SARVAJANIK COLLEGE OF
ENGG. & TECH.
• Patel Tamanna
(130420117043)
• Potla Divya
(130420117045)
• Shah Rajvi
(130420117053)
• Shethnawala Aishwarya
(130420117056)
Branch Name :
Instrumentation & Control
(017)
1
Ch-7 CIRCUIT
ANALYSIS USING
LAPLACE TRANSFORM
2
Defination
3
 Transforms –
A mathematical conversion
from one way of thinking to another to make
a problem easier to solve.
transform
solution
in transform
way of
thinking
inverse
transform
4
The Laplace Transform
Transform Pairs:
f(t)
F(s)  ( t )
F(s)
1
1
u( t ) f (t )
F ( s)
____________________________________ s
1
 st
e
sa
1
t
2
s
n!
n
t
s n 1
The Laplace Transform
Transform Pairs:
f(t)
F(s) at
te
n  at
t e
sin( wt )
cos( wt )
F(s)
1
s  a 2
n!
( s  a )n 1
w
s2  w2
s
s2  w2
The Laplace Transform
Transform Pairs:
f(t)
F(s)
e sin( wt )
 at
e
 at
cos( wt )
sin( wt   )
cos( wt   )
F(s)
w
(s  a)2  w 2
sa
2
2
(s  a)  w
s sin   w cos 
s2  w2
s cos   w sin 
2
2
s w
Yes !
The Laplace Transform
Theorem:
Initial Value Theorem:
If the function f(t) and its first derivative are Laplace transformable and f(t)
Has the Laplace transform F(s),& the lim sF ( s ) exists, then
s
lim sF ( s )  lim f ( t )  f (0)
s
t 0
Initial Value
Theorem
The utility of this theorem lies in not having to take the inverse of F(s)
in order to find out the initial condition in the time domain. This is
particularly useful in circuits and systems.
The Laplace Transform
Theorem:
Final Value Theorem:
If the function f(t) and its first derivative are Laplace transformable and f(t)
lim sF ( s ) exists, then
has the Laplace transform F(s), and the
s
lim sF ( s )  lim f ( t )  f ( )
s0
t 
Final Value
Theorem
Again, the utility of this theorem lies in not having to take the inverse
of F(s) in order to find out the final value of f(t) in the time domain.
This is particularly useful in circuits and systems.
Apply Initial- and Final-Value Theorems to
this Example

Laplace
transform of
the function.

Apply Finalvalue theorem

Apply intialvalue theorem
10
RESISTOR
 Consider the
Ohm’s Law in
time domain 
 Apply the
Laplace
transform 
vR (t )  iR (t ) R
VR (s)  I R (s) R
11
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 )]
12
INDUCTOR
 Inductor’s current
– Rearrange VL(s) equation:

VL ( s) i (0 )
I L ( s) 

sL
s
13
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 )]
14
CAPACITOR
 Capacitor’s voltage
– Rearranged IC(s) equation:
 sC 
Vc(s)  1
 s v ( 0
I c(s)  1
c

)
15
IMPEDANCE
 If we set all initial conditions to zero,
the impedance is defined as:
V
(
s
)
Z ( s) 
I ( s)
[all initial conditions=0]
16
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
17
Ex. 1
 Find vc(t), t>0
 vc (t ) 
0.5 F

v L (t )

1H

v R (t )
3
u (t )

18
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
19
Convert to voltage sourced
s-Domain Circuit
Vc (s) 
2

VL (s )

s
s
I (s)
VR (s) 
3


3
s
20
Find I(s)
2
3


By KVL :  s   3  I ( s )   0
s
s


3
 I (s)  2
s  3s  2
21
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)
22
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
23
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 )
24
Ex. 2
 Find v0(t) for t>0.
25
s-Domain Circuit Elements
Laplace
transform
all
circuit’s
elements
u (t ) 
1
s
1H  sL  s
1
3
F
1
sC

3
s
26
s-Domain Circuit
27
Apply Mesh-Current
Analysis
Loop 1
1  3
3
 1   I1  I 2
s  s
s
Loop 2
3
3

0   I1   s  5   I 2
s
s

1 2
 I1  s  5 s  3 I 2
3


28
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
29
Find V0(s)
V0 ( s )  sI 2
3
 2
s  8s  18
3
2

2
2
2 ( s  4)  ( 2 )
30
Obtain v0(t)
3
2
Vo ( s) 
2
2
2 ( s  4)  ( 2 )
3  4t
v0 (t ) 
e sin 2t
2
31
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
32
TRANSFORM OF CIRCUITSINDUCTOR
 In the time domain:
33
TRANSFORM OF CIRCUITSINDUCTOR
 Inductor’s voltage:
 Inductor’s current:
34
TRANSFORM OF CIRCUITSCAPACITOR
 In the time domain:
35
TRANSFORM OF CIRCUITSCAPACITOR
 Capacitor’s voltage:
 Capacitor’s current:
36
Ex.
 Find v0(t) if the initial voltage is
given as v0(0-)=5 V
37
s-Domain Circuit
38
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
39
Cont’d
10
Vo ( s  2) 
 25
s 1
25s  35
V0 
( s  1)( s  2)
40
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
41
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
vo (t )  (10e  15e
2 t
)u (t )
42