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Thesis: Random Phasor Sum distribution

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Eric Teunis de Boone 2023-05-16 16:16:11 +02:00
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documents/thesis/chapters/beacon_discipline.tex
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@ -591,20 +591,90 @@ opening the way to efficiently measuring the phases in realtime.\Todo{figure?}
% Gaussian noise
The traces will contain noise from various sources, both internal (e.g.~LNA~noise) and external (e.g.~radio~communications) to the detector.
Adding gaussian noise to the traces in simulation gives a simple noise model, associated to many random noise sources.
A simple noise model is given by gaussian noise in the time-domain, associated to many independent random noise sources.
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.
In the following, this aspect is shortly described in terms of two frequency-domain phasors;
the noise phasor written as $\vec{m} = a \, e^{i\pTrue}$ with phase $-\pi < \pTrue \leq \pi$ and amplitude $a \geq 0$,
and the signal phasor written as $\vec{s} = s \, e^{i\pTrue_s}$, but rotated such that its phase $\pTrue_s = 0$.
\Todo{reword; phasor vs plane wave}
Further reading can be found in Ref.~\cite{goodman1985:2.9}.
\\
% Phasor concept
\begin{figure}
\label{fig:phasor}
\caption{
Phasors picture
}
\end{figure}
\bigskip
Phasor concept
\cite{goodman1985:2.9}
Known phasor $\vec{s}$ + random phasor $\vec{m} = a e^{i\pTrue}$ with $-\pi < \pTrue < \pi$ and $a \geq 0$.
% Noise phasor description
The noise phasor is fully described by the joint probability density function
\begin{equation}
\label{eq:random_phasor_pdf}
\label{eq:noise:pdf:joint}
\phantom{,}
p_{A\PTrue}(a, \pTrue; \sigma)
=
\frac{a}{s\pi\sigma^2} e^{-\frac{a^2}{2\sigma^2}}
,
\end{equation}
for $-\pi < \pTrue \leq \pi$ and $a \geq 0$.
\\
Integrating \eqref{eq:noise:pdf:joint} over the amplitude $a$, it follows that the phase is uniformly distributed.
Likewise, the amplitude follows a Rayleigh distribution
\begin{equation}
\label{eq:noise:pdf:amplitude}
\label{eq:pdf:rayleigh}
\phantom{,}
p_A(a; \sigma)
%= p^{\mathrm{RICE}}_A(a; \nu = 0, \sigma)
= \frac{a}{\sigma^2} e^{-\frac{a^2}{2\sigma^2}}
,
\end{equation}
for which the mean is $\bar{a} = \sigma \sqrt{\frac{\pi}{2}}$ and the standard~deviation is given by $\sigma_{a} = \sigma \sqrt{ 2 - \tfrac{\pi}{2} }$.
\begin{figure}
\begin{subfigure}{0.45\textwidth}
\includegraphics[width=\textwidth]{beacon/pdf_noise_phase.pdf}
\caption{
The phase of the noise is uniformly distributed.
}
\label{fig:noise:pdf:phase}
\end{subfigure}
\hfill
\begin{subfigure}{0.45\textwidth}
\includegraphics[width=\textwidth]{beacon/pdf_noise_amplitude.pdf}
\caption{
The amplitude of the noise is Rayleigh distribution \eqref{eq:noise:pdf:amplitude}.
}
\label{fig:noise:pdf:amplitude}
\end{subfigure}
\caption{
Marginal distribution functions of the noise phasor.
Rayleigh and Rice distributions.
\Todo{expand captions}
}
\label{fig:noise:pdf}
\end{figure}
\bigskip
% Random phasor sum
In this work, the addition of the signal phasor to the noise phasor will be named ``Random Phasor Sum''.
The addition shifts the mean in \eqref{eq:noise:pdf:joint}
from $\vec{a}^2 = a^2 {\left( \cos \pTrue + \sin \pTrue \right)}^2$
to ${\left(\vec{a} - \vec{s}\right)}^2 = {\left( a \cos \pTrue -s \right)}^2 + {\left(\sin \pTrue \right)}^2$
,
resulting in a new joint distribution
\begin{equation}
\label{eq:phasor_sum:pdf:joint}
\phantom{.}
p_{A\PTrue}(a, \pTrue; s, \sigma)
= \frac{a}{2\pi\sigma^2}
\exp[ -
@ -615,45 +685,63 @@ Known phasor $\vec{s}$ + random phasor $\vec{m} = a e^{i\pTrue}$ with $-\pi < \p
2 \sigma^2
}
]
.
\end{equation}
requiring $ -\pi < 0 \leq pi $ and $a > 0$, otherwise $p_{A\PTrue} = 0$.
