Time-Domain Analysis and z Transform - Introduction to Digital Signal Processing and Filter Design - page 36

or in another equivalent form

N M

a(k)y(n − k) = b(k)x(n − k) ; a( 0 ) = 1 (2.3)

k = 0 k = 0

Equation (2.2) shows that the output y(n) is determined by the weighted sum
of the previous N values of the output and the weighted sum of the current and
previous M 1 values of the input. Very often the coefficient a( 0 ) as shown in

(2.3) is normalized to unity.
Soon we will introduce the z transform to represent the discrete-time signals
in the set of equations above, thereby generating more models for the system, and
from these models in the z domain, we will derive the transfer function H(z − 1 )
and the unit sample response or the unit impulse response h(n) of the system.
From any one of these models in the z domain, we can derive the other models in
the z domain and also the preceding models given in the time domain. It is very
important to know how to obtain any one model from any other given model
so that the proper tools can be used efficiently, for analysis of the discrete-time
system. In this chapter we will elaborate on the different models of a discrete-
time system and then discuss many tools or techniques for finding the response of
discrete-time systems when they are excited by different kinds of input signals.

+

2.1.2 Recursive Algorithm

Let us consider an example of Equation (2.2) as y(n) = y(n − 1 ) − 0 . 25 y(n −

2 ) + x(n) , where the input sequence x(n) = δ(n) , and the two initial conditions

are y( = 0 . 4.

We compute y( 0 ) , y( 1 ) , y( 2 ) , ... in a recursive manner as follows: y( 0 )

− 1 ) = 1 . 0and y( − 2 )

=

y( − 1 ) − 0 . 25 y( − 2 ) + x( 0 ) .Since x(n) = δ(n) , we substitute x( 0 ) = 1 and get

y( 0 ) = 1 . 0 − 0 . 25 ( 0 . 4 ) + 1 = 1 . 9. Next y( 1 ) = y( 0 ) − 0 . 25 y( − 1 ) + x( 1 ) .We

know y( 0 ) = 1 . 9 from the step shown above, and also that x( 1 ) = 0. So

we get y( 1 ) = 1 . 9 − 0 . 25 ( 1 . 0 ) + 0 = 1 . 65. Next, for n = 2, when we compute

y( 2 ) = y( 1 ) − 0 . 25 y( 0 ) + x( 2 ) . Substituting the known values from above, we

get y( 2 ) = 1 . 65 − 0 . 25 ( 1 . 9 ) + 0 = 1 . 175.

Next, when n = 3, we obtain

y( 3 ) = y( 2 ) − 0 . 25 y( 1 ) + x( 3 )

= 1 . 175 − 0 . 25 ( 1 . 65 ) + 0 = 0 . 760

We can continue to calculate the values of the output y(n) for n = 4 , 5 , 6 , 7 ,... .

This is known as the recursive algorithm , which we use to calculate the output
when we are given an equation of the form (2.2); it can be used when there is
any other input. For a system modeled by an equation of the form (2.2), the
output is infinite in length in general. As a special case, when the input is the
unit impulse function δ(n) , and the initial conditions are assumed to be zero, the