\chapter{Stochastic and Deterministic Processes}

\begin{refsection}

\section{Stochastic Processes}

\begin{itemize}
	\item Stochastic processes $\rightarrow$ random signal
	\item No deterministic description
	\item Description of random parameters (probability, ...)
\end{itemize}

\subsection{Statistic Mean}

Given is family of curves $\vect{x}(t) = \left\{x_1(t), x_2(t), \dots, x_n(t)\right\}$. $\vect{x}(t)$ is called a \index{random vector} \textbf{random vector}.

\begin{figure}[H]
	\centering
	\begin{tikzpicture}
		\begin{axis}[
			height={0.25\textheight},
			width=0.6\linewidth,
			scale only axis,
			xlabel={$t$},
			ylabel={$x(t)$},
			%grid style={line width=.6pt, color=lightgray},
			%grid=both,
			grid=none,
			legend pos=north east,
			axis y line=middle,
			axis x line=middle,
			every axis x label/.style={
				at={(ticklabel* cs:1.05)},
				anchor=north,
			},
			every axis y label/.style={
				at={(ticklabel* cs:1.05)},
				anchor=east,
			},
			xmin=0,
			xmax=11,
			ymin=0,
			ymax=1.7,
			xtick={0, 1, ..., 10},
			ytick={0, 0.5, ..., 1.5},
			xticklabels={0, 1, $t_0$, 3, 4, ..., 10}
		]
			\addplot[black, dashed, smooth, domain=1:10, samples=200] plot (\x,{1.5*abs(sinc((1/(2*pi))*\x))});
			\pgfmathsetseed{100}
			\addplot[red, smooth, domain=1:10, samples=50] plot (\x,{1.5*abs(sinc((1/(2*pi))*\x)) + 0.1*rand});
			\addlegendentry{$x_1$};
			\pgfmathsetseed{200}
			\addplot[blue, smooth, domain=1:10, samples=50] plot (\x,{1.5*abs(sinc((1/(2*pi))*\x)) + 0.1*rand});
			\addlegendentry{$x_2$};
			\pgfmathsetseed{300}
			\addplot[green, smooth, domain=1:10, samples=50] plot (\x,{1.5*abs(sinc((1/(2*pi))*\x)) + 0.1*rand});
			\addlegendentry{$x_3$};
			\addplot[black, very thick, dashed] coordinates {(2,0) (2,2.2)};
		\end{axis}
	\end{tikzpicture}
	\caption{Family of random signals}
\end{figure}

\begin{itemize}
	\item The curves are produced by a random process $\vect{x}(t)$. The random process is time-dependent.
	\item All curves consist of random values, which are gathered around a mean value $\E\left\{\vect{x}(t)\right\}$.
	\item The random process can emit any value $x$. However, each value $x$ has a certain likelihood $p(x, t)$ of being produced. Again, this likelihood is time-dependent like the stochastic process.
\end{itemize}

Let's assume that the values are normally distributed. The \index{probability density function} \textbf{\ac{PDF}} $p(x, t)$ of a \index{normal distribution} \textbf{normal distribution} is:
\begin{equation}
	p(x, t) = \frac{1}{\sigma(t) \sqrt{2 \pi}} e^{-\frac{1}{2} \left(\frac{x - \mu(t)}{\sigma(t)}\right)^2}
\end{equation}
$p(x, t)$ is the likelihood that the stochastic process emits the value $x$ at time instance $t$. Both the mean of the normal distribution $\mu(t)$ and the standard deviation of the normal distribution $\sigma(t)$ are time-dependent.

\begin{attention}
	Do not confuse the mean of the normal distribution $\mu$ and the mean of a series of samples $\E\left\{\cdot\right\}$ (expectation value)!
\end{attention}

\begin{figure}[H]
	\centering
	\begin{tikzpicture}
		\begin{axis}[
			height={0.25\textheight},
			width=0.8\linewidth,
			scale only axis,
			xlabel={$x$},
			ylabel={$p(x, t_0)$},
			%grid style={line width=.6pt, color=lightgray},
			%grid=both,
			grid=none,
			legend pos=north east,
			axis y line=middle,
			axis x line=middle,
			every axis x label/.style={
				at={(ticklabel* cs:1.05)},
				anchor=north,
			},
			every axis y label/.style={
				at={(ticklabel* cs:1.05)},
				anchor=east,
			},
			xmin=-1.2,
			xmax=4.2,
			ymin=0,
			ymax=1.2,
			xtick={-1, 0, 1, 1.47, 2, 3, 4},
			ytick={0, 0.5, ..., 2.0},
			xticklabels={-1, 0, 1, $\E\left\{\vect{x}(t_0)\right\}$, 2, 3, 4}
		]
			% µ = 1.47, simga = 0.5
			\addplot[red, thick, smooth, domain=, samples=200] plot (\x, {(1/(0.5*sqrt(2*pi)))*exp(-0.5*((\x-1.47)/0.5)^2)});
			
