diff -r 7f3760f39bdc -r f8845fc8f38d src/Doc/isac/jrocnik/calulations.tex --- a/src/Doc/isac/jrocnik/calulations.tex Mon Sep 16 12:27:20 2013 +0200 +++ /dev/null Thu Jan 01 00:00:00 1970 +0000 @@ -1,202 +0,0 @@ -%\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