Lecture 5 – 6
Z - Transform
By
Dileep Kumar
1
Frequency domain vs Time domain




Frequency domain is a term used to describe the analysis of
mathematical functions or signals with respect to frequency.
(communications point of view) A plane on which signal strength can
be represented graphically as a function of frequency, instead of a
function of time.
control systems) Pertaining to a method of analysis, particularly useful
for fixed linear systems in which one does not deal with functions of
time explicitly, but with their Laplace or Fourier transforms, which are
functions of frequency.
Speaking non-technically, a time domain graph shows how a signal
changes over time, whereas a frequency domain graph shows how
much of the signal lies within each given frequency band over a range
of frequencies.
2
Cont:



A frequency domain representation can also include information on the
phase shift that must be applied to each sinusoid in order to be able to
recombine the frequency components to recover the original time
signal.
The frequency domain relates to the Fourier transform or Fourier series
by decomposing a function into an infinite or finite number of
frequencies. This is based on the concept of Fourier series that any
waveform can be expressed as a sum of sinusoids (sometimes
infinitely many.)
In using the Laplace, Z-, or Fourier transforms, the frequency spectrum
is complex and describes the frequency magnitude and phase. In many
applications, phase information is not important. By discarding the
phase information it is possible to simplify the information in a
frequency domain representation to generate a frequency spectrum or
spectral density. A spectrum analyser is a device that displays the
spectrum.
3
The Direct Z-Transform
The z-transform of a discrete time signal is defined as the power series
(1)

X(z ) 

x [ n ]z
n
Where z is a complex variable. For convenience, the z-transform of a
signal x[n] is denoted by
X(z) = Z{x[n]}
Since the z-transform is an infinite series, it exists only for those
values of z for which this series converges. The Region of
Convergence (ROC) of X(z) is the set of all values of z for which
this series converges.
We illustrate the concepts by some simple examples.
n  
4
Example 1:
Determine the z-transform of
the following signals
x[n] = [1, 2, 5, 7, 0, 1]
Solution: X(z) = 1 + 2z-1+ 5z-2 + 7z-3 + z-5,
ROC: entire z plane except z = 0
(a)
(b) y[n] = [1, 2, 5, 7, 0, 1]
Solution: Y(z) = z2 + 2z + 5 + 7z-1 + z-3
ROC: entire z-plane except z = 0 and z = .
z[n] = [0, 0, 1, 2, 5, 7, 0, 1]
Solution: z-2 + 2z-3 + 5z-4 + 7z-5 + z-7, ROC: all z except z=0
(c)
5
(d) p[n] = [n]
Solution: P(z) = 1, ROC: entire z-plane.
(e) q[n] = [n – k], k > 0
Solution: Q(z) = z-k, entire z-plane except
z=0.
(f) r[n] = [n+k], k > 0
Solution: R(z) = zk,
ROC: entire z-plane except z = .
6
Example 2: Determine the z-transform of
x[n] = (1/2)nu[n]
Solution:

X(z ) 

x [ n ]z
n
n  
n

1
n
    z

n0  2 
 1
1
z
1
2

 1 1 
2z 

n0 
1 
  z
 2 
n
2
 .......
1

1
1
z
1
2
ROC: |1/2 z-1| < 1, or equivalently |z| > 1/2
7
Example 3: Determine the z-transform of the
x[n] =
signal
n
a u[n]
Solution:

X(z ) 

a z
1

n
n


n0
 1  az

n
 az 
1
n0
 az

1 2
 .......
1
1  az
1
ROC :| z | | a |
8
Properties of z-transform

Linearity
If x1[n]  X1(z)
and x2[[n]  X2(z)
then
a1x1[n] + a2x2[n]  a1X1(z) + a2X2(z)
9
Example:
Determine the z-transform of
the signal x[n] = [3(2n) – 4(3n)]u[n]
Solution:
 z [ a u [ n ]] 
1
n
1  az
1
 z [ 3 ( 2 )  4 3  ]  3
1
n
n
4
1
1  2z
1
1  3z
1
Example 4:
Determine the z-transform of
the signal (cosw0n)u[n]
 cos w 0 n u [ n ] 
1
e
2
 z cos w 0 n u [ n ] 

