When the pmtm MATLAB function is used in the form [f,PSD]=pmtm(x,[],[],Fs) it provides the power spectrum estimation PSD (at frequencies f) for the sampled noisy signal x obtained with sampling frequency Fs. Now if you plot (f,10*log10(PSD)) you do not obtain the "right" PSD in the sense that there is a scaling factor that should be Fs/2. So the correct plot should be obtained with plot (f,10*log10(0.5*PSD*Fs)). Is it correct?
From the documentation
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PSD and Frequency Vector Characteristics with fs Specified
Real/Complex Input Data, nfft Even/Odd, Range of f
Real-valued, Even, [0, fs/2]
Real-valued, Odd, [0, fs/2)
Complex-valued, Even or odd,[0, fs)
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You will only get the expected frequencies for complex-valued signals, otherwise the range will have an upper bound of fs/2
Hi Massimo! Thank you for answering my question!
I read the manual before trying to use pmtm and I was aware about the maximun frequency Fs/2 that it is contained in f if x is real. My problem is with the amplitude of PSD. I don't know whether or not it must be scaled and I have found nothing about it in the documentation. But maybe I have not understood your answer!
Sorry, I didn't understand that your question was related to the amplitude of the spectrum (should have better read the code you posted).
Could you please give an example of the function not returning the expected spectrum?
Hi Massimo! I have find the bug ... not in the pmtm MATLAB implementation but in the data I was using as reference to evaluate whether or not the PSD estimation was right! So, no scaling is need in the estimated PSD, everything is right as it is!
Actually, the PMTM generates a one-sided power spectral density with dB/Hz estimate per (f) frequencies of interest based on the sampling frequency (fs). If a two-sided frequency spectrum is of interest, it is advisable to use:
plot([-f/2 f/2],10*log10(0.5*PSD)). The absolute vale of the estimated signal needs to be considered.
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