Seesion à GEI

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Tournesol02 2026-05-07 15:53:30 +02:00
parent cb01273887
commit 47fac46825
44 changed files with 3511 additions and 964 deletions

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\textbf{Question 1}
En considérant le modèle EDP suivant ainsi que les équations données pour $e_1$ et $e_2$ :
\begin{equation}
\rho(\zeta)\frac{\partial^2 \omega}{\partial t^2}
=
-\frac{\partial^2}{\partial \zeta^2}
\left(
EI(\zeta)\frac{\partial^2 \omega}{\partial \zeta^2}(\zeta,t)
\right)
-q(\zeta,t)
\end{equation}
On obtient comme la représentation d'état :
A partir des equations diferentielles partielles nous avons retrouvé l'espace d'état sous cette forme:
\begin{equation}
\dot{x}
=

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\textbf{Question 10}
Nous avons obtenu une valeur maximale de $0{,}5\,\mathrm{s}$ pour la période d'échantillonnage.
Nous avons trouvé un seuil de $T_s = 0.0356s$ à l'aide d'un boucle qui traverse les
\begin{figure}[H]
\centering
\includegraphics[scale=0.4]{Illustrations/Q10}
\caption{Système à $T_s = 2s$}
\label{fig:Q10_1}
\end{figure}
Manque l'image!!!!!!!!!!!

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\textbf{Question 12}
La Figure~\ref{fig:Q15} montre que le système en boucle fermée suit correctement la référence et converge vers la consigne en régime permanent.
La Figure~\ref{fig:Q15} montre que le système en boucle fermée suit correctement la référence avec un léger depassement et converge vers la consigne en régime permanent.
\begin{figure}[H]
\centering

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\textbf{Question 14}
D'après la Figure~\ref{fig:q14_reponses}, les petites périodes d'échantillonnage permettent de mieux approcher la réponse temporelle continue, comme le montre la Figure~\ref{fig:Q15}.
On choisit donc $T_s = 0{,}01\,\mathrm{s}$, car cette valeur conserve un comportement proche de la réponse continue.
Comme nous avions fait en Q12. Par approche empirique, nous avons trouvé que $T_s = 0.3s$, ce qui convient pour maintenir le comportement temporel voulu. Voir Figure~\ref{fig:q14_reponses}.
\begin{figure}[H]
\centering
\includegraphics[width=0.4\textwidth]{Illustrations/im2} % Remplacez par votre image
\caption{Réponses temporelles du système pour différentes périodes d'échantillonnage.}
\includegraphics[width=0.4\textwidth]{Illustrations/Q14} % Remplacez par votre image
\caption{Réponse du système à $T_s = 0.3s$.}
\label{fig:q14_reponses}
\end{figure}

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\textbf{Question 15}
:o
Après avoir calculé, nous retrouvons la matrice $B_p$ sous la forme:
\begin{equation}
B_p =
\begin{bmatrix}
0 \\ -E^{-1}\int_0^L \Phi(\zeta)d\zeta
\end{bmatrix}
=
\begin{bmatrix}
0 \\ 0 \\ 0 \\ 0 \\ -1 \\ -1 \\ 0 \\ 0
\end{bmatrix}
\end{equation}

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\textbf{Question 16}
La Figure~\ref{fig:q16_perturbation} montre que le système ne corrige pas l'erreur due à la perturbation.
\begin{figure}[H]
\centering

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\textbf{Question 17}
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.
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.
Ainsi, les pôles desirés du système augmenté sont :
\begin{equation}
p_i = -4{,}3805
\end{equation}
Ainsi, les pôles désirés du système augmenté sont :
\begin{equation}
p_{\text{dés}} =
p_{\text{des}} =
\begin{bmatrix}
-84{,}6647\\
-7{,}5585+68{,}7669i\\

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\textbf{Question 18}
En utilisant la fonction \texttt{place} sous MATLAB, nous avons calculé les gains $K_1$ et $K_i$ :
En utilisant la fonction \texttt{place} sous MATLAB, nous avons calculé les gains $K_1$ et $K_i$ à partir du système augmenté:
\begin{equation}
K_{Aug} = \begin{bmatrix}
K_1 & K_i
\end{bmatrix}
\end{equation}
\begin{equation}
K_1 =
\begin{bmatrix}
-3{,}1081 & -18{,}9367 & -0{,}1707 & 2{,}1192 & 19{,}4677 & -2{,}3049 & 2{,}3377 & -1{,}7524
\end{bmatrix}
\end{equation}
\begin{equation}
K_i =
\begin{bmatrix}
26{,}2831
\end{bmatrix}.
\end{bmatrix}
\end{equation}

