Laplace transform - Continuous and Discrete Time Signals and Systems - page 275

Solution
(i) In X 1 ( s ), the characteristic equation is already factorized. In terms of the
partial fractions, X 1 ( s ) can be expressed as follows:

s + 3
s ( s + 1)( s + 2)

≡ k 1
s

k 2
( s + 1)

k 3
( s + 2) ,

X 1 ( s ) = + +

where the partial fraction coefficients k 1 , k 2 , and k 3 are given by

( s + 3)
s ( s + 1)( s + 2)

( s + 3)
( s + 1)( s + 2)

= 3 k 1 = s =

2 ,

s = 0 s = 0

( s + 3)
s ( s + 1)( s + 2)

( s + 3)
s ( s + 2)

k 2 = ( s + 1) = =− 2 ,

s =− 1 s =− 1

and

( s + 3)
s ( s + 1)( s + 2)

( s + 3)
s ( s + 1)

= 1 k 3 = ( s + 2) =

2 .

s =− 2 s =− 2

The partial fraction expansion of the Laplace transform X 1 ( s )isgivenby

s + 3
s ( s + 1)( s + 2)

≡ 3
2 s

2
( s + 1)

1
2( s + 2) ,

X 1 ( s ) = − +

which leads to the following inverse Laplace transform:

3
2

− t + 1

− 2 t

x 1 ( t ) = − 2e

2 e

u ( t ) .

(ii) The characteristic equation of X 2 ( s )isgivenby

D ( s ) = s 3 + 5 s 2 + 17 s + 13 = 0 ,

which has three roots at s =− 1 , − 2 , and j3. The partial fraction expansion

of X 2 ( s )isgivenby

s + 5
( s + 1)( s + 2 + j3)( s + 2 − j3)

k 1
( s + 1)

k 2 s + k 3
( s 2 + 4 s + 13) .

X 2 ( s ) = ≡ +

The partial fraction coefficient k 1 is calculated to be

( s + 5)
( s + 1)( s 2 + 4 s + 13)

( s + 5)
( s 2 + 4 s + 13)

= 2 k 1 = ( s + 1) =

5 .

s =− 1 s =− 1

To compute coefficients k 2 and k 3 , we substitute k 1 = 2 / 5in X 2 ( s ) and expand

s + 5
( s + 1)( s 2 + 4 s + 13)

2
5( s + 1)

k 2 s + k 3
( s 2 + 4 s + 13)

≡ +

as

s + 5 ≡ 0 . 4( s 2 + 4 s + 13) + ( k 2 s + k 3 )( s + 1) .

Comparing the coefficients of s 2

on both sides of the above expression yields

k 2 + 0 . 4 = 0 ⇒ k 2 =− 0 . 4 .