Accelerometer with FFT - strange output

Question

After reading a lot of research and works on subject I still have got problems applying FFT to my accelerometer data. Most of my code is taken from official MATLAB example: FFT for one dimension. After more reading I've found this question: FFT and accelerometer data: why am I getting this output? where there was suggestion to use windowing. So after some more reading I've added hamming window to my code.

My data looks like that on plot:

And this is the code that I am using for FFT:

fs = 1/0.02; %0.02 comes from picking sample each 20ms
m = size(data,1);
w = hanning(m);
yw = w.*data;
n = pow2(nextpow2(yw));
y = fft(yw,size(n,1));
f = (0:size(n,1)-1)*(fs/size(n,1));
power = y.*conj(y)/size(n,1);
figure
plot(f,power)

The problem is that my plot from this code looks like that:

Can someone tell me what is wrong with my code? To be honest I'd excepted it would look better (something like this:https://i.sstatic.net/gNdag.jpg) so that's why I am asking this question.

EDIT: My data can be found here: https://dl.dropboxusercontent.com/u/58774274/exp.txt

Answer 1

Your Fs is 50 so the highest frequency in your data can be Fs/2 = 25Hz.

See if this code helps.

fid = fopen('1.txt','r');
C = textscan(fid, '%f');
fclose(fid);
data = C{1};
fs = 50;
m = length(data);
nfft = 2^nextpow2(m);
y = fft(data,nfft)/m;
f = fs/2 * linspace(0,1,nfft/2+1);
power = abs(y);
subplot(211)
plot(f,power(1:nfft/2+1)) 
t = (0 : m-1)/fs;
s0 = .8*fs : 3.2*fs;  % .8 sec to 3.2 sec
p(s0) = .5*cos(2*pi*3.3*t(s0)+.25*pi); 
p = p + mean(data); 
subplot(212)
plot(t,data);hold on
plot(t,p,'r')

This is your data in frequency domain.

There is a peak at 3.3 Hz.

As a proof I plotted a sinusoidal with frequency of 3.3 Hz along with your data,

As you can see, it totally matches your data.