352 lines
8.1 KiB
TeX
352 lines
8.1 KiB
TeX
\documentclass[12pt,a4paper,austrian]{article}
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\usepackage{graphicx}
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\usepackage[austrian, english]{babel}
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\usepackage[utf8]{inputenc}
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%\usepackage{listings}
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\usepackage{multirow}
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\usepackage{epstopdf}
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\usepackage{amsmath}
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\usepackage{amssymb} % fuer Mengen \N, Q, C, R
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\graphicspath{{./fig/}}
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%% Satzspiegel
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\setlength{\hoffset}{-1in} \setlength{\textwidth}{18cm}
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\setlength{\oddsidemargin}{1.5cm}
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\setlength{\evensidemargin}{1.5cm}
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\setlength{\marginparsep}{0.7em}
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\setlength{\marginparwidth}{0.5cm}
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\setlength{\voffset}{-1.9in}
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\setlength{\headheight}{12pt}
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\setlength{\topmargin}{2.6cm}
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\addtolength{\topmargin}{-\headheight}
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\setlength{\headsep}{3.5cm}
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\addtolength{\headsep}{-\topmargin}
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\addtolength{\headsep}{-\headheight}
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\setlength{\textheight}{27cm}
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%% How should floats be treated?
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\setlength{\floatsep}{12 pt plus 0 pt minus 8 pt}
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\setlength{\textfloatsep}{12 pt plus 0pt minus 8 pt}
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\setlength{\intextsep}{12 pt plus 0pt minus 8 pt}
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\tolerance2000
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\emergencystretch20pt
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%% Text appearence
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% English text
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\newcommand{\eg}[1]%
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{\selectlanguage{english}\textit{#1}\selectlanguage{austrian}}
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\newcommand{\filename}[1]
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{\begin{small}\texttt{#1}\end{small}}
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\newcommand\IFT{\unitlength1mm\begin{picture}(10,2) \put (1,1)
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{\circle{1.7}} \put(2,1){\line(1,0){5}} \put(8,1)
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{\circle*{1.7}}\end{picture}}
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\newcommand\FT{\unitlength1mm\begin{picture}(10,2) \put (1,1)
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{\circle*{1.7}} \put(2,1){\line(1,0){5}} \put(8,1)
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{\circle{1.7}}\end{picture}}
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% A box for multiple choice problems
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\newcommand{\choicebox}{\fbox{\rule{0pt}{0.5ex}\rule{0.5ex}{0pt}}}
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\newenvironment{wahrfalsch}%
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{\bigskip\par\noindent\makebox[1cm][c]{richtig}\hspace{3mm}\makebox[1cm][c]{falsch}
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\begin{list}%
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{\makebox[1cm][c]{\choicebox}\hspace{3mm}\makebox[1cm][c]{\choicebox}}%
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{\setlength{\labelwidth}{2.31 cm}\setlength{\labelsep}{3mm}
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\setlength{\leftmargin}{2.61 cm}\setlength{\listparindent}{0pt}
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\setlength{\itemindent}{0pt}}%
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}
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{\end{list}}
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\newcounter{theaufgabe}\setcounter{theaufgabe}{1}
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\newenvironment{aufgabe}[1]%
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{\bigskip\par\noindent\begin{nopagebreak}
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\textsf{\textbf{\arabic{theaufgabe}.\thinspace Exercise}}\quad
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\textsf{\textit{#1}}\\*[1ex]%
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\stepcounter{theaufgabe}\hspace{2ex}\end{nopagebreak}}
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{\par\pagebreak[2]}
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% Innerhalb der Aufgaben erfolgt die weitere Unterteilung mittels einer
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% enumerate Umgebung, die allerdings a), b),... zaehlen soll.
