depok, october, 2009 laplace transform electric circuit circuit applications of laplace transform...
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![Page 1: Depok, October, 2009 Laplace Transform Electric Circuit Circuit Applications of Laplace Transform Electric Power & Energy Studies (EPES) Department of](https://reader033.vdocument.in/reader033/viewer/2022061618/56649f175503460f94c2e4a1/html5/thumbnails/1.jpg)
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
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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
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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
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Depok, October, 2009 Laplace Transform Electric Circuit
Example 1
For the following circuit, find H(s)=Vo(s)/Vi(s). Assume zero initial conditions.
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Depok, October, 2009 Laplace Transform Electric Circuit
Solution
Transform the circuit into s-domain with zero i.c.:
)(sVs )(sVo
s
s
10
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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
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Depok, October, 2009 Laplace Transform Electric Circuit
Example 2
Obtain the transfer function H(s)=Vo(s)/Vi(s), for the following circuit.
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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
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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
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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).
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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
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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
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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,,
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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
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Depok, October, 2009 Laplace Transform Electric Circuit
)0()()()( LLL LisIsLsV s
i
sL
sVsI LL
L
)0()()(
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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
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Depok, October, 2009 Laplace Transform Electric Circuit
)0(/1
)()( C
CC Cv
sC
sVsI
s
vsI
sCsV C
CC
)0()(
1)(
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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.
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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
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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
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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
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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
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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
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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
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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
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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