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ajout TP2 424

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Pierre-antoine Comby 2019-04-24 00:15:19 +02:00
parent 45383b8f26
commit 3fa29d5ab3
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% Ce script permet de d<EFBFBD>clarer les valeurs des param<EFBFBD>tres du mod<EFBFBD>le
% Il sert aussi <EFBFBD> d<EFBFBD>clarer et <EFBFBD> calculer les diff<EFBFBD>rentes variables
% de commande tout au long du TP (Exemples : Point de fonctionnement,
% trajectoire, commande <EFBFBD> grand gain, ...
clear, clc,
%% Param<EFBFBD>tres :
m = 500; %kg
Mc = 5000; %kg
g = 10; %m/s^2
J = 50; %kg.m^2
b = 0.4; %m
Cd = 20; %kg/s
Cr = 20; %kg.m^2/s
%% Point de fonctionnement
R = 10;
D = 0;
C0 = m*g*b;
F = 0;
%% Trajectoire
Rini = R;
Dini = 0;
Rfin = 5;
Dfin = 20;
zh = 1;
dt = 10;
%% Manip 1
A = [0 0 0 1 0 0 ;
0 0 0 0 1 0 ;
0 0 0 0 0 1 ;
0 0 m*g/Mc -Cd/Mc 0 0 ;
0 0 0 0 -Cr/(b^2*(J/b^2+m)) 0 ;
0 0 -g/R*(1+m/Mc) Cd/(Mc*R) 0 0];
B = [0 0 ;
0 0 ;
0 0 ;
1/Mc 0 ;
0 -1/(b*(J/b^2+m)) ;
-1/(R*Mc) 0 ];
C= eye(6);
Com = [B A*B A^2*B A^3*B A^4*B A^5*B];
rank(Com)
%% Manip 2
vprA = damp(eig(A)); % valeur propres de A
omega0 = vprA(2);
xi = 0.5
i=complex(0,1);
p1 = omega0*(-xi+ i *sqrt(1-xi^2));
p2 = omega0*(-xi- i *sqrt(1-xi^2));
p = [-2 -2.5 -3 -4 p1 p2];
K = place(A,B,p)
C_1 =C(1,:);
B_1 =B(:,1);
eta = -1/(C_1*(A-B*K)^-1*B_1);
%%
sys = ss(A-B*K,B(:,2),C(2,:),D)
H = tf(sys)
bode(H)
% margin(H)
% Avec le correcteur intégral :
Ti = 5e-4
CI = tf(1,[-Ti 0]); % négatif pour avoir une phase >180
%bode(H,CI*H,'grid')
fig =figure();
margin(CI*H)
grid on;
saveas(fig,"manip_5marge.png");
%% pretty figure
% plot(simout)
% legend("d","r","\theta","d d/dt","dr/dt","d\theta/dt");
% grid on;
%% Commande grand gain
Deltat= 10
Dini = 0;
Rini = 10;
Dfin = 20;
Rfin = 5;
ed = omega0/10;
er = ed;
coeff_phi = [6/Deltat^5 -15/Deltat^4 10/Deltat^3 0 0 0]
%%
close all;
fig= figure()
plot(Dc)
hold on
plot(simout.Time, simout.Data(:,1));
xlabel('temps (s)');
grid on;
title('Vérification de la planification de trajectoire');
legend('d_c','d');
fig2= figure()
plot(Rc)
hold on
plot(simout.Time, simout.Data(:,2));
xlabel('temps (s)');
grid on;
title('Vérification de la planification de trajectoire');
legend('r_c','r');

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\documentclass[10pt,a4paper,notitlepage]{article}
\usepackage[utf8]{inputenc}
\usepackage[french]{babel}
\usepackage[T1]{fontenc}
\usepackage{mathtools}
\usepackage[left=2cm,right=2cm,top=2cm,bottom=2cm]{geometry}
\usepackage{graphicx}
\usepackage{float}
\usepackage{subcaption}
\usepackage{tikz}
\usepackage{siunitx}
