applications laplace transform
TRANSCRIPT
1
CIRCUIT ANALYSIS USING LAPLACE TRANSFORM
2
METHODOLOGY
Examples of nonlinear circuits:logic circuits, digital circuits,or any circuits where theoutput is not linearlyproportional to the input.
Examples of linear circuits:amplifiers, lots of OPMcircuits, circuits made ofpassive components (RLCs).
If the circuit is a linear circuit
Laplace transform of the sourcesof excitation: s(t) S(s)
Laplace transform of the all theelements in the circuit
Find the output O(s) in theLaplace freq. domain
Obtain the time response O(t) bytaking the inverse Laplace
transform
Stop or approximatethe circuit into a linear
circuit and continue
NO
YES
3
THE s-DOMAIN CIRCUITS
Equation of circuit analysis: integrodifferential equations.
Convert to phasor circuits for AC steady state.
Convert to s-domain using Laplace transform.
KVL, KCL, Thevenin,etc.
4
KIRCHHOFF’S VOLTAGE LAW
Consider the KVL in time domain:
Apply the Laplace transform:
0)()()()( 4321 tvtvtvtv
0)()()()( 4321 sVsVsVsV
5
KIRCHHOFF’S CURRENT LAW
Consider the KCL in time domain:
Apply the Laplace transform:
0)()()()( 4321 tItItItI
0)()()()( 4321 titititi
6
OHM’S LAW
Consider the Ohm’s Law in time domain
Apply the Laplace transform
RsIsV RR )()(
Rtitv RR )()(
7
INDUCTOR
Inductor’s voltage– In the time domain:
– In the s-domain:
dt
diLtvL )(
)]0()([)( LLL issILsV
8
INDUCTOR
Inductor’s current– Rearrange VL(s) equation:
s
i
sL
sVsI L
L
)0()()(
9
CAPACITOR
Capacitor’s current– In the time domain:
– In the s-domain:
dt
dvCtic )(
)]0()([ ccc vssVC(s)I
10
CAPACITOR
Capacitor’s voltage– Rearranged IC(s) equation:
)(vs(s)IsC(s)V ccc 011
11
RLC VOLTAGE
The voltage across the RLC elements in the s-domain is the sum of a term proportional to its current I(s) and a term that depends on its initial condition.
)]0()([)( LLL issILsV
)(vs(s)IsC(s)V ccc 011
12
CIRCUIT ANALYSIS FOR ZERO INITIAL CONDITIONS (ICs = 0)
13
IMPEDANCE
If we set all initial conditions to zero, the impedance is defined as:
[all initial conditions=0]
)()()( sI
sVsZ
14
IMPEDANCE & ADMITANCE
The impedances in the s-domain are
The admittance is defined as:
sCsZ
sLsZ
RsZ
C
L
R
1)(
)(
)(
sCsYsL
sY
RsY
C
L
R
)(
1)(
1)(
15
Ex.
Find vc(t), t>0
H1
F5.0
3 )(tu
)(tvc
)(tvL
)(tvR
16
Obtain s-Domain Circuit
All ICs are zero since there is no source for t<0
ss
2
3s
1
)(sVc
)(sVL
)(sVR
)(sI
17
Convert to voltage sourced s-Domain Circuit
ss
23
s
3
)(sVc
)(sVL
)(sVR
)(sI
18
23
3)(
03
)(32
2
sssI
ssI
ss :KVL By
Find I(s)
19
Find Capacitor’s Voltage
The capacitor’s voltage:
Rewritten:
)23(
6)(
2)(
2
sss
sIs
sVc
)2)(1(
6
)23(
6)(
2
sssssssVc
20
Using PFE
Expanding Vc(s) using PFE:
Solved for K1, K2, and K3:
21)2)(1(
6)( 321
s
K
s
K
s
K
ssssVc
2
3
1
63
)2)(1(
6)(
ssssss
sVc
21
Find v(t)
Using look up table:
2
3
1
63
)2)(1(
6)(
ssssss
sVc
)(363)( 2 tueetv ttc
22
Ex. Find the Thevenin and Norton
equivalent circuit at the terminal of the inductor.
