From 4d1c4a8b63d257bf2127dd9068a981a95e4e944c Mon Sep 17 00:00:00 2001
From: Nao Pross <np@0hm.ch>
Date: Tue, 21 Dec 2021 15:01:18 +0100
Subject: First half of TA feedback

---
 doc/thesis/chapters/theory.tex | 57 ++++++++++++++++++------------------------
 1 file changed, 24 insertions(+), 33 deletions(-)

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

diff --git a/doc/thesis/chapters/theory.tex b/doc/thesis/chapters/theory.tex
index dfea46d..9e1b563 100644
--- a/doc/thesis/chapters/theory.tex
+++ b/doc/thesis/chapters/theory.tex
@@ -8,7 +8,7 @@
 		\input{figures/tikz/overview}
 	}
 	\caption{
-		Block diagram of our wireless communication system with annotated signal names. Frequency domain representations of signals use the uppercase symbol of their respective time domain name. Amplification constants in the channel (for \(s(t)\) and \(r(t)\)) were omitted throughout the document for readability.
+		Block diagram of the developed wireless communication system with annotated signal names. Frequency domain representations of signals use the uppercase symbol of their respective time domain name. Amplification constants in the channel (for \(s(t)\) and \(r(t)\)) were omitted throughout the document for readability.
 		\label{fig:notation}
 	}
 \end{figure}
@@ -61,7 +61,7 @@ Having analog level signals, it is now possible to mix them with radio frequency
 
 \subsection{Orthogonality of carrier signals}
 
-Before explaining how the two carrier signals are generated, we first need to discuss some important mathematical properties \(\phi_i\) and \(\phi_q\) need to have, in order to modulate two messages over the same frequency \(\omega_c\). The two carriers need to be \emph{orthonormal}\footnote{Actually orthogonality alone would be sufficient, however then the left side of \eqref{eqn:orthonormal-condition} would not equal 1, and an inconvenient factor would be introduced in many later equations \cite{Gallager,Hsu}.} to each other, mathematically this is expressed by the conditions
+Before explaining how the two carrier signals are generated, some important mathematical properties of \(\phi_i\) and \(\phi_q\) have to be discussed, in order to modulate two messages over the same frequency \(\omega_c\). The two carriers need to be \emph{orthonormal}\footnote{Actually orthogonality alone would be sufficient, however then the left side of \eqref{eqn:orthonormal-condition} would not equal 1, and an inconvenient factor would be introduced in many later equations \cite{Gallager,Hsu}.} to each other, mathematically this is expressed by the conditions \cite{Gallager}
 \begin{subequations} \label{eqn:orthonormal-conditions}
 	\begin{align}
 		\langle \phi_i, \phi_q \rangle
@@ -69,7 +69,7 @@ Before explaining how the two carrier signals are generated, we first need to di
 			= 0, \text{ and } \label{eqn:orthogonal-condition} \\
 		\langle \phi_k, \phi_k \rangle
 			&= \int_T \phi_k \phi_k^*  \,dt = 1,
-			\text{ where } k \text{ is either } i \text{ or } q \,\text{\cite{Gallager}}. \label{eqn:orthonormal-condition}
+			\text{ where } k \text{ is either } i \text{ or } q. \label{eqn:orthonormal-condition}
 	\end{align}
 \end{subequations}
 Provided these rather abstract conditions, let's define a new signal 
@@ -140,7 +140,7 @@ It is worth noting that the case of 4-PSK, also known as quaternary phase shift
 
 \section{Multipath fading} \label{sec:multipath-fading}
 
-In the previous section, we discussed how the data is modulated and demodulated at the two ends of the transmission system. This section discusses what happens between the sender and receiver when the modulated passband signal is transmitted wirelessly.
+In the previous section, we discussed how the data is modulated and demodulated at the two ends of the transmission system. This section discusses what happens between the transmitter and receiver when the modulated passband signal is transmitted wirelessly.
 
 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}.
 
