Z-Transforms and Solution of Difference Equations13-9
Proof By definition,
∞∞
−−
==
−
−
−−
−
−
∞
−−
=
−
−
−−
⎛⎞⎛⎞
⋅=
⎜⎟⎜⎟
⎝⎠⎝⎠
=++++
+++++
+++++
=++
++++
=〈+++
+
∑∑
∑
00
1
0001100211
2
2001122
0
01122
0
0
01122
()()
()(
)(
)
(
)
(
nn
nn
nn
nnn
n
mnmn
nnn
n
n
mnmn
nnn
mn
uzvzuzvz
uvuvuvzuvuv
uvzuvuvuv
uvuvz
uvuvuv
uvuvz
Zuvuvuv
uv
bydefinition
=
−
−
=
++〉
⎛⎞
⎜⎟
⎝⎠
∑
0
0
),
mn
m
mnm
m
uv
Zuv
Taking inverse Z-transforms of both sides
−
−
=
⋅=
∑
1
0
(()())
n
mnm
m
Zuzvzuv
(13.60)
Example 13.6
Evaluate
⎛⎞
⎜⎟
+
⎝⎠
1
(1)!
Z
n
. [JNTU 2001]
Solution By definition of Z-transforms
0
()
n
nn
n
Zuuz
∞
−
=
〈〉=
∑
Here
〈〉=
+
=
+
1
(1)!
111
,,,
1!2!(1)!
n
u
n
n
(1)
∞
−−
=
−−
−−−
−+
⎛⎞
∴==+
⎜⎟
++
⎝⎠
++++
+
⎡
=+++
⎢
⎣
⎤
++
⎥
+
⎦
∑
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