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authorPhilipp Le <philipp-le-prviat@freenet.de>2020-05-06 01:03:17 +0200
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WIP: Chapter 2
-rw-r--r--DCS.bib12
-rw-r--r--chapter02/Rainbow.jpgbin0 -> 5197 bytes
-rw-r--r--chapter02/content_ch02.tex269
-rw-r--r--common/settings.tex27
4 files changed, 274 insertions, 34 deletions
diff --git a/DCS.bib b/DCS.bib
index 34dbde5..b47449e 100644
--- a/DCS.bib
+++ b/DCS.bib
@@ -1,6 +1,18 @@
% This file was created with JabRef 2.7.1.
% Encoding: UTF8
+@MISC{Arz2007,
+ author = {"Arz"},
+ title = {Rainbow above Kaviskis Lake, Lithuania},
+ howpublished = {Wikimedia},
+ year = {2007},
+ note = {License: \href{https://creativecommons.org/licenses/by-sa/3.0/deed.en}{CC-BY-SA
+ 3.0}},
+ owner = {ple},
+ timestamp = {2020.05.05},
+ url = {https://en.wikipedia.org/wiki/File:Rainbow_above_Kaviskis_Lake,_Lithuania.jpg}
+}
+
@MISC{Maksim2011,
author = {"Maksim" and "Malyszkz"},
title = {Netowrk topologies},
diff --git a/chapter02/Rainbow.jpg b/chapter02/Rainbow.jpg
new file mode 100644
index 0000000..7c84756
--- /dev/null
+++ b/chapter02/Rainbow.jpg
Binary files differ
diff --git a/chapter02/content_ch02.tex b/chapter02/content_ch02.tex
index 071400c..9dc7df2 100644
--- a/chapter02/content_ch02.tex
+++ b/chapter02/content_ch02.tex
@@ -389,7 +389,7 @@ If the signal $\underline{x_p}(t) = x_p(t)$ is real-valued, i.e., $\Im\left\{\un
\begin{itemize}
\item The amplitude spectrum $|\underline{c}_n|$ is an \underline{even function}. It is symmetric with respect to the $y$-axis.
\item The phase spectrum $\arg\left(\underline{c}_n\right)$ is an \underline{odd function}. It is symmetric with respect to the origin.
- \item As a consequence, the phase of $\arg\left(\underline{c}_0\right)$ at $n = 0$ must be either $0$ or $+\pi$. Note that, $+\pi$ is identical to $-\pi$ in the complex plane. Thus, $-\pi$ is valid, too, but not distinct from $+\pi$. This phase is the sign of the \ac{DC} bias: $\arg\left(\underline{c}_0\right) = 0$ means positive \ac{DC} bias and $\arg\left(\underline{c}_0\right) = \pi$ means negative \ac{DC} bias.
+ \item As a consequence, the phase of $\arg\left(\underline{c}_0\right)$ at $n = 0$ must be either $0$ or $\pm \pi$. Note that, $+\pi$ is identical to $-\pi$ in the complex plane. The phase is the sign of the \ac{DC} bias: $\arg\left(\underline{c}_0\right) = 0$ means positive \ac{DC} bias and $\arg\left(\underline{c}_0\right) = \pi$ means negative \ac{DC} bias.
\end{itemize}
\end{itemize}
These symmetry rules apply for \underline{all} real-valued signals $\underline{x_p}(t) = x_p(t) \in \mathbb{R}$. The symmetry rules ensure that the mono-chromatic components of the Fourier series \eqref{eq:ch02:fourier_series_cmplx} sum up to a real value at each time instance $t \in \mathbb{R}$.
