\documentclass[11pt]{article}

\usepackage{graphicx}
\usepackage{epstopdf}
\usepackage{mathrsfs}
\usepackage{amsmath}


\usepackage{subfigure}
\usepackage{wrapfig}
%\usepackage{mlineno}
%\usepackage{color}
\setlength{\textwidth}{16.58cm}
\setlength{\textheight}{22.94cm}
\setlength{\headheight}{0pt}
\setlength{\headsep}{0pt}
\setlength{\oddsidemargin}{-0.04cm}
\setlength{\topmargin}{-0.04cm}
\renewcommand{\baselinestretch}{1.0}


\includeonly{}
\usepackage{indentfirst}
\usepackage{url}
\usepackage{amsmath}
\usepackage{cite}
%\usepackage[square, comma, sort&compress]{natbib}
\begin{document}

\begin{center}
{\Large
Effect of Phase Center Offsets in LCP/RCP Correlation Maps
}
~\\
~\\
\end{center}

\section{Defining H/V/LCP/RCP Waveforms}
Waveform in the vertical polarization.  The $n$ index represents antenna number and the $i$ index is for time bin.
\begin{equation}
\label{eq:vwaveform}
v_n(t_i) = \sum_k e^{-j \omega_k t_i} V(\omega_k) \Delta \omega
\end{equation}
Now horizontal polarization, giving all of the H-pol phase centers a time offset relative to the V-pol phase centers.
\begin{equation}
\label{eq:hwaveform}
h_n(t_i) = \sum_k e^{-j \left( \omega_k t_i + \omega_k t_0 \right) } H(\omega_k) \Delta \omega
\end{equation}

\begin{equation}
\label{eq:rwaveform}
r_n(t_i) = \dfrac{1}{\sqrt{2}} \left[ v(t_i) + j h(t_i) \right]
\end{equation}

\begin{equation}
\label{eq:lwaveform}
\ell_n(t_i) = \dfrac{1}{\sqrt{2}} \left[ v(t_i) - j h(t_i) \right]
\end{equation}

Substituting Eqs.~\ref{eq:vwaveform} and ~\ref{eq:hwaveform} into Eqs. ~\ref{eq:rwaveform} and ~\ref{eq:lwaveform}:
\begin{equation}
r_n(t_i) = \dfrac{1}{\sqrt{2}}  \sum_k  \Delta \omega e^{-j \omega_kt_i}  \left[ V(\omega_k) + j e^{-j\omega_k t_0}  H(\omega_k) \right]
\end{equation}

\section{Cross-Correlations with LCP/RCP Waveforms}
Now consider two antennas, and antenna 1 has a delay $T$ with respect to 2.  Then,
\begin{equation}
r_1(t_i) = \dfrac{1}{\sqrt{2}}  \sum_k  \Delta \omega e^{-j \omega_k t_i}  \left[ V(\omega_k) + j e^{-j\omega_k t_0}  H(\omega_k) \right]
\end{equation}

And since
\begin{equation}
\label{eq:vwaveform}
v_2(t_i) = \sum_k e^{-j \omega_k (t_i+T)} V(\omega_k) \Delta \omega
\end{equation}
\begin{equation}
\label{eq:hwaveform}
h_2(t_i) = \sum_k e^{-j \left[ \omega_k  (t_i+t_0+T) \right] } H(\omega_k) \Delta \omega
\end{equation}
then
\begin{equation}
r_2(t_i) = \dfrac{1}{\sqrt{2}}  \sum_k  \Delta \omega e^{-j \omega_k (t_i+T)}  \left[ V(\omega_k) + j e^{-j\omega_k t_0}  H(\omega_k) \right]
\end{equation}


