ztransform app

8/8/2019 Ztransform App

1/5

EE406-041 Discrete-Time Signal Processing Applications of the z-Transform

Page - 1

Applications of the z-Transform

One of the major applications of the z-transform is used as an analysis tool for discrete-time

LTI systems. In particular, we will use the z-transform for finding the frequency response

and evaluating the stability of discrete-time LTI systems.

From the convolution property ofz-transform, we have the relationship between the z-

transforms of input and output sequences of a discrete-time LTI system as

( ) ( ) ( )zXzHzY =

where ( )zX , ( )zY and ( )zH are the z-transforms of the system input, output and impulseresponse, respectively. ( )zH is referred as the system function or transfer function of thesystem.

System Function From the Difference Equation Representation

Consider a LTI system for which the input and output satisfy a linear constant-coefficient

difference equation of the form

= =

=N

k

M

k

kk knxbknya0 0

][][

By applying the z-transform to both sides and using the linearity property and the time-shifting property, we obtain

=

=

=

=

=

=

M

k

k

k

N

k

k

k

M

k

k

k

N

k

k

k

zbzXzazY

zXzbzYza

00

00

)()(

)()(

so that

=

=

== N

k

k

k

M

k

k

k

za

zb

zX

zY

zH

0

0

)(

)(

)(

The system function for a system satisfying a linear constant-coefficient difference equation

is always rational.

If we factor the numerator and denominator of the system function H(z), we can rewrite it as:

H zz z z z

z p z p( )

( )( )

( )( )=

1 2

1 2

where the zi are the zeros and the pi are the poles.


2/5


Page - 2

Because we are considering only real coefficients, if any of the roots of the numerator or

denominator are complex then they always occur as complex conjugate pairs (e.g. the roots of

( )z2 1+ are j ). Thus complex poles and zeros will always occur in complex conjugatepairs.

The poles and zeros completely specify the system function and are an alternativerepresentation of it. The pole and zero representation is very useful for checking stability and

a way to visualize the frequency response of the system.

Stability and Causality

Forz evaluated on the unit circle ( )jezei =.,. , ( )zH reduces to the frequency response ofthe system provided that the unit circle is in the ROC. For LTI systems, the BIBO stability is

equivalent to the absolute summability of its impulse response with


3/5


Page - 3

(c) For a = 0.5, plot the pole-zero diagram and shade the region of convergence inz-plane.

(d) Find the impulse response h(n) for this system.

(e) Show that this system is an allpass system, i.e. the magnitude of the frequency response is

a constant. Also, specify the value of the constant.

[Solution]

H za z

az

Y z

X z( )

( )

( ),=

=

1

1

1 1

1

As the system is causal, the ROC is |z| > |a|

(a) Y(z) a z-1

Y(z) = X(z) a-1

z-1

X(z)

Taking the inverse transform, we have

]1[][]1[][ 1 = nxanxnayny

(b) For stability, the ROC must include the unit circle. So, the system is stable for |a| < 1.

(c) a = 0.5

1

1

5.01

21)(

=z

zzH

(d) azaz

za

azzH >

=

,11

1)(

1

11

1

]1[)(1

][][ 1 = nuaa

nuanh nn

1/2 2


4/5


Page - 4

(e) H H za e

aez e

j

jj( ) ( )|

= =

=

1

1

1

H H H a e

ae

a e

ae

j

j

j

j( ) ( ) ( )*

21 11

1

1

1= =

aaa

aa

aaa

aaH

1

cos21

cos211

cos21

cos21)(

2/1

2

22/1

2

12

=

++

=

++

=

As the magnitude response of the system is equal to a constant (1/a), the system is an all-

pass system.

Example 2

A causal LTI system has impulse response h[n], for which the z-transform is

H zz

z z( )

( . )( . )=

+ +

1

1 05 1 0 25

1

1 1

(a) What is the region of convergence ofH(z)?

(b) Is the system stable? Explain.

(c) Find the z-transformX(z) of an inputx[n] that will produce the output

y[n] = (-1/3) (-0.25)nu[n] - (4/3) (2)

nu[-n 1]

(d) Find the impulse response h[n] of the system.

[Solution]

H zz

z z z z( )

( . )( . ) ( . ) ( . )=

+ +

=

+

1

1 05 1 0 25

2

1 0 5

1

1 0 25

1

1 1 1 1

(a) The impulse response ][nh is causal => The ROC is |z| > 0.5

(b) ROC includes the unit circle |z|=1 => The system is stable.

(c) ]1[)2(3

4][)

4

1(

3

1][ = nununy nn

Y zz z

z

z zz( )

/

( / )

/

( ( / ) )( ), /=

+

+

=+

+ <


5/5


Page - 5

Since

X zY z

H z

z

zz( )

( )

( )

( . )

( ),= =

ztransform app

Documents