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\section{Trigger efficiency} |
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\label{sec:trgEff} |
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|
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{\color{red} Here we describe the trigger efficiency calculation. |
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Nothing here yet, since everything is still in flux.} |
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As described in Section~\ref{sec:trigSel} we rely on a |
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mixture of single and double lepton triggers. The trigger |
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efficiency is very high because for most of the phase space |
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we have two leptons each of which can fire a single lepton |
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trigger -- and the single lepton triggers are very efficient. |
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We apply to MC events a simplified model of the trigger efficiency |
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as a function of dilepton species ($ee$, $e\mu$, $\mu\mu$), the $P_T$ |
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of the individual leptons, and, in the case of muons, the $|\eta|$ |
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of the muons. We believe that this model is adequate for |
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the trigger efficiency precision needed for this analysis. |
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The model assumptions are the following |
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|
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\begin{itemize} |
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\item Muon and electron trigger turn-ons as a function of $P_T$ |
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are infinitely sharp. {\color{red} Can we add references?} |
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|
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\item All electron triggers with no ID have 100\% |
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efficiency for electrons passing our analysis cuts. {\color{red} |
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Can we add a reference? Pehaps the top documentation?} |
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|
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\item Electron triggers with (Tight(er))CaloEleId have 100\% |
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efficiency with respect to our offline selection. This we |
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verified via tag-and-probe on $Z\to ee$. |
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|
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\item Electron triggers with EleId have somewhat lower |
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efficiency. This was also measured by tag-and-probe. |
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|
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\item The single muon trigger has zero efficiency for $|\eta|>2.1$. |
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This is conservative, the trigger efficiency here is of order |
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$\approx 40\%$. |
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|
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\item If a muon in $|\eta|>2.1$ fails the single muon trigger, it |
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will also fail the double muon trigger. |
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|
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\item The double muon trigger has efficiency |
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equal to the square of the single muon efficiency. |
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|
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\item The $e\mu$ triggers have no efficiency if the muon has $|\eta|>2.1$. |
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|
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\end{itemize} |
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|
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The model also uses some luminosity fractions and some trigger |
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efficiencies. |
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|
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\begin{itemize} |
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|
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\item $\epsilon_{\mu}$=xx, the single muon trigger efficiency plateau. |
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|
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\item $f9$=xx: fraction of data with the Mu9 trigger unprescaled. |
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(run$\le 147116$). |
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|
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\item $f11$=xx fraction of data with the Mu9 trigger prescaled and |
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the Mu11 trigger unprescaled. |
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(147196 $\leq$ run $\leq$ 148058). |
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|
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\item $e10$=xx: fraction of data with the 10 GeV unprescaled electron triggers. |
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(run$\le 139980$). |
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|
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\item $e15$=xx: fraction of data with the 15 GeV unprescaled electron triggers. |
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(139980 $<$ run $\leq$ 144114). |
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|
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\item $e17$=xx: fraction of data with the 100\% efficient 17 GeV unprescaled electron triggers. |
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(144114 $<$ run $\leq$ 147116). |
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|
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\item $e17b$=xx: fraction of data with 17 GeV unprescaled electron triggers |
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with efficiency $\epsilon_e^b=90\%$ (as measured by tag-and-probe). |
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(147116 $<$ run $\leq$ 148058). |
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|
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\item $emess$=xx: the remainder of the run with several different electron |
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triggers, all of $P_T>17$ GeV. For this period we measure the |
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luminosity-weighted |
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trigger efficiency $\epsilon(P_T)$ via tag and probe to be 99\% |
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($17<P_T<22$, 97\% ($22<P_T<27$), 98\% ($27<P_T<32$) and |
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100\% ($P_T>32$). |
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|
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\end{itemize} |
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|
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The full trigger efficiency model is described separately for |
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$ee$, $e\mu$, and $\mu\mu$. |
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|
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\subsection{$ee$ efficiency model} |
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\label{sec:eemodel} |
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This is the easiest. Throughout the 2010 run we have always |
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had dielectron triggers with thresholds lower than our (20,10) |
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analysis thresholds. Since electron triggers are 100\% efficient, |
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the trigger efficiency for $ee$ is 100\%. We have verified that |
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the efficiency of the dielectron trigger is 100\% with respect |
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to the single electron trigger using $Z \to ee$ data. |
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|
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\subsection{$\mu\mu$ efficiency model} |
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\label{sec:mmmodel} |
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We consider different cases. |
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\subsubsection{Both muons in $|\eta|<2.1$ and with $P_T>15$ GeV} |
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This is the bulk of the $\mu\mu$. |
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|
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\begin{center} |
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$\epsilon = 1 - (1-\epsilon_{\mu})^2$ |
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\end{center} |
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|
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\subsubsection{Both muons in $|\eta|<2.1$, one muon with $11<P_T<15$ GeV} |
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In this case there must be a muon with $P_T>20$ GeV. The single muon |
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trigger is operative for the full dataset on this muon. Some loss |
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of efficiency can be recovered when the 2nd muon fires the trigger. |
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But this can happen only for a fraction of the run. The dimuon trigger |
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cannot fire in our model to recover any of the efficiency lost by |