Cross-correlating the RCP waveforms from antennas 1 and 2 ($r_1$ and $r_2$), and ignoring the normalization factor in the denominator for now, the get the following as a function of delay $\tau$ between the two RCP waveforms:
\begin{equation}
\label{eq:C12rr}
C^{rr}_{12}(\tau) = \sum_{k^{\prime}} \Delta t ~ r_1(t_i) r^*_2 (t_i+\tau)
\end{equation}
where the sum is over the region where the waveforms overlap for a given $\tau$.
Then substituting $r_1(t_i)$ and $r_2(t_i+\tau)$ into Eq.~\ref{eq:C12rr},
\begin{multline}
C^{rr}_{12}(\tau) = \dfrac{1}{2}  \left[  \sum_{k_1}  \Delta \omega e^{-j \omega_{k_1} t_i}  \left[ V(\omega_{k_1}) + j e^{-j\omega_{k_1} t_0}  H(\omega_{k_1}) \right] \right] \times \\
\left[  \sum_{k_2}  \Delta \omega e^{+j \omega_{k_2} (t_i+T+\tau)}  \left[ V(\omega_{k_2}) + j e^{-j\omega_{k_2} t_0}  H(\omega_{k_2}) \right]   \right]
\end{multline}
Collecting terms, we get:
\begin{multline}
C^{rr}_{12}(\tau) = \sum_{k_1}  \sum_{k_2}  (\Delta \omega)^2 e^{-j\left[ \omega_{k_1} t_i-\omega_{k_2} (t_i+T+\tau) \right]} \times \\
\left[ V(\omega_{k_1} )V^* (\omega_{k_2})  -j e^{j \omega_{k_2} t_0 } V(\omega_{k_1} )H^*(\omega_{k_2}) +j e^{-j \omega_{k_1} t_0} H(\omega_{k_1}) V^* (\omega_{k_2}) +H(\omega_{k_1}) H^* (\omega_{k_2}) \right]
\end{multline}

Likewise the LCP waveforms for antennas 1 and 2, where again antenna 1 has a delay $T$ with respect to 2:
\begin{equation}
\ell_1(t_i) = \dfrac{1}{\sqrt{2}}  \sum_k  \Delta \omega e^{-j \omega_k t_i}  \left[ V(\omega_k) - j e^{-j\omega_k t_0}  H(\omega_k) \right]
\end{equation}
\begin{equation}
\ell_2(t_i) = \dfrac{1}{\sqrt{2}}  \sum_k  \Delta \omega e^{-j \omega_k (t_i+T)}  \left[ V(\omega_k) - j e^{-j\omega_k t_0}  H(\omega_k) \right]
\end{equation}
Then, 
\begin{equation}
\label{eq:C12rr}
C^{\ell\ell}_{12}(\tau) = \sum_{k^{\prime}} \Delta t ~ \ell_1(t_i) \ell^*_2 (t_i+\tau)
\end{equation}
\begin{multline}
C^{\ell \ell}_{12}(\tau) = \sum_{k_1}  \sum_{k_2}  (\Delta \omega)^2 e^{-j\left[ \omega_{k_1} t_i-\omega_{k_2} (t_i+T+\tau) \right]} \times \\
\left[ V(\omega_{k_1} )V^* (\omega_{k_2})  +j e^{j \omega_{k_2} t_0 } V(\omega_{k_1} )H^*(\omega_{k_2}) -j e^{-j \omega_{k_1} t_0} H(\omega_{k_1}) V^* (\omega_{k_2}) +H(\omega_{k_1}) H^* (\omega_{k_2}) \right]
\end{multline}



\end{document}

\documentstyle[12pt,epsfig]{article}

\setlength{\textwidth}{6.5in}
\setlength{\textheight}{8.5in}
%\setlength{\footskip}{0.35in}
\setlength{\topmargin}{0.0in}
\setlength{\oddsidemargin}{0in}
\setlength{\parindent}{5ex}
\setlength{\parskip}{.12in}
\newcommand{\detw}{\mbox{$\frac{\Delta E}{2}$}}
\newcommand{\hs}{\hspace*{5ex}}


%\def\decayright#1{\kern#1em\raise1.1ex\hbox{$|$}\kern-.5em\rightarrow}

\begin{document}

\begin{center}
{\Large
Summary of a Simulation of the ANITA Detector
}
~\\
Amy Connolly\\
\today\\

~\\
~\\
{\bf Abstract}\\

\parbox[t]
{5.5in}
{
\hs We outline our code written to simulate the ANITA
detection system.
}
\end{center}

\tableofcontents


\section{Introduction}

The code outlined here is meant to be a simulation of the
ANITA and ANITA-lite detection systems.  It should provide
the collaboration with a tool to study the impact of various
effects on the experiment's sensitivity.  It will also 
provide the energy-dependent sensitivity that will be used
for setting limits on models for UHE neutrinos from
ANITA-lite data and ultimately from the full ANITA flights.
It also serves as a consistency check with the Monte Carlo
simulation being developed by our collaborators at the
University of Hawaii.

