Hello , Im having one matrix which is product of two FFT transforms of one fits image ( astronomical image ). In that matrix you could find 3 peaks. One largest in center, and two around central peak. Main problem is determining those peaks. I could find central ( largest ) peak by searching for max value in matrix. But, when i want to find out size of central peak i`m having problem due to noise. Im not sure where noise begin and where peak end. So, my question is how to find noise value? And when i find it, peaks will be above that value.

  • $\begingroup$ Very few people will be able to understand what you're asking here, I expect. I certainly cannot. $\endgroup$ – Chris Godsil Jun 20 '12 at 14:28
  • 1
    $\begingroup$ I removed some of the tags whose relevance seemed hard to imagine, but I agree that this question as it stands needs significant elaboration and cleanup to be understandable. $\endgroup$ – Noah Stein Jun 20 '12 at 17:04
  • $\begingroup$ I changed question, i hope it is much understandable now $\endgroup$ – viktor.radovic Jun 21 '12 at 9:07
  • $\begingroup$ your question is much much more about data analysis techniques then anything else. I suggest you ask instead at stats.stackexchange.com/faq or scicomp.stackexchange.com/faq (I think the first fits better you needs, but you may decide otherwise). Also, there are established peak fitting softwares available in the wild (Origin, Igor of the general data analysis softwares; R if you want something powerful and used by statisticians). $\endgroup$ – Willie Wong Jun 21 '12 at 10:43
  • $\begingroup$ Math. analysis of such questions would require some (probabilistic or determinant) model, otherwise I am afraid it does not make sense. You should make some assumption - like noise is (un)correlated, peaks have the forms: of gaussians or smth... and so on... But making such model I guess the questions will disappear e.g. if peak is Guassian - then it is reasonable to consider [-3sigma, 3 sigma] interval around the peak.... So on $\endgroup$ – Alexander Chervov Jun 21 '12 at 10:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.