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\documentclass[10pt]{article}
\usepackage{a4}
\usepackage{amsmath}

\oddsidemargin=0pt           % No extra space wasted after first inch.
\evensidemargin=0pt          % Ditto (for two-sided output).
\topmargin=0pt               % Same for top of page.
\headheight=0pt              % Don't waste any space on unused headers.
\headsep=0pt
\textwidth=16cm              % Effectively controls the right margin.
\textheight=24cm

\begin{document}
\centerline{\Large \bf Notes on TPG}
\bigskip
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\centerline{\large \today}

\section{Latency and drain-time limits on signal bandwidth overheads}
In addition to the pileup-generated rate of hits (both for DAQ and TPG),
there are also occasional signal events. Some of these can cause a large number
of hits in a small area, such as a motherboard. The bandwidth for data coming
from the motherboard must include some overhead above the value needed for the
pileup as otherwise these signal events will eventually cause buffer overflows.

There are two issues; latency (particularly important for the TPG) and the
``drain-time'' of the FE buffers. This latter is the time taken to flush out the
extra data from the signal event such that the buffers are back at the steady
state of handling the average pileup. A long drain-time would mean the buffers
are fuller than normal and so there is a risk that another signal event could 
cause an overflow. The effect of latency is clear but it is
harder to put a firm requirement on the drain-time.

Let the maximum signal plus pileup data volume for a BX in a panel be $s$ bits 
and the average pileup data volume per BX be $p$ bits, where clearly $s>p$.
The ``extra'' data due to the signal is obviously $s-p$ bits. A bandwidth $b$
is to be determined, where $b > p$. Clearly $b=s$ would allow any signal event 
to be read out within one BX but would be more bandwidth that really required.
Hence, that the bandwidth is set to be that required for the pileup plus some 
fraction $f$ of the extra data due to the signal, i.e.
\begin{equation} 
b = p+f(s-p) = fs+(1-f)p
\end{equation} 
For this bandwidth then the drain-time $d$ is given by the amount of
extra data, $s-p$, divided by the extra bandwidth above the pileup rate,
$f(s-p)$, i.e.
\begin{equation} 
d = \frac{s-p}{f(s-p)} = \frac{1}{f}
\end{equation} 
The latency $l$ is the time to read out the whole signal event given the total
bandwidth, so 
\begin{equation} 
l = \frac{s}{b} = \frac{s}{fs+(1-f)p} = \frac{1}{f[1-(p/s)]+(p/s)}
\end{equation} 
Note that
\begin{equation} 
\frac{l}{d} = \frac{1}{[1-(p/s)]+(p/fs)} = \frac{1}{1+(p/s)(1/f-1)}<1
\end{equation} 
i.e. the latency is always shorter than the drain-time.

The value of the fraction is obviously limited to $0 < f < 1$ so the limits on
the latency are $(s/p) > l > 1$. The lower limit, which occurs for $f=1$, is
simply adding in enough bandwidth so $b=s$. The upper limit says that even with
zero overhead bandwidth, where $b=p$, the latency for reading out the signal
event is simply $s/b=s/p$.

If this upper latency limit is below the allowable latency, then the drain-time
is the only relevant limit. Otherwise, both limits need to be checked to set the
value of $f$. Take an example; with a latency limit of 10 BX, a drain-time
limit (semi-arbitrarily) of 40 BX (i.e. before the next L1 trigger on average),
and a ratio of $s/p = 20$, then the latency limit means the fraction must be
\begin{equation} 
f \ge \frac{(1/l)-(p/s)}{1-(p/s)} = \frac{0.1-0.05}{1-0.1} = \frac{1}{19}
\end{equation} 
which is stricter that the drain-time limit $f \ge 1/40$ and so sets the 
required fraction. The two limits are equivalent for a particular value
of $p/s$ which is given by
\begin{equation} 
f = \frac{(1/l)-(p/s)}{1-(p/s)} = \frac{1}{d}
\end{equation} 
which gives
\begin{equation} 
\frac{p}{s} = \frac{(d/l)-1}{d-1}
\end{equation} 
and for the example here results in
\begin{equation} 
\frac{p}{s} = \frac{4-1}{40-1} = \frac{3}{39} = \frac{1}{13}
\end{equation} 
Hence, if $s/p \le 13$, then the drain-time limit is stricter.



\section{Hexagonal integer grid}
The centres of a regular array of hexagons can be considered to lie on an
effective Cartesian grid, with the $y$ axis unit equal to $3/2$ of the hexagon
side and the $x$ axis unit equal to $\sqrt{3}/2$ of the hexagonal side, i.e.
the ratio of $y/x$ units is $\sqrt{3}$. The centres of the hexagons then
lie on integer values of $x$ and $y$, $i_x$ and $i_y$, and
occupy a chessboard pattern, here assumed to have the sum of $i_x+i_y$ being
even. The centres are 2 units apart along the $x$ axis but are 1 unit apart
along the $y$ axis. E.g. the valid wafer along the $x$ axis are at $i_y=0$
and $i_x = \dots, -4, -2, 0, 2, 4, \dots$. The next row up is $i_y=1$
and $i_x = \dots, -3, -1, 1, 3, \dots$, while the next row down is $i_y=-1$
with the same values of $i_x$.

