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- 01/29/05 15:01:10 (20 years ago)
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trunk/MagicSoft/TDAS-Extractor/Algorithms.tex
r5993 r6113 16 16 \begin{figure}[htp] 17 17 \includegraphics[width=0.99\linewidth]{ExtractorClasses.eps} 18 \caption{Sketch of the inheritances of three examplary MARS signal extractor classes: MExtractFixedWindow, MExtractTimeFastSpline and MExtractTimeAndChargeDigitalFilter} 18 \caption{Sketch of the inheritances of three examplary MARS signal extractor classes: 19 MExtractFixedWindow, MExtractTimeFastSpline and MExtractTimeAndChargeDigitalFilter} 19 20 \label{fig:extractorclasses} 20 21 \end{figure} … … 402 403 \end{eqnarray} 403 404 404 where the last expression is matricial. $\chi^2$ is a continuous function of $\tau$ and will have to be discretized itself for a 405 desired resolution. $\chi^2$ is also proportional to the auto noise-correlation matrix where increases in the noise level lead to 406 a multiplicative factor for all matrix elements and thus do not affect the position of the maximum of $\chi^2$. 405 where the last expression is matricial. 406 $\chi^2$ is a continuous function of $\tau$ and will have to be discretized itself for a 407 desired resolution. 408 $\chi^2$ is in principle independent from the noise auto-correlation matrix if always the correct noise level is calculated there. 409 In our case however, we decided to use one same matrix $\boldsymbol{B}$ for all levels of night-sky background since increases 410 in the noise level lead only to a multiplicative factor for all matrix elements and thus do not affect the position of the minimum of $\chi^2$. 407 411 The minimum of $\chi^2$ is obtained for: 408 412 … … 417 421 418 422 \begin{eqnarray} 419 0&=&-\boldsymbol{g}^T\boldsymbol{B}^{-1}\boldsymbol{y}+\boldsymbol{g}^T\boldsymbol{B}^{-1}\boldsymbol{g}\overline{E}+\boldsymbol{g}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}\overline{E\tau} 423 0&=&-\boldsymbol{g}^T\boldsymbol{B}^{-1}\boldsymbol{y} 424 +\boldsymbol{g}^T\boldsymbol{B}^{-1}\boldsymbol{g}\overline{E} 425 +\boldsymbol{g}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}\overline{E\tau} 420 426 \\ 421 0&=&-\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\boldsymbol{y}+\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\boldsymbol{g}\overline{E}+\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}\overline{E\tau} \ . 427 0&=&-\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\boldsymbol{y} 428 +\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\boldsymbol{g}\overline{E} 429 +\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}\overline{E\tau} \ . 422 430 \end{eqnarray} 423 431 … … 425 433 426 434 \begin{equation} 427 \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 } \ , 428 \end{equation} 429 430 \begin{equation} 431 \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 } \ . 432 \end{equation} 433 435 \overline{E}(\tau) = \boldsymbol{w}_{\text{amp}}^T (\tau)\boldsymbol{y} \quad \mathrm{with} \quad 436 \boldsymbol{w}_{\text{amp}} 437 = \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}}} 438 {(\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 } \ , 439 \end{equation} 440 441 \begin{equation} 442 \overline{E\tau}(\tau)= \boldsymbol{w}_{\text{time}}^T(\tau) \boldsymbol{y} \quad 443 \mathrm{with} \quad \boldsymbol{w}_{\text{time}} 444 = \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}}} 445 {(\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 } \ . 446 \end{equation} 434 447 435 448 Thus $\overline{E}$ and $\overline{E\tau}$ are given by a weighted sum of the discrete measurements $y_i$ 436 449 with the digital filtering weights for the amplitude, $w_{\text{amp}}(\tau)$, and time shift, $w_{\text{time}}(\tau)$ 437 where the time dependency gets discretized once again leading to a set of weights samples dependingon the450 where the time dependency gets discretized once again leading to a set of weights samples which themselves depend on the 438 451 discretized time $\tau$. 