LAPLACE TRANS FORM
PART-A
UNIT-V
LAPLACE TRANSFORM
The Laplace Transform of a function, f(t), is defined as;

L[ f (t )]  F ( s)   f (t )e dt
 st
0
The Inverse Laplace Transform is defined by
1
L [ F ( s)]  f (t ) 
1
2
  j
F ( s )e

j
 j
ts
ds
The Laplace Transform
We generally do not use Eq B to take the inverse Laplace. However,
this is the formal way that one would take the inverse. To use
Eq B requires a background in the use of complex variables and
the theory of residues. Fortunately, we can accomplish the same
goal (that of taking the inverse Laplace) by using partial fraction
expansion and recognizing transform pairs.
The Laplace Transform
Laplace Transform of the unit step.

 1  st 
L[u(t )]   1e dt  e |
0
s
0
 st
1
L[u(t )] 
s
The Laplace Transform of a unit step is:
1
s
The Laplace Transform
The Laplace transform of a unit impulse:
Pictorially, the unit impulse appears as follows:
(t – t0)
f(t)
t0
0
Mathematically:
(t – t0) = 0 t

t 0 
0
 (t  t )dt 1
0
t0 
 0
The Laplace Transform
The Laplace transform of a unit impulse:
An important property of the unit impulse is a sifting
or sampling property. The following is an important.
t2

t1
t1  t 0  t 2
 f (t 0 )
f (t ) (t  t 0 )dt  
t 0  t1 , t 0  t 2
0
The Laplace Transform
The Laplace transform of a unit impulse:
In particular, if we let f(t) = (t) and take the Laplace

L[ (t )]    (t )e dt  e
0
 st
0 s
1
The Laplace Transform
An important point to remember:
f (t )  F ( s)
The above is a statement that f(t) and F(s) are
transform pairs. What this means is that for
each f(t) there is a unique F(s) and for each F(s)
there is a unique f(t). If we can remember the
Pair relationships between approximately 10 of the
Laplace transform pairs we can go a long way.
The Laplace Transform
Building transform pairs:

L[tu(t )]   te dt
 st
0



 udv  uv |   vdu
0
0
0
tu(t )

1
2
s
u=t
dv = e-stdt
The Laplace Transform
Building transform pairs:

(e jwt  e  jwt )  st
L[cos( wt )]  
e dt
2
0
1 1
1 
 


2  s  jw s  jw 
s
 2
s  w2
cos( wt )u(t )

s
s2  w2
A transform
pair
The Laplace Transform
Time Shift

L[ f (t  a )u(t  a )]   f (t  a )e  st
a
Let x  t  a , then dx  dt and t  x  a
As t  a , x  0 and as t  , x  . So,


0

f ( x )e  s ( x  a ) dx  e as  f ( x )e  sx dx
0
L[ f (t  a)u(t  a)]  e
 as
F ( s)
The Laplace Transform
Frequency Shift
L[e
 at

f (t )]   [e
 at
 st
f (t )]e dt
0


 f ( t )e
( s  a ) t
dt  F ( s  a )
0
L[e
 at
f (t )]  F ( s  a)
The Laplace Transform
Time Integration:
The property is:

 t
  st
L   f (t )dt      f ( x )dx e dt
0
 0 0

Integrate by parts :
t
Let u   f ( x )dx , du  f (t )dt
0
and
 st
dv  e dt ,
1  st
v  e
s
The Laplace Transform
Time Integration:
Making these substitutions and carrying out
The integration shows that

