Time-Domain Analysis and z Transform - Introduction to Digital Signal Processing and Filter Design - page 65

Y 2 (z) H(z) , or the output response of a system when it is excited by a unit

pulse function δ(n) , under zero initial states, is given by the inverse z transform
of H(z) . Thus the unit pulse response denoted by h(n )isgivenby

=

Z − 1 [ H(z) ].
So h(n) is the response of the system due to an excitation δ(n) only. However,
from the general relationship Y 2 (z) H (z)X(z) , we observe that if we know the

transfer function H(z) or if we know the unit pulse response h(n) of the system,
we can find the response due to any other input x(n) . Therefore H(z) or the
unit impulse response h(n) constitutes another model for the system. If we have
derived or have been given H(z) or h(n) ,nextwefindthe z transform X(z)
of the given input, and multiply H(z) and X(z) to get Y(z)

=

H (z)X(z) as the
z transform of the output. Then we find the inverse z transform of Y(z) to get
the output y(n) . For these operations, which are algebraic in nature, finding the
output y(n) as the inverse z transform of H (z)X(z) is an efficient method for
finding the system output. It is this z -transform method that is used extensively
in system analysis, but it depends on the satisfaction of two conditions: (1) we
can find the z transform of the input sequence and (2) we know or can find the
transfer function of the system under investigation. Students should be aware that
in practice, either one or both of these conditions may not be satisfied and other
methods of analysis or design of systems are called for. For example, finding a
closed-form expression for a discrete-time signal obtained by sampling a speech
is not easy. Finding the transfer function of physical systems may not be as easy
and straightforward as the one shown in (2.46). In this topic, we assume that
these conditions are always satisfied.

=

2.6 CONVOLUTION REVISITED

In a previous section on convolution, we had shown that the output y(n) of a
linear, shift-invariant, discrete-time system is obtained by convolution of x(n)
and h(n) , specifically, y(n) = k = 0 x(k)h(n = x(n) ∗ h(n) − k) .

Since Y(z) = H (z)X(z) = X(z)H (z) , we now conclude that convolution sum

operation is commutative:

x(n) ∗ h(n) = h(n) ∗ x(n)

= k = 0 x(k)h(n = k = 0 h(k)x(n Therefore y(n) − k) − k) .

Another way of proving this result is as follows. Let

∞

y(n) = x(k)h(n − k).

k = 0

Then

∞

k) z − n

∞ ∞

y(n)z − n Y(z) = = x(k)h(n −

n = 0 n = 0 k = 0