\documentclass{article}
\usepackage{amssymb, amsmath, geometry, multicol}
\usepackage[europeanresistors]{circuitikz}
\usetikzlibrary{arrows}
\setlength{\tabcolsep}{10pt} % Default value: 6pt
\renewcommand{\arraystretch}{1.5} % Default value: 1
\newcommand{\parspace}{\vspace{0.1cm}}

\linespread{1.25}
\geometry{tmargin=.7in, bmargin=.75in, lmargin=.75in, rmargin = .75in}

\begin{document}
\setlength{\parindent}{0mm}

\title{Electrocinétique}
\author{Aubin SIONVILLE}
\date{MPI Clemenceau - 2021-2023}
\maketitle

\begin{center}
\section*{Lois de Kirchhoff}
\end{center}
\begin{multicols}{2}

\begin{center}
\subsection*{Loi des Nœuds}
\end{center}
\[\boxed{\sum_{k=1}^{n} (\pm)i_k = 0}\]

\centering\subsection*{Loi des Mailles}
\[\boxed{\sum_{k=1}^n (\pm)u_k = 0}\]
\end{multicols}

\begin{center}
\section*{Diviseurs idéaux}
\end{center}
\begin{multicols}{2}
\subsection*{Diviseur idéal de tension}
\begin{multicols}{2}
\begin{circuitikz}
    % Deux résistances en série
    \draw (0,0) to [short, -, i=$i$] (0.5,0)
    to [R, l=$\underline{Z_1}$] (2,0)
    to [R, l=$\underline{Z_2}$] (4,0);

    \draw [->] (2,-0.5) to (1.25,-0.5) node[below] {$u_1$} to (0.5,-0.5);
    \draw [->] (3.75,-0.5) to (3,-0.5) node[below] {$u_2$} to (2.25,-0.5);
    \draw [->] (4, 0.75) to (2,0.75) node[above] {$u$} to (0.25, 0.75);
\end{circuitikz}

\[\boxed{\underline{u_1} = \frac{\underline{Z_1}}{\underline{Z_1}+\underline{Z_2}}\underline{u}}\]

\end{multicols}
\subsection*{Diviseur idéal de courant}
\begin{multicols}{2}
    \begin{circuitikz}
        % Deux résistances en parallèle
        \draw (0,0) to [short, -, i=$\underline{i}$] (1,0)
        to [short, -] (1,0.75)
        to [short, -, i=$\underline{i_1}$] (1.5,0.75)
        to [R, l=$\underline{Z_1}$] (3,0.75)
        to [short, -] (3.25,0.75)
        to [short, -] (3.25,0)
        to [short, -] (3.75,0);
        \draw (1,0) to [short, -] (1,0)
        to [short, -] (1,-0.75)
        to [short, -, i=$\underline{i_2}$] (1.5,-0.75)
        to [R, l=$\underline{Z_2}$] (3,-0.75)
        to [short, -] (3.25,-0.75)
        to [short, -] (3.25,0);

        \draw [->] (3, 0) to (2.25,0) node[above] {$u$} to (1.25, 0);
        
    \end{circuitikz}
    
    \[\boxed{\underline{i_1} = \frac{\frac{1}{\underline{Z_1}}}{\frac{1}{\underline{Z_1}}+\frac{1}{\underline{Z_2}}}\underline{i}}\]
\end{multicols}
\end{multicols}

\centering\section*{Dipôles électriques}
\begin{multicols}{2}
    \subsection*{Condensateur}
        \[\boxed{i = C\frac{\mathrm{d}u}{\mathrm{d}t}}
        \quad\quad\quad\quad
        \boxed{\mathcal{E}_C = \frac{1}{2}Cu^2}\]

    \subsection*{Bobine}
        \[\boxed{u = L\frac{\mathrm{d}i}{\mathrm{d}t}}
        \quad\quad\quad\quad
        \boxed{\mathcal{E}_L = \frac{1}{2}Li^2}\]

