[solved] difference between 2 spectrograms

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

Answer 1

[bryan]

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

[Bjorn Gustavsson]

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.

Answer 2

[Bjorn Gustavsson]

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

Answer 3

[bryan]

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

[Bjorn Gustavsson]

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.

Answer 4

[bryan]

Thank you for your answer, and sorry for the late reply

• for the problem of Z difference, I searched in pspectrogram.m, and it seems to be

    hndl = surf(t, f, 10*log10(abs(Pxx)+eps),'EdgeColor','none');

so I tried it (without eps), and I noticed the original spectrogam plot removes low frequency, so this also explain why the scale is different, because the low frequencies are high on my data.

I will have a look at the time synchro next time

Answer 5

[bryan]

Hello,

for the Z resolution problem (yes it's a late reply), I solved by myself

https://fr.mathworks.com/matlabcentral/answers/482022-difference-between-2-spectrograms#answer_393425

I noticed surf function doesn't plot the last Y value

So I added a dummy column on the matrix, a dummy element on the Y vector and now I see as much lines in the ploted surface as it actually has

[Bjorn Gustavsson]

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

Answer 6

[bryan]

yes, well this is another issue.

this is easy to change, exept I face memory problems (there are lots of data).I focus on the stabilized areas for the moment

[Bjorn Gustavsson]

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]

you are right, nfft=528 is enaught