@@ -159,7 +159,7 @@ The simplest way to understand multipath fading, is to consider it from a geomet
 	}
 \end{figure}
 
-The problem is that, as is geometrically evident, some paths are longer than others. Because of this fact, the signal is seen by the receiver multiple times with different phase shifts \cite{Gallager,Messier}. To mathematically model this effect, we describe the received signal \(r(t)\) as a linear combination of delayed copies of the sent signal \(s(t)\), each with a different attenuation \(c_k\) and phase shift \(\tau_k\):
+The problem is that, as is geometrically evident, some paths are longer than others. Because of this fact, the signal is seen by the receiver multiple times with different phase shifts~\cite{Gallager,Messier}. To mathematically model this effect, we describe the received signal \(r(t)\) as a linear combination of delayed copies of the sent signal \(s(t)\), each with a different attenuation \(c_k\) and phase shift \(\tau_k\):
 \begin{equation} \label{eqn:geom-multipath-rx}
 	r(t) = \sum_k c_k s(t - \tau_k).
 \end{equation}
@@ -177,7 +177,7 @@ A further complication arises, when one end (or both) is not stationary. In that
 	}
 \end{figure}
 
-We have thus observed that the arrangement can be modelled as a linear time-\emph{varying} system (LTV), if the sender or the receiver (or anything else in the channel) is moving, and as a linear time \emph{invariant} (LTI) system if the geometry is stationary. Regardless of which of the two cases, linearity alone is sufficient to approximate the channel as a finite impulse response (FIR) filter \cite{Messier}. We can rewrite an LTV version of equation \eqref{eqn:geom-multipath-rx} using a convolution product as follows:
+Thus the arrangement can be modelled as a linear time-\emph{varying} system (LTV), if the transmitter or the receiver (or anything else in the channel) is moving, and as a linear time \emph{invariant} (LTI) system if the geometry is stationary. Regardless of which of the two cases, linearity alone is sufficient to approximate the channel as a finite impulse response (FIR) filter \cite{Messier}. We can rewrite an LTV version of equation \eqref{eqn:geom-multipath-rx} using a convolution product as follows:
 \begin{align*}
 	r(t) = \sum_k c_k(t) s(t - \tau_k(t)) &= \sum_k c_k(t) \int_\mathbb{R} s(\tau) \delta(\tau - \tau_k(t)) \,d\tau \\
 		&= \int_\mathbb{R} s(\tau) \sum_k c_k(t) \delta(\tau - \tau_k(t)) \,d\tau = s(\tau) * h(\tau, t),
@@ -190,7 +190,7 @@ that describes the \emph{channel impulse response} (CIR). This function depends
 
 \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 \(h(\tau, t)\) 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
+With a continuous time channel model the spectral properties of a fading channel can now be discussed, since the frequency response is the Fourier transform of the impulse response, i.e. \(H(f, t) = \fourier h(\tau, t)\). In this case \(h(\tau, t)\) 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)}.
@@ -213,7 +213,7 @@ Equation \eqref{eqn:multipath-frequency-response} shows that the frequency respo
 
 \subsection{Quantifying dispersion}
 
-Having discussed how multipath fading affects communication systems, the next important step is to be able to quantify its effects to be able to compare different multipath channels to each other.
+Having discussed how multipath fading affects communication systems, the next important step is to be able to quantify its effects in order to 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}
@@ -240,20 +240,20 @@ Finally, another important mean of parametrizing a multipath fading channel is w
 
