Databases Reference
In-Depth Information
T A B L E 12 . 1
Some Z-transform pairs.
{
f
n
}
F
(
z
)
z
a
n
u
[
n
]
z
−
a
Tz
−
1
nTu
[
n
]
z
−
1
2
(
1
−
)
z
−
1
(
sin
α
nT
)
sin
(α
nT
)
1
−
2cos
(α
T
)
z
−
1
+
z
−
2
z
−
1
(
cos
α
nT
)
cos
(α
nT
)
z
−
1
z
−
2
1
−
2cos
(α
T
)
+
δ
n
1
12.9.1 Tabular Method
The inverse Z-transform has been tabulated for a number of interesting cases (see Table.
12.1
).
If we can write
F
(
z
)
as a sum of these functions
F
(
z
)
=
α
i
F
i
(
z
)
then the inverse Z-transform is given by
f
n
=
α
i
f
i
,
n
where
F
i
(
z
)
=
Z
[{
f
i
,
n
}]
.
Example12.9.3:
z
2
z
F
(
z
)
=
5
+
3
From our earlier example we know the inverse Z-transform of
z
z
−
0
.
z
−
0
.
/(
−
)
z
a
. Using that, the inverse
(
)
Z-transform of
F
z
is
5
n
u
n
u
f
n
=
0
.
[
n
]+
2
(
0
.
3
)
[
n
]
12.9.2 Partial Fraction Expansion
In order to use the tabular method, we need to be able to decompose the function of interest to
us as a sum of simpler terms. The partial fraction expansion approach does exactly that when
the function is a ratio of polynomials in
z
.
Suppose
F
(
)
(
)
(
)
z
can be written as a ratio of polynomials
N
z
and
D
z
. For the moment let
us assume that the degree of
D
(
z
)
is greater than the degree of
N
(
z
)
, and that all the roots of
D
(
z
)
are distinct (distinct roots are referred to as simple roots); that is,
N
(
z
)
F
(
z
)
=
(83)
(
z
−
z
1
)(
z
−
z
2
)
···
(
z
−
z
L
)
