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+++ /dev/null	Thu Jan 01 00:00:00 1970 +0000
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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