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\renewcommand{\labelenumi}{\alph{enumi})}
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\renewcommand{\labelenumii}{\arabic{enumii})}
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% A box to tick for everything which has to done
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\newcommand{\abgabe}{\marginpar{$\Box$}}
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% Margin paragraphs on the left side
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\reversemarginpar
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% Language for listings
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%\lstset{language=Vhdl,
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% basicstyle=\small\tt,
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% keywordstyle=\tt\bf,
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% commentstyle=\sl}
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% No indention
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\setlength{\parindent}{0.0cm}
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% Don't number sections
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\setcounter{secnumdepth}{0}
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%% Beginning of the text
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\begin{document}
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\selectlanguage{austrian}
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\pagestyle{plain}
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%=== This is the header section ============================================================
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\thispagestyle{empty}
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\noindent
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\begin{minipage}[b][4cm]{1.0\textwidth}
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\begin{center}
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\begin{bf}
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\begin{large} Digitale Signalverarbeitung WS 2025/26 -- 1.~Aufgabe\end{large} \\
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\vspace{0.3cm}
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\begin{Large} Analog Signals and Systems \end{Large} \\
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\vspace{0.3cm}
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\end{bf}
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\begin{large}
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Gruppennummer 211 \\
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Manuel Illmayer, k12308149 \\
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Quirin Ecker, k12310122 \\
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\end{large}
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\end{center}
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\end{minipage}
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\noindent \rule[0.8em]{\textwidth}{0.12mm}\\[-0.5em]
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%=======================================================================================
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\begin{aufgabe}{}
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\begin{align*}
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c_1 &= -5 + 3j
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= \sqrt{(-5)^2 + 3^2}\, e^{j \arctan\left(\frac{3}{-5}\right)}
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= \sqrt{34}\, e^{-j\,0.54} \\
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\end{align*}
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\begin{align*}
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c_2 &= \frac{\sqrt{2}}{2} e^{-j \frac{3\pi}{4}}
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= \frac{\sqrt{2}}{2} \cos\left(-\frac{3\pi}{4}\right)
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+ j \frac{\sqrt{2}}{2} \sin\left(-\frac{3\pi}{4}\right)
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= -\frac{1}{2} - \frac{1}{2}j \\
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\end{align*}
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\begin{align*}
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c_3 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}j
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= \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}
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\, e^{j \arctan\left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right)}
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= e^{j \frac{\pi}{4}} \\
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\end{align*}
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\begin{align*}
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c_4 &= c_1 + c_2
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= (-5 + 3j) + \left(-\frac{1}{2} - \frac{1}{2}j\right)
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= -\frac{11}{2} + \frac{5}{2}j \\
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\end{align*}
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\begin{align*}
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c_5 &= c_1 \cdot c_2
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= \left(\sqrt{34}\, e^{-j\,0.54}\right)\left(\frac{\sqrt{2}}{2} e^{-j \frac{3 \pi}{4}}\right)
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\approx \sqrt{17} e^{-j\,2.8961} \\
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\end{align*}
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\begin{align*}
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c_6 &= |c_3|^2
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= \left(\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}\right)^2
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= 1 \\
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\end{align*}
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\begin{align*}
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c_7 &= \arg(c_3) = \frac{\pi}{4} \\
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\end{align*}
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\begin{align*}
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c_8 &= \frac{c_1}{c_2}
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= \frac{-5 + 3j}{-\frac{1}{2} - \frac{1}{2}j}
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\cdot \frac{-\frac{1}{2} + \frac{1}{2}j}{-\frac{1}{2} + \frac{1}{2}j} \\
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&= \frac{(-5)(-\frac{1}{2}) + 3(-\frac{1}{2}) + j\left(3(-\frac{1}{2}) - (-5)(-\frac{1}{2})\right)}{\left(-\frac{1}{2}\right)^2 + \left(-\frac{1}{2}\right)^2} \\
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&= 2 - 8j \\
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\end{align*}
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\begin{align*}
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c_9 &= c_1 \cdot c_1^*
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= \sqrt{34}\, e^{-j\,0.54} \cdot \sqrt{34}\, e^{j\,0.54}
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= 34
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\end{align*}
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\end{aufgabe}
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\begin{aufgabe}{}
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\[
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x(t) = \hat{X} \cos(2\pi f_0 t) \xleftrightarrow{} \iff X(f) = \frac{\hat{X}}{2} \delta(f - f_0) + \frac{\hat{X}}{2} \delta(f + f_0) \\
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\]
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\[
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x(t) = \hat{X}( \frac{1}{2} \cdot e^{j2\pi f_0 t} + \frac{1}{2} \cdot e^{-j2\pi f_0 t} )
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\]
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\[
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X(f) = \hat{X}( \frac{1}{2} \cdot \delta (f-f_0) + \frac{1}{2} \cdot \delta (f+f_0) )
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\]
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\[
