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1.4 +%\documentclass[a4paper]{scrartcl}
1.5 +%\usepackage[top=2cm, bottom=2.5cm, left=3cm, right=2cm, footskip=1cm]{geometry}
1.6 +%\usepackage[german]{babel}
1.7 +%\usepackage[T1]{fontenc}
1.8 +%\usepackage[latin1]{inputenc}
1.9 +%\usepackage{endnotes}
1.10 +%\usepackage{trfsigns}
1.11 +%\usepackage{setspace}
1.12 +%
1.13 +%\setlength{\parindent}{0ex}
1.14 +%\def\isac{${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}
1.15 +%\def\sisac{{\footnotesize${\cal I}\mkern-2mu{\cal S}\mkern-5mu{\cal AC}$}}
1.16 +%
1.17 +%\begin{document}
1.18 +%\title{Interactive Course Material for Signal Processing based on Isabelle/\isac}
1.19 +%\subtitle{Problemsolutions (Calculations)}
1.20 +%\author{Walther Neuper, Jan Rocnik}
1.21 +%\maketitle
1.22 +
1.23 +
1.24 +%------------------------------------------------------------------------------
1.25 +%FOURIER
1.26 +
1.27 +\subsection{Fourier Transformation}
1.28 +\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally fragment of the exam to \emph{Signaltransformationen VO} from 04.03.2011. Translated from German.}}
1.29 +\textbf{(a)} Determine the fourier transform for the given rectangular impulse:
1.30 +
1.31 +\begin{center}
1.32 +$x(t)= \left\{
1.33 + \begin{array}{lr}
1.34 + 1 & -1\leq t\geq1\\
1.35 + 0 & else
1.36 + \end{array}
1.37 + \right.$
1.38 +\end{center}
1.39 +
1.40 +\textbf{\noindent (b)} Now consider the given delayed impulse, determine its fourie transformation and calculate phase and magnitude:
1.41 +
1.42 +\begin{center}
1.43 +$x(t)= \left\{
1.44 + \begin{array}{lr}
1.45 + 1 & -1\leq t\leq1\\
1.46 + 0 & else
1.47 + \end{array}
1.48 + \right.$
1.49 +\end{center}
1.50 +
1.51 +\subsubsection{Solution}
1.52 +\textbf{(a)} \textsf{Subproblem 1}
1.53 +\onehalfspace{
1.54 +\begin{tabbing}
1.55 +000\=\kill
1.56 +\texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\
1.57 +\`Insert Condition: $x(t) = 1\;$ for $\;\{-1\leq t\;\land\;t\leq 1\}\;$ and $\;x(t)=0\;$ otherwise\\
1.58 +\texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\
1.59 + \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\
1.60 +\texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{-1}^1$\\
1.61 + \` pbl: integration in $\cal C$\\
1.62 +\texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{-1}^1\right)$\\
1.63 + \` $f\;t\;|_a^b = f\;b-f\;a$\\
1.64 +\texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1} - \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot -1}$\\
1.65 +\texttt{\footnotesize{06}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega} - \frac{1}{-j\cdot\omega}\cdot e^{j\cdot\omega}$\\
1.66 +\` Lift $\frac{1}{j\omega}$\\
1.67 +\texttt{\footnotesize{07}} \> $\frac{1}{j\cdot\omega}\cdot(e^{j\cdot\omega} - e^{-j\cdot\omega})$\\
