fft function seemingly not evaluating correctly

I have a function defined as an inverse fourier transform of a complicated weighting function. Normally I'd go to integral or trapz for such a task but I remembered that fourier transforms can famously be computed fast, so why not give that a shot. Since I've never used fft before, I thought to check how the results were scaled: there's always the confusion over the factor of . So I ran a test on a standard normal and got this:

What am I not understanding here? Those should be the same line, right?

Answer 1

[Sebastian]

That took a lot longer for me to understand that I expected. Technically all the info I neede was in the Matlab documentation, but I really don't think it was clear what the function is actually doing.

If on the off-chance that someone else runs into this page from a complete misunderstanding of FFTs, this Desmos graph I made should help.

Answer 2

[Torsten]

syms x
f = 1/(2*sym(pi))*exp(-0.5*x^2);
fourier(f)

ans = 

The Gaussian Pulse example under

https://uk.mathworks.com/help/matlab/ref/fft.html

might help to understand the difference to fft.

Answer 3

[Matt J]

Pretty close. There are several issues with how you set up your axes, both in the original domain and the Fourier domain. For one thing, you need to sample on a wider time interval than [-5,5] to get good frequency-domain sampling. Also, you need to scale G by dt/sqrt(s*pi).

N=2^14;
T=100; %time sample on [-100,100]

AX=-ceil((N-1)/2):+floor((N-1)/2);  %normalized axis

%Non-Fourier (time) domain
x=-AX*T/AX(1);   
dt=(x(2)-x(1));  %time sampling interval
g = 1/(2*pi)*exp(-0.5*x.^2);

%Fourier domain
X=AX*pi/T;  %frequency axis points (radians/sec)
G=fftshift(abs(fft(g)))*dt/sqrt(2*pi);

%Compare
skip=round(N/2^10);
h=plot(x(1:skip:end),g(1:skip:end),'ro',  X,G,'-b'); 
xlim([-5,5]);
legend g G

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