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META TOPICPARENT |
name="StewartBoogertPhotometry2015" |
Theory |
| The gain measures the amount of photoelectrons you get per ADU.
Since we have N_ADU and we can determine B and find RN from documentations, we should be able to use this to find the actual amount of photon events that occured, and see if the standard deviations from fitting the actual number of photon events still deviates from Poissonian. |
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> > | Last week's analysis
The plot below shows what we did last week, please see the corresponding section.
The plot shows the difference between the standard deviation found from the fit and the standard deviation found from Poisson such that standard deviation is square root of the mean. The negative deviation at short exposure times could be due to a constant offset between N_ADU and N_photoelectrons, whereas higher exposure deviation could be due to electric noise. We wanted to see if we account for these using the formula above and this is done below. |
| Analysis
The bias level was found to be 107.456260541. The read out noise was found from documentations to be 15e- rms for our CCD. |
| This plot for this is not good.This would mean that maybe our assumption of the effects on the readout of ADU is not exactly correct. As an alternative, maybe Poisson assumption is not valid for high exposure times, and this would give the larger difference. We know that if one takes the total amount of data to infinite, the Poisson distribution turns Gaussian.
Some of the corrected means were negative, due to the fact that we took away constant B and RN. In order to compute the Poisson standard deviation, we chose to take the absolute magnitude of the mean. |
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> > | Bias only
If the number of ADUs is corrected only for the bias effects, then the following plot is made.
The plot is not good in comparison to the original plot. We need to explore this more.
Read noise only
If the number of ADUs is corrected only for the read noise effects, then the following plot is made.
This plot is more like the one we found before, as well as the difference is only around 0.5 pixels. However, it is negative, which means that the standard deviation found from fitting is smaller than the one defined by the Poisson distribution. |
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-- ElenaCukanovaite - 27 Oct 2015
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> > |
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