1.68 + \` trick~!\\
1.69 +\texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot(\frac{-e^{j\cdot\omega} + e^{-j\cdot\omega}}{j})$\\
1.70 + \` table\\
1.71 +\texttt{\footnotesize{09}} \> $2\cdot\frac{\sin\;\omega}{\omega}$
1.72 +\end{tabbing}
1.73 +}
1.74 +
1.75 +\noindent\textbf{(b)} \textsf{Subproblem 1}
1.76 +\onehalfspace{
1.77 +\begin{tabbing}
1.78 +000\=\kill
1.79 +\texttt{\footnotesize{01}} \> Definition: $X(j\omega)=\int_\infty^\infty{x(t)\cdot e^{-j\omega t}}$\\
1.80 +\`Insert Condition: $x(t) = 1\;$ for $\;\{1\leq t\;\land\;t\leq 3\}\;$ and $\;x(t)=0\;$ otherwise\\
1.81 +\texttt{\footnotesize{02}} \> $X(j\omega)=\int_{-1}^{1}{1\cdot e^{-j\omega t}}$\\
1.82 + \` $\int_a^b f\;t\;dt = \int f\;t\;dt\;|_a^b$\\
1.83 +\texttt{\footnotesize{03}} \> $\int 1\cdot e^{-j\cdot\omega\cdot t} d t\;|_{1}^3$\\
1.84 + \` pbl: integration in $\cal C$\\
1.85 +\texttt{\footnotesize{04}} \> $\left(\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot t} \;|_{1}^3\right)$\\
1.86 + \` $f\;t\;|_a^b = f\;b-f\;a$\\
1.87 +\texttt{\footnotesize{05}} \> $\frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 3} - \frac{1}{-j\cdot\omega}\cdot e^{-j\cdot\omega\cdot 1}$\\
1.88 +\`Lift $\frac{1}{-j\omega}$\\
1.89 +\texttt{\footnotesize{06}} \> $\frac{1}{j\cdot\omega}\cdot(e^{-j\cdot\omega} - e^{-j\cdot\omega3})$\\
1.90 + \`Lift $e^{j\omega2}$ (trick)\\
1.91 +\texttt{\footnotesize{07}} \> $\frac{1}{j\omega}\cdot e^{j\omega2}\cdot(e^{j\omega} - e^{-j\omega})$\\
1.92 +\`Simplification (trick)\\
1.93 +\texttt{\footnotesize{08}} \> $\frac{1}{\omega}\cdot e^{j\omega2}\cdot(\frac{e^{j\omega} - e^{-j\omega}}{j})$\\
1.94 + \` table\\
1.95 +\texttt{\footnotesize{09}} \> $2\cdot e^{j\omega2}\cdot\frac{\sin\;\omega}{\omega}$\\
1.96 +\noindent\textbf{(b)} \textsf{Subproblem 2}\\
1.97 +\`Definition: $X(j\omega)=|X(j\omega)|\cdot e^{arg(X(j\omega))}$\\
1.98 +\`$|X(j\omega)|$ is called \emph{Magnitude}\\
1.99 +\`$arg(X(j\omega))$ is called \emph{Phase}\\
1.100 +\texttt{\footnotesize{10}} \> $|X(j\omega)|=\frac{2}{\omega}\cdot sin(\omega)$\\
1.101 +\texttt{\footnotesize{11}} \> $arg(X(j\omega)=-2\omega$\\
1.102 +\end{tabbing}
1.103 +}
1.104 +%------------------------------------------------------------------------------
1.105 +%CONVOLUTION
1.106 +
1.107 +\subsection{Convolution}
1.108 +\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the SPSC Problem Class 2, Summer term 2008}}
1.109 +Consider the two discrete-time, linear and time-invariant (LTI) systems with the following impulse response:
1.110 +
1.111 +\begin{center}
1.112 +$h_1[n]=\left(\frac{3}{5}\right)^n\cdot u[n]$\\
1.113 +$h_1[n]=\left(-\frac{2}{3}\right)^n\cdot u[n]$
1.114 +\end{center}
1.115 +
1.116 +The two systems are cascaded seriell. Derive the impulse respinse of the overall system $h_c[n]$.
