Laplace Transform
Frequency-domain methods are extremely useful in the study of signals and LTI systems. However, F.T. has some
limitations in dealing with growing exponentials and with some signals or their derivatives that contains jump
discontinuities.
This is resolved by using the basis function e st instead of e j t , where s    j .
In Fourier transform, F ( j ) 
 f (t )e
 t
1

2





F (  j )e d 
Let,   j  s ,  d 



1 
 f (t ) 
F (  j )e(  j ) t d 


2
1  
 f (t ) 
F ( s)e st ds



2 j
jt


f (t )e jt dt ;  F (  j)   f (t )e(  jt ) dt   [ f (t )e t ]e jt dt
1
ds
j
We define Laplace and inverse Laplace transform as,

 
f (t ) 
L.T .
f (t ) 
 F ( s)
F ( s)  L[ f (t )] and f (t )  L1[ F ( s)]

Symbolically,
1
F (s)   f (t )e st dt and,
or,
2 j 

F ( s)e st ds .
The region of convergence
Let, x(t )  e at u (t ) ;
a real.
Then the Laplace transform of x(t ) is, X ( s) 
If, Re( s  a )  0 ,
Therefore, X ( s ) 
lim e
 ( s  a )t
t 
1
;
sa



e at u (t )e st dt 
1  ( s  a ) t
e
sa

0
 0 , otherwise not.
Re( s  a)  0 or, Re s  a i.e.,   a .
The region of convergence is displayed in the complex plane as shown in Figure below.
Again consider the signal, x(t )  e at u (t ) . Its Laplace transform is,

X ( s )   e  at u (t )e  st dt 

1
; Re s  a . The ROC plot is shown on the next page.
sa
It is seen that the Laplace transform of the two functions are the same except for the ROC. Therefore, in order for
the Laplace transform to be unique for each signal x(t ) , the ROC must be specified as part of the transform.
The ROC of Laplace transform is the set of values of s for which Laplace transform converges. If we restrict our signal
to be causal, the ambiguity does not arise. For this reason we define unilateral Laplace transform,

F ( s)   f (t )e st dt
0
The inverse Laplace transform requires integration in the complex plane. We do not use the formula, rather we shall
use Transformation table to find inverse Laplace transform.
Laplace transform of some basic signals
1. L  (t ) 
2.

L u (t )  

0

0
 (t )e st dt  1
e  st dt 
1  st
e
s

0
1
 ; Re s  0
s
1 1
1 
1
L e j0t u (t )  e  j0t u (t )   

 ; Re( s  j0 )  0
2
2  s  j0 s  j0 
s
; Re s  0
Or, L cos 0tu (t )  2
[Ans]
s  2
3. L cos 0tu (t ) 
Properties of Laplace transform
Differentiation
d 2 x(t )
dx(t )
 sX ( s)  x(0 ) ;
 s 2 X ( s)  s  x(0 )  x(0 )
2
dt
dt
3
d x(t )
 s3 X ( s)  s 2  x(0 )  s  x(0 )  x(0 )
3
dt
Some Laplace transform pairs
Convergence of two-sided Laplace transform
Consider the signal, x(t )  e2t u (t )  et u (t ) . Its Laplace transform is given by,



0


0

X (s)   e2t u(t )e st dt   et u(t )e st dt   e( s 2)t dt   e( s 1)t dt
Or,
Or,
1
1

; Re s  2 and Re s  1
s  2 s 1
3
X ( s)  2
;  2  Re s  1
s s2
X (s) 
Example
Find the response of an LTI system with h(t )  [e2t  et ]u (t ) to the input x(t )  e2t u (t )   (t ) .
1
1
2s  3


Re s  1
s  2 s  1 ( s  2)( s  1)
1
L
x(t ) 
 X (s) 
 1 Re s  2
s2
2s  3
s 1
Y ( s)  H ( s) X ( s) 

, Re s  1
( s  1)( s  2) s  2
2s  3
Or,
Y (s)  H (s) X (s) 
, Re s  2
( s  2)2
2s  3
A
B
Expanding using partial fraction, Y ( s) 


, Re s  2
2
( s  2)
s  2 ( s  2) 2
2s  3
A
2
and
 B  (2s  3) s 2  1
( s  2) s 
L
h(t ) 
 H ( s) 
2
1