\end{multicols}

\vspace{0.5cm}

\centering\section*{Circuits RC, RL, RLC}
\begin{multicols}{3}
\subsection*{Circuit RC}
\begin{circuitikz}
    \draw (0,2) to [R, l=$R$] (3,2);
    \draw (3,2) to [C, l_=$C$] (3,0);
    \draw (3,0) -- (0,0);
    \draw[-triangle 45] (3.8,0) -- (3.8,2) node[right, midway] {$\underline{U}\,_s$};
    % utilisation de la librairie arrow
    \draw[-triangle 45] (-0.2,0) -- (-0.2,2) node[left, midway] {$\underline{U}\,_e$};
\end{circuitikz}

\subsection*{Circuit RL}
\begin{circuitikz}
    \draw (0,2) to [R, l=$R$] (3,2);
    \draw (3,2) to [L, l_=$L$] (3,0);
    \draw (3,0) -- (0,0);
    \draw[-triangle 45] (3.8,0) -- (3.8,2) node[right, midway] {$\underline{U}\,_s$};
    % utilisation de la librairie arrow
    \draw[-triangle 45] (-0.2,0) -- (-0.2,2) node[left, midway] {$\underline{U}\,_e$};
\end{circuitikz}

\subsection*{Circuit RLC}
\begin{circuitikz}
    % R and L on top, C on the right
    \draw (0,2) to [R, l=$R$] (2,2);
    \draw (2,2) to [L, l_=$L$] (4,2);
    \draw (4,2) to [C, l_=$C$] (4,0);
    \draw (4,0) -- (0,0);
    \draw[-triangle 45] (4.8,0) -- (4.8,2) node[right, midway] {$\underline{U}\,_s$};
    % utilisation de la librairie arrow
    \draw[-triangle 45] (-0.2,0) -- (-0.2,2) node[left, midway] {$\underline{U}\,_e$};
\end{circuitikz}
\end{multicols}

\newpage

\section*{Mise en série et en parallèle de composants}

\subsection*{Résistances}

\begin{multicols}{2}
    \begin{circuitikz}
        \draw (0,1.4) to [short] (0,1.4);
        \draw (0,0) to [short] (1,0)
        to [R, l_=$R_1$] (2,0)
        to [short] (3,0)
        to [R, l_=$R_2$] (4,0)
        to [short] (5,0);

    \node at (2.5,-2.75) {$\boxed{\displaystyle R_{\mathrm{eq}}=R_1+R_2}$};

    \end{circuitikz}

    \begin{circuitikz}
        \draw (0,0) to [short] (1,0)
        to [short] (1,1)
        to [short] (2,1)
        to [R, l_=$R_1$] (3,1)
        to [short] (4,1);
        \draw (1,0) to [short] (1,-1)
        to [short] (2,-1)
        to [R, l_=$R_2$] (3,-1)
        to [short] (4,-1);
    
    \node at (2.5,-2.75) {$\boxed{\displaystyle \frac{1}{R_\mathrm{eq}} = \frac{1}{R_1} + \frac{1}{R_2}}$};

    \end{circuitikz}
\end{multicols}

\subsection*{Condensateurs}

\begin{multicols}{2}
    \begin{circuitikz}
        \draw (0,1.4) to [short] (0,1.4);
        \draw (0,0) to [short] (1,0)
        to [C, l_=$C_1$] (2,0)
        to [short] (3,0)
        to [C, l_=$C_2$] (4,0)
        to [short] (5,0);

    \node at (2.5,-2.75) {$\boxed{\displaystyle \frac{1}{C_\mathrm{eq}} = \frac{1}{C_1} + \frac{1}{C_2}}$};

    \end{circuitikz}

    \begin{circuitikz}
        \draw (0,0) to [short] (1,0)
        to [short] (1,1)
        to [short] (2,1)
        to [C, l_=$C_1$] (3,1)
        to [short] (4,1);
        \draw (1,0) to [short] (1,-1)
        to [short] (2,-1)
        to [C, l_=$C_2$] (3,-1)
        to [short] (4,-1);
    
