From fcdd01314d6e7bbc3d5093aaf1d8bd325fb82d17 Mon Sep 17 00:00:00 2001
From: Nao Pross <np@0hm.ch>
Date: Thu, 9 Dec 2021 13:56:50 +0100
Subject: Typos and begin writing "Quantifying Dispersion"

---
 doc/thesis/chapters/theory.tex | 51 +++++++++++++++++++++++++-----------------
 1 file changed, 31 insertions(+), 20 deletions(-)

(limited to 'doc/thesis/chapters/theory.tex')

diff --git a/doc/thesis/chapters/theory.tex b/doc/thesis/chapters/theory.tex
index b01a31f..f1c9eb1 100644
--- a/doc/thesis/chapters/theory.tex
+++ b/doc/thesis/chapters/theory.tex
@@ -15,12 +15,11 @@
 
 \section{Overview}
 
-The first two sections will briefly give the mathematics required by the modulation schemes used in the project. The notation used is summarised in \figref{fig:notation}. For conciseness encoding schemes and (digital) signal processing calculations are left out and discussed later. Thus for this section \(m_e = m\).
+The first two sections will briefly introduce mathematical formulations of the modulation schemes and of the channel models used in the project. The notation used is summarised in \figref{fig:notation}. For conciseness encoding schemes and (digital) signal processing calculations are left out and discussed later. Thus for this section \(m_e = m\).
 \skelpar[4]{Finish overview of the chapter.}
 
 \skelpar[3]{Discuss notation \(m(n) = m(nT)\) in discrete time and some other details.}
 
-
 %% TODO: A section on maths?
 % \section{Signal space and linear operators}
 
@@ -37,7 +36,7 @@ The first two sections will briefly give the mathematics required by the modulat
 	}
 \end{figure}
 
-Quadrature amplitude modulation is a family of modern digital modulation methods, that use an analog carrier signal. The simple yet effective idea behind QAM is to encode extra information into an orthogonal carrier signal, thus increasing the number of bits sent per unit of time \cite{Gallager,Kneubuehler,Mathis,Hsu}. A block diagram of the process is shown in \figref{fig:quadrature-modulation}.
+Quadrature amplitude modulation is a family of modern digital modulation methods, that use an analog carrier signal. The simple yet effective idea behind QAM is to encode extra information into an orthogonal carrier signal, thus increasing the number of bits sent per unit of time (symbol) \cite{Gallager,Kneubuehler,Mathis,Hsu}. A block diagram of the process is shown in \figref{fig:quadrature-modulation}.
 
 %% TODO: Quick par on "we will dicusss M-Ary QAM, M is 2^something"
 
@@ -94,7 +93,7 @@ A graphical way to see what is happening, is to observe a so called \emph{conste
 
 \paragraph{Example}
 
-A concrete example for \(M = 16\): if the message is 1110 the bit splitter creates two values \(\vec{m}_q = 11\) and \(\vec{m}_i = 10\); both are converted into analog amplitudes \(m_q = 3\) and \(m_i = 4\); that are then mixed with their respective carrier, resulting in \(s(t)\) being the point inside the bottom right sub-quadrant of the top right quadrant (blue dot in \figref{fig:qam-constellation}).
+A concrete example for \(M = 16\): if the message is 1110 the bit splitter creates two values \(\vec{m}_q = 11\) and \(\vec{m}_i = 10\); both are converted into analog amplitudes (symbols) \(m_q = 3\) and \(m_i = 4\); that are then mixed with their respective carrier, resulting in \(s(t)\) being the point inside the bottom right sub-quadrant of the top right quadrant (blue dot in \figref{fig:qam-constellation}).
 
 In \figref{fig:qam-constellation} the dots of the constellation have coordinates that begin on the bottom left corner, and are nicely aligned on a grid. Both are not a necessary requirement for QAM, in fact there are many schemes (for example when \(M = 32\)) that are arranged on a non square shape, and place the dots in different orders. The only constraint that most QAM modulators have in common, with regards to the geometry of the constellation, is that between any two adjacent dots (along the axis, not diagonally) only one bit of the represented value changes (gray code). This is done to improve the bit error rate (BER) of the transmission.
 