, Re s  2
s  2 ( s  2) 2
Performing inverse Laplace transform, y(t )  [2e2t  te2t ]u (t ) .
Y ( s) 
Thus,
[Ans]
Poles and zeros of Rational Laplace transforms
Exponential signals have rational Laplace transform. That is, they are ratios of numerator and denominator
polynomials.
N ( s)
.
D( s)
X ( s) 
The roots of N ( s ) are called the zeros of Laplace transform and the roots of D ( s ) are called the poles of Laplace
transform.
Example: X ( s ) 
2s  1
;  2  Re s  1 has poles at p1  1 and p2  2 .
s s2
2
For LTI systems, the roots of the characteristic polynomials are equal to the poles of the Laplace transform of the
impulse response.
The transfer function of LTI system
The general equation of an LTI system is,
Q( D) y (t )  P( D) x(t ) or,  D n  an 1 D n 1 
 a0  y (t )   bn D n  bn 1D n 1 
 b0  x (t )
Performing L.T. assuming all initial conditions to be zero,
s
n
 an 1s
n 1

 a0  Y ( s )   bn s  bn 1s
n
b s  b
 b  X (s ) Or, Y ( s) 
s  a
n
n 1

n
n 1
Y ( s)  bn s  bn 1s   b0  P( s)
or, H ( s) 


X ( s)  s n  an 1s n 1   a0  Q( s)
n 1
n
0
n
n 1
s n 1 
s n 1 
H ( s) is called the transfer function of the system.
Example
1. Find the transfer function of an ideal delay of T, of ideal differentiator, and of ideal integrator.
a) Ideal delay: y (t )  x(t  T )  Y ( s)  e sT X ( s)  H ( s)  e sT .
b) Ideal differentiator: y (t )  dx(t ) / dt  Y ( s)  sX ( s)  H ( s)  s .
c) ) Ideal integrator: y(t ) 
t
 x( )d
0
 Y ( s)  X ( s) / s
 H ( s)  1/ s .
 b0 
 a0 
X (s) ;
2. Find the inverse Laplace transform of F ( s ) 
a) F ( s) 
3
if ROC is, a) 2  Re s  1 , b) Re s  1 , c) Re s  2 .
( s  2)( s  1)
1
1
.

s  2 s 1
Re s  2 corresponds to the causal signal. Re s  1 corresponds to anti-causal signal.
Therefore, f (t )  e2t u (t )  et u (t ) .
b) Both poles lie to the left of the ROC. So both poles correspond to the causal signal.
Therefore, f (t )  e2t u (t )  et u (t ) .
c) Both poles lie to the right of ROC. So both poles correspond to the anti-causal signal.
f (t )  [e2t  et ]u (t ) .
Causality and stability
h(t )  0 for t  0 for a causal system.
The ROC associated with the transfer function of a causal system is a right-half plane. However, a right-sided signal
started at t  10 is not causal, although the ROC is a right-half plane.
1
, Re s  1 is a causal system.
s 1
es
H ( s) 
, Re s  1 is not causal as h(t ) starts at t=-1.
s 1
e s
H ( s) 
, Re s  1 is causal as h(t ) starts at t=1.
s 1
H ( s) 
An LTI system is stable if its impulse response is absolutely integrable. Also a system is considered stable if the zeros
of its characteristic polynomial have a negative real part.


H (s)   h(t )e st dt  H (s)   h(t ) e st dt

Let s  j , then, H (s) 





h(t ) e jt dt   h(t ) dt . If the integration converges the system is stable.

Therefore, a stable system should include the imaginary axis in the ROC for convergence.
Example: Consider a system with transfer function, H ( s) 
below shows the impulse response for the three ROCs.
s 1
. There are three possible ROCs. Table
( s  1)( s  2)
Systems interconnections
Fig: (a) time-domain representation, (b) s-domain representation
Transform circuits
1. Signal sources:
2. Resistance, R:
3. Inductance, L:
4. Capacitance, C:
v(t )  V ( s)
i (t )  I ( s)
v(t )  Ri(t )  V ( s)  RI ( s)
di (t )
v(t )  L
 V ( s )  LsI ( s )  Li (0 )
dt
dv(t )
i (t )  C
 I ( s )  CsV ( s )  Cv(0 )
dt