			\addplot[black, very thick, dashed] coordinates {(1.47,0) (1.47,1)};
		\end{axis}
	\end{tikzpicture}
	\caption{Probability density function for an output value of a stochastic process at time $t_0$ with $\mu(t_0) = 1.47$ and $\sigma(t_0) = 0.5$}
\end{figure}

Given that
\begin{itemize}
	\item We know neither the mean of the normal distribution $\mu(t)$ nor the standard deviation of the normal distribution $\sigma(t)$.
	\item We only have $n$ samples of the curves $x_i(t_0)$ ($i \in \mathbb{N}, 0 \leq i \leq n$) at the time instance $t_0$.
	\item We do know that the random distribution of our samples $x_i(t_0)$ follows the \ac{PDF} $p(x, t_0)$.
\end{itemize}

\paragraph{How do we get the mean of out samples $\E\left\{X(t_0)\right\}$? (Finite case)}

The mean of the samples is the \index{expectation value} \textbf{expectation value} $\E\left\{\vect{x}(t_0)\right\}$. \nomenclature[Se]{$\E\left\{\cdot\right\}$}{Expectation value}

To get an approximation, we can calculate the \index{arithmetic mean} \textbf{arithmetic mean} of out $n$ samples:
\begin{equation}
	\E\left\{\vect{x}(t_0)\right\} \approx \frac{1}{n} \sum\limits_{i = 0}^{n} x_i(t_0)
	\label{eq:ch03:arith_mean}
\end{equation}
The approximation converges to the real $\E\left\{\vect{x}(t_0)\right\}$ for $n \rightarrow \infty$, because the random distribution of our samples $x_i(t_0)$ follows the \ac{PDF} $p(x, t_0)$.

\paragraph{What about an arbitrary \ac{PDF}? (Continuous case)}

\begin{itemize}
	\item We cannot collect an indefinite number of samples.
	\item However, if the \ac{PDF} is known, we can calculate the mean of our samples.
\end{itemize}

Extending, the arithmetic mean \eqref{eq:ch03:arith_mean}, with $n \rightarrow \infty$ and using all $x$ but weighted by their \ac{PDF} $p(x, t_0)$, we can determine the expectation value.
\begin{definition}{Stochastic mean}
	The \index{stochastic mean} \textbf{stochastic mean} of a \ac{PDF} is:
	\begin{equation}
		\E\left\{\vect{x}(t_0)\right\} = \int\limits_{-\infty}^{\infty} x \cdot p(x, t_0) \; \mathrm{d} x
	\end{equation}%
	\nomenclature[Se]{$\E\left\{\vect{x}\right\}$}{Stochastic mean}
\end{definition}

\begin{fact}
	In general, stochastic means are time-dependent.
\end{fact}

\paragraph{Other measures?}

The \index{quadratic stochastic mean} \textbf{quadratic stochastic mean}:
\begin{equation}
	\E\left\{\vect{x}^2(t_0)\right\} = \int\limits_{-\infty}^{\infty} x^2 \cdot p(x, t_0) \; \mathrm{d} x
\end{equation}

\subsection{Temporal Mean}

Given is a random time-domain signal $x_i(t)$ (where $i \in \mathbb{N}$ an arbitrary integer index):

\begin{figure}[H]
	\centering
	\begin{tikzpicture}
		\begin{axis}[
			height={0.25\textheight},
			width=0.6\linewidth,
			scale only axis,
			xlabel={$t$},
			ylabel={$x_i(t)$},
			%grid style={line width=.6pt, color=lightgray},
			%grid=both,
			grid=none,
			legend pos=north east,
			axis y line=middle,
			axis x line=middle,
			every axis x label/.style={
				at={(ticklabel* cs:1.05)},
				anchor=north,
			},
			every axis y label/.style={
				at={(ticklabel* cs:1.05)},
				anchor=east,
			},
			xmin=-5.5,
			xmax=5.5,
			ymin=0,
			ymax=3.2,
			xtick={-5, -4, ..., 5},
			ytick={0, 1, ..., 3},
			xticklabels={-5, -4, -3, -2, $-\frac{T}{2}$, 0, $\frac{T}{2}$, 2, 3, 4, 5}
		]
			\pgfmathsetseed{100}
			\addplot[red, smooth, domain=-5:5, samples=50] plot (\x,{1.5 + 0.8*rand});
			\addplot[black, thick, dashed] coordinates {(-1,0) (-1,3.2)};
			\addplot[black, thick, dashed] coordinates {(1,0) (1,3.2)};
			\addplot[black, dashed] coordinates {(-5,1.5) (5,1.5)};
		\end{axis}
	\end{tikzpicture}
	\caption{Random time-domain signal}
\end{figure}

\textit{Remark:} The signal can be a sample of a family of signals, but it is not required to be.

