diff --git a/latex/contents.tex b/latex/contents.tex index 659242c..049c5c2 100644 --- a/latex/contents.tex +++ b/latex/contents.tex @@ -1,45 +1,31 @@ -\psection{Introduction} - -Blabla - -\newpage - \section{Modelisation} -\begin{equation*} - \dot{x_{1}}, \dot{x_{2}} = f(e_{1},e_{2},q) -\end{equation*} - -\begin{equation*} +\begin{align*} + \dot{x_{1}}, \dot{x_{2}} = f(e_{1},e_{2},q) & & \left\{ \begin{aligned} x_{1}(\zeta,t) &= \frac{\partial^{2}\omega}{\partial\varepsilon^{2}}\\ x_{2}(\zeta,t) &= \rho(\zeta)\frac{\partial\omega}{dt} - \end{aligned} \right. -\end{equation*} - -\begin{equation*} + \end{aligned} \right. & & \left\{ \begin{aligned} e_{1}(\zeta,t) &= EI(\zeta)x_{1}\\ e_{2}(\zeta,t) &= \frac{1}{\rho(\zeta)}x_{2} = \frac{\partial\omega}{\partial t} \end{aligned} \right. -\end{equation*} +\end{align*} \subsection{Question 1} -\begin{equation*} +\begin{align} \left\{\begin{aligned} \dot{x_{1}} &= \frac{\partial x_{1}}{\partial t} = \frac{\partial}{\partial t} (\frac{\partial^2\omega}{\partial \zeta^2}) = \frac{\partial^2}{\partial \zeta^2}(\frac{\partial^2\omega}{\partial t})\\ \dot{x_{2}} &= \rho(\zeta)\frac{\partial^2\omega}{\partial t} = \underbrace{-\frac{\partial^{2}}{\partial \zeta^2}(EI(\zeta)\frac{\partial^2\omega}{\partial\zeta^2})-q(\zeta,t)}_\text{EDP} - \end{aligned}\right. -\end{equation*} - -\begin{equation} - \Rightarrow\boxed{\left\{\begin{aligned} - \dot{x_{1}} &= \frac{\partial^2 e_2}{\partial\zeta^2} \\ - \dot{x_{2}} &= -\frac{\partial^2 e_1}{\partial\zeta^2}-q(\zeta,t) - \end{aligned}\right. } -\end{equation} + \end{aligned}\right. & & + \Rightarrow & & + \boxed{\left\{\begin{aligned} + \dot{x_{1}} &= \frac{\partial^2 e_2}{\partial\zeta^2} \\ + \dot{x_{2}} &= -\frac{\partial^2 e_1}{\partial\zeta^2}-q(\zeta,t) + \end{aligned}\right. } +\end{align} \subsection{Question 2} @@ -58,48 +44,25 @@ Blabla En utilisant la première ligne de l'équation 1, on trouve : \begin{equation*} - \int\phi(\zeta)d\zeta\times\phi^T\dot{x_{1d}} = \int\phi(\zeta)d\zeta\times\frac{\partial^2}{\partial\zeta^2} = \ddot{\phi}(\zeta)^Te_{2*d} + \int\phi(\zeta)d\zeta\times\phi^T\dot{x_{1d}} = \int\phi(\zeta)d\zeta\times\frac{\partial^2}{\partial\zeta^2} = \ddot{\phi}(\zeta)^Te_{2d} \end{equation*} \begin{equation} \underbrace{\int_{0}^{L}\phi(\zeta)\phi(\zeta)^Td\zeta}_\text{E}\times\dot{x_{1d}} = \left(\int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)d\zeta\right)e_{2d} \end{equation} -Intégration par partie (IPP) : - -\begin{equation*} - \int_{0}^{L}u\cdot \dot{v} d\zeta = [u\cdot v]_{0}^{L} - \int_{0}^{L}\dot{u}\cdot v d\zeta -\end{equation*} - -Dans notre cas : - -\begin{align*} - \left\{\begin{aligned} - u&=\phi(\zeta) \\ - \dot{v} &= \ddot{\phi}(\zeta)^T - \end{aligned}\right. & - \Rightarrow\left\{\begin{aligned} - \dot{u} &= \dot{\phi}(\zeta) \\ - v &= \dot{\phi}(\zeta)^T - \end{aligned}\right. -\end{align*} - -On applique plusieurs fois : +On applique plusieurs fois de l’intégration par partie (IPP) : \begin{equation*} \begin{aligned} \int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)^T d\zeta &= [\phi(\zeta)\dot{\phi}(\zeta)^T]_{0}^{L} - \int_{0}^{L}\dot{\phi}(\zeta)\dot{\phi}(\zeta)^T