Difference: 20160111MSciPhotometryAnalysisOb1part5 (1 vs. 11)

Revision 1114 Jan 2016 - ElenaCukanovaite

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"
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Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
  • To generate the positions of stars we uniformly generated x and y pixel positions.
  • To create a background we generated Poisson distributed random variables around the mean of 160, which corresponded to a typical value for images of Location 1.
  • For each position we wanted to create a Gaussian distribution that represents a star centered on the star's x,y position.
\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\} \end{equation} <br />
  • As the standard deviations should be the same for each star we decided to generate only one value for $\sigma$ for all stars. We also assumed that the standard deviations are the same in both directions so we generated only one $\sigma$ for $\sigma_x$ and $\sigma_y$.
  • To experiment we uniformly generated values for $a$ and $\rho$ for each individual star.
  • We then for each pixel in the image calculated the Gaussian value at that particular pixel for each star.
  • Resulting image for 50 stars is shown below:
monte_carlo.png
 

Comparing stars between different filters

  • In order to compare the quality of our data, we looked at the data with the largest signal to noise ratio, in this case data taken with a green filter.
  • For this, we found the largest and smallest stars, as determined by the amount of pixels which constituted them.
  • The coordinates of the stars were recorded, so they can later be compared to the same stars in other filters.
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  • Below fits can be seen for the maximum and minimum size stars for the green filter:
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  • For the fitting the backgrounds were removed, therefore the error on the no background data is the following (assuming error is Poissonian):
\begin{equation} \rm{Data \ with \ background \ removed} = \sqrt{(\Delta \rm{Original \ data})^2 + (\Delta \rm{Background})^2}\end{equation} \begin{equation} \rm{Data \ with \ background \ removed} = \sqrt{ \rm{Original \ data} + \rm{Background}}\end{equation} <br />
  • Below fits can be seen for the maximum and minimum size stars for all filters:
 
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None_1_max.png red_1_max.png
blue_1_max.png green_1_max.png
None_1_min.png red_1_min.png
blue_1_min.png green_1_min.png
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elena_no_br.png elena_r_b.png
elena_b_b.png elena_g_b.png
elena_no_d.png elena_r_d.png
elena_b_d.png elena_g_d.png
  • The plots are projections (they are summed in rows and columns), so to plot the errors the errors were summed in the following way:
\begin{equation} \rm{Error \ projection} = \sqrt{\sum_i^{Number \ of \ pixels} \Delta p_i^2} \end{equation}
  • where the $\Delta p_i$ is the error on the data point (pixel value).
 

Introductory $\chi^2$ estimation

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  • Then set about finding $\chi^2$ for the above stars for no filters and the green filter.
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  • Then set about finding $\chi^2$ for the above stars for all filters.
 
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  Visual Brightest Star Visual Dimmest Star Green Filter Brightest Star Green Filter Dimmest Star
Degrees of Freedom 53 53 53 53
$\chi^2$ 3640000 61100 1930000 67700
reduced $\chi^2$ 68600 1150 36500 1277
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  Visual Brightest Star Visual Dimmest Star Green Filter Brightest Star Green Filter Dimmest Star Red Filter Brightest Star Red Filter Dimmest Star} Blue Filter Brightest Star Blue Filter Dimmest Star
$\rm{Degrees \ of \ Freedom}$ $3593$ $3593$ $3593$ $3593$ $3593$ $3593$ $3593$ $3593$
$\chi^2$ $3250402$ $1746$ $1602258$ $2001$ $869411$ $2080$ $3172054$ $2242$
$\rm{reduced} \chi^2$ $904$ $0.486$ $445$ $0.557$ $241$ $0.579$ $882$ $0.624$
$\rho$ $0.107 \pm 0.00518$ $0.118 \pm 0.0316$ $0.0286 \pm 0.00530$ $0.106 \pm 0.0485$ $0.0631 \pm 0.00473$ $0.0715 \pm 0.0294$ $0.00456 \pm 0.00625$ $0.0423 \pm 0.0358$
$a$ $9399820\pm 43459$ $18450 \pm 441$ $9242014 \pm 42579$ $14155 \pm 522$ $12354155 \pm 51521$ $25730 \pm 573$ $12945440 \pm 70680$ $20255 \pm 547$
$x_0$ $30.1 \pm 0.0201$ $30.3 \pm 0.0867$ $30.7 \pm 0.0248$ $31.0 \pm 0.158$ $30.5 \pm 0.0195$ $30.0 \pm 0.0855$ $30.3 \pm 0.0263$ $30.4 \pm 0.108 $
$y_0$ $31.7 \pm 0.0218$ $29.9 \pm 0.0902$ $31.7 \pm 0.021$ $30.4 \pm 0.130$ $30.8 \pm 0.0204$ $30.4 \pm 0.0904$ $31.6 \pm 0.0246$ $30.1 \pm 0.104 $
$\sigma_x$ $4.18 \pm 0.0162$ $3.76 \pm 0.0870$ $5.16 \pm 0.020$ $4.51 \pm 0.160$ $4.51 \pm 0.0157$ $4.01 \pm 0.0861$ $4.66 \pm 0.0215$ $4.19 \pm 0.109 $
$\sigma_y$ $4.51 \pm 0.0178$ $3.91 \pm 0.0907$ $4.41 \pm 0.0178$ $3.71\pm 0.131$ $4.69 \pm 0.0166$ $4.25 \pm 0.0909$ $4.34 \pm 0.0205$ $4.06 \pm 0.105 $
$d$ $159 \pm 8.36$ $-4.96 \pm 0.373$ $140 \pm 8.68$ $-3.69 \pm 0.416$ $173 \pm 9.505$ $-2.30 \pm 0.448$ $255 \pm 12.2$ $1.52 \pm 0.430 $
 