The valid wafer centres can also be described by plane polar coordinates,
$i_r$ and $i_\phi$, where $i_r \ge 0$ and 
$i_\phi$ is limited to the range 0 to $6i_r-1$,
except the centre $i_x=0$ and $i_y=0$ has $i_r=0$ and $i_\phi=0$.
\bigskip
\begin{center}
\begin{tabular}{c|c}
\hline
$i_x, i_y$ & $i_r, i_\phi$ \cr\hline
$ 0, 0$ & $0, 0$ \cr\hline
$ 2, 0$ & $1, 0$ \cr 
$ 1, 1$ & $1, 1$ \cr 
$-1, 1$ & $1, 2$ \cr 
$-2, 0$ & $1, 3$ \cr 
$-1,-1$ & $1, 4$ \cr 
$ 1,-1$ & $1, 5$ \cr\hline
$ 4, 0$ & $2, 0$ \cr
$ 3, 1$ & $2, 1$ \cr
$ 2, 2$ & $2, 2$ \cr
$ 0, 2$ & $2, 3$ \cr
$-2, 2$ & $2, 4$ \cr
$-3, 1$ & $2, 5$ \cr
$-4, 0$ & $2, 6$ \cr
$-3,-1$ & $2, 7$ \cr
$-2,-2$ & $2, 8$ \cr
$ 0,-2$ & $2, 9$ \cr
$ 2,-2$ & $2,10$ \cr
$ 3,-1$ & $2,11$ \cr
\hline
\end{tabular}
\end{center}

A rotation by $\Delta\phi = 60^\circ$ and $-60^\circ \equiv 300^\circ$ 
in Cartesian coordinates would be
\begin{equation} 
\begin{pmatrix}
x^\prime \\ y^\prime
\end{pmatrix}
= \begin{pmatrix}
1/2 & \mp\sqrt{3}/2 \\ \pm\sqrt{3}/2 & 1/2
\end{pmatrix}
\begin{pmatrix}
x \\ y
\end{pmatrix}
\end{equation}
But the units mean $x^{(\prime)} \propto i_x^{(\prime)}$ while 
$y^{(\prime)} \propto \sqrt{3}i_y^{(\prime)}$ so
\begin{equation} 
\begin{pmatrix}
i_x^\prime \\ \sqrt{3}i_y^\prime
\end{pmatrix}
= \begin{pmatrix}
1/2 & \mp\sqrt{3}/2 \\ \pm\sqrt{3}/2 & 1/2
\end{pmatrix}
\begin{pmatrix}
i_x \\ \sqrt{3}i_y
\end{pmatrix}
\qquad{\rm so}\qquad
\begin{pmatrix}
i_x^\prime \\ i_y^\prime
\end{pmatrix}
= \begin{pmatrix}
1/2 & \mp 3/2 \\ \pm 1/2 & 1/2
\end{pmatrix}
\begin{pmatrix}
i_x \\ i_y
\end{pmatrix}
\end{equation}
A rotation of $120^\circ$ and $-120^\circ \equiv 240^\circ$ is therefore
\begin{equation} 
\begin{pmatrix}
i_x^\prime \\ i_y^\prime
\end{pmatrix}
= \begin{pmatrix}
-1/2 & \mp 3/2 \\ \pm 1/2 & -1/2
\end{pmatrix}
\begin{pmatrix}
i_x \\ i_y
\end{pmatrix}
\end{equation}
while a rotation of $180^\circ$ is simply
\begin{equation} 
\begin{pmatrix}
i_x^\prime \\ i_y^\prime
\end{pmatrix}
= \begin{pmatrix}
-1 & 0 \\ 0 & -1
\end{pmatrix}
\begin{pmatrix}
i_x \\ i_y
\end{pmatrix}
\end{equation}
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\section{Optimisation of layer weights}
Each event $e$ gives some values $d_{e,i}$ of the deposited energy in 
layer $i$; these
can be in any units, e.g. MIPs.
Assume these are to be multiplied by some constant coefficients $a_i$ (which
are approximately the integrated dE/dx values if the $d_{e,i}$ are in MIPs)
to give the estimate of the incoming EM photon or electron energy. 
Hence, the energy
estimation for event $e$ is
\begin{equation} 
E_e = \sum_i a_i d_{e,i}
\end{equation} 
To find the optimal coefficients, then we need to know the truth energy per
event $T_e$. For a given set of coefficients, the RMS$^2$ of the energy 
around the truth value is given by
\begin{equation} 
\mathrm{RMS}^2 = \frac{1}{N} \sum_e (E_e - T_e)^2
= \frac{1}{N} \sum_e \left(\sum_i a_i d_{e,i} - T_e\right)^2
\end{equation} 
This can be thought of as similar to a chi-squared; we want to minimise this
expression. If all the $a_i$ are considered as independent parameters (so 28 for
the EE only), then explicitly
\begin{equation} 
\frac{\partial \mathrm{RMS}^2}{\partial a_j}
= \frac{1}{N} \sum_e 2d_{e,j} \left(\sum_i a_i d_{e,i} - T_e\right)
= \frac{2}{N} \sum_i a_i \left(\sum_e d_{e,j} d_{e,i} \right)
- \frac{2}{N} \sum_e d_{e,j} T_e 
\end{equation} 
Hence, for the minimum, we require
\begin{equation} 
\sum_i \frac{\sum_e d_{e,j} d_{e,i}}{N} a_i
= \frac{\sum_e d_{e,j} T_e}{N} 
\end{equation} 
Writing in matrix notation with $M$ and $v$ defined as
\begin{equation} 
M_{ji} = \frac{\sum_e d_{e,j} d_{e,i}}{N},\qquad
v_j = \frac{\sum_e d_{e,j} T_e}{N}
\end{equation} 
then the requirement is
\begin{equation} 
M a = v\qquad\mathrm{so}\qquad a = M^{-1}v
\end{equation}
Inverting the large matrix $M$ is required to give the solution for the
optimal $a_i$.
Note, $M$ is similar (but not identical) to the error matrix of the $d_i$.