    \node at (2.5,-2.75) {$\boxed{\displaystyle C_{\mathrm{eq}}=C_1+C_2}$};

    \end{circuitikz}
\end{multicols}

\subsection*{Bobines}

\begin{multicols}{2}
    \begin{circuitikz}
        \draw (0,1.4) to [short] (0,1.4);
        \draw (0,0) to [short] (1,0)
        to [L, l_=$L_1$] (2,0)
        to [short] (3,0)
        to [L, l_=$L_2$] (4,0)
        to [short] (5,0);
    
    \node at (2.5,-2.75) {$\boxed{\displaystyle L_{\mathrm{eq}}=L_1+L_2}$};

    \end{circuitikz}

    \begin{circuitikz}
        \draw (0,0) to [short] (1,0)
        to [short] (1,1)
        to [short] (2,1)
        to [L, l_=$L_1$] (3,1)
        to [short] (4,1);
        \draw (1,0) to [short] (1,-1)
        to [short] (2,-1)
        to [L, l_=$L_2$] (3,-1)
        to [short] (4,-1);
    
    \node at (2.5,-2.75) {$\boxed{\displaystyle \frac{1}{L_\mathrm{eq}} = \frac{1}{L_1} + \frac{1}{L_2}}$};

    \end{circuitikz}
\end{multicols}

\newpage

\begin{center}
\section*{Filtrage à l'ordre 1}
\begin{multicols}{2}
\subsection*{Filtre passe-bas}
\begin{circuitikz}
    \draw (0,2) to [R, l=$R$] (3,2);
    \draw (3,2) to [C, l_=$C$] (3,0);
    \draw (3,0) -- (0,0);
    \draw[-triangle 45] (3.8,0) -- (3.8,2) node[right, midway] {$\underline{U}\,_s$};
    % utilisation de la librairie arrow
    \draw[-triangle 45] (-0.2,0) -- (-0.2,2) node[left, midway] {$\underline{U}\,_e$};
\end{circuitikz}

\parspace

Equation différentielle : $\displaystyle\boxed{\frac{\mathrm{d}u}{\mathrm{d}t} + \frac{1}{\tau}u = e(t)}\ (\tau =RC)$

\parspace

Solutions : $\displaystyle \boxed{u(t) = \lambda \mathrm{e}^{-t/\tau}}$

\parspace

Fonction de transfert : $\displaystyle \boxed{\underline{H}(j\omega) = H_0\frac{1}{1+j\frac{\omega}{\omega_0}}}$

\subsection*{Filtre passe-haut}
\begin{circuitikz}
    \draw (0,2) to [R, l=$R$] (3,2);
    \draw (3,2) to [L, l_=$L$] (3,0);
    \draw (3,0) -- (0,0);
    \draw[-triangle 45] (3.8,0) -- (3.8,2) node[right, midway] {$\underline{U}\,_s$};
    % utilisation de la librairie arrow
    \draw[-triangle 45] (-0.2,0) -- (-0.2,2) node[left, midway] {$\underline{U}\,_e$};
\end{circuitikz}

\parspace

Equation différentielle : $\displaystyle\boxed{\frac{\mathrm{d}u}{\mathrm{d}t} + \frac{1}{\tau}u = e(t)}\ \Bigg(\tau = \frac{L}{R}\Bigg)$

\parspace

Solutions : $\displaystyle\boxed{u(t) = \lambda \mathrm{e}^{-t/\tau}}$

\parspace

Fonction de transfert : $\displaystyle\boxed{\underline{H}(j\omega) = H_0\frac{j \frac{\omega}{\omega_0}}{1+j\frac{\omega}{\omega_0}}}$