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X(f) = \frac{\hat{X}}{2} \cdot \delta (f-f_0) + \frac{\hat{X}}{2} \cdot \delta (f+f_0)
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\]
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\begin{center}
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\includegraphics[width=0.95\textwidth]{fig/aufgabe2.png}
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\end{center}
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\end{aufgabe}
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\begin{aufgabe}{}
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\begin{enumerate}
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\item
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\[
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\sin\left(2\pi f_i (t - 0.1)\right) = \sin\left(2\pi f_i t + \phi_i\right)
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\]
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\[
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2\pi f_i (t - 0.1) = 2\pi f_i t + \phi_i
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\]
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\[
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2\pi f_i (t - 0.1) - 2\pi f_i t = \phi_i
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\]
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\[
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2\pi f_i \big((t - 0.1) - t\big) = \phi_i
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\]
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\[
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\phi_i = 2\pi f_i (-0.1) = -0.1 \cdot 2\pi f_i
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\]
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The fourier transform shift theorem states $x(t - T) \;\longleftrightarrow\; X(f)\, e^{-j 2\pi f T}$, where the angle is $2\pi fT$ and $T = 0.1$ in this case. Combining everything, we get $-0.1 \cdot 2\pi f_i$ which is equivalent to the result we got.
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\item Figures:
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\begin{center}
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\includegraphics[width=0.95\textwidth]{fig/aufgabe3_1.png}
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\includegraphics[width=0.95\textwidth]{fig/aufgabe3_2.png}
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\end{center}
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\end{enumerate}
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\end{aufgabe}
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\begin{aufgabe}{}
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\begin{itemize}
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\item $y(t) = (x(t))^2$
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\[
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y(t) = (x(t))^2
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\]
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\[
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y_1(t) = (\alpha x_1(t))^2 + (\beta x_2(t))^2
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\]
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\[
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= \alpha^2 (x_1(t))^2 + \beta^2 (x_2(t))^2
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\]
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\[
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y_2(t) = \alpha (x_1(t))^2 + \beta (x_2(t))^2
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\]
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\[
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y_1(t) \ne y_2(t)
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\]
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\[
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\implies \text{Not linear}
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\]
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\[
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y_1(t) &= \left(x(t - \Delta)\right)^2 \\
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\]
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\[
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y_2(t) &= y_1(t - \Delta) = \left(x(t - \Delta)\right)^2 \\
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\]
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\[
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y_1 &= y_2 \\
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\]
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\[
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\implies \text{Time invariant}
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\]
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\item $y(t) = x(t) sin(\Omega_0 t)$
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\[
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y(t) &= x(t) \sin(\Omega_0 t) \\
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\]
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\[
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y_1(t) &= \left( \alpha \, x(t) \sin(\Omega_0 t) \right) + \left( \beta \, x(t) \sin(\Omega_0 t) \right) \\
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\]
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\[
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y_t(t) &= \alpha \left( x(t) \sin(\Omega_0 t) \right) + \beta \left( x(t) \sin(\Omega_0 t) \right) \\
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\]
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\[
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y_1 &= y_2
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\]
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\[
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\implies \text{linear}
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\]
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\[
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y_1(t) = &\coloneqq x(t - \Delta) \sin(\Omega_0 t - \Delta) \\
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\]
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\[
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y_2(t) &= y_1(t - \Delta) = x(t - \Delta) \sin(\Omega_0 t - \Delta) \\
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\]
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\[
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y_1 &= y_2 \\
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\]
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\[
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\implies \text{Time invariant}
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\]
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\end{itemize}
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\end{aufgabe}
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\begin{aufgabe}
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\begin{enumerate}
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\item System A is linear because there is no difference between the plot $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$.
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You can also see in the error plot that the error is small.
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It is not exactyly zero becuase rounding errors.
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\item Input and output are not proportional in a non linear system and
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therefore the error change is greater than the one of the linear system
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\item The output sound of System B does differ when comparing $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$.
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But is the same when using System A. This matches the differences in the graphs.
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\end{enumerate}
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\end{aufgabe}
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\end{document}
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