Databases Reference
In-Depth Information
Let us call the remaining ratio of polynomials
F
1
(
z
)
. We find the roots of the denominator of
F
1
(
z
)
as
5
z
3
4
z
2
−
+
z
+
1
F
1
(
z
)
=
(106)
2
2
(
z
−
0
.
5
)(
z
−
1
)
Then
5
z
3
4
z
2
F
1
(
z
)
−
+
z
+
1
=
(107)
2
z
2
z
(
z
−
0
.
5
)(
z
−
1
)
A
1
z
+
A
2
A
3
A
4
=
5
+
1
+
(108)
2
z
−
0
.
z
−
(
z
−
1
)
Thus
z
=
0
=−
5
z
3
4
z
2
−
+
+
z
1
A
1
=
(109)
1
2
2
(
z
−
0
.
5
)(
z
−
1
)
z
=
0
.
5
=
5
z
3
4
z
2
−
+
z
+
1
A
2
=
4
.
5
(110)
2
2
z
(
z
−
1
)
z
=
1
=
5
z
3
4
z
2
−
+
z
+
1
A
4
=
3
(111)
2
z
(
z
−
0
.
5
)
To find
A
3
, we take the derivative with respect to
z
, then set
z
=
1:
z
=
1
=−
5
z
3
4
z
2
d
dz
−
+
2
z
+
1
A
3
=
3
(112)
2
z
(
z
−
0
.
5
)
Therefore,
.
4
5
z
3
z
3
z
F
1
(
)
=−
+
5
−
1
+
(113)
z
1
−
.
−
(
−
)
2
z
0
z
z
1
and
n
u
f
1
,
n
=−
δ
n
+
4
.
5
(
0
.
5
)
[
n
]−
3
u
[
n
]+
3
nu
[
n
]
(114)
and
n
u
f
n
=
δ
n
−
1
−
δ
n
+
4
.
5
(
0
.
5
)
[
n
]−
(
3
−
3
n
)
u
[
n
]
(115)
12.9.3 Long Division
If we could write
F
as a power series, then from the Z-transform expression the coefficients
of
z
−
n
would be the sequence values
f
n
.
(
z
)
Example12.9.6:
Let's find the inverse
z
-transform of
z
(
)
=
F
z
z
−
a
