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 functions
2. Convolution integrals
3. RLC circuit with initial conditions
sCC
sLL
RR
1
Depok, October, 2009 Laplace Transform Electric Circuit
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
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
s
10
Depok, October, 2009 Laplace Transform Electric Circuit
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
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 s
2
)(2 sI
s
Depok, October, 2009 Laplace Transform Electric Circuit
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
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 )(sVos
2
Step 2: Find the TF
)(2)(2
2
1)/2(
/2
)(
)()( 21
tuethss
s
sV
sVsH 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
dt
tdiLtv L
L
)()(
)a1.....()0()()()]0()([)( LLLLL LisIsLissILsV
)b1.....()0()(
)(s
i
sL
sVsI LL
L
Depok, October, 2009 Laplace Transform Electric Circuit
)0()()()( LLL LisIsLsV s
i
sL
sVsI LL
L
)0()()(
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
)0(/1
)()( C
CC Cv
sC
sVsI
s
vsI
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
)(sVs
4161
s
80s4
810
)(sV
Depok, October, 2009 Laplace Transform Electric Circuit
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
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)(
ss
s
s
ssV
V)()2sin2302cos5()( 4 tuetttv t
Finding i(t),
ssss
s
s
VI
2
)208(
)96(25.1
4
82
Depok, October, 2009 Laplace Transform Electric Circuit
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
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(,A8.45
24)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
625.0
3
1//625.0
32
101010
ss
s
ssI
ss
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
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
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