Standard Deviation vs Focus/Angle plots

  • Normalised $N_{\mathrm{pe}}$ into $N_{\mathrm{pe}}$ per pixel number per exposure time and replotted graphs of $\sigma$ vs focus distance/camera angle.
  • Converted $\sigma$ from units of pixel number into mm using dimensions of CCD: 765 x 510, 6.9mm x 4.6mm
  • Standard Deviation vs Focus Distance:
FOCUS.png

Plotted Hyperbole to above plot of functional form:

 \begin{equation} f(x; A,B,C) = A\sqrt{1+((x-C)^{2} / B^{2})} \end{equation}<br />

Curve_fit parameters:


Parameter Values
A 0.0253 1.56 x 10^-5
B -4.19 4.60 x 10^-3
C 9.03 3.23 x 10^-3

Gave value $\chi^{2}$ = 668.1

Best focus distance = 19.9mm

  • Standard Deviation vs Camera Angle:
ANGLE.png

Plotted sine function to above plot of functional form:

 \begin{equation} f(x; A, f, p, back) = Asin(f(x+p))+back \end{equation} <br />

Curve_fit parameters:

Parameter Value
Amplitude, A -0.0152 1.74 x 10^-4
Frequency, f 0.911 6.46 x 10^-3
Phase, p -2.25 1.22 x 10^-2
Background, back 0.0366 1.78 x 10^-4

Gives value $\chi^{2}$ = 73.4

Best camera angle = 3.95mm


Fitting the spectral line

Gaussian Fit

First Gaussian fit was fitted to the speactral line, this was done using the gaussian equation

 \begin{equation} f(x; A,\mu,\sigma) = A \frac{1}{\sqrt{2 \pi \sigma^{2}}} e^{-\frac{(x - \mu)^{2}}{2 \sigma^2}} +b\end{equation}

This shows a $\chi^2$ value of

$\chi^2 = 12905$

and the parameters were

Parameter value
Amplitude, A $499636 \pm 6901$
mean, $\mu$ $32.15 \pm 0.04$

width, $\sigma$

$2.574 \pm 0.038$
background, b $2665 \pm 223$

The Gaussian was then convoluted with a Box plot which has the equation

  \begin{equation} f(x; A,a,b) =  A \quad : \quad a&amp;lt;x&amp;lt;b \end{equation}  \begin{equation} f(x; A,a,b) = 0 \quad : \quad {\rm{otherwise}} \end{equation}

This gave a value of $\chi^2$ as

$\chi^2 = 1589$.

The fit parameters were

Parameter Value
Gaussian Amplitude, amp 48.28
Gaussian mean $\mu$ 34.11
Gaussian $\sigma$ 1.203
Box Ampitude, A 1424
Box start, a 29
Box end, b 35
Background, B 2908

All of the covariance matrix elements were infinity?

As the fibres in the spectrometer are circular part of the instrumental line width could be due to this. Therefore the histogram of a uniformly illuminated disk was plotted,

This has 30 bins with 100000 entries. This gave a distribution of a uniform circle, which look roughly semicircular therefore a semicircular distribution was fitted, where

 \begin{equation} f(x; A,r,\mu) = A\frac{2 \sqrt{r^2-(x - \mu)^{2}}}{\pi r^2}. \end{equation}

This gave a $\chi^2$ of

$\chi^2 = 21.39$ and fit parameters of

Parameters Value
Radius, r $14.86 \pm 0.058$
Mean $\mu$ $14.49 \pm 0.012$
Amplitude, A $198473 \pm 4757$
Background, b $34 \pm 155$

This semicircular distribution was then convoluted with the Gaussian distribution and fitted to the spectral line.

This gave a $\chi^2$ of

$\chi^2 = 869$

The fit parameters were

Parameters Value
Gaussian Amplitude, amp $978.8$
Gaussian and Semicircle Mean, $\mu$ $33.031 \pm 0.001$
Gaussian $\sigma$ $1.09 \pm 0.01 $
Semicircle radius, r $4.053 \pm 0.004$
Semicircle Amplitude, A $495$
Background, b $2909 \pm 7$

Ommiting Hot Pixels

To find the positions of the hot pixels the plot was 'differentiated' by subracting the data rolled by one position. i.e

 \begin{equation} \frac{dy}{dx} = \frac{y_{2}-y_{1}}{\delta x} \end{equation}

The plot of this can be seen below for the background summed onto the x-axis:

And also summed on to the y-axis:

By choosing the hot pixels as values greater than one $\sigma$ above the mean and that occurred in more than one frame, the hot pixels positions could be determined.

For the data summed onto the x-axis the hot pixels were found to be:

Frame 1 Frame 2 Frame 3
90 90 90
    179
180 180 180
222 222 222
4077 407 407
  490  
497 497 497
617 617 617

Therefore the hot pixels are the values which fall in all of the frames.

For data summed onto the y-axis the hot pixels were found to be:

Frame 1 Frame 2 Frame 3
24    
  52 52
    61
84 84 84
  90 90
98 98  
  132  
197 197 197
216 216 216
  223 223
273 273 273
    286
312 312 312
348 348  
354 354 354
  381 381
417 417  
  440 440

Again the pixel numbers in all three frames correspond to pixel values.

By then setting the pixel value of these hot pixels to the average of the pixels surrounding it, the hot pixels were emmited. The pixels were average over the 10 surrounding pixels. This is shown for the x-axis projection:

and for the y-axis projection:

In both of these plots the intensity of the background can be seen to rise towards one side, this is due to the electronics in that corner of the CCD, by taking a bias frame of 0s exposure this can be accounted for.

-- AshleaKemp - 08 Nov 2015

Topic attachments
I Attachment Action Size Date Who Comment
PNGpng ANGLE.png manage 37.0 K 10 Nov 2015 - 14:11 AshleaKemp sigma vs angle
PNGpng FOCUS.png manage 36.7 K 10 Nov 2015 - 14:09 AshleaKemp sigma vs focus

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Topic revision: r8 - 02 Feb 2016 - AshleaKemp

 
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