### Initial histogram

The dark frames were analysed by plotting a histogram of all the photoelectron values. However, specifying too many bins will mean that some of the bins inside the peak will have values equal to 0, since the bin width is so small. This can be seen in the figure below. This deviates the Gaussian from the curve. To fix this the values of 0 inside the peak were set to nan. The peak was defined such that it starts above 1000 bin counts. The plot without the zeros in the peak can be seen below. As can be seen the distribution can be described well by the Gaussian distribution. This was done for the various exposure time images. What we wanted to compare was the sigma found from the fitting procedure (sigma_fitting) and the sigma as derived from Poission distribution, i.e. sigma_poisson = square root of the mean pixel value given from the fit. The plot below shows the difference, sigma_fitting - sigma_poisson = delta(sigma). Therefore, as the exposure time increases the background signal deviates from Poissonian. To check this, the mean from fitting was plotted as a function of exposure time below. The mean also increases with the exposure time in approximately linear fashion. This could explain why the delta(sigma) increases with exposure time, because as the mean increases so does the numerical value of its standard deviation, therefore allowing more deviation between values.

-- ElenaCukanovaite - 20 Oct 2015

Topic attachments
I Attachment History Action Size Date Who Comment png difference_between_sigmas.png r1 manage 26.5 K 20 Oct 2015 - 20:28 ElenaCukanovaite png example.png r1 manage 35.1 K 20 Oct 2015 - 20:28 ElenaCukanovaite png mean_as_function_of_time.png r1 manage 27.2 K 20 Oct 2015 - 20:28 ElenaCukanovaite png no_zeros_histogram.png r1 manage 37.1 K 20 Oct 2015 - 20:28 ElenaCukanovaite

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