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Thesis: Random Phasor Sum distribution
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@ -591,20 +591,90 @@ opening the way to efficiently measuring the phases in realtime.\Todo{figure?}
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% Gaussian noise
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The traces will contain noise from various sources, both internal (e.g.~LNA~noise) and external (e.g.~radio~communications) to the detector.
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Adding gaussian noise to the traces in simulation gives a simple noise model, associated to many random noise sources.
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A simple noise model is given by gaussian noise in the time-domain, associated to many independent random noise sources.
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Especially important is that this simple noise model will affect the phase measurement depending on the strength of the beacon with respect to the noise level.
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In the following, this aspect is shortly described in terms of two frequency-domain phasors;
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the noise phasor written as $\vec{m} = a \, e^{i\pTrue}$ with phase $-\pi < \pTrue \leq \pi$ and amplitude $a \geq 0$,
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and the signal phasor written as $\vec{s} = s \, e^{i\pTrue_s}$, but rotated such that its phase $\pTrue_s = 0$.
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\Todo{reword; phasor vs plane wave}
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Further reading can be found in Ref.~\cite{goodman1985:2.9}.
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\\
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% Phasor concept
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\begin{figure}
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\label{fig:phasor}
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\caption{
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Phasors picture
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}
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\end{figure}
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\bigskip
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Phasor concept
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\cite{goodman1985:2.9}
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Known phasor $\vec{s}$ + random phasor $\vec{m} = a e^{i\pTrue}$ with $-\pi < \pTrue < \pi$ and $a \geq 0$.
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% Noise phasor description
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The noise phasor is fully described by the joint probability density function
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\begin{equation}
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\label{eq:random_phasor_pdf}
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\label{eq:noise:pdf:joint}
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\phantom{,}
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p_{A\PTrue}(a, \pTrue; \sigma)
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=
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\frac{a}{s\pi\sigma^2} e^{-\frac{a^2}{2\sigma^2}}
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,
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\end{equation}
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for $-\pi < \pTrue \leq \pi$ and $a \geq 0$.
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\\
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Integrating \eqref{eq:noise:pdf:joint} over the amplitude $a$, it follows that the phase is uniformly distributed.
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Likewise, the amplitude follows a Rayleigh distribution
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\begin{equation}
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\label{eq:noise:pdf:amplitude}
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\label{eq:pdf:rayleigh}
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\phantom{,}
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p_A(a; \sigma)
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%= p^{\mathrm{RICE}}_A(a; \nu = 0, \sigma)
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= \frac{a}{\sigma^2} e^{-\frac{a^2}{2\sigma^2}}
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,
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\end{equation}
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for which the mean is $\bar{a} = \sigma \sqrt{\frac{\pi}{2}}$ and the standard~deviation is given by $\sigma_{a} = \sigma \sqrt{ 2 - \tfrac{\pi}{2} }$.
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\begin{figure}
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\begin{subfigure}{0.45\textwidth}
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\includegraphics[width=\textwidth]{beacon/pdf_noise_phase.pdf}
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\caption{
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The phase of the noise is uniformly distributed.
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}
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\label{fig:noise:pdf:phase}
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\end{subfigure}
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\hfill
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\begin{subfigure}{0.45\textwidth}
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\includegraphics[width=\textwidth]{beacon/pdf_noise_amplitude.pdf}
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\caption{
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The amplitude of the noise is Rayleigh distribution \eqref{eq:noise:pdf:amplitude}.
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}
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\label{fig:noise:pdf:amplitude}
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\end{subfigure}
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\caption{
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Marginal distribution functions of the noise phasor.
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Rayleigh and Rice distributions.
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\Todo{expand captions}
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}
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\label{fig:noise:pdf}
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\end{figure}
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\bigskip
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% Random phasor sum
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In this work, the addition of the signal phasor to the noise phasor will be named ``Random Phasor Sum''.
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The addition shifts the mean in \eqref{eq:noise:pdf:joint}
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from $\vec{a}^2 = a^2 {\left( \cos \pTrue + \sin \pTrue \right)}^2$
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to ${\left(\vec{a} - \vec{s}\right)}^2 = {\left( a \cos \pTrue -s \right)}^2 + {\left(\sin \pTrue \right)}^2$
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,
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resulting in a new joint distribution
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\begin{equation}
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\label{eq:phasor_sum:pdf:joint}
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\phantom{.}
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p_{A\PTrue}(a, \pTrue; s, \sigma)
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= \frac{a}{2\pi\sigma^2}
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\exp[ -
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@ -615,45 +685,63 @@ Known phasor $\vec{s}$ + random phasor $\vec{m} = a e^{i\pTrue}$ with $-\pi < \p
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2 \sigma^2
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}
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]
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.
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\end{equation}
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requiring $ -\pi < 0 \leq pi $ and $a > 0$, otherwise $p_{A\PTrue} = 0$.
