[388] | 1 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 2 | \documentclass[12pt,a4paper]{article} |
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| 3 | |
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| 4 | %-- used packages ------------------------------------------------------ |
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| 5 | |
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| 6 | %%%JEC 3/2/06 does not compile at LAL \usepackage{cite} |
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| 7 | %\usepackage[T1]{fontenc} |
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| 8 | %\usepackage[latin1]{inputenc} |
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| 9 | \usepackage{graphicx} |
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| 10 | \usepackage{epsfig} |
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| 11 | \usepackage{amssymb} |
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| 12 | \usepackage{amsmath} |
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| 13 | \usepackage{latexsym} |
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| 14 | |
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| 15 | \setlength{\textheight}{23cm} |
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| 16 | \setlength{\textwidth}{16cm} |
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| 17 | \setlength{\topmargin}{0cm} |
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| 18 | \setlength{\oddsidemargin}{0cm} |
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| 19 | \setlength{\evensidemargin}{0cm} |
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| 20 | \setlength{\marginparwidth}{0pt} |
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| 21 | |
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| 22 | \def\beq{\begin{equation}} |
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| 23 | \def\eeq{\end{equation}} |
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| 24 | \def\bea{\begin{eqnarray}} |
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| 25 | \def\eea{\end{eqnarray}} |
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| 26 | |
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| 27 | |
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| 28 | |
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| 29 | \begin{document} |
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| 30 | |
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| 31 | \title{Dark Current induced trigger in MEMPHYS\thanks{Preliminary}} |
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| 32 | \author{J.E Campagne - LAL - Orsay} |
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| 33 | \date{\today} |
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| 34 | \maketitle |
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| 35 | |
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| 36 | \begin{abstract} |
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| 37 | Blabla |
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| 38 | \end{abstract} |
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| 39 | |
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| 40 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 41 | \section{Motivations} |
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| 42 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 43 | On 5th of July, J.~Bouchez had demonstrated the need to investigate deeper than expected the dark current induced trigger for a MEMPHYS tank. As a matter of fact, a 3~MeV reactor neutrino produce around 15~p.e, so basically 15 PMTs on a Cerenkov cone are hit, while in the same time at 2~kHz dark current noise per PMT in a 400~ns time window one can expect around 160 PMT hit randomly distributed. Although, it may be not a problem (not yet proved) to recover the PMTs of the signal, the main concern is the trigger. J.~Bouchez has made a preliminary numerical analysis to show that the question is relevant. |
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| 44 | A long this line, I have investigated a toy MC and analytical analysis. |
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| 45 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 46 | \section{A toy MC} |
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| 47 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 48 | The toy MC consists simply of the generation of dark current random noise on each PMT baseline, then it simulates a digital sum of all the PMT baselines, and applies a threshold on a total number of PMT hit during a sliding window. This toy MC can be adapted according to: |
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| 49 | \begin{verbatim} |
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| 50 | NumberOfPMT = 81,000 for MEMPHYS |
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| 51 | DarkCurrentRate = 6.0*kHz more likely for 12" |
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| 52 | TriggerSlidingGate = 425.0*ns for MEMPHYS |
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| 53 | PulseWidth = 5.0*ns this is the time quantum (a pulse width equivalent) |
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| 54 | TimeDurationOfExp = 6000.0*micros (+TriggerSlidingGate) |
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| 55 | as a total time for investigation |
