Digital Signal Processing Reference
In-Depth Information
2.4.2 TABLE LOOKUP
In this method, the
z
-transform is in a form that can simply be looked up in a table of time domain
sequences versus corresponding
z
-transforms.
Example 2.25.
Compute the impulse response that corresponds to a certain causal LTI system that has
the
z
-transform
1
X(z)
=
0
.
9
z
−
1
Since the system is causal (i.e., positive-time), the ROC will include all
z
such that
−
1
|
|
z
>
0
.
9.We
note that the
z
-transform of the sequence
a
n
u
[
n
]
where
|
a
|
<
1 is
1
1
−
az
−
1
=
a
n
u
= 0
.
9
n
u
and hence are able to determine that
h
[
n
]
[
n
]
[
n
]
.
Example 2.26.
Compute the impulse response that corresponds to a certain causal LTI system that has
the
z
-transform
z
−
1
1
+
X(z)
=
0
.
9
z
−
1
Since the system is causal (i.e., positive-time), the ROC will include all
z
such that
1
−
>
0
.
9.
We can exercise a little ingenuity by breaking this
z
-transform into the product of the numerator and
denominator, obtain the impulse response corresponding to each, and then convolve them in the time
domain. We will thus convolve the sequences 0.9
n
u
|
z
|
[
n
]
and [1, 1]. A suitable call to compute the first 30
samples is
ImpResp = conv([0.9.ˆ(0:1:29)],[1,1])
We can check this with the call
ImpRespAlt = filter([1,1],[1,-0.9],[1,zeros(1,29)])
2.4.3 PARTIAL FRACTION EXPANSION
In this method, the
z
-transform is not in a form that can readily be looked up. A
z
-transform of the
general form
b
0
+
b
1
z
−
1
+
...
+
b
M
z
−
M
X(z)
=
(2.13)
1
+
a
1
z
−
1
+
...
+
a
N
z
−
N
can be rewritten as
M
−
N
b
0
+
b
1
z
−
1
+
...
+
b
N
−
1
z
−
(N
−
1
)
C
k
z
−
k
X(z)
=
+
(2.14)
1
+
a
1
z
−
1
+
...
+
a
N
z
−
N
k
=
0
which can be rewritten as a sum of fractions, one for each pole, plus a sum of polynomials if
M
≥
N
.
N
M
−
N
R
n
C
k
z
−
k
X(z)
=
−
p
n
z
−
1
+
(2.15)
1
n
=
1
k
=
0
