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\begin{document}
\centerline{\Large \bf Notes on TPG}
\bigskip
\centerline{\large May 2016}

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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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\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 
data on the link need to be represented in a small number of bits $n$,
typically $\sim 8$.
This could be linear or logarithmic.
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} 
For logarithmic, the general case would be $y=a\log(x)+b$ but with
$c=b/a$ and $x_\mathrm{min}=e^-c$, then
\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} 


\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}