  • In general, a good $\chi^2$ should be around equal to the number of degrees of freedom,

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  • In the above case, they all seem to give very poor agreement between the fit and the data, as the $\chi^2$ values are all very large in comparison to the number of degrees of freedom.

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  • From the plots it is obvious that the Poisson errors are not correct as they are too small for the brightest star.

  • Also the brightest stars are poorly fitted and therefore have high $\chi^2$ in comparison to dimmest stars.
 

Testing curve_fit for better error estimation

  • In order to better obtain errors from curve_fit (as it tends to give outlandish errors), we set out to create a Gaussian of known parameters, and then fit a general Gaussian curve to this.
 

Fractional Error When Guessing Incorectly

 Fractional Error When Guessing Exactly
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 \end{equation}


%ENDLATEX%

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Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
  • To generate the positions of stars we uniformly generated x and y pixel positions.
  • To create a background we generated Poisson distributed random variables around the mean of 160, which corresponded to a typical value for images of Location 1.
  • For each position we wanted to create a Gaussian distribution that represents a star centered on the star's x,y position.
\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\} \end{equation} <br />
  • As the standard deviations should be the same for each star we decided to generate only one value for $\sigma$ for all stars. We also assumed that the standard deviations are the same in both directions so we generated only one $\sigma$ for $\sigma_x$ and $\sigma_y$.
  • To experiment we uniformly generated values for $a$ and $\rho$ for each individual star.
  • We then for each pixel in the image calculated the Gaussian value at that particular pixel for each star.
  • Resulting image for 50 stars is shown below:
monte_carlo.png
  -- ElenaCukanovaite - 11 Jan 2016
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Revision 1014 Jan 2016 - ElenaCukanovaite

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 66 to 66
 %ENDLATEX%
  • We cropped an area of 20x20 pixels around each star and found the standard deviations within the area.
  • Histograms below are of the ratio of $\frac{\sigma_x}{\sigma_y}$ for Location 1 images:
Changed:
<
<		stan_dev_n_t.png
>
>		stan_dev_n_t.png
  • $sigma_x$ for Location 1 images:
sigma_x_loc_1.png
  • Ratio of $\frac{\sigma_x}{\sigma_y}$ for Location 2 images:
location_2.png
  • $sigma_x$ for Location 2 images:
sigma_x_loc_2.png
  • Therefore for both locations we decided to set the requirement that an objects is only a star if its ratio:
\begin{equation} 0.9 &amp;lt; \frac{\sigma_x}{\sigma_y} &amp;lt; 1.1 \end{equation} <br />
  -- ElenaCukanovaite - 11 Jan 2016
Line: 98 to 110
 
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Revision 913 Jan 2016 - DavidHadden

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 36 to 36
 
reduced $\chi^2$ 68600 1150 36500 1277

  • In general, a good $\chi^2$ should be around equal to the number of degrees of freedom,

  • In the above case, they all seem to give very poor agreement between the fit and the data, as the $\chi^2$ values are all very large in comparison to the number of degrees of freedom.

