FFT of a Signal

Hi, I would like to take the FFT of a signal but for some reason I do something wrong.

May someone could help me out?

I would like to see the FFT Spectrum for 2 seconds.

I attached the Excel file and the code.

Thanks!

Answer 1

[Star Strider]

The constant (‘direct current’ or ‘d-c’ 0 Hz) offset is hiding the rest of your spectrum. This is a common problem, and since removing the constant offset does not affect the rest of your signal (and you can easily recover it if needed, since it is the mean of your signal), remove it before calculating the fft to see the rest of your spectrum.

This works:

[d,s] = xlsread('Good State.xls');

Time = d(:,1);
Good = d(:,2);
L = size(d,1);                                              % Vector Length
Ts = mean(diff(Time));                                      % Sampling Interval (Assume ‘seconds’)
Fs = 1/Ts;                                                  % Sampling Frequency (Assume ‘Hz’)
Fn = Fs/2;                                                  % Nyquist Frequency (Assume ‘Hz’)

Good = Good - mean(Good);                                   % Remove D-C (Constant) Offset
Y = fft(Good)/L;                                            % Scaled Two-Sided Discrete Fourier Transform
Fv = linspace(0, 1, fix(L/2)+1)*Fn;                         % Frequency Vector
Iv = 1:length(Fv);                                          % Index Vector For Plot

figure(1)
plot(Fv, 2*abs(Y(Iv)))
grid
xlabel('Frequency')
ylabel('Magnitude')

Answer 2

[Harry]

Brilliant! Thank you!

[Star Strider]

My pleasure!

[Harry]

May I ask you one more question?

I would like to apply feature extraction to a good and bad signal.

The FFT of the two signals is attached as a picture to this question.

How would you proceed?

[Star Strider]

I’ve not done this in a while.

I would use principal components analysis (the pca function in the Statistics and Machine Learning Toolbox) as a first approach. We did that with EEGs a couple decades ago with good results. It will tell you the components of the signal responsible for the maximum variance between the two, so I would begin there.

This is no longer an area of my expertise, so there may be better approaches. Consulting with a statistician would be appropriate.

[Harry]

Nice one!

Much appreciated!

Thank you

[Star Strider]

My pleasure!