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

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zahra 2022-8-10

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评论： dpb 2022-8-10

Hi everyone .I wrote a code in MATLAB. In section 2, the FWHM value is almost twice the value I expected. Where is the problem? run order: 1.section 1 2.simulink 3.section 2

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dpb 2022-8-10

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Looks like it to me,yes...why? Because the Nyquist frequency is

Fmax = Fs/2;

when you compute the FFT, the associated frequency goes from -Fmax : +Fmax; the two-sided FFT, where as above Fmax is 1/2 the signal sampling frequency.

If you're sample at

>> fs2/1E6
ans =
  127.7400
>> 

or 127 MHz, then when you plot the FFT, if your frequency vector is correct, your plot should go from

>> Fmax=ans/2
Fmax =
   63.8700
>> 

-63.87 to +63.87 MHz. If it doesn't, you've messed up; and that you get twice the value you think you should makes me think that's what you did...

I get you should have something like

L=1000;                 % arbitrary signal length
Fs=fs2;                 % your sample rate
f = (0:L/2)*(Fs/L);     % compute the frequency vector (baseband one-sided)
f(end)                  % upper limit --> Fmax = Fs/2
ans =
    63870000
>> 

For your double-sided, you need to go from -L/2:L/2 to get the negative frequency as well, of course, but the maximum is still limited by Nyquist.

zahra 2022-8-10

Thank you so much for your explanation

According to you, the frequency axis is between -63.87 and +63.87 MHz and it is correct, so I don't think the problem is due to the Nyquist frequency.

dpb 2022-8-10

在 MATLAB Online 中打开

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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