0
$\begingroup$

In Welch periodogram Power Spectral Density estimate, we devide the N-long signal into K segments, each of length L with overlapping D, such that N=L+(K-1)*D.

In this paper, just above eqn (7), the authors mention that the step of performing the DFT is to be done to each (windowed) segment (of length L) such that the length of the resulting DFT is L/2. Finally all DFT's will be averaged and normalized.

At the end, you get a PSD of a max. length L/2. Why is that? I would expect the length of the resulting PSD to be related to N, not to L...

Improve this question
$\endgroup$

2 Answers 2

Reset to default
2
$\begingroup$

And that's absolutely correct. As you mentioned, procedure can be described as follows:

  • Take long signal of length $N$ and slice it with some overlap and window into segments of length $L$.
  • For each of these segments calculate the DFT. Assuming no zero padding, you will get spectrum with $L$ points - same as segment length.
  • Because signals are real, then we just care about first half of spectra (squred magnitude to be correct), thus take only $L/2$ points.
  • Do this procedure for each segment and you will end up with lot's of vectors (dependent of overlapping and signal length $N$) of length $L/2$.
  • Now you need to average all the spectra. This is simply average for each frequency bin across number of segments and PSD is calculated.

Let's assume you will end up with 50 segments of length $L$, then you must average 50 spectra of length $L/2$. The longer is your signal, more segments will be averaged (better estimate), but obviously their length is not changing - you've chosen it in the beginning.

Improve this answer
$\endgroup$
7
  • $\begingroup$ @student1: You can also refer here for a recent answer regarding implementation. $\endgroup$
    – jojeck ♦
    Aug 13, 2014 at 14:56
  • 1
    $\begingroup$ Thanks. Is this the case with other PSD estimators? I mean getting a result with length that depends on L? I know for one that Bartlett method have the same thing. Actually, any method that depends on segmenting the data will be the same. $\endgroup$
    – student1
    Aug 13, 2014 at 16:34
  • $\begingroup$ Another related question: when I want to construct the frequency axis. I see that it is constructed as freq[i]*=fs/L, where Fs is the sampling freq., and L is as above. Does this make sense? $\endgroup$
    – student1
    Aug 15, 2014 at 15:11
  • $\begingroup$ Frequency bins are defined as: f = (0:L-1)*(fs/L) $\endgroup$
    – jojeck ♦
    Aug 15, 2014 at 15:32
  • 1
    $\begingroup$ @student1: Apart from tens of answers on DSP SE and almost every single book about DSP, I can think of the common sense... Formula given by me is absolutely correct. Try to substitute the numbers and result will speak for itself. Secondly in answer you are linking to, there is no single word that confirms your equation f = (0:L-1)*fs. So for fs=1000Hz and signal of length L=256 we get upper bin at 255 000 Hz? Nyquist won't be happy... They are mentioning spacing between the bins which is not a frequency axis itself but separation between frequencies, is it not? $\endgroup$
    – jojeck ♦
    Aug 21, 2014 at 23:20
0
$\begingroup$

D in equation N=L+(K-1)*D does not actually represent the length of the overlapping segment but the length of the non-overlapping segments, i.e. D = L-L_{ovelap}

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.