Orthographie bien faite
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latex/Illustrations/NouveauxImages/Q16.png
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@ -8,13 +8,13 @@ Nous avons trouvé un seuil de $T_s = 0.0356s$ à l'aide d'un boucle qui travers
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\begin{subfigure}{0.48\textwidth}
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\begin{subfigure}{0.48\textwidth}
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\centering
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\centering
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q10}
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q10}
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\caption{Système au seuil: $T_s = 0.0356s$}
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\caption{Système sur le seuil: $T_s = 0.0356s$}
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\end{subfigure}
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\end{subfigure}
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\hfill
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\hfill
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\begin{subfigure}{0.48\textwidth}
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\begin{subfigure}{0.48\textwidth}
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\centering
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\centering
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q10_diverge}
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q10_diverge}
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\caption{Système hors seuil: $T_s = 0.0366s$}
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\caption{Système hors le seuil: $T_s = 0.0366s$}
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\end{subfigure}
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\end{subfigure}
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\end{figure}
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\end{figure}
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@ -1,6 +1,6 @@
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\textbf{Question 13}
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\textbf{Question 13}
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Avec le même periode d'echantillonnage comme dans la section precedente: $T_e=0.0356$ notre système est rendu instable.
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Avec la même periode d'echantillonnage comme dans la section precedente: $T_e=0.0356$ notre système se déstabilise.
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\begin{figure}[H]
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\begin{figure}[H]
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\centering
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\centering
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@ -1,19 +1,19 @@
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\textbf{Question 14}
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\textbf{Question 14}
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Nous retrouvons un seuil d'echantillonnage à: $T_e = 0.0107s$. Voici les comportements du système dedans et hors seuil.
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Nous retrouvons un seuil d'echantillonnage à: $T_e = 0.0107s$. Voici les comportements du système sur et hors le seuil.
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\begin{figure}[H]
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\begin{figure}[H]
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\centering
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\centering
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\begin{subfigure}{0.48\textwidth}
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\begin{subfigure}{0.48\textwidth}
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\centering
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\centering
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q14}
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q14}
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\caption{Système au seuil: $T_s = 0.0107s$}
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\caption{Système sur le seuil: $T_s = 0.0107s$}
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\end{subfigure}
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\end{subfigure}
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\hfill
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\hfill
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\begin{subfigure}{0.48\textwidth}
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\begin{subfigure}{0.48\textwidth}
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\centering
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\centering
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q14_instable}
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q14_instable}
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\caption{Système hors seuil: $T_s = 0.0108s$}
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\caption{Système hors le seuil: $T_s = 0.0108s$}
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\end{subfigure}
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\end{subfigure}
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\begin{subfigure}{0.48\textwidth}
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\begin{subfigure}{0.48\textwidth}
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\centering
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\centering
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@ -1,4 +1,14 @@
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\textbf{Question 15}
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\textbf{Question 15}
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\begin{center}
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Soit $q(\zeta, t) = q_0(t)$ et $F_{ext}=-\int_0^L\Phi(\zeta)q(\zeta,t)d\zeta$
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$\Leftrightarrow F_{ext}=-\int_0^L\Phi(t)q_0(t)$\\
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Or $E\dot{x}_{2d}=-D^Tx_{1d}(t)-\Phi(L)u(t)-F_{ext}$
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$\Leftrightarrow$
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$\dot{x}_{2d}=-E^{-1}D^Tx_{1d}(t)-E^{-1}\Phi(L)u(t)+ E^{-1}\int_0^L\Phi(\zeta)d\zeta q_0(t)$\\
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D'où $\dot{x}_d(t)=Ax_d(t)+Bu(t)+ \begin{bmatrix}
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0_{\rm I\!R_{_{4\times1}}} \\ E^{-1}\int_0^L\Phi(\zeta)d\zeta
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\end{bmatrix}q_0(t)$
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\end{center}
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Après avoir calculé, nous retrouvons la matrice $B_p$ sous la forme:
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Après avoir calculé, nous retrouvons la matrice $B_p$ sous la forme:
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\begin{equation}
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\begin{equation}
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B_p =
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B_p =
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@ -7,7 +17,7 @@ B_p =
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\end{bmatrix}
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\end{bmatrix}
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=
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=
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\begin{bmatrix}
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\begin{bmatrix}
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0 & 0 & 0 & 0 & -1 & -1 & 0 & 0
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0 & 0 & 0 & 0 & 1 & 1 & 0 & 0
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\end{bmatrix}^{T}
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\end{bmatrix}^{T}
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\end{equation}
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\end{equation}
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@ -3,7 +3,7 @@
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La Figure~\ref{fig:q16_perturbation} montre que le système ne corrige pas l'erreur due à la perturbation.