@@ -416,15 +416,15 @@ The symmetry rules do \underline{not} apply for complex-valued signals $\underli
ymax=3,
xtick={-3, -2, ..., 3},
ytick={0, 0.5, ..., 2.5},
- axis y line=middle,
+ 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,
+ every axis x label/.style={
+ at={(ticklabel* cs:1.05)},
+ anchor=north,
+ },
+ every axis y label/.style={
+ at={(ticklabel* cs:1.05)},
+ anchor=east,
}
]
\addplot[red, thick] coordinates {(-3, 0) (-3, 2.0)};
@@ -460,17 +460,18 @@ The symmetry rules do \underline{not} apply for complex-valued signals $\underli
ymin=-4,
ymax=4,
xtick={-3, -2, ..., 3},
- ytick={-3.14159, -1.5708, 1.5708, 3.14159},
- yticklabels={$-\pi\hspace{0.30cm}$, $-\frac{\pi}{2}$, $\frac{\pi}{2}$, $\pi\hspace{0.10cm}$},
- axis y line=middle,
+ ytick={-3.14159, -1.5708, 1.5708, 3.14159},
+ yticklabels={$-\pi\hspace{0.30cm}$, $-\frac{\pi}{2}$,
+$\frac{\pi}{2}$, $\pi\hspace{0.10cm}$},
+ 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,
+ every axis x label/.style={
+ at={(ticklabel* cs:1.05)},
+ anchor=north,
+ },
+ every axis y label/.style={
+ at={(ticklabel* cs:1.05)},
+ anchor=east,
}
]
\addplot[red, thick] coordinates {(-3, 0) (-3, 3.14159)};
@@ -488,9 +489,61 @@ The symmetry rules do \underline{not} apply for complex-valued signals $\underli
\label{fig:ch02:FSeries_Phase_Spectrum}
\end{figure}
-\section{Non-Periodic Signals and Fourier Transform}
+\begin{excursus}{Spectra in the nature}
+ The spectrum is no abstract, mathematical theory. You can see spectra with your eye:
+ \begin{figure}[H]
+ \centering
+ \includegraphics[scale=1]{../chapter02/Rainbow.jpg}
+ \caption[A rainbow showing the spectrum of the sunlight]{A rainbow showing the spectrum of the sunlight: The white sunlight is composed of mono-chromatic, electromagnetic waves of all frequencies which are optically visible for humans. When light passes through a dispersive medium (glass prism, raindrop, etc.), it is refracted. Each mono-chromatic component has a different refraction index. The light components are separated by its frequency and become individually visible. An example, is a rainbow as depicted above. \licensequote{\cite{Arz2007}}{``Arz''}{\href{https://creativecommons.org/licenses/by-sa/3.0/deed.en}{CC-BY-SA 3.0}}}
+ \end{figure}
+ The rainbow is a natural example of an visible spectrum of the sunlight.
+\end{excursus}
-\subsection{Derivation of The Fourier Transform}
+\section{Non-Periodic Signals and The Continuous Fourier Transform}
+
+\begin{figure}[H]
+ \centering
+ \begin{tikzpicture}
+ \begin{axis}[
+ height={0.25\textheight},
+ width=0.6\linewidth,
+ scale only axis,
+ xlabel={$t$},
+ ylabel={$x_{np}(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=-3,
+ xmax=3,
+ ymin=0,
+ ymax=2,
+ xtick={-2, -0.5, ..., 2},
+ ytick={0, 0.5, ..., 1.5}
+ ]
+ \addplot[blue, thick] coordinates {(-2, 0) (-0.5, 0)};
+ \addplot[blue, dashed] coordinates {(-0.5, 0) (-0.5, 1)};
+ \addplot[blue, thick] coordinates {(-0.5, 1) (0.5, 1)};
+ \addplot[blue, dashed] coordinates {(0.5, 1) (0.5, 0)};
+ \addplot[blue, thick] coordinates {(0.5, 0) (2, 0)};
+ \end{axis}
+ \end{tikzpicture}
+ \caption{The rectangular function $\mathrm{rect}$ as an example for a non-period signal}
+ \label{fig:ch02:rect_function}
+\end{figure}
+\index{rectangular function}
+
+\subsection{Derivation of The Continuous Fourier Transform}
Non-periodic signals have no repeating pattern. Consequently, there is no period $T_0$. Mathematically, the period is indefinite $T_0 \rightarrow \infty$.
@@ -529,15 +582,15 @@ The outer sum is a Rieman sum. $\frac{1}{T_0}$ is substituted by $\frac{\Delta \
\underline{x_{np}}(t) = \underbrace{\frac{1}{2 \pi} \int\limits_{\omega = -\infty}^{\infty} \underbrace{\left( \int\limits_{t' = -\infty}^{\infty} \underline{x_{np}}(t') \cdot e^{-j \omega t'} \, \mathrm{d} t' \right)}_{\text{Fourier transform}} \cdot e^{j \omega t} \, \mathrm{d} \omega}_{\text{Inverse Fourier transform}}
\end{equation}
-The inner integral is the \index{Fourier transform} \textbf{Fourier transform}.