\end{multicols}

\section*{Filtrage à l'ordre 2}
\textit{NB : Attention à la résonance possible dans le diagramme de Bode}
    \begin{center}
        Equation différentielle : \[\boxed{\frac{\mathrm{d}^2u}{\mathrm{d}t^2} + \frac{\omega_0}{Q} \frac{\mathrm{d}u}{\mathrm{d}t} + \omega_0^2 u = \omega_0^2 e(t)}\ \Bigg(\omega_0 = \frac{1}{\sqrt{LC}}, \displaystyle Q = \frac{1}{R}\sqrt{\frac{L}{C}}\Bigg)\]
    \end{center}
    \begin{multicols}{2}
        \subsection*{Filtre passe-bas}
        \begin{circuitikz}
            % R and L on top, C on the right
            \draw (0,2) to [R, l=$R$] (2,2);
            \draw (2,2) to [L, l_=$L$] (4,2);
            \draw (4,2) to [C, l_=$C$] (4,0);
            \draw (4,0) -- (0,0);
            \draw[-triangle 45] (4.8,0) -- (4.8,2) node[right, midway] {$\underline{U}\,_s$};
            % utilisation de la librairie arrow
            \draw[-triangle 45] (-0.2,0) -- (-0.2,2) node[left, midway] {$\underline{U}\,_e$};
        \end{circuitikz}
        
        Fonction de transfert : \[\boxed{\underline{H}(j\omega) = H_0\frac{1}{1+\frac{1}{Q}j\frac{\omega}{\omega_0}-\Big(\frac{\omega}{\omega_0}\Big)^2}}\]
        
        \subsection*{Filtre passe-haut}
        \begin{circuitikz}
            % R and L on top, C on the right
            \draw (0,2) to [R, l=$R$] (2,2);
            \draw (2,2) to [C, l_=$L$] (4,2);
            \draw (4,2) to [L, l_=$C$] (4,0);
            \draw (4,0) -- (0,0);
            \draw[-triangle 45] (4.8,0) -- (4.8,2) node[right, midway] {$\underline{U}\,_s$};
            % utilisation de la librairie arrow
            \draw[-triangle 45] (-0.2,0) -- (-0.2,2) node[left, midway] {$\underline{U}\,_e$};
        \end{circuitikz}
        
        Fonction de transfert : \[\boxed{\underline{H}(j\omega) = H_0\frac{-\Big(\frac{\omega}{\omega_0}\Big)^2}{1+\frac{1}{Q}j\frac{\omega}{\omega_0}-\Big(\frac{\omega}{\omega_0}\Big)^2}}\]
    \end{multicols}
    
    \subsection*{Filtre passe-bande}
    \begin{multicols}{2}
        \begin{circuitikz}

            % R and L on top, C on the right
            \draw (0,2) to [C, l=$R$] (2,2);
            \draw (2,2) to [L, l_=$L$] (4,2);
            \draw (4,2) to [R, l_=$C$] (4,0);
            \draw (4,0) -- (0,0);
            \draw[-triangle 45] (4.8,0) -- (4.8,2) node[right, midway] {$\underline{U}\,_s$};
            % utilisation de la librairie arrow
            \draw[-triangle 45] (-0.2,0) -- (-0.2,2) node[left, midway] {$\underline{U}\,_e$};
        \end{circuitikz}
        
        Fonction de transfert : \[\boxed{\underline{H}(j\omega) = H_0\frac{\frac{1}{Q}j\frac{\omega}{\omega_0}}{1+\frac{1}{Q}j\frac{\omega}{\omega_0}-\Big(\frac{\omega}{\omega_0}\Big)^2}}\]
    \end{multicols}
\end{center}

\newpage

\section*{Régime libre des filtres à l'ordre 2}
\begin{center}

Equation : $\displaystyle \frac{\mathrm{d}^2u}{\mathrm{d}t^2} + \frac{\omega_0}{Q} \frac{\mathrm{d}u}{\mathrm{d}t} + \omega_0^2 u = 0$