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La Figure~\ref{fig:q16_perturbation} montre que le système ne corrige pas l'erreur due à la perturbation.
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\begin{figure}[H]
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\begin{figure}[H]
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\centering
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\centering
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\includegraphics[scale=0.3]{Illustrations/im3} % Remplacez par votre image
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\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q16}
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\caption{Simulation du système avec une perturbation.}
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\caption{Simulation du système avec une perturbation.}
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\label{fig:q16_perturbation}
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\label{fig:q16_perturbation}
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\end{figure}
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\end{figure}
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@ -1,4 +1,4 @@
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This is pdfTeX, Version 3.141592653-2.6-1.40.25 (TeX Live 2023/Debian) (preloaded format=pdflatex 2026.4.14) 7 MAY 2026 19:39
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This is pdfTeX, Version 3.141592653-2.6-1.40.25 (TeX Live 2023/Debian) (preloaded format=pdflatex 2026.4.14) 7 MAY 2026 23:28
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@ -306,7 +306,7 @@ Overfull \hbox (18.91605pt too wide) in paragraph at lines 3--5
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Overfull \hbox (35.55566pt too wide) in paragraph at lines 3--5
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\OMS/cmsy/m/n/10 ^^@\/cmr/m/n/10 4[]3805$
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\OMS/cmsy/m/n/10 ^^@\/cmr/m/n/10 4\OML/cmm/m/it/10 :\/cmr/m/n/10 3805$
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[]
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[]
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@ -479,11 +479,11 @@ cmsy/m/n/10 ^^@ \/cmr/m/n/10 3\OML/cmm/m/it/10 i \OMS/cmsy/m/n/10 ^^@ \/cmr/m/n
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@ -1,6 +1,6 @@
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\textbf{Question 17}
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\textbf{Question 17}
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À partir des huit pôles obtenus avec la commande par retour d'état de la Question 11, on ajoute un pôle supplémentaire associé à l'action intégrale $p_i = -4{,}3805$, correspondant au pôle du système le plus rapide.
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À partir des huit pôles obtenus avec la commande par retour d'état de la Question 11, on ajoute un pôle supplémentaire associé à l'action intégrale $p_i = -4.3805$, correspondant au pôle du système le plus rapide.
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Ainsi, les pôles desirés du système augmenté sont :
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Ainsi, les pôles desirés du système augmenté sont :
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\begin{center}
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\begin{center}
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$p_{\text{des}} =$
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$p_{\text{des}} =$
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@ -1,11 +1,19 @@
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\textbf{Question 19}
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\textbf{Question 19}
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La Figure~\ref{fig:q19_integrale} montre qu'avec l'action intégrale, la perturbation est rejetée et l'erreur en régime permanent est corrigée.
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La Figure~\ref{fig:q19_integrale} montre qu'avec l'action intégrale, la perturbation est rejetée et l'erreur en régime permanent est corrigée.
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\begin{figure}[H]
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\begin{figure}[H]
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\centering
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\begin{subfigure}{0.48\textwidth}
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\centering
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\centering
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\includegraphics[width=0.4\textwidth]{Illustrations/Q17} % Remplacez par votre image
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\includegraphics[scale=0.5]{Illustrations/NouveauxImages/Q19}
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\caption{Erreur en regime permanente}
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\end{subfigure}
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\hfill
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\begin{subfigure}{0.48\textwidth}
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\centering
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\includegraphics[scale=0.5]{Illustrations/NouveauxImages/Q18}
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\caption{Erreur nulle en regime permanente}
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\end{subfigure}
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\caption{Simulation du système avec correction par action intégrale.}
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\caption{Simulation du système avec correction par action intégrale.}
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\label{fig:q19_integrale}
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\label{fig:q19_integrale}
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\end{figure}
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\end{figure}
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@ -1,73 +1,134 @@
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\textbf{Question 2}
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\textbf{Question 2}
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À partir des informations données dans l'exercice, on peut écrire :
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\begin{equation}
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\begin{equation}
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\dot{x}_1
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\dot{x}_1
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=
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=
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\frac{\partial^2 e_2}{\partial \zeta^2}
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\frac{\partial^2}{\partial \zeta^2} e_2(\zeta,t)
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=
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\phi(\zeta)^T \dot{x}_{1d}(t)
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\end{equation}
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\end{equation}
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En multipliant les deux membres par le vecteur d'approximation $\phi(\zeta)$, on obtient :
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Par la méthode de discrétisation :
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\begin{equation}
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\begin{equation}
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\phi(\zeta)
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\phi^T(\zeta)\dot{x}_{1d}(t)
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\frac{\partial^2 e_2}{\partial \zeta^2}
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=
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=
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\phi(\zeta)\phi(\zeta)^T \dot{x}_{1d}(t)
