Depok, October, 2009 Laplace Transform Electric Circuit

Circuit Applications of Laplace Transform

Electric Power & Energy Studies (EPES)Department of Electrical Engineering

University of Indonesiahttp://www.ee.ui.ac.id/epes

Chairul Hudaya, ST, M.Sc

Depok, October, 2009 Electric Circuit

Depok, October, 2009 Laplace Transform Electric Circuit

Circuit applications

1. Transfer functions2. Convolution integrals3. RLC circuit with initial conditions

sCC

sLLRR

1

Depok, October, 2009 Laplace Transform Electric Circuit

Transfer function

h(t) y(t)x(t)

)()()( txthty

)(Input)(Output

)()()(,functionTransfer

ss

sXsYsH

In s-domain, )()()( sXsHsY In time domain,

Network

System

Depok, October, 2009 Laplace Transform Electric Circuit

Example 1

For the following circuit, find H(s)=Vo(s)/Vi(s). Assume zero initial conditions.

Depok, October, 2009 Laplace Transform Electric Circuit

Solution

Transform the circuit into s-domain with zero i.c.:

)(sVs )(sVo

s

s10

Depok, October, 2009 Laplace Transform Electric Circuit

ss

sso

Vss

Vss

Vs

s

sVs

s

sV

309220

)52)(2(2020

252

2052

20

210//4

10//4

2

Using voltage divider

309220

)()()( 2

sssVsVsH

s

o

Depok, October, 2009 Laplace Transform Electric Circuit

Example 2

Obtain the transfer function H(s)=Vo(s)/Vi(s), for the following circuit.

Depok, October, 2009 Laplace Transform Electric Circuit

Solution

Transform the circuit into s-domain (We can assume zero i.c. unless stated in the question)

)(sVs )(sVo

)(sI s2

)(2 sI

s

Depok, October, 2009 Laplace Transform Electric Circuit

Iss

IsIs

V

IIIV

s

o

9323)3(2

9)2(3

We found that

2939

9329

)()()( 2

ss

s

ss

sVsVsH

s

o

Depok, October, 2009 Laplace Transform Electric Circuit

Example 3

Use convolution to find vo(t) in the circuit ofFig.(a) when the excitation (input) is thesignal shown in Fig.(b).

Depok, October, 2009 Laplace Transform Electric Circuit

Solution

Step 1: Transform the circuit into s-domain (assume zero i.c.)

)(sVs )(sVos2

Step 2: Find the TF

)(2)(2

21)/2(

/2)()()( 21

tuethss

ssVsVsH t

s

o

L

Depok, October, 2009 Laplace Transform Electric Circuit

Step 3: Find vo(t)

dvthtvthtv

sVsHsV

s

t

so

so

)()()()()(

)()()(

0

)(20)1(20

2020

102)(

22

02

0

2

0

)(2

tttt

tttt

t to

eeee

eedee

deetv

For t < 0 0)( tvo

For t > 0

Depok, October, 2009 Laplace Transform Electric Circuit

Circuit element models

Apart from the transformations

we must model the s-domain equivalents of the circuit elements when there is involving initial condition (i.c.)

Unlike resistor, both inductor and capacitor are able to store energy

sCCsLLRR 1,,

Depok, October, 2009 Laplace Transform Electric Circuit

Therefore, it is important to consider the initial current of an inductor and the initial voltage of a capacitor

For an inductor :– Taking the Laplace transform on both sides of eqn gives

or

dttdiLtv L

L)()(

)a1.....()0()()()]0()([)( LLLLL LisIsLissILsV

)b1.....()0()()(s

isL

sVsI LLL

Depok, October, 2009 Laplace Transform Electric Circuit

)0()()()( LLL LisIsLsV s

isL

sVsI LLL

)0()()(

Depok, October, 2009 Laplace Transform Electric Circuit

For a capacitor Taking the Laplace transform on both sides of eqn gives

or

dttdvCti C

C)()(

)a2.....()0(/1

)()]0()([)( CC

CCC CvsCsVvssVCsI

)b2.....()0()(1)(s

vsIsC

sV CCC

Depok, October, 2009 Laplace Transform Electric Circuit

)0(/1

)()( CC

C CvsCsVsI

svsI

sCsV C

CC)0()(1)(

Depok, October, 2009 Laplace Transform Electric Circuit

Example 4

Consider the parallel RLC circuit of the following. Find v(t) and i(t) given that v(0) = 5 V and i(0) = −2 A.

Depok, October, 2009 Laplace Transform Electric Circuit

Solution

Transform the circuit into s-domain (use the given i.c. to get the equivalents of L and C)

)(sI

)(sVs4

161

s80s4

810

)(sV

Depok, October, 2009 Laplace Transform Electric Circuit

Then, using nodal analysis

208)96(516

961616

80)208(

1614

80)8(

48

01614

80//10

2

2

sssV

ss

ssVss

sVs

sV

ss

VI

Depok, October, 2009 Laplace Transform Electric Circuit

Since the denominator cannot be factorized, we may write it as a completion of square:

22222 2)4()2(230

2)4()4(5

4)4()96(5)(

sss

sssV

V)()2sin2302cos5()( 4 tuetttv t

Finding i(t),

sssss

sVI 2

)208()96(25.1

48

2

Depok, October, 2009 Laplace Transform Electric Circuit

A)(])2sin375.112cos6(4[)( 4 tuettti t

Using partial fractions,

sssCBs

sA

ssssssI 2

2082

)208()96(25.1)( 22

It can be shown that 75.46,6,6 CBA

Hence,

22222 2)4()2(375.11

2)4()4(64

20875.4664)(

sss

ssss

ssI

Depok, October, 2009 Laplace Transform Electric Circuit

Example 5

The switch in the following circuit moves from position a to position b at t = 0 second. Compute io(t) for t > 0.

0t

V 42

5

1F 1.0H 625.0

)(tio

a b

Depok, October, 2009 Laplace Transform Electric Circuit

Solution

The i.c. are not given directly. Hence, at firstwe need to find the i.c. by analyzing the circuitwhen t ≤ 0:

V24 )0(Li

)0(Cv

5

V0)0(,A8.4524)0( CL vi

Depok, October, 2009 Laplace Transform Electric Circuit

Then, we can analyze the circuit for t > 0 by considering the i.c.

1025.6625.0)10(3

625.03

1//625.03

210

1010

ss

sss

Iss

3)0( LLi

s625.0

s10 1

)(sIo

Let

I

Depok, October, 2009 Laplace Transform Electric Circuit

Using current divider rule, we find that

)8)(2(48

161048

1025.6625.030

1010

1

2

210

10

ssss

ssI

sII

s

so

Using partial fraction we have

28

88)(

sssIo

A)()(8)( 28 tueeti tto