\documentclass[12pt,a4paper,austrian]{article} \usepackage{graphicx} \usepackage[austrian, english]{babel} \usepackage[utf8]{inputenc} %\usepackage{listings} \usepackage{multirow} \usepackage{epstopdf} \usepackage{amsmath} \usepackage{amssymb} % fuer Mengen \N, Q, C, R \graphicspath{{./fig/}} %% Satzspiegel \setlength{\hoffset}{-1in} \setlength{\textwidth}{18cm} \setlength{\oddsidemargin}{1.5cm} \setlength{\evensidemargin}{1.5cm} \setlength{\marginparsep}{0.7em} \setlength{\marginparwidth}{0.5cm} \setlength{\voffset}{-1.9in} \setlength{\headheight}{12pt} \setlength{\topmargin}{2.6cm} \addtolength{\topmargin}{-\headheight} \setlength{\headsep}{3.5cm} \addtolength{\headsep}{-\topmargin} \addtolength{\headsep}{-\headheight} \setlength{\textheight}{27cm} %% How should floats be treated? \setlength{\floatsep}{12 pt plus 0 pt minus 8 pt} \setlength{\textfloatsep}{12 pt plus 0pt minus 8 pt} \setlength{\intextsep}{12 pt plus 0pt minus 8 pt} \tolerance2000 \emergencystretch20pt %% Text appearence % English text \newcommand{\eg}[1]% {\selectlanguage{english}\textit{#1}\selectlanguage{austrian}} \newcommand{\filename}[1] {\begin{small}\texttt{#1}\end{small}} \newcommand\IFT{\unitlength1mm\begin{picture}(10,2) \put (1,1) {\circle{1.7}} \put(2,1){\line(1,0){5}} \put(8,1) {\circle*{1.7}}\end{picture}} \newcommand\FT{\unitlength1mm\begin{picture}(10,2) \put (1,1) {\circle*{1.7}} \put(2,1){\line(1,0){5}} \put(8,1) {\circle{1.7}}\end{picture}} % A box for multiple choice problems \newcommand{\choicebox}{\fbox{\rule{0pt}{0.5ex}\rule{0.5ex}{0pt}}} \newenvironment{wahrfalsch}% {\bigskip\par\noindent\makebox[1cm][c]{richtig}\hspace{3mm}\makebox[1cm][c]{falsch} \begin{list}% {\makebox[1cm][c]{\choicebox}\hspace{3mm}\makebox[1cm][c]{\choicebox}}% {\setlength{\labelwidth}{2.31 cm}\setlength{\labelsep}{3mm} \setlength{\leftmargin}{2.61 cm}\setlength{\listparindent}{0pt} \setlength{\itemindent}{0pt}}% } {\end{list}} \newcounter{theaufgabe}\setcounter{theaufgabe}{1} \newenvironment{aufgabe}[1]% {\bigskip\par\noindent\begin{nopagebreak} \textsf{\textbf{\arabic{theaufgabe}.\thinspace Exercise}}\quad \textsf{\textit{#1}}\\*[1ex]% \stepcounter{theaufgabe}\hspace{2ex}\end{nopagebreak}} {\par\pagebreak[2]} % Innerhalb der Aufgaben erfolgt die weitere Unterteilung mittels einer % enumerate Umgebung, die allerdings a), b),... zaehlen soll. \renewcommand{\labelenumi}{\alph{enumi})} \renewcommand{\labelenumii}{\arabic{enumii})} % A box to tick for everything which has to done \newcommand{\abgabe}{\marginpar{$\Box$}} % Margin paragraphs on the left side \reversemarginpar % Language for listings %\lstset{language=Vhdl, % basicstyle=\small\tt, % keywordstyle=\tt\bf, % commentstyle=\sl} % No indention \setlength{\parindent}{0.0cm} % Don't number sections \setcounter{secnumdepth}{0} %% Beginning of the text \begin{document} \selectlanguage{austrian} \pagestyle{plain} %=== This is the header section ============================================================ \thispagestyle{empty} \noindent \begin{minipage}[b][4cm]{1.0\textwidth} \begin{center} \begin{bf} \begin{large} Digitale Signalverarbeitung WS 2025/26 -- 1.~Aufgabe\end{large} \\ \vspace{0.3cm} \begin{Large} Analog Signals and Systems \end{Large} \\ \vspace{0.3cm} \end{bf} \begin{large} Gruppennummer 211 \\ Manuel Illmayer, k12308149 \\ Quirin Ecker, k12310122 \\ \end{large} \end{center} \end{minipage} \noindent \rule[0.8em]{\textwidth}{0.12mm}\\[-0.5em] %======================================================================================= \begin{aufgabe}{} \begin{align*} c_1 &= -5 + 3j = \sqrt{(-5)^2 + 3^2}\, e^{j \arctan\left(\frac{3}{-5}\right)} = \sqrt{34}\, e^{-j\,0.54} \\ \end{align*} \begin{align*} c_2 &= \frac{\sqrt{2}}{2} e^{-j \frac{3\pi}{4}} = \frac{\sqrt{2}}{2} \cos\left(-\frac{3\pi}{4}\right) + j \frac{\sqrt{2}}{2} \sin\left(-\frac{3\pi}{4}\right) = -\frac{1}{2} - \frac{1}{2}j \\ \end{align*} \begin{align*} c_3 &= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}j = \sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2} \, e^{j \arctan\left(\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}}\right)} = e^{j \frac{\pi}{4}} \\ \end{align*} \begin{align*} c_4 &= c_1 + c_2 = (-5 + 3j) + \left(-\frac{1}{2} - \frac{1}{2}j\right) = -\frac{11}{2} + \frac{5}{2}j \\ \end{align*} \begin{align*} c_5 &= c_1 \cdot c_2 = \left(\sqrt{34}\, e^{-j\,0.54}\right)\left(\frac{\sqrt{2}}{2} e^{-j \frac{3 \pi}{4}}\right) \approx \frac{\sqrt{70}}{2} e^{-j\,2.8961} \\ \end{align*} \begin{align*} c_6 &= |c_3|^2 = \left(\sqrt{\left(\frac{1}{\sqrt{2}}\right)^2 + \left(\frac{1}{\sqrt{2}}\right)^2}\right)^2 = 1 \\ \end{align*} \begin{align*} c_7 &= \arg(c_3) = \frac{\pi}{4} \\ \end{align*} \begin{align*} c_8 &= \frac{c_1}{c_2} = \frac{-5 + 3j}{-\frac{1}{2} - \frac{1}{2}j} \cdot \frac{-\frac{1}{2} + \frac{1}{2}j}{-\frac{1}{2} + \frac{1}{2}j} \\ &= \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} \\ &= 2 - 8j \\ \end{align*} \begin{align*} c_9 &= c_1 \cdot c_1^* = \sqrt{34}\, e^{-j\,0.54} \cdot \sqrt{34}\, e^{j\,0.54} = 34 \end{align*} \end{aufgabe} \begin{aufgabe}{} \[ x(t) = \hat{X} \cos(2\pi f_0 t) \xleftrightarrow{} \iff X(f) = \frac{\hat{X}}{2} \delta(f - f_0) + \frac{\hat{X}}{2} \delta(f + f_0) \\ \] \[ x(t) = \hat{X}( \frac{1}{2} \cdot e^{j2\pi f_0 t} + \frac{1}{2} \cdot e^{-j2\pi f_0 t} ) \] \[ X(f) = \hat{X}( \frac{1}{2} \cdot \delta (f-f_0) + \frac{1}{2} \cdot \delta (f+f_0) ) \] \[ X(f) = \frac{\hat{X}}{2} \cdot \delta (f-f_0) + \frac{\hat{X}}{2} \cdot \delta (f+f_0) \] \begin{center} \includegraphics[width=0.95\textwidth]{fig/aufgabe2.png} \end{center} \end{aufgabe} \begin{aufgabe}{} \begin{enumerate} \item \[ \sin\left(2\pi f_i (t - 0.1)\right) = \sin\left(2\pi f_i t + \phi_i\right) \] \[ 2\pi f_i (t - 0.1) = 2\pi f_i t + \phi_i \] \[ 2\pi f_i (t - 0.1) - 2\pi f_i t = \phi_i \] \[ 2\pi f_i \big((t - 0.1) - t\big) = \phi_i \] \[ \phi_i = 2\pi f_i (-0.1) = -0.1 \cdot 2\pi f_i \] 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. \item Figures: \begin{center} \includegraphics[width=0.95\textwidth]{fig/aufgabe3_1.png} \includegraphics[width=0.95\textwidth]{fig/aufgabe3_2.png} \end{center} \end{enumerate} \end{aufgabe} \begin{aufgabe}{} \begin{itemize} \item $y(t) = (x(t))^2$ \[ y(t) = (x(t))^2 \] \[ y_1(t) = (\alpha x_1(t))^2 + (\beta x_2(t))^2 \] \[ = \alpha^2 (x_1(t))^2 + \beta^2 (x_2(t))^2 \] \[ y_2(t) = \alpha (x_1(t))^2 + \beta (x_2(t))^2 \] \[ y_1(t) \ne y_2(t) \] \[ \implies \text{Not linear} \] \[ y_1(t) &= \left(x(t - \alpha)\right)^2 \\ \] \[ y_2(t) &= y_1(t - \Delta) = \left(x(t - \alpha)\right)^2 \\ \] \[ y_1 &= y_2 \\ \] \[ \implies \text{Time invariant} \] \item $y(t) = w(t) sin(\Omega_0 t)$ \[ y(t) &= x(t) \sin(n \Omega_0 t) \\ \] \[ y_1(t) &= \left( \alpha \, x(t) \sin(n_1 \omega t) \right) + \left( \beta \, x(t) \sin(n_0 t) \right) \\ \] \[ y_t(t) &= \alpha \left( x(t) \sin(n_1 \omega t) \right) + \beta \left( x(t) \sin(n_0 t) \right) \\ \] \[ y_1 &= y_2 \] \[ \implies \text{linear} \] \[ y_1(t) &\coloneqq x(t - \alpha) \sin(\omega_0 t - \alpha) \\ \] \[ y_2(t) &= y_1(t - \alpha) = x(t - \alpha) \sin(\omega_0 t - \alpha) \\ \] \[ y_1 &= y_2 \\ \] \[ \implies \text{Time invariant} \] \end{itemize} \end{aufgabe} \begin{aufgabe} \begin{enumerate} \item System A is linear because there is no difference between the plot $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$. You can also see in the error plot that the error is small. It is not exactyly zero becuase rounding errors. \item Input and output are not proportional in a non linear system and therefore the error change is greater than the one of the linear system \item The output sound of System B does differ when comparing $T [x_1(t) + x_2(t)]$ and $T [x_1(t)] + T [x_2(t)]$. But is the same when using System A. This matches the differences in the graphs. \end{enumerate} \end{aufgabe} \end{document}