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| 56 | CountThreshold = number of PMTs at least to be fired in a TriggerSlidingGate to |
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| 57 | produce a Trigger |
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| 58 | \end{verbatim} |
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| 59 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 60 | \subsection{Dark current generation} |
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| 61 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 62 | According to the fact that dark current is a random noise, if ones make the assumption that there is no correlation between two noise hit (at least on the scale of \verb|TriggerSlidingGate|), then the time interval between two hit $\delta t$ follow an exponential probability with parameter $\tau = 1/\verb|DarkCurrentRate|$. So, the probability to get 1 hit in the time window $[t_0,t_0+\Delta t]$ reads: |
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| 63 | \bea |
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| 64 | P(1;[t_0,t_0+\Delta t]) & = & \int_{t_0}^{t_0+\Delta t} e^{(t-t_0)/\tau} \frac{dt}{\tau} \nonumber \\ |
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| 65 | & = & 1 - e^{-\Delta t/\tau} = \frac{\Delta t}{\tau} - \frac{1}{2} \left( \frac{\Delta t}{\tau}\right)^2 \dots |
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| 66 | \label{eq:singleProba} |
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| 67 | \eea |
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| 68 | and is independent of $t_0$. |
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| 69 | |
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| 70 | If one piles up $N$ PMTs and if $\Delta t/\tau \rightarrow 0$, then it is well known that the probability to get $n$ PMTs hit in a time window of $\Delta t$ tends to a Poisson distribution: |
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| 71 | \beq |
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| 72 | P(n|N;[t,t+\Delta t]) \stackrel{\Delta t/\tau \rightarrow 0}{\longrightarrow}\mathrm{Poisson}(n;N\Delta t/\tau) |
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| 73 | \label{eq:poissonLimit} |
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| 74 | \eeq |
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| 75 | One can see these properties reproduced by the toy MC on Fig.~\ref{fig:PMTPoisson} for some parameters values used on purpose. |
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| 76 | \begin{figure}[htb] |
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| 77 | \begin{minipage}[c]{0.44\textwidth} |
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| 78 | \includegraphics[width=\textwidth]{deltaT.eps} |
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| 79 | \end{minipage} |
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| 80 | \begin{minipage}[c]{0.05\textwidth} |
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| 81 | \end{minipage} |
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| 82 | \begin{minipage}[c]{0.44\textwidth} |
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| 83 | \includegraphics[width=\textwidth]{NumberPMTs.eps} |
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| 84 | \end{minipage} |
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| 85 | \caption{\label{fig:PMTPoisson} {\it left panel}: exponential generation of $\delta t$ with $1/\tau = 1$~MHz dark current noise; {\it right panel}: number of PMTs fired among 50 PMTs, per $\Delta t = 100$~ns windows. One expects a Poisson distribution with a mean of $N\Delta t/\tau = 5$ as it is shown. |
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| 86 | } |
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| 87 | \end{figure} |
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| 88 | |
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| 89 | In fact, as it is mentioned, the Poisson distribution is an asymptotic distribution. The real probability to get exactly $n$ PMTs fired among a collection of $N$ PMTs in a $\Delta t$ time window, with the hypothesis that each PMT dark current rate has a $1/\tau$ frequency, reads (using Eq.~\ref{eq:singleProba}): |
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| 90 | \bea |
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| 91 | P(n|N;[t,t+\Delta t]) & = & C_N^n P(1;[t,t+\Delta t])^n \left(1-P(1;[t,t+\Delta t])\right)^{N-n} \nonumber \\ |
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| 92 | & = &\mathrm{Binomial}(n,N;P(1,[t,t+\Delta t])) |
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| 93 | \label{eq:BinomialProb} |
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| 94 | \eea |
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| 95 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 96 | \subsection{Trigger simulation} |
