# Time axis of inverse Fourier transform with truncated frequencies

11 views (last 30 days)
Daniel Lalich on 23 Oct 2020
Edited: Swetha Polemoni on 4 Nov 2020
I have a signal in frequency space that I am trying to move into a time series through an inverse fourier transform. My vector of amplitudes spans from 8-10 MHz. My question is how to label the time axis after performing the transform. Normally, the time sampling rate would be twice the highest frequency sampled by the signal, so in this case 20 MHz. However, that's for the general case where your sampled frequencies start at 0 (DC). In this case, my vector starts at 8 MHz, and the bandwidth is 2 MHz (8 to 10), so as far as the transform is concerned my signal is essentially base-banded to 0-2 MHz. Does that make the sampling rate 4 MHz instead? I could pad out my vector with leading zeros to fill in the lower frequencies, but I would prefer not to do that if I can avoid it.

Swetha Polemoni on 4 Nov 2020
Edited: Swetha Polemoni on 4 Nov 2020
Hi Daniel Lalich,
t=0:1/fs:n/fs; % n is FFT length and fs is sampling frequency
Secondly, it is my understanding that you want to calculate sampling frequency.
According to Nyquist sampling theorem ,
• For a base band signal fs (sampling frequency) should be greater than the fmax(maximimum frequency ). Therefore the minimum sampling frequency is 2*fmax.
• On the other hand for band pass signals, the minimum sampling frequency is different .
Let f_H and f_L be the highest and lowest frequencies respectively in the band pass signal
fmin= 2*f_H/k
where k= floor(f_H/(f_H-f_L))
There are many other possible sampling frequencies for band pass signals. You may have to understand Nyquist Band pass sampling theorem to know more.
To conclude, you need prior knowledge of sampling frequency to get the exact time axis labels. On the other hand , you can also use minimum sampling frequency.