The fft function does not change the units of the time-domain argument it is given.
The fft function produces a double-sided Fourier transform, and that divides the energy in the signal equally between each half of the output, so it is necessary to mulitply the amplitude by 2 to approximate the amplitude of the time-domain signal in each half.
If ‘s’ is your signal vector and ‘t’ is your ttime vector (of eaually-spaced correspondiing smapling instants), this will produce an accurate representation of the single-sided Fourier transform of your signal:
Fs = 1/mean(diff(t)); % Sampling Frequency
Fn = Fs/2; % Nyquiat Frequency
L = size(D,1);
FTs = fft(s)/L;
Fv = linspace(0, 1, fix(L/2)+1)*Fn; % Frequency Vector
Iv = 1:numel(Fv); % Index Vector
figure
plot(Fv, abs(FTs(Iv))*2) % Multiply By ‘2’ To Correct For Double-Sided Initial ‘fft’ Output
grid
xlabel('Frequency (Hz)')
ylabel('Power (dB)')
xlim([min(Fv) max(Fv)])
if you want the output to be plotted in decibels (logarithmic power) instead:
plot(Fv, mag2db(abs(FTs(Iv))*2)) % Multiply By ‘2’ To Corredt For Double-Sided Initial ‘fft’ Output
If your signal has a significant D-C (constant) offset, it will likely be necessary to subttract the mean of the signal from the rest of the signal in order to see the other peaks:
FTs = fft(s - mean(s))/L;
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