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the single muon trigger on the high $P_T$ muon. |
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|
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\begin{center} |
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$\epsilon = \epsilon_{\mu} + (1-\epsilon_{\mu})\epsilon_{\mu}(f9+f11)$ |
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\end{center} |
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|
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\subsubsection{Both muons in $|\eta|<2.1$, one muon with $10<P_T<11$ GeV} |
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Same basic idea as above. |
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|
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\begin{center} |
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$\epsilon = \epsilon_{\mu} + (1-\epsilon_{\mu})\epsilon_{\mu}f9$ |
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\end{center} |
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|
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\subsubsection{Both muons with $|\eta|>2.1$} |
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This is a very small fraction of events. In our model they can only be |
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triggered by the dimuon trigger. |
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|
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\begin{center} |
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$\epsilon = \epsilon_{\mu}^2$ |
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\end{center} |
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|
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\subsubsection{First muon with $P_T>15$ and $|\eta|<2.1$; second muon |
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with $|\eta|>2.1$} |
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The single muon trigger is always operative. If it fails the double muon |
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trigger also fails. |
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\begin{center} |
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$\epsilon = \epsilon_{\mu}$ |
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\end{center} |
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|
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\subsubsection{First muon with $11<P_T<15$ and $|\eta|<2.1$; second muon |
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with $|\eta|>2.1$} |
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The single muon trigger is operative only for a fraction of the run. |
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For the remaining fraction, we must rely on the double muon trigger. |
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|
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\begin{center} |
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$\epsilon = (f9+f11)\epsilon_{\mu} + (1-f9-f11)\epsilon_{\mu}^2$ |
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\end{center} |
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|
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\subsubsection{First muon with $10<P_T<11$ and $|\eta|<2.1$; second muon |
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with $|\eta|>2.1$} |
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Same basic idea as above. |
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|
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\begin{center} |
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$\epsilon = f9~\epsilon_{\mu} + (1-f9)\epsilon_{\mu}^2$ |
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\end{center} |
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\subsection{$e\mu$ efficiency model} |
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\label{sec:emumodel} |
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This is the most complicated case. The idea is that the muon trigger |
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is used to get the bulk of the efficiency. Then the single electron |
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trigger(s) and the $e\mu$ triggers are used to get back dome of the |
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efficiency loss. The various cases are listed below. |
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\subsubsection{Muon with $|\eta|<2.1$ and $P_T>15$} |
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This is the bulk of the acceptance. |
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|
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\begin{center} |
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$\epsilon = \epsilon_{\mu} + (1-\epsilon_{\mu})\Delta_1$ |
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\end{center} |
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where $\Delta_1$ is the efficiency from the electron trigger: |
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\begin{itemize} |
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\item $P_T(ele)<15 \to \Delta_1=e10$ |
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\item $15<P_T(ele)<17 \to \Delta_1=e10+e15$ |
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\item $P_T(ele)>15 \to \Delta_1=e10+e15+e17+\epsilon_e^b~e17b+\epsilon(P_T)~emess$ |
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\end{itemize} |
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|
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\subsubsection{Muon with $|\eta|<2.1$ and $11<P_T>15$} |
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This is the similar to the previous case, except that the muon |
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trigger is operative only for a subset of the data taking period. |
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|
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\begin{center} |
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$\epsilon = (f11+f9)\epsilon_{\mu} + \Delta_2 + \Delta_3$ |
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\end{center} |
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Here $\Delta_2$ is associated with the period where the muon |
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trigger was at 15 GeV, in which case we use electron triggers or |
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$e\mu$ triggers. Note that the electron in this case must be |
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above 20 GeV. This can happen only in the latter part of the run, |
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thus we write |
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\begin{center} |
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$\Delta_2 = (1-f11-f9)~(\epsilon_{\mu}~+~ |
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(1-\epsilon_{\mu})\epsilon(P_T))$ |
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\end{center} |
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\noindent where the first term is for the $e\mu$ trigger and the |
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second term corresponds to $e\mu$ trigger failures, in which case we have |
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to rely on the electron trigger. |
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|
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Then, $\Delta_3$ is associated with muon trigger failures in early runs, |
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{\em i.e.}, run $<148819$. In this case the electron trigger picks it |
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up and the $e\mu$ trigger does not help. |
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|
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\begin{center} |
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$\Delta_3 = (f11+f9)(1-\epsilon_{\mu}) \cdot \epsilon_e$ |
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\end{center} |
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|
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\noindent where $\epsilon_e$ is the efficiency of the electron |
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trigger for $P_T>20$. This is 100\% up to run 147716 (fraction |
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$(e10_e15+e17)/(f11+f9)$; then it is somewhat lower up to |
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run 148058, then it becomes very close to 100\% again. |
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For this latter part of the run we approximate it as $\epsilon_e^b$. |
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Thus: |
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|
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\begin{center} |
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$\epsilon_e = (e10_e15+e17)/(f11+f9) + |
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\epsilon_e^b(f11+f9-e10-e15-e17)/(f11+f9)$ |
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\end{center} |
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|
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\subsubsection{Muon with $|\eta|<2.1$ and $9<P_T>11$} |
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Identical to the previous case, but replace $(f11+f9)$ with $f9$ everywhere. |
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|
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\subsubsection{Muon with $|\eta|>2.1$} |
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This is a 10\% effect to start with. We assume no single muon efficiency, |
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no $e\mu$ efficiency. Then we can only ise the single electron trigger. |
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|
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\begin{center} |
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$\epsilon = \Delta_1$ |
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\end{center} |
