fft of given data

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Juan
Juan 2013-11-16
评论: Wayne King 2013-11-17
I got data from a lab, and the FFT I get have a peak in zero Hz. But it is not right, it should have a peak at 50 Hz (european frecuncy). I just don't know what else to try, I'm without new ideas of way to continuo. Please give me an adavise!
The data is xlsx below.
Summering:
% same as in http://www.mathworks.es/es/help/matlab/ref/fft.html but with Fs=150, thus 150/2=75 so close to my 50 Hz, and y and t are vectors given.
y=v;
Fs = 150; % Sampling frequency
% T = 1/Fs; % Sample time
L = length(t); % Length of signal
% t = (0:L-1)*T; % Time vector
NFFT = 2^nextpow2(L); % Next power of 2 from length of y
Y = fft(y,NFFT)/L;
f = Fs/2*linspace(0,1,NFFT/2+1);
plot(f,2*abs(Y(1:NFFT/2+1)))
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')
  2 个评论
Juan
Juan 2013-11-17
Finally semi solved, from
x=i;
%//If youre getting a massive peak at zero that dwarfs everything else, you
%//probably have a large DC offset. Easily removed in the time domain using
%//the following ..
x = x-mean(x);
tAxis = t;
dt = diff(tAxis(1:2)); %//sample period from time axis
fs = 1/dt;%//sample rate from sample period
NFFT = numel(x); %//number of fft bins - change if you like
Y = abs(fft(x, NFFT)).^2; %power spectrum
%//Calculate frequency axis
df = fs/NFFT;
fAxis = 0:df:(fs-df);
%//Plot it all
figure; plot(fAxis(1:NFFT/2), Y(1:NFFT/2))
xlabel('Frequency in Hz')
ylabel('Power')
xlim([0,300]);
Juan
Juan 2013-11-17
Note: for better view of my data
change:
NFFT = 1e6;
add:
xlim([0,300]);
PD: it seems like i didnt need help from here, the stackoverflow-given-an-array-of-data-extract-possible-frequencies-with-fft-how-to would have been enough, but I wont delete this. This may help others.

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回答(1 个)

Wayne King
Wayne King 2013-11-16
编辑:Wayne King 2013-11-16
That indicates that the data has a non-zero mean. First, subtract the mean
y = detrend(y,0);
or simply
y = y-mean(y);
Then execute your commands above.
  2 个评论
Juan
Juan 2013-11-17
编辑:Juan 2013-11-17
Thanks for the answer.
But that doesnt solve the problem. The DC offset or mean is so small that I think is not a problem:
>> mean(y)
ans =
1.3573e-13
>> max(y)
ans =
309.9389
Wayne King
Wayne King 2013-11-17
which column in the excel file is the data you are trying to Fourier transform? what is the title of the column

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