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  • Committer: Chris Rogers
  • Date: 2014-04-16 11:48:45 UTC
  • mfrom: (707 merge)
  • mto: This revision was merged to the branch mainline in revision 711.
  • Revision ID: chris.rogers@stfc.ac.uk-20140416114845-h3u3q7pdcxkxvovs
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doc/doc_src/detectors/tracker/01-Current/12-Appendix-2-Space-point-standard-deviation/12-Appendix-2-Space-point-standard-deviation.tex
1
 
\section{Space-point variance}
2
 
\label{App:SpcPntSD}
3
 
 
4
 
Figure \ref{Fig:SenseArea} shows the arrangement of the fibre channels in
5
 
the tracker.
6
 
The regions in which a space point will be reconstructed are shown by
7
 
the shaded areas.
8
 
The area of the triangular intersection is given by:
9
 
\begin{eqnarray}
10
 
  A & = & 4 \frac{1}{2} \frac{c_p}{\sqrt{3}} \frac{c_p}{2} \\
11
 
    & = & \frac{c_p^{2}}{\sqrt{3}} \, ;
12
 
\end{eqnarray}
13
 
where $c_p$ is the channel pitch.
14
 
Therefore, for the triangular intersection, the mean values of $x$ and
15
 
$y$ are given by:
16
 
\begin{eqnarray}
17
 
  \bar{x} & = & \frac{1}{A} \int \int x dx dy                           \\
18
 
          & = & \frac{1}{A} \int_{0}^{c_p} x dx                         
19
 
                \int_{-\frac{x}{\sqrt{3}}}^{\frac{x}{\sqrt{3}}} dy                \\
20
 
          & = & \frac{1}{A} \frac{2}{\sqrt{3}} \int_{0}^{c_p} x^{2} dx   \\
21
 
          & = & \frac{1}{A} \frac{2}{\sqrt{3}} \frac{c_p^{3}}{3}        \\
22
 
          & = & \frac{\sqrt{3}}{c_p^2} \frac{2}{\sqrt{3}} \frac{c_p^{3}}{3}  \\
23
 
          & = & \frac{2}{3}c_p \, {\rm ; and}
24
 
\end{eqnarray}
25
 
\begin{eqnarray}
26
 
  \bar{y}&=&\frac{1}{A} \int \int y dx dy                            \\
27
 
         &=&\frac{1}{A} \int_{0}^{c_p} dx                            
28
 
            \int_{-\frac{x}{\sqrt{3}}}^{\frac{x}{\sqrt{3}}} y dy               \\
29
 
         &=&\frac{1}{A}  \int_{0}^{c_p} 
30
 
            \left[ \frac{y^{2}}{2} \right]_{-\frac{x}{\sqrt{3}}}^{\frac{x}{\sqrt{3}}} dx \\
31
 
