Gridded convolution

An example grid over a beam     [l] The aim of the output routine is to simulate what a given telescope would see if it were pointed at the cloud. The beamsize depends on the diameter of the telescope ($D$) and is given by ${\rm FWHM}=\frac{1.22 \lambda}{D}$ where $\lambda$ is the wavelength of the transition. Since all real telescopes have a non-point like beam it is necessary to simulate the beam. This is done by placing a grid of points over the beam and calculating the emission at each of those points. These values are then added together with an appropriate weighting. This is demonstrated in figure 4.23, where the circle has a radius equal to the beam's FWHM (ie. is two beamwidths in diameter). The weighting factor for each position is derived by assuming the beam is Gaussian (ie. follows $e^{-d^2}$, where $d$ is the distance from the centre of the beam). Figure 4.24 shows the Gaussian used to model the beam. The input files give the value for $\theta_{\rm FWHM}$ (ie. the width of the beam at half the maximum, for a normalised beam half the maximum is $\frac{1}{2}$). However, the beam shape is defined in terms of $\theta_0$ (which is the half-width at $\frac{1}{e}$ of the power) as
The beam model   [r]

$$\omega=e^{-\frac{\theta^2}{\theta_0^2}}$$ (4.53)

where $w$ is the weighting to be applied to a position a distance $\theta$ from the centre of the beam. It is therefore necessary to convert between $\theta_0$ and $\theta_{\rm FWHM}$. With reference to figure 4.24 it can be seen that $e^{-\frac{\left(\frac{\theta_{\rm FWHM}}{2}\right)^2}{\theta_0^2}}=\frac{1}{2}$ which can be re-arranged to yield

$$\theta_{\rm FWHM}=2 \times \sqrt{\ln 2} \times \theta_0$$

which in the program appears as $\theta_{\rm FWHM}=1.665 \times \theta_0$.

The unit of each step in the grid (ie. the gap between the dots in figure 4.23) is given by $s=\frac{2 \times \theta_{\rm FWHM}}{n_{\rm odd}-1}$ where $n_{\rm odd}$ are the number of gridding points to be used ( $n_{\rm odd}$ must be an odd integer). The factor of 2 is because the radius of the circle was taken to be equal to the FWHM of the beam, however, $n_{\rm odd}$ covers the diameter of the circle. So the distance from the centre of the beam is $\theta=\sqrt{(xs)^2+(ys)^2}$ (where $xs$ and $ys$ are the $x$ and $y$ co-ordinates of the point of interest in say, arcseconds). This can be re-written as $\theta=\frac{2 \times \sqrt{x^2+y^2} \hspace{1mm} \theta_{\rm FWHM}}{n_{\rm odd}-1}$, where the $x$ and $y$are the $x$ and $y$ co-ordinates of the point of interest but this time in units of $s$. Therefore with the inclusion of the factor 1.665 to convert $\theta_{\rm FWHM}$ to $\theta_0$ this can be substituted into equation 4.53 to give the weighting for each position as

$$\omega=e^{-\frac{\left(x^2+y^2\right) \times 1.665^2 \times 4}{\left( n_{\rm odd}-1 \right)^2}}$$ (4.54)

Thus, as an example, the position in figure 4.23 which has co-ordinates (4,2) (marked by a small cross) is given a weighting $e^{-\frac{\left( 4^2+2^2 \right) \times 1.665^2 \times 4}{\left(11-1\right)^2}}=e^{-0.554}= 0.109$. In order to retain the correct absolute emission temperature at the end of the weighting routine it is necessary for the weighting factors used for each position in the grid to be summed and the final temperature (ie. that obtained by adding all the temperatures from each grid position together) are divided by this sum, ie.

$$T_{\rm final}=\frac{\sum_{i=1}^n T_iw_i}{\sum_{i=1}^n w_i}$$ (4.55)

where the $T_i$ are the emission temperatures at the $i^{\rm th}$ position in the grid (with a total of $n$ positions) weighted by $w_i$.