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I am trying to write my own Matlab code to sample a Gaussian function and calculate its DFT, and make a plot of the temporal Gaussian waveform and its Fourier transform.

According to the FT pair:

\$ e^{-at^2} \iff \sqrt{\frac{\pi}{a}} e^{- \pi^2 \nu^2 /a}, \$

The FT of a Gaussian is a Gaussian, and it should also be a real function.

So here is my graph:

enter image description here

So, why is my FT a flat line instead of being Gaussian?

And why is it that when I plot the argument of the DFT, the imaginary part is nonzero? Shouldn't the FT be real?

Any help would be greatly appreciated.

Here is my code:

clear; clf; clc;
a=5;
n=50; % Length of DFT
k=[-n:1:n];

x=exp(-a.*k.^2); % Gaussian function

subplot(221);
plot(k,x);grid;
axis([1.5,n,-max(-x),max(x)]);
xlabel('Data waveform');
hold on; plot(k,x,'o'); hold off;

xx=fft(x)/n;
subplot(222);
plot(k, abs(xx));grid;xlabel('Transformed, magnitude');
axis([1,n,0,1]);
hold on; plot(k,abs(xx),'o'); hold off;

subplot(223);

plot(k, real(xx));grid;xlabel('Transformed, real');
hold on; plot(k, real(xx), 'o'); hold off;
axis([1,n,-1,1]);

subplot(224)
plot(k,imag(xx)); grid;
axis([1,n,-1,1]);
xlabel('Transformed, imaginary');
hold on; plot(k, imag(xx), 'o'); hold off;

figure(gcf);
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  • \$\begingroup\$ Didn't inspect the code, but from your plot you are effectively sampling an impulse function (with a sine FT pair, hence the imaginary oscillations) instead of a gaussian. You should start plots with a lower a coefficient or a higher time resolution ("sampling rate"). \$\endgroup\$ Apr 22, 2016 at 12:14
  • \$\begingroup\$ @VicenteCunha But how could that be? I explicitly defined a Gaussian function x=exp(-a.*k.^2);. I did try lower \$ a\$ values but I am still getting the same features on the graph. \$\endgroup\$
    – Merin
    Apr 22, 2016 at 13:09
  • \$\begingroup\$ My MATLAB's a bit rusty. I'm happy that as k is vector, k.^2 produces a k squared vector, but I'm not sure about the a., as a is a scalar. One way or another however, you've got to get the simple x=a_guassian right before you do the difficult bit of interpreting its FFT. <quick edit>Ah! how about dividing k by n, so that the argument is sigma=5 at the end of the vector, rather than at the first bin away from the middle? <\quick edit> \$\endgroup\$
    – Neil_UK
    Apr 22, 2016 at 13:17
  • \$\begingroup\$ With k=0, x=1, and with k=1, x=0.007; so the x values are very small wrt the one at k=0. Hence the transformed spectrum looks like that of an impulse ie flat. \$\endgroup\$
    – Chu
    Apr 22, 2016 at 13:45

1 Answer 1

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The FFT looks flat, because the value of your a is too high.

Secondly, you should use a different x-axis for the FFT plot. With N+1 time values, you get FFT points for nu values from -N/2 to +N/2.

You should know, that the vector, which the Matlab FFT function returns has these points in the order [0,1,2,...,+N/2,-N/2,...,-1]. This means you have to shift the vector in order to look like a Gaussian.

I would recommend, don't scale the axis manually for a first test.

There are imaginary parts, because of how the Gaussian is represented in the Matlab vector. More info here. You can use x = circshift(x,[0 -50]); to get a FFT without imaginary parts.

clear; clf; clc;
a=5;
n=50; % Length of DFT
k=[-n:1:n];

x=exp(-a.*k.^2); % Gaussian function
xs = circshift(x,[0 -50]); 

subplot(221);
plot(k,x);grid;
%axis([1.5,n,-max(-x),max(x)]);
xlabel('Data waveform');
hold on; plot(k,x,'o'); hold off;

nu=[0:1:2*n];
xx=fft(xs);
subplot(222);
plot(nu, abs(xx));grid;xlabel('Transformed, magnitude');
%axis([1,n,0,1]);
hold on; plot(nu,abs(xx),'o'); hold off;

subplot(223);

plot(nu, real(xx));grid;xlabel('Transformed, real');
hold on; plot(nu, real(xx), 'o'); hold off;
%axis([1,n,-1,1]);

subplot(224)
plot(nu,imag(xx)); grid;
%axis([1,n,-1,1]);
xlabel('Transformed, imaginary');
hold on; plot(nu, imag(xx), 'o'); hold off;

figure(gcf);
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  • \$\begingroup\$ Thank you so much for the explanation and the code, I really do appreciate it. I used the syntax plot(nu, abs(fftshift(xx))) and the final result looked Gaussian. Thank you. \$\endgroup\$
    – Merin
    Apr 23, 2016 at 11:11

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