         &=&0 \, .
32
 
\end{eqnarray}
33
 
The variance of the $x$ and $y$ coodinates are then given by:
34
 
\begin{eqnarray}
35
 
  V_x = \sigma^2_x &=& \frac{1}{A} \int \int (x-\bar{x})^{2} dx dy   \\
36
 
                  &=& \frac{1}{A} \int_{0}^{c_p} dx 
37
 
                      \int_{-\frac{x}{\sqrt{3}}}^{\frac{x}{\sqrt{3}}}  
38
 
                      (x-\bar{x})^{2} dy                            \\
39
 
                  &=& \frac{1}{A} \frac{2}{\sqrt{3}} 
40
 
                      \int_{0}^{c_p} x (x-\bar{x})^{2}              \\
41
 
                  &=& \frac{1}{A} \frac{2}{\sqrt{3}} 
42
 
                      \int_{0}^{c_p}
43
 
                      ( x^{3}-2x^{2}\bar{x}+\bar{x}^{2}x ) dx           \\
44
 
                  &=& \frac{1}{A} \frac{2}{\sqrt{3}} 
45
 
                      \int_{0}^{c_p} ( x^{3}-\frac{4}{3}c_p x^{2}
46
 
                      + \frac{4}{9} c_p^{2} x ) dx                  \\
47
 
                  &=& \frac{1}{A} \frac{2}{\sqrt{3}} 
48
 
                      \left[ \frac{x^{4}}{4} - \frac{4}{9} c_p x^{3}
49
 
                      + \frac{2}{9} c_p^{2} x^{2}\right]_{0}^{c_p}      \\
50
 
                  &=& \frac{c_p^4}{A} \frac{2}{\sqrt{3}} 
51
 
                      \left[ \frac{1}{4}-\frac{4}{9}+\frac{2}{9} \right]  \\
52
 
                  &=& \frac{c_p^4}{A} \frac{2}{\sqrt{3}} 
53
 
                      \left[ \frac{1}{4}-\frac{2}{9} \right]  \\
54
 
                  &=& c_p^4 \frac{\sqrt{3}}{c_p^2} \frac{2}{\sqrt{3}} 
55
 
                           \frac{1}{36}  \\
56
 
                  &=& \frac{1}{18} c_p^{2} \\
57
 
  V_x = \sigma^2_x &=& \left( \frac{c_p}{3\sqrt{2}} \right)^2 \, ;
58
 
\end{eqnarray}
59
 
\begin{eqnarray}
60
 
  V_y = \sigma^2_y 
61
 
        &=& \frac{1}{A} \int_{0}^{c_p} dx
62
 
            \int_{-\frac{x}{\sqrt{3}}}^{\frac{x}{\sqrt{3}}}(y-\bar{y})^{2} dy \\
63
 
        &=& \frac{1}{A} \int_{0}^{c_p} \left[ \frac{y^{3}}{3} 
64
 
            \right]_{-\frac{x}{\sqrt{3}}}^{\frac{x}{\sqrt{3}}} dy            \\
65
 
        &=& \frac{1}{A} \frac{2}{3\sqrt{3}} 
66
 
            \int_{0}^{c_p} x^{3}dx \nonumber                        \\
67
 
        &=& \frac{1}{A} \frac{2}{9\sqrt{3}} 
68
 
            \left[ \frac{x^{4}}{4} \right]_{0}^{c_p}                \\
69
 
        &=& \frac{1}{A} \frac{2}{9\sqrt{3}} \frac{c_p^4}{4}      \\
70
 
        &=& \frac{\sqrt{3}}{c_p^2} \frac{2}{9\sqrt{3}} \frac{c_p^4}{4}  \\
71
 
        &=& \frac{1}{9} \frac{c_p^2}{2}  \\
72
 
        &=& \frac{1}{18} c_p^2  \\
73
 
  V_y = \sigma^2_y 
74
 
        &=& \left( \frac{c_p^{2}}{3\sqrt{2}} \right)^2 \, .
75
 
\end{eqnarray}
76
 
The covariance is given by:
77
 
\begin{eqnarray}
78
 
  V_{xy} &=& \frac{1}{A} \int \int (x-\bar{x}) (y-\bar{y}) dx dy         \\
79
 
        &=& \frac{1}{A} \int_{0}^{c_p} (x-\bar{x}) dx
80
 
            \int_{-\frac{x}{\sqrt{3}}}^{\frac{x}{\sqrt{3}}}(y-\bar{y}) dy     \\
81
 
        &=& \frac{1}{A} \int_{0}^{c_p} (x-\bar{x})
82
 
            \left[ 
83
 
              \frac{1}{2}y^2 - y \bar{y} 
84
 
            \right]_{-\frac{x}{\sqrt{3}}}^{\frac{x}{\sqrt{3}}}                \\
85
 
        &=& 0 \, .
86
 
\end{eqnarray}
87
 
Therefore: 
88
 
\begin{equation}
89
 
  \sigma_{x}=\sigma_{y}=\frac{c_p}{3\sqrt{2}} = 384.4 \mu{\rm m} \, .
90
 
\end{equation}
91
 
