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FFT of Unsteady Temperature Data Resulting a Peak at 0 Hz

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Oguzhan M
Oguzhan M on 25 Apr 2020
Edited: Oguzhan M on 28 Apr 2020
Hi all,
I have a data that was sampled at 50 Hz for 120 seconds. When I apply fft, dominant frequency seems to be 0 Hz which I initially thought it could actually very well be. However, when I apply same fft to different set of data acquired from the experiment, I get the same peak at 0 Hz, which does not seem to be correct as I expect it to show relatively high frequency(ies) ( this is due to nature of experiment in which the unsteady temperature data is acquired, and the second set of data is expected to show some form of periodicity).
As you may see below in the code, DC offset is removed by "detrend" and a low pass filter is applied ( the result was the same even before low pass filtering).
Can you please take a look at the code (data is attached as well) for any possible corrections or recommendations? If everything is okay with the code, I will appreciate some comments on the physical meaning of the frequency analysis result in this particular case.
Best.
Fs = 100; % Sampling frequency
T = 1/Fs; % Sampling period
dt = 0 :T:120-T; % Time vector
nfft= length(S2TR3_0); %Length of FFT
nfft2 = 2.^nextpow2(nfft) ;
S2TR3_0 = detrend(S2TR3_0);
figure
plot(S2TR3_0)
fy = fft(S2TR3_0,nfft2);
fy = fy(1:nfft2/2);
xfft = Fs.*(0:nfft2/2-1)/nfft2;
plot(xfft,abs(fy/max(fy)));
%low pass filter
cut_off = 2/Fs/2;
order = 256;
h = fir1(order,cut_off);
con = conv(S2TR3_0,h);
figure
plot(con);
fh = fft(h,nfft2);
fh = fh(1:nfft2/2);
fh = fh';
mul = fh.*fy;
figure
plot(abs(mul));
  3 Comments
Daniel M
Daniel M on 25 Apr 2020
No, sampling rate is the same as Fs. You are thinking of Nyquist rate, which is half of Fs, and is typically used for filtering (as you did with cut_off = 2 / (Fs/2) ).
For the second comment, he is saying that he can visually identify an underlying 0.1 Hz oscillation in the time-domain data. 0.1 Hz means one oscillation every 10 seconds.

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Accepted Answer

Daniel M
Daniel M on 25 Apr 2020
Edited: Daniel M on 25 Apr 2020
Data looks fine. You shoud plot using semilogy to better visualize the higher frequencies. Your data shows a typical 1/f noise pattern, seen in white and pink noise dominated data. (Fig 1).
You can also try removing this 1/f curve by taking the gradient of the data, this is a rough method. (Fig 2)
Compare these semilogy plots with the periodic data and see if you can spot the differences now.
  3 Comments
Oguzhan M
Oguzhan M on 28 Apr 2020
Hi again Daniel,
I've been looking into details of the code you shared here for some time now, and I have few questions regarding results generated by semilogy and gradient plots.First, please see the attached plots (and data if you like). Also, please bear with me as I try to make sense out of all these.
1- Looking at two gradient plots, the dominant frequency for 250 RPM is 0.75, and for 600 RPM is 0.36. These results make perfect sense considering nature of the physics here. ( At specific measurement point for both measurements, 250 RPM is expected to show more oscillations than 600 RPM, so all good here.)
I wonder how to interpret the number of dominant frequency contents . As you can see, there are few peaks in 250 RPM whereas there are 2-3 distinct peaks in 600 RPM.
2- Why did you call gradient method as tough one?
3- Looking at two semilogy plots, what is the interpretation for the frequency domain plot? I can't see any distinguishable differences between 250 and 600 RPM as both frequency peaks are at 0 Hz.
4- Is there anything I need to pay attention to when using gradient code?
Many thanks.

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