# Why will be obtained FWHM in the frequency domain twice the value expected?

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dpb
on 10 Aug 2022

No -- that's the one I computed -- the one I got from your code as written is double that -- what do your plots of your original code show on the frequency axis?

If I take your code instead of mine above, I get

>> L=1000; % arbitary length; immaterial yours is n2

>> fs2 = 12774e4; % your sample rate

>> n2=L; % set your n2 to something

>> f2 = (0:n2-1)*(fs2/n2); % the way you computed the f vector

>> f2(end) % and your thus-computed Fmax

ans =

127612260

>>

You see that's returning the Fmax value as being (essentially) the same as the sampling frequency, or about double the Nyquist. That's what's doubling your results from what you think should be.

You see the difference in the two? There's an "L/2" in the (0:L/2) portion in mine but your's goes from (0:n2) so the maximum will be twice what mine produces. But, you've still got a 2-sided spectrum at

Signal_fft =fft(Signal_I + 1i*Signal_Q);

plot(f2,real(Signal_fft));

but you've plotted it against a baseband (0:Fs) frequency vector instead of one from (-Fmax:+Fmax). That's why you didn't have a length mismatch with the positive frequency of the FFT being only half as many points as the total -- you plotted both negative and positive frequencies.

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