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@tarokiritani Yes, that is correct. More precisely, the FFT returns energy across the spectrum of positive, negative, and zero frequencies. The folding process takes all the energy from the negative frequencies and places it at the matching positive frequencies. There is a single entry in fft_x for zero frequency, so there is no folding required there.
is the lowest plot equivalent to the PSD?
This notebook is gold ! Thank you very much !
@jedludlow I know you create this gist 8 years ago. Just curious if maybe we could make it do the ifft operation? We have already know that the signal was the combination with 5 * np.sin(2 * np.pi * f * time) and 2 * np.sin(10 * 2 * np.pi * f * time) , what if we don't know the formula before, and we only get the value of x, which is like:
array([ 0.00000000e+00, 2.52877920e+00, 2.41901994e+00, 6.65052259e-01,
5.06655338e-01, 2.93892626e+00, 5.32484856e+00, 5.02813672e+00,
3.04606912e+00, 2.62202223e+00, 4.75528258e+00, 6.81354929e+00,
6.16570415e+00, 3.81456314e+00, 3.00932322e+00, 4.75528258e+00,
6.42624829e+00, 5.39721013e+00, 2.67699571e+00, 1.52062250e+00,
2.93892626e+00, 4.31088140e+00, 3.01619327e+00, 6.78789312e-02,
-1.27544686e+00, -1.83697020e-15, 1.27544686e+00, -6.78789312e-02,
-3.01619327e+00, -4.31088140e+00, -2.93892626e+00, -1.52062250e+00,
-2.67699571e+00, -5.39721013e+00, -6.42624829e+00, -4.75528258e+00,
-3.00932322e+00, -3.81456314e+00, -6.16570415e+00, -6.81354929e+00,
-4.75528258e+00, -2.62202223e+00, -3.04606912e+00, -5.02813672e+00,
-5.32484856e+00, -2.93892626e+00, -5.06655338e-01, -6.65052259e-01,
-2.41901994e+00, -2.52877920e+00, -6.12323400e-15, 2.52877920e+00,
2.41901994e+00, 6.65052259e-01, 5.06655338e-01, 2.93892626e+00,
5.32484856e+00, 5.02813672e+00, 3.04606912e+00, 2.62202223e+00,
4.75528258e+00, 6.81354929e+00, 6.16570415e+00, 3.81456314e+00,
3.00932322e+00, 4.75528258e+00, 6.42624829e+00, 5.39721013e+00,
2.67699571e+00, 1.52062250e+00, 2.93892626e+00, 4.31088140e+00,
3.01619327e+00, 6.78789312e-02, -1.27544686e+00, -1.97217653e-14,
1.27544686e+00, -6.78789312e-02, -3.01619327e+00, -4.31088140e+00,
-2.93892626e+00, -1.52062250e+00, -2.67699571e+00, -5.39721013e+00,
-6.42624829e+00, -4.75528258e+00, -3.00932322e+00, -3.81456314e+00,
-6.16570415e+00, -6.81354929e+00, -4.75528258e+00, -2.62202223e+00,
-3.04606912e+00, -5.02813672e+00, -5.32484856e+00, -2.93892626e+00,
-5.06655338e-01, -6.65052259e-01, -2.41901994e+00, -2.52877920e+00,
-1.22464680e-14, 2.52877920e+00, 2.41901994e+00, 6.65052259e-01,
5.06655338e-01, 2.93892626e+00, 5.32484856e+00, 5.02813672e+00,
3.04606912e+00, 2.62202223e+00, 4.75528258e+00, 6.81354929e+00,
6.16570415e+00, 3.81456314e+00, 3.00932322e+00, 4.75528258e+00,
6.42624829e+00, 5.39721013e+00, 2.67699571e+00, 1.52062250e+00,
2.93892626e+00, 4.31088140e+00, 3.01619327e+00, 6.78789312e-02,
-1.27544686e+00, 5.02600372e-15, 1.27544686e+00, -6.78789312e-02,
-3.01619327e+00, -4.31088140e+00, -2.93892626e+00, -1.52062250e+00,
-2.67699571e+00, -5.39721013e+00, -6.42624829e+00, -4.75528258e+00,
-3.00932322e+00, -3.81456314e+00, -6.16570415e+00, -6.81354929e+00,
-4.75528258e+00, -2.62202223e+00, -3.04606912e+00, -5.02813672e+00,
-5.32484856e+00, -2.93892626e+00, -5.06655338e-01, -6.65052259e-01,
-2.41901994e+00, -2.52877920e+00])And we need to separate it to x1 and x2, I mean x1 and x2 would be the signal with the same basic frequency.
Awesome thanks again for your share.
@raproth, I apologize for not replying to your comment from so long ago. I somehow missed any notification about it.
The plot is not precisely a PSD. The power spectral density expresses signal power distribution as a function of frequency and has the property that the integral under the PSD over some frequency range represents the total signal power in that frequency range.
The plot I'm producing here is really just a plot of the magnitude of the Fourier coefficients. It's useful for reasoning about the amplitude of the various sinusoids that make up the underlying signal.
@pivdets, I'm not sure I fully understand your question, but I'll make an attempt at an answer to the ifft part.
I don't have time right now to extend the notebook to include the ifft case, but I can describe how you can figure it out on your own. You can take the Fourier spectrum results that are plotted in cell 9 at the end of the notebook and try to work backwards through the problem, using the ifft instead of fft in cell 4. If your code reproduces the signal plotted in cell 3 then you know you've got it working correctly. Maybe you can submit a pull request if you get it working. Once you have proven your method you can apply it to other Fourier spectra to recover the underlying signal.
I assume the DC amplitude appears only once in fft_x, and therefore should not be multiplied by 2? Am I mistaken?