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+%\documentclass[a4paper]{scrartcl}
+%\usepackage[top=2cm, bottom=2.5cm, left=3cm, right=2cm, footskip=1cm]{geometry}
+%\usepackage[german]{babel}
+%\usepackage[T1]{fontenc}
+%\usepackage[latin1]{inputenc}
+%\usepackage{endnotes}
+%\usepackage{trfsigns}
+%\usepackage{setspace}
+%
+%\setlength{\parindent}{0ex}
+%\def\isac{${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}
+%\def\sisac{{\footnotesize${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}}
+%
+%\begin{document}
+%\title{Interactive Course Material for Signal Processing based on Isabelle/\isac}
+%\subtitle{Problemsolutions (Calculations)}
+%\author{Walther Neuper, Jan Rocnik}
+%\maketitle
+
+
+%------------------------------------------------------------------------------
+%FOURIER
+
+\subsection{Fourier Transformation}
+\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally fragment of the exam to \emph{Signaltransformationen VO} from 04.03.2011. Translated from German.}}
+\textbf{(a)} Determine the fourier transform for the given rectangular impulse:
+
+\begin{center}
+$x(t)= \left\{
+     \begin{array}{lr}
+       1 & -1\leq t\geq1\\
+       0 & else
+     \end{array}
+   \right.$
+\end{center}
+
+\textbf{\noindent (b)} Now consider the given delayed impulse, determine its fourie transformation and calculate phase and magnitude:
+
+\begin{center}
+$x(t)= \left\{
+     \begin{array}{lr}
+       1 & -1\leq t\leq1\\
+       0 & else
+     \end{array}
+   \right.$
+\end{center}
+
+\subsubsection{Solution}
+\textbf{(a)} \textsf{Subproblem 1}
+\onehalfspace{
+\begin{tabbing}
+000\=\kill
+\texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\
+\`Insert Condition: $x(t) = 1\;$ for $\;\{-1\leq t\;\land\;t\leq 1\}\;$ and $\;x(t)=0\;$ otherwise\\
+\texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\
+      \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\
+\texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{-1}^1$\\
+       \` pbl: integration in $\cal C$\\
+\texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{-1}^1\right)$\\
+      \` $f\;t\;|_a^b = f\;b-f\;a$\\
+\texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1} -  \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot -1}$\\
+\texttt{\footnotesize{06}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega} -  \frac{1}{-j\cdot\omega}\cdot e^{j\cdot\omega}$\\
+\` Lift $\frac{1}{j\omega}$\\
+\texttt{\footnotesize{07}} \> $\frac{1}{j\cdot\omega}\cdot(e^{j\cdot\omega} - e^{-j\cdot\omega})$\\
+      \` trick~!\\
+\texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot(\frac{-e^{j\cdot\omega} + e^{-j\cdot\omega}}{j})$\\
+      \` table\\
+\texttt{\footnotesize{09}} \> $2\cdot\frac{\sin\;\omega}{\omega}$
+\end{tabbing}
+}
+
+\noindent\textbf{(b)} \textsf{Subproblem 1}
+\onehalfspace{
+\begin{tabbing}
+000\=\kill
+\texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\
+\`Insert Condition: $x(t) = 1\;$ for $\;\{1\leq t\;\land\;t\leq 3\}\;$ and $\;x(t)=0\;$ otherwise\\
+\texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\
+      \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\
+\texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{1}^3$\\
+       \` pbl: integration in $\cal C$\\
+\texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{1}^3\right)$\\
+      \` $f\;t\;|_a^b = f\;b-f\;a$\\
+\texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 3} -  \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1}$\\
+\`Lift $\frac{1}{-j\omega}$\\
+\texttt{\footnotesize{06}} \> $\frac{1}{j\cdot\omega}\cdot(e^{-j\cdot\omega} - e^{-j\cdot\omega3})$\\
+      \`Lift $e^{j\omega2}$ (trick)\\
+\texttt{\footnotesize{07}} \> $\frac{1}{j\omega}\cdot e^{j\omega2}\cdot(e^{j\omega} - e^{-j\omega})$\\
+\`Simplification (trick)\\
