Partial fraction expansion - Continuous and Discrete Time Signals and Systems - page 824

Example D.7
Using the partial fraction method, calculate the inverse CTFT of the following
function:

4( j ω ) 2 + 20( j ω ) + 19
( j ω ) 3 + 5( j ω ) 2 + 8( j ω ) + 4 .

X ( ω ) =

(D.36)

Solution
The characteristic equation of X ( ω )isgivenby

( j ω ) 3 + 5( j ω ) 2 + 8( j ω ) + 4 = 0 ,

which has roots at j ω =− 1, − 2, and − 2. The partial fraction expansion of

X ( ω ) is therefore given by

4( j ω ) 2 + 20( j ω ) + 19
( j ω ) 3 + 5( j ω ) 2 + 8( j ω ) + 4

k 1
( j ω + 1)

k 2 , 1
( j ω + 2)

k 2 , 2
( j ω + 2) 2 .

X ( ω ) = ≡ + +

The partial fraction coefficients k 1 and k 2 , 2 are calculated using the Heaviside
formula:

( j ω + 1) 4( j ω ) 2 + 20( j ω ) + 19
( j ω + 1)( j ω + 2) 2

k 1 = = 3

j ω=− 1

and

4( j ω ) 2 + 20( j ω ) + 19
( j ω + 1)(j ω + 2) 2

( j ω + 2) 2 k 2 , 2 = = 5 .

j ω=− 2

The remaining partial fraction coefficient is calculated using Eq. (D.23):

4( j ω ) 2 + 20( j ω ) + 19
( j ω + 1)

1
(2 − 1)!

d
d( j ω )

k 2 , 1 = , (D.37)

j ω=− 2

where the differentiation is with respect to j ω . To simplify the notation for
differentiation, we substitute s = j ω in Eq. (D.37) to obtain:

4 s 2 + 20 s + 19
( s + 1)

1
(2 − 1)!

d
d s

k 2 , 1 =

s =− 2

( s + 1)(8 s + 20) − (4 s 2 + 20 s + 19)
( s + 1) 2

= = 1 .

s =− 2

The partial fraction expansion of X ( ω ) is therefore given by

4( j ω ) 2 + 20( j ω ) + 19
( j ω ) 3 + 5( j ω ) 2 + 8( j ω ) + 4

3
( j ω + 1)

1
( j ω + 2)

5
( j ω + 2) 2 .
(D.38)

X ( ω ) = = + +

Using Table 5.2, the inverse CTFT x ( t )of X ( ω )isgivenby

x ( t ) = [3e − t + e − 2 t + 5 t e − 2 t ] u ( t ) . (D.39)