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I've read from the book "Proakis, J. G., & Manolakis, D. G. (2004). Digital signal processing. PHI Publication: New Delhi, India." Chapter 12 that the number of FFTs required for estimating the PSD from Welch method with 50% overlap is $2N/M$ where $N$ is the amount of data that you have from the signal, $M$ the size of the segments you choose for the method and I guess the $2$ is for the 50% overlap but from the articles I've read they usually do one FFT less than the one described in the equation, why is that? or what am I missing from this interpretation? My theory is that maybe the FFT missing it's supposed to be done with the last 50% of the data and another 50% of data completed with zero padding. Am I correct?

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It should really be 1 less than 2N/M. There's nothing to overlap the last "half-transform" (N/M) worth of samples.

You're also right, one could zero-pad that, but then that wouldn't be an Welch estimate.

So, I'll do something I rarely do: Blame Proakis for this :)

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The source paper is P. Welch: The use of fast Fourier transform for the estimation of power spectra: A method based on time averaging over short, modified periodograms

The use of the fast Fourier transform in power spectrum analysis is described. Principal advantages of this method are a reduction in the number of computations and in required core storage, and convenient application in nonstationarity tests. The method involves sectioning the record and averaging modified periodograms of the sections.

It clearly states the minus-one version:

enter image description here

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