\usepackage{numprint}
\addto{\captionsfrench}{\renewcommand{\abstractname}{Introduction}}
\author{Pierre-Antoine Comby}
\title{TP1 : Commande d'un bras à liaison flexible par bouclage linéarisant}
\input{/home/pac/Scripts/Raccourcis.tex}
\renewcommand{\R}{\mathbb{R}}
\renewcommand{\Z}{\mathbb{Z}}
\renewcommand{\vec}{\overrightarrow}
\usepackage{minted}
\begin{document}
\maketitle
\section{Modélisation}
\paragraph{Prépa.1}
\begin{itemize}
\item On applique le PFD au chariot \{1\}de masse $M_c$:
\begin{equation}
M_c\vec{a_{\{1\}}} = \vec{F}+\vec{T}-C_d\vec{v_{\{1\}}}+M_c\vec{g}
\end{equation}
Soit en projetant:
\begin{equation}
M_c \ddot{d} = F+T\sin(\theta)-C_d\dot{d}
\end{equation}
\item On fait de meme pour la masse pendulaire \{2\} $m$:
\begin{equation}
m \vec{a_{\{2\}}} = -\vec{T} +m \vec{g}
\end{equation}
et il vient directement:
\begin{equation}
\begin{cases}
m\ddot{x} = -T\sin(\theta)\\
m\ddot{z} = -T\cos(\theta)+mg
\end{cases}
\end{equation}
\item et un théorème du moment cinétique sur l'arbre donne (avec $\alpha$ angle du tambour par rapport au chariot):
\begin{equation}
J\ddot{\alpha} = -C +bT-C_r \dot{\alpha}
\end{equation}
soit :
\begin{equation}
J \frac{\ddot{r}}{b} = -C+bT -C_r \frac{\dot{r}}{b}
\end{equation}
\end{itemize}
\paragraph{Prépa.2}
Comme $b\ll r$ on a :
\begin{equation}
\begin{cases}
x = r\sin\theta+d\\
z = r\cos\theta
\end{cases}
\end{equation}
soit :
\begin{equation}
\begin{cases}
\dot{x} = \dot{r}\sin\theta + r\dot{\theta}\cos\theta +\dot{d}\\
\dot{z} = \dot{r}\cos\theta -r\dot{\theta}\sin\theta
\end{cases}
\quad\text{ et }\quad
\begin{cases}
\ddot{x} = \ddot{r}\sin\theta+2\dot{r}\dot{\theta}\cos\theta+r\ddot{\theta}\cos\theta- r\dot{\theta}^2\sin\theta+\ddot{d}\\
\ddot{z} = \ddot{r}\cos\theta-2\dot{r}\dot{\theta}\sin\theta-r\ddot{\theta}\sin\theta-r\dot{\theta}^2\cos\theta
\end{cases}
\end{equation}
\paragraph{Prépa.3} en remplacant dans le système (1) donné dans l'énoncé on trouve:
\begin{equation}
\begin{cases}
m(
\ddot{r} \sin\theta+2\dot{r}\dot{\theta}\cos\theta+r\ddot{\theta}\cos\theta-r\dot{\theta}^2\sin\theta+\ddot{d}
) &=-T\sin\theta\\
m(
\ddot{r}\cos{\theta}-2\dot{r}\dot{\theta}\sin\theta-r\ddot{\theta}\sin\theta-r\dot{\theta}^2\cos\theta
) &= - T\cos\theta+ mg
\end{cases}
\end{equation}
A partir des identités trigonométrique on a :
\begin{equation}
\begin{cases}
m\ddot{r} &= - m\ddot{d}\sin\theta+mr\dot{\theta}^2 -T + mg \cos{\theta}\\
r \ddot{\theta} &= -\ddot{d}\cos\theta-2\dot{r}\dot{\theta}-g\sin\theta
\end{cases}
\end{equation}
\paragraph{Prépa.4}
Des équations précédentes on en déduit le modèle d'état( en remplacant l'expression de $T$ par celle déduite de la prépa.3):