\section{Overall Strategy}
\label{sec:strategy}
\hs The simulation models interactions from neutrinos
of a specified energy.
The Askaryan pulse is parameterized by the current
theoretical model.
We require an interaction to occur within
the Antarctic ice volume within the balloon's horizon.  
We generate two different types of events.  The first is where the
ray seen by the balloon is direct; the ray is emitted from the
interaction upward.  The second is where the ray is emitted downward
and is reflected from the ice-rock interface before being reaching
the surface.  For half of the neutrinos, we force the ray to be the
first type and for the other half, the second type.
For a given type of ray, we find the unique path along which
an RF signal would travel from the interaction to the balloon, snelled
through ice layers 
at the ice-air surface.  Next, we pick a direction for the neutrino
path, only considering directions such that the Cerenkov cone is
close enough to the unique ray from interaction to balloon that
the signal is still detectable under a best-case scenario.
Depth-dependent attenuation lengths and indices of refraction
in the ice are based on recent South Pole measurements.  Surface
slopeyness is taken into account under a simple model.
Frequency-dependent 
antenna response is based on the manufacturer's specifications.
The trigger configuration is based on the most
current design.
The restrictions in neutrino phase space 
that come about in our selection of neutrino interaction positions
and neutrino directions 
are corrected for in the
final calculation (see Section~\ref{sec:volstr}).


\subsection{Coordinate System}


\hs Where Cartesian coordinates are used, we define $x$,$y$ and $z$ to 
be zero at the center of the earth.  The $+z$ axis 
runs from the center of the earth
to the south pole (so, our coordinate system is defined so that
the earth is ``upside-down'').  The $x$ and $y$ axes lie in the
0$^{\circ}$ latitude plane, with $+x$ pointing to
the 90$^{\circ}$ E longitudinal line, and $+y$ points to
0$^{\circ}$ longitude.  This coordinate system is right-handed.

The $\phi$ coordinate is defined as usual, zero along the $+x$
axis, and increasing moving counter-clockwise from the view 
looking down on the
$+z$ axis.  The $\theta$ coordinate is measured relative to the
$+z$ axis.

Bedrock and ice surface elevations are measured radially from
the center of the earth, and are quoted relative to sea level.
Sea level $r(\theta)$ is latitude-dependent and is quoted as a
distance from the center of the earth.  
We use a geoid shape given by~\cite{geoid}:
\begin{equation}
r(\theta)=\frac{r_{min}\cdot r_{max}}{\sqrt{r_{min}^2-(r_{min}^2-r_{max}^2) \cos^2{\theta}}}  
\end{equation}
where $r_{min}=6356.752$ km and $r_{max}=6378.137$ km.

To accommodate the global crust model 2.0 \cite{crust2}
for thicknesses of sediment layers through the crust as well as ice
thicknesses, which is on a $2^{\circ} \times 2^{\circ}$ grid, we bin
the earth's  surface into 180 bins in longitude and
90 bins in latitude.  Note that the resulting bins are not uniform
in area.
 The bins in latitude are
numbered such that the 0$^{th}$ bin is closest to the $+z$ axis (at the
south pole), and the 89$^{th}$ bin is closest to the north pole ($-z$).


\subsection{CSEDI Crust 2.0}
\label{sec:crust2}
\hs Crust 2.0 is the latest model of the Earth's interior near the 
surface published as the result of
an initiative called Cooperative Studies of the Earth's Deep
Interior (CSEDI).  It is based on seismological data, and
the model gives thicknesses and densities of each of seven
layers in $2^{\circ} \times 2^{\circ}$ bins:  ice, water, soft sediments, hard sediments, upper crust,
middle crust, and lower crust.  The ice thicknesses in Crust 2.0 are 
claimed to be within 250 m of the true ice thickness. Sediment thicknesses 
in each cell are to within 1.0 km of the true sediment thickness 
and crustal thickness are within 5 km of the true crustal thicknesses. 




%\hs  The user may input a list of
%bedrock and ice surface altitudes along with their coordinates
%in longitude and latitude, and the gaps are filled in with a 
%smoothing process.  In other words, for the bins where no data
%is given,  the altitudes for the bedrock and ice surface are
%interpolated.  The data that we are using now for bedrock and 
%ice surface elevations comes from: \\
%\centerline{http://classroomantarctica.aad.gov.au/materials/Elevations.pdf }
%which gives rockbed and ice surface elevations only along a single
%line across the continent. 