d\zeta \\ - &= \phi(L)\dot{\phi}(L)^T - \underbrace{\phi(0)\dot{\phi}(0)^T }_\text{=0} - \int_{0}^{L}\dot{\phi}(\zeta)\dot{\phi}(\zeta)^T d\zeta \\ - &= \phi(L)\dot{\phi}(L)^T - \dot{\phi}(L)\phi(L)^T + \underbrace{\dot{\phi(0)}\phi(0)^T }_\text{=0} + \int_{0}^{L} \ddot{\phi}(\zeta)\phi(\zeta)^T d\zeta \\ + &= \phi(L)\dot{\phi}(L)^T - \underbrace{\phi(0)\dot{\phi}(0)^T }_\text{=0} - \dot{\phi}(L)\phi(L)^T + \underbrace{\dot{\phi(0)}\phi(0)^T }_\text{=0} + \int_{0}^{L} \ddot{\phi}(\zeta)\phi(\zeta)^T d\zeta \\ &= \phi(L)\dot{\phi}(L)^T - \dot{\phi}(L)\phi(L)^T + \int_{0}^{L} \ddot{\phi}(\zeta)\phi(\zeta)^T d\zeta = D \end{aligned} \end{equation*} -On continue avec ça en remplaçant dans l'équation 2 : +On continue avec ça en remplaçant dans l'équation 2 : $\Rightarrow \boxed{E\dot{x}_{1d} = De_{2d}}$ \\ -\begin{equation*} - \Rightarrow \boxed{E\dot{x}_{1d} = De_{2d}} -\end{equation*} Puis on a dans la deuxième ligne de l'équation 1 : @@ -114,47 +77,35 @@ Puis on trouve D : \begin{equation*} \begin{aligned} - D &= \phi(L)\dot{\phi}(L)^T - \phi(L)\dot{\phi}(L)^T + \int_{0}^{L}\ddot{\phi}(\zeta)\phi(\zeta)^Td\zeta \\ - \Rightarrow D^T &= \dot{\phi}(L)\phi(L)^T - \dot{\phi}(L)\phi(L)^T + \int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)d\zeta \\ - \Rightarrow \boxed{\int_{0}^{L}\underbrace{\phi(\zeta)}\ddot{\phi}(\zeta)^Td\zeta} &= D^T - \dot{\phi}(L)\phi(L)^T + \phi(L)\dot{\phi}(L)^T + & D = \phi(L)\dot{\phi}(L)^T - \phi(L)\dot{\phi}(L)^T + \int_{0}^{L}\ddot{\phi}(\zeta)\phi(\zeta)^Td\zeta \\ + \Rightarrow & D^T = \dot{\phi}(L)\phi(L)^T - \dot{\phi}(L)\phi(L)^T + \int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)d\zeta \\ + \Rightarrow & \boxed{\int_{0}^{L}\underbrace{\phi(\zeta)}\ddot{\phi}(\zeta)^Td\zeta} = D^T - \dot{\phi}(L)\phi(L)^T + \phi(L)\dot{\phi}(L)^T \\ + \Rightarrow & E\dot{x}_{2d} = -e_{1d}D^T + \overbrace{e_{1d}\dot{\phi}(L)\phi(L)^T}^{=0} - e_{1d}\phi(L)\dot{\phi}(L)^T - F_{ext} \end{aligned} \end{equation*} -Donc : - -\begin{equation*} - E\dot{x}_{2d} = -e_{1d}D^T + \overbrace{e_{1d}\dot{\phi}(L)\phi(L)^T}^{=0} - e_{1d}\phi(L)\dot{\phi}(L)^T - F_{ext} -\end{equation*} - Or on sait que : -\begin{equation} +\begin{align*} \Rightarrow\left\{\begin{aligned} - \frac{de_1}{d\zeta} = u(t) &\Rightarrow \dot{\phi}(L)^T e_{1d} = u(t) \\ - e_1 (L,t) = 0 &\Rightarrow \phi(L)^T e_{1d} = 0 - \end{aligned}\right. -\end{equation} - -Donc : - -\begin{equation*} + \frac{de_1}{d\zeta} = u(t) &\Rightarrow \dot{\phi}(L)^T e_{1d} = u(t) \\ + e_1 (L,t) = 0 &\Rightarrow \phi(L)^T e_{1d} = 0 + \end{aligned}\right. & & + \Rightarrow & & \boxed{E\dot{x}_{2d} = - e_{1d}D^T - \phi(L)u(t) - F_{ext}} -\end{equation*} +\end{align*} -Puis on a : +Puis on a ceci : -\begin{equation*} +\begin{align*} \left\{\begin{aligned} y(t) &= -e_2 (L,t) \\ e_2 (L,t) &\approx \phi(L)^T e_{2d} - \end{aligned}\right. -\end{equation*} - -Ce qui nous donne l'approximation : - -\begin{equation*} + \end{aligned}\right. & & + \Rightarrow & & \boxed{y(t) = -\phi(L)^T e_{2d}} -\end{equation*} +\end{align*} + \subsection{Question 3} On a : @@ -201,7 +152,7 @@ Pour : 0 \\ 0 \end{pmatrix} - \end{aligned}\right. \Rightarrow