1

200

200)(

network Overdamped :1 :1 Case

network dUnderdampe :1 :2 Case 2

002,1 1 js :poles)1cos()( 2 tKeth o

to

network damped Critically :1 :3 Case

LEARNING EXAMPLE)()()(

sVsVsH

i

o functiontransfer the Determine

Transform the circuit to the Laplacedomain. All initial conditions set to zero

Mesh analysis

212)( IIsVi

21110 I

sCsI

)(sVi

)(1)( 2 sIsC

sVo

CssCsVo /1)2/1(

)2/1()( 2

25.025.02,1 js :poles 8FC a)

25.02,1 s :poles 16FC b)

073.0,427.02,1 s :poles 32FC c)

LEARNING EXTENSION

8410)( 2

ss

ssH

Determine the pole-zero plot, the type of damping and the unit step response

10-z :zero

22084 2,12 jsss

:poles

jx

x

2j

2O10

842 ss2o

o222

)84(101)()( 2

sss

ss

sHsY

2222)(

*221

jsK

jsK

sKsY

)22)(22(842 jsjsss

810|)( 01 sssYK

)4)(22(28|)()22( 222 jj

jsVjsK jso

)()cos(||2)()( 11

*11 tuKteK

jsK

jsK t

21173.0

90413583.21425.8

2K

210( ) 1.46 cos(2 211 ) ( )8

tov t e t u t

Second order networks: variation of poles with damping ratio

Normalized second order system

200

2

20

2)(

sssH

12002,1 s :poles

network dUnderdampe :1 :2 Case 2

002,1 1 js :poles

cos

LEARNING EXAMPLE

RLsCs

CssVsVsG

in

ov

1

1

)()()(

LCs

LRs

LC1

1

2

LR

LC oo 2,12

2000o Usepoles. of Variation

POLE-ZERO PLOT/BODE PLOT CONNECTIONBode plots display magnitude and phase information of jssG |)(

They show a cross section of G(s)

52)( 2

2

ssssG

Cross sectionshown by Bode

If the poles get closer toimaginary axis the peaksand valleys are morepronounced

LCs

LRs

LCsVsVsG

in

o1

1

)()()(

2

Cross section

Front view

Due to symmetryshow only positivefrequencies

Amplitude Bode plot

Uses log scales

STEADY STATE RESPONSE)()()( sXsHsY Response when all initial conditions are zero

Laplace uses positive time functions. Even for sinusoids the response containstransitory termsEXAMPLE ))(][cos)(()(,

11)( 22 tuttx

sssX

ssH

jsK

jsK

sK

jsjssssY

*221

1))()(1()(

)()cos(||2)( 22 tuKtKKety t

transient Steady state response

If interested in the steady state responseonly, then don’t determine residuesassociated with transient terms

termstransient

oo

x

o

M

o

M

jsK

jsK

jsX

jsXsHsY x

*

21)()(

)(21|)()( oMjsox jHXsYjsK

o

termstransient )cos(||2)( 2KtKty ox

))(cos(|)(|)( oooMss jHtjHXty

o

M

o

MtjtjMM js

Xjs

XsXeeXtutX

21)(

2)(cos

For the general case

If ( ) cos( ) ( )( ) | ( ) | cos( ( ) )

oM

ss o o oM

x t X t u ty t X H j t H j

LEARNING EXAMPLEDetermine the steady state response

Transform the circuit to the Laplace domain.Assume all initial conditions are zero

1 1 11KCL@V : 022 1

iV V V Vs

s

112

1 V

s

Vo

:divider Voltage

443)()(

443)( 2

2

2

2

ssssHsV

ssssV io

10,2 Mo X

45354.04)2(4)2(3

)2()2( 2

2

jjjjH

If ( ) cos( ) ( )( ) | ( ) | cos( ( ) )

oM

ss o o oM

x t X t u ty t X H j t H j

Vttys )452cos(54.3)(

LEARNING EXTENSION 0),( ttvoss DetermineIf ( ) cos( ) ( )

( ) | ( ) | cos( ( ) )oM

ss o o oM

x t X t u ty t X H j t H j

12,2 Mo X

Transform circuit to Laplace domain.Assume all initial conditions are zero

s

s1

)(sVi

Thevenin

)(1

1)(11

1

)( sVs

sV

s

ssV iiOC

11

11||1,1||)(

2

s

sss

ss

ssZTh

)()(2

2)( sVsZ

sV OCTh

o

)(33

2)( 2 sVss

sV io

)(sH

46.9908.62

612

3642)2(

jjjH

)(1

1

112

2)( 2 sVs

sss

sV io

)46.992cos(08.6212)( ttvoss

APPLICATIONLAPLACE