+The inner integral is the \textbf{continuous Fourier transform}, also called only \index{Fourier transform} \emph{Fourier transform}.
\begin{definition}{Fourier Transform}
- The \index{Fourier transform} \textbf{Fourier transform} of the function $\underline{x}(t)$ is:
+ The \index{continuous Fourier transform} \textbf{continuous Fourier transform} of the function $\underline{x}(t)$ is:
\begin{equation}
\underline{X}(j \omega) = \mathcal{F} \left\{\underline{x}(t)\right\} = \int\limits_{t = -\infty}^{\infty} \underline{x}(t) \cdot e^{-j \omega t} \, \mathrm{d} t
\end{equation}
- The \index{inverse Fourier transform} \textbf{inverse Fourier transform} is:
+ The \index{inverse Fourier transform} \index{inverse continuous Fourier transform} \textbf{inverse (continuous) Fourier transform} is:
\begin{equation}
\underline{x}(t) = \mathcal{F}^{-1} \left\{\underline{X}(j \omega)\right\} = \frac{1}{2 \pi} \int\limits_{\omega = -\infty}^{\infty} \underline{X}(j \omega) \cdot e^{+j \omega t} \, \mathrm{d} \omega
\end{equation}
@@ -545,7 +598,7 @@ The inner integral is the \index{Fourier transform} \textbf{Fourier transform}.
\subsection{Amplitude and Phase Spectra}
-The value-continuous complex frequency variable $j \omega$ in the Fourier transforms replaced the value-discrete index $n$ of the Fourier series. Due to their similarity, the constraints for all signals and the \index{spectrum!symmetry rules} \textbf{symmetry rules} for real-valued signals apply analogously.
+The value-continuous complex frequency variable $j \omega$ in the continuous Fourier transforms replaced the value-discrete index $n$ of the Fourier series. Due to their similarity, the constraints for all signals and the \index{spectrum!symmetry rules} \textbf{symmetry rules} for real-valued signals apply analogously.
\begin{itemize}
\item The Fourier transform $\underline{X}(j \omega) \in \mathbb{C}$ is always complex-valued, for both real-valued $\underline{x}(t) = x(t) \in \mathbb{R}$ and complex-valued $\underline{x}(t) \in \mathbb{C}$ signals.
@@ -555,13 +608,175 @@ The value-continuous complex frequency variable $j \omega$ in the Fourier transf
\begin{itemize}
\item The amplitude spectrum $|\underline{X}(j \omega)|$ is an \underline{even function}. It is symmetric with respect to the $y$-axis.
\item The phase spectrum $\arg\left(\underline{X}(j \omega)\right)$ is an \underline{odd function}. It is symmetric with respect to the origin.
- \item As a consequence, the phase of $\arg\left(\underline{X}(0)\right)$ at $j \omega = 0$ must be either $0$ or $+\pi$. Note that, $+\pi$ is identical to $-\pi$ in the complex plane. Thus, $-\pi$ is valid, too, but not distinct from $+\pi$. This phase is the sign of the \ac{DC} bias: $\arg\left(\underline{X}(0)\right) = 0$ means positive \ac{DC} bias and $\arg\left(\underline{X}(0)\right) = \pi$ means negative \ac{DC} bias.
+ \item As a consequence, the phase of $\arg\left(\underline{X}(0)\right)$ at $j \omega = 0$ must be either $0$ or $\pm \pi$. Note that, $+\pi$ is identical to $-\pi$ in the complex plane. The phase is the sign of the \ac{DC} bias: $\arg\left(\underline{X}(0)\right) = 0$ means positive \ac{DC} bias and $\arg\left(\underline{X}(0)\right) = \pi$ means negative \ac{DC} bias.
\end{itemize}
\end{itemize}
+Let's investigate the \index{rectangular function} rectangular function from Figure \ref{fig:ch02:rect_function}. It is defined as:
+\begin{equation}
+ \mathrm{rect}(t) = \begin{cases}
+ 0 & \qquad \text{if } \; |t| > \frac{1}{2}, \\
+ 1 & \qquad \text{if } \; |t| < \frac{1}{2}
+ \end{cases}
+\end{equation}
+The function is undefined for $t = \pm \frac{1}{2}$. The function is now transformed, i.e., $\underline{x}(t) = \mathrm{rect}(t)$.
+
+\begin{equation}
+ \underline{X}\left(j \omega\right) = \int\limits_{t = -\infty}^{\infty} \mathrm{rect}(t) \cdot e^{-j \omega t} \, \mathrm{d} t = \mathrm{sinc}\left(\frac{\omega}{2 \pi}\right)
+\end{equation}
+where $\mathrm{sinc}(t)$ is the \emph{normalized} sinc function.
