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\begin{document}

\title{Dark Current induced trigger in MEMPHYS\thanks{Preliminary}}
\author{J.E Campagne - LAL - Orsay}
\date{\today}
\maketitle

\begin{abstract}
Blabla
\end{abstract}

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\section{Motivations}
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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. 
A long this line, I have investigated a toy MC and analytical analysis.
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\section{A toy MC}
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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: 
\begin{verbatim}
   NumberOfPMT  = 81,000 for MEMPHYS
   DarkCurrentRate = 6.0*kHz  more likely for 12"
   TriggerSlidingGate = 425.0*ns for MEMPHYS
   PulseWidth = 5.0*ns this is the time quantum (a pulse width equivalent)
   TimeDurationOfExp = 6000.0*micros (+TriggerSlidingGate) 
                       as a total time for investigation  
   CountThreshold = number of PMTs at least to be fired in a TriggerSlidingGate to 
                    produce a Trigger
\end{verbatim}
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\subsection{Dark current generation}
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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:
\bea
P(1;[t_0,t_0+\Delta t]) & = & \int_{t_0}^{t_0+\Delta t} e^{(t-t_0)/\tau} \frac{dt}{\tau} \nonumber \\
	& = & 1 - e^{-\Delta t/\tau} = \frac{\Delta t}{\tau} - \frac{1}{2} \left( \frac{\Delta t}{\tau}\right)^2 \dots
\label{eq:singleProba}
\eea
and is independent of $t_0$. 

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:
\beq
P(n|N;[t,t+\Delta t]) \stackrel{\Delta t/\tau \rightarrow 0}{\longrightarrow}\mathrm{Poisson}(n;N\Delta t/\tau)
\label{eq:poissonLimit}
\eeq
One can see these properties reproduced by the toy MC on Fig.~\ref{fig:PMTPoisson} for some parameters values used on purpose.
\begin{figure}[htb]
 \begin{minipage}[c]{0.44\textwidth}
\includegraphics[width=\textwidth]{deltaT.eps}
\end{minipage}
 \begin{minipage}[c]{0.05\textwidth}
\end{minipage}
 \begin{minipage}[c]{0.44\textwidth}
\includegraphics[width=\textwidth]{NumberPMTs.eps}
\end{minipage}
\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.
}
\end{figure}

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}):
\bea
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 \\
											& = &\mathrm{Binomial}(n,N;P(1,[t,t+\Delta t]))
\label{eq:BinomialProb}
\eea
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\subsection{Trigger simulation}
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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 $n(t)=n_{Th}+1 \rightarrow n(t+dt)=n_{Th}$. 
\begin{figure}[htb]
 \begin{minipage}[c]{0.44\textwidth}
\includegraphics[width=\textwidth]{SumBaseline.eps}
\end{minipage}
 \begin{minipage}[c]{0.05\textwidth}
\end{minipage}
 \begin{minipage}[c]{0.44\textwidth}
\includegraphics[width=\textwidth]{SumBaselineZoom.eps}
\end{minipage}
\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.
}
\end{figure}

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 ($v = 22,6~\mathrm{cm/ns}$). 
\begin{table}
\begin{center}
\begin{tabular}{cccc}\hline\hline
DC per PMT (kHz) & Mean of PMTs fired & Threshold (at 3$\sigma$)  & Trigger (kHz) & Th (kHz)\\
1                 &      34           &  52                     & 412             & 57\\
2                 &      69           &  93                     & 378              & 64.9\\
3                 &      103           &  133                     & 198            & 2.8 10^3\\
4                 &      138           &  172                     & 378            & 7.1\\
5                 &      172           &  211                     & 268            & 3.1\\
6                 &      206          &  249                     & 416             & 1.5\\ \hline
DC per PMT (kHz) & Mean of PMTs fired & Threshold  & Trigger \& rel. err (kHz) \\
1                 &      34           & 40 (1$\sigma$)                 &   30 811 (0.7\%) & 1475.5 \\
1                 &      34           & 46 (2$\sigma$)                 &    4580 (2\%)    & 363.4\\
1                 &      34           & 52 (3$\sigma$)                 &     412 (7\%)     & 41.2   \\
1                 &      34           & 57 (4$\sigma$)                 &      28 (12\%)    & 3.9\\
\hline\hline
\end{tabular}
\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.}
\end{center}
\end{table} 
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\subsection{Analytical formula}
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\end{document}
