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

Wed May 28 16:57:52 EDT 2008

Hi

I see that I cann't be inaccurate when discussing with.
Obviously using 'in a formula using ...' was still not sufficient
to make it fuzzy enough.

But I hope that everyone got the crucial difference I wanted
to make between MEAN and TOTAL.

Clear skies
 Wolfgang

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

----- Original Message ----- 
From: "Michael Newberry"
To: "Wolfgang Renz"; "Jim Roe"
Cc: "AAVSO-PHOTOMETRY" <aavso-photometry at mira.aavso.org>
Sent: Wednesday, May 28, 2008 9:25 PM
Subject: Re: [Aavso-photometry] CCD 'fainter-than' question

> We define a 3-sigma detection as having SNR = 3.0. Then the issue becomes 
> how the SNR is calulated at a given magnitude so one can determine what 
> magnitude corresponds to SNR=3. Computing the SNR simply by dividing (Star 
> Signal - Sky Signal) by the Noise from the sky annulus is *not* the correct 
> way to compute the SNR because it hide certain values which must be added to 
> make the statistics correct. First, there are 2 general strategies for 
> estimating the noise value (I'll call it "sigma"). One is to use the scatter 
> of values in the sky annulus and the other is to go back to theory and use 
> the readout noise, gain, and exposure time to estimate it. Both methods have 
> their supporting arguments. But in either case, you would not just divide 
> the measured or calulated sigma into the net signal to get the SNR of a 
> photometric measurement. I introduced the correct formulae in a 1991 paper, 
> here:
> http://adsabs.harvard.edu/abs/1991PASP..103..122N
> About half way down the page, click [Send PDF].
> 
> There's lots of information in that paper. I hope it helps clarify the 
> issues.
> 
> Michael
> 
> ----- Original Message ----- 
> From: "Wolfgang Renz" <w_renz at onlinehome.de>
> To: "Jim Roe" <jroe at jamesroe.com>
> Cc: "AAVSO-PHOTOMETRY" <aavso-photometry at mira.aavso.org>
> Sent: Wednesday, May 28, 2008 11:44 AM
> Subject: Re: [Aavso-photometry] CCD 'fainter-than' question
> 
>> 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]
>>
>>
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>> 
> 
> 
>