d\acute{e}cla \ A, \ B, \ C \ en \ MATLAB + \end{aligned}\right. \end{equation*} \begin{equation*} @@ -444,147 +395,3 @@ et à garantir une erreur nulle en régime permanent. \subsection{Question 20} - -\newpage - -% VIEUX PROJET À RETIRER MAIS ON LE GARDE TEMPORAIREMENT POUR COPIER-COLLER -\section{VIEUX PROJET À RETIRER MAIS ON LE GARDE TEMPORAIREMENT POUR COPIER-COLLER} - -\vspace{0.25cm} -Le bille sur rail est une manipulation où le but est de stabiliser une bille sur un rail. Le rail est commandé par une tension, et les données lues sont l'angle du rail et la position de la bille. La position est achevé à l'aide d'un lecture d'impedance. -\vspace{0.5cm} -\\ \textbf{Le schèma de forces de la bille sur rail:} - -\includegraphics{./Illustrations/Schema_Forces.png} - -\newpage - -\subsection{Analyse du schèma bloc et setup} -Nous avons remarqué que l'identification du système se fait en bouclé fermé. Voici le schèma bloc désignant le système que nous pouvons manipuler: \{Sett inn bilde av schèma bloc, système rail\} - -\subsection{Mise en oeuvre de N4SID} -On a utilisé la fonction n4sid() du GIT de Mr. Poussot. Nous avons fait une experiènce temporel, frequentiel et avec Loewner. -Voici le comportement des differents modèles obtenu: \{Sett inn bilde av n4sid\} - -Cela nous avait mené à résumer le systeme du rail à la fonction de transfert suivante : -$$G(p) = \frac{NUM}{DEN}$$ - -\subsection{Fonction transfert du système: Rail} -Après avoir trouvé un modèle qui nous va, nous avons ensuite retrouvé la vraie fonction transferte du rail. Avec la relation qui suit: -\\ - - -%%Lånt av disse her, smarte folk! -% Source - https://tex.stackexchange.com/q/175969 -% Posted by student1, modified by community. See post 'Timeline' for change history -% Retrieved 2026-04-02, License - CC BY-SA 3.0 - - -% Source - https://tex.stackexchange.com/a/175970 -% Posted by Peter Grill, modified by community. See post 'Timeline' for change history -% Retrieved 2026-04-02, License - CC BY-SA 3.0 - - - -\tikzstyle{block} = [draw, fill=white, rectangle, - minimum height=3em, minimum width=6em] -\tikzstyle{sum} = [draw, fill=white, circle, node distance=1cm] -\tikzstyle{input} = [coordinate] -\tikzstyle{output} = [coordinate] -\tikzstyle{pinstyle} = [pin edge={to-,thin,black}] - -\begin{tikzpicture}[auto, node distance=2cm,>=latex] - - \node [input, name=input] {}; - \node [sum, right of=input] (sum) {}; - %%\node [block, right of=sum] (controller) {}; - \node [block, right of=sum, - node distance=3cm] (system) {$G_{Rail}(s)$}; - - \draw [->] (sum) -- node[name=u] {$u$} (system); - \node [output, right of=system] (output) {}; - %\node [block, below of=u] (measurements) {Measurements}; - \coordinate [below of=u] (measurements) {}; - - \draw [draw,->] (input) -- node {$r$} (sum); - %\draw [->] (sum) -- node {$e$} (system); - \draw [->] (system) -- node [name=y] {$y$}(output); - %\draw [->] (y) |- (measurements); - \draw [-] (y) |- (measurements); - %\draw [->] (measurements) -| node[pos=0.99] {$-$} - \draw [->] (measurements) -| %node[pos=1.00] {$-$} - node [near end] {$y_m$} (sum); - -\coordinate [below=1.7cm of