Image Processing Reference
In-Depth Information
The result is
lim
z !1
X ( z ) ¼ x (
0
) þ
0
þ
0
þ¼ x (
0
)
(
3
:
65
)
(f) Final Value Theorem :Ifx ( n ) ¼
0 for n <
0 and X ( z )
does not have any
poles on the boundary of unit circle, then
x (1) ¼ lim
z ! 1
( z
1
) X ( z )
(
3
:
66
)
To derive this property, we use the delay and linear properties of the
z-transform, mainly
( z
1
) X ( z ) ¼ zX ( z ) X ( z ) ¼ z-
transform of [ x ( n þ
1
) x ( n )]
(
3
:
67
)
Therefore,
1
) x ( n )] z n
( z
1
) X ( z ) ¼
1 [ x ( n þ
1
(
3
:
68
)
n ¼
Taking the limit as z !
1 results in
1
z! 1 ( z
lim
1
) X ( z ) ¼
1 [ x ( n þ
1
) x ( n )]
n ¼
¼ x (
0
) x (
1
) þ x (
1
) x (
0
) þ x (
2
) x (
1
) þ
(
3
:
69
)
Since x (
1
) ¼
0, then:
lim
z ! 1 ( z
1
) X ( z ) ¼ x (1)
(
3
:
70
)
Example 3.18
Find the initial and the
final values of the one-sided signal x(n), if
z(z
0
:
7)
X(z)
¼
z 2
1
:
25z þ
0
:
25
S OLUTION
Using the initial and
final value theorems, we have
z(z
0
:
7)
x(0)
¼
lim
z!1
X(z)
¼
lim
z!1
25 ¼
1
z 2
1
:
25z þ
0
:
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