• 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}

  • 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
$b_0$ 0.0372 0.0015 0.0867 0.0031
$b_1$ -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 $z = v/c $
  • 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


  • 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,

  • 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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