Signals and Systems - Electrical Engineering

9.5 One-Sided Z-Transform Inverse 545

where the second term is proper rational as the denominator is of a higher degree in powers ofz−^1

than the numerator. The inverse Z-transform ofX(z)will then be

x[n]=δ[n]+Z−^1

[

1 − 2 z−^1

1 + 2 z−^1 +z−^2

]

The inverse of the proper rational term is done as indicated in this section. n

nExample 9.15

Find the inverse Z-transform of

X(z)=

1 +z−^1

( 1 +0.5z−^1 )( 1 −0.5z−^1 )

=

z(z+ 1 )

(z+0.5)(z−0.5)

|z|>0.5

by using the negative and the positive powers ofzexpressions.

Solution

Clearly,X(z)is proper if it is considered a function of negative powersz−^1 (inz−^1 , the numerator

is of degree 1 and the denominator of degree 2), but it is not proper if it is considered a function

of positive powersz(the numerator and the denominator are both of degree 2). It is, however,

unnecessary to perform long division to makeX(z)proper when it is considered as a function of

z. One simple approach is to considerX(z)/zas the function. We wish to find its partial fraction

expansion—that is,

X(z)

z

=

z+ 1

(z+0.5)(z−0.5)

(9.32)

which is proper. Thus, wheneverX(z), as a function ofzterms, is not proper it is always possible to

divide it by some power inzto make it proper. After obtaining the partial fraction expansion then

thezterm is put back.

Consider then the partial fraction expansion inz−^1 terms,

X(z)=

1 +z−^1

( 1 +0.5z−^1 )( 1 −0.5z−^1 )

=

A

1 +0.5z−^1

+

B

1 −0.5z−^1

Given that the poles are real—one at z=−0.5 and the other at z=0.5—from the

Z-transform table, we get that a general form of the inverse is

x[n]=[A(−0.5)n+B0.5n]u[n]