Discrete Fourier Transform and Signal Spectrum - Digital Signal Processing: Fundamentals and Applications - page 96

We can define the frequency resolution as the frequency step between two consecutive DFT
coefficients to measure how fine the frequency domain presentation is and obtain

D u ¼ u s
N

ð radians per second Þ (4.14)

or in terms of Hz, it follows that

D f ¼ f s

N ð Hz Þ

(4.15)

Let us study the following example.

EXAMPLE 4.4
In Example 4.2, given a sequence xðnÞ for 0 n 3, where xð0Þ¼1, xð1Þ¼2, xð2Þ¼3, and xð3Þ¼4, we
computed 4 DFT coefficients X ðkÞ for 0 k 3asX ð0Þ¼10, X ð1Þ¼2 þ j2, X ð2Þ¼2, and

X ð3Þ¼2 j2. If the sampling rate is 10 Hz,
a. determine the sampling period, time index, and sampling time instant for a digital sample xð3Þ in the time
domain;
b. determine the frequency resolution, frequency bin, and mapped frequencies for the DFT coefficients X ð1Þ and
X ð3Þ in the frequency domain.
Solution:
a. In the time domain, the sampling period is calculated as

T ¼ 1=f s ¼ 1=10 ¼ 0:1 second

For xð3Þ, the time index is n ¼ 3 and the sampling time instant is determined by

t ¼ nT ¼ 3$0:1 ¼ 0:3 second

b. In the frequency domain, since the total number of DFT coefficients is four, the frequency resolution is
determined by

f s

N ¼ 10

D f ¼

¼ 2:5Hz

4

The frequency bin for X ð1Þ should be k ¼ 1 and its corresponding frequency is determined by

kf s

N ¼ 1 10

f ¼ ¼ 2:5Hz

4

Similarly, for X ð3Þ and k ¼ 3,

kf s

N ¼ 3 10

f ¼ ¼ 7:5Hz

4

Note that from Equation (4.4) , k ¼ 3 is equivalent to k N ¼ 3 4 ¼1; and f ¼ 7.5 Hz is also equivalent to
the frequency f ¼ð1 10Þ=4 ¼2:5 Hz, which corresponds to the negative side spectrum. The amplitude
spectrum at 7.5 Hz after folding should match the one at f s f ¼ 10:0 7:5 ¼ 2:5 Hz. We will apply these
developed notations in the next section for amplitude and power spectral estimation.