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\\
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\bigskip
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Noise only Amplitude:
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Rayleigh distribution
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Integrating \eqref{eq:phasor_sum:pdf:joint} over $\pTrue$ one finds
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a Rice (or Rician) distribution for the amplitude,
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\begin{equation}
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\label{eq:amplitude_pdf:rayleigh}
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p_A(a; s=0, \sigma)
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= p^{\mathrm{RICE}}_A(a; \nu = 0, \sigma)
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= \frac{a}{\sigma^2} e^{-\frac{a^2}{2\sigma^2}}
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\end{equation}
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with $\sigma = \frac{\mu_1}{\sqrt{\frac{\pi}{2}}}$ and $\mu_2 = \frac{ 4 - \pi }{2}\sigma^2$.
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\bigskip
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Gaussian distribution
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\begin{equation}
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\label{eq:amplitude_pdf:gauss}
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p_A(a; \sigma) = \frac{1}{\sqrt{2\pi}} \exp[-\frac{{\left(a + s\right)}^2}{2\sigma^2}]
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\end{equation}
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Rician distribution ( 2D Gaussian at $\nu$ with $\sigma$ spread)
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\begin{equation}
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\label{eq:amplitude_pdf:rice}
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p^{\mathrm{RICE}}_A(a; s, \sigma)
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\label{eq:phasor_sum:pdf:amplitude}
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\label{eq:pdf:rice}
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\phantom{,}
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p_A(a; s, \sigma)
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= \frac{a}{\sigma^2}
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\exp[-\frac{a^2 + s^2}{2\sigma^2}]
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\;
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I_0\left( \frac{a s}{\sigma^2} \right)
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,
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\end{equation}
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with $I_0(z)$ the modified Bessel function of the first kind with order zero.\\
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No signal $\mapsto$ Rayleigh ($s = 0$);\\
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Large signal $\mapsto$ Gaussian ($s \gg a$)
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where $I_0(z)$ is the modified Bessel function of the first kind with order zero.
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For the Rician distribution, two extreme cases can be highlighted (as can be seen in Figure~\ref{fig:phasor_sum:pdf:amplitude}).
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In the case of a weak signal ($s \ll a$), \eqref{eq:phasor_sum:pdf:amplitude} behaves as a Rayleigh distribution~\eqref{eq:noise:pdf:amplitude}.
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Meanwhile, it approaches a gaussian distribution around $s$ when a strong signal ($s \gg a$) is presented.
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\begin{equation}
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\label{eq:strong_phasor_sum:pdf:amplitude}
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p_A(a; \sigma) = \frac{1}{\sqrt{2\pi}} \exp[-\frac{{\left(a - s\right)}^2}{2\sigma^2}]
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\end{equation}
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\begin{figure}
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\begin{subfigure}{0.45\textwidth}
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\includegraphics[width=\textwidth]{beacon/pdf_phasor_sum_phase.pdf}
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\caption{
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The Random Phasor Sum phase distribution \eqref{eq:phasor_sum:pdf:phase}.
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}
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\label{fig:phasor_sum:pdf:phase}
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\end{subfigure}
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\hfill
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\begin{subfigure}{0.45\textwidth}
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\includegraphics[width=\textwidth]{beacon/pdf_phasor_sum_amplitude.pdf}
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\caption{
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The Random Phasor Sum amplitude distribution \eqref{eq:phasor_sum:pdf:amplitude}.
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}
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\label{fig:phasor_sum:pdf:amplitude}
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\end{subfigure}
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\caption{
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A signal phasor's amplitude in the presence of noise will follow a Rician distribution.
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For strong signals, this approximates a gaussian distribution, while for weak signals, this approaches a Rayleigh distribution.
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\Todo{expand captions}
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}
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\label{fig:phasor_sum:pdf}
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\end{figure}
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\bigskip
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Random Phasor Sum phase distribution: uniform (low $s$) + gaussian (high $s$)
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Like the amplitude distribution \eqref{eq:phasor_sum:pdf:amplitude}, the marginal phase distribution of \eqref{eq:phasor_sum:pdf:joint} results in two extremes cases;
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weak signals correspond to the uniform distribution for \eqref{eq:noise:pdf:joint}, while strong signals are well approximated by a gaussian distribution.
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The analytic form takes the following complex expression,
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\begin{equation}
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\label{eq:phase_pdf:random_phasor_sum}
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p_\PTrue(\pTrue; s, \sigma) =
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@ -667,23 +755,20 @@ Random Phasor Sum phase distribution: uniform (low $s$) + gaussian (high $s$)
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\right)}{2}
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\cos{\pTrue}
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\end{equation}
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with
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where
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\begin{equation}
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\label{eq:erf}
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\phantom{,}
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\erf{\left(z\right)} = \frac{2}{\sqrt{\pi}} \int_0^z \dif{t} e^{-t^2}
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,
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\end{equation}
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.
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\bigskip
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Phase distribution: gaussian
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\begin{equation}
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\label{eq:phase_pdf:gaussian}
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p_\PTrue(\pTrue; s, \sigma) = \frac{1}{\sqrt{2} \sigma} \exp\left(- \frac{s^2}{2\sigma^2} \right)
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\end{equation}
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is the error function.
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\begin{figure}
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\includegraphics[width=0.5\textwidth]{beacon/time_res_vs_snr.pdf}
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\caption{Measured Time residuals vs Signal to Noise ration}
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\caption{
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Measured Time residuals vs Signal to Noise ratio
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}
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\label{fig:time_res_vs_snr}
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\end{figure}
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