+
+\begin{attention}
+ Mathematics and engineering use a slightly different definition of the sinc function.
+
+ In mathematics, it is \index{sinc function!unnormalized} \textbf{\textit{unnormalized} sinc function}:
+ \begin{equation*}
+ \mathrm{sinc}(t) = \frac{\sin\left(t\right)}{t}
+ \end{equation*}
+
+ In the context of signal processing and information theory, it is the \index{sinc function!normalized} \textbf{\textit{normalized} sinc function}:
+ \begin{equation*}
+ \mathrm{sinc}(t) = \frac{\sin\left(\pi t\right)}{\pi t}
+ \end{equation*}
+
+ In either case, the value at $t = 0$ is defined to:
+ \begin{equation*}
+ \mathrm{sinc}(t = 0) = \lim\limits_{t \rightarrow 0} \frac{\sin\left(t\right)}{t} = 1
+ \end{equation*}
+\end{attention}
+
+The resulting spectra of $\underline{X}\left(j \omega\right)$ can now be drawn. The rectangular function is special. The imaginary part $\Im\left\{\underline{X}\left(j \omega\right)\right\} = 0$ is zero. Thus, the phase can only be $0$ or $\pm \pi$. However, this is a special property of the sinc function, but not of every function.
+
+\begin{figure}[H]
+ \centering
+ \begin{tikzpicture}
+ \begin{axis}[
+ height={0.25\textheight},
+ width=0.6\linewidth,
+ scale only axis,
+ xlabel={$\omega$},
+ ylabel={$|\underline{X}\left(j \omega\right)|$},
+ %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=-52,
+ xmax=52,
+ ymin=0,
+ ymax=1.2,
+ xtick={-50, -40, ..., 50},
+ ytick={0, 0.25, ..., 1.0}
+ ]
+ \addplot[red, thick, smooth, domain=-50:50, samples=200] plot (\x,{abs(sinc((1/(2*pi))*\x))});
+ \end{axis}
+ \end{tikzpicture}
+ \caption{Amplitude spectrum of the rectangular function}
+ \label{fig:ch02:rect_function_ampl_spectrum}
+\end{figure}
+
+\begin{figure}[H]
+ \centering
+ \begin{tikzpicture}
+ \begin{axis}[
+ height={0.25\textheight},
+ width=0.6\linewidth,
+ scale only axis,
+ xlabel={$\omega$},
+ ylabel={$\arg\left(\underline{X}(j \omega)\right)$},
+ %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=-52,
+ xmax=52,
+ ymin=-4,
+ ymax=4,
+ xtick={-50, -40, ..., 50},
+ ytick={-3.14159, -1.5708, 1.5708, 3.14159},
+ yticklabels={$-\pi\hspace{0.30cm}$, $-\frac{\pi}{2}$,
+ $\frac{\pi}{2}$, $\pi\hspace{0.10cm}$},
+ ]
+ \addplot[red, thick] coordinates {(-50, 0) (-39.48, 0)};
+ \addplot[red, dashed] coordinates {(-39.48, 0)(-39.48, -3.14159)};
+ \addplot[red, thick] coordinates {(-39.48, -3.14159) (-19.74, -3.14159)};
+ \addplot[red, dashed] coordinates {(-19.74, -3.14159) (-19.74, 0)};
+ \addplot[red, thick] coordinates {(-19.74, 0) (19.74, 0)};
+ \addplot[red, dashed] coordinates {(19.74, 0) (19.74, 3.14159)};
+ \addplot[red, thick] coordinates {(19.74, 3.14159) (39.48, 3.14159)};
+ \addplot[red, dashed] coordinates {(39.48, 3.14159)(39.48, 0)};
+ \addplot[red, thick] coordinates {(39.48, 0)(50, 0)};
+ \end{axis}
+ \end{tikzpicture}
+ \caption[Phase spectrum of the rectangular function]{Phase spectrum of the rectangular function. Please note that $- \pi$ is equivalent to $+ \pi$.}
+ \label{fig:ch02:rect_function_phase_spectrum}
+\end{figure}
+
\subsection{Time Domain and Frequency Domain}
-\section{Properties of The Fourier Transform}
+You have learnt two representations of a signal, so far.
