[Aavso-photometry] CCD 'fainter-than' question

[Wolfgang Renz]

Hi Jim

That part was just the basic general math/stat part and I adapted my
language away from pure math/stat to make it easier to understand.
But you are right that it lost accuracy by doing so not well enough.

To continue your example:

How is the SNR of a star calculated in aperture photometry ?

Most programs use (and output) the net ADU counts (which is the
sum of the actual ADU counts in the star aperture minus the number
of pixel in the star aperture times the mean/median/mode of the
background level in the sky annulus).

I never assumed that the SNR isn't integrated on a pixel by pixel
base over all pixel in the star aperture.
But long way back I incorrectly thought that the SNR is calculated
in a formula using the MEAN net ADU count per pixel  divided by
the ADU stddev in the sky annulus. But this is not the case.
The SNR is actually calculated in a formula using the total net ADU
count in the star aperture divided by the ADU stddev in the sky
annulus.

Clear skies
 Wolfgang

-- 
Wolfgang Renz, Karlsruhe, Germany
Rz.BAV = WRe.vsnet = RWG.AAVSO



----- Original Message ----- 
From: "Jim Roe"
Cc: "AAVSO-PHOTOMETRY" <aavso-photometry at mira.aavso.org>
Sent: Wednesday, May 28, 2008 2:23 PM
Subject: Re: [Aavso-photometry] CCD 'fainter-than' question

> Wolfgang Renz wrote:
>> Hi Bob
>> 
>> If one assumes that the noise in the sky annulus is normal distributed,
>> then the following applies for the probability that a star measurement
>> is still just background noise:
>> <http://en.wikipedia.org/wiki/68-95-99.7_rule>
>> <http://en.wikipedia.org/wiki/Standard_deviation#Rules_for_normally_distributed_data>
>> <http://upload.wikimedia.org/wikipedia/commons/8/8c/Standard_deviation_diagram.svg>
>> <http://upload.wikimedia.org/wikipedia/commons/thumb/8/8c/Standard_deviation_diagram.svg/400px-Standard_deviation_diagram.svg.png>
>> 
>> So the probability for a measurement to lie outside the following
>> ranges and to be still just a background detection are as follows:
>> +/- 1 sigma   100% - 68.27% = 31.73%
>> +/- 2 sigma   100% - 95.45% = 4.55%
>> +/- 3 sigma   100% - 99.73% = 0.27%
>> +/- 4 sigma   100% - 99.9934% = 0.0066%
>> +/- 5 sigma   100% - 99.99994% = 0.00006%
>> +/- 6 sigma   100% - 99.9999998% = 0,0000002%
>> +/- 7 sigma   100% - 99.99999999974 % = 0.00000000026%
>> But as there are no 'darker than the background' stars, one must
>> half the above probabilities for just 'brighter than the background'
>> stars.
> 
> A precise definition, Wolfgang, but I would try to amplify it with a 
> numerical example.  First off, these results must be applied pixel by pixel.
> 
> Thus, consider a background average of 100 and a sigma of 10.  The 
> probability that a pixel in the measuring aperture that measures 130 (+ 
> 3 sigma) is just a background outlier is, as stated above, 0.0027/2. 
> But how many pixels are in your measuring aperture?  The math is pi r 
> squared where r is the radius of the aperture in pixels.  Suppose r = 6 
> then the number of pixels is 113.  Thus, the probability that at least 
> one of the background pixels in the measurement aperture is as bright as 
> 130 is only 15%.
> 
> But one should never claim a "detection" based upon one pixel.  There 
> should be several contiguous pixels at or above the threshold before 
> claiming such.  Even then, a bit of caution is warranted as cosmic ray 
> strikes can produce some very star-like images.
> 
> Jim Roe [ROE]