Commit 87bcf303 by dauncey

### Various studies

parent e046d9e4
 ... ... @@ -537,6 +537,38 @@ Representation & Exponent & Mantissa & Value \cr\hline \end{tabular} \end{center} To get an unbiased decoded value, it is necessary to decode to a value a factor of two larger than the original value. To do this, all values stored with the exponent greater than 1 have 0b01111\dots appended to the mantissa where the number of 1 bits is set to be one less than the original LSB. Expanding the example table above, this gives \bigskip \begin{center} \begin{tabular}{c|c|c|c|c|c|c|c} \hline Representation & Exponent & Mantissa & Value & $2\times$Rep & Appended & $2\times$Value & Value\cr\hline 0 & 0b00 & 0b00 & 0b00000 = \phantom{2}0 & \phantom{2}0 & & \phantom{2}0 & 0 \cr 1 & 0b00 & 0b01 & 0b00001 = \phantom{2}1 & \phantom{2}2 & & \phantom{2}2 & 1 \cr 2 & 0b00 & 0b10 & 0b00010 = \phantom{2}2 & \phantom{2}4 & & \phantom{2}4 & 2\cr 3 & 0b00 & 0b11 & 0b00011 = \phantom{2}3 & \phantom{2}6 & & \phantom{2}6 & 3\cr 4 & 0b01 & 0b00 & 0b00100 = \phantom{2}4 & \phantom{2}8 & & \phantom{2}8 & 4\cr 5 & 0b01 & 0b01 & 0b00101 = \phantom{2}5 & 10 & & 10 & 5\cr 6 & 0b01 & 0b10 & 0b00110 = \phantom{2}6 & 12 & & 12 & 6\cr 7 & 0b01 & 0b11 & 0b00111 = \phantom{2}7 & 14 & & 14 & 7 \cr 8 & 0b10 & 0b00 & 0b01000 = \phantom{2}8 & 16 & 0b0001 & 0b010001 = 17 & 8.5 \cr 9 & 0b10 & 0b01 & 0b01010 = 10 & 20 & 0b0101 & 0b010101 = 21 & 10.5 \cr 10 & 0b10 & 0b10 & 0b01100 = 12 & 24 & 0b1001 & 0b011001 = 25 & 12.5\cr 11 & 0b10 & 0b11 & 0b01110 = 14 & 28 & 0b1101 & 0b011101 = 29 & 14.5 \cr 12 & 0b11 & 0b00 & 0b10000 = 16 & 32 & 0b00011 & 0b0100011 = 35 & 17.5\cr 13 & 0b11 & 0b01 & 0b10100 = 20 & 40 & 0b01011 & 0b0101011 = 43 & 21.5\cr 14 & 0b11 & 0b10 & 0b11000 = 24 & 48 & 0b10011 & 0b0110011 = 51 & 25.5\cr 15 & 0b11 & 0b11 & 0b11100 = 28 & 56 & 0b11011 & 0b0111011 = 59 & 29.5\cr \hline \end{tabular} \end{center} \section{Template fit of energy in depth} Assume a template shape for a photon of $P_l$ per photon energy GeV for layer $l$. ... ... @@ -731,6 +763,10 @@ a factor of $\sin\theta$, which is given by \sin\theta = \sin\left(\tan^{-1}\rho_s\right) = \sin\left[\tan^{-1}\left(\frac{1}{\sinh\eta}\right)\right] \qquad\mbox{Note also}\qquad \sin\theta = \frac{1}{\cosh\eta},\quad \tan\theta = \frac{1}{\sinh\eta},\quad \cos\theta = \tanh\eta % \begin{figure}[ht!] ... ... @@ -1189,4 +1225,130 @@ which in terms of rate in kHz are R_2 = 20.768 \end{eqnarray} \section{Photon model} For photons at shower maximum (around layer 10, counting from 0), the cumulative 0.683 and 0.900 radii are around 1.2\,cm and 2.8\,cm, respectively. \section{Gaussian} Assume a Gaussian radial distribution for the shower shape in a given layer with a width parameter $a$, so having a density \begin{eqnarray} \rho = \frac{1}{2\pi a^2} e^{-r^2/2a^2}\,rdr\,d\phi \end{eqnarray} Check the normalisation: integrating to some radius $R$ gives \begin{eqnarray} \int_0^R \int_0^{2\pi} \rho\,rdr\,d\phi = \int_0^R \frac{r}{a^2} e^{-r^2/2a^2}\,dr = \left[-e^{-r^2/2a^2}\right]_0^R = 1 - e^{-R^2/2a^2} \end{eqnarray} This goes to 1 as $R\rightarrow\infty$ and so is normalised. For a cumulative amount $C<1$, then the radius needed is the solution of \begin{eqnarray} e^{-R^2/2a^2} = 1-C, \qquad\mbox{so}\qquad \frac{R}{a} = \sqrt{-2\ln(1-C)} \end{eqnarray} For the standard values, $C=0.683$ requires $R/a = 1.52$ while $C=0.900$ requires $R/a = 2.15$. These imply parameter values of $a = 1.2/1.52 = 0.79$\,cm and $a = 2.8/2.15 = 