Matlab FFT for gaussian function

问题

I am trying to obtain gaussian curve by using matlab's fft. The problem is that in one case my attemp to reduce noise by dividing F=dt.*F./exp(-1i.*nu.*T/2) is not working (img 1) and in the second case if I am trying to take absolute value of fft result's I dont have decent scale in graph (img 2).

N=512;
T=10;
dt=T/(N-1);
t=linspace(-5,5,N);
f=exp(-t.^2);
F=fft(f);

F1=F(1:N/2+1);
F2=F(N/2+1:N);
F=[F2,F1];

dnu=(N-1)/(N*T);
nuNyq=1/(2*dt);
nu=-nuNyq+dnu*(0:N);
F=dt.*F;
%F=dt.*F./exp(-1i.*nu.*T/2);


y=linspace(-5,5,N);
F2=pi.^(1/2).*exp(-y.^2/4);

hold on
plot(y,F2); 
%plot(nu,real(F),'r');
plot(nu,abs(F),'r');
legend('analiticFT','FFT')
xlim([-5 5])
hold off

img 1

img2

回答1:

It seems the scaling in your formula for the analytic Fourier Transform is not quite correct. According to this Fourier Transform table on Wikipedia, the transform of the continuous time-domain signal

where in your case a=1. Correspondingly, you should compare the FFT of the time domain signal

t=linspace(-5,5,N);
f=exp(-t.^2);

with the analytic Fourier Transform

F2 = sqrt(pi)*exp(-(pi*y).^2);

So, plotting the comparison with:

hold off;
plot(y,F2); 
hold on;
plot(nu,abs(F),'r');
legend('analiticFT','FFT')
xlim([-5 5])

yields:

Now that we have establish a proper basis for comparison, we can look at why you are getting the oscillations in img 1. Simply put, the reference Gaussian pulse f=exp(-t.^2); you have generated has a peak at t=0. The corresponding "zero" discrete time instant is naturally the first index in the array (index 1). However in your array, this peak appears at index N/2. Under the Shift theorem, this causes an additional exp(-pi*j*k) term in the frequency domain, responsible for the oscillations you are seeing. To fix this, you should shift back your Gaussian pulse with ifftshift:

F=fftshift(fft(ifftshift(f)));

来源：https://stackoverflow.com/questions/34367889/matlab-fft-for-gaussian-function

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matlab

fft