Digital Signal Processing Reference
In-Depth Information
Partial-fraction expansions are used in finding the inverse transform of the ra-
tional function on the right side of (F.1). We will present a procedure for expanding
a rational function as partial fractions. For the general case,
Á
b
m
s
m
+
+ b
1
s + b
0
N(s)
D(s)
,
F(s) =
=
m 6 n,
(F.2)
+
Á
+ a
1
s + a
0
s
n
+ a
n- 1
s
n- 1
where is the numerator polynomial and is the denominator polynomial.
To perform a partial-fraction expansion, first we must find the roots of the denomi-
nator polynomial,
N(s)
D(s)
D(s).
Then we can express the denominator polynomial as
n
D(s) = (s - p
1
)
(s - p
2
)
Á
(s - p
n
) =
q
(s - p
i
),
(F.3)
i = 1
where
ß
indicates the product of terms and the
p
i
are called the
poles
of
F(s)
[the
values of
s
for which is unbounded].
We first consider the case that the roots of
F(s)
D(s)
are distinct (i.e., there are no
repeated roots). The function
F(s)
in (F.2) can then be expressed as
N(s)
D(s)
=
N(s)
k
1
s - p
1
k
2
s - p
2
k
n
s - p
n
.
Á
F(s) =
=
+
+
+
(F.4)
n
(s - p
i
)
q
i = 1
This partial-fraction expansion is completed by calculating the constants
To calculate
k
1
, k
2
, Á , k
n
.
k
j
, 1 F j F n,
first we multiply
F(s)
by the term
(s - p
j
):
k
1
(s - p
j
)
s - p
1
k
n
(s - p
j
)
s - p
n
Á
Á
(s - p
j
)F(s) =
+
+ k
j
+
+
.
(F.5)
If we evaluate this equation at
s = p
j
,
all terms on the right are zero except the term
k
j
.
Therefore,
k
j
= (s - p
j
)F(s) ƒ
s =p
j
,
j = 1, 2, Á , n,
(F.6)
which is the desired result.
Next, we consider the case that the denominator has repeated roots. For ex-
ample, suppose that the rational function is given by
N(s)
(s - p
1
)
(s - p
2
)
r
F(s) =
(F.7)
k
1
s - p
1
k
21
s - p
2
k
22
(s - p
2
)
2
+
k
2r
(s - p
2
)
r
.
Á
=
+
+
+
