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+\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}}
+
+\begin{document}
+%\header{BlattNr}{Tutor}{Abgabedatum}{Vorlesungsname}{Namen}{Semester}{Anzahl Aufgaben}
+\header{4}{}{2015-11-11}{Kommunikationsnetze}{\textit{Jonas Jaszkowic, 3592719}\\\textit{Jan-Peter Hohloch, 3908712}}{WS 15/16}{4}
+\vspace{1cm}
+\Aufgabe{Scrambling}{5+5+10+10}
+\begin{enumerate}
+  \item To be perfectly in sync, the transmitter and receiver must have access to very accurate and identical time signals.
+  Due to technical difficulties this is often not the case, so that transmitter and receiver have slightly different time signals.
+  This results in incorrectly assigned bits, e.g. when the receiver reads a bit short time later than it is supposed to do.
+  \item Synchronization issues can be fixed by introducing synchronization bits in the signal. Usually the signal must
+  be chunked into frames, at the beginning of each frame a synchronization bit is added. Subsequent synchronization bits form
+  a specific binary pattern which repeats at a regular interval in the stream. This is used to identify the current
+  synchronization state (in sync vs. out of sync).
+  \item The bit sequence $011000000000100000000110$ is given.
+    \begin{itemize}
+      \item[a)] AMI (Alternate Mark Inversion, 0 is encoded as zero,
+      1 is alternately encoded as positive or negative voltage):
+      \begin{align*}
+          0+-000000000+00000000-+0
+      \end{align*}
+       B8ZS substitutes eight consecutive zeros with $000VB0VB$: ($+00000000$ is encoded as $+000+-0-+$, $-00000000$ is encoded as $-000-+0+-0$):
+       \begin{align*}
+         0+\mathbf{-000-+0+-0}\mathbf{+000+-0-+}
+       \end{align*}
+      \item[b)] HDB3 ($0$ is encoded as $0$, $1$ is alternately encoded as positive or negative voltage,
+      $0000$ as $000V$, where $V$ is a violation pulse, $0000 0000$ as $B00V B00V$, where $B$ is a balancing pulse):
+      \begin{align*}
+        0+-\mathbf{B00V B00V}0+\mathbf{B00V B00V}-+0
+      \end{align*}
+    \end{itemize}
+\end{enumerate}
+\Aufgabe{Pulse-Code Modulation}{15+15+5+5}
+\begin{enumerate}
+  \item We have $8$ available levels from $0$ to $7$. The sampling points lead to the level sequence $3,6,0,2,6,4,1,2,3$ and thus to
+  the word sequence $011,110,000,010,110,100,001,010,011$.
+  \item The word sequence $011,110,000,010,110,100,001,010,011$ modulated using the given constellation diagram and a sine as the carrier.
+  The amplitude modulations are $-1, 1, -2, -1, 1, 2, -2, 1$ and $-2$. The phase modulations are $10/8, 2/8, 7/8, 6/8, 2/8, 1/8, 9/8, 6/8$ and $10/8$ of $\pi$.
+  Two periods of each word are shown: \par
+  \begin{minipage}{\linewidth}
+      \centering
+      \includegraphics[width=\linewidth]{QAM8.png}
+  \end{minipage}
+  \item According to the Nyquist theorem, the sampling rate $f_s$ must be at least 2 times the highest frequency contained in the signal.
+  Thus, a sampling rate of $2 \cdot 4kHz = 8kHz$ is necessary for human voice.
+  \item The minimum bandwidth for transmission of PCM modulated signals is given as:
+  \begin{align*}
+    B_{PCM} = c \cdot 2 \cdot B_{analog} \cdot n_{b} \cdot \frac{1}{r}
+  \end{align*}
+  where $c$ is the average probability, that the next bit changes ($c = 0.5$), $r$ is the number of data per signal (here $r = 2$).
+  Therefore the required bandwidth is:
+  \begin{align*}
+      B = 0.5 \cdot 8000Hz \cdot 8bit \cdot \frac{1}{2} = 16kHz
+  \end{align*}
+\end{enumerate}
+\Aufgabe{Modulation}{10+10+10}
+\begin{enumerate}
+  \item Binary Amplitude Shift Keying (BASK): \par
+  \begin{minipage}{\linewidth}
+      \centering
+      \includegraphics[width=\linewidth]{BASK.png}
+  \end{minipage}
+  \item Binary Frequency Shift Keying (BFSK): \par
+  \begin{minipage}{\linewidth}
+      \centering
+      \includegraphics[width=\linewidth]{BFSK.png}
+  \end{minipage}
+  \item Binary Phase Shift Keying (BPSK): \par
+  \begin{minipage}{\linewidth}
+      \centering
+      \includegraphics[width=\linewidth]{BPSK.png}
+  \end{minipage}
+\end{enumerate}
+\end{document}