Added:
>
>

Testing curve_fit for better error estimation

  • In order to better obtain errors from curve_fit (as it tends to give outlandish errors), we set out to create a Gaussian of known parameters, and then fit a general Gaussian curve to this.
 

Fractional Error When Guessing Incorectly

 Fractional Error When Guessing Exactly
Correlation Coefficient $0.015$ 0
Amplitude $1.87 \times 10^{-10}$ 0
X Centre $0.09$ 0
Y Centre $0.01$ 0
$\sigma_x$ $1.25 \times 10^{-10}$ 0
$\sigma_y$ $1.25 \times 10^{-10}$ 0
Offset $1.66 \times 10^{-10}$ 0
 

Setting the threshold

  • We decided to use our mode image to set the threshold.
Line: 82 to 92
 
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Revision 813 Jan 2016 - ElenaCukanovaite

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 55 to 55
  %ENDLATEX%
  • We cropped an area of 20x20 pixels around each star and found the standard deviations within the area.
Changed:
<
<
>
>
  • Histograms below are of the ratio of $\frac{\sigma_x}{\sigma_y}$ for Location 1 images:
stan_dev_n_t.png
 -- ElenaCukanovaite - 11 Jan 2016

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Line: 78 to 80
 
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Added:
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Revision 713 Jan 2016 - DavidHadden

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 27 to 27
 
blue_1_max.png green_1_max.png
None_1_min.png red_1_min.png
blue_1_min.png green_1_min.png
Added:
>
>

Introductory $\chi^2$ estimation

  • Then set about finding $\chi^2$ for the above stars for no filters and the green filter.

  Visual Brightest Star Visual Dimmest Star Green Filter Brightest Star Green Filter Dimmest Star
Degrees of Freedom 53 53 53 53
$\chi^2$ 3640000 61100 1930000 67700
reduced $\chi^2$ 68600 1150 36500 1277

  • In general, a good $\chi^2$ should be around equal to the number of degrees of freedom,

  • In the above case, they all seem to give very poor agreement between the fit and the data, as the $\chi^2$ values are all very large in comparison to the number of degrees of freedom.

 

Setting the threshold

  • We decided to use our mode image to set the threshold.
  • The mode image is shown below:
Line: 67 to 77
 
META FILEATTACHMENT attachment="None_1_max.png" attr="" comment="" date="1452690774" name="None_1_max.png" path="Y:\Major Project\20151113_DH_EC\Analysis\Loaction_1\max_min_tests\None_1_max.png" size="115849" user="zxap012" version="1"
META FILEATTACHMENT attachment="None_1_min.png" attr="" comment="" date="1452690783" name="None_1_min.png" path="Y:\Major Project\20151113_DH_EC\Analysis\Loaction_1\max_min_tests\None_1_min.png" size="112431" user="zxap012" version="1"
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Added:
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Revision 613 Jan 2016 - ElenaCukanovaite

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 16 to 16
 
  • To experiment we uniformly generated values for $a$ and $\rho$ for each individual star.
  • We then for each pixel in the image calculated the Gaussian value at that particular pixel for each star.
  • Resulting image for 50 stars is shown below:
Added:
>
>		monte_carlo.png
 

Comparing stars between different filters

  • In order to compare the quality of our data, we looked at the data with the largest signal to noise ratio, in this case data taken with a green filter.
  • For this, we found the largest and smallest stars, as determined by the amount of pixels which constituted them.
  • The coordinates of the stars were recorded, so they can later be compared to the same stars in other filters.
  • Below fits can be seen for the maximum and minimum size stars for the green filter:
Changed:
<
<
None_1_max.png red_1_max.png
blue_1_max.png green_1_max.png
None_1_min.png red_1_min.png
blue_1_min.png green_1_min.png
>
>
None_1_max.png red_1_max.png
blue_1_max.png green_1_max.png
None_1_min.png red_1_min.png
blue_1_min.png green_1_min.png
 

Setting the threshold

  • We decided to use our mode image to set the threshold.
  • The mode image is shown below:
Line: 43 to 44
 \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}

%ENDLATEX%

Changed:
<
<
  • We found the following:
>
>
  • We cropped an area of 20x20 pixels around each star and found the standard deviations within the area.
 -- ElenaCukanovaite - 11 Jan 2016