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The resulting RMS using the best fit values is given by
\begin{eqnarray*}
\mathrm{RMS}^2_\mathrm{min}
&=& \frac{1}{N} \sum_e \left[
\left(\sum_i a_i d_{e,i} \right)^2
- 2 T_e \sum_i a_i d_{e,i} + T_e^2 \right] \\
&=& \frac{1}{N} \sum_j \sum_i a_j a_i \sum_e d_{e,j} d_{e,i}
- \frac{2}{N} \sum_i a_i \sum_e T_e d_{e,i} + \frac{1}{N} \sum_e T_e^2 \\
&=& \sum_j \sum_i a_j a_i M_{ji}
- 2 \sum_i a_i v_i + \frac{1}{N} \sum_e T_e^2 
= a^T M a - 2 a^T v + \frac{1}{N} \sum_e T_e^2 
\end{eqnarray*}
But since the solution is defined by $Ma=v$, then $a^T M a = a^T v$. Hence
\begin{equation} 
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\mathrm{RMS}^2_\mathrm{min} = \frac{1}{N} \left(\sum_e T_e^2\right) - a^T M a
= \frac{1}{N} \left(\sum_e T_e^2\right) - v^T M^{-1} v
= \frac{1}{N} \left(\sum_e T_e^2\right) - a^T v
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\end{equation}
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The above can be extended slightly, which may improve the energy response 
linearity as well as the RMS. The energy estimation for the event (i.e. the 
first equation in this section) can be generally considered to be a polynomial 
in the $d_{e,i}$, but with the quadratic and higher terms neglected. However,
it also neglects any constant term. A more general expression would then be
to add another coefficient $b$ to give
\begin{equation} 
E_e = b + \sum_i a_i d_{e,i}
\end{equation} 
The easiest way to handle this is to allow the index $i$ to go one higher than
previously, specifically change from $i=0,27$ to $i=0,28$ and then define
$a_{28}=b$ and $d_{e,28}=1$. This means the expression simplifies to
\begin{equation} 
E_e = \sum_{i=0}^{28} a_i d_{e,i}
\end{equation} 
and so an identical calculation to previously holds, simply with the index 
running
over a larger range. Explicitly,
the matrix $M$ is now $29\times 29$ with the extra elements being
\begin{equation} 
M_{i,28} = M_{28,i} = \frac{1}{N} \sum_e d_{e,28}d_{e,i} 
= \frac{1}{N} \sum_e d_{e,i}
\end{equation} 
and
\begin{equation} 
M_{28,28} = \frac{1}{N} \sum_e d_{e,28}d_{e,28} = 1
\end{equation} 
while the extra element in $v$ is
\begin{equation} 
v_{28} = \frac{1}{N} \sum_e T_e d_{e,28} = \frac{1}{N} \sum_e T_e
\end{equation} 

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\section{Units}
Keeping quantities to 16-bit integers.

The FE ASIC works in fC with an overall LSB of 0.1\,fC and upper range of
10\,pC $= 10^4$\,fC. This requires 17 bits total (although represented as
a 10-bit and a 12-bit pair of values.

Reconstructed energy (not deposited energy) with an LSB of 10\,MeV and 16-bit 
unsigned representation gives a maximum energy of 655\,GeV. These are 
initially MIPS $\times \int (dE/dx)\,dx$ for each layer
until after forming the 3D clusters
when the total energy is set more exactly.

Position in $x$ and $y$ with an LSB of 100\,$\mu$m and a 16-bit signed 
representation gives a range of $\pm 328$\,cm (with $\pm 190$\,cm required).
If needed, $z$ can be represented in a 16-bit unsigned representation with the 
same LSB, giving a range up to 655\,cm (with 408\,cm required).
Note, the endcaps are handled
separately so the negative $z$ endcap can be treated like the positive
$z$ one.

