FFT code on time series?

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Sam
Sam 2013-2-19
I have run through the example used here: http://www.mathworks.co.uk/products/matlab/examples.html;jsessionid=20a658a2dcd514f152ada895e6f8?file=/products/demos/shipping/matlab/fftdemo.html and am now trying to do the same with my data. I have a time series that consists of 5127 data points which were collected every 12 hours for just over 7 years. It is of ocean current water movement and I am trying to look at cycles within it.
I have tried to do this and the graph comes up blank. I suspect it is something to do with this part f = 1000/251*(0:127);
I don't know why they have used 1000 or 127.
Would someone be able to explain this? Thanks
Ok sorry, This is what I used. I have a time variable and a measurement of water volume. By following the example above I used this, but have clearly done something wrong:
t=0:.5:2578;
>> plot (Thermocline(1:5157))
>> Y=fft(Thermocline,5157);
>> Pyy=Y.*conj(Y)/5127;
>> f=1000/5127*(0:2563);
plot(f,Pyy(1:2564))
Pyy - 1x5157
Max ans =
1.6713e+06
ans =
4.4727e-07
I am unsure of the range as I don't fully understand what Pyy is. If its time then its half days. If its volume then the range is 18.7 Sv and if its periodicity then I expect to see a strong annual signal (720).
  2 个评论
Oleg Komarov
Oleg Komarov 2013-2-19
Without a code example is hard to understand what you're plotting against what.
Image Analyst
Image Analyst 2013-2-19
编辑：Image Analyst 2013-2-19
Tell us this:
whos Pyy
max(Pyy(:))
min(Pyy(:))
Also what range is Pyy in and is that the range you expected it to be in?

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采纳的回答

Youssef Khmou
Youssef Khmou 2013-2-20
hi Sam,
in case you have verified the properties of the signal in the above comment,we can proceed as the following :
We have a signal of length L=3950 taken every 12 hours for 2563.5 days. So formally the sampling rate is Fs= 3950/2563 =~1.54 Hz .
For these types of times series , the range is dynamically changing therefore we plot the frequency in dB to expand the low values and compress the high ones :
1)Solution 1:
Fs=1.55; %
L=length(Thermocline);
N=ceil(log2(L));
FFTherm=fft(Thermocline,2^N);
f=(Fs/2^N)*(0:2^(N-1)-1);
Power=FFTherm.*conj(FFTherm);
figure,
semilogy(f,Power(1:2^(N-1)))
xlabel(' Frequency (Hz)'), ylabel(' Magnitude (w)'),
title(' Power Spectral Density'), grid on;
2) Solution : 2
% Directly you type in the C.P :
psd(Thermocline)
So the frequency is in the range [0,...,0.1] Hz. To see if this result, physically, makes sens, try to check :
"Stochastic climate models, Part I1, Application to sea-surface temperature anomalies and thermocline variability" by CLAUDE FRA NKIG NOUL :
And another paper: "MEAN AND EDDY DYNAMICS OF THE MAIN THERMOCLINE" by GEOFFREY K. VALLIS
http://www.princeton.edu/~gkv/papers/thermocline.pdf
These two papers give a general idea on how much an estimate of the frequency must be .
I hope this helps .
KHMOU Youssef.
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Sam
Sam 2013-2-20
Yes it does, thank you very much for all of your help, I understand it now, thanks for thaking the time for such detailed answers!

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更多回答（1 个）

Youssef Khmou
Youssef Khmou 2013-2-19
hi Sam,
In the demo, they used 1000 and 127 because :
t and y are of length 251, and they computed FFT resulting in same number of points which is 251, so the spectrum is "TWO" sided ( Symmetric) and they created a frequency axis based on a number N satisfying the relation 2^N>=length(y) which is N=8 and this is for adjusting the frequency axis to match the exact frequency .
for a value 1000: its the sampling frequency Fs=1000 Hz , you can anderstand that by looking at the vector t :
t = 0:.001:.25; % t=0:1/Fs......
So to adjust the frequency axis you have to do :
F=Fs/Length(vector)*(0:2^(N-1)-1) ;
Second Part : concerning your data , i think you have to increase the sampling rate to 10 or even 20, to make sure that the Nyquist condition is met.
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Sam
Sam 2013-2-19
I imagine there was a better way to do that but couldn't see one.
Youssef Khmou
Youssef Khmou 2013-2-19
编辑：Youssef Khmou 2013-2-20
hi Sam it is ok, i tried to copy the vector, so to make sure the process of copying was correct verify these properties :
1.Length=3950.
2. Expectation=-17.7831.
3. Variance = 10.6303 .

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