%The smoothing procedure is as follows.  The longitude and
%latitude coordinates for the inputted data points are translated
%to their appropriate $\phi$ and $\cos{\theta}$ bins and the
%bedrock and ice surface altitudes are set for these bins according
%to the data.  Also, the bins along the edge of the circle that
%defines our coordinate system (latitude=-60$^{\circ}$ and all bins
%in $\phi$) are set to be at sea level (altitudes set to zero
%relative to the earth radius).  All of the bins that have been
%set up to this point are called \emph{reference bins}.

%For a given bin that is not a reference bin, we need to
%set its bedrock and ice surface altitude.  First, we find
%the closest reference bin that is non-trivial.  A trivial reference
%bin is one set to sea level on the outer edge of the circle.
%Next, we find the closest reference bin (this time allowing trivial
%reference bins) that is on the other side of the bin of interest.
%So we want to be able to draw a straight line whose endpoints are
%reference points (at least one non-trivial) and that includes the
%bin of interest somewhere along the line.  Then, the bedrock altitude
%for the bin of interest is set to a value between those of the
%two reference points, determined by the bin's relative distance 
%from each of the reference points.  The same is done for the ice
%surface altitude.

Figures~\ref{fig:upperlayers} and~\ref{fig:lowerlayers} show the 
elevations of the seven crustal layers included in Crust 2.0.
\begin{figure}
\begin{center}
\epsfig{figure=upperlayers.eps,height=3in}
\caption{Altitude of the upper four layers given in Crust 2.0 along the
the $75^{\circ}$ S latitude line.  The horizontal axis is degrees in
longitude.
\label{fig:upperlayers}}
\end{center}
\end{figure}
\begin{figure}
\begin{center}
\epsfig{figure=lowerlayers.eps,height=3in}
\caption{Altitude of the lower three layers given in Crust 2.0 along the
the $75^{\circ}$ S latitude line.
\label{fig:lowerlayers}}
\end{center}
\end{figure}

We compute
the total Antarctic ice volume by summing the product of ice
thickness and surface area for each bin within the Antarctic continent.
For the area of each bin, we use:
\begin{equation}
\int_{\phi_1}^{\phi_2}\int_{\theta_1}^{\theta_2}\sin{\theta}~d\theta~d\phi=\left(\phi_1-\phi_2\right)\times\left(\cos{\theta_1}-\cos{\theta_2}\right)
\end{equation}
Where the limits of the integrals define the edges of the bin in
latitude and longitude.
We find $2.7 \times 10^{16}$ m$^2$ of Antarctic ice in this model.
Compare this to the $3.01098 \times 10^{16}$ m$^3$ volume of ice
in Antarctica reported by the US Geological Survey~\cite{usgs};
they are different by 10\%.
Unfortunately, Crust 2.0 calls any ice that sits above water (ice
shelves) just water.  Therefore, the Ross Ice Shelf needs to be
added into the model by hand.  Ice shelves make up 2.4\% of the
ice volume in Antarctica, so it won't account for the entire
$10\%$ difference. 

%\begin{center}
%\begin{tabular}{|l|l|l|l|}
%\hline\hline
%term      &   symbol &  with arguments & units for neutrinos\\
%\hline\hline
%neutrinos &
%  $N_{\nu}$    & 
%  $N_{\nu}$ &
%  (unitless) \\
%\hline
%neutrino intensity & 
%  $I$ &
%  $I(\theta,\phi)$  &
%  [length]$^{-2}$ [sr]$^{-1}$ [time]$^{-1}$\\
%\hline
%neutrino brightness, {\it or} & 
%  $I_{E}$, &
%  $I_{E}(E, \theta, \phi)$ &
%  [length]$^{-2}$ [sr]$^{-1}$ [time]$^{-1}$ [energy]$^{-1}$\\
%neutrino specific intensity  &
%  &
%  &
%  \\
%\hline
%net flux & 
%  $F_{E}$ &
%  $F_{E} (E)$ &
%  [length]$^{-2}$ [time]$^{-1}$ [energy]$^{-1}$\\
%\hline 
%total integrated flux &
%  $F$ &
%  $F$ &
%  [length]$^{-2}$ [time]$^{-1}$\\
%\hline\hline
%\end{tabular}
%\end{center}