Sine and cosines can be represented in a 16-bit signed representation 
where they are multiplied by $2^{15}$. Hence, the result of a multiplication
by this value needs to be stored in up to 31 bits and then bitshifted by 15.
Note, this does not allow a representation of exactly $+1$, 
i.e. for angles of 0
or $\pi/2$. Neither of these should occur in the HGC. Similarly, if needed,
$\tan(\theta)$ is in the appproximate range $\pm 0.1$ to $\pm 0.5$ and so can
be represented in the same way (and hence is similar to $\sin\theta$ for small
angles). Hence, the scaled variables $x/z = \tan\theta \cos\phi$ and
$y/z = \tan\theta \sin\phi$ can also have the same representation.

\section{FE ASIC TOT non-linearity}
Modelled as a response $r$ for an input charge $q$ given by
\begin{equation} 
r = 0\quad\mathrm{for}\ q<100\,\mathrm{fC},
\qquad r = q - \frac{100(100-q_0)}{q-q_0} 
= q \left[1 - \frac{100(100-q_0)}{q(q-q_0)}\right]
\quad \mathrm{otherwise}
\end{equation} 
where value of the
parameter is chosen to be $q_0=90$\,fC.
For $q(q-q_0) \gg 100(100-q_0)=1000$\,fC$^2$,
the non-linear term becomes negligible. 
E.g. for $q=200$\,fC, then $q(q-q_0) = 22000$ and so is a 5\% correction,
while for $q=400$\,fC, then $q(q-q_0) = 124000$ and so is a 0.8\% correction.

Inverting the above response function for $q \ge 100$\,fC, then
\begin{equation} 
r(q-q_0) = q(q-q_0) - 100(100-q_0)
\qquad\mathrm{so}\qquad
q^2-q(q_0+r)+rq_0 -100(100-q_0)
\end{equation} 
Hence
\begin{equation} 
q = \frac{1}{2}\left[q_0+r \pm \sqrt{(q_0+r)^2-4rq_0+400(100-q_0)}\right]
=\frac{q_0+r}{2} \pm \sqrt{\left(\frac{q_0-r}{2}\right)^2 + 100(100-q_0)}
\end{equation} 
where the positive sign is required for $q>100$\,fC.

\section{Link data representation}
The selected trigger cell 
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data are calculated to a large number of bits, typically 16-18.
On the links, they need to be represented in a small number of bits $n$,
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typically $\sim 8$.
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This could be linear or logarithmic or floating.

\subsection{Linear representation}
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For linear, then in general it can be linear betwen $x_\mathrm{min}$ and
$x_\mathrm{max}$ and 0 or $2^n-1$ outside this range. This can be represented
within the range by
\begin{equation}
y = \frac{(2^n-1)(x-x_\mathrm{min})}{x_\mathrm{max}-x_\mathrm{min}}
\end{equation} 
which can be inverted to give
\begin{equation}
x = x_\mathrm{min} + \frac{y(x_\mathrm{max}-x_\mathrm{min})}{2^n-1}
\end{equation} 
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\subsection{Logarithmic representation}
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For logarithmic, the general case would be $y=a\log(x)+b$ but with
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$c=b/a$ and $x_\mathrm{min}=e^{-c}$, then
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\begin{equation}
y = a\log(x)+b = a\log(x)+ac = a(\log(x)+c) = a(\log(x)-\log(x_\mathrm{min}))
= a \log(x/x_\mathrm{min})
\end{equation} 
Therefore
\begin{equation}
y = (2^n-1) \frac{\log(x/x_\mathrm{min})}{\log(x_\mathrm{max}/x_\mathrm{min})}
\end{equation} 
This can be inverted to give
\begin{equation}
x = x_\mathrm{min}\left(\frac{x_\mathrm{max}}{x_\mathrm{min}}\right)^{y/(2^n-1)}
\end{equation} 

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\subsection{Float representation}
Here, the $2^n$ values are split into an exponent of $E$ bits and a mantissa of
$M$ bits. The naive approach is simply to take the actual value as the
mantissa shifted up by $E$ bits. For example, for $E=2$ and $M=2$, then the
16 possible values would give the table below.

\bigskip
\begin{center}
\begin{tabular}{c|c|c|c}
\hline
Representation & Exponent & Mantissa & Value \cr\hline
 0 & 0b00 & 0b00 & 0b00000 = \phantom{2}0\cr
 1 & 0b00 & 0b01 & 0b00001 = \phantom{2}1\cr
 2 & 0b00 & 0b10 & 0b00010 = \phantom{2}2\cr
 3 & 0b00 & 0b11 & 0b00011 = \phantom{2}3\cr
 4 & 0b01 & 0b00 & 0b00000 = \phantom{2}0\cr
 5 & 0b01 & 0b01 & 0b00100 = \phantom{2}2\cr
 6 & 0b01 & 0b10 & 0b01000 = \phantom{2}4\cr
 7 & 0b01 & 0b11 & 0b01100 = \phantom{2}6\cr
 8 & 0b10 & 0b00 & 0b00000 = \phantom{2}0\cr
 9 & 0b10 & 0b01 & 0b00100 = \phantom{2}4\cr
10 & 0b10 & 0b10 & 0b01000 = \phantom{2}8\cr
11 & 0b10 & 0b11 & 0b01100 = 12 \cr
12 & 0b11 & 0b00 & 0b00000 = \phantom{2}0\cr
13 & 0b11 & 0b01 & 0b01000 = \phantom{2}8\cr
14 & 0b11 & 0b10 & 0b10000 = 16 \cr
15 & 0b11 & 0b11 & 0b11000 = 24 \cr
\hline
\end{tabular}
\end{center}

It is clear this is neither monotonic nor efficient, as the same values appear
for several representations.
A better representation is made by realising that for all but the lowest
exponent representations, there is always a leading bit. Hence, this does not
have to be stored explicitly. This means this leading bit must be added to the
mantissa before the bit shift, and since this increments the length by one bit,
then the shift up needed is only $E-1$. The table below shows this improved
representation. It is monotonic, there are no duplicates, and the lowest two
exponent ranges give an exact representation.