1.117 +\subsubsection*{Solution}
1.118 +
1.119 +\doublespace{
1.120 +\begin{tabbing}
1.121 +000\=\kill
1.122 +\texttt{\footnotesize{01}} \> $h_c[n]=h_1[n]*h_2[n]$\\
1.123 +\texttt{\footnotesize{02}} \> $h_c[n]=\left(\left(\frac{3}{5}\right)^n\cdot u[n]\right)*\left(\left(-\frac{2}{3}\right)^n\cdot u[n]\right)$\\
1.124 +\`Definition: $a^n\cdot u[n]\,*\,b^n\cdot u[n]=\sum\limits_{k=-\infty}^{\infty}{a^k\cdot u[k]\cdot b^{n-k}\cdot u[n-k]}$\\
1.125 +\texttt{\footnotesize{03}} \> $h_c[n]=\sum\limits_{k=-\infty}^{\infty}{\left(\frac{3}{5}\right)^k\cdot u[n]\,\cdot \,\left(-\frac{2}{3}\right)^{n-k}\cdot u[n-k]}$\\
1.126 +\`$u[n]= \left\{
1.127 + \begin{array}{lr}
1.128 + 1 & for\ n>=0\\
1.129 + 0 & else
1.130 + \end{array}
1.131 + \right.$\\
1.132 +\`We can leave the unitstep through simplification.\\
1.133 +\`So the lower limit is 0, the upper limit is n.\\
1.134 +\texttt{\footnotesize{04}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n-k}}$\\
1.135 +\`Expand\\
1.136 +\texttt{\footnotesize{05}} \> $h_c[n]=\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{n}\cdot \left(-\frac{2}{3}\right)^{-k}}$\\
1.137 +\`Lift\\
1.138 +\texttt{\footnotesize{06}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{\infty}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{2}{3}\right)^{-k}}$\\
1.139 +\texttt{\footnotesize{07}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(\frac{3}{5}\right)^k\cdot \left(-\frac{3}{2}\right)^{k}}$\\
1.140 +\texttt{\footnotesize{08}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\sum\limits_{k=0}^{n}{\left(-\frac{9}{10}\right)^{k}}$\\
1.141 +\`Geometric Series: $\sum\limits_{k=0}^{n}{q^k}=\frac{1-q^{n+1}}{1-q}$\\
1.142 +\`Now we have to consider the limits again.\\
1.143 +\`It is neccesarry to put the unitstep in again.\\
1.144 +\texttt{\footnotesize{09}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n+1}}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\
1.145 +\texttt{\footnotesize{10}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{1-\left(-\frac{9}{10}\right)}\cdot u[n]$\\
1.146 +\texttt{\footnotesize{11}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot\frac{1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)}{\left(\frac{19}{10}\right)}\cdot u[n]$\\
1.147 +\texttt{\footnotesize{12}} \> $h_c[n]=\left(-\frac{2}{3}\right)^{n}\cdot \left(1-\left(-\frac{9}{10}\right)^{n}\cdot\left(-\frac{9}{10}\right)\right)\cdot\left(\frac{10}{19}\right)\cdot u[n]$\\
1.148 +\`Lift $u[n]$\\
1.149 +\texttt{\footnotesize{13}} \> $\left(\frac{10}{19}\cdot\left(-\frac{2}{3}\right)^n+\frac{9}{19}\cdot\left(\frac{3}{5}\right)^n\right)\cdot u[n]$\\
1.150 +\end{tabbing}
1.151 +}
1.152 +
1.153 +%------------------------------------------------------------------------------
1.154 +%Z-Transformation
1.155 +
1.156 +\subsection{Z-Transformation\label{sec:calc:ztrans}}
1.157 +\subsubsection*{Problem\endnote{Problem submitted by Bernhard Geiger. Originally part of the signal processing problem class 5, summer term 2008.}}
1.158 +Determine the inverse $\cal{z}$ transform of the following expression. Hint: applay the partial fraction expansion.
