Changeset 5944

Timestamp:

01/22/05 18:30:19 (20 years ago)

Author:

gaug

Message:

*** empty log message ***

File:

• trunk/MagicSoft/TDAS-Extractor/Algorithms.tex (modified) (5 diffs)

trunk/MagicSoft/TDAS-Extractor/Algorithms.tex

@@ -397 +397 @@
 
 \begin{eqnarray}
-\chi^2(E, E\tau) &=& \sum_{i,j}(y_i-E g_i-E\tau \dot{g}_i) B^{-1}_{ij} (y_j - E g_j-E\tau \dot{g}_j) \\
+\chi^2(E, E\tau) &=& \sum_{i,j}(y_i-E g_i-E\tau \dot{g}_i) (\boldsymbol{B}^{-1})_{ij} (y_j - E g_j-E\tau \dot{g}_j) \\
 &=& (\boldsymbol{y} - E
 \boldsymbol{g} - E\tau \dot{\boldsymbol{g}})^T \boldsymbol{B}^{-1} (\boldsymbol{y} - E \boldsymbol{g}- E\tau \dot{\boldsymbol{g}}) \ ,
@@ -403 +403 @@
 
 where the last expression is matricial. $\chi^2$ is a continuous function of $\tau$ and will have to be discretized itself for a 
-desired resolution.
+desired resolution.  $\chi^2$ is also proportional to the auto noise-correlation matrix where increases in the noise level lead to 
+a multiplicative factor for all matrix elements and thus do not affect the position of the maximum of $\chi^2$.
 The minimum of $\chi^2$ is obtained for:
 
@@ -409 +410 @@
 \frac{\partial \chi^2(E, E\tau)}{\partial E} = 0 \qquad \text{and} \qquad \frac{\partial \chi^2(E, E\tau)}{\partial(E\tau)} = 0 \ .
 \end{equation}
+
 
 Taking into account that $\boldsymbol{B}$ is a symmetric matrix, this leads to the following 
@@ -423 +425 @@
 
 \begin{equation}
-\overline{E}= \boldsymbol{w}_{\text{amp}}^T \boldsymbol{y} \quad \mathrm{with} \quad \boldsymbol{w}_{\text{amp}} = \frac{ (\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}) \boldsymbol{B}^{-1} \boldsymbol{g} -(\boldsymbol{g}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}})  \boldsymbol{B}^{-1} \dot{\boldsymbol{g}}}  {(\boldsymbol{g}^T \boldsymbol{B}^{-1} \boldsymbol{g})(\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}) -(\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\boldsymbol{g})^2 } \ ,
-\end{equation}
-
-\begin{equation}
-\overline{E\tau}= \boldsymbol{w}_{\text{time}}^T \boldsymbol{y} \quad \mathrm{with} \quad \boldsymbol{w}_{\text{time}} = \frac{ ({\boldsymbol{g}}^T\boldsymbol{B}^{-1}{\boldsymbol{g}}) \boldsymbol{B}^{-1} \dot{\boldsymbol{g}} -(\boldsymbol{g}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}})  \boldsymbol{B}^{-1} {\boldsymbol{g}}}  {(\boldsymbol{g}^T \boldsymbol{B}^{-1} \boldsymbol{g})(\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}) -(\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\boldsymbol{g})^2 } \ .
+\overline{E}(\tau) = \boldsymbol{w}_{\text{amp}}^T (\tau)\boldsymbol{y} \quad \mathrm{with} \quad \boldsymbol{w}_{\text{amp}} = \frac{ (\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}) \boldsymbol{B}^{-1} \boldsymbol{g} -(\boldsymbol{g}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}})  \boldsymbol{B}^{-1} \dot{\boldsymbol{g}}}  {(\boldsymbol{g}^T \boldsymbol{B}^{-1} \boldsymbol{g})(\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}) -(\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\boldsymbol{g})^2 } \ ,
+\end{equation}
+
+\begin{equation}
+\overline{E\tau}(\tau)= \boldsymbol{w}_{\text{time}}^T(\tau) \boldsymbol{y} \quad \mathrm{with} \quad \boldsymbol{w}_{\text{time}} = \frac{ ({\boldsymbol{g}}^T\boldsymbol{B}^{-1}{\boldsymbol{g}}) \boldsymbol{B}^{-1} \dot{\boldsymbol{g}} -(\boldsymbol{g}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}})  \boldsymbol{B}^{-1} {\boldsymbol{g}}}  {(\boldsymbol{g}^T \boldsymbol{B}^{-1} \boldsymbol{g})(\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}) -(\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\boldsymbol{g})^2 } \ .
 \end{equation}
 
 
 Thus $\overline{E}$ and $\overline{E\tau}$ are given by a weighted sum of the discrete measurements $y_i$ 
-with the digital filtering weights for the amplitude, $w_{\text{amp}}(t)$, and time shift, $w_{\text{time}}(t)$.
+with the digital filtering weights for the amplitude, $w_{\text{amp}}(\tau)$, and time shift, $w_{\text{time}}(\tau)$
+where the time dependency gets discretized once again leading to a set of weights samples depending on the 
+discretized time $\tau$.
+
+\par
+\ldots {\textit{\bf IS THIS CORRECT LIKE THIS???}} \ldots
+\par
+
 Note the remaining time dependence of the two weights which follow from the dependency of $\boldsymbol{g}$ and 
 $\dot{\boldsymbol{g}}$ on the position of the pulse with respect to the FADC bin positions.
@@ -459 +468 @@
 
 For the MAGIC signals, as implemented in the MC simulations, a pedestal RMS of a single FADC slice of 4 FADC counts introduces an error in the reconstructed signal and time of:
+
 
 \begin{equation}\label{of_noise}