diff -r 7f3760f39bdc -r f8845fc8f38d doc-isac/jrocnik/calulations.tex --- /dev/null Thu Jan 01 00:00:00 1970 +0000 +++ b/doc-isac/jrocnik/calulations.tex Tue Sep 17 09:50:52 2013 +0200 @@ -0,0 +1,202 @@ +%\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