The temporal mean is calculated as the arithmetic mean with following differences to \eqref{eq:ch03:arith_mean}:
\begin{itemize}
	\item The mean is calculation over the time, not over a number of samples.
	\item For a time-continuous signal, the sum extends to an integral.
	\item The arithmetic mean is calculated over the time interval $[-\frac{T}{2}, \frac{T}{2}]$. Let's make the interval indefinite.
\end{itemize}

\begin{definition}{Temporal mean}
	The \index{temporal mean} \textbf{temporal mean} of time-domain signal $x_i(t)$ is:
	\begin{equation}
		\overline{x_i} = \E\left\{x_i(t)\right\} = \lim\limits_{T \rightarrow \infty} \frac{1}{T} \int\limits_{-\frac{T}{2}}^{\frac{T}{2}} x_i{t} \; \mathrm{d} t
	\end{equation}%
	\nomenclature[Sx]{$\overline{x}$, $\E\left\{x_i(t)\right\}$}{Temporal mean}
\end{definition}

The temporal mean is not time-dependent.

\begin{fact}
	In general, temporal means are sample-dependent.
\end{fact}

Actually $x_i(t)$ would not need the index $i$ if there is only one sample. Nevertheless, it was kept here, to emphasize the dependency on the sample, in contrast to the dependency on the time of the stochastic mean.

\paragraph{Other measures?}

The \index{quadratic temporal mean} \textbf{quadratic temporal mean}:
\begin{equation}
	\overline{x^2_i} = \E\left\{x^2_i(t)\right\} = \lim\limits_{T \rightarrow \infty} \frac{1}{T} \int\limits_{-\frac{T}{2}}^{\frac{T}{2}} |x_i{t}|^2 \; \mathrm{d} t
\end{equation}

\subsection{Ergodic Processes}

\begin{definition}{Ergodic process}
	\index{ergodic process} A process is \textbf{ergodic} if:
	\begin{enumerate}
		\item The stochastic means are equal at all times.
		\begin{equation}
			\E\left\{\vect{x}(t_0)\right\} = \E\left\{\vect{x}(t_1)\right\} = \dots = \E\left\{\vect{x}\right\}
		\end{equation}
		\item The temporal means of all samples are equal.
		\begin{equation}
			\overline{x_1} = \overline{x_2} = \dots = \overline{x}
		\end{equation}
		\item The stochastic mean equals the temporal mean.
		\begin{equation}
			\E\left\{\vect{x}\right\} = \overline{x} = \mu_x
		\end{equation}
	\end{enumerate}
\end{definition}

As a consequence:
\begin{itemize}
	\item One single, sufficiently long, random sample of the process is enough to deduct the statistical properties of an ergodic process.
	\item The ergodic process is in steady state (\ac{WSS}), i.e., it does not erratically change its behaviour and properties.
\end{itemize}

\begin{figure}[H]
	\centering
		\begin{tikzpicture}
			\begin{axis}[
			height={0.25\textheight},
			width=0.6\linewidth,
			scale only axis,
			xlabel={$t$},
			ylabel={$x_i(t)$},
			%grid style={line width=.6pt, color=lightgray},
			%grid=both,
			grid=none,
			legend pos=north east,
			axis y line=middle,
			axis x line=middle,
			every axis x label/.style={
				at={(ticklabel* cs:1.05)},
				anchor=north,
			},
			every axis y label/.style={
				at={(ticklabel* cs:1.05)},
				anchor=east,
			},
			xmin=-5.5,
			xmax=5.5,
			ymin=0,
			ymax=3.2,
			xtick={-5, -4, ..., 5},
			ytick={0, 1, ..., 3},
			xticklabels={-5, -4, -3, -2, $-\frac{T}{2}$, 0, $\frac{T}{2}$, 2, 3, 4, 5}
			]
			\pgfmathsetseed{100}
			\addplot[red, smooth, domain=-5:5, samples=50] plot (\x,{1.5 + 0.8*rand});
			\addlegendentry{$x_1$};
			\pgfmathsetseed{200}
			\addplot[blue, smooth, domain=-5:5, samples=50] plot (\x,{1.5 + 0.8*rand});
			\addlegendentry{$x_2$};
			\pgfmathsetseed{300}
			\addplot[green, smooth, domain=-5:5, samples=50] plot (\x,{1.5 + 0.8*rand});
			\addlegendentry{$x_3$};
			\addplot[black, dashed] coordinates {(-5,1.5) (5,1.5)};
			\addlegendentry{$\mu_x$};
		\end{axis}
		\end{tikzpicture}
	\caption{Three samples of the same ergodic process}
\end{figure}