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| 97 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 98 | One may wish to know the occurrence of "how many times per second, $n_{Th}$ PMTs are fired among $N$ PMTs?". In fact, J.~Bouchez noticed that this question is not complete as one may trigger only if the number of PMTs fired in a $\Delta t$ sliding windows just exceeds $n_{Th}$. That is to say, one should consider the stepping function of the number of PMTs fired in the sliding window $\Delta t$ examined at $t$, noted $n(t)$, and consider as "trigger" the passage $n(t)=n_{Th} \rightarrow n(t+dt)=n_{Th}+1$ in an infinitesimal time duration $dt$. In fact, looking at Fig.~\ref{fig:DigitalSum}, where is displayed the digital sum of $N=81,000$ PMts with 6~kHz dark current noise rate in sliding window of $425$~ns (computed every $dt=5$~ns), one notices that it is true that the threshold\footnote{mean number of PMTS is 206 and the threshold is set 3 sigma above} at $250$ is reached by steps, but one may also specify that if there is a trigger at $t_1$, before to re-trigger the $n(t>t_1)$ function may pass through a sequence $n(t)=n_{Th}+1 \rightarrow n(t+dt)=n_{Th}$. |
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| 99 | \begin{figure}[htb] |
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| 100 | \begin{minipage}[c]{0.44\textwidth} |
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| 101 | \includegraphics[width=\textwidth]{SumBaseline.eps} |
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| 102 | \end{minipage} |
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| 103 | \begin{minipage}[c]{0.05\textwidth} |
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| 104 | \end{minipage} |
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| 105 | \begin{minipage}[c]{0.44\textwidth} |
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| 106 | \includegraphics[width=\textwidth]{SumBaselineZoom.eps} |
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| 107 | \end{minipage} |
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| 108 | \caption{\label{fig:DigitalSum} {\it left panel}: Digital sum of the time evolution of $N=81,000$ PMTs with a dark current of $6$~kHz, computed for a time sliding window of $425$~ns every $5$~ns; {\it right panel}: zoom around the location of a trigger threshold at $250$ PMTs. |
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| 109 | } |
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| 110 | \end{figure} |
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| 111 | |
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| 112 | So in practice, one considers as a trigger the passage $n(t)=n_{Th} \rightarrow n(t+dt)=n_{Th}+1$ and groups together the following $n(t)$ values contiguous in time for which $n(t)>=n_{Th}$. Using this definition, one gets the results for one MEMPHYS tank (Tab.\ref{tab:TriggerRate}): $N=81,000$ PMTs, a sliding window of $425$~ns which corresponds to the maximum time of flight of \v{C}erenkov light in the $\phi: 65~\mathrm{m} \times H:65~\mathrm{m}$ cylinder water volume ($v = 22,6~\mathrm{cm/ns}$). |
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| 113 | \begin{table} |
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| 114 | \begin{center} |
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| 115 | \begin{tabular}{cccc}\hline\hline |
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| 116 | DC per PMT (kHz) & Mean of PMTs fired & Threshold (at 3$\sigma$) & Trigger (kHz) & Th (kHz)\\ |
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| 117 | 1 & 34 & 52 & 412 & 57\\ |
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| 118 | 2 & 69 & 93 & 378 & 64.9\\ |
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| 119 | 3 & 103 & 133 & 198 & 2.8 10^3\\ |
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| 120 | 4 & 138 & 172 & 378 & 7.1\\ |
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| 121 | 5 & 172 & 211 & 268 & 3.1\\ |
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| 122 | 6 & 206 & 249 & 416 & 1.5\\ \hline |
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| 123 | DC per PMT (kHz) & Mean of PMTs fired & Threshold & Trigger \& rel. err (kHz) \\ |
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| 124 | 1 & 34 & 40 (1$\sigma$) & 30 811 (0.7\%) & 1475.5 \\ |
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| 125 | 1 & 34 & 46 (2$\sigma$) & 4580 (2\%) & 363.4\\ |
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| 126 | 1 & 34 & 52 (3$\sigma$) & 412 (7\%) & 41.2 \\ |
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| 127 | 1 & 34 & 57 (4$\sigma$) & 28 (12\%) & 3.9\\ |
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| 128 | \hline\hline |
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| 129 | \end{tabular} |
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| 130 | \caption{\label{tab:TriggerRate} Mean number of PMTs ($m$)fired among $N=81,000$ PMTs on a time sliding window of $\Delta t=425$~ns corresponding to the dark current (DC) rate ($f$) per PMT (this is $N\Delta t * f$). Then, setting a threshold to $m+3\sqrt m$, one reads the trigger rate (see text for the definition of a trigger). The relative errors is also reported as it is not negligeable.} |
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| 131 | \end{center} |
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| 132 | \end{table} |
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| 133 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 134 | \subsection{Analytical formula} |
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| 135 | %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% |
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| 136 | |
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| 137 | |
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| 138 | \end{document} |
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