Laplace Transform Circuit Application.ppt - Website Staff...
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Transcript of Laplace Transform Circuit Application.ppt - Website Staff...
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 ==∴
Depok, October, 2009 Laplace Transform Electric Circuit
)(p)(
Example 1p
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 withTransform the circuit into s-domain with zero i.c.:
s
)(sVs )(sVos10
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
Example 2p
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 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
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
Example 3p
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-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
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
)(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
Depok, October, 2009 Laplace Transform Electric Circuit
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 +=
Depok, October, 2009 Laplace Transform Electric Circuit
ssL
)0()()()( LLL LisIsLsV −=isVsI LL
L)0()()( +=)0()()()( LLL LisIsLsV
ssLsIL )( +
Depok, October, 2009 Laplace Transform Electric Circuit
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 +=
Depok, October, 2009 Laplace Transform Electric Circuit
)0(/1
)()( CC
C CvCsVsI −=
vsIC
sV CCC
)0()(1)( += )(/1
)( CC sCssC CC )()(
Depok, October, 2009 Laplace Transform Electric Circuit
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.
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
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 −+++
=−
=
Depok, October, 2009 Laplace Transform Electric Circuit
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−+−=∴
)()(
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
)(,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
Depok, October, 2009 Laplace Transform Electric Circuit
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
Depok, October, 2009 Laplace Transform Electric Circuit
A)()(8)( 28 tueeti tto
−− −=∴