Signal Processing Stack Exchange is a question and answer site for practitioners of the art and science of signal, image and video processing. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am specifically working on a categorical time series of best n number of parallel threads (for n=1..16) for a multicore workstation. I constructed the Walsh-Fourier periodogram for size 2^10 and 2^11 data points. Both of the images are with the first value removed to allow a scaled image. The difference is dramatic as you can see.

My naive interpretation is that there is something stationary going on with the $2^{11}$ range that isn't happening in the $2^{10}$ range. In particular my data covers minute intervals and if I am looking for background programs that will effect the the best thread policy, if they are scheduled nightly (antivirus, back-up etc) then in terms of 1440 minutes=1 day they will fall within $2^{11}$ minutes because $2^{10}<1440<2^{11}$. Am I making the correct assumption here?

Because I am too new on this particular stackexchange I am not allowed to post the periodograms but they are hosted here.

share|improve this question
When you say 2^10 or 2^11, you are talking about the number of samples pre-FFT correct, or are you talking about FFT size? Also, can you please scale the images properly on both axes. – Mohammad Dec 6 '12 at 6:16
I'm a little concerned that I am in over my head but by 2^10 and 2^11, I meant number of samples. However I have a much larger sample size, but since I thought I might use a fast Walsh-Hadamard transform I got used to sub-selecting sample sizes as powers of two to fit the algorithm. Since I'm not yet using that, and my computer is fast enough, I should stick to a 1440 sample size (1 day). Also I'm embarrassed to say, I don't know what is wrong with my axes scales. – Meadowlark Bradsher Dec 6 '12 at 7:10
I realized I had neglected to average my periodograms. When they were averaged they both looked very similar. However I still don't know the explanation for the difference between those two un-averaged periodograms. It doesn't appear adequately random. – Meadowlark Bradsher Dec 6 '12 at 8:52
Yes, the averaging is crucial to a smoothed periodogram, or Bartletts Method. Make sure you do that. I am still not clear as to the sizes in question here. Can you put some images of the total signal, its length, the FFT size, etc etc. This will make the problem more easily attacked by everyone. – Mohammad Dec 6 '12 at 15:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.