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How to make the frequency vector to analyze a complex signal?

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I have this example and I'm puzzled with the frequency vector (hz) if I change any parameter in it, the result won't be the frequency components. How to calculate the frequency vector, generally?
srate = 1000;
time = -2: 1/srate : 2;
pnts = length(time);
hz = linspace(0,srate/2,(pnts-1)/2);
freq = [2 10 25 40];
amp = [3 5 7 9];
phas = [-pi/4 pi/2 pi 2*pi];
signal = zeros(size(time));
for i = 1:length(freq)
signal = signal + amp(i)*sin( 2*pi*freq(i)*time + phas(i));
end
subplot(211)
plot(time,signal,'k-','linew',0.5)
title("Time Domain")
y = fft(signal)/length(signal);
pwr = abs( y ).^2;
pwr(1) = [];
subplot(212)
plot(hz,pwr(1:length(hz)),'r-')
title("Frequency Domain")

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Muhamed Sewidan
Muhamed Sewidan on 24 Dec 2020
Thank you for your response.
I mean by changing srate/2 or (pnts-1)/2, I don't get in the end the values of the frequencies.
Pardon me, but my second question why should I use pwr = 2*abs(y) and not pwr = abs(y).^2 ?
Thank you
Michael Soskind
Michael Soskind on 24 Dec 2020
I see what you mean, you are referring to the small errors in frequency value compared to the ideal frequency values. If you increase the sample rate, you will find that the approximaton of the DFT should improve. If I increase srate to srate = 10000, I get the pairs:
freq: 2, amp: 3
freq: 10, amp: 5
freq: 25, amp: 6.99
freq: 40, amp: 9
With respect to the second question, pwr = 2*abs(y) should be used since this is really the folding of the FFT on itself in a single sided Fourier Transform. In order to get the correct signal amplitude as you had built in your composite signal, you should just double the magnitude, which will get you the correct signal amplitudes.

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Accepted Answer

Cris LaPierre
Cris LaPierre on 24 Dec 2020
Edited: Cris LaPierre on 24 Dec 2020
I think I would explain it this way. Due to Nyquist theory, the highest frequency you can detect is half your sample rate (must have at least 2 points in one period). The lowest is obviously 0. The assumption here is that the detectable frequencies increase linearly from 0 to srate/2.
If you inspect the peaks in your plot, you'll notice this approach is an approximation, as the peaks are about 1 point off.

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Muhamed Sewidan
Muhamed Sewidan on 24 Dec 2020
Yes I know but rather than using (pnts-1)/2 which equals 2000. Can I use 1000 for example rather than 2000?
Cris LaPierre
Cris LaPierre on 25 Dec 2020
Not without having to change some other things. The number of points is set by your sample rate. Changing it directly impacts the results, as you've observed. To get the frequencies to line up, you'd have to half your sample rate, which halves the frequencies you can resolve in the signal.
Better to zoom in on your axis if you want to focus on the peaks. Use xlim.
srate = 1000;
time = -2: 1/srate : 2;
pnts = length(time);
hz = linspace(0,srate/2,(pnts-1)/2);
freq = [2 10 25 40];
amp = [3 5 7 9];
phas = [-pi/4 pi/2 pi 2*pi];
signal = zeros(size(time));
for i = 1:length(freq)
signal = signal + amp(i)*sin( 2*pi*freq(i)*time + phas(i));
end
y = fft(signal)/length(signal);
pwr = abs( y ).^2;
pwr(1) = [];
plot(hz,pwr(1:length(hz)),'r-')
title("Frequency Domain")
xlim([0 250])

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