1
1
V ( s )  i (0 )
Ls
s
1
1
 V (s) 
I ( s )  v(0 )
Cs
s
 I (s) 
Example
1. Find the Laplace transform X ( s) and sketch the pole-zero plot with the ROC for the following signal.
x(t )  e3t u(t )  e2t u(t ) .
The ROC is shown on the left. X ( s) has no zeros and two poles at
s  2 and s  3 .
2. Find the Laplace transform of the signal, x(t )  e2t [u(t )  u(t  5)] .
x(t )  e2t u(t )  e2t u (t  5)  e2t u (t )  e10  e2(t 5)u (t  5)
1
1
1
 e 10  e 5 s 

[1  e 5( s  2) ] , Re s  2
 X (s) 
s2
s2 s2
3. Derive the Laplace transform of the signal, x(t )  e  at cos 0tu (t ) .
1
1
1
1
s
cos 0tu (t )  
 
 2
2 s  j0 2 s  j0 s  02
Applying the shifting in the s-domain property,
sa
.
( s  a) 2  02
2s  4
,  3  Re s  1 .
4. Find the inverse L.T. of X ( s )  2
s  4s  3
c
c
2s  4
X (s)  2
 1  2
 2s  4  c1 (s  3)  c2 (s  1)  c1  1, c2  1 .
s  4s  3 s  1 s  3
2s  4
1
1


, 3  Re s  1
 X ( s)  2
s  4s  3 s  1 s  3
Hence, x(t )  e3t u (t )  et u (t ) .
[Ans]
5s  13
5. Find the inverse L.T. of X ( s) 
, Re s  0 .
2
s( s  4s  13)
c s  c3
c
5s  13
 1 2 2
 5s  13  c1 ( s 2  4s  13)  (c2 s  c3 ) s
2
s( s  4s  13) s s  4s  13
Equating coefficients of s 2 :
0  c1  c2
e at cos 0tu (t ) 
Equating coefficients of s :
5  4c1  c3
Solving, c1  1 , c2  1 and c3  1 .
Equating constant coefficients : 13  13c1
1 ( s  2)  3
1
s 1
 2
 
s s  4s  13
s ( s  2) 2  32
 x(t )  u(t )  e2t cos3tu(t )  e2t sin 3tu(t ) = [1  e2t (cos3t  sin 3t )]u(t )
Thus, X ( s ) 
[Ans]
6. Find the impulse response of the system y(t )  2 y (t )  x(t )  x(t ) using Laplace transform.
From system equation we get, H ( s ) 
s 1
1
 1
s2
s2
[Ans]
 h(t )   (t )  e2t u (t ) .
7. Solve the 2nd order system y(t )  5 y(t )  6 y (t )  x(t ) for x(t )  et u (t ) with y (0)  2 and y(0)  1 .
Performing one-sided L.T.: s 2Y (s)  sy(0)  y(0)  5sY ( s)  5 y(0)  6Y ( s)  X ( s)
Or, Y (s)(s 2  5s  6)  2s  1  10  X (s)
 Y ( s)( s 2  5s  6) 
2s 2  13s  12
s 1
2s 2  13s  12
1/ 2
6
9/ 2



2
(s  1)(s  5s  6) s  1 s  2 s  3
9
1

Thus, y (t )   e  t  6e 2t  e 3t  u (t )
[Ans]
2
2

Or, Y ( s) 
8. Consider the RC circuit shown in Figure below. The switch is closed at t  0 . The initial capacitor voltage is v0 .
Find i (t ) and vc (t ) using transform network technique.
The transform circuit is shown in Figure on the right side. Applying KVL,
v0 V
1 

R
 I (s)  
Cs 
s
s

Thus, i (t ) 
V  v0
V  v0
1
1



s
R  1/ Cs
R
s  1/ RC
v
1
1
1
 v0
 Vc ( s ) 
I ( s )  0  (V  v0 )  

Cs
s
 s s  1/ RC  s
 I ( s) 
V  v0  t / RC
e
u (t ) and vc (t )  V (1  e t / RC )u (t )  v0e t / RC u (t )
R
[Ans]