\begin{equation}
\begin{cases}
M_c\ddot{d} &= F +T \sin\theta-C_d\dot{d}\\
r\ddot{\theta} &= -\ddot{d}\cos\theta-2\dot{r}\dot{\theta}-g\sin(\theta)\\
(m+\frac{J}{b^2})\ddot{r} = -\frac{C}{b}-\frac{C_r\dot{r}}{b^2}-m\ddot{d}\sin\theta-mr\dot{\theta}^2+mg\cos{\theta}
\end{cases}
\end{equation}
\paragraph{Prépa.5}
On linéarise autour de $d=D$,$r=R$,$\theta=0$, $T=mg$,$F=0$,$C=mbg$:
\begin{equation}
\begin{cases}
M_{c}(\Delta d+D) &=(\Delta F)+T \sin (\Delta \theta)-C_{d}(\Delta d+D) \\(\Delta r+R)(\Delta \theta) &=-(\Delta d+D) \cos (\Delta \theta)-2(\Delta r+R)(\Delta \theta)-g \sin (\Delta \theta) \\\left(m+\frac{J}{b^{2}}\right)(\Delta r+R) &=-\frac{\Delta C+m g b}{b}-\frac{C_{r}(\Delta r+R)}{b^{2}}-m(\Delta d+D) \sin (\Delta \theta)-m(\Delta r+R)(\Delta \theta)^{2}+m g \cos (\Delta \theta)
\end{cases}
\end{equation}
En faisant une approximation au 1er ordre on a:
\begin{equation}
\begin{cases}
M_{c} \Delta d &=\Delta F+m g \Delta \theta-C_{d} \Delta d \\
R \Delta \theta+\Delta d &=-g \Delta \theta \\
\left(m+\frac{J}{b^{2}}\right) \ddot{\Delta r} &=-\frac{\Delta C+m g b}{b}-\frac{C_{r} \Delta r}{b^{2}}-m g \Delta \theta
\end{cases}
\implies
\begin{cases} M_{c} \ddot{\Delta} d &=\Delta F+m g \Delta \theta-C_{d} \dot{\Delta} d \\
R \Delta \theta+\Delta d &=-g \Delta \theta \\
\left(m+\frac{J}{b^{2}}\right) \ddot{\Delta r} &=-\frac{\Delta C}{b}-\frac{C_{r} \Delta r}{b^{2}}
\end{cases}
\end{equation}
On en deduit le modèle d'état linéarisé:
\[
\deriv{t}\vect{\Delta d \\ \Delta r \\ \Delta\theta \\ \dot{\Delta d}\\ \dot{\Delta r}\\ \dot{\Delta \theta}} =
\left[ \begin{array}{cccccc}
{0} & {0} & {0} & {1} & {0} & {0} \\
{0} & {0} & {0} & {0} & {1} & {0} \\
{0} & {0} & {0} & {0} & {0} & {1} \\
{0} & {0} & {\frac{m g}{M_{c}}} & {-\frac{C_{d}}{M_{c}}} & {0} & {0} \\
{0} & {0} & {0} & {0} & {-\frac{C_{r}}{J+m b^{2}}} & {0} \\
{0} & {0} & {-\left(1+\frac{m}{M c}\right) \frac{g}{R}} & {\frac{C_{d}}{R M_{c}}} & {0} & {0}
\end{array}\right] \cdot
\left[ \begin{array}{c}{\Delta d} \\ {\Delta r} \\ {\Delta \theta} \\ {\Delta d} \\ {\Delta r} \\ {\Delta \theta}\end{array}\right]+
\left[ \begin{array}{ccc}{0} & {0} \\ {0} & {0} \\ {0} & {0} \\ {\frac{1}{M_{c}}} & {0} \\ {0} & {-\frac{1}{\frac{J}{b}+m b^{2}}} \\ {-\frac{1}{R M_{c}}} & {0}\end{array}\right]
\cdot \left[ \begin{array}{c}{\Delta F} \\ {\Delta C}\end{array}\right]
\]
\section{Commande linéaire}
\paragraph{Manip.1}
Dans le cas linéaire on construit la matrce de Kallman :
\[
\mathcal{C} = \vect{B & AB & A^2B & A^5B }
\]
avec la fonction matlab \texttt{rank(Com)} on vérifie que le système est bien commandable.