sum] (u1) {}; -\coordinate [below=1.88cm of y] (u2) {}; -\draw[ -decorate, -decoration={brace, mirror, amplitude=8pt} -] -(u1.south west) -- (u2.south east) -node[midway, below=10pt] {$H(s)$}; - - - %\draw [->] -\end{tikzpicture} -\bigskip - -\begin{equation} - H(s)=\frac{G(s)}{1+G(s)}\Rightarrow G(s)=\frac{H(s)}{1+H(s)} -\end{equation} - - - -\subsection{Calcul du correcteur du système: P} -On a testé plusieurs valeurs, et conclue que juste un correcteur proportionnel du gain 1 fonctionne très bien. -\\ - - -\begin{tikzpicture}[auto, node distance=2cm,>=latex] - - \node [input, name=input] {}; - \node [sum, right of=input] (sum) {}; - \node [block, right of=sum] (controller) {Controleur: P}; - \node [block, right of=controller, - node distance=3cm] (system) {$G_{Rail}(s)$}; - - \draw [->] (controller) -- node[name=u] {$u$} (system); - \node [output, right of=system] (output) {}; - %\node [block, below of=u] (measurements) {Measurements}; - \coordinate [below of=u] (measurements) {}; - - \draw [draw,->] (input) -- node {$r$} (sum); - \draw [->] (sum) -- node {$e$} (controller); - \draw [->] (system) -- node [name=y] {$y$}(output); - %\draw [->] (y) |- (measurements); - \draw [-] (y) |- (measurements); - %\draw [->] (measurements) -| node[pos=0.99] {$-$} - \draw [->] (measurements) -| %node[pos=1.00] {$-$} - node [near end] {$y_m$} (sum); - -\coordinate [below=1.7cm of sum] (u1) {}; -\coordinate [below=1.88cm of y] (u2) {}; -\draw[ -decorate, -decoration={brace, mirror, amplitude=8pt} -] -(u1.south west) -- (u2.south east) -node[midway, below=10pt] {$H_C(s)$}; - - - %\draw [->] -\end{tikzpicture} - -\subsection{Theorie de la loi de commande} -Faut mettre des choses ici. Je push! - -\subsection{Une sous section} -\subsubsection{Une sous sous section} -Un mot compliqué\footnote{Une note de bas de page} - - -\newpage - -\psection{Conclusion} - -Une conclusion diff --git a/latex/contents/annexes.tex b/latex/contents/annexes.tex index 116f261..0fb729c 100644 --- a/latex/contents/annexes.tex +++ b/latex/contents/annexes.tex @@ -1,25 +1,25 @@ -\newpage -\appendix -\thispagestyle{empty} -\psection{Table des annexes} -\addtocontents{toc}{\protect\setcounter{tocdepth}{0}} % Désactivation de la table des matières +%\newpage +%\addtocontents{toc}{\protect\setcounter{tocdepth}{0}} % Désactivation de la table des matières % Personnalisation de la table des annexes -\renewcommand{\stctitle}{} % Titre (issue with previous subsection showing up) -\renewcommand\thesubsection{A\arabic{subsection}} % Numérotation -\renewcommand{\stcSSfont}{} % Police normale, pas en gras -\mtcsetrules{secttoc}{off} % Désactivation des lignes en haut et en bas de la table +%\renewcommand{\stctitle}{} % Titre (issue with previous subsection showing up) +%\renewcommand\thesubsection{A\Alph{section}} % Numérotation +%\renewcommand{\stcSSfont}{} % Police normale, pas en gras +%\mtcsetrules{secttoc}{off} % Désactivation des lignes en haut et en bas de la table % Affichage de la table des annexes -\secttoc - +%\secttoc \newpage -% Annexe 1 -\subsection{Annexe A} -Contenu de l'annexe A. +\pagenumbering{Alph} +\section{Annexe} + +\subsection{Question 