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\textbf{Question 19}
La Figure~\ref{fig:q19_integrale} montre qu'avec l'action intégrale, la perturbation est rejetée et l'erreur en régime permanent est annulée.
La Figure~\ref{fig:q19_integrale} montre qu'avec l'action intégrale, la perturbation est rejetée et l'erreur en régime permanent est corrigée.
\begin{figure}[H]
\centering

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n remplaçant ensuite $e_2$ par son approximation, on obtient :
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mm/m/it/10 ^^^\/cmr/m/n/10 (\OML/cmm/m/it/10 ^^P\/cmr/m/n/10 )[] \OML/cmm/m/it/
10 d^^P []\/cmr/m/n/10 (\OML/cmm/m/it/10 t\/cmr/m/n/10 ) = [][] \OML/cmm/m/it/1
0 ^^^\/cmr/m/n/10 (\OML/cmm/m/it/10 ^^P\/cmr/m/n/10 )[] \OML/cmm/m/it/10 d^^P e
[]\/cmr/m/n/10 (\OML/cmm/m/it/10 t\/cmr/m/n/10 )
[]
! LaTeX Error: Missing \begin{document}.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.49 A
près avoir fait une integration par partie:
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If that doesn't work, type X <return> to quit.
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[]\/cmr/m/n/10 ^^Re
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! LaTeX Error: Missing \begin{document}.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.53 O
n identifie alors les matrices $E$ et $D$. Ainsi, les deux équations ...
You're in trouble here. Try typing <return> to proceed.
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\OML/cmm/m/it/10 E$
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\OML/cmm/m/it/10 D$
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! LaTeX Error: Missing \begin{document}.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.55 \begin{equation}
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! LaTeX Error: Environment aligned undefined.
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Type H <return> for immediate help.
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l.56 \begin{aligned}
Your command was ignored.
Type I <command> <return> to replace it with another command,
or <return> to continue without it.
! Misplaced alignment tab character &.
l.57 E\dot{x}_{1d}(t) &
= D e_{2d}(t),\\
I can't figure out why you would want to use a tab mark
here. If you just want an ampersand, the remedy is
simple: Just type `I\&' now. But if some right brace
up above has ended a previous alignment prematurely,
you're probably due for more error messages, and you
might try typing `S' now just to see what is salvageable.
! Misplaced alignment tab character &.
l.58 E\dot{x}_{2d}(t) &
= -D^T e_{1d}(t)-\phi(L)u(t)-F_{ext}.
I can't figure out why you would want to use a tab mark
here. If you just want an ampersand, the remedy is
simple: Just type `I\&' now. But if some right brace
up above has ended a previous alignment prematurely,
you're probably due for more error messages, and you
might try typing `S' now just to see what is salvageable.
! LaTeX Error: \begin{equation} on input line 55 ended by \end{aligned}.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.59 \end{aligned}
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Type I <command> <return> to replace it with another command,
or <return> to continue without it.
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\OML/cmm/m/it/10 E[]\/cmr/m/n/10 (\OML/cmm/m/it/10 t\/cmr/m/n/10 ) = \OML/cmm/m
/it/10 De[]\/cmr/m/n/10 (\OML/cmm/m/it/10 t\/cmr/m/n/10 )\OML/cmm/m/it/10 ; []
E[]\/cmr/m/n/10 (\OML/cmm/m/it/10 t\/cmr/m/n/10 ) = \OMS/cmsy/m/n/10 ^^@\OML/cm
m/m/it/10 D[]e[]\/cmr/m/n/10 (\OML/cmm/m/it/10 t\/cmr/m/n/10 ) \OMS/cmsy/m/n/10
^^@ \OML/cmm/m/it/10 ^^^\/cmr/m/n/10 (\OML/cmm/m/it/10 L\/cmr/m/n/10 )\OML/cmm
/m/it/10 u\/cmr/m/n/10 (\OML/cmm/m/it/10 t\/cmr/m/n/10 ) \OMS/cmsy/m/n/10 ^^@ \
OML/cmm/m/it/10 F[]:
[]
! LaTeX Error: Missing \begin{document}.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.62 E
nfin, on obtient :
You're in trouble here. Try typing <return> to proceed.
If that doesn't work, type X <return> to quit.
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! LaTeX Error: Missing \begin{document}.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
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l.64 \begin{equation}
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[]
! Undefined control sequence.
l.65 \boxed
{
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.
! LaTeX Error: Environment aligned undefined.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.66 \begin{aligned}
Your command was ignored.
Type I <command> <return> to replace it with another command,
or <return> to continue without it.
! Misplaced alignment tab character &.
l.67 E\dot{x}_{1d}(t) &
= D e_{2d}(t),\\
I can't figure out why you would want to use a tab mark
here. If you just want an ampersand, the remedy is
simple: Just type `I\&' now. But if some right brace
up above has ended a previous alignment prematurely,
you're probably due for more error messages, and you
might try typing `S' now just to see what is salvageable.
! Misplaced alignment tab character &.
l.68 E\dot{x}_{2d}(t) &
= -D^T e_{1d}(t)-\phi(L)u(t)-F_{ext},\\
I can't figure out why you would want to use a tab mark
here. If you just want an ampersand, the remedy is
simple: Just type `I\&' now. But if some right brace
up above has ended a previous alignment prematurely,
you're probably due for more error messages, and you
might try typing `S' now just to see what is salvageable.
! Misplaced alignment tab character &.
l.69 y(t) &
= -\phi(L)^T e_{2d}(t).
I can't figure out why you would want to use a tab mark
here. If you just want an ampersand, the remedy is
simple: Just type `I\&' now. But if some right brace
up above has ended a previous alignment prematurely,
you're probably due for more error messages, and you
might try typing `S' now just to see what is salvageable.
! LaTeX Error: \begin{equation} on input line 64 ended by \end{aligned}.
See the LaTeX manual or LaTeX Companion for explanation.
Type H <return> for immediate help.
...
l.70 \end{aligned}
Your command was ignored.
Type I <command> <return> to replace it with another command,
or <return> to continue without it.
Overfull \hbox (356.01259pt too wide) detected at line 72
[]
[]
)
! Emergency stop.
<*> Q2.tex
*** (job aborted, no legal \end found)
Here is how much of TeX's memory you used:
41 strings out of 474222
714 string characters out of 5748733
1925975 words of memory out of 5000000
22394 multiletter control sequences out of 15000+600000
558377 words of font info for 37 fonts, out of 8000000 for 9000
1141 hyphenation exceptions out of 8191
23i,3n,32p,163b,91s stack positions out of 10000i,1000n,20000p,200000b,200000s
! ==> Fatal error occurred, no output PDF file produced!