\parspace
D'où : $\displaystyle z^2 + \frac{\omega_0}{Q}z + \omega_0^2 = 0$

    
    \vspace{1cm}
    %Tableau à 3 colonnes\renewcommand*{\arraystretch}{1.4}}
    \begin{tabular}{|c|c|c|}
        \hline
        \textbf{Régime apériodique} & \textbf{Régime critique} & \textbf{Régime Pseudo-périodique}\\
        \hline
        \rule{0pt}{4ex}  \text{Racines : } & \text{Racine double } & \text{Racines : }\\
        \text{Deux racines réelles négatives} & \text{Une racine réelle négative} & \text{Deux racines complexes conjugées}\\
        \rule{0pt}{4ex} $\displaystyle r = \frac{1}{2}\Big(-\frac{\omega_0}{Q} \pm \sqrt{\Delta}\Big)$ & $\displaystyle r = -\frac{\omega_0}{2Q}$ & $\displaystyle r = \frac{1}{2}\Big(-\frac{\omega_0}{Q} \pm j\sqrt{|\Delta|}\Big)$\\
        \hline
        \multicolumn{3}{|c|}{Les racines sont homogènes à des temps}\\
        \hline
        \rule{0pt}{4ex} On note $\displaystyle r_1 = -\frac{1}{\tau_1}, r_2 = -\frac{1}{\tau_2}$ & On note $\displaystyle r = -\frac{1}{\tau} = -\frac{\omega_0}{2Q}$ & On note $\displaystyle r = -\frac{1}{\tau}\pm j \omega_p$\\
        \rule{0pt}{4ex} $\displaystyle s(t) = A\mathrm{e}^{-t/\tau_1} + B\mathrm{e}^{-t/\tau_2}$ & $\displaystyle s(t) = (At+B)\mathrm{e}^{-t/\tau}$ & $\displaystyle s(t) = (A\cos(\omega_pt) + B\sin(\omega_pt))\mathrm{e}^{-t/\tau}$\\
        \hline
    \end{tabular}

\end{center}

\vspace{1cm}

\begin{center}
    \section*{Filtrage sur un signal complet}

    \subsection*{Décomposition en Série de Fourier}
    \[\boxed{v(t) = V_0 + \sum_{n=1}^{+\infty}C_n\cos(n\omega_e t + \varphi_n)}\]
    $V_0$ : valeur moyenne (composante continue), $\displaystyle V_0 \triangleq \frac{1}{T}\int_{t_0}^{t_0+T}v(t)\mathrm{d}t$

    \parspace
    \parspace

    Pour filtrer le signal, on applique le filtre à chaque composante de Fourier.\\
    Stratégie : déssiner le diagramme de Bode du filtre par dessus le spectre du signal
\end{center}

\section*{Résonance}
\subsection*{Bande passante}
Bande passante : Intervalle vérifiant $\displaystyle\boxed{I(\omega) \geq \frac{I_\text{max}}{\sqrt{2}}}$

\begin{multicols}{2}

    \subsection*{Résonance en intensité}
    \[\boxed{\underline{I}(\omega) = I\exp(j\varphi) = \frac{I_\text{max}}{1+jQ\Big(\frac{\omega}{\omega_0} - \frac{\omega_0}{\omega}\Big)}}\]
    $I$ max à la puslation de résonance : $\boxed{\omega_r = \omega_0}$\\
    Existence pour tout $Q$\\
    Acuité : $\displaystyle \frac{\omega_0}{\Delta\omega}$
    Bande passante faible $\iff$ acuité élevée

    \subsection*{Résonance en tension}
    \[\boxed{\underline{U}(\omega) = U\exp(j\varphi) = \frac{A_\text{m}}{1+\frac{j\omega}{Q\omega_0}-\Big(\frac{\omega}{\omega_0}\Big)^2}}\]
    Existence pour $\boxed{\displaystyle Q > \frac{1}{\sqrt{2}}}$\\
    Pulsation de résonance : $\boxed{\displaystyle \omega_r = \omega_0\sqrt{1-\frac{1}{2Q^2}}}$
\end{multicols}

\end{document}