# FFT result looks nothing like analytic result

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James Kirk am 29 Nov. 2015
Bearbeitet: Rick Rosson am 14 Mär. 2016
Forgive me if I am missing understanding something simple here but I am confused by exactly what the FFT function returns in Matlab. I have code to compare the result of FFT to the analytic result of the fourier transform of a gaussian:
steps = 2^10; lim = 4;
x = linspace(-lim, lim, steps);
w = lim/4; A = 1;
func = A*exp((-1).*(x./w).^2); % Gaussian
Analytic = sqrt(pi)*w*exp(-(w*pi.*x).^2); % Analytic Transform
Func = fftshift(fft(func)); % FFT transform
When I plot this, the fft function looks nothing like the analytic result:
If anyone could shed some light on why this is and how I can adapt my use of the FFT function to give me back the analytic result I would greatly appreciate it!
I am asking this question as I have been developing some code to model the propagation of various beams through a turbulent medium using phase screens. For the code to be valid it is important that the Fourier Transforms produced agree with Physics.
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### Antworten (1)

Rick Rosson am 6 Jan. 2016
Bearbeitet: Rick Rosson am 14 Mär. 2016
steps = 2^10; lim = 4;
dx = 2*lim/steps;
x = -lim:dx:lim-dx;
% x = linspace(-lim, lim, steps);
Fs = 1/dx;
dF = Fs/steps;
f = -Fs/2:dF:Fs/2-dF;
w = lim/4; A = 1;
func = A*exp((-1).*(x./w).^2); % Gaussian
% Analytic = sqrt(pi)*w*exp(-(w*pi.*x).^2); % Analytic Transform
Analytic = sqrt(pi)*w*exp(-(w*pi.*f).^2);
% Func = fftshift(fft(func)); % FFT transform
Func = dx*fftshift(fft(ifftshift(func))); % FFT transform
diff = abs(Func) - Analytic;
figure;
subplot(2,1,1);
plot(f,abs(Func),f,Analytic);
xlim([-4 4]);
subplot(2,1,2);
plot(f,diff);
xlim([-4 4]);
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Rick Rosson am 14 Mär. 2016
They are the same to within +/- 3 x 10^-8.

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