\begin{minted}{matlab}
A = [0 0 0 1 0 0 ;
0 0 0 0 1 0 ;
0 0 0 0 0 1 ;
0 0 m*g/Mc -Cd/Mc 0 0 ;
0 0 0 0 -Cr/(b^2*(J/b^2+m)) 0 ;
0 0 -g/R*(1+m/Mc) Cd/(Mc*R) 0 0];
B = [0 0 ;
0 0 ;
0 0 ;
1/Mc 0 ;
0 -1/(b*(J/b^2+m)) ;
-1/(R*Mc) 0 ];
Com = [B A*B A^2*B A^3*B A^4*B A^5*B];
rank(Com)
\end{minted}
On a un rang de 6, le système linéaire est bien commandable.
\paragraph{Manip.2}
avec la fonction \texttt{damp(eig(A))} on trouve les valeurs propres et constantes de temps de la matrice d'état
\begin{table}[H]
\centering
\begin{tabular}[t]{llll}
\hline
{\bf Pole} & {\bf Damping} & {\bf Time Constant} & \\
\hline
& & {(rad/TimeUnit)} & {(TimeUnit)} \\
0.00e+00 & -1.00e+00 & 0.00e+00 & Inf \\
-1.82e-04 + 1.48e+00i & 1.23e-04 & 1.48e+00 & 5.50e+03 \\
-1.82e-04 - 1.48e+00i & 1.23e-04 & 1.48e+00 & 5.50e+03 \\
-3.64e-03 & 1.00e+00 & 3.64e-03 & 2.75e+02 \\
0.00e+00 & -1.00e+00 & 0.00e+00 & Inf \\
-1.54e-01 & 1.00e+00 & 1.54e-01 & 6.50e+00 \\
\hline
\end{tabular}
\caption{Valeur propre et pulsation caractéristique}
\end{table}
On a donc la pulsation propre du système : $\omega_0$= \SI{1.48}{rad/s}. On choisi alors d'imposer les poles suivant, (d'après le cahier des charges) en ammortissant le pole double conjugué principal:
\[
p_{1,2} = \omega_0(\xi\pm j \sqrt{1-\xi^2}) = 1.48(0.5\pm \sqrt{1-0.5^2})
\]
\begin{minted}{matlab}
vprA = damp(eig(A)); % valeur propres de A
omega0 = vprA(2);
xi = 0.5
i=complex(0,1);
p1 = omega0*(xi+ i *sqrt(1-xi^2));
p2 = omega0*(xi- i *sqrt(1-xi^2));
p = [-2 -2.5 -3 -4 p1 p2];
K = place(A,B,p)
\end{minted}
\[K = 10^{5}\cdot
\begin{bmatrix}
0.4288 & 0.1360 & 2.2614 & 0.0215 & 0.0506 & -0.9539 \\
-0.0035 & -0.0261 & -0.0026 & 0.0019 & -0.0182 & 0.0195 \\
\end{bmatrix}\]
\paragraph{Manip.3}
Pour le terme de précommande, si l'on veux un gain statique unitaire, il faut\footnote{cf UE 421}:
\[
\eta =\frac{-1}{C_1(A-BK)^{-1}B_1} = 4.1039. 10^{4}
\]
Avec $C_1$ et $B_1$ lignes et colonnes des matrices $C$ et $B$ propre à $\Delta d$
\paragraph{Manip.4} L'ajout d'une composante intégrale va rendre notre système précis. On construit alors la fonction de transfert grace à MATLAB:
\begin{minted}{matlab}
sys = ss( A-B*K,B(:,2),C(2,:), 0 );
H = tf(sys);
bode(H);
\end{minted}
On obtient le diagramme de bode suivant:
\begin{figure}[ht]
\centering
\includegraphics[width=0.9\textwidth]{manip_5_bode.png}
\caption{Diagramme de bode système en boucle ouverte}
\label{fig:label}
\end{figure}
\paragraph{Manip.5}
La marge de phase est nulle, on met en place un correcteur intégrale pur qui apporte stabilité et précission au système:
\begin{minted}{matlab}
Ti = 4e-4
CI = tf(1,[-Ti 0]); % négatif pour avoir une phase >180
%bode(H,CI*H,'grid')
figure();
margin(CI*H)
