APPLICATION OF THE LAPLACE TRANSFORM TO CIRCUIT ANALYSIS LEARNING GOALS Laplace circuit solutions Showing the usefulness of the Laplace transform Circuit.

APPLICATION OF THE LAPLACE TRANSFORMTO CIRCUIT ANALYSIS

LEARNING GOALS

Laplace circuit solutionsShowing 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 SOLUTIONS

We compare a conventional approach to solve differential equations with a technique using the Laplace transform

)()()( tdt

diLtRitvS :KVL

tCC

CC eKtit

dt

diLtRi )(0)()(

equationary Complement

pC iii

L

ReLKeRK t

Ct

C 0)(

pp Kti )(

case thisfor solution Particular

pS 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

dt

diLsRIsVS L)()(

)()0()( ssIissIdt

di

L

)()(1

sLsIsRIs

)(

1)(

LsRssI

LRs

K

s

K

sLRs

LsI

/)/(

/1)( 21

RsILRsK

RssIK

LRs

s

1|)()/(

1|)(

/2

01

0;11

)(

te

Rti

tLR

)()()( tdt

diLtRitvS

LInitial conditionsare automaticallyincluded

No need tosearch forparticularor comple-mentarysolutions

Only algebrais needed

Particular

Complementary

LEARNING BY DOING 0),( ttv Find

KCL using Model

dt

dvC

R

vv S

Sv

0

R

vv

dt

dvC S

Svvdt

dvRC

L)()( sVsV

dt

dvRC S

L

)()0()( ssVvssVdt

dv

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)(

RCss

RC

RCsssV

Use partial fractions to determine inverse

RCs

K

s

K

RCss

RCsV

/1)/1(

/1)( 21

1|)()/1(

1|)(

/12

01

RCs

s

sVRCsK

ssVK

0,1)(

tetv RCt

CIRCUIT ELEMENT MODELS

The 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

)()(

)()(

sIti

sVtv

SS

SS

sourcest Independen

...

)()()()(

)()()()(

sBVsItBvti

sAIsVtAitv

CDCD

CDCD

sources Dependent

Resistor

Capacitor: Model 1

)0()(1

)(0

vdxxiC

tvt

)0()()( CvsCsVsI

Source transformation

)0(1

)0(

Cv

Cs

sv

Ieq

s

vsI

CssV

)0()(

1)(

Impedance in serieswith voltage source

Capacitor: Model 2

Impedance in parallelwith current source

s

sIdxxi

t )()(

0

L

Inductor Models

))0()(()()()( issILsVtdt

diLtv

)0()( issIdt

di

L

s

i

Ls

sVsI

)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

1

1)()(1)(

ssVsIsV

Law sOhm'

)()1(1 sIs :KVL

Steady state for t<0

ANALYSIS TECHNIQUES

All 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

1

3)(

ssIS

)(1

||)( sICs

RsV So

1

103

/1

/1)(

1)(

3

sRCs

CsI

CsR

CsR

sV So

25.0)1025)(1010( 63 RC

14)1)(4(

120)( 21

s

K

s

K

sssVo

40|)()1(

40|)()4(

12

41

so

so

sVsK

sVsK

)(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

sVsV

ss

sV

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)(

s

ssVo

Could haveused voltagedivider here

s

sV1

2

)(1

Loop Analysis

ssI

4)(1

1 Loop

ssIsI

ssIsIs

12)(2)(

1))()(( 2212

2 Loop

22)1(

)3(4)(

s

ssI

22)1(

)3(8)(2)(

s

ssIsVo

Source Superposition Applying current source

Current divider

'2I

' 4( ) 2

12

os

V sss

s

Voltage divider

sss

sVo12

12

2)("

2"'

)1(

)3(8)()()(

s

ssVsVsV

ooo

Applying voltage source

Source Transformation

The resistance is redundant

Combine the sources and use currentdivider

2

124

21

2)(ss

ss

ssVo

2)1(

)3(8)(

s

ssVo

Using Thevenin’s Theorem

Reduce this part

s

s

ss

ssVOC

124412)(

Only independent sources

s

ss

sZTh

11 2

s

s

ss

sVo124

12

2)( 2

Voltagedivider

2)1(

)3(8)(

s

ssVo

Using Norton’s Theorem

2

124/124)(

s

s

s

s

ssISC

Reduce this part

sZTh

2

124

21

2)(s

s

ss

ssVo

2)1(

)3(8)(

s

ssVo

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

s

sVsI

s

sVsV :supernode KCL@

2

)()( 1 sV

sI : variablegControllin

)(1

1)( 20 sV

ssV

:divider Voltage

ssVsV /12)()( 21 :algebra the Doing

ssVsI /62/)()( 2

0)1/()(

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

2

22

ssV

ssVssVs

)54(

)3)(1(12)( 22

sss

sssV

)54(

)3(12)( 2

sss

ssVo

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

/12

2

/12

sVI

Is

sVsV

OC

OCOC

ssVOC

12)(

' 0I

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

s/12

sI /6"

s

sISC

)3(6 3

2

)(

)(

ssI

sVZ

SC

OCTH

ss

ssVo

12

32

1

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

sss

ssVo

)12)(12(542 jsjsss

)12()12()12)(12(

)3(12)(

*11

js

K

js

K

s

K

jsjss

ssV o

o

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

*11 tuKteK

js

K

js

K t

5

36|)( 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 t

o

1)2( 2 s

1)2(1)2(

)2(

12

)3(12)( 2

22

12

s

C

s

sC

s

C

ss

ssV o

o

)(]sincos[)()(

)(2122

222

1 tutCtCes

C

s

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 tt

o

One can also usequadratic factors...

