Databases Reference
In-Depth Information
Taking the derivative of both sides with respect to
z
, we get
2
2
d
dz
F
(
z
)(
z
−
z
2
)
2
A
1
(
z
−
z
2
)(
z
−
z
1
)
−
A
1
(
z
−
z
2
)
=
+
A
2
(98)
z
(
z
−
z
1
)
2
If we now evaluate the expression at
z
=
z
2
, we get
z
=
z
2
2
d
dz
F
(
z
)(
z
−
z
2
)
A
2
=
(99)
z
has a root of order
m
at some
z
k
, that
portion of the partial fraction expansion can be written as
Generalizing this approach, we can show that if
D
(
z
)
F
(
z
)
A
1
A
2
A
m
=
z
k
+
2
+···+
(100)
m
z
z
−
(
z
−
z
k
)
(
z
−
z
k
)
and the
l
th coefficient can be obtained as
z
=
z
k
d
(
m
−
l
)
dz
(
m
−
l
)
m
(
)(
−
z
k
)
1
F
z
z
A
l
=
(101)
(
−
)
!
m
l
z
Finally, let us drop the requirement that the degree of
D
(
z
)
be greater or equal to the degree
of
N
(
z
)
. When the degree of
N
(
z
)
is greater than the degree of
D
(
z
)
, we can simply divide
N
(
z
)
by
D
(
z
)
to obtain
N
(
z
)
R
(
z
)
F
(
z
)
=
=
Q
(
z
)
+
(102)
D
(
z
)
D
(
z
)
where
Q
(
z
)
is the quotient and
R
(
z
)
is the remainder of the division operation. Clearly,
R
(
z
)
will have degree less than
D
.
To see how all this works together, consider the following example.
(
z
)
Example12.9.5:
Let us find the inverse Z-transform of the function
2
z
4
+
1
F
(
z
)
=
(103)
2
z
3
5
z
2
−
+
4
z
−
1
The degree of the numerator is greater than the degree of the denominator, so we divide once
to obtain
5
z
3
4
z
2
−
+
z
+
1
F
(
z
)
=
z
+
(104)
2
z
3
−
5
z
2
+
4
z
−
1
The inverse Z-transform of
z
is
δ
n
−
1
, where
δ
n
is the discrete delta function defined as
1
n
0
0 otherwise
=
δ
n
=
(105)