1} + + + + + + + -\newpage -% Annexe 2 -\subsection{Annexe B} -Contenu de l'annexe B. \ No newline at end of file diff --git a/latex/main.pdf b/latex/main.pdf index 2919c88..875138d 100644 Binary files a/latex/main.pdf and b/latex/main.pdf differ diff --git a/latex/main.tex b/latex/main.tex index 4c30b07..b7748a1 100644 --- a/latex/main.tex +++ b/latex/main.tex @@ -65,7 +65,7 @@ %\input{contents/figures} % annexes - %\input{contents/annexes} + \input{contents/annexes} \input{template/derniere_page} % ne pas toucher \end{document} diff --git a/latex/template/derniere_page.tex b/latex/template/derniere_page.tex index bf054da..d63e081 100644 --- a/latex/template/derniere_page.tex +++ b/latex/template/derniere_page.tex @@ -19,8 +19,8 @@ }; % Des logos cliquables - \node at (5.53, -15.15) {\href{https://www.facebook.com/INSAToulouse/}{\includegraphics[width=0.8cm]{template/assets/carre.png}}}; - \node at (6.64, -15.15) {\href{https://www.instagram.com/insatoulouse/}{\includegraphics[width=0.8cm]{template/assets/carre.png}}}; - \node at (7.75, -15.15) {\href{https://www.linkedin.com/school/institut-national-des-sciences-appliqu%C3%A9es-de-toulouse/}{\includegraphics[width=0.8cm]{template/assets/carre.png}}}; - \node at (8.89, -15.15) {\href{https://www.youtube.com/user/insatoulouse}{\includegraphics[width=0.8cm]{template/assets/carre.png}}}; + \node at (6.45, -15.15) {\href{https://www.facebook.com/INSAToulouse/}{\includegraphics[width=0.8cm]{template/assets/carre.png}}}; + \node at (7.5, -15.15) {\href{https://www.instagram.com/insatoulouse/}{\includegraphics[width=0.8cm]{template/assets/carre.png}}}; + \node at (8.65, -15.15) {\href{https://www.linkedin.com/school/institut-national-des-sciences-appliqu%C3%A9es-de-toulouse/}{\includegraphics[width=0.8cm]{template/assets/carre.png}}}; + \node at (9.8, -15.15) {\href{https://www.youtube.com/user/insatoulouse}{\includegraphics[width=0.8cm]{template/assets/carre.png}}}; \end{tikzpicture} \ No newline at end of file diff --git a/latex/template/preambule.tex b/latex/template/preambule.tex index 44e6525..78ef4f6 100644 --- a/latex/template/preambule.tex +++ b/latex/template/preambule.tex @@ -4,8 +4,8 @@ % marges du document \geometry{ - left=3cm, - right=3cm, + left=2cm, + right=2.5cm, top=2cm, bottom=3.5cm } diff --git a/latex/template/premiere_page.tex b/latex/template/premiere_page.tex index 2cfc2f7..0d53ea5 100644 --- a/latex/template/premiere_page.tex +++ b/latex/template/premiere_page.tex @@ -13,9 +13,9 @@ \fill[fill=couleurcarre]([xshift=-4.7cm, yshift=-4.5cm]current page.north east) rectangle ++(3.5cm, 3.5cm); % Ajouter du texte à une position spécifique - \node at (3.25, -3.5) {\LARGE \bfseries \MakeUppercase{\sffamily \titre}}; - \node at (6.8, -6.2) {\large \firstcouverture}; - \node at (6.8, -19) {\secondcouverture}; + \node at (4.25, -3.5) {\LARGE \bfseries \MakeUppercase{\sffamily \titre}}; + \node at (8.5, -6.2) {\large \firstcouverture}; + \node at (8.5, -19) {\secondcouverture}; %\node at (11.5, -12.2) {\includegraphics[height=3cm]{\imagecouverture}}; %\node at (15.05, -0.3) {\includegraphics[width=3cm]{\imagecouverture}}; \end{tikzpicture}