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@ -46,6 +46,11 @@ Comme $e_{2d}(t)$ ne dépend pas de $\zeta$, il peut être sorti de l'intégrale
e_{2d}(t)
\end{equation}
Après avoir fait une integration par partie:
à inserer:\\
On identifie alors les matrices $E$ et $D$. Ainsi, les deux équations d'état discrétisées s'écrivent :
\begin{equation}

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@ -0,0 +1,3 @@
\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.

1406
latex/Questions/Q3.log Normal file

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@ -24,6 +24,10 @@ Et, on obtient comme valeurs de $A$, $B$ et $C$ :
\begin{equation}
A =
\begin{bmatrix}
0 & E^{-1}D \\
-E^{-1}D^T & 0
\end{bmatrix} =
\begin{bmatrix}
0 & 0 & 0 & 0 & 114 & 6 & 12 & -2 \\
0 & 0 & 0 & 0 & -54 & -6 & -2 & 4 \\
0 & 0 & 0 & 0 & -1188 & -12 & -114 & 6 \\
@ -38,6 +42,10 @@ A =
\begin{equation}
B=
\begin{bmatrix}
0 \\ -E^{-1}\Phi(L)
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
@ -50,6 +58,10 @@ B=
\qquad
C=
\begin{bmatrix}
0 & -\Phi^{T}(L,t)
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 0 & 0 & 0 & -1 & 0 & 0
\end{bmatrix}
\end{equation}