@@ -159,9 +158,9 @@ In the previous section, we discussed how the data is modulated and demodulated
 
 In theory because wireless transmission happens through electromagnetic radiation, to model a wireless channel one would need to solve Maxwell's equations for either the electric or magnetic field, however in practice that is not (analytically) possible. Instead what is typically done, is to model the impulse response of the channel using a geometrical or statistical model, parametrized by a set of coefficients that are either simulated or measured experimentally \cite{Gallager}.
 
-In our model we are going to include an additive white Gaussian noise (AWGN) and a Rician (or Rayleighan) fading; both are required to model physical effects of the real world. The former in particular is relevant today, as it mathematically describes dense urban environments.
+In our relatively simple model we are going to include an additive white Gaussian noise (AWGN) and a Rician (or Rayleighan) fading; both are required to model physical effects of the real world. The former in particular is relevant today, as it mathematically describes dense urban environments.
 
-\subsection{Geometric multipath fading model}
+\subsection{Geometric model}
 
 The simplest way to understand the multipath fading, is to consider it from a geometrical perspective. \figref{fig:multipath-sketch} is a sketch a wireless transmission system affected by multipath fading. The sender's antenna radiates an electromagnetic wave in the direction of the receiver (red line), however even under the best circumstances a part of the signal is dispersed in other directions (blue lines).
 
@@ -179,9 +178,9 @@ The problem is that, as is geometrically evident, some paths are longer than oth
 	r(t) = \sum_k c_k s(t - \tau_k).
 \end{equation}
 
-The linearity of the model is justified by the assumption that the underlying electromagnetic waves behave linearly (superposition holds) \cite{Gallager}. How many copies of \(s(t)\) (usually referred to as ``taps'' or ``rays'') should be included in the formula, depends on the precision requirements of the model.
+The linearity of the model is justified by the assumption that the underlying electromagnetic waves behave linearly (superposition holds) \cite{Gallager}. How many copies of \(s(t)\) (usually referred to as ``taps'' or ``rays'') should be included in \eqref{eqn:geom-multipath-rx}, depends on the precision requirements of the model.
 
-A further complication arises, when one end (or both) is not stationary. In that case the lengths of the paths change over time, and as a result both the delays \(\tau_k\) as well as the attenuations \(c_k\) become functions of time: \(\tau_k(t)\) and \(c_k(t)\) respectively \cite{Gallager,Messier}. Even worse is when the velocity at which the device is moving is high, because then Doppler shifts of the electromagnetic wave frequency become non negligible \cite{Gallager}.
+A further complication arises, when one end (or both) is not stationary. In that case the lengths of the paths change over time, and as a result both the delays \(\tau_k\) as well as the attenuations \(c_k\) become functions of time: \(\tau_k(t)\) and \(c_k(t)\) respectively \cite{Gallager,Messier}. Even worse when the velocity at which the device is moving is high, then Doppler shifts of the electromagnetic wave frequency become non negligible \cite{Gallager}.
 
 \begin{figure}
 	\centering
@@ -201,23 +200,23 @@ obtaining a new function
 \begin{equation} \label{eqn:multipath-impulse-response}
 	h(\tau, t) = \sum_k c_k(t) \delta(\tau - \tau_k(t)),
 \end{equation}
-that describes the \emph{channel impulse response} (CIR). This function depends on two time parameters: actual time \(t\) and convolution time \(\tau\). To better understand \(h(\tau, t)\), consider an example shown in figure \ref{fig:multipath-impulse-response}. Each stem represents a weighted Dirac delta, so each series of stems of the same color, along the convolution time \(\tau\) axis, is a channel response at some specific time \(t\). Along the other \(t\) axis we see how the entire channel response changes over time\footnote{In the figure only a finite number of stems was drawn, but actually \(h(\tau, t)\) is continuous in \(t\), i.e. the weights \(c_k(t)\) of the Dirac deltas change continuously.}. Notice that the stems are not quite aligned to the \(\tau\) time raster (dotted lines), that is because in \eqref{eqn:multipath-impulse-response} not only the weights \(c_k\) but also the delays \(\tau_k\) are time dependent.
+that describes the \emph{channel impulse response} (CIR). This function depends on two time parameters: actual time \(t\) and convolution time \(\tau\). To better understand \(h(\tau, t)\), consider an example shown in figure \ref{fig:multipath-impulse-response}. Each stem represents a weighted Dirac delta, so each series of stems of the same color, along the convolution time \(\tau\) axis, is a channel response at some specific time \(t\). Along the other \(t\) axis we see how the entire channel response changes over time\footnote{In the figure only a finite number of stems was drawn, but actually the weights \(c_k(t)\) of the Dirac deltas change continuously.}. Notice that the stems are not quite aligned to the \(\tau\) time raster (dotted lines), that is because in \eqref{eqn:multipath-impulse-response} not only the weights \(c_k\) but also the delays \(\tau_k\) are time dependent.
 