+\texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot e^{j\omega2}\cdot(\frac{e^{j\omega} - e^{-j\omega}}{j})$\\
+      \` table\\
+\texttt{\footnotesize{09}} \> $2\cdot e^{j\omega2}\cdot\frac{\sin\;\omega}{\omega}$\\
+\noindent\textbf{(b)} \textsf{Subproblem 2}\\
+\`Definition: $X(j\omega)=|X(j\omega)|\cdot e^{arg(X(j\omega))}$\\
+\`$|X(j\omega)|$ is called \emph{Magnitude}\\
+\`$arg(X(j\omega))$ is called \emph{Phase}\\
+\texttt{\footnotesize{10}} \> $|X(j\omega)|=\frac{2}{\omega}\cdot sin(\omega)$\\
+\texttt{\footnotesize{11}} \> $arg(X(j\omega)=-2\omega$\\
+\end{tabbing}
+}
+%------------------------------------------------------------------------------
+%CONVOLUTION
+
+\subsection{Convolution}
+\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the SPSC Problem Class 2, Summer term 2008}}
+Consider the two discrete-time, linear and time-invariant (LTI) systems with the following impulse response:
+
+\begin{center}
+$h_1[n]=\left(\frac{3}{5}\right)^n\cdot u[n]$\\
+$h_1[n]=\left(-\frac{2}{3}\right)^n\cdot u[n]$
+\end{center}
+
+The two systems are cascaded seriell. Derive the impulse respinse of the overall system $h_c[n]$.
+\subsubsection*{Solution}
+
+\doublespace{
+\begin{tabbing}
+000\=\kill
+\texttt{\footnotesize{01}} \> $h_c[n]=h_1[n]*h_2[n]$\\
+\texttt{\footnotesize{02}} \> $h_c[n]=\left(\left(\frac{3}{5}\right)^n\cdot u[n]\right)*\left(\left(-\frac{2}{3}\right)^n\cdot u[n]\right)$\\
+\`Definition: $a^n\cdot u[n]\,*\,b^n\cdot u[n]=\sum\limits_{k=-\infty}^{\infty}{a^k\cdot u[k]\cdot b^{n-k}\cdot u[n-k]}$\\
+\texttt{\footnotesize{03}} \> $h_c[n]=\sum\limits_{k=-\infty}^{\infty}{\left(\frac{3}{5}\right)^k\cdot u[n]\,\cdot \,\left(-\frac{2}{3}\right)^{n-k}\cdot u[n-k]}$\\
+\`$u[n]= \left\{
+     \begin{array}{lr}
+       1 & for\ n>=0\\
+       0 & else
+     \end{array}
+   \right.$\\
+\`We can leave the unitstep through simplification.\\
+\`So the lower limit is 0, the upper limit is n.\\
+\texttt{\footnotesize{04}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n-k}}$\\
+\`Expand\\
+\texttt{\footnotesize{05}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n}\cdot \left(-\frac{2}{3}\right)^{-k}}$\\
+\`Lift\\
+\texttt{\footnotesize{06}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{\infty}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{-k}}$\\
+\texttt{\footnotesize{07}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{3}{2}\right)^{k}}$\\
+\texttt{\footnotesize{08}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(-\frac{9}{10}\right)^{k}}$\\
+\`Geometric Series: $\sum\limits_{k=0}^{n}{q^k}=\frac{1-q^{n+1}}{1-q}$\\
+\`Now we have to consider the limits again.\\
+\`It is neccesarry to put the unitstep in again.\\
+\texttt{\footnotesize{09}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n+1}}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\
+\texttt{\footnotesize{10}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\
+\texttt{\footnotesize{11}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{\left(\frac{19}{10}\right)}\cdot u[n]$\\
+\texttt{\footnotesize{12}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot \left(1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)\right)\cdot\left(\frac{10}{19}\right)\cdot u[n]$\\
+\`Lift $u[n]$\\
+\texttt{\footnotesize{13}} \> $\left(\frac{10}{19}\cdot\left(-\frac{2}{3}\right)^n+\frac{9}{19}\cdot\left(\frac{3}{5}\right)^n\right)\cdot u[n]$\\
+\end{tabbing}
+}
+
+%------------------------------------------------------------------------------
+%Z-Transformation
+
+\subsection{Z-Transformation\label{sec:calc:ztrans}}
+\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the signal processing problem class 5, summer term 2008.}}
+Determine the inverse $\cal{z}$ transform of the following expression. Hint: applay the partial fraction expansion.
