Databases Reference
In-Depth Information
Using the approach described above, we obtain
z
=
0
.
5
(
6
z
−
9
)(
z
−
0
.
5
)
A
1
=
(90)
(
z
−
0
.
5
)(
z
−
2
)
=
4
(91)
z
=
2
(
−
)(
−
)
6
z
9
z
2
A
2
=
(92)
(
z
−
0
.
5
)(
z
−
2
)
=
2
(93)
Therefore,
4
z
2
z
F
(
z
)
=
5
+
z
−
0
.
z
−
2
and
n
n
f
n
=[
4
(
0
.
5
)
+
2
(
2
)
]
u
[
n
]
The procedure becomes slightly more complicated when we have repeated roots of
D
(
z
)
.
Suppose we have a function
(
)
N
z
F
(
z
)
=
(
−
z
1
)(
−
z
2
)
2
z
z
The partial fraction expansion of this function is
F
(
z
)
A
1
A
2
A
3
=
z
1
+
z
2
+
2
z
z
−
z
−
(
z
−
z
2
)
The values of
A
1
and
A
3
can be found as shown previously:
z
=
z
1
F
(
z
)(
z
−
z
1
)
A
1
=
(94)
z
z
=
z
2
2
F
(
z
)(
z
−
z
2
)
A
3
=
(95)
z
However, we run into problems when we try to evaluate
A
2
. Let's see what happens when
we multiply both sides by
(
z
−
z
2
)
:
F
(
z
)(
z
−
z
2
)
A
1
(
z
−
z
2
)
A
3
=
+
A
2
+
(96)
z
z
−
z
1
z
−
z
2
If we now evaluate this equation at
z
z
2
, the third term on the right-hand side becomes
undefined. In order to avoid this problem, we first multiply both sides by
=
2
and take
(
z
−
z
2
)
the derivative with respect to
z
prior to evaluating the equation at
z
=
z
2
:
2
2
F
(
z
)(
z
−
z
2
)
A
1
(
z
−
z
2
)
=
+
A
2
(
z
−
z
2
)
+
A
3
(97)
z
z
−
z
1
