265 lines
12 KiB
TeX
265 lines
12 KiB
TeX
\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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\usepackage{url}
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\setlength{\topmargin}{10mm}
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\setlength{\topmargin}{-13mm}
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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 RRF Fits}%
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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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\section{Rotating Reference Frame Fits}
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High transverse field \musr (HTF-\musr) experiments will typically lead to
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rather large data sets since it is necessary to follow the high frequencies
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present in the positron decay histograms. Currently the HAL-9500 instrument
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at PSI \cite{HAL9500} is operated with $2\time 8$ positron detector, with a
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typical number of $\sim 4\times 10^5$ histogram bins. To analyze HTF-\musr
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data on rather slugish computer hardware is a challenge. In the last millennium
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the people invented the rotating reference frame transformation (RRF) \cite{Riseman90}
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to reduce to data sets to be handled.
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Here I will shortly describe the ways how it is implemented in \textsc{Musrfit},
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and why it should be avoided to be used altogether. The starting point of all
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is given by the positron decay spectrum which formally takes the form
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\begin{equation}\label{eq:positron_decay_spectrum}
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N^{(j)}(t) = N_0^{(j)} \exp(-t / \tau_\mu) \, \left[ 1 + A^{(j)}(t) \right] + N_{\rm bkg}^{(j)},
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\end{equation}
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\noindent where $(j)$ is the index of the positron counter, $N_0$ gives the scale
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of recorded positrons, $\tau_\mu$ is the muon lifetime, $A(t)$ the asymmetry, and $N_{\rm bkg}$
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describes the background due to uncorrelated events.
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The idea behind the RRF is twofolded: (i) try to extract $A(t)$, and (ii) shift the high frequency
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data set $A(t)$ to lower frequencies such that the number of necessary bins needed can be
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reduced (packing / rebinning), and hence the overall number of bins is much smaller.
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As I will try to explain, this is not for free, and there are problems arising from
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this kind of data treatment.
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\subsection{Single Histogram RRF Implementation}
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In a first step the asymmetry needs to be determined. This is done the following way:
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\begin{enumerate}
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\item Determine the background, $N_{\rm bkg}$, at times before $t_0$ ($t_0$ is the time of the muon
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implantation). Hopefully the background before and after $t_0$ is equal, which is not always
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the case.
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\item Multiple the background corrected histogram with $\exp(+t / \tau_\mu)$, this is leading to
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\end{enumerate}
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\begin{equation}\label{eq:Mt}
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M(t) \equiv \left[ N(t) - N_{\rm bkg} \right]\,\exp(+t / \tau_\mu) = N_0 \left[ 1 + A(t) \right].
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\end{equation}
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\begin{enumerate}
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\setcounter{enumi}{2}
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\item In order to extract $A(t)$ from $M(t)$, $N_0$ needs to be determined, which is almost the
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most tricky part here. The idea is simple: since $A(t)$ is dominated by high frequency signals,
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proper averaging over $M(t)$ should allow to determine $N_0$, assuming that $\langle A(t) \rangle = 0$.
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Is this assumption always true? \emph{No!} For instance it is \emph{not} true if the averaging is preformed
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over incomplete periodes of a single assumed signal. Another case where it will fail is if multiple signals
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with too far apart frequencies is present, as \eg in the case of muonium. Said all this, let's come
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back and try to determine $N_0$:
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\end{enumerate}
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\begin{equation}\label{eq:N0estimate}
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N_0 = \sum_{k=0}^{N_{\rm avg}} w_k M(t_k),
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\end{equation}
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\noindent where $N_{\rm avg}$ is determined such that $N_{\rm avg} \Delta t \simeq 1 \mu$s ($\Delta t$ being the
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time resolution. $1 \mu$s means averaging over many cyles). In order to get a good estimate for $N_{\rm avg}$, $N(t)$ is Fourier
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transformed, and from this power spectrum the frequency with the largest amplitude is determined, $\nu_{\rm 0}$. From $\nu_0$,
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$\Delta t$, the number of cycles fitting into $1 \mu$s can be determined, and from this and the time resolution $N_{\rm avg}$
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can be calculated. The weight $w_k$ is given by:
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\begin{equation}
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w_k = \frac{\left[\Delta M(t_k)\right]^{-2}}{\sum_{j=0}^{N_{\rm avg}} \left[\Delta M(t_j)\right]^{-2}},
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\end{equation}
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\noindent where
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\begin{equation}
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\Delta M(t) = \left[ \left(\frac{\partial M}{\partial N} \Delta N\right)^2 +
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\left(\frac{\partial M}{\partial N_{\rm bkg}} \Delta N_{\rm bkg}\right)^2 \right]^{1/2}
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\simeq \exp(+t/\tau_\mu) \sqrt{N(t)}.