439 440 \par 441 \ldots {\textit{\bf IS THIS CORRECT LIKE THIS???}} \ldots 442 \par 443 444 Note the remaining time dependence of the two weights which follow from the dependency of $\boldsymbol{g}$ and 452 \par 453 Note the remaining time dependency of the two weights samples which follow from the dependency of $\boldsymbol{g}$ and 445 454 $\dot{\boldsymbol{g}}$ on the position of the pulse with respect to the FADC bin positions. 446 455 \par … … 452 461 453 462 \begin{equation} 454 \left(\boldsymbol{V}^{-1}\right)_{i,j}=\frac{1}{2}\left(\frac{\partial^2 \chi^2(E, E\tau)}{\partial \alpha_i \partial \alpha_j} \right) \quad \text{with} \quad \alpha_i,\alpha_j \in \{E, E\tau\} \ . 463 \left(\boldsymbol{V}^{-1}\right)_{i,j} 464 =\frac{1}{2}\left(\frac{\partial^2 \chi^2(E, E\tau)}{\partial \alpha_i \partial \alpha_j} \right) \quad 465 \text{with} \quad \alpha_i,\alpha_j \in \{E, E\tau\} \ . 455 466 \end{equation} 456 467 … … 458 469 459 470 \begin{equation}\label{of_noise} 460 \sigma_E^2=\boldsymbol{V}_{E,E}=\frac{\dot{\boldsymbol{g}}^T\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} \ . 471 \sigma_E^2=\boldsymbol{V}_{E,E} 472 =\frac{\dot{\boldsymbol{g}}^T\boldsymbol{B}^{-1}\dot{\boldsymbol{g}}} 473 {(\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} \ . 461 474 \end{equation} 462 475 … … 464 477 465 478 \begin{equation}\label{of_noise_time} 466 E^2 \cdot \sigma_{\tau}^2=\boldsymbol{V}_{E\tau,E\tau}=\frac{{\boldsymbol{g}}^T\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} \ . 479 E^2 \cdot \sigma_{\tau}^2=\boldsymbol{V}_{E\tau,E\tau} 480 =\frac{{\boldsymbol{g}}^T\boldsymbol{B}^{-1}{\boldsymbol{g}}} 481 {(\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} \ . 467 482 \end{equation} 468 483 … … 506 521 \includegraphics[totalheight=7cm]{noise_autocorr_AB_36038_TDAS.eps} 507 522 \end{center} 508 \caption[Noise autocorrelation.]{Noise autocorrelation matrix for open camera including the noise due to night sky background fluctuations.} \label{fig:noise_autocorr_AB_36038_TDAS} 523 \caption[Noise autocorrelation.]{Noise autocorrelation matrix for open camera including the noise due to night sky background fluctuations.} 524 \label{fig:noise_autocorr_AB_36038_TDAS} 509 525 \end{figure} 510 526 … … 546 562 \end{figure} 547 563 548 549 550 In the current implementation a two step procedure is applied to reconstruct the signal. The weight functions $w_{\mathrm{amp}}(t)$ and $w_{\mathrm{time}}(t)$ are computed numerically with a resolution of $1/10$ of an FADC slice. In thefirst step the quantities $e_{i_0}$ and $e\tau_{i_0}$ are computed using a window of $n$ slices:564 In the current implementation a two step procedure is applied to reconstruct the signal. The weight functions $w_{\mathrm{amp}}(t)$ 565 and $w_{\mathrm{time}}(t)$ are computed numerically with a resolution of $1/10$ of an FADC slice. 566 In the first step the quantities $e_{i_0}$ and $e\tau_{i_0}$ are computed using a window of $n$ slices: 551 567 552 568 \begin{equation} … … 554 570 \end{equation} 555 571 556 for all possible signal start slices $i_0$. Let $i_0^*$ be the signal start slice with the largest $e_{i_0}$. Then in a second step the timing offset $\tau$ is calculated: 572 for all possible signal start slices $i_0$. Let $i_0^*$ be the signal start slice with the largest $e_{i_0}$. 573 Then in a second step the timing offset $\tau$ is calculated: 557 574 558 575 \begin{equation} … … 563 580 564 581 \begin{equation} 565 E=\sum_{i=i_0^*}^{i_0^*+n-1} w_{\mathrm{amp}}(t_i - \tau)y(t_{i+i_0^*}) \qquad E \theta=\sum_{i=i_0^*}^{i_0^*+n-1} w_{\mathrm{time}}(t_i - \tau)y(t_{i+i_0^*}) \ . 582 E=\sum_{i=i_0^*}^{i_0^*+n-1} w_{\mathrm{amp}}(t_i - \tau)y(t_{i+i_0^*}) \qquad 583 E \theta=\sum_{i=i_0^*}^{i_0^*+n-1} w_{\mathrm{time}}(t_i - \tau)y(t_{i+i_0^*}) \ . 566 584 \end{equation} 567 585
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