 1
L  f (t )dt    f (t )e  st dt
0
 s0
1
 F ( s)
s
The Laplace Transform
Time Differentiation:
If the L[f(t)] = F(s), we want to show:
df (t )
L[
]  sF ( s )  f (0)
dt
Integrate by parts:
u  e , du   se dt and
 st
 st
df ( t )
dv 
dt  df ( t ), so v  f ( t )
dt
The Laplace Transform
Time Differentiation:
Making the previous substitutions gives,
 df 
L    f ( t )e
 dt 

|  f (t ) se dt
 st 
0
0

 0  f (0)  s  f (t )e  st dt
0
So we have shown:
 df (t ) 
L

sF
(
s
)

f
(
0
)
 dt 
 st
The Laplace Transform
Time Differentiation:
We can extend the previous to show;
 df (t ) 2  2
L
 s F ( s )  sf (0)  f ' (0)
2 
 dt 
 df (t ) 3 
3
2
L

s
F
(
s
)

s
f (0)  sf ' (0)  f ' ' (0)
3 
 dt 
general case
 df (t ) n 
n
n 1
n2
L

s
F
(
s
)

s
f
(
0
)

s
f ' (0)
n 
 dt 
 ...  f ( n 1) (0)
The Laplace Transform
Transform Pairs:
f(t)
F(s)
 (t )
1
1
u( t ) ____________________________________
s
1
 st
e
sa
1
t
s2
n!
n
t
s n 1
f (t )
F ( s)
The Laplace Transform
Transform Pairs:
f(t)
te
 at
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
Common Transform Properties:
f(t)
F(s)
f ( t  t 0 )u( t  t 0 ), t 0  0
f ( t )u( t  t 0 ), t  0
e  at f ( t )
d n f (t )
dt n
e
e
 to s
 to s
F ( s)
L[ f ( t  t 0 )
F (s  a)
s n F ( s )  s n  1 f ( 0)  s n  2 f ' ( 0)  . . .  s 0 f n  1 f ( 0)
dF ( s )
ds
tf ( t )

t
1
F ( s)
s
 f ( )d
0
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), and 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)
has the Laplace transform F(s), and the lim sF ( s ) exists, then
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.
The z-Transform
UNIT-V PART B
Content
•
•
•
•
•
•
•
Introduction
z-Transform
Zeros and Poles
Region of Convergence
Important z-Transform Pairs
Inverse z-Transform
z-Transform Theorems and Properties
The z-Transform
Introduction
Why z-Transform?
• A generalization of Fourier transform
• Why generalize it?
– FT does not converge on all sequence
– Notation good for analysis
– Bring the power of complex variable theory
deal with the discrete-time signals and systems
Definition
• The z-transform of sequence x(n) is defined

by
n
X ( z) 
 x ( n) z
n  

Let z = ej.
j
X (e ) 

 x ( n )e
n 
Fourier
Transform
 j n
z-Plane
X ( z) 

 x ( n) z
n
Im
z = ej
n  
j
X (e ) 

 x ( n )e
n 

 j n
Re
z-Plane
Im
X(z)
z = ej

Re
Im
Re
Periodic Property of FT
X(ej)
X(z)

Im
Re


The z-Transform
Zeros and Poles
Definition
• Give a sequence, the set of values of z for
which the z-transform converges, i.e.,
|X(z)|<, is called the region of convergence.
| X ( z ) |

n
x
(
n
)
z


n  

n
|
x
(
n
)
||
z
|


n  
Example: Region of Convergence
| X ( z ) |

n
x
(
n
)
z


n  

n
|
x
(
n
)
||
z
|


n  
Im
r
Re
Rx  | z | Rx 
j
ROC  {z  re | Rx   r  Rx  }
Stable Systems
• A stable system requires that its Fourier
transform is uniformly convergent.
Im

1

Re
Fact: Fourier transform is to
evaluate z-transform on a unit
circle.
A stable system requires the
ROC of z-transform to include
the unit circle.
Example: A right sided Sequence
For convergence of X(z), we
require that
x ( n)  a u ( n )
n
X ( z) 


 a u (n)z
n
n  
n
1
|
az
|

n 0
| z || a |

  a n z n
n 0

  (az 1 ) n
n 0
| az 1 | 1

1
z
X ( z )   (az ) 

