Digital Signal Processing Reference
In-Depth Information
10z
z
2
z þ 1
c.
X ðzÞ¼
z
4
z 1
þ z
6
z
3
z þ 0:5
d.
X ðzÞ¼
þ
Solution:
xðnÞ¼2Z
1
ð1Þþ4Z
1
z
z 1
Z
1
z
z 0:5
a.
x
n
¼ 2d
n
þ 4u
n
ð0:5Þ
n
u
n
xðnÞ¼Z
1
!
Z
1
!
¼ 5Z
1
z
ðz 1Þ
!
Z
1
!
5z
ðz 1Þ
2z
ðz 0:5Þ
2
0:5
0:5z
ðz 0:5Þ
b.
2
2
2
2
Then
xðnÞ¼5nu
n
4nð0:5Þ
n
u
n
10
sinðaÞ
10z
z
2
z þ 1
¼
sinðaÞz
z
2
2zcosðaÞþ1
c.
X ðzÞ¼
By coefficient matching, we have
2cosðaÞ¼1
Hence, cosðaÞ¼0:5, and a ¼ 60
0
. Substituting a ¼ 60
0
into the sine function leads to
sin
a
¼ sin
60
0
¼ 0:866
Finally, we have
xðnÞ¼
10
sinðaÞ
Z
1
sinðaÞz
z
2
2zcos
a
þ 1
10
0:866
sinðn,60
0
Þ¼11:547sinðn,60
0
Þ
¼
xðnÞ¼Z
1
z
5
þ Z
1
z
6
,1
þ Z
1
z
4
d.
z
z 1
z
z þ 0:5
Using
Table 5.1
and the shift property, we get
n4
uðn 4Þ
xðnÞ¼uðn 5Þþdðn 6Þþð0:5Þ
Now, we are ready to deal with the inverse z-transform using the partial fraction expansion and
lookup table. The general procedure is as follows:
1.
Eliminate the negative powers of
z
for the z-transform function
XðzÞ
.
2.
Determine the rational function
XðzÞ=z
(assuming it is proper), and apply the partial fraction
3.
Multiply the expanded function
XðzÞ=z
by
z
on both sides of the equation to obtain
XðzÞ
.
4.
Apply the inverse z-transform using
Table 5.1
.
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