The z-Transform - Digital Signal Processing: Fundamentals and Applications - page 152

where B ¼ 4; p 1 ¼ 1;

A ¼1:5 0:5j ; p ¼ 0:5 þ 0:5j ;

A ¼1:5 þ 0:5j;

and p ¼ 0:5 0:5j

c. Similarly, if we use
» D [ conv(conv([1 L 1], [1 L 0.5]),[1 L 0.5])
D [
1.0000 L 2.0000 1.2500 L 0.2500

then

z 2
ðz 1Þðz 0:5Þ 2 ¼

z 2
z 3 2z 2 þ 1:25z 0:25

X ðzÞ¼

and we yield

X ð z z ¼ z

z 3 2z 2 þ 1:25z 0:25

From MATLAB, we obtain
» [R,P,K] [ residue([1 0],[1 L 2 1.25 L 0.25])
R [
4.0000
L 4.0000
L 1.0000
P [
1.0000
0.5000
0.5000
K [
[]
»

Using the previous results leads to

X ðzÞ¼ 4z

4z
z 0:5

z
ðz 0:5Þ 2

z 1

5.4 SOLUTION OF DIFFERENCE EQUATIONS USING THE Z-TRANSFORM
To solve a difference equation with initial conditions, we have to deal with time-shifted sequences such
as yðn 1 Þ , yðn 2 Þ , . , yðn mÞ , and so on. Let us examine the z-transform of these terms. Using the
definition of the z-transform, we have