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App. of LT in Elec. Circuit.pptx
1. Application of Laplace
Transform to Electric Circuit
S.R. Balaji
Associate Professor & Head
PG Department of Physics
V.O.Chidambaram College (SF)
Tuticorin – 8
Email: prambala@gmail.com
Blog site: www.srbphysics.blogspot.com
2. 2
Contents
Introduction to Laplace Transform (L.T)
Properties and Important Results of L.T
Application – Simple Electric Circuit
Problems – An RLC Circuit & RL Circuit
Conversion of Simple Circuit to 2nd Order Ordinary diff. equation
Calculation of Charge and Current in the Circuits Using Laplace
Transform Method.
Results
Application of Laplace Transform to Electric Circuit
3. 3
Introduction to Laplace Transform
If the integral exists and is equal to F(s), then F(s) is
called the Laplace transform of F(t) and it is denoted by the symbol
L {F(t)}.
i.e.,
Conditions for existence of Laplace Transform are:
1. the function f(t) should be an arbitrary piecewise continuous
function in every finite interval.
2. the function f(t) should be of exponential order.
Application of Laplace Transform to Electric Circuit
dt
)
t
(
f
e st
0
)
s
(
F
dt
)
t
(
f
e
)
t
(
f
L st
0
4. 4
Properties of Laplace Transform
Linearity property:
Change of scale property:
First shifting property:
Second shifting property:
Derivative of Laplace Transform:
Integral of Laplace Transform:
Application of Laplace Transform to Electric Circuit
)}
(
{
)}
(
{
)
(
)
( 2
2
1
1
2
2
1
1
t
f
L
a
t
f
L
a
t
f
a
t
f
a
L
a
s
f
a
)}
at
(
f
{
L
1
a
s
f
)}
t
(
f
e
{
L
,
a
s
f
)}
t
(
f
e
{
L at
at
;
s
f
e
)}
t
(
G
{
L as
a
t
if
)
a
t
(
F
a
t
if
)
t
(
G
where
0
n
n
n
n
ds
)
s
(
f
d
)
(
)}
t
(
f
t
{
L 1
s
ds
)
s
(
f
t
)
t
(
f
L
5. 5
Properties of Inverse Laplace Transform
Linearity property:
Change of scale property:
First shifting property:
Derivative of Laplace Transform:
Convolution Theorem:
Application of Laplace Transform to Electric Circuit
)}
(
{
)}
(
{
)
(
)
( 2
1
2
1
1
1
2
2
1
1
1
s
f
L
a
s
f
L
a
s
f
a
s
f
a
L
a
t
f
a
as
f
L
1
)}
(
{
1
)}
(
{
)}
(
{ 1
1
s
f
L
e
a
s
f
L at
)}
(
{
)
(
)}
(
'
{ 1
1
s
f
L
t
s
f
L
0
1
2
0
2
1
2
1
1
)
(
)
(
)
(
)
(
)}
(
)
(
{ du
u
t
f
u
f
du
u
t
f
u
f
s
f
s
f
L
)}
(
{
)}
(
{ 1
1
s
f
L
e
a
s
f
L at
6. 6
Important Results of Laplace Transform
Application of Laplace Transform to Electric Circuit
F(t) L{F(t)}
eat
e-at
cosh at
sinh at
cos at
sin at
F(t) L{F(t)}
1
K
t
tn
t ½
t -½
a
s
1
a
s
1
2
2
a
s
s
2
2
a
s
a
2
2
a
s
s
2
2
a
s
a
s
1
s
.
K
1
2
1
s
1
n
s
!
n
2
3
2 /
s
s
7. 7
Application of Laplace Transform
The Laplace transform can be used to solve differential equations
and is used extensively in electrical engineering.
Two types of approaches are avail in literature, they are:
1. Derive the circuit (differential) equations in the time domain,
then transform these ODEs to the s-domain;
2. Transform the circuit to the s-domain, then derive the circuit
equations in the t-domain (using the concept of "impedance").
Of it, we will use the first approach. i.e., we derive the system
equations(s) in the t-plane, then transform it to the s-plane.
Application of Laplace Transform to Electric Circuit
8. 8
Problem for Study
1. A inductor L, a resistor R and a capacitor C are connected in
series with an emf E. At t = 0, the charge of the capacitor and
current in the circuit are zero. Find the current and charge at any
time t > 0.
