[solved] difference between 2 spectrograms
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Hello,
Here is the situation:
I have a dynamic system equipped with a vibration sensor.
this system has 2 configurations; I have 1 record of 10 minutes in each configuration
I plot spectrogram for each. So far, everything is fine
[s,f,t]=spectrogramspectrogram(data,hamming(wdw),overlap,[],10240);
[s2,f,t]=spectrogram(data2,hamming(wdw),overlap,[],10240);
Now I want to compare (make a difference) of these spectrograms, but
I can do s-s2, but before
I can't plot back the spectrogram from s
I tried several thing such as
im=imagesc(h/1000,(t/60),(abs((s')))) ; % tested with imag real and with image
colorbar
ylabel('Time (mins)');
xlabel('Frequency (kHz)');
but the result is different from
spectrogram(data,hamming(wdw),overlap,[],10240);
thank you for your answer
采纳的回答
Hello,
thank you for your answer, this is great help
here is the result
this is the best result I got so far. :D
some possible improvments
-the last time index is deleted
-Z data is different (but I will make comparisons, si not very important)
-the pscolor give a flat image, when spectrogram give a 3D I can rotate
1 个评论
Bjorn Gustavsson
2019-9-25
For the last point you can simply change pcolor to surf. For the point about the colour-scale, it might be that the top plot is something like:
pcolor(f,t,10*log10(abs(s)/max(abs(s(:))))), shading flat
that is display the spectrogram in dB relative to the strongest component.
更多回答（5 个）
Bjorn Gustavsson
2019-9-25
Spectrogram called without input arguments seems to plot log10 of the power. try this:
im = pcolor(h/1000,(t/60),log10(abs((s'))));
shading flat
colorbar
ylabel('Time (mins)');
xlabel('Frequency (kHz)');
HTH
bryan
2019-9-25
0 个投票
thank you again
surf work fine (I think I got a parenthesis problem)
I will try this suggestion for amplitude
but for the time, I have a synchronisation problem visible in the rounded area
1 个评论
Bjorn Gustavsson
2019-9-25
That has something to do with what your red curve is, and not that much with the spectrogram-plot, I think.
A completely different question is if you need this spectral resolution - you might get more useful results if you reduce the length of your window. That would reduce the frequency-resolution but will give you improved time-resolution.
1 个评论
Bjorn Gustavsson
2019-10-25
Yes, that is true, but the "red curve" of yours varies at shorter time-scales than your time-resolution of your spectrogram - to try to resolve that you could try to run spectrogram with a shorter window, wdw, and overlap (this will reduce the spectral resolution, but you have plenty of that already.)
HTH
bryan
2019-10-28
0 个投票
2 个评论
Bjorn Gustavsson
2019-10-28
You will not get more data in your spectrogram if you increase the time-resolution and reduce the spectral resolution - this is Heisenberg! As your spectrogram looks now you have "too high" spectral resolution and too low temporal resolution.
bryan
2020-2-28
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