Depok, October, 2009 Laplace Transform Electric Circuit Circuit Applications of Laplace Transform Electric Power & Energy Studies (EPES) Department of

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


Circuit applications

1. Transfer functions

2. Convolution integrals

3. RLC circuit with initial conditions

sCC

sLL

RR

1


Transfer function

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

)()()( txthty

)(Input

)(Output

)(

)()(,functionTransfer

s

s

sX

sYsH

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

Network

System


Example 1

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


Solution

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

)(sVs )(sVo

s

s

10


ss

sso

Vss

Vss

Vs

s

sVs

s

sV

3092

20

)52)(2(20

20

252

2052

20

210

//4

10//4

2

Using voltage divider

3092

20

)(

)()(

2

sssV

sVsH

s

o


Example 2

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


Solution

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

)(sVs )(sVo

)(sI s

2

)(2 sI

s


Iss

IsIs

V

IIIV

s

o

932

3)3(2

9)2(3

We found that

293

9

932

9

)(

)()(

2

ss

s

ss

sV

sVsH

s

o


Example 3

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


Solution

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

)(sVs )(sVos

2

Step 2: Find the TF

)(2)(2

2

1)/2(

/2

)(

)()( 21

tuethss

s

sV

sVsH t

s

o

L


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


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,,


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

dt

tdiLtv L

L

)()(

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

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

)(s

i

sL

sVsI LL

L


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

i

sL

sVsI LL

L

)0()()(


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

or

dt

tdvCti C

C

)()(

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

)()]0()([)( C

CCCC Cv

sC

sVvssVCsI

)b2.....()0(

)(1

)(s

vsI

sCsV C

CC


)0(/1

)()( C

CC Cv

sC

sVsI

s

vsI

sCsV C

CC

)0()(

1)(


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.


Solution

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

)(sI

)(sVs

4161

s

80s4

810

)(sV


Then, using nodal analysis

208

)96(516

96

16

16

80

)208(

16

14

80

)8(

4

8

016

1480

//10

2

2

ss

sV

s

s

ss

Vss

s

Vs

s

V

ss

VI


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)(

ss

s

s

ssV

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

Finding i(t),

ssss

s

s

VI

2

)208(

)96(25.1

4

82


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

Using partial fractions,

sss

CBs

s

A

ssss

ssI

2

208

2

)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

208

75.4664)(

ss

s

sss

s

ssI


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


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(,A8.45

24)0( CL vi


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

1025.6625.0

)10(3

625.0

3

1//625.0

32

101010

ss

s

ssI

ss

3)0( LLi

s625.0

s10 1

)(sIo

Let

I


Using current divider rule, we find that

)8)(2(

48

1610

48

1025.6625.0

30

10

10

1

2

210

10

ssss

ssI

sII

s

so

Using partial fraction we have

2

8

8

8)(

sssIo

A)()(8)( 28 tueeti tto

Documents

Depok, October, 2009 Laplace Transform Electric Circuit Circuit Applications of Laplace Transform Electric Power & Energy Studies (EPES) Department of