Problem with frequency axis during Fourier Transform of Gaussian Pulse

Hello, I noticed that whenever I change dt(sampling time) the bandwidth of FFT Gaussian changes. Smaller dt will have wider bandwidth.

• Why is this happening? Doesn't smaller dt just allows more points for better signal sampling in time domain?

clc 
clear all
close all

dt=15e-12;
t=0:dt:2e-9;
tao = .5e-9;
nfft = length(t);

%frequency spectrum
fs = 1/dt;
df = fs/(nfft-1);
f=(-fs/2):df:(fs/2);

sigma1=.1E-9;
x1=((t-tao)/sigma1).*((t-tao)/sigma1);
G1=(1/(sqrt(6.28)*sigma1)).*exp(.5.*(-x1));

%Fourier Transform
G7 = fftshift(abs(fft(G1)));
plot(f/1E9,G7)
xlim([-50 50]);

Thanks,

[Rick Rosson]

Please post screen shots of the spectrum for each of two different values of dt.

Answer 1

[Rick Rosson]

A few minor tweaks:

   t = 0:dt:2e-9-dt;

   df = fs/nfft;

   f = -fs/2:df:fs/2-df;

[Vivian]

Thank you for looking into this. Actually, I wasn't quite clear on what I was asking. I was wondering about why if changed:

t=0:dt:2e-9;

to

t=0:dt*4:2e-9;

The bandwidth of G7 widened. I think it is incorrect to change dt to dt*4, as this changes the time spacing between the number of samples in time domain without reflecting the changes of it in the frequency domain. The proper way is to:

dt=4*15e-12;

To show the correct fourier transform of the sampling frequency. Please correct me if I'm wrong.

Thank you!

[Rick Rosson]

Yes, that is correct. If you change dt, then that also changes fs, and that will change the range of frequencies over which the Fourier spectrum is defined. So it is crucial that when you change dt, you recompute fs and everything else that depends on dt.