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Circuit Applications of Laplace Transform

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

University of Indonesia

Chairul Hudaya, ST, M.Sc

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

Depok, October, 2009 Laplace Transform Electric CircuitDepok, October, 2009 Electric Circuit

Circuit applicationspp

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

sLLRR

1→→

sCC 1

→

Depok, October, 2009 Laplace Transform Electric Circuit

Transfer function

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

Network

)()()( txthty ∗=In time domain,

System

)()()(y

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

)(Input)(Output

)()()(,functionTransfer

ss

sXsYsH ==∴


)(p)(

Example 1p

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


Solution

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

s

)(sVs )(sVos10


Using voltage divider

2010//4 ⎟⎠⎞

⎜⎝⎛

Using voltage divider

sso Vs

sVs

sV2

5220

52

210//4 ++

+=++⎟

⎠⎞

⎜⎝⎛

⎟⎠

⎜⎝=

VV

ss2020

52

==

+⎟⎠

⎜⎝

ss Vss

Vss 3092)52)(2(20 2 +++++

309220

)()()( 2 ++==∴

sssVsVsH

s

o


Example 2p

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


Solution

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

2)(sI s

2

s

)(sVs )(sVo)(2 sI


We found that

IIIVo =+= 9)2(3

We found that

Iss

IsIs

Vs

o

⎟⎠⎞

⎜⎝⎛ ++=++= 9323)3(2

)(

ss ⎠⎝

99)( ssV293

9

9329

)()()( 2 ++

=++

==∴ss

s

ss

sVsVsH

s

o

s


Example 3p

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

)(sVs )(sVo2s

Step 2: Find the TFStep 2: Find the TF)(2)(

22

1)/2(/2

)()()( 21

tuethss

ssVsVsH to −=⎯→⎯

+=

+==

−L


21)/2()( sssVs ++

Step 3: Find v (t)Step 3: Find vo(t)

sVsHsV so )()()( =

λλλ dvthtvthtv s

t

so )()()()()(0∫ −=∗=

For t < 0 0)( =tvo

[ ]102)(

0

)(2

tt

t to deetv −−− ⋅=

∫

∫ λλ λFor t > 0

[ ])(20)1(20

202022

02

0

2

tttt

tttt

eeee

eedee−−−

−− == ∫ λλ λ


)(20)1(20 eeee −=−=

Circuit element models

Apart from the transformations1

t d l th d i i l t f th i itsC

CsLLRR 1,, →→→

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 , pstore energy


Therefore it is important to consider the initial current ofTherefore, it is important to consider the initial current of an inductor and the initial voltage of a capacitorFor an inductorTaking the Laplace transform on both sides of eqn gives

tdi )(ordt

tdiLtv LL

)()( =

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

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

isL

sVsI LLL +=


ssL

)0()()()( LLL LisIsLsV −=isVsI LL

L)0()()( +=)0()()()( LLL LisIsLsV

ssLsIL )( +


For a capacitortdvCti C )()( =For a capacitor

Taking the Laplace transform on both sides of eqn givesdtCtiC )( =

or)a2.....()0(

/1)()]0()([)( C

CCCC Cv

sCsVvssVCsI −=−=

/1 sC

)b2()0()(1)( vsIsV C+ )b2.....()()(s

sIsC

sV CCC +=


)0(/1

)()( CC

C CvCsVsI −=

vsIC

sV CCC

)0()(1)( += )(/1

)( CC sCssC CC )()(


Example 4p

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


Solution

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

)(sI4 80s4

)(sV

)(sVs4

161

s80s4

8−10


Then using nodal analysisThen, using nodal analysis

014=−−+

VI 01680//10

=−−⎟⎠⎞

⎜⎝⎛

+s

s

I

1614

80)8(

48

+=+

+−

sVs

sV

1696

1616

80)208( 2 +

=+=++

ss

ssVss

208)96(5161680

2 +++

=ss

sV

sss


208 ++ ss

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

)2(230)4(5)96(522222 2)4(

)2(2302)4(

)4(54)4(

)96(5)(++

+++

+=

+++

=ss

ss

ssV

V)()2sin2302cos5()( 4 tuetttv t−+=∴

Finding i(t),V 2)96(2518 +

sssss

sVI 2

)208()96(25.1

48

2 −+++

=−

=


Using partial fractionsUsing partial fractions,CBsAssI 22)96(25.1)( −

++=−

+=

sssssssssI

208)208()( 22 ++

+=++

=

It b h th t 754666 CBAIt 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

A)(])2sin375.112cos6(4[)( 4 tuettti t−+−=∴

)()(


Example 5p

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

Ω 5 a b

0=t )(tio

V 42 Ω1F 1.0H 625.0


Solution

The i c are not given directly Hence at firstThe i.c. are not given directly. Hence, at firstwe need to find the i.c. by analyzing the circuitwhen t ≤ 0: Ω5when t ≤ 0:

+

Ω5

V24 )0(Li

−)0(Cv

V0)0(,A8.424)0( ===∴ CL vi


)(,5

)( CL

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

s625.010 1

)(sIo

3)0( =LLi

s10 1I

Let

( ) 1025.6625.0)10(3

625.03

1//625.03

210

1010 +++−

=+

−=

+−

=+ ss

sss

Iss


Using current divider rule we find thatUsing current divider rule, we find that 3010

2

10 −=== III s

o

48481025.6625.0101 210

−=

−=

++++ sssso

)8)(2(16102 ++=

++=

ssss

Using partial fraction we have88)(I

28)(

+−

+=

sssIo

A)()(8)( 28 tt


A)()(8)( 28 tueeti tto

−− −=∴

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