3C273

We initially took images of the quasar to locate it, the transmission grating was then places in front of the CCD to obtain its spectrum.
As 3c273 is an object with a magnitude of 12.9, many frames were taken to try and increase the signal to noise.
A total of 30 4x4 binned frames were taken with an exposure of 120s
This is a single frame of 3c273 and its surrounding stars,
3c273_Spec_1_Image.jpg.png:
This is then a stacked image of 30 frames with 3c273 highlighted:
Full_Bin2_Image.jpg.png:

The area highlighted on the plot was sliced from the array, and an area just above was also sliced to be used as the background.
The plot below shows the initial slice without the zeroth order star, the background slice and the slice with the background subtracted,
Slices.png:
The final slice was the projected onto the x axis:

The first image shows 10 frames stacked:
3c273_2.png:
The second image shows all 30 frames stacked:
3c273_1.png:
This plot then shows 30 of the slices stacked in python,
PythonStack.png:

To then calibrate the wavelength, a reference spectrum was used from the star Denebola

Deneb_Star.png:

The zeroth order star was then fit with a Gaussian to find the centre.

$\begin{equation} G(x;V) = \frac{1}{\sqrt{2\pi \sigma^2}} e^{\frac{-(x - \mu)^2}{2 \sigma^2}} \end{equation}$

GaussianFit.png:

This was repeated for the reference stars Denebola and Aldebaran and 3c273, the data was then sliced so that the centre of the star (Gaussian) was at zero, and plotted,

two_Spectrums_Positions.png:

To then map the pixel position to the wavelength the reference of a line within Denebola was used,
The $H_{\beta}$ line was used which has a wavelength of 486.1 nm,
To find the position of the line in Denebola it was fit with a voigtian,
The line was found to fall at the pixel value of 73, however the pixel value needed to be converted into mm.
The size of a pixel was 0.009 mm however as we used 4x4 binning this was then 0.036 mm.
The slit density of the grating was set at 200 however this needs to be checked,
The grating equation was used to find the distance of the grating from the CCD,

$\begin{equation} \lambda k n = \sin{(\theta_{m})}, \end{equation}$

where k is the order = 1, m is the slit density and $\theta_m$ is the angle of diffraction.

The angle $\theta_m$ was related to the distance of the grating from the CCD by,

$\begin{equation} \tan{(\theta_m)} = \frac{0.036p}{l}, \end{equation}$

where p is the pixel position and l is the distance of the grating from the CCD.
from these l = 26.9 mm

This could then be used to map the wavelength position from the pixel number,

$\begin{equation} \lambda = \frac{1}{n} \sin{\left[ \tan^{-1}{\left( \frac{0.036p}{l}\right)}\right]}, \end{equation}$

This then gave a plot of

two_Spectrums.png:

the spectral line of 3c273 spectrum were then fit with a Voigtian,

Hbeta.png:

Halpha.png:

Which gave the parameters:

Parameter	$H_{\beta}$	$H_{\alpha}$
Amp	0.924 ± 0.1201	1.065 ± 0.6452
$\mu$	565.89 ± 0.0295	753.456± 1.7232
$\sigma$	0.5508 ± 5.4785	0.0219 ± 6.9023
$\gamma$	20.3071 ± 4.1472	19.6402 ± 5.9619
	0.0372 ± 0.0015	0.0867 ± 0.0031
	-0.0003 ± 0.0	0.0001 ± 0.0
b_2	0.0 ± 0.0	-0.0 ± 0.0

The important parameter was the mean, $\mu$ , which give the wavelength of the line,
This could then be compared to the emitted wavelength of the line to find the redshift: * The redshift was found using:

$\begin{equation} z = \frac{\lambda_{\rm{obs}} - \lambda_{\rm{em}}}{\lambda_{\rm{em}}} \end{equation}$

The velocity was found using
And the distance with $v = H_{0}d$

Parameter	$H_{\beta}$	$H_{\alpha}$
Observed	565.89 ± 0.0295	753.456± 1.7232
Emitted	486.1	656.3
z	0.16	0.148
v [ms^-1]	4.80*10^7	4.44*10^7
d [Mpc]	707	654
real z	0.158	0.158
real d	750	750

Vega

The width of the Gaussian from the fit was plotted against the number of points within the convolution:
where

Gaussian:

$\begin{equation} G(\lambda;\lambda_0,\sigma) = \frac{1}{\sqrt{2\pi} \sigma^2} e^{\frac{-(\lambda - \lambda_0)^2}{2 \sigma^2}} \end{equation}$

Lorentz:

$\begin{equation} L(\lambda;\gamma) = \frac{1}{\pi} \frac{\gamma/2}{(\lambda)^2 + (\gamma / 2)^2} \end{equation}$

Circle:

$\begin{equation} C(\lambda; r) = \frac{2 \sqrt{r^2-(\lambda)^{2}}}{\pi r^2}. \end{equation}$

Background:

$\begin{equation} B(\lambda) = b_{0}+b_{1}(\lambda-\lambda_0)+b_{2}(\lambda-\lambda_0)^{2} \end{equation}$

$\begin{equation}f(\lambda;A,\lambda_0,\sigma,\gamma,b_0,b_1,b_2) = A((G \mathop{*} L) \mathop{*} C)(\lambda) + B(\lambda)\end{equation}$

Discrete Convolution:

$\begin{equation} (G \mathop{*} L \mathop{*}C) (\lambda) = \sum_j{\sum\limits_{i}{G(\lambda-j,\lambda_0,\sigma)L(j-i,\gamma)C(i,r)}} \end{equation}$

sig_vs_its.png:

Our spectrum was attempted to be normalised to the elodie spectrum by fitting an exponential to the elodie spectrum and setting the background of our data to that,

VegaElodie.png:

The exponential was not a great fit so not sure how else to do this.

-- JosephBayley - 18 Feb 2016

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Topic revision: r2 - 19 Feb 2016 - JosephBayley

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