\end{minted}
\begin{figure}[H]
\centering
\includegraphics[width=0.7\textwidth]{manip_5marge.png}
\caption{Marge de Phase et Gain pour $T_i=4.10^{-4}$}
\label{fig:margin}
\end{figure}
sur la figure\ref{fig:margin} on relève une marge de phase de 50$^o$ , le cahier des charge est respecté.
\begin{figure}[ht]
\centering
\includegraphics[width=0.9\textwidth]{bouclage_modele_L.png}
\caption{Modèle Linéaire compléter par le correcteru par retour d'état.}
\label{fig:label}
\end{figure}
\paragraph{Manip.6}
Avec une commande $\Delta d_c = 20$m on obtient la figure suivante:
\begin{figure}[ht]
\centering
\includegraphics[width=0.7\textwidth]{manip6_20.png}
\caption{sortie du système pour une commande de 20m}
\label{fig:gain20}
\end{figure}
On remarque que le système n'est plus précis, il faut recaler le gain, ici manuellement, pour obtenir un système précis.
On atteint les limites du modèle linéaire, pour des commandes plus grande ou plus complexe il va falloir tenir compte des non-linéarités du système.
\paragraph{Manip.7}
On applique le correcteur au modèle non linéaire, en ayant recentrer le
retourd'état autour de son point de fonctionnement.
On obtient une sortie chaotique, mais douce. comme on peux le voir sur la figure \ref{fig:dc_var}
\paragraph{Manip.8}
Un changement de consigne apporte un comportement différents :
\begin{figure}[H]
\centering
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\textwidth]{NL_correcL10.png}
\caption{$d_c=10$}
\label{fig:label}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\linewidth]{NL_correcL100.png}
\caption{$d_c=100$}
\label{fig:label}
\end{subfigure}\\
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\linewidth]{NL_correcL1.png}
\caption{$d_c=1$}
\label{fig:label}
\end{subfigure}
\caption{différentes sorties pour des consignes différentes}
\label{fig:dc_var}
\end{figure}
\section{Commande non linéaire hiérarchisante}
\subsection{commande à grand gain}
\paragraph{Prépa.6} On planifie la trajectoire avec:
\begin{equation}
\begin{aligned} D_{c}(t) &=(1-\phi(t)) D_{i n i}+\phi(t) D_{f i n} \\
R_{c}(t) &=(1-\phi(t)) R_{i n i}+\phi(t) R_{f i n}
\end{aligned}
\end{equation}
Avec les conditions initiales et finales sur les positions
$\phi(0)=0 \quad$ et $\quad \phi(\Delta t)=1$
et celles sur la vitesse et l'accélération
$\forall i=1,2\left.\frac{d^{i} \phi(t)}{d t^{i}}\right|_{t=0}=0 \quad$ et $\quad\left.\frac{d^{i} \phi(t)}{d t^{i}}\right|_{t=\Delta t}=0$
On a 6 conditions à respecter, donc on choisit un polynôme de degré 5:
\[
\phi(t) = a_5 t^5 + a_4 t^4 +a_3 t^3 + a_2 t^2+a_1 t+ a_0
\]
Les conditions initiales imposent $a_0=a_1=a_2 = 0$ on a donc le polynome : $ \phi(t) = a_5 t^5 + a_4 t^4 +a_3 t^3$.