1 z
1  2z
1
1
jw 0 n

1
e
 jw 0 n
2
1
1
2 1e
jw 0
z
1

1
2 1e
1
 jw 0
z
1
cos w 0
cos w 0  z
2
10
Time Shifting Property:
If x[n]  X(z) then x[n-k]  z-kX(z)
Proof:
since

z [ x [ n  k ]] 
 x[ n  k ] z
n
n  
then the change of variable m = n-k
produces

z [ x [ n  k ]] 

x [ m ]z
(m k )
m  
 z
k


x [ m ]z
m
 z
k
X(z )
m  
11
Example: Find the z-transform of a unit step
function. Use time shifting property to find ztransform of u[n] – u[n-N].
The z-transform of u[n] can be found as

z [ u [ n ]] 

u [ n ]z
n


n  
 1 z
1
 z
2

z
n
n0
 ....... 
1
1 z
1
Now the z-transform of u[n]-u[n-N] may be
found as follows:
z [ u [ n ]  u [ n  N ]] 

1 z
N
1 z
1
1
1 z
1
 z
N
1
1 z
1
12
Scaling in the z-domain
If x[n]  X(z)
Then anx[n]  X(a-1z)
For any constant a, real or complex.
Proof:


  a x [ n ]z
z a x[ n ] 
n
n
n


n  


1
x[ n ] a z

n

1
Xa z

n  
Example 5: Determine the z-transform of the signal
an(cosw0n)u[n].
Solution: since
z [cos( w 0 n )u [ n ] 
1 z
1  2z
1
 z [ a cos w 0 n u [ n ]] 
n
1
cos w 0
cos w 0  z
1  az
1  2 az
1
2
1
cos w 0
cos w 0  a z
2
2
13
Time reversal
If x[n]  X(z) then x[-n]  X(z-1)
Proof:

z [ x[  n ]] 
 x [  n ]z
n  
n



m  

x [ m ]z
m

 x[ m ]z 
1  m
1
 X(z )
m  
Example 6: Determine the z-transform of
u[-n].
Solution: since z[u[n]] = 1/(1 – z-1)
Therefore,
Z[u[-n]] = 1/(1-z)
14
Differentiation in the z - Domain
x[n]  X(z) then nx[n] = -z(dX(z)/dz)
Tutorial 4: Q1: Prove the differentiation
property of z – transform.
Example 7: Determine the z-transform of the
signal x[n] = nanu[n].
Solution:
 z [ a u [ n ]] 
n
1
1  az
 z [ na u [ n ]]   z
n
1
d
1
dz 1  az
1

az
1
1  az 
1
2
15
Convolution and Correlation




To study the LTI systems, convolution plays important
role. Shifting multiplications and summation are operations
in computation of convolution.
Correlation which is very much similar to convolution
provides information about the similarity between the two
sequences.
It is used in Radars, digital communication and mobile
communication etc.
The main application of correlation is that the
incoming/received signal is correlated with standard
signals and signal of this set which has maximum
correlation with the incoming/received signal is detected.
16
Convolution of two sequences
If x1[n]  X1(z) and x2[n]  X2(z) then
x1[n]*x2[n] = X1(z)X2(z)
Proof:
The convolution of x1[n] and x2[n] is defined as

x [ n ]  x1 [ n ] * x 2 [ n ] 
 x [k ]x
1
2
[n  k ]
k  
The z-transform of x[n] is

X(z ) 
 x [ n ]z
n  
n



n  
 
 n
  x 1 [ k ]x 2 n  k  z
 n  

Upon interchanging the order of the summation
and applying the time shifting property, we obtain


 
n 
k
X ( z )   x 1 k   x 2 n  k z   X 2  z   x 1 [ k ]z  X 2  z X 1  z 
k  
k  
 n  