 % TODO: discuss the "bins" of discrete time
 
-Since in practice signal processing is done digitally, it is meaningful to discuss the properties of a discrete-time model. To keep the complexity of the model manageable some assumptions are necessary, thus the sent discrete signal\footnote{This is an abuse of notation. The argument \(n\) is used to mean the \(n\)-th digital sample of \(s\), whereas \(s(t)\) is used for the analog waveform. A more correct but longer notation is \(s(nT)\), where \(T\) is the sample time.} \(s(n)\) is assumed to have a finite single sided bandwidth \(W\). This implies that in the time-domain the signal is a series of sinc-shaped pulses each shifted from the previous by a time interval \(T = 1 / (2W)\) (Nyquist rate):
+Since in practice signal processing is done digitally, it is meaningful to discuss the properties of a discrete-time model. To keep the complexity of the model manageable some assumptions are necessary, thus the sent discrete signal\footnote{This is an abuse of notation. The argument \(n\) is used to mean the \(n\)-th digital sample of \(s\), whereas \(s(t)\) is used for the analog waveform. A more correct but longer notation is \(s(nT)\), where \(T\) is the sample time.} \(s(n)\) is assumed to have a finite single sided bandwidth \(W\). This implies that the time-domain the signal is a series of sinc-shaped pulses each shifted from the previous by a time interval \(T = 1 / (2W)\) (Nyquist rate) \cite{Messier}:
 \begin{equation}
-	s(t) = \sum_n s(n) \sinc(t/T - n)
+	s(t) = \sum_n s(n) \sinc(t/T - n).
 \end{equation}
 The waveform \(s(t)\) is then convolved with the CIR function \(h(\tau, t)\) (with respect to \(\tau\)) from the continuous time model resulting in the waveform at the receiver
 \begin{align*}
 	r(t) &= \int_ \mathbb{R} \sum_n s(n) \sinc(\tau / T - n) \sum_k c_k(t) \delta(\tau - \tau_k(t)) \,d\tau \\
 	&= \sum_n s(n) \sum_k c_k(t) \sinc(t/T - \tau_k(t)/T - n),
 \end{align*}
-which is then sampled at the Nyquist rate of \(2W = 1/T\), resulting in a set of samples\footnote{Again, the abusing notation \(r(m)\) means the \(m\)-th digital sample of \(r(t)\), i.e. \(r(mT)\).}:
+which is then sampled at the Nyquist rate of \(2W = 1/T\), resulting in a set of samples\footnote{Again, the abusing notation \(r(m)\) means the \(m\)-th digital sample of \(r(t)\), i.e. \(r(mT)\).} \cite{Messier}:
 \[
 	r(m) = \sum_n s(n) \sum_k c_k(mT) \sinc(m - \tau_k(mT)/T - n).
 \]
-Finally the substitution \(l = m - n\) eliminates the sender's sample counter \(n\) (unknown to the receiver) and reformulates \(r(m)\) as a discrete convolution product of with a discrete CIR function \(h_l(m)\):
+Finally the substitution \(l = m - n\) eliminates the sender's sample counter \(n\) (unknown to the receiver) and reformulates \(r(m)\) as a discrete convolution product of with a discrete CIR function \(h_l(m)\) \cite{Messier}:
 \begin{equation}
 	r(m) = \sum_l s(m - l) \sum_k c_k(mT) \sinc(l - \tau_k(mT)/T) 
 	= \sum_l s(m - l) h_l (m).
@@ -273,7 +273,7 @@ is different from \eqref{eqn:multipath-impulse-response} consider again the plot
 	}
 \end{figure}
 
-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 \(\tau_k(t)\).
+From a signal processing perspective \eqref{eqn:discrete-multipath-impulse-response} can be interpreted as a simple tapped delay line \cite{Messier, Gallager}, 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 \(\tau_k(t)\) \cite{Messier}.
 
 \subsection{Simulating multipath CIR with FIR filters} \label{sec:fractional-delay}
 
@@ -294,17 +294,17 @@ From a signal processing perspective \eqref{eqn:discrete-multipath-impulse-respo
 	}
 \end{figure}
 
-As mentioned in \ref{sec:discrete-time-model} a FIR filter can be used to simulate discrete-time models of multipath fading. But with FIR filters the delays can only be integer multiples of the sample rate. When the delays are non integer an approximation needs to be done \cite{Valimaki1995}, that is because FIR filters have a transfer function of the form
+As mentioned in \ref{sec:discrete-time-model} a FIR filter can be used to simulate discrete-time models of multipath fading. But with FIR filters the delays can only be integer multiples of the sample rate. When the delays are non integer an approximation needs to be done, that is because FIR filters have a transfer function of the form
 \begin{equation} \label{eqn:transfer-function-fir}
 	H(j\omega) = \sum_{n = 0}^{N} h(n) e^{-j\omega nT}
 	\quad \text{commonly written as} \quad
 	H(z) = \sum_{n = 0}^{N} h(n) z^{-n},
 \end{equation}
-but a non integer delay of \(\tau\) in the frequency domain is \(H_\tau(j\omega) = e^{-j\omega \tau}\). There are multiple ways to find coefficients \(h(n)\) that approximate \(H_\tau\), in this case the least squares method was used by minimizing the error function
+but a non integer delay of \(\tau\) in the frequency domain is \(H_\tau(j\omega) = e^{-j\omega \tau}\). There are multiple ways to find coefficients \(h(n)\) that approximate \(H_\tau\) \cite{Valimaki1995}, in this case the least squares method was used by minimizing the error function
 \begin{equation}
 	E(j\omega) = H(j\omega) - H_\tau(j\omega).
 \end{equation}
-The least square method plus the assumption of finite bandwidth and the requirement of causality gives the following rule for computing the FIR filter coefficients:
+The least square method plus the assumption of finite bandwidth and the requirement of causality gives the following rule for computing the FIR filter coefficients \cite{Valimaki1995}:
 \begin{equation}
 	h(n)= \begin{cases}
 		\sinc (n - \tau) & 0 \leq n \leq N \\ 
@@ -315,14 +315,14 @@ where the odd order of the filter \(N\) should satisfy the condition
 \begin{equation} \label{eqn:fractional-fir-length}
 	N = 2 \lfloor \tau \rfloor + 1
 \end{equation}
-for a minimal error in the approximation. It is worth mentioning that it is also possible to build FIR filters of even length with a different condition, or that do not satisfy \eqref{eqn:fractional-fir-length}, in which cases more consideration is required. An example of a fractional delay FIR filter is shown in \figref{fig:fractional-delay-sinc-plot}.
+for a minimal error in the approximation \cite{Valimaki1995}. It is worth mentioning that it is also possible to build FIR filters of even length with a different condition, or that do not satisfy \eqref{eqn:fractional-fir-length}, in which cases more consideration is required. An example of a fractional delay FIR filter is shown in \figref{fig:fractional-delay-sinc-plot}.
 