Alors :
\begin{equation}
\begin{aligned} \alpha(\Delta t)=1 & \Rightarrow \quad a_5 \Delta t^{5} \quad+a_4 \Delta t^{4} \quad+a_3 \Delta t^{3}=1 \\
\left.\frac{d \alpha(t)}{d t}\right|_{t=\Delta t}=0 & \Rightarrow 5 a_5 \Delta t^{4} \quad+4 a_4 \Delta t^{3}+3 a_3 \Delta t^{2}=0 \\
\left.\frac{d^{2} \alpha(t)}{d t^{2}}\right|_{t=\Delta t}=0 & \Rightarrow 20 a_5 \Delta t^{4}+12 a_4 \Delta t^{3}+6 a_3 \Delta t^{2}=0
\end{aligned}
\end{equation}
En résolvant le système on obtient:
\[
a_5 = \frac{6}{\Delta t^5} ,\quad a_4 = \frac{-15}{\Delta t^4}, \quad a_3 = \frac{10}{\Delta t^3}
\]
\paragraph{Prépa.7} à partir de l'équation (2.4) de l'énoncé on a :
\begin{equation}
\frac{1}{\omega_{0}^{2}}=\frac{R}{g} \Rightarrow \omega_{0}=\sqrt{\frac{g}{R}}
\end{equation}
\paragraph{Prépa.8} on a les commandes :
\begin{equation}
F=\frac{M_{c}}{\epsilon_{d}}\left(\dot{D}_{c}-\dot{d}\right) \text{ et } C=-\frac{J / b+m b}{\epsilon_{r}}\left(\dot{R}_{c}-\dot{r}\right)
\end{equation}
soit le modèle :
\begin{equation}
\dot{x}=\left[ \begin{array}{c}{\Delta d} \\ {\Delta r} \\ {\frac{m g}{M_{c}} \Delta \theta-\frac{C_{d}}{M_{c}} \Delta d+\frac{1}{\epsilon_{d}}\left(\dot{D}_{c}-\dot{d}\right)} \\ {\frac{-C_{r}}{J+m b^{2}}+\frac{F_{c}-\dot{r}}{\epsilon_{r}}} \\ {\Delta \theta}\end{array}\right]
\end{equation}
D'où :
\begin{equation}
\begin{aligned}
\ddot{\Delta d} &=-\left(\frac{C_{d}}{M_{c}}+\frac{1}{\epsilon_{d}}\right) \Delta d+\frac{\dot{D}_{c}}{\epsilon_{d}}+\frac{m g}{M_{c}} \Delta \theta \\
\ddot{\Delta r} &=-\left(\frac{C_{r}}{J+m b^{2}}+\frac{1}{\epsilon_{r}}\right) \Delta r+\frac{\dot{R}_{c}}{\epsilon_{r}} \\
\text { On peut poser } \tau_{d} &=\frac{1}{\frac{C_{d}}{M_{c}}+\frac{1}{\epsilon_{d}}} \\
\text { et } \tau_{r} &=\frac{1}{\frac{C_{r}}{J+m b^{2}}+\frac{1}{\epsilon_{r}}}
\end{aligned}
\end{equation}
En considérant $\epsilon_d \ll 1 $ et $\epsilon_r <<1 $ on a $\tau_d \simeq \epsilon_d$ et $\tau_r \simeq \epsilon_r$
\paragraph{Prépa.9}
On doit prendre $\epsilon_d$ et $\epsilon_r$ suffisamment petit devant la pulsation du sytème $\omega_0$, pour avoir une commande qui puisse compenser suffisament vite les oscillations du système.
