1.1 --- a/src/Doc/isac/jrocnik/calulations.tex Mon Sep 16 12:27:20 2013 +0200
1.2 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000
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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