 \subsection{Spectrum of a multipath fading channel}
 
-With a continuous time channel model we can now discuss the spectral properties of a fading channel since the frequency response is the Fourier transform of the impulse response, mathematically \(H(f, t) = \fourier h(\tau, t)\). In this case however \(h(\tau, t)\) depends on two time variables, but that is actually not an issue, it just means that the frequency response is also changing with time. Hence we perform the Fourier transform with respect to the channel (convolution) time variable \(\tau\) to obtain
+With a continuous time channel model we can now discuss the spectral properties of a fading channel since the frequency response is the Fourier transform of the impulse response, mathematically \(H(f, t) = \fourier h(\tau, t)\). In this case however \(h(\tau, t)\) depends on two time variables, but that is actually not an issue, it just means that the frequency response is also changing over time. Hence we perform the Fourier transform with respect to the channel (convolution) time variable \(\tau\) to obtain
 \begin{equation} \label{eqn:multipath-frequency-response}
 	H(f, t) = \int_\mathbb{R} \sum_k c_k(t) \delta(\tau - \tau_k(t)) e^{-2\pi jf\tau} \, d\tau
 	= \sum_k c_k(t) e^{-2\pi jf \tau_k(t)}.
 \end{equation}
 
-Equation \eqref{eqn:multipath-frequency-response} shows that the frequency response is a periodic complex exponential, which has some important implications. Notice that if there is only one tap (term), the magnitude of \(H(f, t)\) is a constant (with respect to \(f\)) since \(|e^{j\alpha f}| = 1\). This means that the channels attenuates all frequencies by the same amount, therefore it is said to be a \emph{frequency non-selective} channel. Whereas in the case when there is more than one tap, the taps interfere destructively at certain frequencies and the channel is called \emph{frequency selective}. Plots of the frequency response of a two tap channel model are shown in \figref{fig:multipath-frequency-response-plots}. On the left is the magnitude of \(H(f, t)\), which presents periodic ``dips'', and on the right complex loci for the two taps (red and blue), as well as their sum (magenta), over the frequency range near the first dip (2 to 2.5 MHz) are shown.
+Equation \eqref{eqn:multipath-frequency-response} shows that the frequency response is a periodic complex exponential, which has some important implications. Notice that if there is only one tap (term), the magnitude of \(H(f, t)\) is a constant (with respect to \(f\)) since \(|e^{j\alpha f}| = 1\). This means that the channels attenuates all frequencies by the same amount, therefore it is said to be a \emph{frequency non-selective} or \emph{flat fading} channel. Whereas in the case when there is more than one tap, the taps interfere destructively at certain frequencies and the channel is called \emph{frequency selective}. To illustrate how this happens, plots of the frequency response of a two tap channel model are shown in \figref{fig:multipath-frequency-response-plots}. On the left is the magnitude of \(H(f, t)\), which presents periodic ``dips'', and on the right complex loci for the two taps (red and blue), as well as their sum (magenta), over the frequency range near the first dip (2 to 2.5 MHz) are shown.
 
 \begin{figure}
 	\centering
 	\resizebox{\linewidth}{!}{
-		% \input{figures/tikz/multipath-frequency-response-plots}
-		\skelfig[width = .8 \linewidth, height = 3cm]{}
+		\input{figures/tikz/multipath-frequency-response-plots}
+		% \skelfig[width = .8 \linewidth, height = 3cm]{}
 	}
 	\caption{
 		Frequency response of a multipath fading channel.
@@ -225,7 +224,19 @@ Equation \eqref{eqn:multipath-frequency-response} shows that the frequency respo
 	}
 \end{figure}
 
