diff --git a/ex05/Ex-05.pdf b/ex05/Ex-05.pdf new file mode 100644 index 0000000..a9883cb --- /dev/null +++ b/ex05/Ex-05.pdf Binary files differ diff --git a/ex05/kn05.pdf b/ex05/kn05.pdf new file mode 100644 index 0000000..3e34477 --- /dev/null +++ b/ex05/kn05.pdf Binary files differ diff --git a/ex05/kn05.tex b/ex05/kn05.tex new file mode 100644 index 0000000..0a1d227 --- /dev/null +++ b/ex05/kn05.tex @@ -0,0 +1,148 @@ +\documentclass[a4paper,12pt]{scrartcl} +\usepackage[ngerman]{babel} +\usepackage{graphicx} %BIlder einbinden +\usepackage{amsmath} %erweiterte Mathe-Zeichen +\usepackage{amsfonts} %weitere fonts +\usepackage[utf8]{inputenc} %Umlaute & Co +\usepackage{hyperref} %Links +\usepackage{ifthen} %ifthenelse +\usepackage{enumerate} + +\usepackage{color} +\usepackage{algpseudocode} %Pseudocode +\usepackage{dsfont} % schöne Zahlenräumezeichen +\usepackage{amssymb, amsthm} %noch stärker erweiterte Mathe-Zeichen +\usepackage{tikz} %TikZ ist kein Zeichenprogramm +\usetikzlibrary{trees,automata,arrows,shapes} +\usepackage{pgfplots} + +\pagestyle{empty} + + +\topmargin-50pt + +\newcounter{aufgabe} +\def\tand{&} + +\newcommand{\makeTableLine}[2][0]{% +\setcounter{aufgabe}{1}% +\whiledo{\value{aufgabe} < #1}% +{% +#2\tand\stepcounter{aufgabe}% +} +} + +\newcommand{\aufgTable}[1]{ +\def\spalten{\numexpr #1 + 1 \relax} +\begin{tabular}{|*{\spalten}{p{1cm}|}} +\makeTableLine[\spalten]{A\theaufgabe}$\Sigma$~~\\ \hline +\rule{0pt}{15pt}\makeTableLine[\spalten]{}\\ +\end{tabular} +} + +\def\header#1#2#3#4#5#6#7{\pagestyle{empty} +\begin{minipage}[t]{0.47\textwidth} +\begin{flushleft} +{\bf #4}\\ +#5 +\end{flushleft} +\end{minipage} +\begin{minipage}[t]{0.5\textwidth} +\begin{flushright} +#6 \vspace{0.5cm}\\ +% Number of Columns Definition of Columns second empty line +% \begin{tabular}{|*{5}{C{1cm}|}}\hline A1&A2&A3&A4&$\Sigma$\\\hline&&&&\\\hline\end{tabular}\\\vspace*{0.1cm} +\aufgTable{#7} +\end{flushright} +\end{minipage} +\vspace{1cm} +\begin{center} +{\Large\bf Blatt #1} + +{(Abgabe #3)} +\end{center} +} + + + +%counts the exercisenumber +\newcounter{n} + +%Kommando für Aufgaben +%\Aufgabe{AufgTitel}{Punktezahl} +\newcommand{\Aufgabe}[2]{\stepcounter{n} +\textbf{Aufgabe \arabic{n}: #1} (#2 Punkte)} + +\newcommand{\textcorr}[1]{\textcolor{red}{#1}} +\newenvironment{corr}{\color{red}}{\color{black}\newline} +\newcommand{\ok}{\begin{corr} + $\checkmark$ + \end{corr}} + +\newcommand{\enot}[2]{#1 \cdot 10^{#2}} + +\begin{document} +%\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben} +\header{5}{}{2015-11-18}{Kommunikationsnetze}{\textit{Jonas Jaszkowic, 3592719}\\\textit{Jan-Peter Hohloch, 3908712}}{WS 15/16}{3} +\vspace{1cm} +\Aufgabe{Frequencies and Wavelenghts}{48}\\\\ +The formula for converting frequency $f$ to wavelength $\lambda$ given speed $c$ is: +\begin{align} + \lambda = \frac{c}{f} +\end{align} +Therefore, the resulting wavelengths and position in the electromagnetic spectrum are:\\ +\begin{center} + $ + \begin{array}{c|c|c|c} + \text{Frequency $f$} & \text{Frequency $f$ in Hz} & \text{Wavelength $\lambda$} & \text{Electromagnetic Spectrum} \\ \hline + 33 ~EHz & \enot{3.3}{19} ~Hz & 9.09 ~pm & \text{Gamma-rays} \\ + 4 ~EHz & \enot{4}{18} Hz & 75 ~pm & \text{X-rays, Gamma-rays} \\ + 348 ~THz & \enot{3.48}{14} ~Hz & 862 ~nm & \text{Infrared} \\ + 428 ~THz & \enot{4.28}{14} ~Hz & 701 ~nm & \text{Visible red} \\ + 517 ~THz & \enot{5.17}{14} ~Hz & 580 ~nm & \text{Visible yellow} \\ + 652 ~THz & \enot{6.52}{14} ~Hz & 460 ~nm & \text{Visible blue} \\ + 750 ~THz & \enot{7.5}{14} ~Hz & 400 ~nm & \text{Visible violet} \\ + 6 ~GHz & 600000000 ~Hz & 50 ~cm & \text{Ultra High Frequency UHF} \\ + 600 ~MHz & \enot{6.0}{9} ~Hz & 5 ~cm & \text{Super High Frequency SHF} \\ + 96 ~MHz & 96000000 ~Hz & 3.12 ~m & \text{Very High Frequency VHF} \\ + 420 ~MHz & \enot{6.42}{9} ~Hz & 4.67 ~cm & \text{Super High Frequency SHF} \\ + 900 ~MHz & 900000000 ~Hz & 33.33 ~cm & \text{Ultra High Frequency UHF} \\ + 1820 ~MHz & \enot{1.82}{9} ~Hz & 16.48 ~cm & \text{Super High Frequency SHF} \\ + 2140 ~MHz & \enot{2.14}{9} ~Hz & 14.02 ~m & \text{High Frequency HF} \\ + 6190 ~kHz & 6190000 ~Hz & 48.46 ~m & \text{Medium Frequency MF} \\ + 549 ~kHz & 549000 ~Hz & 546.45 ~m & \text{Low Frequency LF} + \end{array} + $ +\end{center} \vspace{1cm} +\Aufgabe{Frequency-Division Multiplexing (FDM)}{6+6+6+6} +The available analogue satellite channel with bandwidth $B_{STA} = 4MHz$ is divided into the required $5$ +channels, of which each has now $5MHz / 5 = 0.8MHz$ for transferring $N = 6Mbit/s$. Each digital channel is +modulated by QAM and sent over the analogue satellite link using the reserved bandwidth. At the receiver, the signal +is demultiplexed and again demodulated with QAM. This leads to 5 output channel with the original 6Mbit/s. +\begin{enumerate} + \item The required bandwidth for each channel is $B_{QAM} = (1 + d) \cdot S$, with $d = 0$ and $S = 0.8MHz$ this leads to + $B_{QAM} = 0.8MHz$. + \item The nyquist theorem states, that the maximum bit rate $R_{max}$ is given as $R_{max} = 2 \cdot \log{2}{L}$. + Since $\log{2}{L}$ is the number of bits per Baud, the maximum Baud per second $Bd_{max}$ is: + \begin{align*} + Bd_{max} = 2 \cdot B_{channel} = 2 \cdot 0.8 MHz = 1.6 MHz + \end{align*} + \item The required number of bits per symbol is $r = \frac{N}{S} = \frac{\enot{6}{6}bit/s}{\enot{0.8}{6}1/s} = 7.5 Bits$ + \item The required number of symbols for QAM is $L = 2^r = 2^8 = 256$. +\end{enumerate} \vspace{1cm} +\Aufgabe{Analog and Digital Hierarchy}{14+14} +\begin{enumerate} + \item Both, analog and digital hierarchy are used to carry many individual data streams over a single link. This link usually has a + large bandwidth. The individual channels are subsequently multiplexed into larger groups, until the final bandwidth is reached. The + receiver has to know how to demultiplex this giant data stream into the original channels. In analog hierarchy, a bunch of + \textbf{analog} signals are multiplexed into a single link using \textbf{Frequency Division Multiplexing}. In digital hierarchy, a bunch + of \textbf{digital} signals are multiplexed into a single link using \textbf{Time Division Multiplexing}. It is also to be noted, that + the digital signals can carry analog information (e.g. speech signals). + \item + \begin{enumerate} + \item STS-1 + \item OC-1 + \item DS-0 + \end{enumerate} +\end{enumerate} +\end{document}