Partial fraction expansion - Continuous and Discrete Time Signals and Systems - page 817

which has roots at s = 0, − 2 + j3, and − 2 − j3. The partial fraction expansion
of X ( s ) is therefore given by

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 .

≡ + + (D.11)

Note that in this case, there are two complex-conjugate poles at s =− 2 j3.
Using the Heaviside formula, the residues k r are given by

6 s 2 + 11 s + 26
s ( s + 2 + j3)( s + 2 − j3)

k 1 = s = 2 ,

s = 0

6 s 2 + 11 s + 26
s ( s + 2 + j3)( s + 2 − j3)

= 2 − j 5 k 2 = ( s + 2 + j3)

6 ,

s =− 2 − j3

and

6 s 2 + 11 s + 26
s ( s + 2 + j3)( s + 2 − j3)

= 2 + j 5 k 3 = ( s + 2 + j3)

6 .

s =− 2 + j3

Substituting the values of the partial fraction coefficients k 1 , k 2 , and k 3 ,we
obtain

2 − j 6
s + 2 + j3

2 + j 6

2
s

X ( s ) = + +

s + 2 − j3 .

(D.12)

(ii) Assuming the function x ( t ) to be causal or right-sided, we use Table 6.1
to determine the inverse Laplace transform x ( t )ofthe X (s) as follows:

2 − j 5
6

2 + j 5
6

e − (2 + j3) t e − (2 − j3) t x ( t ) = 2 + + u ( t )

2 − j 5
6

2 + j 5
6

2 + e − 2 t e − j3 t e j3 t = + u ( t )

j5
6

2 + e − 2 t e j3 t

+ e − j3 t

e j3 t

− e − j3 t = +

2 u ( t )

5
3 sin(3 t )

2 + e − 2 t

= 4 cos(3 t ) − u ( t )

5
3 e − 2 t sin(3 t )

2 + 4e − 2 t cos(3 t ) −

= u ( t ) . (D.13)

In Example D.2, the complex-valued poles of the Laplace transform X ( s ) occur
in conjugate pairs. This is true, in general, for any polynomial with real-valued
coefficients. Although the Heaviside formula may be used to determine the
values of the partial fraction residues corresponding to the complex poles, the
procedure is often complicated due to complex algebra. Below, we present
another procedure, which expresses such complex-valued and conjugate poles
in terms of a quadratic term in the partial fraction expansion.