FFT results Matlab VS Numpy (Python) : not the same results

Question

I have a Matlab script to compute the DFT of a signal and plot it:

(data can be found here)

clc; clear; close all;

fid = fopen('s.txt');
txt = textscan(fid,'%f'); 

s = cell2mat(txt);

nFFT = 100;
fs = 24000;
deltaF = fs/nFFT;
FFFT = [0:nFFT/2-1]*deltaF;
win = hann(length(s));

sw = s.*win;
FFT = fft(sw, nFFT)/length(s);
FFT = [FFT(1); 2*FFT(2:nFFT/2)];
absFFT = 20*log10(abs(FFT));

plot(FFFT, absFFT)
grid on

I am trying to translate it to Python and can't get the same result.

import numpy as np
from matplotlib import pyplot as plt

x = np.genfromtxt("s.txt", delimiter='  ')

nfft = 100
fs = 24000
deltaF = fs/nfft;
ffft = [n * deltaF for n in range(nfft/2-1)]
ffft = np.array(ffft)
window = np.hanning(len(x))

xw = np.multiply(x, window)
fft = np.fft.fft(xw, nfft)/len(x)
fft = fft[0]+ [2*fft[1:nfft/2]]
fftabs = 20*np.log10(np.absolute(fft))

plt.figure()
plt.plot(ffft, np.transpose(fftabs))
plt.grid()

The plots I get (Matlab on the left, Python on the right):

What am I doing wrong?

Maybe not your main problem, but your window function should be the same size as the FFT, i.e. nfft, not len(x) (applies to both MATLAB and Python code). — Paul R
– Paul R, Commented Jun 21, 2017 at 13:51
@Paul R Interesting, where can I find more information about that ? — hibol
– hibol, Commented Jun 21, 2017 at 14:18
There are quite a few questions and answers right here on StackOverflow which cover window functions and FFTs. The bottom line though is that you apply a window function to the input data of an FFT in order to reduce spectral leakage. The size of this window function therefore needs to match the size of the FFT. — Paul R
– Paul R, Commented Jun 21, 2017 at 14:22
unlikely to be the issue, but be aware that matlab's [0:49] is not the same as python's range(49). In python you're off by one. (i.e. range(49) is equivalent to [0:48]) — Tasos Papastylianou
– Tasos Papastylianou, Commented Jun 21, 2017 at 14:28
There are at least two points i found incorrect in your python code. For instance the line ffft = [n * deltaF for n in range(nfft/2-1)]. 1) you need to use integer division nfft//2; 2) By range(nfft//2-1) you have missed 1. The correct shall be range(nfft//2). — MathArt
– MathArt, Commented Jul 30, 2024 at 11:59

ymmx · Accepted Answer · 2017-06-21 14:25:52Z

3

Both codes are different in one case you concatenate two lists

FFT = [FFT(1); 2*FFT(2:nFFT/2)];

in the matlab code

in the other you add the first value of fft with the rest of the vector

fft = fft[0]+ [2*fft[1:nfft/2]]

'+' do not concatenate here because you have numpy array

In python, it should be:

fft = fft[0:nfft/2]
fft[1:nfft/2] =  2*fft[1:nfft/2]

edited Jun 21, 2017 at 14:25

answered Jun 21, 2017 at 14:07

ymmx

5,2077 gold badges40 silver badges78 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Dr. Paul Kenneth Shreeman · Accepted Answer · 2017-06-21 13:22:00Z

0

I am not a Mathlab user so I am not sure but there are few things I'd ask to see if I can help you.

You called np.array after array has been made (ffft). That probably will not change the nature of array as well as you hoped, perhaps it would be better to try to define it inside np.array(n * deltaF for n in range(nfft/2-1)) I am not sure of formatting but you get the idea. The other thing is that the range doesn't seem right to me. You want it to have a value of 49?

Another one is the fft = fft[0]+ [2*fft[1:nfft/2]] compared to FFT = [FFT(1); 2*FFT(2:nFFT/2)]; I am not sure if the comparsion is accurate or not. It just seemed to be a different type of definition to me?

Also, when I do these type of calculations, I 'print' out the intermediate steps so I can compare the numbers to see where it breaks.

Hope this helps.

answered Jun 21, 2017 at 13:22

Dr. Paul Kenneth Shreeman

3141 gold badge2 silver badges10 bronze badges

1 Comment

hibol Over a year ago

Thanks. I changed the definition of ffft that way : ffft = np.array(n * deltaF for n in range(nfft/2-1)). Not sure either about the second portion of code you mention. I found a solution and will post it below. And I don't reaaly care about the number of points I get. I'll probably do some tests with different values.

hibol · Accepted Answer · 2017-06-21 14:27:46Z

I found out that using np.fft.rfft instead of np.fft.fft and modifying the code as following does the job :

import numpy as np
from matplotlib import pyplot as pl

x = np.genfromtxt("../Matlab/s.txt", delimiter='  ')

nfft = 100
fs = 24000
deltaF = fs/nfft;
ffft = np.array([n * deltaF for n in range(nfft/2+1)])
window = np.hanning(len(x))

xw = np.multiply(x, window)
fft = np.fft.rfft(xw, nfft)/len(x)
fftabs = 20*np.log10(np.absolute(fft))

pl.figure()
pl.plot(np.transpose(ffft), fftabs)
pl.grid()

The resulting plot : right result with Python

I can see that the first and the last points, as well as the amplitudes are not the same. It isn't a problem for me (I am more interested in the general shape), but if someone can explain, I'd be happy.

Collectives™ on Stack Overflow

FFT results Matlab VS Numpy (Python) : not the same results

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

Comments

1 Comment

Comments

Your Answer

Sign up or log in

Post as a guest

Related