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\frac{\partial^2 \phi^T(\zeta)}{\partial \zeta^2} e_{2d}(t) \Leftrightarrow \int_0^L \phi(\zeta)\phi^T(\zeta)\,d\zeta \; \dot{x}_{1d}(t)
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\end{equation}
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En remplaçant ensuite $e_2$ par son approximation, on obtient :
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\begin{equation}
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\phi(\zeta)
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\frac{\partial^2}{\partial \zeta^2}
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\left(
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\phi(\zeta)^T e_{2d}(t)
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\right)
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=
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\phi(\zeta)\phi(\zeta)^T \dot{x}_{1d}(t)
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\end{equation}
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Comme $e_{2d}(t)$ ne dépend pas de $\zeta$, il peut être sorti de l'intégrale. En intégrant sur $[0,L]$, on obtient :
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\begin{equation}
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\int_0^L
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\phi(\zeta)\phi(\zeta)^T
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\, d\zeta \,
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\dot{x}_{1d}(t)
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=
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=
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\underbrace{
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\int_0^L
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\int_0^L
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\phi(\zeta)
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\phi(\zeta)
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\frac{\partial^2 \phi(\zeta)^T}{\partial \zeta^2}
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\frac{\partial^2 \phi^T(\zeta)}{\partial \zeta^2}
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\, d\zeta \,
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\,d\zeta
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e_{2d}(t)
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e_{2d}(t)
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}_{J}
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\end{equation}
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\end{equation}
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Après avoir fait une integration par partie:
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Par intégration par parties deux fois :
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à inserer:\\
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\begin{equation}
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J
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On identifie alors les matrices $E$ et $D$. Ainsi, les deux équations d'état discrétisées s'écrivent :
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=
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\left[
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\phi(\zeta)
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\frac{\partial \phi^T(\zeta)}{\partial \zeta}
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\right]_0^L
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-
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\int_0^L
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\frac{\partial \phi(\zeta)}{\partial \zeta}
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\frac{\partial \phi^T(\zeta)}{\partial \zeta}
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\,d\zeta
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\end{equation}
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\begin{equation}
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\begin{equation}
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\begin{aligned}
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\begin{aligned}
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E\dot{x}_{1d}(t) &= D e_{2d}(t),\\
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J
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E\dot{x}_{2d}(t) &= -D^T e_{1d}(t)-\phi(L)u(t)-F_{ext}.
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&=
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\phi(L)\frac{\partial \phi^T}{\partial \zeta}(L)e_{2d}(t)
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-0
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-
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\left[
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\frac{\partial \phi}{\partial \zeta}(\zeta)
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\phi^T(\zeta)e_{2d}(t)
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\right]_0^L
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+
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\int_0^L
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\frac{\partial^2 \phi}{\partial \zeta^2}(\zeta)
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\phi^T(\zeta)\,d\zeta \; e_{2d}(t)
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\\[0.2cm]
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&=
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\phi(L)\frac{\partial \phi^T}{\partial \zeta}(L)e_{2d}(t)
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-
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\frac{\partial \phi}{\partial \zeta}(L)
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\phi^T(L)e_{2d}(t)
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+
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\frac{\partial \phi}{\partial \zeta}(0)
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e_{2d}(0,t)
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+
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\int_0^L
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\frac{\partial^2 \phi}{\partial \zeta^2}(\zeta)
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\phi^T(\zeta)\,e_{2d}(t)d\zeta \;
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\end{aligned}
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\end{aligned}
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\end{equation}
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\end{equation}
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Enfin, on obtient :
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On a $e_2(0,t)=0 \Leftrightarrow E\dot{x}_1 = De_{2d}(t)$
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Comme $D =
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\phi(L)\frac{d\phi^T}{d\zeta}(L)
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-
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\frac{d\phi}{d\zeta}(L)\phi^T(L)
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+
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\int_0^L
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\frac{d^2\phi}{d\zeta^2}(\zeta)\phi^T(\zeta)\,d\zeta
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$. Et $(AB)^T = B^T A^T$, on a :
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\begin{equation}
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\begin{equation}
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\boxed{
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-D^T =
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\begin{aligned}
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-
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E\dot{x}_{1d}(t) &= D e_{2d}(t),\\
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\frac{d\phi}{d\zeta}(L)\phi^T(L)
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E\dot{x}_{2d}(t) &= -D^T e_{1d}(t)-\phi(L)u(t)-F_{ext},\\
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+
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y(t) &= -\phi(L)^T e_{2d}(t).