-\subsection{Discrete-time model}\label{sec:Discrete-time-model}
+\subsection{Quantifying dispersion}
+
+Having discussed how multipath fading affects communication systems, the next important step is to be able quantify its effects to be able to compare different multipath channels to each other.
+
+An intuitive parameter to quantify how dispersive channel is, is to take the time difference between the fastest and slowest paths with significant energy. What in the literature is called \emph{delay spread}, and is denoted here by \(T_d\). Consequently, a low delay spread means that all paths have more or less the same length, while a high delay spread implies that there is a large difference in length among the paths. Thus \(T_d\) could be be defined as
+\begin{equation}
+	T_d = \max_{k} (\tau_k(t)) -  \min_{k} (\tau_k(t)),
+\end{equation}
+as is done in \cite{Gallager}. However since in reality some paths get more attenuated than others (\(c_k(t)\) parameters) it also not uncommon to define the delay spread as a weighted mean or even as a statistical second moment (RMS value), where mean tap power \(\expectation\{|c_k(t)|^2\}\) is taken into account \cite{Mathis,Messier}. More sophisticated definitions of delay spread will be briefly mentioned later in section \ref{sec:statistical-model}.
+
+Another important parameter for quantifying dispersion is \emph{coherence bandwidth}, a measure how 
+
+\subsection{Discrete-time model} \label{sec:discrete-time-model}
 
 % TODO: discuss the "bins" of discrete time
 
@@ -265,24 +276,24 @@ is different from \eqref{eqn:multipath-impulse-response} consider again the plot
 From a signal processing perspective \eqref{eqn:discrete-multipath-impulse-response} can be interpreted as a simple tapped delay line, schematically drawn in \figref{fig:tapped-delay-line}, which confirms that the presented mathematical model is indeed a FIR filter. Simple multipath channels can be simulated with just a few lines of code, for example the data for the static fading channel in \figref{fig:multipath-frequency-response-plots} is generated in just four lines of Python. The difficulty of fading channels in practice lies in the estimation of the constantly changing parameters \(c_k(t)\) and \(\tau_k(t)\).
 
 
-\subsection{Fractional Delay}\label{sec:fractional-delay}
+\subsection{Fractional Delay} \label{sec:fractional-delay}
 % TO Do quelle:  http://users.spa.aalto.fi/vpv/publications/vesan_vaitos/ch3_pt1_fir.pdf
 
 \begin{figure}
 	\centering
 	\begin{subfigure}{.45\linewidth}
 		\includegraphics[width=\linewidth]{./figures/screenshots/Fractional_delay_6}
-		\caption{sinc function shifted by the delay = 6.0, with the sample points}
+		\caption{Pulse with an integer delay of 6 samples.}
 	\end{subfigure}
 	\hskip 5mm
 	\begin{subfigure}{.45\linewidth}
 		\includegraphics[width=\linewidth]{./figures/screenshots/Fractional_delay_637}
-		\caption{sinc function shifted by the delay = 6.37, with the sample points}
+		\caption{Pulse with a fractional delay of 6.37 samples.}
 	\end{subfigure}
-	\label{fig:fractional-delay-sinc-plot}	
+	\caption{\label{fig:fractional-delay-sinc-plot}}
 \end{figure}
 
-As in \ref{sec:Discrete-time-model} mentioned a FIR filter can be used to cheat a discrete time model of multiparty fading. But with a FIR filter, the delays are set at the sample rate, so the delays are integer. When the delays are noninteger a approximation had to be done. 
+As in \ref{sec:discrete-time-model} mentioned a FIR filter can be used to cheat a discrete time model of multiparty fading. But with a FIR filter, the delays are set at the sample rate, so the delays are integer. When the delays are noninteger a approximation had to be done. 
 
 In the example shown in \figref{fig:fractional-delay-sinc-plot}. For a integer delays in the sinc function all sample values are zero except the one by the delayed sample, which is the amplitude value, here one. When the delay is a fractional number all samples are non-zero. In theory this filter is notrealizable because its noncasual and the impulse respond is infinity long. This problem can't be solve by adding them because of the imaginary part.
 
@@ -323,7 +334,7 @@ The first non-zero sample can be find out with the help of the index M in \eqref
 
 \skelpar{Discrete frequency response. Discuss bins, etc.}
 
-\subsection{Statistical model}
+\subsection{Statistical model} \label{sec:statistical-model}
 
 Because as mentioned earlier it is difficult to estimate the time-dependent parameters of \(h_l(m)\) in many cases it is easier to model the components of the CIR as stochastic processes, thus greatly reducing the number of parameters. This is especially effective for channels that are constantly changing, because by the central limit theorem the cumulative effect of many small changes tends to a normal statistical distribution.
 
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