Digital Signal Processing Reference
In-Depth Information
where
z¼0:5
¼3 and
z¼1
¼ 4
z
þ 1
z 1
z þ 1
z 0:5
A ¼
B ¼
The z-transform step response is therefore
Y ðzÞ¼
3z
z 0:5
þ
4z
z 1
Applying the inverse z-transform table yields the step response as
n
uðnÞþ4uðnÞ
yðnÞ¼3ð0:5Þ
c. To determine the system output response, we first find the z-transform of the input xðnÞ,
z
z 0:25
n
uðnÞg ¼
X ðzÞ¼Zfð0:25Þ
Then Y ðzÞ can be obtained via Equation
(6.10)
, that is,
z þ 1
z 0:5
$
z
z 0:25
¼
zðz þ 1Þ
ðz 0:5Þðz 0:25Þ
Y ðzÞ¼HðzÞX ðzÞ¼
Using the partial fraction expansion, we have
Y
ð
z
z
¼
A
z 0:5
þ
ðz þ 1Þ
ðz 0:5Þðz 0:25Þ
¼
B
z 0:25
2
1
0
0
1
2
3
4
5
6
7
8
9
10
Sample number
4
2
0
0
1
2
3
4
5
6
7
8
9
10
Sample number
2
1
0
0
1
2
3
4
5
6
7
8
9
10
Sample number
FIGURE 6.4
Impulse, step, and system response in Example 6.7.
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