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\end{equation}
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\noindent The error estimate on $N_0$ is then
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\begin{equation}\label{eq:N0-rrf}
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\Delta N_0 = \sigma_{N_0} = \sqrt{\sum_k w_k^2 \Delta M(t_k)^2}.
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\end{equation}
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\noindent Having estimated $N_0$, the asymmetry can be extracted as:
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\begin{equation}
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A(t) = M(t) / N_0 - 1.
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\end{equation}
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\begin{enumerate}
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\setcounter{enumi}{3}
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\item Now the actual RRF transformation can take place: $A_{\rm rrf}(t) = 2 \times A(t) \cos(\omega_{\rm rrf} t + \phi_{\rm rrf})$.
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The factor of 2 is introduced to conserve the asymmetry amplitude. The idea is the following: Fourier transform theory tells as that
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\end{enumerate}
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\begin{equation}
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{\cal F}\left\{2 \times A(t) \cos(\omega_{\rm rrf} t + \phi_{\rm rrf}) \right\} =
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{\cal F}\left\{A(t)\right\}(\omega-\omega_{\rm rrf}) + {\cal F}\left\{A(t)\right\}(\omega+\omega_{\rm rrf}),
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\end{equation}
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\noindent \ie that the Fourier spectrum of $A(t)$ is shifted down and up by $\omega - \omega_{\rm rrf}$ and $\omega + \omega_{\rm rrf}$, respectively.
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In order to get rid of the high frequency part (${\cal F}\left\{A(t)\right\}(\omega+\omega_{\rm rrf})$), $A_{\rm rrf}(t)$ will be heavily over-binned,
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\ie
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\begin{enumerate}
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\setcounter{enumi}{4}
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\item Do the rrf packing: $A_{\rm rrf}(t) \rightarrow \langle A_{\rm rrf}(t) \rangle_p$. Packing itself is a filtering of data! Especially this
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kind of filter is dispersive \cite{King89}, \ie that it potentially is leading to line shape distortions. For symmetric, rather narrow
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lines, this is unlikely to be a problem. However, this might be quite different for complex line shapes as in the case of vortex lattices.
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\end{enumerate}
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\noindent The property $\langle A_{\rm rrf}(t) \rangle_p$ is what is fitted. The error on this property is estimated the following way: (i) the
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unpacked error of $A(t)$ is:
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\begin{equation}
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\Delta A(t) \simeq \frac{\exp(+t/\tau_\mu)}{N_0}\,\left[ N(t) + \left( \frac{N(t)-N_{\rm bkg}}{N_0} \right)^2 \Delta N_0^2 \right]^{1/2},
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\end{equation}
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\noindent and form this the packed $A_{\rm rrf}(t)$ error can be calculated.
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\subsection{Asymmetry RRF Implementation}
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\begin{enumerate}
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\item In order to circumvent the difficulties to estimate $N_0$ the asymmetry of the starting positron histograms is formed. For details
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see \cite{musrfit_userManual15}. For this, positron detectors geometrically under $180^\circ$ are used. However, due the the spiraling
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of the positron in sufficiently high magnetic fields, and the uncertainties of the $t_0$'s, the geometrical phase might \emph{not}
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correspond to the positron signal phase! At $B=9$T the uncertainty in $t_0$ by one channel leads to a phase shift of
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$\gamma_\mu B \Delta t \cdot (180 / \pi) = 1.7^\circ$. Fig.\ref{fig:hal-9500-t0} shows the $t_0$-region of a typical HAL-9500 spectrum.
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It shows that it is very hard to get the absolut value of $t_0$ right.
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\item Carry out the RRF transformation $A_{\rm rrf}(t) = 2 \times A(t) \cos(\omega_{\rm rrf} t + \phi_{\rm rrf})$.
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\item Do the rrf packing.
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\end{enumerate}
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.5\textwidth]{HAL-9500-t0.pdf}
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\caption{The $t_0$ region of a typical HAL-9500 spectrum. The broad black hump with the green line, is the ``prompt'' peak.
