Partial fraction expansion - Continuous and Discrete Time Signals and Systems - page 821

Solution
(i) The partial fraction expansion of Eq. (D.24) is given by
X ( s ) = s 3 + 10 s 2 + 27 s + 20
( s + 1 )( s + 2 ) 3

k 1
s + 1

k 2 , 1
s + 2

k 2 , 2
( s + 2 ) 2

+ k 2 , 3

( s + 2 ) 3 .
The partial fraction coefficient k 1 is calculated using the Heaviside formula,
Eq. (D.6), as follows:

≡ + +

s 3 + 10 s 2 + 27 s + 20
( s + 2) 3

2
1

k 1 = = = 2 .

s =− 1

The partial fraction coefficient k r , 3 is calculated using Eq. (D.22) as follows:

s 3 + 10 s 2 + 27 s + 20
s + 1

− 8 + 40 − 54 + 20
− 1

k 2 , 3 = = = 2 .

s =− 2

The remaining partial fraction coefficients are calculated using Eq. (D.22) as
follows:

s 3 + 10 s 2 + 27 s + 20
s + 1

1
(3 − 2)!

d
d s

k 2 , 2 =

s =− 2

1
( s + 1) 2

d
d s ( s 3 + 10 s 2 + 27 s + 20)

= ( s + 1)

d
d s ( s + 1)

− ( s 3 + 10 s 2 + 27 s + 20)

s =− 2

1
( s + 1) 2 [( s + 1)(3 s 2 + 20 s + 27) − ( s 3 + 10 s 2 + 27 s + 20)]

=

s =− 2

1
( s + 1) 2 [2 s 3 + 13 s 2 + 20 s + 7]

= = 3

s =− 2

and

d 2
d s 2

s 3 + 10 s 2 + 27 s + 20
s + 1

1
(3 − 1)!

k 2 , 1 =

s =− 2

1
2

d
d s

2 s 3 + 13 s 2 + 20 s + 7
( s + 1) 2

=

s =− 2

1
2

1
( s + 1) 4

d
d s (2 s 3 + 13 s 2 + 20 s + 7)

( s + 1) 2

=

d
d s ( s + 1) 2

− (2 s 3 + 13 s 2 + 20 s + 7)

s =− 2

1
2

1
( s + 1) 4

( s + 1) 2
= 1

( 6 s 2 + 2 6 s + 20 )

=

=− 8

= 1

− ( 2 s 3 + 13 s 2 + 20 s + 7 ) ( 2 s + 2 )

=− 2 = 3

s =− 2

1
2 − 8 + 6 =− 1 .

=