1 H
0 .5 F
3 u (t )
23
Obtain s-domain circuit
s
2 /s
3 1 /s
24
Find ZTH
2 /s
3
sZTH
23
25
Find VTH or Voc
2 /s
3 1 /s+V T H
-
ssVTH
313
26
Draw The Thevenin Circuit
Using ZTH and VTH:
2 /s 3+- 3 /s
27
Obtain The Norton Circuit The norton current is:
2 /s
33 /(3s + 2)
23
323
3
ss
sZ
VI
TH
THN
28
Ex.
Find v0(t) for t>0.
29
s-Domain Circuit Elements
Laplace transform all circuit’s elements
ssC
s
F
ssLH
tu
3131
1
1
)(
30
s-Domain Circuit
31
Apply Mesh-Current Analysis
21
3
5
31
1I
sI
s
Loop 1
Loop 2
22
1
21
353
1
35
30
IssI
Is
sIs
32
Substitute I1 into eqn loop 1
sssI
Isss
Is
Isss
188
3
1883
335
3
1
5
31
1
232
223
222
33
Find V0(s)
22
2
20
)2()4(
2
2
3
188
3
)(
s
ss
sIsV
34
Obtain v0(t)
tetv t 2sin2
3)( 4
0
22 )2()4(
2
2
3)(
ssVo
35
Ex. The input, is(t) for the circuit below is
shown as in Fig.(b). Find i0(t)
1
0 2 t(s)
is(t)
(b)
)(tis
1H1
)(tio
(a)
36
s-Domain Circuit
)(sIs1s
)(sIo
37
Using current divider:
)1()(1
)(
sI
s
ssIo
38
1
0 2 t(sec)
is(t)
Derive Input signal, Is
0t
is1(t)
0 2t
is2(t)
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Obtain Is(t) and Is(s)
Expression for is(t):
Laplace transform of is(t):
)2()()( tututis
)2(1111
)( 22 sss e
sse
ssI
40
Substitute eqn. (2) into (1):
11
1
)1(
)1()(
2
2
0
s
e
s
ss
essI
s
s
41
)2()()( )2( tuetueti tto
Inverse Laplace transform
42
CIRCUIT ANALYSIS FOR NON-ZERO INITIAL CONDITION (ICs ≠ 0)
43
TIME DOMAIN TO s-DOMAIN CIRCUITS
s replaced t in the unknown currents and voltages.
Independent source functions are replaced by their s-domain transform pair.
The initial condition serves as a second element, the initial condition generator.
44
THE ELEMENTS LAW OF s-DOMAIN
)0(1
)(1
)(
)0()()(
)()(
CCC
LLL
RR
vsC
sIsC
sV
LissLIsV
sRIsV
45
THE ELEMENTS LAW OF s-DOMAIN
)0()()(
)0()()(
)()(
CC
LLL
RR
CvssCVsI
si
sLsVsI
RsVsI
46
TRANSFORM OF CIRCUITS- RESISTOR
In the time domain:
In the s-domain:
i (t ) + v (t )-
R v (t )= i (t )R
I (s ) + V (s )-
R
V (s )= I (s )R
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TRANSFORM OF CIRCUITS- INDUCTOR
In the time domain:
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TRANSFORM OF CIRCUITS- INDUCTOR
Inductor’s voltage: Inductor’s current:
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TRANSFORM OF CIRCUITS- CAPACITOR
In the time domain:
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TRANSFORM OF CIRCUITS- INDUCTOR
Capacitor’s voltage: Capacitor’s current:
51
Ex.
Find v0(t) if the initial voltage is given as v0(0-)=5 V
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s-Domain Circuit
53
Apply nodal analysis method
5.21
1)2(
10
1
5.210101
1
10
5.021010
0
10
)1(10
0
ssV
sVV
s
V
VVV
o
oo
s
oos
54
Cont’d
)2)(1(
3525
251
10)2(
0
ss
sV
ssVo
55
Using PFE
Rewrite V0(s) using PFE:
Solved for K1 and K2:
21)2)(1(
3525 21
s
K
s
K
ss
sVo
15;10 21 KK
56
Obtain V0(s) and v0(t)
Calculate V0(s):
Obtain V0(t) using look up table:
2
15
1
10)(
sssVo
)()1510()( 2 tueetv tto