fft: significant peak in 0 Hz component

I'm recently dealing with a problem about finding the frequencies of a data vector using fft. I've read in some sources that the 0 Hz component comes from the mean so I need to detrend the data.but there is no visible change in my output. the 0 Hz component still dominates significantly. If it helps, the vector contains velocity values over time and MY SAMPLE FREQUENCY IS 1HZ. (I've attached my sample data)

load file.txt;
Data=file(:,2);
Time=file(:,1);
N = length(Data(:,1));
Ts = 1;
W = 0:2*pi/N:2*pi*(1-1/N);
X = W / Ts;
Data=Data-mean(Data);
Fx = fft(Data);
plot(X,abs(Fx));ylabel('fft of V_x');xlabel('Ferq(HZ)')

Answer 1

[Greg Heath]

One of the basic assumptions is that that your data is a finite window sample of an infinitely periodic function. Very often there is a polynomial trend to the data which violates this assumption. The function detrend can, in general, remove linear trends

 help detrend
 doc detrend

However, sometimes I find it necessary to remove quadratic trends (especially in the acceleration data of a tumbling and spinning radar target which is undergoing gravitational acceleration).

Sometimes you can plot and "see" a polynomial trend. Sometimes looking at the differences in the plots and spectra of x, diff(x) and diff(diff(x) are helpful.

I have never dealt with a trend higher than quadratic.

The importance of removing significant trends is that their spectral sidelobes contaminate, often overwhelmingly, the low frequency spectra of the true periodic components.

Hope this helps.

Greg

Answer 2

[Carlos]

From your code and the file uploded I can see ypur sampling frequency is 1000 Hz and you are trying to represent frequencies up to 6000Hz.

>> NFFT = 2^nextpow2(N); % Next power of 2 from length of y
Y = fft(Data,NFFT)/N;
f = (1/Ts)/2*linspace(0,1,NFFT/2+1);

% Plot single-sided amplitude spectrum.
plot(f,2*abs(Y(1:NFFT/2+1))) 
title('Single-Sided Amplitude Spectrum of y(t)')
xlabel('Frequency (Hz)')
ylabel('|Y(f)|')

The DC component has disappeared as you can see in the attached file

[Ashkan]

you're right, but as you see there is huge peak somewhere very close to 0Hz. I tried detrending the data but the peak remained somewhere close to 0 Hz which doesn't sound logical according to my data

[Carlos]

Yes, there is a peak at f=3Hz. I cannot figure out where the problem might be then. At least now your fft representation is correct.

[Ashkan]

Thanks for your time Carlos. can it be an aliasing problem ?

[Carlos]

I have thought about aliasing too, but aliasing tends to appear in spectrum as distortion at frequencies close to the maximum frequency of the signal.

Aliasing appears because the effects of sampling in frequency spectrum are scaling and periodic repetition of the spectra. The problem comes when the periodic repetitions intersect. That happens if the the maximum frequency of the signal W satisfies W<fs-W.From this we can infere that the lowest aliased frequency is fs-W, which is not close to 0.

[Ashkan]

Today I found out my sample frequency is 1Hz. do you think it helps ?

Answer 3

[Carlos]

You are right, the 0 Hz component is DC mean component

Try something like

x=x-mean(x);

y = detrend(x,'constant')

And see what happens with your fft.

Are you sure you are plotting and calculating your fft correctly?

[Ashkan]

Thanks for your answer Alex. I've tried both codes you mentioned but no visible change happened. Here is my code: (I've attached my sample data)

 load file.txt;
Data=file(:,2);
Time=file(:,1);
N = length(Data(:,1));
Ts = 10^-3;
W = 0:2*pi/N:2*pi*(1-1/N);
X = W / Ts;
Data=Data-mean(Data);
Fx = fft(Data);
plot(X,abs(Fx));ylabel('fft of V_x');xlabel('Ferq(HZ)')