You can calculate your fft as follows:

Specify your signal length, such as L= 2200

NFFT= 2^nextpow2 (L) ;

% next higher power of 2

Y= fft(a, NFFT)/ L;

% where “a” is your signal and “Y” is its fft

f= Fs/ 2* linspace (0, 1, NFFT/2+1);

% calculating the frequency where Fs is the sampling frequency

% to plot:

f1= f(1:513); % choose your points to plot.

ft= 2*abs(Y(1:NFFT/2+1));

% multiply by to for: folding: keeping only positive magnitude, so it multiplied by 2 to remove negative mirror component

Plot (f1(1:513), ft(1:513));

The answer is divided in two sections! Sorry!

Thank you very much for your answer.

I have some questions so I can understand:

- you mentioned that “a” is my signal, the things is I have net acceleration from the three axis, how can I go from acceleration to signal?

- How can I determine signal length and sampling frequency?

You have analog signal which varies with time. We call this waveform. So you have a waveform. Say the the time duration of your record is T. Then you can discretize it by sampling it with a sampling frequency fs and then quantize it in amplitude by an n bit for every sample. Both processes are the A/D conversion of your acceleration a(t).

The sampling frequency must be greater equal the 2 the highest frequency in a(t) according to the Nyquest theorem. So, if you have a vibration then you know how much is the highest frequency in the vibration. Otherwise you can limit your signal frequency by an antialiasing filter which is a low pass filter to cut off the signal above the highest required frequency.

After sampling then you will have a data in form a sequence a[i] , with i from 0 to N with N is the highest index of the last sample.

Continue as described above by Yahya.

The lowest signal length is determined by the lowest frequency in the waveform fl= fs/ N, then N = fs/fl

Best wishes