Difference: 20160201MSciPhotometryAnalysisOb1part8 (1 vs. 6)

Revision 604 Feb 2016 - ElenaCukanovaite

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

Checking fit validity

  • p values give a probability, so should be normalised to 1.
Line: 46 to 46
 
724_479_hist.png 1328_235_hist.png
  • As one would expect, the bigger the $\chi^2$, the further away the histogram is from a normalised Gaussian.

Parameter errors generated with Monte Carlo

Changed:
<
<
  • Generated 40 images (100x100 pixels) of the same star.
>
>
  • Generated 40 images (100x100 pixels) of the same BRIGHT star.
 
  • As before generated a star according to:
%BEGINLATEX%
Line: 59 to 59
 
  • Added a background of 160 pixel values.
  • Then for each pixel, its pixel value was taken as the mean value of the Poisson distribution and a new pixel value was then generated according to that distribution for the given pixel.
  • The 40 images were fitted using our maximum likelihood fitting algorithm.
Added:
>
>
bright.png
 
  • For each parameter fitted the following was calculated:
%BEGINLATEX%
Line: 90 to 91
 
$\sigma_y$ $6.97281 \pm 0.0243$
$x_0$ $30.008308 \pm 0.0256$
$y_0$ $29.998781 \pm 0.0215$
Added:
>
>
  • Generated dimmer star to see how errors scaled:
dim.png
  • For a dim star, fractional errors:
$\alpha$ $\pm 0.484$
$\beta_1$ $\pm 0.0949$
$\beta_2$ $\pm 0.245$
$\rho$ $\pm 0.113$
$\sigma_x$ $\pm 0.0374$
$\sigma_y$ $\pm 0.0604$
$x_0$ $\pm 0.0108$
$y_0$ $\pm 0.0121$
  • An example with fitted params:
$\alpha$ $0.038647 \pm 0.0187$
$\beta_1$ $0.868236 \pm 0.0824$
$\beta_2$ $-1.137944 \pm -0.279$
$\rho$ $-0.200749 \pm -0.0138$
$\sigma_x$ $4.297991 \pm 0.1609$
$\sigma_y$ $5.313384 \pm 0.321$
$x_0$ $29.971642 \pm 0.324$
$y_0$ $29.935687 \pm 0.363$
 

Observation 2

  • Took a series of bias, dark and flat frames to check for variabilities and to figure out how long it took the cooling ring to disappear.
  • Here is the progression of the cooling ring:
Line: 176 to 197
 
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Revision 504 Feb 2016 - ElenaCukanovaite

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

Checking fit validity

  • p values give a probability, so should be normalised to 1.
Line: 92 to 92
 
$y_0$ $29.998781 \pm 0.0215$

Observation 2

  • Took a series of bias, dark and flat frames to check for variabilities and to figure out how long it took the cooling ring to disappear.
Added:
>
>
  • Here is the progression of the cooling ring:
1_ring.png 3.png
5.png
  • The ring took 1hr50mins to dissapear.
 
  • Took a selection of master frames for later analysis porpoises.
Added:
>
>
  • We had a major problem, for some reason the average pixel value for the bias frame, the dark frame exposure time 30s and 60s were all around 900.
bias_1.png bias_100.png
master_bias.png

master_dark_30s.png

master_flat.png

  • We tried some initial removal of bias and flat frames:
  • Original:
original.png original_bg.png
  • Bias master frame removed (substracted bias):

bias_removed.png

bias_removed_bg.png
  • Master flat removed (divided by master flat):
flat_removed.png flat_removed_bg.png
 -- DavidHadden - 01 Feb 2016
$\sigma_x$
Line: 143 to 162
 
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Revision 404 Feb 2016 - ElenaCukanovaite

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

Checking fit validity

  • p values give a probability, so should be normalised to 1.
Line: 45 to 45
 
118_219_hist.png 972_394_hits.png
724_479_hist.png 1328_235_hist.png
  • As one would expect, the bigger the $\chi^2$, the further away the histogram is from a normalised Gaussian.
Added:
>
>

Parameter errors generated with Monte Carlo

  • Generated 40 images (100x100 pixels) of the same star.
  • As before generated a star according to:
  \begin{equation} f_G(x,y,\mu_x,\mu_y,\sigma_x,\sigma_y,\rho, \beta_1, \beta_2) = \frac{1}{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\} \times \Bigg[1 + \rm{erf} \frac{\beta_1 x + \beta_2 y}{\sqrt{2}}\Bigg]  \end{equation}
  • Therefore, each pixel had a distinct pixel value.
  • Added a background of 160 pixel values.
  • Then for each pixel, its pixel value was taken as the mean value of the Poisson distribution and a new pixel value was then generated according to that distribution for the given pixel.
  • The 40 images were fitted using our maximum likelihood fitting algorithm.
  • For each parameter fitted the following was calculated:
  \begin{equation} \rm{fractional \ error \ on \ parameter} = \frac{\rm{fitted \ parameter} - \rm{generated \ parameter})}{\rm{fitted \ parameter}}  \end{equation}
  • Bellow are the histograms for the above calculation:

alpha_err.png beta1_err.png
rho_err.png sigmax_err.png
  • Decided to take the error on the parameter as the difference between the biggest and smalles fractional errors.
  • Here is the fractional errors:
$\alpha$ $\pm 0.000738$
$\beta_1$ $\pm 0.0217$
$\beta_2$ $\pm 0.00842$
$\rho$ $\pm 0.113$
$\sigma_x$ $\pm 0.00349$
$\sigma_y$ $\pm 0.000738$
$x_0$ $\pm 0.000853$
$y_0$ $\pm 0.000717$
  • An example with fitted parameters:
$\alpha$ $0.921682 \pm 0.000680$
$\beta_1$ $0.400025 \pm 0.00866$
$\beta_2$ $-1.519446 \pm -0.0127$
$\rho$ $-0.043813 \pm -0.00493$
$\sigma_x$ $4.368815 \pm 0.00818$
$\sigma_y$ $6.97281 \pm 0.0243$
$x_0$ $30.008308 \pm 0.0256$
$y_0$ $29.998781 \pm 0.0215$
 

Observation 2

  • Took a series of bias, dark and flat frames to check for variabilities and to figure out how long it took the cooling ring to disappear.
  • Took a selection of master frames for later analysis porpoises.
-- DavidHadden - 01 Feb 2016
Added:
>
>
$\sigma_x$
 
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Revision 304 Feb 2016 - ElenaCukanovaite

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

Checking fit validity

  • p values give a probability, so should be normalised to 1.
Line: 4 to 4
 
  • p values give a probability, so should be normalised to 1.
  • Found p value for each pixel, and summed over all pixels
  • Found summed values as: 0.999999876234048 $ \approx $ 1
Deleted:
<
<
 
  • Found previously around 80% signal and 20% background,
  • Performed back of the envelope calculations to validate this, found s = 80%, b = 20% as was fitted.
Added:
>
>

Skewed Gaussian

  • Since our stars looked skewed, decided to fit skewed Gaussians.
  • The new total function fitted, $f$:
%BEGINLATEX%

\begin{equation} f_G(x,y,\mu_x,\mu_y,\sigma_x,\sigma_y,\rho, \beta_1, \beta_2) = \frac{1}{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\} \times \Bigg[1 + \rm{erf} \frac{\beta_1 x + \beta_2 y}{\sqrt{2}}\Bigg]

\end{equation}

\begin{equation} f_{bg}(N,M) = \frac{1}{N \times M}

\end{equation}

\begin{equation} f(\alpha) = \alpha f_G + (1 - \alpha) f_{bg}

 
Added:
>
>
\end{equation}


%ENDLATEX%

  • Below are examples of fitted stars from one of our images:
1124_513.png 630_844.png
118_219.png 972_394.png
724_479.png 1328_235.png
  • Well fitted stars seemed to have $\chi^2$ at around 1.
  • Therefore, we could set our threshold for the boundary between 1 star and 2 stars as $\chi^2 = 2$.
  • It appears that the most bright stars have been fitted terribly and their $\chi^2$ are even bigger than 2. It seems that although the tails fit nicely, the fit cannot reach the required amplitude.
  • Below are histograms for the same stars in the same order of
  \begin{equation} \rm{x \ axis} = \frac{\rm{data \ value} - \rm{fitted \ data \ value}}{\sqrt{\rm{fitted \ data \ value}}}  \end{equation}  <br />
  • This should have produced a Gaussian with mean 0 and standard deviation of 1.

1124_513_hist.png 630_844_hist.png
118_219_hist.png 972_394_hits.png
724_479_hist.png 1328_235_hist.png
  • As one would expect, the bigger the $\chi^2$, the further away the histogram is from a normalised Gaussian.
 

Observation 2

  • Took a series of bias, dark and flat frames to check for variabilities and to figure out how long it took the cooling ring to disappear.
  • Took a selection of master frames for later analysis porpoises.
Line: 15 to 51
 -- DavidHadden - 01 Feb 2016

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Revision 203 Feb 2016 - DavidHadden

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

Checking fit validity

  • p values give a probability, so should be normalised to 1.
Line: 7 to 7
 
  • Found previously around 80% signal and 20% background,
  • Performed back of the envelope calculations to validate this, found s = 80%, b = 20% as was fitted.
Added:
>
>

Observation 2

  • Took a series of bias, dark and flat frames to check for variabilities and to figure out how long it took the cooling ring to disappear.
  • Took a selection of master frames for later analysis porpoises.
 -- DavidHadden - 01 Feb 2016

META FILEATTACHMENT attachment="latexf67d9f51527949b615d8ef71ed72d610.png" attr="h" comment="" date="1454335148" name="latexf67d9f51527949b615d8ef71ed72d610.png" user="zxap012" version="1"

Revision 101 Feb 2016 - DavidHadden

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

Checking fit validity

  • p values give a probability, so should be normalised to 1.
  • Found p value for each pixel, and summed over all pixels
  • Found summed values as: 0.999999876234048 $ \approx $ 1

  • Found previously around 80% signal and 20% background,
  • Performed back of the envelope calculations to validate this, found s = 80%, b = 20% as was fitted.
-- DavidHadden - 01 Feb 2016

META FILEATTACHMENT attachment="latexf67d9f51527949b615d8ef71ed72d610.png" attr="h" comment="" date="1454335148" name="latexf67d9f51527949b615d8ef71ed72d610.png" user="zxap012" version="1"
 
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