1.159 +
1.160 +\begin{center}
1.161 +$X(z)=\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}},\ \ x[n]$ is absolute summable
1.162 +\end{center}
1.163 +
1.164 +\subsubsection*{Solution}
1.165 +\onehalfspace{
1.166 +\begin{tabbing}
1.167 +000\=\kill
1.168 +\textsf{Main Problem}\\
1.169 +\texttt{\footnotesize{01}} \> $\frac{3}{z-\frac{1}{4}-\frac{1}{8}z^{-1}}$ \\
1.170 +\`Divide through z, neccesary for z-transformation\\
1.171 +\texttt{\footnotesize{02}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}$ \\
1.172 +\`Start with partial fraction expansion\\
1.173 +\texttt{\footnotesize{03}} \> $\frac{3}{z^2-\frac{1}{4}z-\frac{1}{8}}=\frac{A}{z-z_1}+\frac{B}{z-z_2}$ \\
1.174 +\`Eliminate Fractions\\
1.175 +\texttt{\footnotesize{04}} \> $3=A(z-z_2)+B(z-z_1)$ \\
1.176 +\textsf{Subproblem 1}\\
1.177 +\`Setup a linear equation system by inserting the zeros $z_1$ and $z_2$ for $z$\\
1.178 +\texttt{\footnotesize{05}} \> $3=A(z_1-z_2)$ \& $3=B(z_2-z_1)$\\
1.179 +\texttt{\footnotesize{06}} \> $\frac{3}{z_1-z_2}=A$ \& $\frac{3}{z_2-z_1}=B$\\
1.180 +\textsf{Subproblem 2}\\
1.181 +\`Determine $z_1$ and $z_2$\\
1.182 +\texttt{\footnotesize{07}} \> $z_1=\frac{1}{8}+\sqrt{\frac{1}{64}+\frac{1}{8}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{1}{64}+\frac{1}{8}}$\\
1.183 +\texttt{\footnotesize{08}} \> $z_1=\frac{1}{8}+\sqrt{\frac{9}{64}}$ \& $z_2=\frac{1}{8}-\sqrt{\frac{9}{64}}$\\
1.184 +\texttt{\footnotesize{09}} \> $z_1=\frac{1}{8}+\frac{3}{8}$ \& $z_2=\frac{1}{8}-\frac{3}{8}$\\
1.185 +\texttt{\footnotesize{10}} \> $z_1=\frac{1}{2}$ \& $z_2=-\frac{1}{4}$\\
1.186 +\textsf{Continiue with Subproblem 1}\\
1.187 +\`Get the coeffizients $A$ and $B$\\
1.188 +\texttt{\footnotesize{11}} \> $\frac{3}{\frac{1}{2}-(-\frac{1}{4})}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\
1.189 +\texttt{\footnotesize{12}} \> $\frac{3}{\frac{1}{2}+\frac{1}{4}}=A$ \& $\frac{3}{-\frac{1}{4}-\frac{1}{2}}=B$\\
1.190 +\texttt{\footnotesize{13}} \> $\frac{3}{\frac{3}{4}}=A$ \& $\frac{3}{-\frac{3}{4}}=B$\\
1.191 +\texttt{\footnotesize{14}} \> $\frac{12}{3}=A$ \& $-\frac{12}{3}=B$\\
1.192 +\texttt{\footnotesize{15}} \> $4=A$ \& $-4=B$\\
1.193 +\textsf{Continiue with Main Problem}\\
1.194 +\texttt{\footnotesize{16}} \> $\frac{A}{z-z_1}+\frac{B}{z-z_2}$\\
1.195 +\texttt{\footnotesize{17}} \> $\frac{4}{z-\frac{1}{2}}+\frac{4}{z-\left(-\frac{1}{4}\right)}$ \\
1.196 +\texttt{\footnotesize{18}} \> $\frac{4}{z-\frac{1}{2}}-\frac{4}{z+\frac{1}{4}}$ \\
1.197 +\`Multiply with z, neccesary for z-transformation\\
1.198 +\texttt{\footnotesize{19}} \> $\frac{4z}{z-\frac{1}{2}}-\frac{4z}{z+\frac{1}{4}}$ \\
1.199 +\texttt{\footnotesize{20}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}$ \\
1.200 +\`Transformation\\
1.201 +\texttt{\footnotesize{21}} \> $4\cdot\frac{z}{z-\frac{1}{2}}+(-4)\cdot\frac{z}{z+\frac{1}{4}}\ \Ztransf\ 4\cdot\left(-\frac{1}{2}\right)^n\cdot u[n]+(-4)\cdot\left(\frac{1}{4}\right)^n\cdot u[n]$\\
1.202 +\end{tabbing}
1.203 +}
1.204 +\theendnotes
1.205 +%\end{document}
1.206 \ No newline at end of file