\bigskip
\begin{center}
\begin{tabular}{c|c|c|c}
\hline
Representation & Exponent & Mantissa & Value \cr\hline
 0 & 0b00 & 0b00 & 0b00000 = \phantom{2}0\cr
 1 & 0b00 & 0b01 & 0b00001 = \phantom{2}1\cr
 2 & 0b00 & 0b10 & 0b00010 = \phantom{2}2\cr
 3 & 0b00 & 0b11 & 0b00011 = \phantom{2}3\cr
 4 & 0b01 & 0b00 & 0b00100 = \phantom{2}4\cr
 5 & 0b01 & 0b01 & 0b00101 = \phantom{2}5\cr
 6 & 0b01 & 0b10 & 0b00110 = \phantom{2}6\cr
 7 & 0b01 & 0b11 & 0b00111 = \phantom{2}7\cr
 8 & 0b10 & 0b00 & 0b01000 = \phantom{2}8\cr
 9 & 0b10 & 0b01 & 0b01010 = 10 \cr
10 & 0b10 & 0b10 & 0b01100 = 12 \cr
11 & 0b10 & 0b11 & 0b01110 = 14 \cr
12 & 0b11 & 0b00 & 0b10000 = 16 \cr
13 & 0b11 & 0b01 & 0b10100 = 20 \cr
14 & 0b11 & 0b10 & 0b11000 = 24 \cr
15 & 0b11 & 0b11 & 0b11100 = 28 \cr
\hline
\end{tabular}
\end{center}

In this improved representation, the mantissa has $M+1$ bits (except in
the lowest exponent range). The exponent can represent numbers up to
$2^E-1$ and hence will bit shift by a maximum of $2^E-2$ bits.
Hence, the number of bits in the representation is $M+E$ bits, while
the maximum number represented has $M+1+2^E-2 = M+2^E-1$ bits, i.e. is
is less than $2^{M+2^E-1}$. The reduction is $2^E-E-1$ bits.
For the example of $M=2$, $E=2$ 
in the table above, this gives $2-1+4=5$ bits, i.e.
numbers up to $2^5=32$ as shown and the reduction is 1 bit.

For the extreme values of $E$, then $E=0$ and $E=1$ both give an exact
representation as they only use the lowest range or two lowest ranges,
respectively. The reduction is $2^0-0-1=0$ and $2^1-1-1=0$ bits in both cases. 
Explicitly, for $E=0$, then the representation has $M$ bits
while the value represented has $M$
bits also, i.e. the reduction is 0 bits.
For $E=1$, the representation has $M+1$ bits, while the 
value represented has $M+1$ bits also, again with a reduction of 0 bits.

For the extreme value of $M=0$, then the representation is
just the number of bits in the input word. E.g. for $M=0$, $E=4$, then the
table is given below. The value range is less thn $2^{15}=32768$.

\bigskip
\begin{center}
\begin{tabular}{c|c|c|c}
\hline
Representation & Exponent & Mantissa & Value \cr\hline
 0 & 0b0000 & 0 & 0b000000000000000 = \phantom{1222}0\cr
 1 & 0b0001 & 0 & 0b000000000000001 = \phantom{1222}1\cr
 2 & 0b0010 & 0 & 0b000000000000010 = \phantom{1222}2\cr
 3 & 0b0011 & 0 & 0b000000000000100 = \phantom{1222}4\cr
 4 & 0b0100 & 0 & 0b000000000001000 = \phantom{1222}8\cr
 5 & 0b0101 & 0 & 0b000000000010000 = \phantom{122}16\cr
 6 & 0b0110 & 0 & 0b000000000100000 = \phantom{122}32\cr
 7 & 0b0111 & 0 & 0b000000001000000 = \phantom{122}64\cr
 8 & 0b1000 & 0 & 0b000000010000000 = \phantom{12}128\cr
 9 & 0b1001 & 0 & 0b000000100000000 = \phantom{12}256 \cr
10 & 0b1010 & 0 & 0b000001000000000 = \phantom{12}512 \cr
11 & 0b1011 & 0 & 0b000010000000000 = \phantom{1}1024 \cr
12 & 0b1100 & 0 & 0b000100000000000 = \phantom{1}2048 \cr
13 & 0b1101 & 0 & 0b001000000000000 = \phantom{1}4096 \cr
14 & 0b1110 & 0 & 0b010000000000000 = \phantom{1}8192 \cr
15 & 0b1111 & 0 & 0b100000000000000 = 16384 \cr
\hline
\end{tabular}
\end{center}

The maximum bit lengths of the value, i.e. $M+2^E-1$, for various values of 
$M$ and $E$ are shown in the table below.

