Discrete Fourier Transform and Signal Spectrum - Digital Signal Processing: Fundamentals and Applications - page 90

We note the following points:

a. As displayed in Figure 4.3 , only the line spectral portion between the frequency f s = 2 and
frequency f s = 2 (folding frequency) represents frequency information of the periodic signal.
b. Notice that the spectral portion from f s = 2to f s is a copy of the spectrum in the negative frequency
range from f s = 2 to 0 Hz due to the spectrum being periodic for every Nf 0 Hz. Again, the amplitude
spectral components indexed from f s = 2to f s can be folded at the folding frequency f s = 2 to match the
amplitude spectral components indexed from 0 to f s = 2 in terms of f s f Hz, where f is in the range
from f s = 2to f s . For convenience, we compute the spectrum over the range from 0 to f s Hz with
nonnegative indices, that is,

N 1
n¼ 0 xðnÞe j

1
N

2 pkn
N

c k ¼ ; k ¼ 0 ; 1 ; / ; N 1 (4.5)

We can apply Equation (4.4) to find the negative indexed spectral values if they are required.

c. For the k th harmonic, the frequency is

f ¼ kf 0 Hz (4.6)

The frequency spacing between the consecutive spectral lines, called the frequency resolution, is f 0 Hz.

EXAMPLE 4.1
The periodic signal

xðtÞ¼sinð2ptÞ

is sampled using the sampling rate f s ¼ 4 Hz.
a. Compute the spectrum c k using the samples in one period.
b. Plot the two-sided amplitude spectrum jc k j over the range from 2 to 2 Hz.
Solution:
a. From the analog signal, we can determine the fundamental frequency u 0 ¼ 2p radians per second and
f 0 ¼ u 0

2p ¼ 2p

2p ¼ 1 Hz, and the fundamental period T 0 ¼ 1 second.
Since using the sampling interval T ¼ 1=f s ¼ 0:25 second, we get the sampled signal as
xðnÞ¼xðnTÞ¼sinð2pnTÞ¼sinð0:5pnÞ

and plot the first eight samples as shown in Figure 4.4 .

x ()

x (1

1

x () 2

x (0

n

0

x (3

N 4

FIGURE 4.4

Periodic digital signal.