Discrete Fourier transform - Continuous and Discrete Time Signals and Systems - page 545

Table 12.1. Comparison between the DFT and DTFT coefficients in Example 12.9

DTFT frequency,
DFT index, r Ω r

= 2 π r / N DFT coefficients, X [ r ] DTFT coefficients, X ( Ω )

− 5 −π

0.6212 0.6250

− 4 − 0 . 8 π

0 . 6334 + j0.1504 0 . 6373 + j0.1513

− 3 − 0 . 6 π 0 . 6807 + j0.3277 0 . 6849 + j0.3297

− 2 − 0 . 4 π 0 . 8185 + 0 . 5734 0 . 8235 + j0 . 5769

− 1 − 0 . 2 π 1 . 3142 + j0.9007 1 . 3222 + j0 . 9062

0 0 2.4848 2.5000

1 0.2 π 1 . 3142 − j0.9007 1 . 3222 − j0 . 9062

2 0.4 π 0 . 8185 − j0.5734 0 . 8235 − j0 . 5769

3 0.6 π 0 . 6807 − j0.3277 0 . 6849 − j0 . 3297

4 0.8 π 0 . 6334 − j0.1504 0 . 6373 − j0 . 1513

Step 3: DFT computation The M ATLAB code for computing the DFT is as
follows:

>> N = 10; k = 0:N-1; % set sequence length
%toN=10
>> x = 0.6.ˆk; % compute the DT sequence
>> X = fft(x); % calculate the 10-point DFT
>> X = fftshift(X); % shift the DFT coefficients
>> w = -pi:2*pi/N:pi-2*pi/N; % compute DTFT frequencies

Table 12.1 compares the computed DFT coefficients with the corresponding
DTFT coefficients obtained from the following DTFT pair:

←−−→ 1

DTFT

0 . 6 k u [ k ]

1 − 0 . 6e − j Ω .
We observe that the values of the DFT coefficients are fairly close to the DTFT
values.

Example 12.10
Calculate the DTFT of the aperiodic sequence x [ k ] = [2, 1, 0, 1] for 0 ≤ k ≤
3.

Solution
Using Eq. (12.6), the DFT coefficients are given by

X [ r ] = [4 , 2 , 0 , 2] for 0 ≤ r ≤ 3 .

Mapping in the DTFT domain, the corresponding DTFT coefficients are given
by

X ( Ω r ) = [4 , 2 , 0 , 2] for

Ω r = [0 , 0 . 5 π, π, 1 . 5 π ] radians / s .