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It is \emph{not} straight forward how to define $t_0$.}\label{fig:hal-9500-t0}
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\end{figure}
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\section{Discussion}
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Both RRF transformation sketched above have weak points which makes it hard to estimate systematic errors. Both methods will fail at too
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low fields of $\lesssim 1$T. The only and single purpose of the RRF transformation is slughish computer power! We developed GPU based fitting
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which overcomes \emph{all} this uncontrolled weaknesses and henceforth RRF could be omitted altogether. I still added it for the time being,
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since strong GPU/CPU hardware is still a bit costly and therefore not affordable to everyone.
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In order to give a feeling about what might go ``wrong'' with the RRF, I was running a couple of test cases. The chosen asymmetry is
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\begin{equation}\label{eq:asymmetry-simulation}
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A^{(j)}(t) = A_0^{(j)} \sum_{k=1}^3 f_k \exp\left[-0.5\cdot (\sigma_k t)^2 \right] \cos(\gamma_\mu B_k t + \phi^{(j)}),
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\end{equation}
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\noindent with values found in Tab.\ref{tab:asymmetry-parameters}. For the simulation 4 positron detector signals were generated with
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$A_0^{(j)} = \left\{ 0.2554, 0.2574, 0.2576, 0.2566 \right\}$. The further ingredients were: $N_0^{(j)} = \left\{ 27.0, 25.3, 25.6, 26.9 \right\}$,
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$N_{\rm bkg}^{(j)} = \left\{ 0.055, 0.060, 0.069, 0.064 \right\}$, and $ \phi^{(j)} = \left\{ 5.0, 95.0, 185.0, 275.0 \right\}$.
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\begin{table}[h]
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\centering
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\begin{tabular}{c|c|c|c}
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$k$ & $f_k$ & $\sigma_k$ & $B_k$ \\
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& & ($1/\mu$s) & (T) \\ \hline\hline
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1 & 0.5 & 7 & $1$ or $9$ \\
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2 & 0.2 & 0.75 & $1.02$ or $9.02$ \\
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3 & 0.3 & 0.25 & $1.06$ or $9.06$
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\end{tabular}
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\caption{Parameters used in Eq.(\ref{eq:asymmetry-simulation}).}\label{tab:asymmetry-parameters}
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\end{table}
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\begin{figure}[h]
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\centering
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\includegraphics[width=0.45\textwidth]{08011-Fourier-Averaged.pdf} \quad
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\includegraphics[width=0.45\textwidth]{08011-RRF-Histo-Fourier-Averaged-Ghost.pdf} \\
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\includegraphics[width=0.45\textwidth]{08011-RRF-Asym-Fourier-Averaged-Ghost.pdf}
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\caption{Averaged Fourier power spectra. Top left: from single histogram fit for the 1T data set.
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Top right: from single histogram RRF fit. Bottom: from asymmetry RRF fit. Both RRF
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sets show ghost lines.}\label{fig:08011-Fourier}
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\end{figure}
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Figure.\ref{fig:08011-Fourier} shows the averaged Fourier power spectra for the simulated data sets at 1T. Both RRF transformation are
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showing ghost lines, even for optimally chosen RRF rebinning. At higher fields this is less pronounced. The ghost lines have various origins
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such as aliasing effects due to the RRF packing not prefectly suppressing the high frequency part of $A_{\rm rrf}(t)$, leakage of the RRF frequency
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for not sufficently precise known $N_0$ (see Eq.(\ref{eq:N0-rrf})) for single histogram RRF fits, etc.
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Fits of simulated data as described above (see Eq.(\ref{eq:asymmetry-simulation}), with fields 0.5, 1.0, 3.0, 5.0, 7.0, and 9.0T) show that above
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about 1T the model parameters of the RRF fits are acceptable, but the error bars are typically about a factor 3 larger compared to single histogram
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fits. The asymmetries of the RRF fits are ``substantially'' too small. The $\chi^2$ values are close to meaningless for the RRF fits, since they
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are strongly depending on the RRF packing, time interval chosen, etc.
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To summaries: RRF fits can be used for online analysis if no GPU accelerator is available, but \emph{must not} be used for any final analysis!
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\bibliographystyle{amsplain}
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\bibliography{rrf}
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\end{document} |