Download CIRCUIT ANALYSIS USING LAPLACE TRANSFORM

CIRCUIT ANALYSIS
USING LAPLACE
TRANSFORM
METHODOLOGY
If the circuit is a linear circuit
YES
Laplace transform of the sources
of excitation: s(t)  S(s)
Laplace transform of the all the
elements in the circuit
Find the output O(s) in the
Laplace freq. domain
Obtain the time response O(t) by
taking the inverse Laplace
transform
NO
Stop or approximate
the circuit into a linear
circuit and continue
Examples of nonlinear circuits:
logic circuits, digital circuits,
or any circuits where the
output is not linearly
proportional to the input.
Examples of linear circuits:
amplifiers, lots of OPM
circuits, circuits made of
passive components (RLCs).
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.
KIRCHHOFF’S VOLTAGE LAW
 Consider the KVL in time domain:
v1 (t )  v2 (t )  v3 (t )  v4 (t )    0
 Apply the Laplace transform:
V1 (s)  V2 (s)  V3 (s)  V4 (s)    0
KIRCHHOFF’S CURRENT LAW
 Consider the KCL in time domain:
i1 (t )  i2 (t )  i3 (t )  i4 (t )    0
 Apply the Laplace transform:
I1 (t )  I 2 (t )  I 3 (t )  I 4 (t )    0
OHM’S LAW
 Consider the
Ohm’s Law in
time domain 
 Apply the
Laplace
transform 
vR (t )  iR (t ) R
VR (s)  I R (s) R
INDUCTOR
 Inductor’s voltage
– In the time domain:
di
vL (t )  L
dt
– In the s-domain:

VL (s)  L[sI L (s)  iL (0 )]
INDUCTOR
 Inductor’s current
– Rearrange VL(s) equation:

VL ( s) i(0 )
I L ( s) 

sL
s
CAPACITOR
 Capacitor’s current
– In the time domain:
dv
ic (t )  C
dt
– In the s-domain:

I c(s)  C[ sVc ( s)  vc (0 )]
CAPACITOR
 Capacitor’s voltage
– Rearranged IC(s) equation:
 
Vc(s)  1
 

1
I c(s) 
vc( 0 )
sC
s
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.

VL (s)  L[sI L (s)  iL (0 )]
 
Vc(s)  1
 

1
I c(s) 
vc( 0 )
sC
s
CIRCUIT ANALYSIS FOR
ZERO INITIAL
CONDITIONS (ICs = 0)
IMPEDANCE
 If we set all initial conditions to zero,
the impedance is defined as:
V
(
s
)
Z (s) 
I (s)
[all initial conditions=0]
IMPEDANCE & ADMITANCE
 The impedances in
the s-domain are
Z R (s)  R
Z L ( s )  sL
1
Z C ( s) 
sC
 The admittance is
defined as:
1
YR ( s ) 
R
1
YL ( s ) 
sL
YC ( s )  sC
Ex.
 Find vc(t), t>0
 vc (t ) 
0.5 F

v L (t )

1H

v R (t )

3
u (t )
Obtain s-Domain Circuit
 All ICs are zero since there is no
source for t<0
Vc (s) 
2

VL (s )

s
s

VR (s )
I (s)

3
1
s
Convert to voltage sourced
s-Domain Circuit
Vc (s) 
2

VL (s )

s
s
I (s)
VR (s) 
3


3
s
Find I(s)
2
3


By KVL :  s   3  I ( s )   0
s
s


3
 I ( s)  2
s  3s  2
Find Capacitor’s Voltage
 The capacitor’s voltage:
2
6
Vc ( s)   I ( s) 
2
s
s( s  3s  2)
 Rewritten:
6
6
Vc ( s) 

2
s( s  3s  2) s( s  1)( s  2)
Using PFE
 Expanding Vc(s) using PFE:
K3
6
K1 K 2
Vc ( s) 



s( s  1)( s  2) s s  1 s  2
 Solved for K1, K2, and K3:
6
3
6
3
Vc ( s) 



s( s  1)( s  2)
s s 1 s  2
Find v(t)
6
3
6
3
Vc ( s) 



s( s  1)( s  2)
s s 1 s  2
 Using look up table:

vc (t )   3  6e
t
 3e
2 t
 u (t )
Ex.
 Find the Thevenin and Norton
equivalent circuit at the terminal of
the inductor.
0.5 F
1H
3
u(t)
Obtain s-domain circuit
2/s
s
3
1/s
Find ZTH
2/s
3
Z TH
2
 3
s
Find VTH or Voc
+
VTH
2/s
3
-
VTH
1 3
 3 
s s
1/s
Draw The Thevenin Circuit
 Using ZTH and VTH:
2/s
3
+
-
3/s
Obtain The Norton Circuit
 The norton current is:
3
VTH
3
s
IN 