Pictorial representation:
(OR)
Application of Laplace Transform to Electric Circuit
E E
9. 9
Conversion of RLC Circuit to ODE Equation
Let Q and I be instantaneous charge and current respectively at
time t. We know that, using Krichoff’s Loop Law the equation for an
RLC circuit can be written as:
Dividing by L, we get,
which can also be written as,
[1]
Application of Laplace Transform to Electric Circuit
E
C
Q
R
dt
dQ
L
dt
Q
d
2
2
L
E
LC
Q
dt
dQ
L
R
dt
Q
d
2
2
L
E
Q
LC
'
Q
L
R
'
'
Q
1
10. 10
Solution:
Taking Laplace transform on both side of equ (1), we get,
and
the above equation can be written as,
Application of Laplace Transform to Electric Circuit
L
E
L
Q
LC
1
L
'
Q
L
R
L
'
'
Q
L
0
x'
0
sx
x
s
dt
x
d
'
x' 2
2
2
0
x
sx
dt
dx
x'
1
L
L
E
Q
L
LC
1
L
L
R
0
Q
sQ
0
Q'
0
sQ
Q
s
L 2
]
2
[
1
L
L
E
Q
L
LC
1
L
R
sL
L
R
s
0
Q
Q
0
Q'
0
Q
Q
L
s2
11. 11
Solution:
Substituting L{Q} = q, the above equation becomes,
Since Q(0) = 0, Q’(0) must also be 0, then we get,
We know that
[3]
Application of Laplace Transform to Electric Circuit
1
L
L
E
q
LC
1
L
R
sq
L
R
s
0
Q
0
Q'
0
Q
q
s2
1
L
L
E
q
LC
1
sq
L
R
q
s2
s
1
1
L
Ls
E
q
LC
1
sq
L
R
q
s2
Ls
E
LC
1
s
L
R
2
s
q
12. 12
Solution:
Then,
[4]
Using the method of partial fractions, the above equation becomes,
Then,
Application of Laplace Transform to Electric Circuit
)
LC
1
(
s
)
L
R
(
Ls
E
2
s
q
)
LC
1
(
s
)
L
R
(
s
L
E
2
s
q
]
5
[
)
LC
1
(
s
)
L
R
(
C
Bs
s
A
)
LC
1
(
s
)
L
R
(
s
L
E
2
2
s
s
q
)
s
(
)
C
Bs
(
)
LC
1
(
s
)
L
R
(
A
2
s
L
E
13. 13
Solution:
[6]
Substituting s = 0, in equation (6), we get,
Equating the coefficient of s2 on equation (6), we get,
Application of Laplace Transform to Electric Circuit
)
LC
1
(
0
A
0
L
E
A
.
LC
1
L
E
EC
L
ELC
A
LC
/
1
L
/
E
EC
A
s
C
s
B
)
LC
1
(
A
s
A
)
L
R
(
A 2
2
s
L
E
B
A
0
EC
A
B
EC
B
14. 14
Solution:
Equating the coefficient of s on equation (6), we get,
Substituting the values A, B & C on equation (5), we get,
Application of Laplace Transform to Electric Circuit
C
L
AR
C
A
)
L
R
(
0
L
R
)
EC
(
L
AR
C
L
ECR
C
)
LC
1
(
s
)
L
R
(
L
ECR
s
)
EC
(
s
EC
2
s
q
15. 15
Solution:
Application of Laplace Transform to Electric Circuit
)
LC
1
(
s
)
L
R
(
L
ECR
s
)
EC
(
s
EC
q
2
s
LC
1
s
L
R
s
L
ECR
ECS
s
EC
2
2
2
2
L
2
R
L
2
R
LC
1
s
L
R
s
L
ECR
ECS
s
EC
16. 16
Solution:
Application of Laplace Transform to Electric Circuit
2
2
2
L
2
R
LC
1
s
L
R
L
2
R
s
L
ECR
ECS
s
EC
2
2
2
L
4
R
LC
1
L
2
R
s
L
2
ECR
L
2
ECR
ECS
s
EC
2
2
2
L
4
R
LC
1
L
2
R
s
L
2
ECR
L
2
R
s
EC
s
EC
17. 17
Solution:
Taking Inverse Laplace Transform on both sides we get,
Application of Laplace Transform to Electric Circuit
2
2
2
2
2
2
L
4
R
LC
1
L
2