\subsection{Cross-Correlation}

\begin{itemize}
	\item Imagine you have two random processes.
	\item They produce the (complex) random vectors $\cmplxvect{x}(t)$ and $\cmplxvect{y}(t)$.
	\item The random processes can be somehow related (correlated) to each other. But they can also be independent instead.
	\item How can we find this out?
\end{itemize}

We need a similarity measure. The cross-correlation is such a measure.

\begin{definition}{Cross-correlation of stochastic processes}
	The \index{cross-correlation!stochastic process} \text{cross-correlation of two stochastic processes} $\cmplxvect{x}(t_1)$ and $\cmplxvect{y}(t_2)$ between the times $t_1$ and $t_2$ is:
	\begin{equation}
		\mathrm{R}_{XY}(t_1, t_2) = \E\left\{ \cmplxvect{x}(t_1) \cmplxvect{y}^{*}(t_2) \right\}
	\end{equation}%
	\nomenclature[Sr]{$\mathrm{R}_{XY}$}{Cross-correlation of two random vectors}%
	\nomenclature[Na]{$\left(\cdot\right)^{*}$}{Complex conjugate of $\left(\cdot\right)$}
	where $\left(\cdot\right)^{*}$ denotes the complex conjugate.
\end{definition}

The expectation value can be expressed as:
\begin{equation}
	\mathrm{R}_{XY}(t_1, t_2) = \E\left\{ \cmplxvect{x}(t_1) \cmplxvect{y}*(t_2) \right\} = \int\limits_{y = -\infty}^{\infty} \int\limits_{x = -\infty}^{\infty} x y \cdot p(x, y, t_1, t_2) \; \mathrm{d} x \mathrm{d} y
\end{equation}
$p(x, y, t_1, t_2)$ is the joint \ac{PDF} of the two random processes. It defines the likelihood that $x$ is produced at time $t_1$ \textbf{and} $y$ is produced at time $t_2$.

Let's derive a special case for \textbf{ergodic} processes:
\begin{itemize}
	\item The time difference is $\tau = t_2 - t_1$.
	\item Because of the ergodicity of the two processes, only one sample of each $x_i(t)$ and $y_i(t)$ needs to be taken.
	\item An estimation for the cross-correlation is averaging the products of the time-shifted samples $x_i(t) \cdot y_i(t+\tau)$. This resembles
\end{itemize}
Extending this to complex number yields:
\begin{equation}
	\mathrm{R}_{XY}(\tau) = \E\left\{ \cmplxvect{x}(t) \cmplxvect{y}*(t+\tau) \right\} \approx \lim\limits_{T \rightarrow \infty} \frac{1}{T} \int\limits_{t = -\frac{T}{2}}^{\frac{T}{2}} \underline{x}_i^{*}(t) \cdot \underline{x}_i(t+\tau) \; \mathrm{d} t
\end{equation}

This resembles the cross-correlation of deterministic signals
\begin{definition}{Cross-correlation of deterministic signals}
	The \index{cross-correlation!deterministic signals} \text{cross-correlation of two deterministic signals} $\underline{f}(t_1)$ and $\underline{g}(t_2)$ between the times $\tau = t_2 - t_1$ is:
	\begin{equation}
		\left(f \star g\right)(\tau) = \int\limits_{t = -\infty}^{\infty} \underline{f}^{*}(t) \cdot \underline{g}(t+\tau) \; \mathrm{d} t
	\end{equation}%
	\nomenclature[N]{$\left(f \ast g\right)(\tau)$}{Cross-correlation of two signals}
\end{definition}

\begin{attention}
	You must not confuse the operators for the convolution $*$ and correlation $\star$.
\end{attention}

For the random signals $x(t)$ and $y(t)$, the cross-correlation can not be determined analytically, but numerically.
\begin{equation}
	\mathrm{R}_{XY}(\tau) \approx \left(x \star y\right)(\tau)
\end{equation}

\paragraph{What's the use?}

\begin{itemize}
	\item The cross-correlation ``scans'' the two signals for common features.
	\item The cross-correlation $\mathrm{R}_{XY}(\tau)$ will show a peak at the time shift $\tau$, if
	\begin{itemize}
		\item The signals are correlated, i.e., have a common feature.
		\item The common feature is time-shifted by $\tau$.
	\end{itemize}
	\item A flat near $0$ cross-correlation means that the signals are uncorrelated.
\end{itemize}

\section{Spectral Density}

\subsection{Autocorrelation}

\subsection{Power Spectral Density}

\subsection{Decibel}

\section{Noise}

\subsection{Types of Noise}

\subsection{Thermal Noise}

\subsection{White Noise}

\subsection{Noise Floor and Noise Figure}

\phantomsection
\addcontentsline{toc}{section}{References}
\printbibliography[heading=subbibliography]
\end{refsection}

