Digital Signal Processing Reference
In-Depth Information
x
2
[
k
]
x
1
[
k
] =(2
k
− 1)
u
[
k
]
63
1
31
x
3
[
k
]
15
346
219
7
≈
≈
≈
0.4
0.2
0.11
76
3
≈
≈
≈
1
6
56
k
k
k
−2
−1
0
1
3
−2
−1
0
1
3
−2
−1
0
1
3
≈
≈
−7297
(a)
(b)
(c)
−49 239
Fig. 13.3. DT sequences
obtained in Example 13.4.
13.3.4 Power series method
When
X
(
z
) is a rational function of the form in Eq. (13.9), the partial fraction
expansion is a convenient method of calculating the inverse z-transform. At
times, however, it may be difficult to expand
X
(
z
) as partial fractions, especially
when
X
(
z
) is not a rational function. In such cases, we use the power series
method. Alternatively, we may be interested in determining a few values of
x
[
k
]
for
k
≥
0. The power series method is easy to apply since it does not require
the evaluation of the complete inverse z-transform.
In the power series method, the transform
X
(
z
) is expanded by long division
as follows:
N
(
z
)
D
(
z
)
=
a
+
bz
−
1
+
cz
−
2
+
dz
−
3
+ .
X
(
z
)
=
(13.15a)
Taking the inverse z-transform of Eq. (13.15), we obtain
x
[
k
]
=
a
δ
[
k
]
+
b
δ
[
k
−
1]
+
c
δ
[
k
−
2]
+
d
δ
[
k
−
3]
+ ,
(13.15b)
which implies that
x
[0]
=
a
,
x
[1]
=
b
,
x
[2]
=
c
, and
x
[3]
=
d
. Additional
samples of
x
[
k
] can be obtained by determining additional terms in the quotient
of Eq. (13.15a). We now illustrate the application of the power series method
with an example.
Example 13.5
Calculate the first four non-zero values of the following right-sided sequences
using the power series approach:
z
z
2
−
3
z
+
2
;
X
1
(
z
)
=
(i)
1
(
z
−
0
.
1)(
z
−
0
.
5)(
z
+
0
.
2)
;
(ii)
X
2
(
z
)
=
2
z
(3
z
+
17)
(
z
−
1)(
z
2
−
6
z
+
25)
.
(iii)
X
3
(
z
)
=
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