Partial fraction expansion - Continuous and Discrete Time Signals and Systems - page 825

D.3 Discrete-time Fourier transform

To illustrate the partial fraction expansion of the DTFT, consider the following
rational function:

= b m e j m Ω + b m − 1 e j( m − 1) Ω ++ b 1 e j Ω + b 0
a n e j n Ω + a n − 1 e j( n − 1) Ω ++ a 1 e j Ω + a 0

X ( Ω ) = N ( Ω )
D ( Ω )

, (D.40)

where the numerator N ( Ω ) is a polynomial of degree m and the denominator
D ( Ω ) is a polynomial of degree n . An alternative representation for Eq. (D.40)
is obtained by dividing both the numerator and the denominator by e j n Ω as
follows:

− j Ω ++ b 1 e − j( m − 1) Ω + b 0 e − j m Ω

X ( Ω ) = N ( Ω )
D ( Ω )

= e j( m − n ) Ω b m + b m − 1 e .

a n + a n − 1 e − j Ω ++ a 1 e − j( n − 1) Ω + a 0 e − j n Ω

X ′ ( ω )

(D.41)
We need to express Eq. (D.41) in simpler terms using the partial fraction expan-
sion with respect to e − j Ω . To simplify the factorization process, we substitute
z = e j Ω :

X ( z ) = z ( m − n ) b m + b m − 1 z − 1 ++ b 1 z − ( m − 1) + b 0 z − m
a n + a n − 1 z − 1 ++ a 1 z − ( n − 1) + a 0 z − n

. (D.42)

The process for the partial fraction expansion of Eq. (D.41) is the same as for
the CTFT and Laplace transform, except that the expansion is performed with
respect to z − 1 . Below we illustrate the process with an example.

Example D.8
Using the partial fraction method, calculate the inverse CTFT of the following
function:

2e j2 Ω − 5e j Ω
e j2 Ω − (4 / 9)e j Ω + (1 / 27) .

N ( Ω )
D ( Ω )

X ( Ω ) = = (D.43)

Solution
Dividing both the numerator and the denominator of Eq. (D.43) by e j2 Ω yields

2 − 5e − j Ω
1 − (4 / 9)e − j Ω + (1 / 27)e − 2j Ω .

X ( Ω ) =

Substitute z = e j Ω in the above equation to obtain

2 − 5 z − 1
1 − (4 / 9) z − 1 + (1 / 27) z − 2 ,
with the characteristic equation

X ( z ) =

4
9 z − 1 +

1
27 z − 2

4
9 z +

1
27

= 0or z 2 −

1 − = 0 ,