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Partial Fractions

where θ = arctan(b /a ) and A = a / cos θ . When the proper fraction contains
k
k
k
k
k
k
a multiple pole of order r, the coefficients in the partial-fraction expansion A ,
p1
A , . . . , A that are involved in the terms
pr
p2
A p1 A p2 A pr
+ + · · · +
(s − s ) (s − s ) 2 (s − s ) r
p
p
p
must be evaluated. A simple application of (2.15) is not adequate. Now the
r
procedure is to multiply both sides of (2.14) by (s − s ) , which gives
p
[ ]
A 1 A 2 A k
r
(s − s ) F(s) = (s − s ) r + + · · · + +
p
p
s − s 1 s − s 2 s − s k
+A (s − s ) r−1 + · · · + A p(r−1) (s − s ) + A . (2.18)
p
p1
p
pr
In the limit as s = s all terms on the right vanish with the exception of A .
pr
p
Suppose now that this equation is differentiated once with respect to s. The
constant A will vanish in the differentiation but A p(r−1) will be determined by
pr
setting s = s . This procedure will be continued to find each of the coefficients
p
A . Specifically, the procedure is specified by
pk
{ }
1 d r−k
A pk = (r − k)! ds r−k F(s)(s − s ) r , k = 1, 2, . . . , r. (2.19)
p
. s=s p
Example 2.34, Find the inverse transform of the following
2
3
function: F(s) = s +2s +3s+1 .
2
s (s+1)
. . . . .
.
Solution. This is not a proper fraction. The numerator polynomial is
divided by the denominator polynomial by simple long division. The result
is
2
s + 3s + 1
F(s) = 1 + .
2
s (s + 1)
The proper fraction is expanded into partial fraction forms

2
s + 3s + 1 A 11 A 12 A 2
F (s) = = + + .
p
2
s (s + 1) s s 2 s + 1
The value of A is deduced using (2.15)
2
2
s + 3s + 1

A = [(s + 1)F (s)] s=−1 = = −1.
2
p
s 2
s=−1
To find A 11 and A 12 we proceed as specified in (4.11)

2
s + 3s + 1
2
A 12 = [s F (s)] s=0 = = 1
p
. . . . s + 1 s=0
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