add N0 estimate to single histogram fit. Some improvements in musredit (see ChangLog)
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doc/examples/test-histo-PSI-BIN.msr
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doc/examples/test-histo-PSI-BIN.msr
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FeSe 9p4 TF100 p107apr09_sample*1p02
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###############################################################
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FITPARAMETER
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# Nr. Name Value Step Pos_Error Boundaries
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1 Asy 0.2622 -0.0014 0.0014 0 0.33
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2 Rate 0.3188 -0.0044 0.0044
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3 Field 97.853 -0.056 0.056 0 200
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4 Phase_L 178.95 -0.41 0.41
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5 Phase_R 1.75 -0.40 0.39
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6 N0_L 1097.90 -1.00 1.00
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7 N0_R 1159.7 -1.0 1.0
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8 Bkg_L 54.47 -0.20 0.20
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9 Bkg_R 46.70 -0.19 0.19
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###############################################################
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THEORY
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asymmetry 1
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simplExpo 2 (rate)
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TFieldCos map1 fun1 (phase frequency)
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###############################################################
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FUNCTIONS
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fun1 = par3 * gamma_mu
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###############################################################
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RUN data/deltat_pta_gpd_0423 PIE1 PSI PSI-BIN (name beamline institute data-file-format)
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fittype 0 (single histogram fit)
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norm 6
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backgr.fit 8
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lifetimecorrection
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map 4 0 0 0 0 0 0 0 0 0
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forward 1
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#background 30 152 # estimated bkg: 49.2393
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data 165 7965
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t0 162.0
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fit 0 8.2
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packing 25
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RUN data/deltat_pta_gpd_0423 PIE1 PSI PSI-BIN (name beamline institute data-file-format)
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fittype 0 (single histogram fit)
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norm 7
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backgr.fit 9
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lifetimecorrection
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map 5 0 0 0 0 0 0 0 0 0
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forward 2
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#background 30 152 # estimated bkg: 43.1545
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data 205 7965
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t0 202.0
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fit 0 8.2
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packing 25
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###############################################################
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COMMANDS
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SCALE_N0_BKG TRUE
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MINIMIZE
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MINOS
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SAVE
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###############################################################
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PLOT 0 (single histo plot)
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runs 1 2
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range 0 7 -0.3 0.3
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###############################################################
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FOURIER
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units Gauss # units either 'Gauss', 'MHz', or 'Mc/s'
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fourier_power 12
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apodization NONE # NONE, WEAK, MEDIUM, STRONG
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plot POWER # REAL, IMAG, REAL_AND_IMAG, POWER, PHASE
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phase 8.5
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#range_for_phase_correction 50.0 70.0
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range 0.0 200.0
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###############################################################
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STATISTIC --- 2013-02-12 13:15:32
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chisq = 663.9, NDF = 515, chisq/NDF = 1.289169
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BIN
doc/memos/estimateN0/PSI_Logo_narrow_blau.jpg
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doc/memos/estimateN0/PSI_Logo_narrow_blau.jpg
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doc/memos/estimateN0/PSI_Logo_wide_blau.pdf
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doc/memos/estimateN0/PSI_Logo_wide_blau.pdf
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doc/memos/estimateN0/estimateN0.tex
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doc/memos/estimateN0/estimateN0.tex
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% $Id$
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\documentclass[twoside]{article}
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\usepackage[english]{babel}
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\usepackage{a4}
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\usepackage{amssymb}
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\usepackage{graphicx}
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\usepackage{fancyhdr}
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\usepackage{array}
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\usepackage{float}
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\usepackage{hyperref}
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\usepackage{xspace}
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\usepackage{rotating}
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\usepackage{dcolumn}
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\setlength{\topmargin}{10mm}
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\setlength{\topmargin}{-13mm}
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% \setlength{\oddsidemargin}{0.5cm}
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% \setlength{\evensidemargin}{0cm}
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\setlength{\oddsidemargin}{1cm}
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\setlength{\evensidemargin}{1cm}
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\setlength{\textwidth}{14.5cm}
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\setlength{\textheight}{23.8cm}
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\def\mathbi#1{\textbf{\em #1}}
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\pagestyle{fancyplain}
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\addtolength{\headwidth}{0.6cm}
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\fancyhead{}%
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\fancyhead[RE,LO]{\bf Estimate $\mathbi{N}_0$}%
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\fancyhead[LE,RO]{\thepage}
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\cfoot{--- Andreas Suter -- \today ---}
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\rfoot{\includegraphics[width=6.4cm]{PSI_Logo_wide_blau.pdf}}
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\DeclareMathAlphabet{\bi}{OML}{cmm}{b}{it}
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\newcommand{\mean}[1]{\langle #1 \rangle}
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\newcommand{\lem}{LE-$\mu$SR\xspace}
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\newcommand{\musr}{$\mu$SR\xspace}
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\newcommand{\etal}{\emph{et al.\xspace}}
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\newcommand{\ie}{\emph{i.e.\xspace}}
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\newcommand{\eg}{\emph{e.g.\xspace}}
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\newcolumntype{d}[1]{D{.}{.}{#1}}
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\begin{document}
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% Header info --------------------------------------------------
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\thispagestyle{empty}
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\noindent
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\begin{tabular}{@{\hspace{-0.7cm}}l@{\hspace{6cm}}r}
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\noindent\includegraphics[width=3.4cm]{PSI_Logo_narrow_blau.jpg} &
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{\Huge\sf Memorandum}
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\end{tabular}
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%
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\vskip 1cm
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%
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\begin{tabular}{@{\hspace{-0.5cm}}ll@{\hspace{4cm}}ll}
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Datum: & \today & & \\[3ex]
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Von: & Andreas Suter & An: & \\
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Telefon: & +41\, (0)56\, 310\, 4238 & & \\
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Raum: & WLGA / 119 & cc: & \\
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e-mail: & \verb?andreas.suter@psi.ch? & & \\
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\end{tabular}
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%
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\vskip 0.3cm
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\noindent\hrulefill
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\vskip 1cm
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%
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\noindent The \musr decay histogram can be described by
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\begin{equation}\label{eq:decay}
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N(t) = N_0\, e^{-t/\tau_\mu} \left[ 1 + A P(t) \right] + \mathrm{Bkg}
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\end{equation}
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\noindent For single histogram fits a good initial estimate for $N_0$ is needed
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in order that \textsc{Minuit2} has a chance to converge.
