Discrete-time Fourier series and transform - Continuous and Discrete Time Signals and Systems - page 485

cosine waveform, yet its DTFT can be expressed mathematically. A similar
behavior is exhibited by most periodic sequences. So how do we determine
the DTFT for a periodic sequence? We cannot use the definition of the DTFT,
Eq. (11.28b), since the procedure will lead to infinite DTFT values. In such
cases, an alternative procedure based on the DTFS is used; this is explained in
Section 11.4.

11.4 DTFT of perio dic functions

Consider a periodic function x [ k ] with fundamental period K 0 . The DTFS
representation for x [ k ]isgivenby

n = K 0 D n e j n Ω 0 k ,

x [ k ] = (11.2831)

where Ω 0 = 2 π/ K 0 and the DTFS coefficients are given by

k = K 0 x [ k ]e

1
K 0

− j n Ω 0 k .

D n = (11.2832)

Calculating the DTFT of both sides of Eq. (11.2831), we obtain

n = K 0 D n e j n Ω 0 k

X ( Ω ) =ℑ .

Since the DTFT satisfies the linearity property, the above equation can be
expressed as follows:

n = K 0 D n ℑ e j n Ω 0 k ,

X ( Ω ) = (11.2833)

where the DTFT of the complex exponential sequence is given by

m =−∞ δ ( Ω

∞

ℑ e j n Ω 0 k = 2 π

− n Ω 0 − 2 π m ) .

Using the above value for the DTFT of the complex exponential, Eq. (11.2833)
takes the following form:

n = K 0 D n 2 π m =−∞ δ ( Ω

∞

X ( Ω ) = − n Ω 0 − 2 π m ) .

By changing the order of summation in the above equation and substituting
Ω 0

= 2 π/ K 0 ,wehave

∞

n = K 0 D n δ

− 2 n π
K 0

X ( Ω ) = 2 π − 2 π m .

Ω

m =−∞

Since the DTFT is periodic with a period of 2 π , we determine the DTFT in
the range Ω

= [0 , 2 π ] and use the periodicity property to determine the DTFT
values outside the specified range. Taking n = 0 , 1 , 2 ,..., K 0 − 1 and m = 0,