Partial fraction expansion - Continuous and Discrete Time Signals and Systems - page 816

Solution
(i) The characteristic equation of X ( s )isgivenby

s 3 + 3 s 2 − 6 s − 8 = 0 ,

which has roots at s =− 1, 2, and − 4. The partial fraction expansion of X ( s )

is therefore given by

4 s 2 + 20 s − 2
s 3 + 3 s 2 − 6 s − 8

k 1
s + 1

k 2
s − 2

k 3
s + 4 .

X ( s ) = ≡ + +

Using the Heaviside formula, the residues k r are given by

4 s 2 + 20 s − 2
( s − 2)( s + 4)

4 − 20 − 2
− 9

k 1 = = = 2 ,

s =− 1

4 s 2 + 20 s − 2
( s + 1)( s + 4)

16 + 40 − 2
3 6

k 2 = = = 3 ,

s = 2

and

4 s 2 + 20 s − 2
( s + 1)( s − 2)

64 − 80 − 2
( − 3) ( − 6)

− 18
18

k 3 = = = =− 1 .

s =− 4

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

2
s + 1

3
s − 2

1
s + 4 .

X ( s ) = + − (D.8)

(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:

2e − t + 3e 2 t − e − 4 t

x ( t ) = u ( t ) . (D.9)

Example D.2
For the function

6 s 2 + 11 s + 26
s 3 + 4 s 2 + 13 s ,

X ( s ) =

(D.10)

(i) calculate the partial fraction expansion;
(ii) based on your answer to (i), calculate the inverse Laplace transform of
X ( s ).

Solution
(i) The characteristic equation of X ( s )isgivenby

s 3 + 4 s 2 + 13 s = 0 ,