Partial fraction expansion - Continuous and Discrete Time Signals and Systems - page 818

Example D.3
Repeat Example D.2 by expressing the complex-valued poles as a quadratic
term.

Solution
(i) Combining the complex-valued terms in Eq. (D.11),

X ( s ) = 6 s 2 + 11 s + 26
s 3 + 4 s 2 + 13 s

= k 1
s

k 2
s + 2 + j3

k 3
s + 2 − j3 ,

+ +

= k 1
s

+ ( k 2

+ k 3 ) s + 2 ( k 2 + k 3 )

( s + 2) 2 − ( j3) 2

.

Since k 2 and k 3 are constants, their linear combinations can be replaced with
other constants. Substituting k 2

+ k 3 = A 1 and k 2 + k 3 = A 2 , we obtain

X ( s ) = 6 s 2 + 11 s + 26
s 3 + 4 s 2 + 13 s

k 1
s

A 1 s + A 2
s 2 + 4 s + 13 .

≡ + (D.14)

It may be noted that the above expression could have been obtained directly by
factorizing the denominator,

s 3 + 4 s 2 + 13 s = s ( s 2 + 4 s + 13) ,

and writing the partial fraction expansion of X ( s ) in terms of two terms, one
with a linear polynomial s in the denominator and the other with a quadratic
polynomial ( s 2 + 4 s + 13).
The partial fraction coefficient k 1 of the linear polynomial denominator is
obtained using the Heaviside formula as follows:

s 6 s 2 + 11 s + 26
s ( s 2 + 4 s + 13)

k 1 = = 2 .

s = 0

In order to calculate the remaining coefficients A 1 and A 2 , we substitute k 1 = 2
in Eq. (D.14). Cross-multiplying and equating the numerators in Eq. (D.5), we
obtain

6 s 2 + 11 s + 26 = 2( s 2 + 4 s + 13) + ( A 1 s + A 2 ) s

or

( A 1 + 2) s 2 + ( A 2 + 8) s + 26 = 6 s 2 + 11 s + 26 .

Equating the coefficients of the polynomials of the same degree on both sides
of the above equation, we obtain:

coefficient of s 2 ( A 1 + 2) = 6 , A 1 = 4;

coefficient of s ( A 2 + 8) = 11 , A 2 = 3 .