\section{The Askaryan Signal}

\subsection{Magnitude of the Pulse}

We use the parameterization outlined in~\cite{jaime} for the peak of the
Askaryan signal:
\begin{equation}
\label{eq:vmmhz}
\frac{\cal E^{(\mathrm{@ 1m})} }{\mathrm{V/m/MHz}}=2.53\times 10^{-7}\cdot \frac{\sin{\theta_v}}{\sin{\theta_c}}\cdot  \frac{E_{em}}{\mathrm{TeV}} \cdot \frac{\nu}{\nu_0} \cdot \frac{1}{1.+\left( \frac{\nu}{\nu_0} \right)^{1.44}}
\end{equation}
where $\nu$ is the frequency, $\nu_0=1.15$ GHz, $E_{em}$ is the shower energy, $\theta_v$ is the viewing angle and $\theta_c$ is the Cerenkov angle. 

\subsection{Electromagnetic and Hadronic Components of the Shower}
 We assume flavor democracy; that by the time the neutrinos get
here, the flavors are fully mixed.  We also keep track of each
flavor individually so the sensitivity to each flavor may be
quoted separately.

We pick an inelasticity $y$ according to~\cite{ghandi},
which we have approximated by a double-exponential.
If we call the electromagnetic component $f_e$ and the 
hadronic component $f_h$, then the dependence of $f_e$ and
$f_h$ on $y$ depends on the whether it is a charged current
interaction that occurred or a neutral current interaction.
We choose
$70.64\%$ of the events at random to be charged-current
and the remainder to be neutral current.
If the incident neutrino was a $\nu_e$ and it is a charged current
event, then $f_e=1-y$ and $f_h=y$.
If the incident neutrino was $\nu_\tau$ or $\nu_\mu$, or
if the $\nu_e$ interacts through a neutral current, then
the electromagnetic component is for now treated as negligible, and
$f_h=y$.  A future version of the code will include $\mu/\tau$
bremsstrahlung and photo-nuclear interactions.


\subsection{Width of Cerenkov Cone}
\label{sec:cerenkov_width}

The user can select one of two different parameterizations of the
width of the Cerenkov cone in inputs.txt.  Here we refer to the
two parameterizations as the ``old'' parameterization and 
``new'' parameterization, for lack of a better terminology.
The user should select 
0 (1) for the old (new) parameterizations in the appropriate line in

\documentstyle[12pt,epsfig]{article}

\setlength{\textwidth}{6.5in}
\setlength{\textheight}{8.5in}
%\setlength{\footskip}{0.35in}
\setlength{\topmargin}{0.0in}
\setlength{\oddsidemargin}{0in}
\setlength{\parindent}{5ex}
\setlength{\parskip}{.12in}
\newcommand{\detw}{\mbox{$\frac{\Delta E}{2}$}}
\newcommand{\hs}{\hspace*{5ex}}



%\def\decayright#1{\kern#1em\raise1.1ex\hbox{$|$}\kern-.5em\rightarrow}

\begin{document}


\begin{center}
{\Large
Summary of a Variables in tsignals Tree
}
~\\
Amy Connolly\\
\today\\
\end{center}
~\\
~\\

\section{The tsignals Tree}

The tree called ``tsignals'' is stored in outputs/signals.root, which
is written by \texttt{icemc}.  Only events that survive any preselection cuts
to reach the Anita trigger are written to this tree.
Remember that as with the \texttt{passing\_events} tree, each event
has a weight given by \texttt{weight\_test/dnutries} to account for earth absorption and
selection of the interaction position within the horizon of the payload.
This is only filled for the antenna that is closest to the interaction, which
is always on the lowest of the three layers of the antennas (there are no
drop-downs implemented yet).  For all waveforms, there are 512 samples separated
by (1/2.6)~ns.

Where I refer to the first polarization, I mean LCP for Anita~I and vertical polarization for
Anita~II.  And for second polarization I mean RCP for Anita~I and horizontal polarization for
Anita~II.