1
1  az
za
n 0
1 n
| z || a |
Example: A left sided Sequence
x(n)  a u(n  1)
n

X ( z )    a u (n  1)z
n
n  
1
   a n z n
n
For convergence of X(z), we
require that

1
|
a
 z|
| a 1 z | 1
n 0
| z || a |
n  

  a  n z n
n 1

 1   a n z n
n 0

1
z
X ( z )  1   (a z )  1 

1
1 a z z  a
n 0
1
n
| z || a |
The z-Transform
Region of
Convergence
Represent z-transform as a
Rational Function
P( z )
X ( z) 
Q( z )
where P(z) and Q(z) are
polynomials in z.
Zeros: The values of z’s such that X(z) = 0
Poles: The values of z’s such that X(z) = 
Example: A right sided Sequence
z
X ( z) 
,
za
x ( n)  a n u ( n )
| z || a |
Im
a
Re
ROC is bounded by the
pole and is the exterior
of a circle.
Example: A left sided Sequence
z
X ( z) 
,
za
x(n)  a nu(n  1)
| z || a |
Im
a
Re
ROC is bounded by the
pole and is the interior
of a circle.
Example: Sum of Two Right Sided Sequences
x(n)  ( 12 ) n u (n)  ( 13 ) n u (n)
z
z
2 z ( z  121 )
X ( z) 


1
1
z2 z3
( z  12 )( z  13 )
Im
ROC is bounded by poles
and is the exterior of a circle.
1/12
1/3
1/2
Re
ROC does not include any pole.
Example: A Two Sided Sequence
x(n)  ( 13 ) n u (n)  ( 12 ) n u (n  1)
z
z
2 z ( z  121 )
X ( z) 


1
1
z3 z2
( z  13 )( z  12 )
Im
ROC is bounded by poles
and is a ring.
1/12
1/3
1/2
Re
ROC does not include any pole.
Example: A Finite Sequence
x(n)  a n ,
0  n  N 1
N 1
X ( z)   a z
n
n 0
n
Im
N 1
  ( az )
1 n
n 0
N-1 zeros
1  (az 1 ) N

1  az 1
1 zN  aN
 N 1
z
za
ROC: 0 < z < 
ROC does not include any pole.
N-1 poles
Re
Properties of ROC
• A ring or disk in the z-plane centered at the origin.
• The Fourier Transform of x(n) is converge absolutely iff the ROC
includes the unit circle.
• The ROC cannot include any poles
• Finite Duration Sequences: The ROC is the entire z-plane except
possibly z=0 or z=.
• Right sided sequences: The ROC extends outward from the outermost
finite pole in X(z) to z=.
• Left sided sequences: The ROC extends inward from the innermost
nonzero pole in X(z) to z=0.
More on Rational z-Transform
Consider the rational z-transform
with the pole pattern:
Im
Case 1: A right sided Sequence.
a b
c
Re
More on Rational z-Transform
Consider the rational z-transform
with the pole pattern:
Im
Case 2: A left sided Sequence.
a b
c
Re
More on Rational z-Transform
Consider the rational z-transform
with the pole pattern:
Im
Case 3: A two sided Sequence.
a b
c
Re
More on Rational z-Transform
Consider the rational z-transform
with the pole pattern:
Im
Case 4: Another two sided Sequence.
a b
c
Re
Z-Transform Pairs
Sequence
z-Transform
(n)
1
(n  m)
z m
ROC
All z
All z except 0 (if m>0)
or  (if m<0)
u (n)
1
1  z 1
| z | 1
 u (n  1)
1
1  z 1
| z | 1
a u (n)
1
1  az 1
| z || a |
 a nu (n  1)
1
1  az 1
| z || a |
n
Z-Transform Pairs
Sequence
z-Transform
[cos 0 n]u(n)
1  [cos 0 ]z 1
1  [2 cos 0 ]z 1  z 2
| z | 1
[sin 0 n]u (n)
[sin 0 ]z 1
1  [2 cos 0 ]z 1  z 2
| z | 1
[r n cos 0 n]u (n)
1  [r cos 0 ]z 1
1  [2r cos 0 ]z 1  r 2 z 2
| z | r
[r n sin 0 n]u (n)
[r sin 0 ]z 1
1  [2r cos 0 ]z 1  r 2 z 2
| z | r
1 aN zN
1  az 1
| z | 0
a n

