obtain the fourier trasnform doing the fft. Do I need to normalize??

Hi,

I have a signal in time domain, actually, specific kinetic energy was calculated in time domain and its medium value is 10 m2/s2 (with some peak at 60 m2/s2). I wanted to obtain the Fourier transform of this signal in time domain. Then I did an FFT on the signal and converted to frequency domain; but the amplitude of this kinetic energy in frequency domain is at the magnitude order of 10e6. Why this difference? Should I normalize this FFT in some way to obtain the Fourier trasform? If I do the fft the unit lenght is m2/s2, while if I do the Fourier trasform the unit lenght should be m2/s, so I thought that my problem could be due to a normalization

This is the code (the tket.txt contain the signal, which can be read at the 34th column) can you help me?

tket1=readtable('tket.txt');
tket=table2array(tket1);
Tinc=0.001;
% Tinc is the time step
Fs=1/Tinc;
L=length(tket);
f = Fs*(0:(L/2))/L;
tkef1= fft(tket(:,34:34),L);
fl=length(f);
tkef=tkef1(1:fl,:);

[dpb]

Yes, divide by L.

NB: however, the time domain max values won't match the values in the frequency domain for anything excepting a pure sine wave. Also note the FFT itself is a complex quantity; and is also double-sided in frequency. See the example at

doc fft

on computing PSD from the FFT--it shows how to normalized a one-sided PSD to match the input magnitude. Again note that the match will only be exact the value of a single frequency bin if the input signal frequency EXACTLY matches a frequency bin in the FFT and there is no noise. Otherwise, energy will be "smeared" around the nearest bin and the total will be the integral over the peak.

[SYML2nd]

Ok. Thank you very much for your detailed answer. I tried to solve the fact that is a complex quantity by doing

y=abs(tkef)

Do you think it is a proper way to do that??

[dpb]

As Matt J says, "it all depends" on what you're actually trying to compute.

Answer 1

[Matt J]

To approximate a continuous Fourier transform integral, you need to multiply by the time step,

tkef1= fft(tket(:,34:34),L) * Tinc;

Depending on whether you are an engineer or a physicist, you may also need to multiply by 1/sqrt(2*pi), as different professions define the Fourier Transform scaling differently.

[Matt J]

Francesco's comment moved here

Sorry I am confused. Should I divide by L or multiply by the time step?

[Matt J]

My advice hasn't changed. I think you want to multiply by Tinc. But you need to be absolutely sure of what you're trying to compute. Here is one guy's idea of the formula for the continuous Fourier transform that I found online,

and here is how Matlab defines the FFT,

As you can see, Matlab's FFT after multiplying by dx=Tinc becomes a discretized version of the continuous Fourier Transform integral, assuming you agree with that definition.

[SYML2nd]

Ok, I tried to study and this solution of multiplying by Tinc seems to me have sense.

Indeed using this method is verified the fact that, if take the frequency 0 of the fourier trasform I should obtain the integral of the time history, as stated below:

But I am still confused about the fact that here https://it.mathworks.com/help/matlab/ref/fft.html it is recommended to divide by L. Can you explain me why?

[Matt J]

Dviding by L gives the Discrete Time Fourier Series coefficients. Possibly, that is what they were trying to compute.