\bigskip
\begin{center}
\begin{tabular}{r||c|c|c|c|c|c|c|c|c}
\hline
$M=$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \cr\hline
$E=0$ & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \cr
1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 \cr
2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 \cr
3 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \cr
4 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \cr
5 & 31 & 32 & 33 & 34 & 35 & 36 & 37 & 38 & 39 \cr
6 & 63 & 64 & 65 & 66 & 67 & 68 & 69 & 70 & 71 \cr
7 & 127 & 128 & 129 & 130 & 131 & 132 & 133 & 134 & 135 \cr
8 & 255 & 256 & 257 & 258 & 259 & 260 & 261 & 262 & 263 \cr
\hline
\end{tabular}
\end{center}


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\section{Template fit of energy in depth}
Assume a template shape for a photon of $P_l$ per photon energy GeV 
for layer $l$.
The minimum bias gives a template shape of $M_l$ for layer $l$ in some
arbitrary units. The total expected per layer will then be
$E_l = E_pP_l + E_m M_l$ for a photon energy $E_p$ and some scaling $E_m$ of the
minimum bias template. Hence, the chi-squared compared to the observed
energy $O_l$ will be
\begin{equation} 
\chi^2 = \sum_l \frac{(E_p P_l + E_m M_l - O_l)^2}{\sigma_l^2}
\end{equation}
The $\sigma_l$ are given by the photon and minimum bias shower
fluctuations around
the average of the template. As such, the errors will depend on $E_p$ and
$E_m$. However, ``expected'' values can be used initially to fix the
$\sigma_l$ so that the problem remains linear. It could then be iterated 
several times with improved values to get a better fit.

Minimising the chi-squared requires
\begin{equation} 
\frac{d\chi^2}{dE_p} = \sum_l \frac{2P_l(E_p P_l + E_m M_l - O_l)}{\sigma_l^2}=0,\qquad
\frac{d\chi^2}{dE_m} = \sum_l \frac{2M_l(E_p P_l + E_m M_l - O_l)}{\sigma_l^2}=0
\end{equation}
such that
\begin{equation} 
E_p \sum_l \frac{P_l^2}{\sigma_l^2} + E_m\sum_l \frac{P_lM_l}{\sigma_l^2}
= \sum_l \frac{O_l P_l}{\sigma_l^2},
\qquad
E_p \sum_l \frac{P_l M_l}{\sigma_l^2} + E_m\sum_l \frac{M_l^2}{\sigma_l^2}
= \sum_l \frac{O_l M_l}{\sigma_l^2},
\end{equation}
This can be written as a matrix equation
\begin{equation} 
\begin{pmatrix}
\sum_l \frac{P_l^2}{\sigma_l^2} & \sum_l \frac{P_l M_l}{\sigma_l^2} \\
\sum_l \frac{P_l M_l}{\sigma_l^2} & \sum_l \frac{M_l^2}{\sigma_l^2}
\end{pmatrix}
\begin{pmatrix}
E_p \\ E_m
\end{pmatrix} =
\begin{pmatrix}
\sum_l \frac{O_l P_l}{\sigma_l^2} \\
\sum_l \frac{O_l M_l}{\sigma_l^2}
\end{pmatrix}
\end{equation}
As long as the $P_l$ and $M_l$ are not proportional to each other, the
matrix on the left can be inverted to solve for $E_p$ (and $E_m$).
This matrix is a constant for all events and so can be precalculated and
inverted once, offline. The vector on the right must be calculated per
event. However, explicitly the matrix determinant is
\begin{equation} 
\Delta = \left(\sum_l \frac{P_l^2}{\sigma_l^2}\right)
\left(\sum_l \frac{M_l^2}{\sigma_l^2}\right)
- \left(\sum_l \frac{P_l M_l}{\sigma_l^2} \right)^2
\end{equation}
so the inverse is
\begin{equation} 
\frac{1}{\Delta}
\begin{pmatrix}
\sum_l \frac{M_l^2}{\sigma_l^2} & -\sum_l \frac{P_l M_l}{\sigma_l^2} \\
-\sum_l \frac{P_l M_l}{\sigma_l^2} & \sum_l \frac{P_l^2}{\sigma_l^2}
\end{pmatrix}
\end{equation}
and hence
\begin{equation} 
\begin{pmatrix}
E_p \\ E_m
\end{pmatrix} =
\frac{1}{\Delta}
\begin{pmatrix}
\sum_{l^\prime} \frac{M_{l^\prime}^2}{\sigma_{l^\prime}^2} & -\sum_{l^\prime} \frac{P_{l^\prime} M_{l^\prime}}{\sigma_{l^\prime}^2} \\
-\sum_{l^\prime} \frac{P_{l^\prime} M_{l^\prime}}{\sigma_{l^\prime}^2} & \sum_{l^\prime} \frac{P_{l^\prime}^2}{\sigma_{l^\prime}^2}
\end{pmatrix}
\begin{pmatrix}
\sum_l \frac{O_l P_l}{\sigma_l^2} \\
\sum_l \frac{O_l M_l}{\sigma_l^2}
\end{pmatrix}
\end{equation}
which means
\begin{equation} 
\begin{pmatrix}
E_p \\ E_m
\end{pmatrix} =
\begin{pmatrix}
\sum_l O_l 
\left[\frac{P_l}{\Delta \sigma_l^2}
\left( \sum_{l^\prime} \frac{M_{l^\prime}^2}{\sigma_{l^\prime}^2}\right)
- \frac{M_l}{\Delta \sigma_l^2}
\left( \sum_{l^\prime} \frac{P_{l^\prime} M_{l^\prime}}{\sigma_{l^\prime}^2}\right) \right]\\
\sum_l O_l 
\left[\frac{M_l}{\Delta \sigma_l^2}
\left(\sum_{l^\prime} \frac{P_{l^\prime}^2}{\sigma_{l^\prime}^2}\right)
-\frac{P_l}{\Delta \sigma_l^2}
\left(\sum_{l^\prime} \frac{P_{l^\prime} M_{l^\prime}}{\sigma_{l^\prime}^2}\right)\right]
\end{pmatrix}
\end{equation}
Hence
\begin{equation} 
E_p = \sum_l O_l A_l,\qquad
E_m = \sum_l O_l B_l
\end{equation}
where $A_l$ and $B_l$ correspond to the quantities in the square brackets
and can be precalculated, except for any subtleties with the errors.