0
0  n  N 1
otherwise
ROC
Linearity
Z[ x(n)]  X ( z ),
z  Rx
Z [ y (n)]  Y ( z ),
z  Ry
Z [ax(n)  by (n)]  aX ( z )  bY ( z ),
z  Rx  R y
Overlay of
the above two
ROC’s
Shift
Z[ x(n)]  X ( z ),
z  Rx
Z[ x(n  n0 )]  z X ( z )
n0
z  Rx
Multiplication by an Exponential Sequence
Z[ x(n)]  X ( z ),
Rx- | z | Rx 
1
Z [a x(n)]  X (a z )
n
z | a | Rx
Differentiation of X(z)
Z[ x(n)]  X ( z ),
dX ( z )
Z [nx(n)]   z
dz
z  Rx
z  Rx
Conjugation
Z[ x(n)]  X ( z ),
Z[ x * (n)]  X * ( z*)
z  Rx
z  Rx
Reversal
Z[ x(n)]  X ( z ),
1
Z[ x(n)]  X ( z )
z  Rx
z  1 / Rx
Real and Imaginary Parts
Z[ x(n)]  X ( z ),
z  Rx
Re[ x(n)]  12 [ X ( z )  X * ( z*)]
Im[ x(n)]  21j [ X ( z )  X * ( z*)]
z  Rx
z  Rx
Initial Value Theorem
x(n)  0,
for n  0
x(0)  lim X ( z )
z 
Convolution of Sequences
Z[ x(n)]  X ( z ),
Z [ y (n)]  Y ( z ),
z  Rx
z  Ry
Z[ x(n) * y (n)]  X ( z )Y ( z )
z  Rx  R y
Convolution of Sequences
x ( n) * y ( n) 

 x(k ) y (n  k )
k  
 
 n
Z[ x(n) * y (n)]     x(k ) y (n  k )  z
n    k  





k  
n  
 x(k )  y(n  k )z
 X ( z )Y ( z )
n



k  
x(k ) z  k

n
y
(
n
)
z

n  
The z-Transform
System Function
Shift-Invariant System
y(n)=x(n)*h(n)
x(n)
h(n)
X(z)
H(z)
Y(z)=X(z)H(z)
Shift-Invariant System
H(z)
X(z)
Y(z)
Y ( z)
H ( z) 
X ( z)
Stable and Causal Systems
Causal Systems : ROC extends outward from the outermost pole.
Im
M
H ( z) 
A (1  cr z 1 )
r 1
N
1
(
1

d
z
 r )
k 1
Re
Stable and Causal Systems
Stable Systems : ROC includes the unit circle.
Im
M
H ( z) 
A (1  cr z 1 )
r 1
N
1
(
1

d
z
 r )
k 1
1
Re
Determination of Frequency Response from
pole-zero pattern
• A LTI system is completely characterized by
its pole-zero pattern.
Im
Example:
z  z1
H ( z) 
( z  p1 )( z  p2 )
j0
e
 z1
j0
H (e )  j0
(e  p1 )(e j0  p2 )
p1
e j 0
z1
Re
p2
Determination of Frequency Response from
pole-zero pattern
• A LTI system is completely characterized by
its pole-zero pattern.
Im
Example:
|H(ej)|
=
|
|
2
|
||
|
H(ej) = 1(2+ 3 )
z1
p1
e j 0
1
p2
3
Re