A channel passes the trigger if the output of the tunnel diode\\
(\texttt{timedomain\_output\_1\_inanita[j]} or 
\texttt{timedomain\_output\_1\_inanita[j]})
goes lower than the (negative) 
threshold within the time window that lies between the 
232$^{\mathrm{nd}}$ and 284$^{\mathrm{th}}$ bins, or between about
89~ns and 109~ns.  Notice that this window
is centered about 15~ns later than the center of the 
signal waveforms since under my
tunnel diode model it imposes a delay.
The thresholds for ANITA~I are -1.251E-14 (Lo),-1.13106E-14 (Mid1),
-7.98583E-15 (Mid2),-1.40634E-14 (Hi) and -3.55069E-11 (full when used).
For ANITA~II, they are -1.15638e-14 (Lo), -1.42725e-14 (Mid),
-2.20845e-14 (Hi), -2.71869e-14 (Full).


\begin{center}
  \begin{tabular}{p{2.2in}p{4.0in}}
\texttt{double dangle} & Viewing angle - Cerenkov angle \\
\texttt{double emfrac} & EM fraction of the shower \\
\texttt{double hadfrac} & Hadronic fraction of the shower \\
\texttt{int inu} & \texttt{icemc}'s event number \\
\multicolumn{2}{p{6.0in}}{The event weight is the quotient \texttt{weight\_test/dnutries}}\\
\texttt{double dnutries} &  The number of ``tries'' for the interaction position across Antarctica before it would have landed within the horizon of the balloon.\\
\texttt{double weight\_test} & Weight due to absorption in the earth.  Is it the probability that the neutrino would have survived its trip through the earth.  When we fill this tree, we haven't stepped through the Earth using the full geographical model to calculate the absorption yet, so instead we just assume the 
neutrino only traverses ice on its way to the interaction.\\
\texttt{double peak\_e\_rx} & Peak of the signal waveform at the antenna receiver, first polarization.  No RFCM amplification or any attenuation of any sort imposed.  Just antenna effective heights imposed on the incident signal.\\
\texttt{double peak\_h\_rx} & Same as previous, but second polarization.\\
\texttt{double volts\_rx\_rfcm\_lab\_e[512]} & Time domain {\it signal} waveform read by the surf in the first polarization\\
\texttt{double peak\_e\_rx\_rfcm\_lab} & Peak of the previous waveform \\
\texttt{double volts\_rx\_rfcm\_lab\_h} & Time domain {\it signal} waveform read by the surf in the second polarization\\
\texttt{double peak\_h\_rx\_rfcm\_lab} & Peak of the previous waveform \\
\end{tabular}
\end{center}

\begin{center}
  \begin{tabular}{p{3.0in}p{3.2in}}
\texttt{double signal\_vpol\_inanita[5][512]} & Time domain {\it signal} waveforms (voltage v. time) for each of the five channels and each of the 512 samples in the vertical polarization only, before coverting to LCP, RCP for ANITA~I. \\
\texttt{double noise\_vpol\_inanita[5][512]} & Same as previous, but just the {\it noise} waveform in the vertical polarization only, with no signal.  This is before converting to LCP, RCP for ANITA~I. \\
\texttt{double total\_vpol\_inanita[5][512]} & Sum of the previous two waveforms. \\
\texttt{double total\_diodeinput\_1\_inanita[5][512]} & Waveform that is input to the tunnel diode for the first polarization.\\
\texttt{double total\_diodeinput\_2\_inanita[5][512]} & Waveform that is input to the tunnel diode for the second polarization.\\
\texttt{double peak\_e[5]} & Peak voltage of the signal waveform in the first polarization \\
\texttt{double peak\_h[5]} & Peak voltage of the noise waveform in the second polarization \\
\texttt{double timedomain\_output\_1\_inanita[5][512]} & Output waveforms of the tunnel diodes for each channel (voltage v. time) for the first polarization \\
\texttt{double timedomain\_output\_2\_inanita[5][512]} & Output waveforms of the tunnel diodes for each channel (voltage v. time) for the second polarization \\
\texttt{int flag\_e\_inanita[5][512]} & 1 when the channel has fired and 0 otherwise, for the first polarization.  This flag is held high for a L1 trigger coincidence window of 11.19~ns. \\
\texttt{int flag\_h\_inanita} & same as previous, but for the second polarization. \\
\texttt{double bwslice\_vrms[5]} & RMS noise voltage for each channel \\
\texttt{double ston[5]} & This is the peak signal voltage divided by rms noise voltage for each channel (\texttt{peak\_e[i]/bwslice\_vrms[i]}).  Not that this is not what is input to the discriminator, it is just one measure of signal strength. \\
\texttt{int channels\_passing\_e[5]} & Which channels in the first polarization pass the single-channel trigger at some time.  1 for yes, 0 for no. \\
\texttt{int channels\_passing\_h[5]/I} & Same as previous, but second polarization. \\
\texttt{int l1\_passing} & Whether this antenna passes the L1 trigger. 1 for yes, 0 for no. \\
 \texttt{double integral\_vmmhz} & Integral of the electric field in V/m/MHz incident on the closest antenna.  For the older frequency domain simulation, this was the measure of signal strength. \\
\end{tabular}
\end{center}