ZTH 3  2
3s  2
s
2/s
3/(3s+2)
3
Ex.
 Find v0(t) for t>0.
s-Domain Circuit Elements
Laplace
transform
all
circuit’s
elements
u (t ) 
1
s
1H  sL  s
1
3
F
1
sC

3
s
s-Domain Circuit
Apply Mesh-Current
Analysis
Loop 1
Loop 2
1  3
3
 1   I1  I 2
s  5
s
3
3

0   I1   s  5   I 2
s
s

1 2
 I1  s  5 s  3 I 2
3


Substitute I1 into eqn loop 1


1  31 2
 3
 1   s  5 s  3 I 2    I 2
s  53
s


3  s  8s  18s I 2
3
2
3
 I2  3
2
s  8s  18s
Find V0(s)
V0 ( s )  sI 2
3
 2
s  8s  18
3
2

2
2
2 ( s  4)  ( 2 )
Obtain v0(t)
3
2
Vo ( s ) 
2
2
2 ( s  4)  ( 2 )
3  4t
v0 (t ) 
e sin 2t
2
CIRCUIT ANALYSIS FOR
NON ZERO INITIAL
CONDITION (ICs ≠ 0)
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.
THE ELEMENTS LAW OF sDOMAIN
 VR ( s)  RI R ( s)

 VL ( s)  sLI L ( s)  LiL (0 )
1
1

 VC ( s) 
I C ( s) 
vC (0 )
sC
sC
THE ELEMENTS LAW OF sDOMAIN
V
(
s
)
R
 I R ( s) 
V
(
s
)
L
 I L (s) 
R

i
(
0
)
L

sL
s

 I C ( s )  sCV ( s )  CvC (0 )
TRANSFORM OF CIRCUITSRESISTOR
 In the time
domain:
i(t)
+ v(t)R
v(t)=i(t)R
 In the s-domain:
I(s)
+ V(s)R
V(s)=I(s)R
TRANSFORM OF CIRCUITSINDUCTOR
 In the time domain:
TRANSFORM OF CIRCUITSINDUCTOR
 Inductor’s voltage:
 Inductor’s current:
TRANSFORM OF CIRCUITSCAPACITOR
 In the time domain:
TRANSFORM OF CIRCUITSINDUCTOR
 Capacitor’s voltage:
 Capacitor’s current:
Ex.
 Find v0(t) if the initial voltage is
given as v0(0-)=5 V
s-Domain Circuit
Apply nodal analysis
method
V0  10 ( s 1) Vo Vo
   2  0.5
10
10 10 s
V0
Vo sVo
1
 
 
 2.5
10 s  1 10 10
1
1
 Vo ( s  2) 
 2.5
10
s 1
Cont’d
10
Vo ( s  2) 
 25
s 1
25s  35
V0 
( s  1)( s  2)
Using PFE
 Rewrite V0(s) using PFE:
25s  35
K1
K2
Vo 


( s  1)( s  2) s  1 s  2
 Solved for K1 and K2:
K1  10; K2  15
Obtain V0(s) and v0(t)
 Calculate V0(s):
10
15
Vo ( s ) 

s 1 s  2
 Obtain V0(t) using look up table:
t
2 t
vo (t )  (10e  15e )u (t )
Ex.
 The input, is(t) for the circuit below is
shown as in Fig.(b). Find i0(t)
is(t)
io (t )
is (t )
1H
(a)
1
1
2
0
(b)
t(s)
s-Domain Circuit
I o (s)
I s (s)
s
1
 Using current divider:
 s 
I o (s)  
 I ( s)  (1)
 s 1
Derive Input signal, Is
is1(t)
is(t)
t
1
0

is2(t)
0
2
t(sec)
t
0
2
Obtain Is(t) and Is(s)
 Expression for is(t):
is (t )  u(t )  u(t  2)
 Laplace transform of is(t):

1 2 s 1 1
2 s
I s ( s)   e
 1 e
s
s s

 (2)
 Substitute eqn. (2) into (1):
2 s
s (1  e )
I 0 (s) 
s ( s  1)
2 s
1
e


s 1 s 1
Inverse Laplace
transform
t
io (t )  e u(t )  e
(t 2)
u(t  2)