17
Example
of
the
 1,
x 2 [n ]  
0,
8:
signals
Compute
x1[n] =
the
convolution
[1, -2, 1] and
0n  5
elsewhere
Solution:
X1(z) = 1 – 2z-1 + z-2
X2(z) = 1 + z-1 + z-2 + z-3 + z-4 + z-5
Now X(z) = X1(z)X2(z) = 1 – z-1 – z-6 + z-7
Hence x[n] = [1, -1, 0, 0, 0, 0, -1, 1]
Note: You should verify this result from the
definition of the convolution sum.
18
Exercise

Find the convolution
of sequences?
x1  {1,  3 , 2} and x 2  {1, 2 , 1}
19
Correlation of two sequences
If x1[n]  X1(z) and x2[n]  X2(z)

rx1 x 2 ( m ) 


x1(n)x 2 (n  m) 
  X 1(z)X 2 (z
z
1
)
n  
Proof:
The follow ing is the
correlati on of two sequences

rx 1 x 2 ( m ) 
x
1
( n ) x 2 [( n  m )]
(1)
n  
Arranging the term x 2 (n  m) as x
above equa tion, we g et
x 1(n)x 2 (n) :
2
[  (m  n)] in

rx 1 x 2 ( m ) 
x
1
( n ) x 2 [  ( m  n )] (2)
n  
Therefore, the RHS o f Eq. ( 2 ) represen ts the con volution
x1(n) and x 2 (  m) can be written as
rx 1 x 2 ( m )  x1 ( m )  x 2 (  m )
20
Continue:
Z [ rx 1 x 2 ( m )]  Z [ x1 ( m )  x 2 (  m )]
Z [ rx 1 x 2 ( m )]  Z [ x1 ( m )]  Z [ x 2 (  m )]
1
Z [ x1 ( m )]  X 1 ( z ) and Z [ x 2 (  m )]  X 2 ( z )
1
Z [ rx 1 x 2 ( m )]  X 1 ( z ) X 2 ( z )
21
Correlation of two sequences
If x1[n]  X1(z) and x2[n]  X2(z)
then rx1x2[k] = X1(z)X2(z-1)

Tutorial 4 Q2: Prove this property.
The Initial Value Theorem:
X(z )
If x[n] is causal then x [ 0 ]  lim
z 
Proof:

X(z ) 

x [ n ]z
n
 x [ 0 ]  x [1 ]z
1
 x [ 2 ]z
2
 ....
n0
Obviously, as z  , z-n  0 since n >0, this
proves the theorem.
22
Final Value Theorem
1


x
[

]

lim
1

z
X(z )
If x[n]  X(z), then
z1

Tutorial 4 Q3: Prove the Final Value
Theorem
Example 9: Find the final value of
2z
X(z ) 
Solution:
1  1 .8 z
1
1  z X ( z )  1  z 
1

 1 z
1
 0 .8 z
2z
1
1

2z
1
1  1 .8 z
1
 0 .8 z
1
1  z 1  0 . 5 z 
1
1

2
2z
2
1
1  0 .5 z
1
The final value theorem yields
y [  ]  lim
z1
2z
1
1  0 .8 z
1

2
0 .2
 10
23
Inverse z-transform
In general, the inverse z-transform may be
found by using any of the following
methods:
 Power series method
 Partial fraction method
24
Power Series Method
Example 2: Determine the z-transform of
X(z ) 
1
1
1  1 .5 z  0 .5 z
2
By dividing the numerator of X(z) by its
denominator, we obtain the power series
1
1
3
2
z
1

1
2
z
2
 1
3
2
z
1

7
4
z
2

15
8
z
3

31
16
z
4
 ...
 x[n] = [1, 3/2, 7/2, 15/8, 31/16,…. ]
25
Power Series Method
Example 2:Determine the z-transform of
X(z ) 
4z
1
1
2  2z  z
2
By dividing the numerator of X(z) by its
denominator, we obtain the power series
 x[n] = [2, 1.5, 0.5, 0.25, …..]
26
Partial Fraction Method:
Example 1: Find the signal corresponding to
the z-transform
X(z ) 
z
2  3z
Solution:X ( z ) 
X(z )
z
3
1
z
z
2  3z