\end{document}

\documentstyle[12pt,epsfig]{article}

\setlength{\textwidth}{6.5in}
\setlength{\textheight}{8.5in}
%\setlength{\footskip}{0.35in}
\setlength{\topmargin}{0.0in}
\setlength{\oddsidemargin}{0in}
\setlength{\parindent}{5ex}
\setlength{\parskip}{.12in}
\newcommand{\detw}{\mbox{$\frac{\Delta E}{2}$}}
\newcommand{\hs}{\hspace*{5ex}}



%\def\decayright#1{\kern#1em\raise1.1ex\hbox{$|$}\kern-.5em\rightarrow}

\begin{document}


\begin{center}
{\Large
Summary of a Variables in passing\_events Tree
}
~\\
Amy Connolly\\
\today\\
\end{center}
~\\
~\\

\section{The Tree}

The tree called ``passing\_events'' is stored in outputs/icefinal.root, which
is written by \texttt{icemc}.  Only events that pass all three levels
of the Anita trigger are written to the tree.  When using this tree for
plotting, it is important to weight the events by \texttt{weight} (described
below).

All distances are in meters unless otherwise stated.

\begin{center}
  \begin{tabular}{p{2.2in}p{4.0in}}
\multicolumn{2}{p{6.0in}}{The next four variables use a 2d coordinate system 
from the view looking down on the South Pole, 
with the +x axis pointing in the direction of 
+90 longitude and +y in the 0$^{\circ}$ longitude direction }\\
\texttt{double horizcoord} & Position of the neutrino interaction in km along the +x axis \\
\texttt{double vertcoord} & Position of the neutrino interaction in km along the +y axis \\
\texttt{double horizcoord\_bn} & Position of the balloon in km along the +x axis \\
\texttt{double vertcoord\_bn} & Position of the balloon in km along the +y axis \\
\multicolumn{2}{p{6.0in}}{}\\
\texttt{double weight1} & Weight due to absorption in the earth.  Is it the probability that the neutrino would have survived its trip through the earth. \\
\texttt{double weight2} & A phase space factor.  For each balloon position, we choose the neutrino interaction position and direction such that the Cerenkov signal is visible at the balloon.  This phase space factor is supposed to remove the bias that this selection introduces. \\
\texttt{double weight} & Total weight for the neutrino interaction, which is \texttt{weight1$\times$ weight2}\\  
\texttt{double logweight} & log$_{10}$ of \texttt{weight} \\
\end{tabular}
\end{center}

