Hi Daniel,
Let's plot the first function
Bx = 10;
A = sqrt(log(2))/(2*pi*Bx);
fs = 500; %sampling frequency
dt = 1/fs; %time step
T=1; %total time window
t = -T/2:dt:T/2-dt; %time grids
x = exp(-t.^2/2/A^2);
plot(t,x)
Two things that are very important to remember about the Discrete Fourier Transfrom (DFT), as is computed by fft.
One: there is an underlying assumption that the time-domain sequence is periodic.
Two: the fft implemenation of the DFT operates on one period of the sequence starting at t = 0.
The periodic sequence in item one is the so-called periodic extension of x, with period equal to the time duration of x. We can graph a couple of periods of this periodic extension by
figure
plot([t, t+T , t + 2*T],[x x x])
Now, in order for fft to work the way we want based on point 2, we need to provide it the elements of this periodic extension between 0 and 1. Because of the symmetry in the time vector around t = 0 by construction, we can use ifftshift (ifftshift should be used because numel(t) is even and the first point is at -T/2 and the last at T/2-dt; for aperiodic symmetric signals, it's better (in principle) to define the time vector with an odd number of points symmetric around the point of symmetry, which is t = 0 in this case)
figure
plot((0:numel(t)-1)*dt,ifftshift(x)) % note the last point is at T-dt
We see that ifftshift produces a single period of the periodic extension of the signal over the interval starting at t = 0
Let's check the second case
clear
Bx = 10;
A = sqrt(log(2))/(2*pi*Bx);
fs = 500; %sampling frequency
dt = 1/fs; %time step
T = 1; %total time window
t = 0:dt:T-dt; %time grids starting from t=0
% Numerical evaluations
x = exp(-(t - T/2).^2 / (2*A^2)); % Adjusted the time grid
figure
plot(t,x)
Here, x is defined over the desired interval, so there is no need for additional shifting.
Finally, it's important to remember that bolded clause above. ifftshift (and fftshift) assumes a particular symmetry (depending on if the sequence is even or odd length). If that symmetry is not present in the original time vector, then ifftshift (and fftshift) won't give the correct result. However, in the general case, we can use circshift to map an input sequence defined over any interval to the interval expected by fft.
