\documentclass[a4paper,french,11pt]{article}

\title{
Théorie du signal --- TD1
\\ \large Décomposition en Série de Fourier
}
\author{}
\date{Dernière compilation~: \today{} à \currenttime}

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

\begin{document}

\maketitle

\section{I}

\subsection{Signal carré}

    Soit le signal carré $x(t)$, $T_0$-périodique tel que~:
    \begin{align*}
        x(t) =
        \left\{
        \begin{array}{ll}
            -1 \quad \forall\, t \in [-\frac{T_0}{2};0] \\
            1 \quad \forall\, t \in [0;\frac{T_0}{2}] \\
        \end{array}
        \right.
    \end{align*}

    \subsubsection{Tracer le signal $x(t)$}

        \begin{center}
        \begin{tikzpicture}

            \draw[help lines, dashed] (-7,-2) grid (7,2);
            \draw[-latex] (-7,0) -- (7,0) node[below]{$t$};
            \draw[-latex] (0,-2) -- (0,2) node[left]{$x(t)$};

            \foreach \i in {-6, -4, -2, 0, 2, 4, 6}{
                \draw[very thick, teal]
                    (\i-1,1) -- (\i-1,-1)
                    plot[domain=\i-1:\i]({\x}, {-1})
                    (\i,1) -- (\i,-1)
                    plot[domain=\i:\i+1]({\x}, {1})
                    ;
            }

        \end{tikzpicture}
        \end{center}

    \subsubsection{Calculer les coefficients de Fourier réels $a_0, a_n, b_n$ du signal $x(t)$}

        $x$ est impaire donc $a_0 = a_n = 0$.

        \begin{tabularx}{\linewidth}{XX}
            {
            \begin{equation*}
                \text{formule~:} \quad b_n = \frac{2}{T_0} \int_{(T_0)} x(t) \sin(n\omega_0 t) \dif t
            \end{equation*}
            \vspace{4cm}
            \begin{equation*}
                x(t) = \sum_{n=1}^{+\infty} \frac{2}{n\pi}(1 - (-1)^n) \sin(n\omega_0 t)
            \end{equation*}
            } &
            {
            \begin{align*}
                b_n &= \frac{4}{T_0} \int_0^{T_0/2} 1 \sin(n\omega_0 t) \dif t \\
                    &= \frac{4}{T_0} \left[\frac{-\cos(n\omega_0 t)}{n\omega_0}\right]_0^{T_0/2} \\
                    &= \frac{4}{T_0} \left(\frac{-\cos(n2\pi f_0 \frac{T_0}{2}) + 1}{n2\pi f_0}\right) \\
                    &= 2 \left(\frac{-\cos(n\pi) + 1}{n\pi}\right) \\
                    &= \frac{2}{n\pi} (-\cos(n\pi) + 1) \\
                    &= \frac{2}{n\pi} (-(-1)^n + 1) \\
                b_n &= \boxed{\frac{2}{n\pi} (1 -(-1)^n)}
            \end{align*}
            } \\
        \end{tabularx}


    \subsubsection{Tracer la DSP du signal $x(t)$}

        \begin{align*}
            |c_n|^2 = |-j\frac{1}{2}b_n|^2 = \frac{1}{4}b_n^2 = \frac{1}{4}(\frac{2}{n\pi})^2(1 - (-1)^n)^2
        \end{align*}
        \begin{align*}
            S_x(f) = \sum_{-\infty}^{+\infty} |c_n|^2 \implies
            \left\{
            \begin{array}{l}
                \text{si } n \text{ est pair, } b_n = 0 \implies |c_n|^2 = 0 \\
                \text{si } n \text{ est impair, } b_n = \frac{4}{n\pi} \implies |c_n|^2 = \frac{4}{(n\pi)^2} \\
            \end{array}
            \right.
        \end{align*}

\subsection{Signal en dent de scie}

    Soit le signal $x(t)$, $T_0$-périodique tel que~:
    \begin{align*}
        x(t) = A \times \frac{1}{T_0}t \quad \forall\, t \in [0;T_0] \\
    \end{align*}

    \subsubsection{Tracer le signal $x(t)$}

        \begin{center}
        \begin{tikzpicture}

            \draw[help lines, dashed] (-8,-2) grid (8,2);
            \draw[-latex] (-8,0) -- (8,0) node[below]{$t$};
            \draw[-latex] (0,-2) -- (0,2) node[left]{$x(t)$};

            \foreach \i in {-8, -6, -4, -2, 0, 2, 4, 6}{
                \draw[very thick, teal]
                    plot[domain=\i:\i+2]({\x}, {1.7*(\x-\i)/2})
                    (\i+2,0) -- (\i+2,1.7)
                    ;
            }

        \end{tikzpicture}
        \end{center}

\subsection{Signal porte}

    Soit le signal $x(t)$, de largeur $T>0$ tel que~:
    \begin{align*}
        x(t) = A\Pi_r(t) =
        \left\{
        \begin{array}{ll}
            A \quad \forall\, t \in [-\frac{T}{2};\frac{T}{2}] \\
            0 \text{ sinon} \\
        \end{array}
        \right.
    \end{align*}

    \subsubsection{Calculer $X(f)$, la TF de $x(t)$. Représenter la DSE de $x(t)$}
        \begin{equation*}
            X(f) = \int_{\mathbb{R}} x(t) e^{-j2\pi ft} \dif t
        \end{equation*}
        \begin{align*}
            X(f) &= \int_{-T/2}^{T/2} A e^{-j2\pi ft} \dif t \\
                 &= A \left[\frac{e^{-j2\pi ft}}{-j2\pi f}\right]_{-T/2}^{T/2} \\
                 &= A \left(\frac{-e^{-j\pi fT} - e^{j\pi fT}}{-j2\pi f}\right) \\
                 &= A \left(\frac{e^{j\pi fT} - e^{-j\pi fT}}{-j2\pi f}\right) \\
                 &= A \frac{\sin(\pi fT)}{\pi f \color{red}{T}}\color{red}{T} \\
            X(f) &= \boxed{AT \text{sinc}(fT)}
        \end{align*}

        \paragraph{DSE}

            \begin{align*}
                S_x(f) &= |X(f)|^2 \\
                       &= A^2T^2\text{sinc}^2(fT)
            \end{align*}

            \begin{align*}
                \text{Rappel~: } &\text{sinc}(0) = 1 \\
                                 &\text{sinc}(fT) = \frac{\sin(\pi fT)}{\pi fT}
            \end{align*}

            \begin{center}
            \begin{tikzpicture}
                %TODO plot this
                \draw[help lines, dashed] (-7,-2) grid (7,2);
                \draw[-latex] (-7,0) -- (7,0) node[below]{$t$};
                \draw[-latex] (0,-2) -- (0,2) node[left]{$x(t)$};

            \end{tikzpicture}
            \end{center}

    \subsubsection{Que vaut $\int_{\mathbb{R}} S_x(f) \dif f$~?}

        Calculer l'intégrale dans le domaine fréquentiel serait compliqué.
        Mais d'après le théorême de Parseval~:

        \begin{align*}
            E = \int_{\mathbb{R}} S_x(f) \dif f
            &= \int_{\mathbb{R}} |x(t)|^2 \dif t \\
            &= \int_{-\pi/2}^{\pi/2} A^2 \dif t \\
            &= A^2 [t]_{-T/2}^{T/2} \\
            &= A^2 T
        \end{align*}

\section{II}

\end{document}