\begin{center}
  \begin{tabular}{p{2.2in}p{4.0in}}
\multicolumn{2}{p{6.0in}}{The following arrays are defined by a coordinate system whose origin is at the center of the earth, +z is pointed towards the South Pole, +x is parallel to 90$^{\circ}$E at the South Pole and +y is parallel to 0$^{\circ}$ longitude at the South Pole.}\\
\texttt{double posnu[3]} & Position of the neutino interaction point \\
\texttt{double r\_bn[3]} & Position of the balloon \\
\texttt{double n\_bn[3]} & Normalized array pointing from center of the earth to the balloon \\
\texttt{double costheta\_nutraject} & $\cos{\theta_{\nu}}$ where $\theta_{\nu}$ is the zenith angle of the neutrino's incident trajectory \\
\texttt{double phi\_nutraject} & $\phi$ of the neutrino's incident trajectory\\
\texttt{double nnu[3]} & Normalized array pointing in the direction of the neutrino's incident trajectory\\
\texttt{double n\_exit2bn[5][3]} & Five 3-element arrays, each a unit vector pointing from the exit point to the balloon for the ray seen by the balloon.  We find the exit point in three iterations, and each array is the result of a different iteration.  So what you want is \texttt{n\_exit2bn[2]} because that is the third iteration.  The last two are zeros.\\
\texttt{double rfexit[5][3]} & Five 3-element arrays, each a unit vector pointing from the center of the earth to the point where the RF seen at the balloon exits the ice.  \\
\texttt{double n\_exit\_phi} & $\phi$ of \texttt{n\_exit2bn[2]} \\
\multicolumn{2}{p{6.0in}}{}\\
\texttt{double pnu} & Energy of the neutrino in eV\\
\texttt{double altitude\_int} & ``altitude'' of the interaction point relative to the ice surface (it is a negative number because the interaction is always below the surface)\\
\texttt{double elast\_y} & Inelasticity of the interaction (fraction of the neutrino's energy that goes into the hadronic shower) \\
\texttt{double emfrac} & Fraction of neutrino energy that goes into the EM component of the shower \\
\texttt{double hadfrac} & Fraction of the neutrino energy that goes into the hadronic component of the shower \\
\texttt{double sumfrac} & \texttt{emfrac+hadfrac}.  This can be less than unity since some of the energy can be carried away by a lepton.\\
\texttt{int nuflavor} & Neutrino flavor. 1=electron, 2=muon, 3=tau \\
\texttt{int current} & Type of interaction. 1=charged current, 2=neutral current. \\
\texttt{double viewangle} & Angle in the ice between the ray seen by the balloon and the neutrino direction in radians (actually in the firn just below the surface) \\
\texttt{double changle} & Cerenkov angle in the ice or firn, depending on where the interaction occured in radians\\
\texttt{double offaxis} & $|\texttt{viewangle-changle}|$ \\
\end{tabular}
\end{center}

\begin{center}
  \begin{tabular}{p{2.2in}p{4.0in}}
\texttt{int l1trig[3]} & For each trigger layer, which antennas pass L1.  16 bit, 16 bit and 8 bit for trigger layers 1 (top), 2 (bottom) and nadirs\\
\texttt{int l2trig[3][400]} & For each trigger layer, which set of three neighboring antennas pass L2.  16 bit, 16 bit and 8 bit for layers 1 (top) \& 2 (bottom) and nadirs.  Indexed according to the number of the centre antenna.  It is 400 long in the 2nd dimension so that the same array can be used for EeVA.  Only the relevant elements are filled. \\
\texttt{int l3trig[400]} & 16 bit number which says which phi sectors pass L3  \\
\texttt{double arrivaltimes[40]} & Time of arrival of the signal at each antenna, relative to the time the signal hits the first antenna \\
\texttt{double e\_component,h\_component} & Projection of the polarization of the signal along the horizontal (vertical) polarization of the antenna \\
\multicolumn{2}{p{6.0in}}{Each of the three arrays below are filled for the
antenna that is closest to the interaction (with the minimum arrival time).  The first index is the polarization (0=LCP, 1=RCP)}\\
\texttt{double rx0\_signal\_eachband[2][5]} & The signal strength in each band, including noise fluctutions.  Here the signal strength is $\int{V(f)} df$ where 
V(f) is the signal spectrum in Volts/Hz read by each channel 
and df is the frequency bin width. \\
\texttt{double rx0\_noise\_eachband[2][5]} & The average noise voltage that we 
use for each channel.\\
\texttt{double rx0\_threshold\_eachband[2][5]} & The threshold a signal
must exceed for a channel to pass,
in terms of a multiple of the average noise.\\
\texttt{double vmmhz\_max} & Electric field per frequency in V/m/MHz at FREQ\_HIGH=1200~MHz of the signal as it is incident on the payload\\
\texttt{double vmmhz\_min} & Electric field per frequency in V/m/MHz at FREQ\_LOW=200~MHz of the signal as it is incident on the payload\\
\texttt{volts\_rx\_rfcm\_lab\_e\_all[48][512]} & \\
 & Filled voltages in Volts for vertical polarization, 48 antennas and 512 samples each, with the
samples being every 1/2.6 ns so that the waveforms are 196.9 ns long\\
\texttt{volts\_rx\_rfcm\_lab\_h\_all[48][512]} & \\
 & Filled voltages in Volts for horizontal polarization, 48 antennas and 512 samples each, with the
samples being every 1/2.6 ns so that the waveforms are 196.9 ns long\\


\end{tabular}
\end{center}



\end{document}