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\phi(L)\frac{d\phi^T}{d\zeta}(L)
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\end{aligned}
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-
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}
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\int_0^L
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\phi(\zeta)
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\frac{d^2\phi^T}{d\zeta^2}(\zeta)
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\,d\zeta.
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\end{equation}
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\end{equation}
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\begin{equation}
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E\dot{x}_{2d}(t)
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=
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-
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\int_0^L
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\phi(\zeta)
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\frac{d^2\phi^T}{d\zeta^2}(\zeta)
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\,d\zeta \; e_{1d}(t)
|
||||||
|
-
|
||||||
|
F_{ext}.
|
||||||
|
\end{equation}
|
||||||
|
|
||||||
|
Donc,
|
||||||
|
\begin{equation}
|
||||||
|
E\dot{x}_{2d}(t)
|
||||||
|
=
|
||||||
|
-D^T e_{1d}(t)
|
||||||
|
+
|
||||||
|
\frac{d\phi}{d\zeta}(L)e_1(L,t)
|
||||||
|
-
|
||||||
|
\phi(L)\frac{\partial e_1}{\partial \zeta}(L,t)
|
||||||
|
-
|
||||||
|
F_{ext}.
|
||||||
|
\end{equation}
|
||||||
|
|
||||||
|
Comme $e_1(L,t)=0$, $\frac{\partial e_1}{\partial \zeta}(L,t)=u(t)$ on obtient bien :
|
||||||
|
|
||||||
|
\begin{equation}
|
||||||
|
E\dot{x}_{2d}(t)
|
||||||
|
=
|
||||||
|
-D^T e_{1d}(t)
|
||||||
|
-
|
||||||
|
\phi(L)u(t)
|
||||||
|
-
|
||||||
|
F_{ext}.
|
||||||
|
\end{equation}
|
||||||
|
|
||||||
|
|
|
||||||
|
|
@ -1,3 +1,9 @@
|
||||||
\textbf{Question 20}
|
\textbf{Question 20}
|
||||||
|
|
||||||
Nous avons testé avec 6 decimals, mais la valeur absolue d'un des pôles z: $|z|>1$, alors nous n'arrivons jamais à stabiliser le système. Aussi, nous considerons avec $T_s \rightarrow 0.00001$ est un temps continu, la commande n'est pas implemantable sur un controleur numérique.
|
Nous avons testé avec 6 decimals, mais npous avons un pôle: $|z|>1$, alors nous n'arrivons jamais à stabiliser le système. Aussi, nous considerons avec $T_s \rightarrow 0.00001$ est un temps continu, la commande n'est pas donc implemantable sur un controleur numérique.
|
||||||
|
\begin{center}
|
||||||
|
%\begin{figure}
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||||||
|
\includegraphics[scale=0.6]{Illustrations/NouveauxImages/Q_20}
|
||||||
|
%\caption{Valeurs propres du système avec action intégrale}
|
||||||
|
%\end{figure}
|
||||||
|
\end{center}
|
||||||
|
|
@ -1,6 +1,6 @@
|
||||||
\textbf{Question 4}
|
\textbf{Question 4}
|
||||||
|
|
||||||
Voici le lieu des racines du système en continu et en temps discret. $T_s = {0.01 ; 0.02 ; 0.04}$ avec le cercle unitaire. Comme tous nos pôles sont à $|z|=1$, nous ne pouvons pas conclure sur la stabilité ni l'instabilité. En temps continu nous avons instabilité, à cause des pôles: $Re\{\lambda\}>0$
|
Voici le lieu des racines du système en continu et en temps discret. $T_s = {0.01 ; 0.02 ; 0.04}$ avec le cercle unitaire. Comme tous nos pôles sont à $|z|=1$, nous ne pouvons pas conclure sur la stabilité ni l'instabilité. En temps continu le système est instable, à cause des pôles: $Re\{\lambda\}>0$
|
||||||
|
|
||||||
\begin{figure}[H]
|
\begin{figure}[H]
|
||||||
\centering
|
\centering
|
||||||
|
|
|
||||||
|
|
@ -1,10 +1,9 @@
|
||||||
\newpage
|
|
||||||
\textbf{Question 5}
|
\textbf{Question 5}
|
||||||
|
|
||||||
La Figure 2 montre que la sortie $y(t)$ oscille après l’échelon, confirmant l’instabilité du système en boucle ouverte.
|
La Figure 2 montre que la sortie $y(t)$ oscille après l’échelon, confirmant l’instabilité du système en boucle ouverte.
