Partial fraction expansion - Continuous and Discrete Time Signals and Systems - page 823

the inverse CTFT of Eq. (D.30) is given by

x 1 ( t ) = ( k 1 e p 1 t + k 2 e p 2 t ++ k n e p n t ) u ( t ) . (D.32)

Similarly, the complex roots and repeated roots may be expanded in partial
fractions by following the procedure outlined for the Laplace transform.

Example D.6
Using the partial fraction method, calculate the inverse CTFT of the following
function:

2( j ω ) + 7
( j ω ) 3 + 10( j ω ) 2 + 31( j ω ) + 30 .

X ( ω ) = (D.33)

Solution
The characteristic equation of X ( ω )isgivenby

(j ω ) 3 + 10(j ω ) 2 + 31(j ω ) + 30 = 0 ,

which has roots at j ω =− 2, − 3, and − 5. The partial fraction expansion of

X ( ω ) is therefore given by

2( j ω ) + 7
( j ω + 2)( j ω + 3)( j ω + 5)

k 1
j ω + 2

k 2
j ω + 3

k 3
j ω + 5 .

X ( ω ) = ≡ + +

The partial fraction coefficients are calculated using the Heaviside formula:

2( j ω ) + 7
( j ω + 2)( j ω + 3)( j ω + 5)

k 1 = ( j ω + 2) = 1 ,

j ω=− 2

2( j ω ) + 7
( j ω + 2)( j ω + 3)( j ω + 5)

=− 1

k 2 = ( j ω + 3)

2 ,

j ω=− 3

and

2( j ω ) + 7
( j ω + 2)( j ω + 3)( j ω + 5)

=− 1 k 3 = ( j ω + 5)

2 .

j ω=− 5

Therefore, the partial fraction expansion of X ( ω )isgivenby

1
j ω + 2

1
2

1
( j ω + 3)

1
2

1
( j ω + 5) .

X ( ω ) = − − (D.34)

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

1
2 e − 3 t

1
2 e − 5 t

e − 2 t x ( t ) =

− −

u ( t ) . (D.35)