\paragraph{Manip.9}
On réalise la commande suivante :
\begin{figure}[ht]
\centering
\includegraphics[width=0.9\textwidth]{boucleNL.png}
\caption{Élaboration de la commande pour la porsuite de trajectoire}
\label{fig:label}
\end{figure}
\begin{figure}[ht]
\centering
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\linewidth]{traj_plan_d}
\caption{poursuite sur $d$}
\label{fig:label}
\end{subfigure}%
\begin{subfigure}{0.5\textwidth}
\centering
\includegraphics[width=\linewidth]{traj_plan_r}
\caption{Poursuite sur $r$}
\label{fig:label}
\end{subfigure}
\caption{Commande en poursuite de trajectoire}
\label{fig:pours_traj}
\end{figure}
sur la figure \ref{fig:pours_traj} on remarque que la poursuite est plutot bien respecté pour $d$, mais celle sur $r$ conduit à un écart final non nul.
\paragraph{Manip.10}
\paragraph{Manip.11}
\paragraph{Manip.12}
\paragraph{Manip.13}
\subsection{Platitude}
\paragraph{Prépa.10}
Pour montrer que le systeme est plat, ayant pour sortie plates $(x, z),$ il faut montrer que les états,$d, r, \theta$ et les commandes $F, C$ ne dépendent que des sorties $x, z$ et de leurs derivées.
\begin{equation}
\left\{
\begin{aligned}
x &=r \sin \theta+d \\
m \ddot{x} &=-T \sin \theta
\end{aligned}
\right. \Rightarrow d=x+r \frac{m \ddot{x}}{T}
\end{equation}
Or on a:
\begin{equation}
\begin{cases}
m \ddot{x}=-T \sin \theta \\
m \ddot{z}=-T \cos \theta+m g
\end{cases} \Rightarrow T=m \sqrt{\ddot{x}^{2}+(g-\ddot{z})^{2}}
\end{equation}
\begin{equation}
\begin{cases}
x=r \sin \theta+d \\
z=r \cos \theta
\end{cases} \Rightarrow r^{2}=(x-d)^{2}+z^{2}
\end{equation}
Soit :
\begin{equation}
\begin{cases}
x-d &=-r \frac{m \ddot{x}}{T} \\
r^{2} &=(x-d)^{2}+z^{2}
\end{cases} \Rightarrow r^{2}=\frac{z^{2}}{1-\frac{\ddot{x}^{2}}{\vec{x}^{2}+(g-\ddot{z})^{2}}}
\end{equation}
Or $\theta = \arccos{\frac{z}{r}}$ on a donc :
\begin{equation}
r=\sqrt{\frac{z^{2}}{1-\frac{\ddot{x}^{2}}{\ddot{x}^{2}+(g-\ddot{z})^{2}}},}, d=x+\sqrt{\frac{z^{2}}{(g-\ddot{z})^{2}}} \ddot{x}, \quad \text { et } \theta=\arccos \sqrt{1-\frac{\ddot{x}^{2}}{\ddot{x}^{2}+(g-\ddot{z})^{2}}}
\end{equation}
Les variables d'états de dépendent donc que des sorties et de leurs dérivées, de même pour la commande
\begin{equation}
\begin{aligned}
C &=b T-C_{r} \frac{\dot{r}}{b}-J \frac{\ddot{r}}{b} \\
F &=M_{c} \ddot{d}-T \sin \theta+C_{d} \dot{d}
\end{aligned}
\end{equation}
Le système est plat ,avec le sorties plates $x,z$
\paragraph{Prépa.11}
Les conditions imposées sont :
\begin{equation}
\begin{cases}
d(t=0) = D_{ini} \\
r(t=0) = R_{ini} \\