1.30$\,cm, respectively. Hence, they are not really compatible, but an average value of 1.05\,cm is probably adequate. % \begin{figure}[ht!] \begin{center} \includegraphics[width=9cm]{GaussianCumulative.pdf} \label{fig:ExponentialCumulative} \end{center} \end{figure} \section{Exponential} Assume an exponential radial distribution for the shower shape in a given layer with an exponential parameter $a$, so having a density \begin{eqnarray} \rho = \frac{1}{2\pi a^2} e^{-r/a}\,rdr\,d\phi \end{eqnarray} Check the normalisation: integrating to some radius $R$ gives \begin{eqnarray} \int_0^R \int_0^{2\pi} \rho\,rdr\,d\phi = \int_0^R \frac{r}{a^2} e^{-r/a}\,dr \end{eqnarray} Consider \begin{eqnarray} \frac{d}{dr}\left(\frac{r}{a}e^{-r/a}\right) = \frac{1}{a}e^{-r/a} - \frac{r}{a^2}e^{-r/a} \end{eqnarray} Hence \begin{eqnarray} \int_0^R \frac{r}{a^2}e^{-r/a}\,dr = \int_0^R \frac{1}{a}e^{-r/a}\,dr - \left[\frac{r}{a}e^{-r/a}\right]_0^R = \left[-e^{-r/a}\right]_0^R - \frac{R}{a}e^{-R/a} = 1 - \left(1 + \frac{R}{a}\right) e^{-R/a} \end{eqnarray} This goes to 1 as $R\rightarrow\infty$ and so is normalised. For a cumulative amount $C<1$, then the radius needed is the solution of \begin{eqnarray} \left(1 + \frac{R}{a}\right) e^{-R/a} = 1-C \end{eqnarray} which can be solved numerically. For the standard values, $C=0.683$ requires $R/a = 2.36$ while $C=0.900$ requires $R/a = 3.89$. These imply parameter values of $a = 1.2/2.36 = 0.508$\,cm and $a = 2.8/3.89 = 0.720$\,cm, respectively. Hence, they are not really compatible, but an average value of 0.61\,cm is probably adequate. % \begin{figure}[ht!] \begin{center} \includegraphics[width=9cm]{ExponentialCumulative.pdf} \label{fig:ExponentialCumulative} \end{center} \end{figure} \section{Exponential square-root} Assume a radial distribution of an exponential with a square root for the shower shape in a given layer with a exponential parameter $\sqrt{a}$, so having a density \begin{eqnarray} \rho = \frac{1}{24\pi a^2} e^{-\sqrt{r/a}}\,rdr\,d\phi \end{eqnarray} Check the normalisation: integrating to some radius $R$ gives \begin{eqnarray} \int_0^R \int_0^{2\pi} \rho\,rdr\,d\phi = \int_0^R \frac{r}{12a^2} e^{-\sqrt{r/a}}\,dr \end{eqnarray} Let $x=r/a$ and consider \begin{eqnarray} \frac{d}{dx}\left[(2x^{3/2}+6x+12x^{1/2}+12)e^{-\sqrt{x}}\right] &=& (3x^{1/2}+6+6x^{-1/2})e^{-\sqrt{x}} -(x^{3/2}+3x+6x^{1/2}+6)x^{-1/2} e^{-\sqrt{x}}\\ &=& (3x^{1/2}+6+6x^{-1/2} - x-3x^{1/2}-6-6x^{-1/2})e^{-\sqrt{x}}\\ &=& -x e^{-\sqrt{x}} \end{eqnarray} Hence \begin{eqnarray} \int_0^R \frac{r}{12a^2} e^{-\sqrt{r/a}}\,dr = \left[-\left(1+x^{1/2}+\frac{x}{2}+\frac{x^{3/2}}{6}\right)e^{-\sqrt{x}}\right]_0^{R/a} = 1 - \left[1+(R/a)^{1/2}+\frac{R/a}{2}+\frac{(R/a)^{3/2}}{6} \right] e^{-\sqrt{R/a}} \end{eqnarray} This goes to 1 as $R\rightarrow\infty$ and so is normalised. For a cumulative amount $C<1$, then the radius needed is the solution of \begin{eqnarray} \left[1+(R/a)^{1/2}+\frac{R/a}{2}+\frac{(R/a)^{3/2}}{6} \right] e^{-\sqrt{R/a}} = 1-C \end{eqnarray} which can be solved numerically. For the standard values, $C=0.683$ requires $R/a = 21.7$ while $C=0.900$ requires $R/a = 44.6$. These imply parameter values of $a = 1.2/21.7 = 0.0553$\,cm and $a = 2.8/44.6 = 0.0628$\,cm, respectively. Hence, they are quite compatible, and an average value of 0.059\,cm is probably adequate. % \begin{figure}[ht!] \begin{center} \includegraphics[width=9cm]{ExponentialSqrtCumulative.pdf} \label{fig:ExponentialSqrtCumulative} \end{center} \end{figure} \end{document}