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Line: 64 to 66
 
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Added:
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META FILEATTACHMENT attachment="monte_carlo.png" attr="" comment="" date="1452692332" name="monte_carlo.png" path="monte_carlo.png" size="284643" user="zxap014" version="1"

Revision 513 Jan 2016 - DavidHadden

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 21 to 21
 
  • For this, we found the largest and smallest stars, as determined by the amount of pixels which constituted them.
  • The coordinates of the stars were recorded, so they can later be compared to the same stars in other filters.
  • Below fits can be seen for the maximum and minimum size stars for the green filter:
Deleted:
<
<		green_1_max.png
 
Changed:
<
<		green_1_min.png
>
>
None_1_max.png red_1_max.png
blue_1_max.png green_1_max.png
None_1_min.png red_1_min.png
blue_1_min.png green_1_min.png
 

Setting the threshold

  • We decided to use our mode image to set the threshold.
  • The mode image is shown below:
Line: 50 to 52
 
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<
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Revision 413 Jan 2016 - ElenaCukanovaite

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 24 to 24
 green_1_max.png

green_1_min.png

Added:
>
>

Setting the threshold

  • We decided to use our mode image to set the threshold.
  • The mode image is shown below:
mode_sec.png
  • We took the Poissonian error (the square root) of each section.
  • We used the error as our threshold by setting the threshold:
\begin{equation} \rm{threshold} = 3 \times \rm{Poisson \ error}\end{equation} <br />
 

Standard deviation of the stars

  • We previously tried to show that all stars in the image have the same standard deviations.
  • To find our standard deviations we did the following:
Line: 44 to 53
 
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Revision 311 Jan 2016 - ElenaCukanovaite

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 21 to 21
 
  • For this, we found the largest and smallest stars, as determined by the amount of pixels which constituted them.
  • The coordinates of the stars were recorded, so they can later be compared to the same stars in other filters.
  • Below fits can be seen for the maximum and minimum size stars for the green filter:
Changed:
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<		green_1_max.png
>
>		green_1_max.png
 
Changed:
<
<		green_1_min.png
>
>		green_1_min.png

Standard deviation of the stars

  • We previously tried to show that all stars in the image have the same standard deviations.
  • To find our standard deviations we did the following:
%BEGINLATEX% \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}
 
Added:
>
>		\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}

%ENDLATEX%

  • We found the following:
 -- ElenaCukanovaite - 11 Jan 2016

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Revision 211 Jan 2016 - DavidHadden

Line: 1 to 1
 
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
Line: 16 to 16
 
  • To experiment we uniformly generated values for $a$ and $\rho$ for each individual star.
  • We then for each pixel in the image calculated the Gaussian value at that particular pixel for each star.
  • Resulting image for 50 stars is shown below:
Added:
>
>

Comparing stars between different filters

  • In order to compare the quality of our data, we looked at the data with the largest signal to noise ratio, in this case data taken with a green filter.
  • For this, we found the largest and smallest stars, as determined by the amount of pixels which constituted them.
  • The coordinates of the stars were recorded, so they can later be compared to the same stars in other filters.
  • Below fits can be seen for the maximum and minimum size stars for the green filter:
green_1_max.png

green_1_min.png

 -- ElenaCukanovaite - 11 Jan 2016

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Revision 111 Jan 2016 - ElenaCukanovaite

Line: 1 to 1
Added:
>
>
META TOPICPARENT name="StewartBoogertPhotometry2015"

Monte Carlo simulation of a 2D star field

  • We have done some prelimenary work on the simulation.
  • To generate the positions of stars we uniformly generated x and y pixel positions.
  • To create a background we generated Poisson distributed random variables around the mean of 160, which corresponded to a typical value for images of Location 1.
  • For each position we wanted to create a Gaussian distribution that represents a star centered on the star's x,y position.
\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\} \end{equation} <br />
  • As the standard deviations should be the same for each star we decided to generate only one value for $\sigma$ for all stars. We also assumed that the standard deviations are the same in both directions so we generated only one $\sigma$ for $\sigma_x$ and $\sigma_y$.
  • To experiment we uniformly generated values for $a$ and $\rho$ for each individual star.
  • We then for each pixel in the image calculated the Gaussian value at that particular pixel for each star.
  • Resulting image for 50 stars is shown below:
-- ElenaCukanovaite - 11 Jan 2016

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