-- AriannaSaba - 11 Dec 2018


I investigated how the background value changes across a real image with stars in it.

The first approach I used was to keep the row numbers fixed and select 6 adjacent boxes:

box1 = image[0:251,0:251]
box2 = image[0:251,252:503]
box3 = image[0:251,504:755]
box4 = image[0:251,756:1007]
box5 = image[0:251,1008:1259]
box6 = image[0:251,1260:1530]

with size 251x251. However, the last box had a bigger y size (270) because of the shape of the image.

I found the median of each box with np.median:

median values = [937.0, 937.0, 937.0, 938.0, 939.0, 944.0]

and I plotted them against the central pixel number for each box. The result obtained suggested an exponential growth, so I fitted the data with an exponential curve getting the following:


The other approach I tried was to first define a 251x251 box. Then I moved the box one column at time up to the point where the end of the box matched the end of the frame.

At first, the algorithm was taking into account also the boxes with size smaller than 251x251, because as we approach the end of the frame there are not 251 columns for each box anymore, but their number will decrease until no columns are left in the count. I managed to discard those boxes and to take into account only the 251x251 boxes.

For each of those boxes I calculated the median with np.median and took the central pixel i index. This is what I got:


Now, I am not sure if the x axis should contain the index of the central pixel in the box or the first ith value in the box, such that x starts from 0.



JC: Have also been investigating background. Specifically, I have looked at background values as a function of boxsize (of both the 4 boxes surrounding each star and of the 1000 randomly distributed boxes used to determine the threshold):

- For the boxes used to determine the threshold, I investigated background vs iteration of algorithm for 3 different box sizes due to the random nature of the algorithm.


- For larger boxes, the standard deviation of the mean background value averaged over number of iterations of the algorithm is observed to decrease. This is due to the fact that larger boxes are more homogeneous on average. Despite this, larger boxes are more likely to include a star(s) and hence the average background value is greater.

-On the other hand, smaller boxes are less likely to include sources however, if one does include a source, it will result in a larger average background. This is seen in the plot above; the smallest box size has the largest standard deviation but the lowest average background value.

-Subsequently, for the 4 corner boxes per star used to determine source-specific background values, I investigated the median of the summed backgrounds as a function of box half width:back_v_n_plot.png

-Taking a closer look!:


- Observe that the smallest box sizes are more strongly influenced by neighbour starlight.

Topic attachments
I Attachment History Action Size Date Who Comment
PNGpng Schermata_2018-12-13_alle_15.08.11.png r1 manage 144.2 K 13 Dec 2018 - 15:10 AriannaSaba  
PNGpng Schermata_2018-12-13_alle_15.08.37.png r1 manage 158.1 K 13 Dec 2018 - 15:12 AriannaSaba  
PDFpdf Sigma_Background_estimation.pdf r1 manage 416.4 K 11 Dec 2018 - 16:02 AriannaSaba  
PNGpng back_v_i_plot.png r1 manage 53.9 K 13 Dec 2018 - 14:12 JacobChambers  
PNGpng back_v_n_plot.png r1 manage 52.8 K 13 Dec 2018 - 14:25 JacobChambers  
PNGpng back_v_n_plot2.png r1 manage 84.0 K 13 Dec 2018 - 14:25 JacobChambers  
PNGpng backdatadelete.png r1 manage 36.8 K 13 Dec 2018 - 15:03 JacobChambers  
PNGpng backgrounds.png r1 manage 29.3 K 13 Dec 2018 - 14:38 JacobChambers  
PNGpng plot1.png r1 manage 24.3 K 11 Dec 2018 - 15:49 AriannaSaba  
PNGpng plot2.png r1 manage 25.7 K 11 Dec 2018 - 15:58 AriannaSaba  
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Topic revision: r4 - 13 Dec 2018 - AriannaSaba

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