\theta(t=0) = 0
\end{cases} \quad\text{ et }\quad
\begin{cases}
d(t=\Delta t ) = D_{fin} \\
r(t=\Delta t ) = R_{fin} \\
\theta(t=\Delta t ) = 0
\end{cases}
\end{equation}
Ce qui se transpose aux coordonnées:
\begin{equation}
\begin{cases}
x(t=0) = D_{ini}\\
z(t=0) = R_{ini} \\
\end{cases} \quad\text{ et }\quad
\begin{cases}
x(t=\Delta t) = D_{fin}\\
z(t=\Delta t) = R_{fin} \\
\end{cases}
\end{equation}
On pose donc :
\begin{equation}
\begin{aligned}
x_{c}(t) &=(1-\alpha(t)) D_{i n i}+\alpha(t) D_{f i n} \\
z_{c}(t) &=(1-\alpha(t)) R_{i n i}+\alpha(t) R_{f i n}
\end{aligned}
\end{equation}
De manière analogue à la préparation 6 on a les conditions initiales et finales sur les positions et leur dérivées:
\[
\alpha(0) = 0 ,\quad \alpha(\Delta t) =1, \quad \forall i = 1,2,3\left. \deriv[^i\alpha(t)]{t^i}\right|_{t=0}= 0 \text{ et } \left.\deriv[^i\alpha(t)]{t^i}\right|_{t=\Delta t}= 0
\]
Pour satisfaire les 8 conditions on choisit un polynome d'ordre 7 :
\begin{equation}
\alpha(t)=a_{7} t^{7}+a_{6} t^{6}+a_{5} t^{5}+a_{4} t^{4}+a_{3} t^{3}+a_{2} t^{2}+a_{1} t+a_{0}
\end{equation}
Les conditions initiales imposent $a_0=a_1=a_2=a_3 =0$ et les conditions finales donnent :
\begin{equation}
\begin{array}{lllllll}
\alpha(\Delta t) & =1 & a_7\Delta t^7 & + a_6 \Delta t^6 & + a_5 \Delta t^5 & + a_4 \Delta t^4 & =1 \\
\frac{d \alpha(t)}{d t}|t=\Delta t & =0 & 7a_7\Delta t^6 & + 6a_6 \Delta t^5 & 5a_5\Delta t^4 & 4a_4 \Delta t^3 & =0 \\
\frac{d^{2} \alpha(t)}{d t^{2}}|_{t=\Delta t} & =0 & 42a_7\Delta t^5 & +30a_6 \Delta t^4 & 20a_5 t^3 & 12a_4\Delta t^2 & =0 \\
\frac{d^{3} \alpha(t)}{d t^{3}} |_{t=0} & t=\Delta & 210a_z\Delta t^4 & +120a_6\Delta t^3 & 60a_5 \Delta t^2 & 24a_4\Delta t & =0
\end{array}
\end{equation}
On a donc :
\begin{equation}
a_{7}=\frac{-20}{\Delta t^{7}}, \quad a_{6}=\frac{70}{\Delta t^{6}}, \quad a_{5}=\frac{-84}{\Delta t^{5}}, \quad a_{4}=\frac{35}{\Delta t^{4}}
\end{equation}
\paragraph{Prépa.12}
On impose une trajectoire parabolique ainsi,au niveau niveau de l'obstacle :$x_c=x_H \implies z_c=z_H$. On pose donc :
\[
z_c = a(x_c-x_h)^2+z_h
\]
En évaluant cette expression à la position initiale on a:
\[
a = \frac{R_{ini}-z_H}{(D_{ini}-x_H)^2}
\]
Pour déterminer $x_H$ on utilise la position finale et on a:
\begin{equation}
x_{H}=\frac{D_{i n i}+\sqrt{\frac{R_{i n i}-z_{H}}{R_{i n i}-z_{H}}} D_{f i n}}%
{1+\sqrt{\frac{R_{i n i}-z_{H}}{R_{f i n}-z_{H}}}}
\end{equation}
\end{document}
%%% Local Variables:
%%% mode: latex
%%% TeX-master: t
%%% End:

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