R
s
L
2
ECR
L
4
R
LC
1
L
2
R
s
L
2
R
s
EC
s
EC
]
7
[
L
4
R
LC
1
L
2
R
s
L
2
ECR
L
4
R
LC
1
L
2
R
s
L
2
R
s
EC
s
EC
}
Q
{
L
2
2
2
2
2
2
2
2
2
2
2
2
1
L
4
R
LC
1
L
2
R
s
L
2
ECR
L
4
R
LC
1
L
2
R
s
L
2
R
s
EC
s
EC
L
)
t
(
Q
18. 18
Solution:
Application of Laplace Transform to Electric Circuit
2
2
2
1
2
2
2
1
1
L
4
R
LC
1
L
2
R
s
L
2
ECR
L
L
4
R
LC
1
L
2
R
s
L
2
R
s
EC
L
s
EC
L
2
2
2
1
2
2
2
1
1
L
4
R
LC
1
L
2
R
s
1
L
L
2
ECR
L
4
R
LC
1
L
2
R
s
L
2
R
s
L
EC
s
1
L
EC
19. 19
Solution:
We know that, and using the First Shifting Property of
Inverse Laplace Transform the above equation reduces to,
Application of Laplace Transform to Electric Circuit
2
2
2
1
t
L
2
R
2
2
2
1
t
L
2
R
L
4
R
LC
1
s
1
L
e
L
2
ECR
L
4
R
LC
1
s
s
L
e
EC
EC
1
s
1
L 1
2
2
2
2
2
L
4
R
LC
1
a
then
,
L
4
R
LC
1
a
If
2
2
2
2
2
1
2
2
t
L
2
R
2
2
t
L
2
R
L
4
R
LC
1
s
L
4
R
LC
1
L
L
4
R
LC
1
1
e
L
2
ECR
t
L
4
R
LC
1
cos
e
EC
EC
20. 20
Solution:
Substituting
[8]
The above is the charge at the circuit at any time.
Application of Laplace Transform to Electric Circuit
t
L
4
R
LC
1
sin
e
L
4
R
LC
1
L
2
ECR
t
L
4
R
LC
1
cos
e
EC
EC
2
2
t
L
2
R
2
2
2
2
t
L
2
R
2
2
L
4
R
LC
1
t
sin
e
L
2
ECR
t
cos
e
EC
EC t
L
2
R
t
L
2
R
t
sin
e
L
2
ECR
t
cos
e
EC
EC
)
t
(
Q t
L
2
R
t
L
2
R
21. 21
Solution:
To determine the Current in the Circuit:
Application of Laplace Transform to Electric Circuit
dt
dQ
I
t
e
L
ECR
t
e
EC
EC
dt
d
dt
dQ t
L
R
t
L
R
sin
2
cos 2
2
)
(
cos
2
sin
2
2
sin
cos
2
0
2
2
2
2
t
e
L
ECR
t
e
L
R
L
ECR
t
e
EC
t
e
L
R
EC
t
L
R
t
L
R
t
L
R
t
L
R
22. 22
Solution:
Application of Laplace Transform to Electric Circuit
t
e
L
ECR
t
e
L
ECR
t
e
EC
t
e
L
ECR
t
L
R
t
L
R
t
L
R
t
L
R
cos
2
sin
4
sin
cos
2
2
2
2
2
2
2
t
e
L
ECR
t
e
EC t
L
R
t
L
R
sin
4
sin 2
2
2
2
2
2
2
4
sin
L
ECR
EC
t
e t
L
R
2
2
2
4
sin
L
R
EC
t
e t
L
R
2
2
2
2
4
sin
L
R
EC
t
e t
L
R
23. 23
Solution:
The above is the current at the circuit at any time.
Application of Laplace Transform to Electric Circuit
2
2
2
2
4
sin
L
R
EC
t
e t
L
R
2
2
2
2
2
4
1
,
4
1
L
R
LC
L
R
LC
2
2
2
2
2
4
4
1
sin
L
R
L
R
LC
EC
t
e t
L
R
LC
EC
t
e t
L
R
sin
2
t
e
L
E
I t
L
R
sin
2
24. 24
Result
If a inductor L, a resistor R and a capacitor C are connected in
series with an emf E. The charge and current in the circuit at any
time can be expressed as:
Application of Laplace Transform to Electric Circuit
t
e
L
ECR
t
e
EC
EC
t
Q t
L
R
t
L
R
sin
2
cos
)
( 2
2
t
e
L
E
I t
L
R
sin
2
25. 25
Problem for Study
2. A inductor of 3 Henry is in series connected with a resistance of 30
Ohms and an e.m.f of 150 Volts. Assuming that the current the
current is zero at t = 0. Find the current at time t > 0.
