What's the number of FFTs in Welch-based PSD? I've read from the book &quot;Proakis, J. G., &amp; Manolakis, D. G. (2004). Digital signal processing. PHI Publication: New Delhi, India.&quot; 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?
It should really be 1 less than 2N/M. There's nothing to overlap the last &quot;half-transform&quot; (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 :)