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Circuit Applications of Laplace Transform. Chairul Hudaya, ST, M.Sc. Electric Power & Energy Studies (EPES) Department of Electrical Engineering University of Indonesia http://www.ee.ui.ac.id/epes. - PowerPoint PPT Presentation
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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
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