wup.tex 23.6 KB
 vpalladi committed May 13, 2016 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 \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 \centerline{\large May 2016}  dauncey committed Jun 02, 2016 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 \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 E_e = \sum_i a_i d_{e,i} 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 \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 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 \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 Hence, for the minimum, we require \sum_i \frac{\sum_e d_{e,j} d_{e,i}}{N} a_i = \frac{\sum_e d_{e,j} T_e}{N} Writing in matrix notation with $M$ and $v$ defined as M_{ji} = \frac{\sum_e d_{e,j} d_{e,i}}{N},\qquad v_j = \frac{\sum_e d_{e,j} T_e}{N} then the requirement is M a = v\qquad\mathrm{so}\qquad a = M^{-1}v 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 $d_i$.  dauncey committed Jun 07, 2016 64 65 66 67 68 69 70 71 72 73 74 75 76 77 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  dauncey committed Aug 16, 2016 78 79 80 \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  dauncey committed Jun 07, 2016 81   dauncey committed Jun 17, 2016 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 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 E_e = b + \sum_i a_i d_{e,i} 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 E_e = \sum_{i=0}^{28} a_i d_{e,i} 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 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} and M_{28,28} = \frac{1}{N} \sum_e d_{e,28}d_{e,28} = 1 while the extra element in $v$ is v_{28} = \frac{1}{N} \sum_e T_e d_{e,28} = \frac{1}{N} \sum_e T_e  dauncey committed Jun 02, 2016 114   vpalladi committed May 13, 2016 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129 130 131 132 133 134 135 136 137 138 139 140 141 142 143 144 145 146 147 148 149 150 151 152 153 154 155 156 157 158 159 160 161 162 163 164 165 166 167 168 169 170 171 172 173 174 175 176 \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 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} 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 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) Hence 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)} where the positive sign is required for $q>100$\,fC. \section{Link data representation} The selected trigger cell  dauncey committed Nov 14, 2016 177 178 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 $n$,  vpalladi committed May 13, 2016 179 typically $\sim 8$.  dauncey committed Nov 14, 2016 180 181 182 This could be linear or logarithmic or floating. \subsection{Linear representation}  vpalladi committed May 13, 2016 183 184 185 186 187 188 189 190 191 192 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 y = \frac{(2^n-1)(x-x_\mathrm{min})}{x_\mathrm{max}-x_\mathrm{min}} which can be inverted to give x = x_\mathrm{min} + \frac{y(x_\mathrm{max}-x_\mathrm{min})}{2^n-1}  dauncey committed Nov 14, 2016 193 194  \subsection{Logarithmic representation}  vpalladi committed May 13, 2016 195 For logarithmic, the general case would be $y=a\log(x)+b$ but with  dauncey committed Nov 14, 2016 196 $c=b/a$ and $x_\mathrm{min}=e^{-c}$, then  vpalladi committed May 13, 2016 197 198 199 200 201 202 203 204 205 206 207 208 209  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}) Therefore y = (2^n-1) \frac{\log(x/x_\mathrm{min})}{\log(x_\mathrm{max}/x_\mathrm{min})} This can be inverted to give x = x_\mathrm{min}\left(\frac{x_\mathrm{max}}{x_\mathrm{min}}\right)^{y/(2^n-1)}  dauncey committed Nov 14, 2016 210 211 212 213 214 215 216 217 218 219 220 221 222 223 224 225 226 227 228 229 230 231 232 233 234 235 236 237 238 239 240 241 242 243 244 245 246 247 248 249 250 251 252 253 254 255 256 257 258 259 260 261 262 263 264 265 266 267 268 269 270 271 272 273 274 275 276 277 278 279 280 281 282 283 284 285 286 287 288 289 290 291 292 293 294 295 296 297 298 299 300 301 302 303 304 305 306 307 308 309 310 311 312 313 314 315 316 317 318 319 320 321 322 323 324 325 326 327 328 329 330 331 332 333 334 335 336 337 338 339 340 341 342 343 344 345 \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}  vpalladi committed May 13, 2016 346 347 348 349 350 351 352 353 354 355 356 357 358 359 360 361 362 363 364 365 366 367 368 369 370 371 372 373 374 375 376 377 378 379 380 381 382 383 384 385 386 387 388 389 390 391 392 393 394 395 396 397 398 399 400 401 402 403 404 405 406 407 408 409 410 411 412 413 414 415 416 417 418 419 420 421 422 423 424 425 426 427 428 429 430 431 432 433 434 435 436 437 438 439 440 441 442 443 444 445 446 447 448 449 450 451 452 453 454 455 456 457 458 459 460 461 462 463 464 465 466 467 468 469 470 471 472 473 474 475 476 477 478 479 480 481 482 483 484 485 486 487 488 489 490 491 492 493 494 495 496 497 498 499 500 501 502 503 504 505 506 507 508 509 510 511 512 513 514 515 516 517 518 519 520 521 522 523 524 525 526 527 528 529 530 531 532 533 534 535 536 537 538 539 540 541 542 543 544 545 546 547 548 549 550 551 552 553 554 555 556 557 558 559 560 561 562 563 564 565 566 567 568 569 570 571 572 573 574 575 576 577 578 579 580 581 582 583 584 585 586 587 588 589 590 591 592 593 594 595 596 597 598 599 600 601 602 603 604 605 606 607 608 609 610 611 612 613 614 615 616 617 618 619 620 621 622 623 624 625 626 627 628 629 630  \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 \chi^2 = \sum_l \frac{(E_p P_l + E_m M_l - O_l)^2}{\sigma_l^2} 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 \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 such that 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}, This can be written as a matrix 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} 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 \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 so the inverse is \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} and hence \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} which means \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} Hence E_p = \sum_l O_l A_l,\qquad E_m = \sum_l O_l B_l 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 \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) 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. \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) Using the expression for $\Delta\rho$ above, then \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} so that \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) Expressing the right hand side in plane polars gives \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) 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 \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 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 1 - \cos\Delta\phi = 2\sin^2\left(\frac{\Delta\phi}{2}\right) then \overline{\rho}^2\Delta\phi^2 =4\rho_1\rho_2 \sin^2\left(\frac{\Delta\phi}{2}\right) and so \overline{\rho}\Delta\phi =2\sqrt{\rho_1\rho_2} \sin\left(\frac{\Delta\phi}{2}\right) 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 M= \begin{pmatrix} M_{00} & M_{01} & M_{02} \\ M_{01} & M_{11} & M_{12} \\ M_{02} & M_{12} & M_{22} \end{pmatrix} 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 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} Note, if variable 2 becomes uncorrelated with variables 0 and 1, then $M_{02}=M_{12}=0$ so \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 and 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} 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 M_{00}^{-1} = \frac{M_{11}}{M_{00}M_{11}-M_{01}^2} 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}