ERROR MATRIX

\section{Comparing coordinates in plane polars}
On a given layer, then comparing e.g. a track extrapolation to a cluster 
position requires a difference of the two points in 2D;
$x_1$, $y_1$ and $x_2$, $y_2$.
This should be done
in coordinates which preserve the cylindrical (i.e. plane polar for a layer)
geometry. The obvious two are
\begin{equation} 
\Delta\rho = \rho_2-\rho_1 = \sqrt{x_2^2+y_2^2}-\sqrt{x_1^2+y_1^2},
\qquad
\Delta\phi = \phi_2 - \phi_1 = \tan^{-1}(y_2/x_2) - \tan^{-1}(y_1/x_1)
\end{equation}
However, $\Delta\phi$ has two issues; firstly is that it is not a length 
variable
and so makes comparison with $\Delta\rho$ difficult, and secondly that there is
a discontinuity in $\phi$ which needs to be handled.

A length variable can be formed using some radius value $\overline{\rho}$
to give $\overline{\rho}\Delta\phi$
but there is an ambiguity about which radius to use; $\rho_1$, $\rho_2$ or
some average of these. One desirable property is that the two
variables should preserve the total length, i.e.
\begin{equation} 
\Delta\rho^2 + \overline{\rho}^2\Delta\phi^2
= \Delta x^2 + \Delta y^2 = (x_2-x_1)^2 + (y_2-y_1)^2
= x_2^2+y_2^2+x_1^2+y_1^2 - 2(x_1x_2+y_1y_2)
\end{equation}
Using the expression for $\Delta\rho$ above, then
\begin{equation} 
\Delta\rho^2 = x_2^2+y_2^2+x_1^2+y_1^2 - 2\sqrt{x_2^2+y_2^2}\sqrt{x_1^2+y_1^2}
\end{equation}
so that
\begin{equation} 
\overline{\rho}^2\Delta\phi^2
=2\sqrt{x_2^2+y_2^2}\sqrt{x_1^2+y_1^2}- 2(x_1x_2+y_1y_2)
\end{equation}
Expressing the right hand side in plane polars gives
\begin{equation} 
\overline{\rho}^2\Delta\phi^2
=2\rho_1\rho_2 - 2\rho_1\rho_2 (\cos\phi_1\cos\phi_2 + \sin\phi_1\sin\phi_2)
=2\rho_1\rho_2 [1-\cos(\phi_2-\phi_1)]=2\rho_1\rho_2 (1-\cos\Delta\phi)
\end{equation}
This effectively defines $\overline{\rho}$ and hence the second variable
directly. Note there is no issue with the $\phi$ discontinuity as this is
handled automatically by the cosine.
For small $\Delta\phi$, then the above expression is approximated by
\begin{equation} 
\overline{\rho}^2\Delta\phi^2
\approx 2\rho_1\rho_2 \frac{\Delta\phi^2}{2}
\approx \rho_1\rho_2 \Delta\phi^2
\end{equation}
so that $\overline{\rho} \approx \sqrt{\rho_1\rho_2}$, i.e. the geometric
mean.

Note that the sign of
the second variable is not defined by the above; it should be the same as 
the sign of $\Delta\phi$. However, since
\begin{equation} 
1 - \cos\Delta\phi = 2\sin^2\left(\frac{\Delta\phi}{2}\right)
\end{equation}
then
\begin{equation} 
\overline{\rho}^2\Delta\phi^2
=4\rho_1\rho_2 \sin^2\left(\frac{\Delta\phi}{2}\right)
\end{equation}
and so
\begin{equation} 
\overline{\rho}\Delta\phi
=2\sqrt{\rho_1\rho_2} \sin\left(\frac{\Delta\phi}{2}\right)
\end{equation}
where the positive sign for the square-root is taken to agree with $\Delta\phi$.
Again, for small $\Delta\phi$, then $\overline{\rho}$
clearly approximates to the geometric
mean of the two radii, as before.