For the hexagonal case, the area of the overlapping region (shaded
92
 
zone in the right panel of figure \ref{Fig:SenseArea}) is givn by:
93
 
\begin{eqnarray}
94
 
  A & = & 6 \frac{1}{2} \frac{c_p}{\sqrt{3}} \frac{c_p}{2} \\
95
 
    & = & \frac{\sqrt{3}}{2}c_p^{2} \, .
96
 
\end{eqnarray}
97
 
By symmetry, $\bar{x} = \bar{y} = 0$.
98
 
The variance of the $x$ and $y$ coordinates are given by:
99
 
\begin{eqnarray}
100
 
  V_x = \sigma^2_x = \sigma^2_y =
101
 
        &=& \frac{1}{A} \int \int (x-\bar{x})^{2} dx dy \\
102
 
        &=& \frac{1}{A} \int \int x^{2} dx dy \\
103
 
        &=& \frac{2}{A}  \int_{-\frac{c_p}{2}}^{0}x^{2} dx 
104
 
            \int_{-\frac{x}{\sqrt{3}}-\frac{c_p}{\sqrt{3}}}
105
 
                ^{\frac{x}{\sqrt{3}}+\frac{c_p}{\sqrt{3}}} dy \nonumber \\
106
 
        &=& \frac{2}{A}\int_{-\frac{c_p}{2}}^{0}x^{2} 
107
 
            \left[ 2 \left( \frac{x}{\sqrt{3}}+\frac{c_p}{\sqrt{3}}  
108
 
            \right) \right] dx \\
109
 
        &=& \frac{2}{A}\frac{2}{\sqrt{3}} \int_{-\frac{c_p}{2}}^{0} 
110
 
            \left( x^{3}+ x^{2}c_p \right)dx \nonumber \\
111
 
        &=& \frac{2}{A}\frac{2}{\sqrt{3}} \left[ \frac{1}{4}
112
 
            x^4 +\frac{1}{3} x^3 c_p\right]_{-\frac{c_p}{2}}^{0} \\
113
 
        &=& \frac{2}{A}\frac{2}{\sqrt{3}} \left[ -\frac{1}{4}
114
 
            \frac{c_p^{4}}{16}+\frac{1}{3}\frac{c_p^{4}}{8}\right] \\
115
 
        &=& \frac{2}{A}\frac{2}{\sqrt{3}}
116
 
            \left[ \frac{1}{8} \left( \frac{1}{3}-\frac{1}{8} 
117
 
            \right) c_p^4 \right]\\
118
 
        &=& \frac{2}{A}\frac{2}{\sqrt{3}} \frac{1}{8} \frac{5}{24} c_p^4
119
 
         =  \frac{1}{A} \frac{5}{48\sqrt{3}} c_p^4 \\
120
 
        &=& \frac{2}{\sqrt{3}c_p^{2}} \frac{5}{48\sqrt{3}} c_p^4 
121
 
         =  \frac{2}{\sqrt{3}} \frac{5}{48\sqrt{3}} c_p^2 \\
122
 
        &=& \left( \sqrt{\frac{5}{2}}\frac{c_p}{6} \right)^2 \, .
123
 
\end{eqnarray}
124
 
As before, the covariance is given by:
125
 
\begin{eqnarray}
126
 
  V_{xy} &=& \frac{1}{A} \int \int (x-\bar{x}) (y-\bar{y}) dx dy         \\
127
 
        &=& \frac{2}{A}  \int_{-\frac{c_p}{2}}^{0} x dx 
128
 
            \int_{-\frac{x}{\sqrt{3}}-\frac{c_p}{\sqrt{3}}}
129
 
                ^{\frac{x}{\sqrt{3}}+\frac{c_p}{\sqrt{3}}} y dy \nonumber \\
130
 
        &=& \frac{2}{A}  \int_{-\frac{c_p}{2}}^{0} x dx 
131
 
            \left[
132
 
             \frac{1}{2} y^2
133
 
            \right]_{-\frac{x}{\sqrt{3}}-\frac{c_p}{\sqrt{3}}}
134
 
                   ^{\frac{x}{\sqrt{3}}+\frac{c_p}{\sqrt{3}}}   \nonumber \\
135
 
        &=& 0 \, .
136
 
\end{eqnarray}
137
 
Therefore:
138
 
\begin{equation}
139
 
  \sigma_{x}=\sigma_{y}=\sqrt{\frac{5}{2}}\frac{c_p}{6} = 429.8 \mu{\rm m} \, .
140
 
\end{equation}
141
 
\begin{figure}
142
 
  \begin{center}
143
 
    \includegraphics[width=0.7\linewidth]%
144
 
      {12-Appendix-2-Space-point-standard-deviation/Figures/drawing2.eps}
145
 
  \end{center}
146
 
  \caption{
147
 
      Right panel: Fibre arrangement in station 5 of tracker 1. 
148
 
      Left panel: Fibre arrangement in the rest of the stations. 
149
 
      The shaded region shows the intersection of the three channels
150
 
      is triangle for every station other than station 5, where it
151
 
      will be an hexagon.
152
 
  }
153
 
  \label{Fig:SenseArea}
154
 
\end{figure}