\\
\bigskip
Noise only Amplitude:
Rayleigh distribution
Integrating \eqref{eq:phasor_sum:pdf:joint} over $\pTrue$ one finds
a Rice (or Rician) distribution for the amplitude,
\begin{equation}
\label{eq:amplitude_pdf:rayleigh}
p_A(a; s=0, \sigma)
= p^{\mathrm{RICE}}_A(a; \nu = 0, \sigma)
= \frac{a}{\sigma^2} e^{-\frac{a^2}{2\sigma^2}}
\end{equation}
with $\sigma = \frac{\mu_1}{\sqrt{\frac{\pi}{2}}}$ and $\mu_2 = \frac{ 4 - \pi }{2}\sigma^2$.
\bigskip
Gaussian distribution
\begin{equation}
\label{eq:amplitude_pdf:gauss}
p_A(a; \sigma) = \frac{1}{\sqrt{2\pi}} \exp[-\frac{{\left(a + s\right)}^2}{2\sigma^2}]
\end{equation}
Rician distribution ( 2D Gaussian at $\nu$ with $\sigma$ spread)
\begin{equation}
\label{eq:amplitude_pdf:rice}
p^{\mathrm{RICE}}_A(a; s, \sigma)
\label{eq:phasor_sum:pdf:amplitude}
\label{eq:pdf:rice}
\phantom{,}
p_A(a; s, \sigma)
= \frac{a}{\sigma^2}
\exp[-\frac{a^2 + s^2}{2\sigma^2}]
\;
I_0\left( \frac{a s}{\sigma^2} \right)
,
\end{equation}
with $I_0(z)$ the modified Bessel function of the first kind with order zero.\\
No signal $\mapsto$ Rayleigh ($s = 0$);\\
Large signal $\mapsto$ Gaussian ($s \gg a$)
where $I_0(z)$ is the modified Bessel function of the first kind with order zero.
For the Rician distribution, two extreme cases can be highlighted (as can be seen in Figure~\ref{fig:phasor_sum:pdf:amplitude}).
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}.
Meanwhile, it approaches a gaussian distribution around $s$ when a strong signal ($s \gg a$) is presented.
\begin{equation}
\label{eq:strong_phasor_sum:pdf:amplitude}
p_A(a; \sigma) = \frac{1}{\sqrt{2\pi}} \exp[-\frac{{\left(a - s\right)}^2}{2\sigma^2}]
\end{equation}
\begin{figure}
\begin{subfigure}{0.45\textwidth}
\includegraphics[width=\textwidth]{beacon/pdf_phasor_sum_phase.pdf}
\caption{
The Random Phasor Sum phase distribution \eqref{eq:phasor_sum:pdf:phase}.
}
\label{fig:phasor_sum:pdf:phase}
\end{subfigure}
\hfill
\begin{subfigure}{0.45\textwidth}
\includegraphics[width=\textwidth]{beacon/pdf_phasor_sum_amplitude.pdf}
\caption{
The Random Phasor Sum amplitude distribution \eqref{eq:phasor_sum:pdf:amplitude}.
}
\label{fig:phasor_sum:pdf:amplitude}
\end{subfigure}
\caption{
A signal phasor's amplitude in the presence of noise will follow a Rician distribution.
For strong signals, this approximates a gaussian distribution, while for weak signals, this approaches a Rayleigh distribution.
\Todo{expand captions}
}
\label{fig:phasor_sum:pdf}
\end{figure}
\bigskip
Random Phasor Sum phase distribution: uniform (low $s$) + gaussian (high $s$)
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;
weak signals correspond to the uniform distribution for \eqref{eq:noise:pdf:joint}, while strong signals are well approximated by a gaussian distribution.
The analytic form takes the following complex expression,
\begin{equation}
\label{eq:phase_pdf:random_phasor_sum}
p_\PTrue(\pTrue; s, \sigma) =
@ -667,23 +755,20 @@ Random Phasor Sum phase distribution: uniform (low $s$) + gaussian (high $s$)
\right)}{2}
\cos{\pTrue}
\end{equation}
with
where
\begin{equation}
\label{eq:erf}
\phantom{,}
\erf{\left(z\right)} = \frac{2}{\sqrt{\pi}} \int_0^z \dif{t} e^{-t^2}
,
\end{equation}
.
\bigskip
Phase distribution: gaussian
\begin{equation}
\label{eq:phase_pdf:gaussian}
p_\PTrue(\pTrue; s, \sigma) = \frac{1}{\sqrt{2} \sigma} \exp\left(- \frac{s^2}{2\sigma^2} \right)
\end{equation}
is the error function.
\begin{figure}
\includegraphics[width=0.5\textwidth]{beacon/time_res_vs_snr.pdf}
\caption{Measured Time residuals vs Signal to Noise ration}
\caption{
Measured Time residuals vs Signal to Noise ratio
}
\label{fig:time_res_vs_snr}
\end{figure}