The z-Transform - Digital Signal Processing: Fundamentals and Applications - page 148

We first find B :

z¼1 ¼ z¼1 ¼

Y ð z Þ
z

zðz þ 1Þ
z 2 z þ 0:5

1 ð1 þ 1Þ
1 2 1 þ 0:5 ¼ 4

B ¼ðz 1Þ

Notice that A and A are a complex conjugate pair. We determine A as follows:

z¼0:5þj0:5 ¼ z¼0:5þj0:5

Y ð z Þ
z

zðz þ 1Þ
ðz 1Þðz 0:5 þ j0:5Þ

A ¼ðz 0:5 j0:5Þ

ð0:5 þ j0:5 1Þð0:5 þ j0:5 0:5 þ j0:5Þ ¼ ð0:5 þ j 0:5Þð1:5 þ j 0:5Þ

ð0:5 þ j0:5Þð0:5 þ j0:5 þ 1Þ

¼

ð0:5 þ j0:5Þj

Using the polar form, we get

0:707 : 45 0 1:58114 : 18:43 0
0:707 : 135 0 1 : 90 0

¼ 1:58114 : 161:57 0

A ¼

A ¼ A ¼ 1:58114 : 161:57 0

Assume that a first-order complex pole takes the form

P ¼ 0:5 þ 0:5j ¼ jPj : q ¼ 0:707 : 45 0 and P ¼ jPj : q ¼ 0:707 : 45 0

We have

A z
ðz P Þ

Y ðzÞ¼ 4z

Az
ðz PÞ þ z 1 þ

Applying the inverse z-transform from Line 15 in Table 5.1 leads to
yðnÞ¼4Z 1 z
z 1

þ Z 1 Az

A z
ðz P Þ ðz PÞ þ

Using the previous formula, the inversion and subsequent simplification yield
y n ¼ 4u n þ 2jAjðjPjÞ cos nq þ 4 u n

¼ 4u n þ 3:1623ð0:7071Þ

n

cos 45 0 n 161:57 0 u n

n

The situation dealing with real repeated poles is presented next.

EXAMPLE 5.11
Find xðnÞ if

z 2
z 1 z 0:5 2

X ðzÞ¼