+
+\begin{center}
+$X(z)=\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}},\ \ x[n]$ is absolute summable
+\end{center}
+
+\subsubsection*{Solution}
+\onehalfspace{
+\begin{tabbing}
+000\=\kill
+\textsf{Main Problem}\\
+\texttt{\footnotesize{01}} \> $\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}}$ \\
+\`Divide through z, neccesary for z-transformation\\
+\texttt{\footnotesize{02}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}$ \\
+\`Start with partial fraction expansion\\
+\texttt{\footnotesize{03}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}=\frac{A}{z-z_1}+\frac{B}{z-z_2}$ \\
+\`Eliminate Fractions\\
+\texttt{\footnotesize{04}} \> $3=A(z-z_2)+B(z-z_1)$ \\
+\textsf{Subproblem 1}\\
+\`Setup a linear equation system by inserting the zeros $z_1$ and $z_2$ for $z$\\
+\texttt{\footnotesize{05}} \> $3=A(z_1-z_2)$ \& $3=B(z_2-z_1)$\\
+\texttt{\footnotesize{06}} \> $\frac{3}{z_1-z_2}=A$ \& $\frac{3}{z_2-z_1}=B$\\
+\textsf{Subproblem 2}\\
+\`Determine $z_1$ and $z_2$\\
+\texttt{\footnotesize{07}} \> $z_1=\frac{1}{8}+\sqrt{\frac{1}{64}+\frac{1}{8}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{1}{64}+\frac{1}{8}}$\\
+\texttt{\footnotesize{08}} \> $z_1=\frac{1}{8}+\sqrt{\frac{9}{64}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{9}{64}}$\\
+\texttt{\footnotesize{09}} \> $z_1=\frac{1}{8}+\frac{3}{8}$ \& $z_2=\frac{1}{8}-\frac{3}{8}$\\
+\texttt{\footnotesize{10}} \> $z_1=\frac{1}{2}$ \& $z_2=-\frac{1}{4}$\\
+\textsf{Continiue with Subproblem 1}\\
+\`Get the coeffizients $A$ and $B$\\
+\texttt{\footnotesize{11}} \> $\frac{3}{\frac{1}{2}-(-\frac{1}{4})}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\
+\texttt{\footnotesize{12}} \> $\frac{3}{\frac{1}{2}+\frac{1}{4}}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\
+\texttt{\footnotesize{13}} \> $\frac{3}{\frac{3}{4}}=A$ \& $\frac{3}{-\frac{3}{4}}=B$\\
+\texttt{\footnotesize{14}} \> $\frac{12}{3}=A$ \& $-\frac{12}{3}=B$\\
+\texttt{\footnotesize{15}} \> $4=A$ \& $-4=B$\\
+\textsf{Continiue with Main Problem}\\
+\texttt{\footnotesize{16}} \> $\frac{A}{z-z_1}+\frac{B}{z-z_2}$\\
+\texttt{\footnotesize{17}} \> $\frac{4}{z-\frac{1}{2}}+\frac{4}{z-\left(-\frac{1}{4}\right)}$ \\
+\texttt{\footnotesize{18}} \> $\frac{4}{z-\frac{1}{2}}-\frac{4}{z+\frac{1}{4}}$ \\
+\`Multiply with z, neccesary for z-transformation\\
+\texttt{\footnotesize{19}} \> $\frac{4z}{z-\frac{1}{2}}-\frac{4z}{z+\frac{1}{4}}$ \\
+\texttt{\footnotesize{20}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}$ \\
+\`Transformation\\
+\texttt{\footnotesize{21}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}\ \Ztransf\ 4\cdot\left(-\frac{1}{2}\right)^n\cdot u[n]+(-4)\cdot\left(\frac{1}{4}\right)^n\cdot u[n]$\\
+\end{tabbing}
+}
+\theendnotes
+%\end{document}
\ No newline at end of file




