efrei/theorie-signal/exercices/tp1.tex

\documentclass[a4paper,french,12pt]{article}

\title{
Théorie du signal --- TP1
\\ \large Décomposition en Série de Fourier
}
\author{Alexandre CHEN et Tunui FRANKEN}
\date{Dernière compilation~: \today{} à \currenttime}

\usepackage{../../cours}
\usepackage{enumitem}
\usepackage{xfrac}
\usepackage{tikz}

\begin{document}

\maketitle

\section{Partie théorique}

\subsection{Coefficients de Fourier}
    Donner l'expression de la DSF réelle des signaux suivants $T_0$-périodiques~:
    \begin{align*}
        x_1(t) = 1 - \frac{2}{T_0} |t| \quad \forall t \in \left[-\frac{T_0}{2}; \frac{T_0}{2}\right]
    \end{align*}
    avec $T_0 = 0.5s$.

    $x_1$ est paire, donc les $b_n$ sont nuls.
    \begin{align*}
        a_0 &= \frac{1}{T_0} \int_{-\sfrac{T_0}{2}}^{\sfrac{T_0}{2}} 1 - \frac{2}{T_0}|t| \dif t \\
            &= \frac{2}{T_0} \int_0^{\sfrac{T_0}{2}} 1 \dif t - \frac{2}{T_0} \int_0^{\sfrac{T_0}{2}} \frac{2}{T_0}|t| \dif t \\
            &= \frac{2}{T_0} [t]_0^{\sfrac{T_0}{2}} - \frac{4}{T_0^2} \left[\frac{t^2}{2}\right]_0^{\sfrac{T_0}{2}} \\
            &= 1 - \frac{4T_0^2}{8T_0^2} \\ \\
        a_0 &= \frac{1}{2}
    \end{align*}

    \begin{align*}
        a_n &= \frac{2}{T_0} \int_{(T_0)} \left(1-\frac{2}{T_0}|t|\right)\cos(n\omega_0 t) \dif t \\
        \text{IPP avec }
            &\left\{
        \begin{array}{l}
            u = 1 - \frac{2}{T_0}|t| \quad\implies
            u' = -\frac{2}{T_0} \\ \\
            v' = \cos(n\omega_0 t) \quad\implies
            v = \frac{\sin(n\omega_0 t)}{n\omega_0}
            \\
        \end{array}
        \right.\\
        a_n &= \frac{4}{T_0}\left(\left[
            1 - \frac{2}{T_0}|t|
            \frac{\sin(n\omega_0 t)}{n\omega_0}
        \right]_0^{T_0/2}
        - \int_0^{\sfrac{T_0}{2}} -\frac{2}{T_0}\frac{\sin(n\omega_0 t)}{n\omega_0}\dif t \right) \\
            &= \frac{4}{T_0}\left(
                \left[
                    \left(1 - \frac{2T_0}{T_0 2}\frac{\sin(n\pi)}{\frac{n2\pi}{T_0}}\right) - (1 - 0)
                \right]
                + \frac{2}{2n\pi}\left[
                    \left(
                        \frac{-\cos(n\pi)}{\frac{2n\pi}{T_0}}
                    \right) - \frac{1}{\frac{2n\pi}{T_0}}
                \right]
            \right) \\
            &= \frac{4}{T_0}\frac{2}{2n\pi}\left(
                \frac{-\cos(n\pi) - 1}{\frac{2n\pi}{T_0}}
            \right) \\
            &= \frac{2}{(n\pi)^2}(-\cos(n\pi) + 1)
    \end{align*}

    \begin{align*}
        S_x(t) &= a_0 + \sum_{n=0}^{+\infty} a_n\cos(n\omega_0 t) \\
               &= \frac{1}{2} + \sum_{n=0}^{+\infty} \frac{2}{(n\pi)^2}(-\cos(n\pi) + 1) \cos(n\omega_0 t)
    \end{align*}

    \begin{align*}
        x_2(t) =
        \left\{
        \begin{array}{l}
            1 \quad \forall t \in \left[-\frac{T_0 r}{2}; \frac{T_0 r}{2}\right] \\
            0 \quad \text{sinon}
        \end{array}
        \right.
    \end{align*}
    avec $r$ le rapport cyclique tel que $r < 1$ et $T_0 = 0.5s$.

    $x_2$ est paire, donc les $b_n$ sont nuls.
    \begin{align*}
        a_0 &= \frac{1}{T_0}\int_{(T_0)}1\dif t \\
            &= \frac{2}{T_0}\int_0^{\sfrac{rT_0}{2}} 1\dif t \\
            &= \frac{2}{T_0}[t]_0^{\sfrac{rT_0}{2}} \\
            &= \frac{2}{T_0}\frac{rT_0}{2} \\
        a_0 &= r
    \end{align*}

    \begin{align*}
        a_n &= \frac{2}{T_0}\int_{(T_0)}1\cdot\cos(n\omega_0 t)\dif t \\
            &= \frac{4}{T_0}\left[
                \frac{\sin(n\omega_0 t)}{n\omega_0}
            \right]_0^{\sfrac{rT_0}{2}} \\
            &= \frac{4}{T_0}\frac{\sin(n\pi r)}{n2\pi/T_0} \\
            &= \frac{4\sin(n\pi r)}{n2\pi} \\
        a_n &= \frac{2\sin(n\pi r)}{n\pi}
    \end{align*}

    \begin{align*}
        S_x(t) &= a_0 + \sum_{n=0}^{+\infty} a_n\cos(n\omega_0 t) \\
               &= r + \sum_{n=0}^{+\infty} \frac{2\sin(n\pi r)}{n\pi} \cos(n\omega_0 t)
    \end{align*}