2
3
1
z

2
3
2
z  z  1  z  0 . 5 
2
1

z
X(z )  3  z
1


z  1 .5 z  0 .5 z
0 .5
X(z )  3 
or
0 .5
3

z
z
z1
z
2

 (4)
1
1 z
1

1
4
0 .5
z  z  1  z  0 . 5 
1
z 1

4
z  0 .5
z
z  0 .5
1
1  0 .5 z
1
 x [ n ]  3  [ n ]   [ n  1 ]  u [ n ]  4 0 . 5  u [ n ]
n
27
Partial Fraction Method:
Example 2: Find the signal corresponding to the ztransform
Y(z ) 
1
1  0 .2 z 1  0 .2 z 
1 2
1
Solution:
Y(z ) 
Y(z )

z
Y (z ) 
 0 . 25
z
z
3
 0 . 2  z  0 . 2 
z
2
z  0 . 2 z  0 . 2 2
0 . 25 z
z  1
1
1  0 .2 z
1

 0 . 75
0 . 75 z
z  0 .2
1
1  0 .2 z
1

0 . 25
z  0 .2


2
0 .1
0 .2

0 . 75
z  0 .2

0 .1
 z  0 . 2 2
0 .1 z
z
 0 .2 
0 .2 z
2
1
1  0 .2 z 
1 2
 y [ n ]  0 . 25  0 . 2  u [ n ]  0 . 75 0 . 2  u [ n ]  0 . 5 n 0 . 2  u28[ n ]
n
n
n
Z-Transform Solution of Linear
Difference Equations




We can use z-transform to solve the difference
equation that characterizes a causal, linear, time
invariant system. The following expressions are
especially useful to solve the difference
equations:
z[y[(n-1)T] = z-1Y(z) +y[-T]
Z[y(n-2)T] = z-2Y(z) + z-1y[-T] + y[-2T]
Z[y(n-3)T] = z-3Y(z) + z-2y[-T] + z-1y[-2T] +
y[-3T]
29
Example:
Consider the following difference
equation:
y[nT] –0.1y[(n-1)T] – 0.02y[(n-2)T] = 2x[nT] –
x[(n-1)T]
where the initial conditions are y[-T] = -10 and y[2T] = 20. Y[nT] is the output and x[nT] is the unit
step input.
Solution:
Computing the z-transform of the difference
equation gives
Y(z) – 0.1[z-1Y(z) + y[-T]] – 0.02[z-2Y(z) + z-1y[-T]
+ y[-2T]] = 2X(z) – z-1X(z)
Substituting the initial conditions we get
Y(z) – 0.1z-1Y(z) +1 – 0.02z-2Y(z) – 0.2z-1 –0.4 =
30
-1
(2 – z )X(z)
1  0 .1 z
1
 0 . 02 z

Y ( z ) 1  0 .2 z
Y (z ) 
Y ( z )  2  z 
1
1  z 1  0 . 1 z
1
3
 0 .6 z
2
1
1
 0 . 02 z
1 .4  0 .6 z
1 .4 z

1
2

2
2z
1
1 z
1
 0 .2 z
1
1 z
 0 .2 z
2
 0 . 02 z
2
1
1
1
 0 .2 z  0 .6
 0 .6
1 .4  0 .6 z

1
 0 .2 z
2
 1  z 1  0 . 2 z 1  0 . 1 z 
1
1
1
 0 .2 z
 z  1  z  0 . 2  z  0 . 1 
Y (z )
z

1 . 136
z1
Y ( z )  1 . 136

 0 . 567
z  0 .2
1
1 z
1
 0 . 567

0 . 830
z  0 .1
1
1  0 .2 z
1
 0 . 830
1
1  0 .1 z
1
and the output signal y[nT] is
31
y [ nT ]  1 . 136 u[ nT ]  0 . 567 ( 0 . 2 ) u[ nT ]  0 . 830 (  0 . 1 ) u[ nT
]
n
n