+\begin{itemize}
+ \item \index{time domain} \textbf{Time domain} -- A signal is a function $\underline{x}(t)$ of the time.
+ \item \index{frequency domain} \textbf{Frequency domain} -- A signal is a function $\underline{X}(j \omega)$ of the frequency.
+\end{itemize}
+Both $\underline{x}(t)$ and $\underline{X}(j \omega)$ refer to the same signal.
+
+The frequency domain is obtained from the time domain by a transform. For time-continuous signals, these transforms one of:
+\begin{itemize}
+ \item Fourier series
+ \item continuous Fourier transform
+\end{itemize}
+The time domain is obtained by the respective inverse transform.
+
+\begin{definition}{Transform operator}
+ The operation of a transform between time and frequency domain is written as:
+ \begin{equation}
+ \underline{x}(t) \TransformHoriz \underline{X}(j \omega)
+ \end{equation}
+ for the transform from time to frequency domain, and vice versa:
+ \begin{equation}
+ \underline{X}(j \omega) \InversTransformHoriz \underline{x}(t)
+ \end{equation}
+\end{definition}
+
+\textbf{But what is the purpose of the transforms?}
+
+\begin{figure}[H]
+ \centering
+ \begin{tikzpicture}
+ \node[align=center, minimum width=2.5cm, minimum height=1.5cm] (ProbTD) {\textbf{Problem}\\ in time domain};
+ \node[align=center, minimum width=2.5cm, minimum height=1.5cm, right=5cm of ProbTD] (ProbFD) {\textbf{Problem}\\ in frequency domain};
+ \node[align=center, minimum width=2.5cm, minimum height=1.5cm, below=3cm of ProbTD] (SolTD) {\textbf{Solution}\\ in time domain};
+ \node[align=center, minimum width=2.5cm, minimum height=1.5cm, below=3cm of ProbFD] (SolFD) {\textbf{Solution}\\ in frequency domain};
+
+ \draw[-latex, thick] (ProbTD.south) -- node[midway, left, align=right]{Hard to solve} (SolTD.north);
+ \draw[-latex, thick] (ProbTD.east) -- node[midway, above, align=center]{Transform} (ProbFD.west);
+ \draw[-latex, thick] (ProbFD.south) -- node[midway, right, align=left]{Easy to solve} (SolFD.north);
+ \draw[-latex, thick] (SolFD.west) -- node[midway, above, align=center]{Inverse Transform} (SolTD.east);
+ \end{tikzpicture}
+ \caption{Explanation of the purpose of transforms}
+\end{figure}
+
+\section{Properties of The Continuous Fourier Transform}
\subsection{Energy Signals and Power Signals}
diff --git a/common/settings.tex b/common/settings.tex
index 94f97d6..7ba155b 100644
--- a/common/settings.tex
+++ b/common/settings.tex
@@ -59,6 +59,7 @@
\usetikzlibrary{shapes.arrows}
\usetikzlibrary{shapes,arrows}
\usetikzlibrary{decorations.pathreplacing}
+\usetikzlibrary{math}
\usepackage{pgf-umlsd}
% Circuits
@@ -88,6 +89,17 @@
%\renewcommand{\floatpagefraction}{0.66}
%\renewcommand{\textfraction}{0.10}
+% Custom functions
+\tikzmath{
+ function sinc(\x) {
+ if abs(\x) < .001 then { % (|x| < .001) ~ (x = 0)
+ return 1;
+ } else {
+ return sin(\x r)/\x;
+ };
+ };
+}
+
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Symbols
@@ -97,6 +109,7 @@
\usepackage{amsmath}
\usepackage{amssymb}
\usepackage{bm}
+\usepackage{trsym}
% Quotes
\usepackage{csquotes}
@@ -216,14 +229,14 @@
% Exercises
\usepackage{exsheets}
-\SetupExSheets{
- headings = block-subtitle,
- headings-format = \bfseries\sffamily,
- subtitle-format = \bfseries\sffamily,
- counter-within = chapter,
- counter-format = ch.qu\IfQuestionSubtitleT{:},
+\SetupExSheets{
+ headings = block-subtitle,
+ headings-format = \bfseries\sffamily,
+ subtitle-format = \bfseries\sffamily,
+ counter-within = chapter,
+ counter-format = ch.qu\IfQuestionSubtitleT{:},
toc-level = {subsection},
- % solution/print = true % uncomment for tutors
+ % solution/print = true % uncomment for tutors
}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%