 \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 \cite{Messier,Mathis}. 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 distribution.
 
 % \skelpar[3]{Assumptions of the model}
-Before discussing the models themselves, their underlying statistical assumptions need to be considered. In the literature the so called WSSUS assumptions are made, which for a LTV CIR \(h(\tau, t)\) can be formulated as
+Before discussing the models themselves, their underlying statistical assumptions need to be considered. In the literature the so called WSSUS assumptions are made \cite{Messier, Gallager}, which for a LTV CIR \(h(\tau, t)\) can be formulated as
 \begin{subequations}
 	\begin{align}
 		R(\tau') &= \E{h(\tau, t) h^*(\tau + \tau', t)}, \text{ and } \label{eqn:stat-wss} \\
@@ -346,16 +346,16 @@ Multipath fading is a form of multiplicative noise, as mathematically confirmed
 	= e^{j\omega_c \tau} \sum_k c_k(t) e^{-j\omega_c \tau_k(t)}
 	= e^{j\omega_c \tau} \cdot f(t).
 \end{equation}
-If there is no line of sight (NLOS), it is reasonable to assume that all path have more or less the same attenuation, i.e. all \(c_k\) are the same. Another reasonable assumption in this case is that all paths are equally likely to be taken, or in other words the delays \(-\omega_c \tau_k\) can be replaced with random variables \(\vartheta_k\) that are uniformly distributed on \([0,2\pi)\) \cite{Hoher2013,Mathis}; physically this can be imagined as a ``ring of scattering objects'' around the receiver\cite{Messier}, as shown in \figref{fig:ring-of-scattering-objects} (without the red line of sight signal). Finally, assuming that there are infinitely many paths the random variable for the multiplicative fading noise becomes
+If there is no line of sight (NLOS), it is reasonable to assume that all path have more or less the same attenuation, i.e. all \(c_k\) are the same. Another reasonable assumption in this case is that all paths are equally likely to be taken, or in other words the delays \(-\omega_c \tau_k\) can be replaced with random variables \(\vartheta_k\) that are uniformly distributed on \([0,2\pi)\) \cite{Hoher2013,Mathis}; physically this can be imagined as a ``ring of scattering objects'' around the receiver \cite{Messier} as sketched in \figref{fig:ring-of-scattering-objects} (without the red line of sight signal). Finally, assuming that there are infinitely many paths the random variable for the multiplicative fading noise becomes
 \begin{equation} \label{eqn:mult-fading-nlos}
 	f = \lim_{N\rightarrow\infty} \frac{1}{\sqrt{N}}
 		\sum_{k=1}^{N} e^{j \vartheta_k },
 \end{equation}
-where the \(c_k\) where omitted, since they are assumed to be all equal. The factor \(1/\sqrt{N}\) is introduced such that \(\expectation \{|f|^2\} = 1\). It then can be shown that the probability density function of \(|f|\) is
+where the \(c_k\) where omitted, since they are assumed to be all equal \cite{Hoher2013}. The factor \(1/\sqrt{N}\) is introduced such that \(\expectation \{|f|^2\} = 1\). It then can be shown that the probability density function of \(|f|\) is
 \begin{equation}
 	p(a)= 2a e^{-a^2}, \text{ or } |f| \sim \mathcal{R},
 \end{equation}
-i.e. the amplitude of \(f\) is \emph{Raileigh} distributed. The probability density function of a Rayleigh distributed random variable is shown in \figref{fig:rayleigh-rice-pdf}.
+i.e. the amplitude of \(f\) is \emph{Raileigh} distributed \cite{Hoher2013}. The probability density function of a Rayleigh distributed random variable is shown in \figref{fig:rayleigh-rice-pdf}.
 