Pictorial representation:
Application of Laplace Transform to Electric Circuit
Vin = 150V
R = 30Ω
L = 3 Henry
Vout
26. 26
Conversion of RL Circuit to ODE Equation
It is an RL circuit. The differential equation of the given circuit
can be written as:
[1]
Substituting,
[3]
Sub. the values of L, R & E as 3, 30 & 150 respectively, we get,
[4]
Application of Laplace Transform to Electric Circuit
E
V
dt
dI
L
IR in
dt
dQ
I
E
dt
dQ
R
dt
Q
d
L
2
2
150
30
3 2
2
dt
dQ
dt
Q
d
]
2
[
27. 27
Solution:
The above equation can also be written as,
Taking Laplace Transform on both side, we get,
and
Application of Laplace Transform to Electric Circuit
150
'
30
'
'
3
Q
Q
150
'
30
'
'
3 L
Q
Q
L
150
'
30
'
'
3 L
Q
L
Q
L
1
150
'
30
'
'
3 L
Q
L
Q
L
s
Q
L
Q
L
1
150
'
30
'
'
3
0
x'
0
sx
x
s
dt
x
d
'
x' 2
2
2
0
x
sx
dt
dx
x'
28. 28
Solution:
and
[5]
At time t = 0,
Application of Laplace Transform to Electric Circuit
0
Q'
0
sQ
Q
s
dt
Q
d
'
Q' 2
2
2
0
Q
sQ
dt
dQ
Q'
s
150
0
Q
sQ
L
30
0
Q'
0
sQ
Q
s
L
3 2
s
150
0
Q
30
sQ
L
30
0
Q'
3
0
sQ
3
Q
s
L
3 2
s
0
Q
Q
L
s
0
Q'
0
sQ
Q
L
s
3 2 150
30
30
3
3
then
,
0
be
also
must
(0)
Q'
therefore
0,
Q(0)
s
150
Q
L
s
30
Q
L
s
3 2
29. 29
Solution:
Substituting x = L{Q}, we get,
[6]
[7]
Application of Laplace Transform to Electric Circuit
s
x
s
x
s
3 2 150
30
s
s
3s
x 2 150
30
s
3s
s
x 2
30
150
10
3
150
30
150
}
{
s
s
s
s
3s
s
Q
L 2
10
50
10
3
150
2
2
s
s
s
s
30. 30
Solution:
Taking Inverse Laplace Transform, we get,
[8]
Application of Laplace Transform to Electric Circuit
10
1
50
10
50
)
( 2
1
2
1
s
s
L
s
s
L
t
Q
10
1
.
1
50 1
s
s
s
L
t
s
s
L
0
1
10
1
50
t t
s
L
0 0
1
10
1
50
31. 31
Solution:
[9]
Application of Laplace Transform to Electric Circuit
t t
s
L
0 0
1
10
1
50
t
t
t
e
0 0
10
10
50
t
t
e
0
10
1
10
1
50
t t
t
e
0 0
10
50
t t
e
e
0
0
10
10
10
50
t t
e
0
10
10
1
10
50
32. 32
Solution:
[10]
The above eqn. represents the charge in the circuit at any time t.
Application of Laplace Transform to Electric Circuit
t
t
e
0
10
1
10
50
t
t
e
t
0
10
10
5
10
0
10
5
0
10
e
e
t
t
10
1
10
5
10t
e
t
)
1
(
10
1
5
)
( 10t
e
t
t
Q
33. 33
Solution:
To determine the Current in the Circuit:
The above eqn. represents the current in the circuit at any time t.
Application of Laplace Transform to Electric Circuit
dt
dQ
I
1
10
1
5 10t
e
t
dt
d
dt
dQ
1
10
1
.
5 10t
e
t
dt
d
0
)
10
(
10
1
.
5
10
1
10
.
5
10
10 t
t
e
e
t
dt
d
t
e
t
I 10
1
.
5
)
(
34. 34
Result
If A inductor of 3 Henry is in series connected with a
resistance of 30 Ohms and an e.m.f of 150 Volts. Then the charge
and current in the circuit at any time can be expressed as:
Application of Laplace Transform to Electric Circuit
)
1
(
10
1
5
)
( 10t
e
t
t
Q
t
e
t
I 10
1
.
5
)
(
