Rescaling the x-axis of FFT transformed histogram

ROOT
1

I am trying to understand how to use TH1::FFT.

As a test, I am trying to compute the real component of the Fourier transform of a cosine function:
f(t) = cos(2pix)

I expect a result like this:
F(w) = (1/2) * [delta(w+1)+delta(w-1)]
Re(F(w)) = (1/2) * [delta(w+1)+delta(w-1)]

When I do the transform, the output histogram has a range on the x-axis from 0 to 100. The shape of the output looks correct: all bins have zero entries except for two bins with content = 0.5. But the x-axis should range from -1 to +1, not 0 to 100.

What is the prescription for rescaling the x-axis of the “real” transformed histogram? I see a prescription in the tutorial for “magnitude” and “phase” transformed histograms, but not for “real” or “imaginary” transformed histograms:
root.cern.ch/root/html/tutorials/fft/FFT.C.html

My code is below:

  int    n_bins = 100;
  double x_min  = 0.;
  double x_max  = 5.;
  
  TH1F * hist = new TH1F("hist","hist", n_bins, x_min, x_max);
  TF1  * func = new TF1("func", "cos(2.0*TMath::Pi()*x)", x_min, x_max);
  
  for (int bin = 1; bin <= n_bins; ++bin){
    double bin_center = hist -> GetBinCenter(bin);
    double func_value = func -> Eval (bin_center);
    hist -> SetBinContent(bin, func_value);
  }
  
  TH1 * hist_transform;
  hist_transform = hist -> FFT(hist_transform, "RE");
  hist_transform -> Scale (1.0 / float(n_bins));
  hist_transform -> Draw();

Thanks,
Edmund

2

The answer can be found in the fftw documentation:
fftw.org/doc/The-1d-Discrete … _0029.html

Here is what I believe the answer is:

  • Say your input histogram has a sampling frequency of “f”, where f = 1 / (input histogram bin width)
  • Say your output histogram has “nFFT” bins
  • Then for k <= (nFFT / 2), the kth bin of your output histogram corresponds to the frequency k * f / nFFT
  • If you only want to plot positive frequencies, then you’re done. Just plot the first nFFT / 2 bins and ignore the following bins of the FFT output.
  • If you do want to plot negative frequencies, the values corresponding to the negative frequencies are stored in backwards order in the second half of the output [k > (nFFT / 2)].

As an aside: in addition to my question on this topic, I noticed a few very similar questions on this forum that had not been answered. Here are just two examples:

Is this feature simply no longer supported? That would be a shame.

Best,
Edmund

1 Like