PRACTICAL MATLAB® FOR ENGINEERS PRACTICAL MATLAB

492 Practical MATLAB® Applications for Engineers

For the signal f(t) = cos(2πt) just considered (R.5.98), the power spectrum for the
positive frequency range is illustrated as follows:

MATLAB Solution

>> p = abs(fft(fn))/4;

>> power = (p(1:4).^2)’

power =

0.0000

1.0000

0.0000

0.0000

Note that the power content is 1 at the frequency of 1.

R.5.103 Let us use the script fi le sample_cos to illustrate the process of obtaining the power-
spectrum plot (Figure 5.16) of a practical case consisting of 3000 samples taken
from a sinusoidal signal with a frequency of 25 Hz, over a time interval of 2.3 s.

MATLAB Solution

% Script file: sample _ cos

N = 3000; T = 2.3;

t = [0:N-1]/N;

t = T*t;

fn = cos(2*pi*25*t);

p = abs(fft(fn))/(N/2);

power = p(1:N/2).^2;

freq = [0:N/2-1]/T;

plot(freq(1:100),power(1:100));

xlabel(‘frequency in Hertz’), ylabel(‘power’)

title(‘Power spectrum’)

>> sample _ cos

The script fi le sample_cos is executed and the result is shown in Figure 5.16.

25

frequency in Hertz

30 35 40 45

0.45

0.4

0.35

0.3

0.25

power0.2

0.15

0.1

0.05

(^005101520)
Power spectrum
FIGURE 5.16
Power-spectrum plot of R. 5.103.