Image Processing Reference
In-Depth Information
Taking inverse z-transform yields
2)
n
2n(0
:
2)
n
u(n)
3)
n
u(n)
x(n)
¼
30(0
:
u(n)
þ
30(0
:
0
:
2
or
2)
n
2)
n
3)
n
]u(n)
x(n)
¼
[
30(0
:
10n(0
:
þ
30(0
:
Example 3.24
Consider a SISO system described by the second-order DE:
y(n þ
2)
þ
0
:
4y(n þ
1)
þ
0
:
03y(n)
¼ x(n)
Find the output of this system if the input signal is a unit step function x(n)
¼ u(n)
with the initial conditions y(1)
¼ y(0)
¼
0.
S
OLUTION
Taking z-transform from both sides of the above DE yields
z
z
z
2
Y(z)
þ
0
:
4zY(z)
þ
0
:
03Y(z)
¼ X(z)
¼
1
Therefore,
z
z
Y(z)
¼
1)
¼
(z
2
þ
0
:
4z þ
0
:
03)(z
(z þþ
0
:
1)(z þ
0
:
3)(z
1)
Partial fraction of
Y
(
z
)
z
results in
Y(z)
z
¼
1
0
6993
z
:
4
5454
z þ
:
3
8461
z þ
:
1)
¼
1
1
þ
(z þþ
0
:
1)(z þ
0
:
3)(z
0
:
0
:
3
0
:
6993z
z
4
5454z
z þ
:
3
8461z
z þ
:
Y(z)
¼
1
þ
1
0
:
0
:
3
Hence,
1)
n
3)
n
]u(n)
y(n)
¼
[0
:
6993
4
:
5454(
0
:
þ
3
:
8461(
0
:
3.5.3 R
ELATION BETWEEN THE Z
-T
RANSFORM AND THE
L
APLACE
T
RANSFORM
Consider analog signal x
a
(
t
)
1
T
as shown in Figure 3.7.
The output of the sampler could be considered to be either the discrete signal
being sampled at the rate of f
s
¼
x
(
n
)
or the continuous signal x*
(
t
)
de
ned as
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