How to know about FFT bin frequencies
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Hi guys Im not used to computing fft on matlab and I have some doubts so if you could tell me if im right or wrong... My signal is h : N = 16384 Sampling frequency = 65536Hz
if I do : fft(h,65536) I should have a frequency step of 1Hz, right???? Like first bin stands for 0Hz, second for 1 Hz...
I would like to compute a FFT in the frequency range [0 - 6000 Hz] but with a 1 Hz frequency step. Could you help me please??
I would be very thankful for all answers and explainations ! Bye
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Risposte (5)
Star Strider
il 1 Dic 2015
The easiest way to find out is to do the experiment:
Fs = 65536; % Sampling Frequency (Hz)
Ts = 1/Fs; % Sampling Interval (sec)
Fn = Fs/2; % Nyquist Frequency (Hz)
L = 16384; % Signal Length (samples)
t = [0:(L-1)]*Ts; % Time Vector (sec)
s = cos(2*pi*100*t); % Signal (100 Hz)
FTs = fft(s,Fs)/L; % Experiment #1
Fv = linspace(0, 1, fix(Fs/2)+1)*Fn;
Iv = 1:length(Fv);
[PeakFTs,Idx] = max(abs(FTs(Iv))*2); % Find MAximum and Index
figure(1)
plot(Fv, abs(FTs(Iv))*2)
axis([0 500 ylim])
grid
text(Fv(Idx)+10, PeakFTs, sprintf('Peak at %.3f Hz, Index %d', Fv(Idx), Idx), 'HorizontalAlignment','left', 'VerticalAlignment','top')
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Adam
il 1 Dic 2015
Modificato: Adam
il 1 Dic 2015
A 65536-pt FFT would, in your case give 1Hz bins for a sample frequency of 65536 Hz.
If you only want the FFT from 0 to 6000 then just clip the result of the FFT to
y = fft( h, 65536 );
y = y(1:6001);
and this should give you the frequencies from 0 to 6000 at 1Hz intervals.
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himanshu tripathi
il 8 Giu 2019
Please have a look at the code and plot of 1kHz frequency (file attached here).
There is a problem, why the reversed V-shaped plot is obtained in the frequency domain.
Note: Sample 1kHz signal data is attached.
And why Fv = linspace(0,1,fix(LF/2)+1)*Fn;
is used in the fft analysis and what is the concept of the nyquist frequency.
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