Digital Signal Processing

Module 4

System Function

Objective:

1. To understand the system function and analyze it.

2. To understand the concept of stability and frequency response of Stable systems

Introduction:

Given a linear shift-invariant system with a unit sample response h(n), the input and output

are related by a convolution sum

This relationship implies that Y( e jω

)= X( e jω

) H( e jω

) where H( e jω

), the frequency

response of the system, is the discrete-time Fourier transform of h(n). This relationship

between x(n) and y(n) may also be expressed in the z-transform domain as

where H(z), the z-transform of h(n), is the system function of the LSI system. The system

function is very useful in the description and analysis of LSI systems.

Description:

System Function

The frequency response of a linear shift-invariant system is the discrete-time Fourier

transform of the unit sample response, and the system function is the z-transform of the unit

sample response:

The frequency response may be derived from the system function by evaluating H(z) around

the unit circle:

If the z-transform of the input to a linear shift-invariant system with a system function H(z) is

X(z), the z-transform of the output is

For linear shift-invariant systems that are described by a linear constant coefficient difference

equation,

the system function is a rational function of z:

Therefore, the system function is defined, to within a scale factor, by the location of its poles,

αk and zeros, βk. Note that each term in the numerator

contributes a zero to the system function at z = βk and a pole to the system function at z = 0.

Similarly, each term in the denominator contributes a pole at z = αk and a zero at z = 0.

Therefore, including the poles and zeros that may lie at z = 0 or z =∞, the number of zeros in

H (z) is equal to the number of poles.

Stability

The unit sample response of a stable system must be absolutely summable:

Note that because this is equivalent to the condition that

for |z| = 1, the region of convergence of the system function must include the unit circle if the

system is stable.

Frequency Response for Rational System Functions

A linear shift-invariant system with a rational system function may be written in

factored form as follows:

The frequency response of a linear shift-invariant system may be found from the

system function by evaluating H(z) on the unit circle. For a rational function of z, the

frequency response may be found geometrically from the poles and zeros of H(z). With H(z)

written in factored form as in the above equation, the frequency response is

Illustrative Examples:

Problem 1: If the input to a linear shift-invariant system is

the output is

Find the system function, H(z) , and determine whether or not the system is stable and/or

causal.

Solution:

In order to find the system function, recall that H ( z ) = Y ( z ) / X ( z ) . Because we are given

both x ( n ) and y ( n ) , all that is necessary to find H ( z ) is to evaluate the z-transform of

x ( n ) and y ( n ) and divide. With

For the region of convergence of H ( z ) , we have two possibilities. Either |z|>3/4 or |z|<3/4.

Because the region of convergence of Y ( z ) is |z|>3/4 and includes the intersection of the

regions of convergence of X ( z ) and H(z), the region of convergence of H ( z ) must be

|z|>3/4 . Because the region of convergence of H ( z ) includes the unit circle, h ( n ) is stable,

and because the region of convergence is the exterior of a circle and includes z = ∞, h ( n ) is

causal.

Problem 2: Determine H(z) and its poles and zeros for the system described by difference

equation

Solution:

Taking z-transform

Hence solving numerator and denominator polynomials

The zero locations are z=0 and z= -1

The pole locations are z= - ½ and z= -1/4

Problem 3: Determine the frequency response for the three point moving average system

described by 𝑦 𝑛 =1

3 𝑥 𝑛 + 1 + 𝑥 𝑛 + 𝑥(𝑛 − 1) . Plot the magnitude and phase

functions for 0 < ω < π

Solution:

Since ℎ 𝑛 = 1

3,1

3,1

3

It follows that 𝐻 𝑒𝑗𝜔 =1

3 𝑒𝑗𝜔 + 1 + 𝑒−𝑗𝜔 =

1

3 1 + 2𝑐𝑜𝑠 𝜔

Hence magnitude response is 𝐻 𝑒𝑗𝜔 =1

3|1 + 2𝑐𝑜𝑠 𝜔 |

Phase response is 𝜃 𝜔 = 0 0 ≤ 𝜔 ≤ 2𝜋/3𝜋 2𝜋/3 ≤ 𝜔 ≤ 𝜋

Summary:

Hence knowing the transfer function of the LSI system we can find the location of

poles and zeros and thereby characterizing the system as stable or causal. Analyzing the

frequency response of system we can understand how the system behaves with different

frequencies.

Assignment:

Problem 1: Consider an LTI system, initially at rest, described by the difference equation

Determine the system function and hence impulse response

Problem 2: Determine the impulse response of the causal system given below by finding the

homogeneous and particular solutions and discuss on stability

Problem 3: Consider an LTI system, initially at rest, described by the difference equation

Determine the impulse response of the system and sketch the frequency response of this system

Simulation:

% Pole – Zero Map in Z- Domain clc; clear all; close all; %z-domain LTI system %y(n)=(3/8)y(n-1)+(2/3)y(n-2)+x(n)+(1/4)x(n-1) a= input('Enter the Numerator coefficients'); b = input('Enter the Denominator coefficients'); [z1,p1,k]=tf2zp(a,b); disp('pole locations are');p1 disp('zero locations are');z1 figure, zplane(a,b)

OUTPUT:

pole locations are

p1 = 1.0252

-0.6502

zero locations are

z1 = -0.2500

% To verify the stability of the System num = input (' type the numerator vector '); den = input (' type the denominator vector '); [z,p,k] = tf2zp(num,den); disp ('Gain constant is '); disp(k); disp (' Zeros are at '); disp(z) disp ('radius of Zeros ') ; radzero = abs(z) disp ('Poles are at '); disp(p) disp ('radius of Poles ') ; radpole = abs(p) if max(radpole) >= 1 disp (' ALL the POLES do not lie within the Unit Circle '); disp (' The given LTI system is NOT a stable system '); else disp (' ALL the POLES lie WITHIN the Unit Circle '); disp (' The given LTI system is a REALIZABLE and STABLE system '); end; zplane(num,den) title ( ' Pole-Zero plot of the LTI system ' );

Input:

type the numerator vector [1,1/4]

type the denominator vector [1,-3/8,-2/3]

Output:

Gain constant is

1

Zeros are at

-0.2500

radzero = 0.2500

Poles are at 1.0252

-0.6502

radius of Poles

radpole = 1.0252

0.6502

ALL The POLES Do Not Lie Within The Unit Circle

The Given LTI System Is NOT A Stable System

References:

1. Digital Signal Processing, Principles, Algorithms and Applications – John G Proakis, Dimitris G Manolakis,

Pearson Education / PHI, 2007

2. Discrete Time Signal Processing – A V Oppenheim and R W Schaffer, PHI, 2009

3. Digital Signal Processing – Monson H.Hayes – Schaum’s Outlines, McGraw-Hill,1999

4. Fundamentals of Digital Signal Processing using Matlab – Robert J Schilling, Sandra L Harris, Thomson

2007.

5. Digital Signal processing – A Practical Approach, Emmanuel C Ifeachor and Barrie W Jervis, 2nd

Edition, PE

2009

6. Digital Signal Processing – A Computer Based Approach, Sanjit K.Mitra, McGraw Hill,2nd

Edition, 2001