Maximizing S/N in Angular Power Spectrum Signals

In summary, the signal-to-noise ratio for the angular power spectrum signal Cl, which is a function of multipole l, under theoretical noise Nl is given by (S/N)^2= \sum (2l+1) (Cl/Nl)^2. To increase the S/N, it is recommended to bin the power spectrum signal with a bin width of \Delta l, which decreases Nl by a factor of 1/sqrt(\Delta l). This is discussed in Section 3.2.5 of the Planck 2018 results V. CMB power spectra and likelihoods paper. However, binning may not necessarily lead to an increase in the S/N ratio as the cumulative S/N ratio may remain
  • #1
SherLOCKed
13
1
The signal-to-noise ratio for angular power spectrum signal Cl under theoretical noise Nl, where Cl and Nl are functions of multipole l, is given as

(S/N)^2= \sum (2l+1) (Cl/Nl)^2To increase the S/N we bin the power spectrum signal, if bin width \Delta l, this in principle decreases Nl by a factor of 1/sqrt(\Delta l).

Now, in (S/N)^2 should we replace the sum over multipoles with the sum over bin centers?
 
Space news on Phys.org
  • #3
Thanks for the response. I checked the paper, it talks about the power spectrum binning. Suppose I bin the power spectrum as described in the paper.
The confusion I had is, if I just sum over the binned multipoles, I will end with the similar cummulative signal-to-noise ratio as before I started binning. So, binning is not necessarily helping to increase the signal-to-noise ratio.
 
  • Like
Likes vanhees71
  • #4
@SherLOCKed I guess one thing that confuses me is why there is a summation over 2 + 1 in this case. That would make sense whenever averaging over all the m modes within a given multipole ℓ. But C is the same for every m mode at a given by assumption of statistical isotropy. So summing over m modes doesn't make sense to me. What does the summation do, and why isn't S/N just quantified as C/N at every multipole?

I agree that because we're considering power, not just amplitude, random noise produces an N that enters your spectrum as a bias, not just as variance. You can't get rid of it by binning multipoles. But for any actual measurement, the noise also causes multipole-to-multipole variance in the estimation of C that would average down through binning.
 

Similar threads

Replies
1
Views
789
Replies
9
Views
970
Replies
1
Views
807
Replies
2
Views
1K
Replies
3
Views
1K
Replies
1
Views
785
Replies
8
Views
3K
  • Set Theory, Logic, Probability, Statistics
Replies
1
Views
837
Replies
1
Views
3K
Back
Top