Generally, we know that if we have this relation between Fourier transforms of three functions in frequency domain as:

$$\mathfrak{F}\{\mathsf{P}(t)\} = \mathfrak{F}\{\mathsf{Z}(t)\}\mathfrak{F}\{\mathsf{Q}(t)\}$$

We should have this relationship in the time domain as:

$$\mathsf{P}(t) = \int \mathsf{Z}(\tau) \mathsf{Q}(t-\tau) d\tau$$

Now I have this in Python:

import numpy as np

time = np.genfromtxt('time-data.txt',delimiter=',').T

fft = np.genfromtxt('fft-data.txt',delimiter=',',dtype=np.complex_).T

Z_t = time[0]
Q_t = time[1]

Z_fft = fft[0]
Q_fft = fft[1]
P_fft = fft[2]

#Shows that indeed Z_fft is the Fourier transform of Z_t
assert (np.fft.irfft(Z_fft,n=len(Q_t)) == Z_t).all()

#Shows that indeed Q_fft is the Fourier transform of Q_t
assert (np.fft.irfft(Q_fft,n=len(Q_t)) == Q_t).all()

#Shows that indeed Z_fft = P_fft / Q_fft
assert (np.divide(P_fft,Q_fft) == Z_fft).all()

P_t = np.fft.irfft(P_fft,n=len(Q_t))

P_t_from_convolve = np.convolve(Z_t,Q_t,mode='same')

#Expect to see that P_t = P_t_convolve but it's not the case!
assert (P_t == P_t_from_convolve).all()

Is there any reason that $\int \mathsf{Z}(\tau) \mathsf{Q}(t-\tau) d\tau \neq \mathfrak{F^{-1}}\{\mathfrak{F} \{ \mathsf{P}(t) \} \}$?

These are the data files here:

  • 4
    $\begingroup$ Generally it's necessary to 0-pad a discrete time series in order for periodic FFT based convolution to work in the same way as conventional convolution. It's also extremely numerically unstable to deconvolve by spectral division. $\endgroup$ – Brian Borchers Sep 11 '20 at 3:16
  • $\begingroup$ @MaximUmansky NumPy convolve is pretty useless. In fact , I used scipy.ndimage.convolve1d with mode=wrap and solved the problem. $\endgroup$ – Alone Programmer Sep 11 '20 at 15:10

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