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Here I summaries how $N_0$ can be reasonably well estimates. For all estimates,
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it is assumed that rather to take Eq.(\ref{eq:decay}),
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the following ansatz will be used
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\begin{equation}\label{eq:N0ansatz}
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T(t) = N_0\, e^{-t/\tau_\mu} + \mathrm{Bkg},
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\end{equation}
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\noindent \ie the asymmetry is ignored all together.
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\section*{Estimate $\mathbi{N}_0$ with free Background}
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We would like to minimize $\chi^2$, which is
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\begin{equation}\label{eq:chisq}
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\chi^2 = \sum_i \frac{(D_i - T_i)^2}{\sigma_i^2} = \sum_i \frac{(D_i -
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T_i)^2}{D_i},
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\end{equation}
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\noindent where $D_i$ is a histogram entry of the measured data, $T_i$ the
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evaluation of Eq.(\ref{eq:N0ansatz}) at $t_i = \Delta t \cdot i$, with $\Delta
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t$ the histogram time resolution. In the last step Poisson statistics is used,
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\ie $\sigma_i^2 = D_i$.
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In order to minimize Eq.(\ref{eq:N0ansatz}), it is sufficient that $\nabla
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\chi^2 = 0 \Longrightarrow$
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\begin{eqnarray*}
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\frac{\partial \chi^2}{\partial N_0} &=& \sum_i \left[ \frac{2 (D_i -
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T_i)}{D_i}\, e^{-t_i/\tau_\mu} \right] = 0 \\
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\frac{\partial \chi^2}{\partial \mathrm{Bkg}} &=& \sum_i \left[ \frac{2 (D_i -
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T_i)}{D_i} \right] = 0
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\end{eqnarray*}
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\noindent With the following abbreviations
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\begin{eqnarray}\label{eq:abbrv}
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x_i &=& e^{-t_i/\tau_\mu} \\
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\Delta &=& \sum_i 1/D_i \nonumber \\
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S &=& \sum_i 1 \nonumber
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\end{eqnarray}
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\noindent ($S$ is the number of bins in the histogram) the background can be
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written as
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\begin{equation}\label{eq:bkgEstimate}
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\mathrm{Bkg} = \frac{1}{\Delta}\, \sum_i \left[ 1 - \frac{N_0}{D_i}\, x_i
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\right]
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\end{equation}
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\noindent and using this result, $N_0$ is
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\begin{equation}\label{eq:N0estimate}
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N_0 = \frac{\displaystyle\sum_i \left[x_i \left(1-\frac{S}{D_i
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\Delta}\right)\right]}{\displaystyle\sum_j \left[\frac{x_j}{D_j}\left(x_j -
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\frac{1}{\Delta} \sum_k \frac{x_k}{D_k}\right)\right]}.
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\end{equation}
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\noindent Eqs.(\ref{eq:bkgEstimate}) \& (\ref{eq:N0estimate}) allow to estimate
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$N_0$ and Bkg. However, the results are mostly unsatisfactory since typically
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$N_0$ is underestimate whereas Bkg is grossly overestimated. The reason is the
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missing asymmetry term.
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\section*{Estimate $\mathbi{N}_0$ with linked Background}
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A much more robust ansatz is to link the background to $N_0$, \ie
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\begin{equation}\label{eq:N0ansatz_with_linked_bkg}
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T(t) = N_0\, e^{-t/\tau_\mu} + \mathrm{Bkg} = N_0\, e^{-t/\tau_\mu} + \alpha
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N_0,
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\end{equation}
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\noindent where $\alpha$ is typically $0.01$ are smaller. The practical tests
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show that it is mostly save to set $\alpha=0$. From
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$$ \frac{\partial \chi^2}{\partial N_0} = 0 $$
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\noindent it follows
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\begin{equation}\label{eq:N0estimate_with_linked_bkg}
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N_0 = \frac{\displaystyle\sum_i (\alpha +
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x_i)}{\displaystyle\sum_i\frac{(\alpha + x_i)^2}{D_i}}.
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\end{equation}
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\noindent Eq.(\ref{eq:N0estimate_with_linked_bkg}) with $\alpha < 0.01$ leads typically to
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very good results.
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A closer look at Eq.(\ref{eq:N0estimate_with_linked_bkg}) reveals that there is
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a principle problem. How should one deal with $D_i = 0$? There are two
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possibilities:
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\begin{enumerate}
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\item if $D_i = 0$, set it to $D_i = 1$. This implicitly assumes that an empty bin has an uncertainty of 1.
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\item if $D_i = 0$, ignore it!
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\end{enumerate}
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\noindent Currently \texttt{musrfit} ignores empty bins.
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\end{document}
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