# Effect of number of data segments in Periodogram-based PSD

In Welch or Bartlett method, original signal is divided into L segments, of D overlapping (D=0 in Bartlett method). Then FFT is done for each segment, and the result is the average of all segments' FFTs. Averaging reduces the variance. But apart from this effect, is there anything else to take into account when choosing the number of segments for a given signal?

If you increase the number of segments for a fixed total signal length, each segment becomes shorter. This means that you'll have a trade-off between the estimate's variance and the frequency resolution. Averaging many segments will reduce the variance but by using shorter windows, the ability to resolve closely spaced narrow band components is decreased. Shorter windows also decrease the peak amplitudes of any narrow band component. Shorter windows basically average out any details in the spectrum.

• OK, so if, for the sake of argument, I do not care about variance, then I should not segment my data at all (i.e. L = N = total # points). Right? – student1 Aug 24 '14 at 18:54
• @student1: Yes, but your estimate will be worthless. – Matt L. Aug 24 '14 at 19:50
• Here's my problem. So, you see, if I partitioned my data into, say, 32 partitions, then I have a time series of 32 points. I cannot afford having small values of L because then I get almost meaningless number of points. So what to do? – student1 Aug 24 '14 at 23:30
• @student1: So your total signal is only $32\times 32=1024$ samples? If you only have segments of 32 samples then your resolution of course quite limited. Anyway, you can always zero-pad your data to get more densely spaced FFT values (which of course does not increase resolution, but it gives you a more detailed picture of the spectrum with the given resolution). – Matt L. Aug 25 '14 at 15:46
• Only segments of 32 samples. I know this is too little to get anything meaningful, but that's what I have for now. I will try zero padding, thanks. – student1 Aug 25 '14 at 16:34