THE z-TRANSFORM - Signals, Systems, and Transforms - page 570

It is common when specifying digital filters to give the filter-transfer function
or the difference equation. For example, the block-diagram specification for
the -filter of the last example can be any of the four forms illustrated in Figure 11.5.
The representation of the time delay by the transfer function

H(z)

a

z -1

is based on the

property

z [y[n - 1]] = z -1 z [y[n]] = z -1 Y(z).

Inverse z -Transform

Three procedures for finding the inverse z -transform will now be discussed. The
first procedure involves the use of the complex inversion integral, from (11.2):

1
2p j C ≠

F(z)z n- 1 dz. f[n] =

This integration is in the complex plane and is usually too complicated to be of prac-
tical value; hence, we will not use this approach. (See Ref. 3 for applications.)
The second method for finding the inverse z -transform is by partial-fraction
expansion, in which a function that does not appear in the z -transform tables is ex-
pressed as a sum of functions that do appear in the tables. Partial-fraction expan-
sions are presented in Appendix F; those readers unfamiliar with this topic should
read this appendix. We now illustrate the use of partial fractions to find inverse
z -transforms.

Inverse z -transform by partial-fraction expansion EXAMPLE 11.9

We solve for the time response of the a -filter of Example 11.8, for the unit step input. From

Table 11.2, X(z) = z/(z - 1). From (11.45), the transformed output is given by

0.1z
z - 0.9

z
z - 1 .

Y(z) = H(z)X(z) =

We expand Y(z) / z in partial fractions:

Y(z)
z

k 1
z - 0.9 +

k 2
z - 1 .

0.1z
(z - 0.9)(z - 1) =

=

Then (see Appendix F)

0.1z
(z - 0.9)(z - 1)

0.1z
z - 1

B R

k 1 = (z - 0.9)

z= 0.9 = z= 0.9 =-0.9

and

0.1z
(z - 0.9)(z - 1)

0.1z
z - 0.9

B R k 2 = (z - 1)

z= 1 = z= 1 = 1.