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Roughly speaking, the methods of power-spectra estimation can be divided in two categories : parametric and non-parametric.

From what I understood, parametric methods assumes that the signal can be more or less modeled (with an AR model, or a behavior + noise model etc..) and try to use this to perform the estimation.

I have noticed that when I use parametric methods, the result I obtain usually looks "nicer" that if I use a non-parametric method, let's say Welch. The thing is, I don't know anything about my signal and do not know if it can be modelized and if it therefore makes sense to trust parametric methods.

This is where my question comes in : What are the know "problems" which are associated with parametric methods ? Can we trust a spectrogram obtained this way even though the signal can't be modelized or doesn't fit the correct framework of the used method ?

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Can we trust a spectrogram obtained this way even though the signal can't be modelized or doesn't fit the correct framework of the used method ?

Short answer: anything based on a model will work only when applied to the model. Not quite sure which other answer you were expecting here!

Of course, some "model mismatches" are more benign than others: For example, some classes of estimators are very robust against noise not fitting the original model, but break down when the assumed number of signal components isn't what was specified (example: ESPRIT). Others fail the other way around (Bartlett's method will not like you having negative SNR).

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  • $\begingroup$ Hi: generally speaking, ( not just with respect to spectral estimation ), a non-parameteric estimation method will be more robust than a parameteric model when the underlying parametric assumptions are not true. At rhe same time, it won't be as efficient ( greater variance ), when the assumptions underlying the parametric model are true. So, it's key whether the assumptions hold in the parametric case. It's like median and mean. Id distribution is symmetric, mean is way better. If not, median is. $\endgroup$ – mark leeds May 11 at 1:10

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