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The z-transform
Part 2
Dr. Ali Hussein Muqaibel
Dr. Ali Muqaibel
The material to be covered in this lecture is
as follows:
Properties of the z-transform
Linearity
Initial and final value theorems
Time-delay
z-transform table
Inverse z-transform
Application of z-transform to discrete-time systems
Dr. Ali Muqaibel
After finishing this lecture you should be
able to:
Find the z-transform for a given signal utilizing the ztransform tables
Utilize the z-transform properties like the initial and final
value theorems
Find the inverse z-transform.
Utilize z-transform to perform convolution for discretetime systems.
Dr. Ali Muqaibel
Derivation of the z-Transform
The ztransform is defined as follows:
The coefficient denote the sample value and denotes that
the sample occurs n sample periods after the t=0 reference.
Rather than starting form the given definition for the ztransform, we may build a table for the popular signals and
another table for the z-transform properties.
Like the Fourier and Laplace transform, we have two options
either to start from the definition or we may utilize the
tables to find the proper transform.
The next slide illustrates a few z-transform pairs.
Then we will investigate some of the z-transform properties:
Linearity
Time-shifting property
Initial and final value theorems
Dr. Ali Muqaibel
Table of z-transform pairs
F(z)
1
(t k T )
z k
u (t )
z
z 1
Tz
( z 1) 2
t2
e
at
te
at
a nu [n ]
T 2 z ( z 1)
( z 1)3
z
z e a T
Tze a T
( z e a T ) 2
z
z a
Dr. Ali Muqaibel
Linearity of the z-Transform
with region of convergence,
If
And
Then If
with
This follows directly from the definition of the z-transform because
the summation operator is linear.
It is easily extended to a linear combination of an arbitrary number
of signals.
This property includes the multiplication by constant property
which states that if the signal is scaled by a constant its z-transform
will be scaled by the same constant.
with region of convergence,
Dr. Ali Muqaibel
1.
2.
Time-Shifting property for the z-Transform
If
Then
Proof
with
ROC=R
Z { x [ n 1]}
x [ n 1] z n
x [ n 1] z ( n 1 )
x [ m ] z m z 1 Z { x [ n ]}
This property will be very important for producing the ztransform transfer function of a difference
equation which uses the property:
Z
x [n 1] z 1X (z )
Dr. Ali Muqaibel
Example 1: Properties of the z-transform
Find the z-transform for the input signal
x [n ] 7(1/ 3) n 2 u [n 2] 6(1/ 2) n 1u [n 1]
Solution:
We know that
So
z
z
6 z 1
z 1/ 3
z 1/ 2
1
1
7 2
6
z 1/ 3 z
z 1/ 2
X ( z ) 7 z 2
Dr. Ali Muqaibel
Initial and Final Value Theorems
If
then
has a z-transform
and if lim
exists,
lim n 0 x [n ] x [0] lim z X (z )
This theorem can be easily proven by the definition of the ztransform
As we take the limit all terms will be zero except the first
term
n
1
2
X (z ) x [n ]z
x [0] x [1]z
x [2]z
...
n 0
The final value theorem which is given by
lim n x [n ] lim z 1 1 z 1 X (z )
Dr. Ali Muqaibel
Example 2: Application of the initial and
final value theorems
Find the initial and final values for the following signal
expressed in its z-transform
0.792z 2
F (z )
Solution:
(z 1)(z 2 0.416z 0.208)
Initial-value
Final value
0.792z 2
F (z )
0
3
z
f (n )
0.792
1
(1 0.416 0.208)
These answers can be justified by looking at the
expansion of the given expression
F ( z ) 0.792z 1 1.12z 2 1.091z 3 1.01z 4 0.983z 5 0.989z 6 0.99z 7 ....
The coefficient for is zero which is the initial value.
The coefficient converges to one as the negative power
of z increases which corresponds to the final value.
Dr. Ali Muqaibel
Tables of z-transform properties
Dr. Ali Muqaibel
Inverse z-Transform
x [n ]z
X (z )
x [n ]
Find the inverse ztransform for
by long division .
The inverse operation for the z-transform my be accomplished by:
Long division
Partial fraction expansion
The z-transform of a sample sequence can be written as
X (z ) x (0) x (T )z 1 x (2T )z 2 ....
If we can write X(z) into this form, the sample values can be determined by
inspection.
When X(z) is represented in a ratio of polynomials in z, this can be easily
achieved by long division.
Before carrying out the division, it is convenient to arrange both the
numerator and the denominator in ascending powers of z-1.