LEARNING EXTENSION equations node using Find )(tio

supernode

oVSo VV

Assume zero initial conditionsImplicit circuit transformation to s-domain

SV

KCL at supernode

( ) ( )2( ( ) ( )) 0

2o o

o S

V s V sCs V s V s

s s

2

)()(,

12)(

sVsI

ssV o

oS

1615

41

61

15.0

61)( 22

s

s

ss

ssIo

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

js

K

js

K t

415

2

415

41

61

|)(4

15

4

1

415

411

j

j

sIjsKjs

o

72.15653.6

9097.0

72.6633.61K

72.156

4

15cos06.13)( 4 teti

t

o

LEARNING EXTENSION equations loop using Find )(tvo

supermesh

122

IIs

source to due constraint

0212

21

2211 Is

sIIIs

supermesh onKVL

)73.3)(27.0(

216

)14(

216)( 22

sss

s

sss

ssI

73.327.0)( 210

2

s

K

s

K

s

KsI

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

titv

tueeti

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

)2

1( 21 s

Is

ssI

232

12)( 22

ss

ssI

ssI

ssVo

1)(

2)( 2

232

72)( 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 )(4

7

4

3

js

o sVjsK

5.7614.2

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

*11 tuKteK

js

K

js

K t

34 7

( ) 4.28 cos( 76.5 )4

t

ov t e t

Circuit for t>0

LEARNING EXTENSION 0),(1 tti Determine

Initial current through inductor

Aii LL 1)0()0(

12

6 s2s

1)(1 sI

ss

ssI

1

182

2)(1

Current

divider

91 1

1( ) ( ) ( )9

tI s i t e u ts

LEARNING EXTENSION 0),( ttvo Determine

Determine initial current through inductor)0(Li

Use sourcesuperposition

Ai V 212

Ai V 3

24

AiL 3

4)0(

s2

V3

8

)(sVo

divider) (voltage

3

812

24

2)(

sssVo

)2(3

)368()(

ss

ssVo 2

21

s

K

s

K

6|)( 01 so ssVK

3

10|)()2( 22 so sVsK

)(3

86)( 2 tuetv t

o

TRANSFER FUNCTION

)(sX )(sYSystem with allinitial conditionsset to zero

)(

)()(

sX

sYsH

xadt

dxa

dt

xda

dt

xda

ybdt

dyb

dt

ydb

dt

ydb

om

m

mm

m

m

on

n

nn

n

n

11

1

1

11

1

1

...

...

equation

aldifferenti 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 sXasasa

bsbsbsY m

m

nn

01

01

...

...)(

asasa

bsbsbsH 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 Determine

In the Laplace domain, Y(s)=H(s)X(s)

)()()( sVsHsV io

1

1)()()(

ssHtueth t

2

10)()(10)( 2

ssVtuetv i

ti

)2)(1(

10)(

sssVo 21

21

s

K

s

K

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

Keths

KsH

)(

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 oto

network damped Critically :1 :3 Case

LEARNING EXAMPLE)(

)()(

sV

sVsH

i

o functiontransfer the Determine

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

Mesh analysis

212)( IIsVi

211

10 IsC

sI

)(sVi

)(1

)( 2 sIsC

sVo

Css

CsVo

/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

84

10)( 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

o22

2

)84(

101)()( 2

sss

s

ssHsY

2222)(

*221

js

K

js

K

s

KsY

)22)(22(842 jsjsss

8

10|)( 01 sssYK

)4)(22(

28|)()22( 222 jj

jsVjsK jso

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

*11 tuKteK

js

K

js

K t

21173.0

90413583.2

1425.82K

210( ) 1.46 cos(2 211 ) ( )

8t

ov 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

CssV

sVsG

in

ov

1

1

)(

)()(

LCs

LR

s

LC1

1

2

L

R

LC oo 2,12

2000o Use

poles. of Variation

POLE-ZERO PLOT/BODE PLOT CONNECTION

Bode plots display magnitude and phase information of jssG |)(

They show a cross section of G(s)

52)( 2

2

ss

ssG

Cross sectionshown by Bode

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

LCs

LR

s

LCsV

sVsG

in

o

1

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 terms

EXAMPLE ))(][cos)(()(,1

1)( 22 tuttx

s

ssX

ssH

js

K

js

K

s

K

jsjss

ssY

*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

js

K

js

K

js

X

js

XsHsY x

*

2

1)()(

)(2

1|)()( oMjsox jHXsYjsK

o

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

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

o

M

o

MtjtjMM js

X

js

XsXee

XtutX

2

1)(

2)(cos

For the general case

If ( ) cos( ) ( )

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

ss o o oM

x t X t u t

y 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 : 0

22 1

iV V V Vs

s

1

12

1V

s

Vo

:divider Voltage

443)()(

443)( 2

2

2

2

ss

ssHsV

ss

ssV io

10,2 Mo X

45354.04)2(4)2(3

)2()2( 2

2

jj

jjH

If ( ) cos( ) ( )

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

ss o o oM

x t X t u t

y t X H j t H j

Vttys )452cos(54.3)(

LEARNING EXTENSION 0),( ttvoss Determine

If ( ) cos( ) ( )

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

ss o o oM

x t X t u t

y t X H j t H j

12,2 Mo X

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

s

s

1

)(sVi

Thevenin

)(1

1)(

11

1

)( sVs

sV

s

ssV iiOC

1

1

1

1||

1,1||)(

2

s

ss

ss

sssZTh

)()(2

2)( sV

sZsV OC

Tho

)(33

2)( 2 sV

sssV io

)(sH

46.9908.6

2

61

2

364

2)2(

jjjH

)(1

1

11

2

2)( 2 sV

s

sss

sV io

)46.992cos(08.6

212)( ttvoss

APPLICATIONLAPLACE