Mode Substraction

  • We substracted the split mode image from the original image.
  • If one performs mode split on the resulting image, one gets 0 for the mode everywhere, since the most common element has been substracted from itself and therefore 0 becomes the most common element.
mde.png
  • We substracted the mode from the original image and set the threshold above 100 photoelectrons to find 64 stars:
more_stars.png

Further algorithm improvements: shape and size of a star candidate

  • Previously saw that some possible star candidates were definetly not stars, appearing very angular and non-circular, likely bled hot pixels
  • In order to attempt to fix this, we imposed a limit on the 'eccentricity' of the objects, i.e. the ratios of their widths.
\begin{equation} \frac{\sigma_x}{\sigma_y} > 1.4 \end{equation} \begin{equation} \frac{\sigma_y}{\sigma_x} > 1.4 \end{equation}
  • Having imposed this limit, several suspect stars were removed from the image.
  • In addition to the above, we also increased the threshold for an objec to considered a star to amount of pixels = 20.
  • Below can be seen the results.
final_removal.png
  • Where the red dots now show stars which were removed from the image.
  • This appears to remove several problem candidates, such as those seen below.
removed_edge.png

  • Here you can see that stars appearing half on the frame, which we would not be able to fit properly, will not be counted now, as they do not appear circular.
  • Below it can be seen that several apparent 'hot pixels' have been removed, however a concern is objects which could be stars (such as the object on the right) may also be removed.
removed_ratio_1_3.png

Standard deviations of stars

  • We known that the standard deviations of the stars in the image should be the same in x and y directions.
  • We defined our standard deviations as:
\begin{equation} \sigma_x = \sqrt{\frac{\sum_{i = 1}^N (x_i - x_{\rm center})^2 p_i}{\sum_{i=1}^N p_i}}\end{equation} \begin{equation} \sigma_y = \sqrt{\frac{\sum_{i = 1}^N (y_i - y_{\rm center})^2p_i}{\sum_{i=1}^N p_i}}\end{equation} <br />
  • The plots show the histrograms of the standard deviations in x and y directions for the image:
x.png

y.png

Preliminary fitting

  • We started by fitting standard multivariate Gaussian distributions to the identified stars. For this we took areas 60 by 60 pixels around the identified centres of the stars and projected into the x and y, then fitted the following:
\begin{equation} f(x,y,\mu_x,\mu_y,\sigma_x,\sigma_y,\rho,a,d) = \frac{a}{2\pi\sigma_x\sigma_y\sqrt{(1-\rho^2)}} exp\Bigg\{-\frac{1}{2(1-\rho^2)}\Bigg[\bigg(\frac{x-\mu_x}{\sigma_x}\bigg)^2 + \bigg(\frac{y-\mu_y}{\sigma_y}\bigg)^2 -2\rho\bigg(\frac{x-\mu_x}{\sigma_x}\bigg)\bigg(\frac{y-\mu_y}{\sigma_y}\bigg)\Bigg]\Bigg\} + d \end{equation}

b_r.png

  • The above graph is an example of a fitted star for summed rows. This appeared to be a good fit, although a slight skew appears to exist in this data, so we modified our fitting function in order to account for this, by changing the y term in the formula to a polynomial in y. (visible skew in y, not so much in x, left it for now).
\begin{equation} y \rightarrow y_0 + y_1y + y_2y^2 \end{equation}
  • This was then performed, and a comparison can be seen below:
skewed_example.png
  • From this, one can see that this seems to follow the points better, but not perfectly, and highlights the skew of the curve.
  • It appears however that most stars summed columnly are not skewed as can be seen below:
equal.png
  • However, not all fits were so successful.

Fitting problems

  • A recurring issue seen in our projections is the presence of more than 1 star in the cropped image, which obviously does not fit properly.
  • This is currently a major issue for such scenarios, which must be resolved.
bad.png

-- ElenaCukanovaite - 08 Dec 2015

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PNGpng b_r.png r1 manage 66.8 K 08 Dec 2015 - 16:07 ElenaCukanovaite  
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Topic revision: r13 - 11 Jan 2016 - ElenaCukanovaite

 
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