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@ -1,10 +1,6 @@
\textbf{Question 4}
La Figure 1a correspond aux valeurs propres en temps continu, tandis que les figures 1b, 1c et 1d montrent les valeurs propres discrètes obtenues avec les méthodes de Tustin et ZOH pour différents temps d'échantillonnage.
En temps continu, le système présente deux pôles à partie réelle positive, ce qui montre son instabilité.
Après discrétisation, les pôles obtenus avec les méthodes de Tustin et ZOH se trouvent sur ou proches du cercle unité. Le système discret est donc à la limite de stabilité.
La Figure 1a correspond aux valeurs propres en temps continu, tandis que les figures \textit{1b, 1c et 1d} montrent les valeurs propres discrètes obtenues avec les méthodes de Tustin et ZOH pour différents temps d'échantillonnage.
\begin{figure}[h]
\centering
@ -37,3 +33,8 @@ Après discrétisation, les pôles obtenus avec les méthodes de Tustin et ZOH s
\caption{Comparaison des valeurs propres du système en temps continu et en temps discret.}
\end{figure}
En temps continu, le système présente deux pôles à partie réelle positive, ce qui montre son instabilité.
Après discrétisation, les pôles obtenus avec les méthodes de Tustin et ZOH se trouvent sur ou proches du cercle unité. Comme tous nos pôles sont à $|z|=1$, nous ne pouvons pas conclure sur la stabilité ni l'instabilité.

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@ -1,9 +1,10 @@
\newpage
\textbf{Question 5}
La Figure 2 montre que la sortie $y(t)$ oscille après léchelon, confirmant linstabilité du système en boucle ouverte.
\begin{figure}[H]
\centering
\includegraphics[width=0.4\textwidth]{Illustrations/Q5stepReponse} % Remplacez par votre image
\includegraphics[scale=0.4]{Illustrations/Q5stepReponse} % Remplacez par votre image
\caption{Response du système en boucle ouverte.}
\end{figure}

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@ -1,28 +1,12 @@
\textbf{Question 7}
Nous avons implementé la loi de commande: $u(t) = -Ky(t) + Hy_{c}(t)$
Nous avons utilisé une boucle \texttt{for} sous MATLAB afin de tester les valeurs de $k$ entre 0 e 100.
Le gain $H$ est ensuite calculé à partir d'un système en boucle fermée de la forme :
\begin{equation}
\dot{x}(t) = (A-BK)x(t) + BHr(t),
\qquad
y(t)=Cx(t),
\end{equation}
on impose que, en régime permanent, la sortie suive la consigne. On utilise donc :
\begin{equation}
H = -\frac{1}{C(A-BK)^{-1}B}.
\end{equation}
Le gain $H$ est ensuite calculé à partir d'un système en boucle fermée sous la forme:
\begin{center}
$H=-\frac{1}{C(A-BKC)^{-1}B}$
\end{center}
Ainsi, les valeurs retenues pour la commande en boucle fermée sont :
\begin{equation}
\boxed{
k = 1{,}85,
\qquad
H = -3.
}
\end{equation}
$K = 1{,}85$ \textit{\&} $H = -3$

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@ -1,9 +1,10 @@
\textbf{Question 8}
La figure montre que $w(L,t)$ suit la consigne $w_c(t)$ et converge vers 1, avec une erreur nulle en régime permanent.
La Figure 3 montre que $w(L,t)$ suit la consigne $w_c(t)$ unitaire, avec une erreur nulle en régime permanent, et un temps caractéristique à $\tau = 1s$.
\begin{figure}[H]
\centering
\includegraphics[width=0.4\textwidth]{Illustrations/Q8} % Remplacez par votre image
\caption{Suivi de la consigne $wc(t)$ par la position $w(L,t)$.}
\includegraphics[scale=0.4]{Illustrations/Q8} % Remplacez par votre image
\caption{Suivi de la consigne $w_c(t)$ par la position $w(L,t)$.}
\label{fig:Q8_1}
\end{figure}

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@ -9,6 +9,7 @@
\subsection{Discrétisation par éléments finis}
\input{Questions/Q2}
\input{Questions/Q3}
\newpage
\input{Questions/Q4}
\input{Questions/Q5}
\input{Questions/Q6}
@ -23,6 +24,7 @@
%%newpage pour les imgs q9
\newpage
\input{Questions/Q9}
\newpage
\input{Questions/Q10}

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@ -2,41 +2,45 @@
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\textbf{Brage JOHNSEN}\\
\textbf{Gabriela DONZELLI}\\
\textbf{Gabriela DONZELLI}\\
\textbf{Thibaut ALETRUT}\\
Département GEI\\
4A-AE\\

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