Converting Matlab code into Python - FFT

Question

I need to convert a piece of MATLAB code to Python and I'm bad at both. The code in MATLAB uses fft and fftshift. I tried to use NumPy in Python. The code runs but when I compare the outcome they are not matching. I appreciate your help.

Here is the MATLAB code:

h(1,1:Modes_number) = -1i*S;
hfft = fft(h);
hft0 = fftshift(hfft);

and here is the Python code which I wrote:

h = np.zeros((1,self.cfg.Modes_number+1),dtype=complex) 
for i in range(0, self.cfg.Modes_number+1):
   h[0,i] = -1j*S;

hfft = np.fft.fft(h)
hft0 = np.fft.fftshift(hfft)

Here is the values for S and Modes_number:

S = 12.5022214424;
Modes_number = 200;

Here is also an example of the results I get in MATLAB and Python:

MATLAB:
hfft(1,1)
ans =
   1.1857e-13 - 2.5129e+03i

Python:
hfft[0]
0. -2.52544873e+03j

Cheers.

I'm not good at Python either, but I would keep that purely 1D, rather than 2D with a singleton dimension as is standard in MATLAB: h=np.zeros(self.cfg.Modes_number,dtype=complex) — Cris Luengo
– Cris Luengo, Commented May 8, 2018 at 22:54
Did the above comment help? If not, please let us know the value of Modes_number and S, without those we cannot reproduce what you see. — Cris Luengo
– Cris Luengo, Commented May 14, 2018 at 6:31
Hi Cris, Thanks for you reply and follow up. Unfortunately, I couldn't fix it using your hint. Here is the values for S and Modes_number: code S=12.5022214424 code Modes_number = 200 Here is also an example of the results I get in MATLAB and Python: MATLAB hfft(1,1) ans = 1.1857e-13 - 2.5129e+03i Python hfft[0] = 0. -2.52544873e+03j — Alireza Hassani
– Alireza Hassani, Commented May 15, 2018 at 2:33

Cris Luengo · Accepted Answer · 2018-05-15 04:49:31Z

The error in your Python code is that you define h to be of size Modes_number+1, which is one more than the size in the MATLAB code. The first value in hfft is the sum of all input values. In MATLAB this is -1j*S*200 = -2500.4j, and in your Python code this is -1j*S*201 = -2512.9j. These are the values that you are seeing.

This bit of Python code produces the same as your bit of MATLAB code, up to numerical precision (I see some values like -1.68388521e-15 +6.55829989e-15j in Python, which are forced to 0 by MATLAB's algorithms). I am creating h as a one-dimensional vector, rather than a 2D array with one dimension of size 1.

import numpy as np

S = 12.5022214424
Modes_number = 200
h = np.zeros(Modes_number,dtype=complex)
for i in range(0,Modes_number):
   h[i] = -1j*S;
hfft = np.fft.fft(h)
hft0 = np.fft.fftshift(hfft)

Python:

>>> hfft[0]
-2500.4442884800001j

MATLAB:

>> hfft(1)
ans = 
   0.000000000000000e+00 - 2.500444288480000e+03i`

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