Changeset 21455
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- Feb 11, 2009, 6:46:07 PM (17 years ago)
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trunk/doc/psphot/psphot.tex (modified) (1 diff)
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trunk/doc/psphot/psphot.tex
r11599 r21455 389 389 likely that the `peak' was identified in a region of flat flux 390 390 distribution or many saturated or edge pixels. 391 392 \subsubsection{Determination of the Peak Coordinates and Errors} 393 394 We use the 9 pixels which include the source peak to fit for the 395 position and position errors. We model the peak of the sources as a 396 2D quadratic polynomial, and use a very simple bi-quadratic fit to 397 these pixels. We use the following function to describe the peak 398 399 \[ f(x,y) = C_{00} + C_{10}x + C_{01} y + C_{11} x y + C_{20} x^2 + C_{02} y^2 \] 400 401 and write the Chi-Square equation: 402 403 \[ \Chi^2 = \sum_{i,j} (F_{i,j} - f(x,y))^2 / \sigma_{i,j}^2 \] 404 405 By approximating the error per pixel as the error on just the peak, 406 and pulling that term out of the above equation, and recognizing that 407 the values x,y in the 3x3 grid centered on the peak pixel have values 408 of only 0 or 1, we can greatly simplify the chi-square equation to a 409 square matrix equation with the following values: 410 411 %% fix this: 412 \begin{verbatim} 413 | 9 0 0 0 6 6 | C_00 | = \sum F_{i,j} 414 | 0 6 0 0 0 0 | C_10 | = \sum F_{i,j} x 415 | 0 0 6 0 0 0 | C_01 | = \sum F_{i,j} y 416 | 0 0 0 6 0 0 | C_11 | = \sum F_{i,j} x y 417 | 6 0 0 0 6 4 | C_20 | = \sum F_{i,j} x^2 418 | 6 0 0 0 4 6 | C_02 | = \sum F_{i,j} y^2 419 \end{verbatim} 420 421 The inverse of the 3x3 matrix terms for C_{00}, C_{20}, and C_{02} is: 422 \begin{verbatim} 423 | +5/9 -1/3 -1/3 | 424 | -1/3 +1/2 0 | 425 | -1/3 0 +1/2 | 426 \end{verbatim} 427 428 which can be used to determine the errors on the coefficients: 429 430 \begin{eqnarray} 431 \sigma^2_{00} & = & \sigma^2 (5/9) \\ 432 \sigma^2_{10} & = & \sigma^2 (1/6) \\ 433 \sigma^2_{01} & = & \sigma^2 (1/6) \\ 434 \sigma^2_{11} & = & \sigma^2 (1/6) \\ 435 \sigma^2_{20} & = & \sigma^2 (1/2) \\ 436 \sigma^2_{02} & = & \sigma^2 (1/2) \\ 437 \end{eqnarray} 438 439 The location of the peak is determined from the minimum of the 440 bi-quadratic function above, and is given by: 441 442 \begin{eqnarray} 443 Det & = & 4 C_{20} C_{02} - C_{11}^2 \\ 444 x_{min} & = & (C_{11} C_{01} - 2 C_{02} C_{10}) / Det \\ 445 y_{min} & = & (C_{11} C_{10} - 2 C_{20} C_{01}) / Det \\ 446 \end{eqnarray} 447 448 Applying error propagation to the above, we find: 449 450 \begin{eqnarray} 451 \sigma_{Det}^2 & = & \sigma_{11}^2 (4 C_{11}^2) + \sigma_{20}^2 (16 C_{02}^2) + \sigma_{02}^2 (16 C_{20}^2) \\ 452 \sigma_{xn}^2 & = & \sigma_{11}^2 C_{01}^2 + \sigma_{01}^2 C_{11}^2 + \sigma_{02}^2 (4 C_{10}^2) + \sigma_{10}^2 (4 C_{02}^2) \\ 453 \sigma_{yn}^2 & = & \sigma_{11}^2 C_{10}^2 + \sigma_{10}^2 C_{11}^2 + \sigma_{20}^2 (4 C_{01}^2) + \sigma_{01}^2 (4 C_{20}^2) \\ 454 \sigma_{x}^2 & = & x^2 (\sigma_{xn}^2 / xn^2 + \sigma_{Det}^2 / Det^2) \\ 455 \sigma_{y}^2 & = & y^2 (\sigma_{yn}^2 / yn^2 + \sigma_{Det}^2 / Det^2) \\ 456 \end{eqnarray} 391 457 392 458 \subsection{PSF Determination}
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