Digital Signal Processing Reference
In-Depth Information
The values
R
n
are called residues,
p
n
is the
n
-th pole of
X(z)
, and it is assumed that the poles are
all distinct, i.e., no duplicate poles. In such a case, the residues can be computed by:
b
0
+
b
1
z
−
1
b
N
−
1
z
−
(N
−
1
)
+
+
p
n
z
−
1
)
z
=
p
n
...
R
n
=
−
(
1
a
1
z
−
1
a
N
z
−
N
1
+
+
...
+
The above expressions are true for distinct
p
n
.If
p
n
consists of
S
duplicate poles, the partial fraction
expansion is
S
R
n,s
z
−
(s
−
1
)
(
1
R
n,
2
z
−
1
R
n,s
z
−
(S
−
1
)
(
1
R
n,
1
=
p
n
z
−
1
+
p
n
z
−
1
)
2
+
...
+
−
p
n
z
−
1
)
s
−
−
−
p
n
z
−
1
)
S
1
(
1
s
=
1
The time domain impulse response can be written from the partial fraction expansion as
R
n
Z
−
1
N
M
−
N
1
x
[
n
]=
+
C
k
δ
[
n
−
k
]
p
n
z
−
1
1
−
n
=
1
k
=
0
For causal sequences, the inverse
z
-transform of
1
p
n
z
−
1
1
−
is
p
n
u
[
k
]
The function
residuez(b,a)
provides the vector of residues
R
, the corresponding vector of poles
p
, and, when the order of the
numerator is equal to or greater than that of the denominator, the coefficients
C
k
[
R, p, C
k
]=
of the polynomial in
z
−
1
.
Example 2.27.
Construct the partial fraction expansion of the z-transform below; use the function
residuez to obtain the values of
R
,
p
, and
C
k
. Determine the impulse response from the result.
0
.
5
z
−
1
0
.
1
z
−
2
0
.
1
+
+
=
X(z)
1
.
2
z
−
1
0
.
81
z
−
2
1
−
+
We make the call
[R,p,Ck] = residuez([0.1,0.5,0.1],[1,-1.2,0.81])
and receive results
R = [(-0.0117 -0.4726i),(-0.0117 + 0.4726i)]
p = [(0.6 + 0.6708i),(0.6 - 0.6708i)]
Ck = [0.1235]