 \begin{figure}
 	\centering
@@ -387,7 +387,7 @@ i.e. the amplitude of \(f\) is \emph{Raileigh} distributed. The probability dens
 
 \paragraph{LOS case}
 
-Extending the previous NLOS case, if there is a line of sight (LOS) path (red signal in \figref{fig:ring-of-scattering-objects}), the statistical model has to be extended by defining the so called Rice factor \(K\). This \(K\) factor is the ratio between the power from the LOS path and the average power of the NLOS paths (often also referred to as distributed components). Hence, by taking \(K\) into account \eqref{eqn:mult-fading-nlos} becomes
+Extending the previous NLOS case, if there is a line of sight (LOS) path (red signal in \figref{fig:ring-of-scattering-objects}), the statistical model has to be extended by defining the so called Rice factor \(K\). This \(K\) factor is the ratio between the power from the LOS path and the average power of the NLOS paths (often also referred to as distributed components) \cite{Hoher2013}. Hence, by taking \(K\) into account \eqref{eqn:mult-fading-nlos} becomes \cite{Hoher2013}:
 \begin{equation} \label{eqn:mult-fading-los}
 	f = \sqrt{\frac{K}{K+1}} + 
 	\frac{1}{\sqrt{K+1}} \left(
@@ -395,17 +395,8 @@ Extending the previous NLOS case, if there is a line of sight (LOS) path (red si
 		\frac{1}{\sqrt{N}}\sum_{n=1}^{N} e^{j \vartheta_k}
 	\right).
 \end{equation}
-Notice that by letting \(K = 0\), that is no power in the LOS path, \eqref{eqn:mult-fading-los} becomes \eqref{eqn:mult-fading-nlos} or Rayleigh distributed (as expected). Conversely when \(K \to \infty\), i.e. no power in the NLOS paths, then \(f \to 1\) and the fading disappears. The new amplitude density in this case is:
+Notice that by letting \(K = 0\), that is, no power in the LOS path, \eqref{eqn:mult-fading-los} becomes \eqref{eqn:mult-fading-nlos} or Rayleigh distributed (as expected). Conversely when \(K \to \infty\), i.e. no power in the NLOS paths, then \(f \to 1\) and the fading disappears. The new amplitude density in this case is:
 \begin{equation}
 	p(a)= 2a(1+K) \exp{\left(-K -a^2 (K+1) \right)} I_0 \left(2a\sqrt{K(1+K)} \right),
 \end{equation}
-where \(I_0\) the zeroth order modified Bessel function. Random variables with this probability density function are said to have a Rice or Rician distribution.
-
-% The Phase for the strait line component has no influences for the Random process therefore there set to zero. In the case when \(K = 0\).
-% the Rician distribution becomes a Rayleight distribution on the other hand when \(K\rightarrow \infty \) the distribution becomes an AWGN-channel model (additive white Gaussian noise). When \(K > 0 \) is the phase not equally distributed.
-% For this distribution model the expectation value for the real part is \(\E{\Re{f(t)}}=\sqrt{\frac{K}{K+1}} \) and for the imaginary part \(\E{\Im{f(t)}}=0\) 
-
-%\begin{equation}
-%	\Re{h_l(n)}, \Im{h_l(n)}
-%	\sim \mathcal{N} \left( \frac{A_l}{\sqrt{2}}, \frac{1}{2} \sigma_l^2 \right)
-%\end{equation}
+where \(I_0\) the zeroth order modified Bessel function. Random variables with this probability density function are said to have a \emph{Rice} or \emph{Rician} distribution \cite{Hoher2013}.
-- 
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