\section{Shower position and direction fit}

\section{Motion in a magnetic field}

\section{Inverting matrices}
A symmetric $3\times 3$ matrix can be written as
\begin{equation} 
M=
\begin{pmatrix}
M_{00} & M_{01} & M_{02} \\
M_{01} & M_{11} & M_{12} \\
M_{02} & M_{12} & M_{22} 
\end{pmatrix}
\end{equation}
Its determinant is then
\begin{eqnarray} 
\Delta
&=&
M_{00} \begin{vmatrix}
M_{11} & M_{12} \\
M_{12} & M_{22} 
\end{vmatrix}
-M_{01} \begin{vmatrix}
M_{01} & M_{12} \\
M_{02} & M_{22} 
\end{vmatrix}
+M_{02} \begin{vmatrix}
M_{01} & M_{11} \\
M_{02} & M_{12} 
\end{vmatrix}\\
&=& M_{00}M_{11}M_{22}-M_{00}M_{12}M_{12}
 -M_{01}M_{01}M_{22}+M_{01}M_{12}M_{02}
 +M_{02}M_{01}M_{12}-M_{02}M_{11}M_{02}\\
&=& M_{00}M_{11}M_{22}+2M_{01}M_{12}M_{02}
-M_{00}M_{12}^2 -M_{11}M_{02}^2 -M_{22}M_{01}^2
\end{eqnarray}
This can be written is several ways
\begin{eqnarray} 
\Delta
&=& M_{00}(M_{11}M_{22}-M_{12}^2)
+M_{01}(M_{12}M_{02}-M_{22}M_{01})
+M_{02}(M_{01}M_{12}-M_{11}M_{02})\\
&=& M_{01}(M_{12}M_{02}-M_{22}M_{01})
+M_{11}(M_{00}M_{22}-M_{02}^2)
+M_{12}(M_{01}M_{02}-M_{00}M_{12})\\
&=&M_{02}(M_{01}M_{12}-M_{11}M_{02})
+M_{12}(M_{01}M_{02}-M_{00}M_{12})
+M_{22}(M_{00}M_{11}-M_{01}^2)
\end{eqnarray}
which means the inverse must be
\begin{equation} 
M^{-1}=
\frac{1}{\Delta}
\begin{pmatrix}
M_{11}M_{22}-M_{12}^2 & M_{12}M_{02}-M_{22}M_{01} & M_{01}M_{12}-M_{11}M_{02} \\
M_{12}M_{02}-M_{22}M_{01} & M_{00}M_{22}-M_{02}^2 & M_{01}M_{02}-M_{00}M_{12} \\
M_{01}M_{12}-M_{11}M_{02} & M_{01}M_{02}-M_{00}M_{12} & M_{00}M_{11}-M_{01}^2
\end{pmatrix}
\end{equation}
Note, if variable 2 becomes uncorrelated with variables 0 and 1, then 
$M_{02}=M_{12}=0$ so
\begin{equation} 
\Delta
=M_{00}M_{11}M_{22}-M_{22}M_{01}^2= M_{22}(M_{00}M_{11}-M_{01}^2) = M_{22}\Delta_2
\end{equation}
and
\begin{equation} 
M^{-1}=
\frac{1}{\Delta}
\begin{pmatrix}
M_{11}M_{22} & -M_{22}M_{01} & 0 \\
-M_{22}M_{01} & M_{00}M_{22} & 0 \\
0 & 0 & M_{00}M_{11}-M_{01}^2
\end{pmatrix}
=\frac{1}{\Delta_2}
\begin{pmatrix}
M_{11} & -M_{01} & 0 \\
-M_{01} & M_{00} & 0 \\
0 & 0 & \Delta_2/M_{22}
\end{pmatrix}
=
\begin{pmatrix}
M_2^{-1} & 0 \\
0 & 1/M_{22}
\end{pmatrix}
\end{equation}
as expected. The inverse of the $2\times 2$ matrix is not the submatrix in
the inverse of the $3\times 3$ matrix. E.g. for the first element in the
inverse of the $2\times 2$ matrix, this is
\begin{equation} 
M_{00}^{-1} 
= \frac{M_{11}}{M_{00}M_{11}-M_{01}^2}
\end{equation}
while in the $3 \times 3$ case, this is 
\begin{eqnarray}
M_{00}^{-1} 
&=& \frac{M_{11}M_{22}-M_{12}^2}{M_{00}M_{11}M_{22}+2M_{01}M_{12}M_{02}
-M_{00}M_{12}^2 -M_{11}M_{02}^2 -M_{22}M_{01}^2}\\
&=& \frac{M_{11}-(M_{12}^2/M_{22})}{M_{00}M_{11}-M_{01}^2
+(2M_{01}M_{12}M_{02} -M_{00}M_{12}^2 -M_{11}M_{02}^2)/M_{22}}
\end{eqnarray}

\end{document}