From e54fed45d15199fe1c1c5f289671b28f99c71c2c Mon Sep 17 00:00:00 2001 From: quirinecker Date: Sat, 9 May 2026 13:23:38 +0200 Subject: [PATCH] initial commit --- .envrc | 1 + Assignment1_211.tex | 351 ++++++++++++++++++++++++++++++++++++++++++++ flake.nix | 22 +++ 3 files changed, 374 insertions(+) create mode 100644 .envrc create mode 100644 Assignment1_211.tex create mode 100644 flake.nix diff --git a/.envrc b/.envrc new file mode 100644 index 0000000..3550a30 --- /dev/null +++ b/.envrc @@ -0,0 +1 @@ +use flake diff --git a/Assignment1_211.tex b/Assignment1_211.tex new file mode 100644 index 0000000..7f14d13 --- /dev/null +++ b/Assignment1_211.tex @@ -0,0 +1,351 @@ +\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 \sqrt{17} 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 $y(t) = (x(t))^2$ + +\[ +y(t) = (x(t))^2 +\] + +\[ +y_1(t) = (\alpha x_1(t))^2 + (\beta x_2(t))^2 +\] + +\[ += \alpha^2 (x_1(t))^2 + \beta^2 (x_2(t))^2 +\] + +\[ +y_2(t) = \alpha (x_1(t))^2 + \beta (x_2(t))^2 +\] + +\[ +y_1(t) \ne y_2(t) +\] + +\[ +\implies \text{Not linear} +\] + + +\[ + y_1(t) &= \left(x(t - \Delta)\right)^2 \\ +\] + +\[ + y_2(t) &= y_1(t - \Delta) = \left(x(t - \Delta)\right)^2 \\ +\] + +\[ + y_1 &= y_2 \\ +\] + +\[ +\implies \text{Time invariant} +\] + \item $y(t) = x(t) sin(\Omega_0 t)$ + +\[ + y(t) &= x(t) \sin(\Omega_0 t) \\ +\] + +\[ + y_1(t) &= \left( \alpha \, x(t) \sin(\Omega_0 t) \right) + \left( \beta \, x(t) \sin(\Omega_0 t) \right) \\ +\] + +\[ + y_t(t) &= \alpha \left( x(t) \sin(\Omega_0 t) \right) + \beta \left( x(t) \sin(\Omega_0 t) \right) \\ +\] + +\[ + y_1 &= y_2 +\] + +\[ + \implies \text{linear} +\] + + + + +\[ + y_1(t) = &\coloneqq x(t - \Delta) \sin(\Omega_0 t - \Delta) \\ +\] + +\[ + y_2(t) &= y_1(t - \Delta) = x(t - \Delta) \sin(\Omega_0 t - \Delta) \\ +\] + +\[ + y_1 &= y_2 \\ +\] + +\[ + \implies \text{Time invariant} +\] +\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} diff --git a/flake.nix b/flake.nix new file mode 100644 index 0000000..b452254 --- /dev/null +++ b/flake.nix @@ -0,0 +1,22 @@ +{ + description = "LaTeX paper build environment"; + + inputs = { + nixpkgs.url = "github:NixOS/nixpkgs"; + }; + + outputs = + { self, nixpkgs }: + let + system = "x86_64-linux"; + pkgs = import nixpkgs { inherit system; }; + in + { + devShells.${system}.default = pkgs.mkShell { + name = "latex"; + buildInputs = with pkgs; [ + texliveFull + ]; + }; + }; +}