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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