How to find period of a signal using fft?

I have a strange signal which looks like this:

I got it in a form of a matrix (a=[value,value,value,.....])

The x-axis represents time per hour for one year (8760 hours)

As you can see, this is a periodical function with a lot of noise. I need to find the period of that function!

I've seen that using fourier transformation is the best way, but I can not find out how to use it correctly.

If anybody has an idea, i would be greatful to know it! Using fourier transformation isn't necessary.

Thank you in advance.

Answer 1

[Star Strider]

Prototype fft code:

t = ...;                                            % Time Vector
s = ...;                                            % signal Vector

L = numel(t);                                       % Signal Length
Ts = mean(diff(t));                                 % Sampling Interval
Fs = 1/Ts;                                          % Sampling Frequency
Fn = Fs/2;                                          % Nyquist Frequency
FTs = fft(s)/L;                                     % Normalised Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                 % Frequency Vector
Iv = 1:numel(Fv);                                   % Index Vector

figure
plot(Fv, abs(FTs(Iv))*2)
grid
title('Fourier Transform')
xlabel('Frequency (units)')
ylabel('Amplitude (units)')

To calculate the period (the inverse of frequency) of a signal:

[pks,locs] = findpeaks(abs(FTs(Iv)));
Perd = 1./Fv(locs);

I did not test this, however that should work. The units are the inverse of the frequency units, so if the frequency is in Hz, the period will be in seconds/cycle.

EDIT — (12 Jan 2021 at 18:40)

Corrected typographical error (replaced ‘-’ in findpeaks call line with ‘=’).

.

[Chatzistylianos Evangelos]

Actually, I was waiting for a number, or something like that. For example, the "main" period of the signal is 2 days. The results i get, though, is:

The signal is the capacity of a storage technology. So, I want to figure out the period of the cycles of that technology. For example, my cell phone's battery would have a period of 1 day (1 day is need for the battery to be fully discharged and then it is charged again and so on). Did I make it more understandable now?

However, your previous answer was really helpful on understanding FFT.

[Star Strider]

As always, my pleasure!

One other refinement on my code is:

sm = s - mean(s);                                   % Subtract Constant Offset
FTs = fft(sm)/L;                                    % Normalised Fourier Transform

Subtracting the constant offset first makes any other peaks much more visible.

[Qifa]

I couldn't run this line due to unknown variable of pks. I wanted to find period of the signal.

[pks,locs] - findpeaks(abs(FTs(Iv)));

[Sam Pasti]

Code returns period as Perd as a vector, what then is the actually period? sorry if this is obvious. Thanks