|
||||||
|
|
||||||
\begin{figure}[H]
|
\begin{figure}[H]
|
||||||
\centering
|
\centering
|
||||||
\includegraphics[scale=0.3]{Illustrations/Q5stepReponse} % Remplacez par votre image
|
\includegraphics[scale=0.4]{Illustrations/Q5stepReponse} % Remplacez par votre image
|
||||||
\caption{Response du système en boucle ouverte.}
|
\caption{Response du système en boucle ouverte.}
|
||||||
\end{figure}
|
\end{figure}
|
||||||
|
|
@ -1,4 +1,13 @@
|
||||||
\textbf{Question 6}
|
\textbf{Question 6}
|
||||||
|
\begin{center}
|
||||||
:o
|
Nous savons que: $ x_1(\zeta, t) = \frac{d^2 w}{d^2\zeta}(\zeta,t)$
|
||||||
|
$\Rightarrow w(\zeta,t)=\int\int_0^L x_1(\mu , t)d\mu $\\
|
||||||
|
comme $x_1(\mu,t)=\Phi^T(\mu)x_{1d}(t)$\\
|
||||||
|
d'où
|
||||||
|
\begin{equation}
|
||||||
|
w(\zeta,t) =
|
||||||
|
\begin{bmatrix}
|
||||||
|
\int\int_0^\zeta \Phi(\mu)d\mu & 0_{{\rm I\!R}_{1\times4}}
|
||||||
|
\end{bmatrix}x_d(t)
|
||||||
|
\end{equation}
|
||||||
|
\end{center}
|
||||||
|
|
@ -2,9 +2,8 @@
|
||||||
\textbf{Question 9}
|
\textbf{Question 9}
|
||||||
|
|
||||||
La Figure~\ref{fig:q9_eigenvalues} compare les valeurs propres du système en boucle fermée en temps continu avec celles obtenues après discrétisation par ZOH, pour les trois périodes d'échantillonnage proposées: $T_e = \{0.01s ; 0.02s ; 0.04s\}$.
|
La Figure~\ref{fig:q9_eigenvalues} compare les valeurs propres du système en boucle fermée en temps continu avec celles obtenues après discrétisation par ZOH, pour les trois périodes d'échantillonnage proposées: $T_e = \{0.01s ; 0.02s ; 0.04s\}$.
|
||||||
|
|
||||||
\begin{figure}[H]
|
\begin{figure}[H]
|
||||||
\centering
|
\centering
|
||||||
\begin{subfigure}{0.48\textwidth}
|
\begin{subfigure}{0.48\textwidth}
|
||||||
\centering
|
\centering
|
||||||
\includegraphics[scale=0.3]{Illustrations/Q9continu}
|
\includegraphics[scale=0.3]{Illustrations/Q9continu}
|
||||||
|
|
@ -27,8 +26,9 @@ La Figure~\ref{fig:q9_eigenvalues} compare les valeurs propres du système en bo
|
||||||
\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q9_TE004}
|
\includegraphics[scale=0.4]{Illustrations/NouveauxImages/Q9_TE004}
|
||||||
\caption{$T_s = 0.04\,\mathrm{s}$.}
|
\caption{$T_s = 0.04\,\mathrm{s}$.}
|
||||||
\end{subfigure}
|
\end{subfigure}
|
||||||
|
\caption{Valeurs propres dans le cas d'un retour de sortie}
|
||||||
\label{fig:q9_eigenvalues}
|
\label{fig:q9_eigenvalues}
|
||||||
\end{figure}
|
\end{figure}
|
||||||
|
|
||||||
Comme nous pouvons voir, les pôles du système discret restent à l'interieur du cercle unitaire, mais dès que nous passons à $T_e=0.04s$ ça rend le système instable avec un pôle hors le cercle.
|
Comme nous pouvons voir, les pôles du système discret restent à l'interieur du cercle unitaire, mais prendre $T_e=0.04s$ rend le système instable avec un pôle hors du cercle.
|
||||||
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\subsection{Discrétisation par éléments finis}
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\subsection{Discrétisation par éléments finis}
|
||||||
\input{Questions/Q2}
|
\input{Questions/Q2}
|
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\input{Questions/Q3}
|
\input{Questions/Q3}
|
||||||
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\newpage
|
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\input{Questions/Q4}
|
\input{Questions/Q4}
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||||||
\input{Questions/Q5}
|
\input{Questions/Q5}
|
||||||
\input{Questions/Q6}
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|
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\newlabel{fig:Q8_1}{{3}{4}{Suivi de la consigne $w_c(t)$ par la position $w(L,t)$}{figure.caption.3}{}}
|
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|
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Reference in a new issue