\subsection{Densité Spectrale de Puissance}

    \subsubsection{Pour $x_1$}
        \begin{equation*}
            c_0 = a_0 = \frac{1}{2}
        \end{equation*}
        \begin{align*}
            c_n &= \frac{1}{2}\frac{2}{(n\pi)^2}(-\cos(n\pi) + 1) \\
                &= \frac{-\cos{n\pi} + 1}{(n\pi)^2}
        \end{align*}
        \begin{align*}
            S_{x_1}(f) &= \sum_{n=-\infty}^{+\infty} \left|\frac{-\cos(n\pi) + 1}{(n\pi)^2}\right|^2 \\
                   &= \sum_{n=-\infty}^{+\infty} \left|\frac{-(-1)^n + 1}{(n\pi)^2}\right|^2 \\
                   &=
                   \left\{
                   \begin{array}{l}
                       0 \text{ pour $n$ pair} \\
                       \frac{4}{(n\pi)^2} \text{ pour $n$ impair} \\
                   \end{array}
                   \right.
        \end{align*}

        \begin{center}
        \begin{tikzpicture}
            \draw[help lines, dashed] (-6,-1) grid (6,3);
            \draw[-latex] (-6,0) -- (6,0) node[below]{$f$};
            \draw[-latex] (0,-1) -- (0,3) node[left]{$S_x(f)$};
            % j'ajoute 2 pour la visibilité
            \draw[-latex,very thick, teal] (-5, 0) -- (-5, 2+0.016211389382774045);
            \draw[-latex,very thick, teal] (-3, 0) -- (-3, 2+0.04503163717437234);
            \draw[-latex,very thick, teal] (-1, 0) -- (-1, 2+0.4052847345693511);
            \draw[-latex,very thick, teal] (0, 0) -- (0, 2+0.5);
            \draw[-latex,very thick, teal] (1, 0) -- (1, 2+0.4052847345693511);
            \draw[-latex,very thick, teal] (3, 0) -- (3, 2+0.04503163717437234);
            \draw[-latex,very thick, teal] (5, 0) -- (5, 2+0.016211389382774045);
        \end{tikzpicture}
        \end{center}

    \subsubsection{Pour $x_2$}
        \begin{equation*}
            c_0 = a_0 = r
        \end{equation*}
        \begin{align*}
            c_n &= \frac{1}{2}\frac{2\sin(n\pi r)}{n\pi} \\
                &= \frac{\sin(n\pi r)}{n\pi}
        \end{align*}
        \begin{align*}
            S_{x_2}(f) &= \sum_{n=-\infty}^{+\infty} \left|\frac{\sin(n\pi r)}{n\pi}\right|^2 \\
            \text{pour } r=\frac{1}{2} &\implies
            \sum_{n=-\infty}^{+\infty} \left|\frac{\sin(\frac{n\pi}{2})}{n\pi}\right|^2 \\
                                       &= \sum_{n=-\infty}^{+\infty}\frac{1}{(n\pi)^2} \\
            \text{pour } r=\frac{1}{3} &\implies
            \sum_{n=-\infty}^{+\infty} \left|\frac{\sin(\frac{n\pi}{3})}{n\pi}\right|^2 \\
            \text{pour } r=\frac{1}{4} &\implies
            \sum_{n=-\infty}^{+\infty} \left|\frac{\sin(\frac{n\pi}{4})}{n\pi}\right|^2 \\
        \end{align*}

        \begin{center}
        \begin{tikzpicture}
            \draw[help lines, dashed] (-6,-1) grid (6,3);
            \draw[-latex] (-6,0) -- (6,0) node[below]{$f$};
            \draw[-latex] (0,-1) -- (0,3) node[left]{$S_x(f)$};

            \draw[-latex,very thick, teal] (-6, 0) -- (-6, 3.700508937884072e-32);
            \draw[-latex,very thick, teal] (-5, 0) -- (-5, 0.3947841760435743);
            \draw[-latex,very thick, teal] (-4, 0) -- (-4, 3.700508937884072e-32);
            \draw[-latex,very thick, teal] (-3, 0) -- (-3, 1.0966227112321507);
            \draw[-latex,very thick, teal] (-2, 0) -- (-2, 3.700508937884072e-32);
            \draw[-latex,very thick, teal] (-1, 0) -- (-1, 9.869604401089358);
            \draw[-latex,very thick, teal] (0, 0) -- (0, 0.5);
            \draw[-latex,very thick, teal] (1, 0) -- (1, 9.869604401089358);
            \draw[-latex,very thick, teal] (2, 0) -- (2, 3.700508937884072e-32);
            \draw[-latex,very thick, teal] (3, 0) -- (3, 1.0966227112321507);
            \draw[-latex,very thick, teal] (4, 0) -- (4, 3.700508937884072e-32);
            \draw[-latex,very thick, teal] (5, 0) -- (5, 0.3947841760435743);
            \draw[-latex,very thick, teal] (6, 0) -- (6, 3.700508937884072e-32);
        \end{tikzpicture}
        \end{center}
\end{document}