Dr. Ali Muqaibel
Inverse z-transform using Partial Fraction
Expansion
Alternatively, we may avoid the long division by partial
fraction expansion. The idea is similar to the method used
for inverse Laplace transform.
The objective is to manipulate X(z) into a form that can
be inverse z-transformed by using z-transform tables.
Because of the forms of transforms,
it is usually best to perform partial fraction expansion of H(z)/z.
As an alternative z-1 can be treated as the variable in the partial
fraction expansion.
Important: before doing partial-fraction expansion, make
sure the z-transform is in proper rational function of !
Dr. Ali Muqaibel
Example 3: Inverse z-Transform Using
Partial Fraction Expansion
Find the inverse z-transform using both partial fraction
expansion and long division
Solution:
If we treat
as the variable in the partial fraction
expansion, we can write
Utilizing Heavisides Expansion Method:
1
1
0.2
1.25
0.25
Dr. Ali Muqaibel
1 0.2 z 1 0 ( z 0.2 )
Continue Solution of Example 3:
X ( z)
1.25
0.25
n
x
nT
(
)
(1.25
0.25(0.2)
)u[n]
1
1
1 z
1 0.2 z
From which we may find that 0
1,
1.2, 2
1.24, 3
1.248
We may get to the same answer using long division. X(z) is written as
which is, after multiplying numerator and denominator by z-2
Now, it is left for you to show that the long division will result in the same
answer given by
1.2
1.24
1.248
Dr. Ali Muqaibel
Difference Equation
For continuous-time systems, differential equation may be
solved using Laplace transform
Similarly discrete-time systems result in Difference
Equations which may be solved using z-transform
Recall that discrete-time systems process a discrete-time
input signal to produce a discrete-time output signal.
The general symbolic notation for Discrete-Time System:
Dr. Ali Muqaibel
Transfer Function in the z-Domain
There is a simple relationship for a signal time-shift
1 z
This is fundamental for deriving the transfer function of a
difference equation which is expressed in terms of the inputoutput signal delays
The transfer function of a discrete time LTI system is the ztransform of the systems impulse response
The transfer function is a rational polynomial in the complex
number z.
Convolution is expressed as multiplication
X z
and this can be solved for particular signals and systems
Dr. Ali Muqaibel
Example 4: Discrete-Time Convolution
Calculate the output of a first order difference equation of a input signal
0.5
System transfer function (z-transform of the impulse response)
0.8
1
Taking the z-transform of the input signal
Taking the z-transform of the difference equation
0.8
1 0.8
0.5
The (z-transform of the) output is therefore the product:
ROC | | 0.8
z2
Y (z )
(z 0.5)(z 0.8)
1 0.8z
0.5z
0.3 (z 0.5) (z 0.8)
y [n ] (0.8*0.5n u [n ] 0.5*0.8n u [n ]) / 0.3
Dr. Ali Muqaibel
Self Test
Question 1:
If
Answer:
Question 2:
)?
Find the z-transform for
Answer:
5
1.25 6.25 0.2
If we compare with Example 3 we conclude that the answer should be
1.25 0.25 0.2
2
Which is the same . Try for
0, 1,2,3, . . The values are 0,0,1, 1.2,
n
Question 3:
, whats
Find
Answer :
1 1
1
h( nT ) u( n 5)
3 3
3
n
1
1
x ( nT ) u( n 3)
4
4
1 1
1
1
x (nT ) * h (nT ) ( )3 ( )5 [4( ) n 8 3( ) n 8 )]u (n 8)
4 3
3
4
Dr. Ali Muqaibel
1
4
n 5
u( n 5)
n 3
u( n 3)
Continue Self Test
Question 4 :
Calculate the step response to the system describe by the
following difference equation
6
5
1
1
2
Answer
H (z )
1
6 5z 1 1z 2
1
(2 z 1 )(3 z 1 )
u [n ] X (z )
1
(3 z 1 )(2 z 1 )(1 z 1 )
1
1
1
0.5
0.5
(3 z 1 ) (2 z 1 )
(1 z 1 )
1
1
1
0.167
0.5
0.5
(1 1/ 3z 1 )
(1 1/ 2z 1 )
(1 z 1 )
Y (z )
y [n ] (0.167(1/ 3) n 0.5(1/ 2) n 0.5)u [n ]
Dr. Ali Muqaibel
1
1 z 1
Continue Self Test
Q5: Perform the following
Convolution using z-transform and
sketch the final answer
2z
2 1.5
1
2 3.5
0.5
1.5z
0.5z
0.5
2z
Dr. Ali Muqaibel
