\section{Modelisation}

\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. & &
	\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{align*}


\subsection{Question 1}

\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. & &
	\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}

\begin{align*}
		\left\{\begin{aligned}
			x_1 &\approx \phi^Tx_{1d}(t) \\
			x_2 &\approx \phi^Tx_{2d}(t)
		\end{aligned}\right. & &
		\left\{\begin{aligned}
			e_1 &\approx \phi^Te_{1d}(t) \\
			e_2 &\approx \phi^Te_{2d}(t)
		\end{aligned}\right.
\end{align*}

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_{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}

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} - \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 : $\Rightarrow \boxed{E\dot{x}_{1d} = De_{2d}}$ \\


Puis on a dans la deuxième ligne de l'équation 1 : 

\begin{equation*}
	\begin{aligned}
			\phi^T(\zeta)\dot{x}_{2d} &= - \frac{\partial^2}{\partial\zeta^2} (\phi(\zeta)^T e_{1d}) - q(\zeta, t) \\
			\int_{0}^{L} \phi^T(\zeta)d\zeta\times\dot{x}_{2d} &= -e_{1d}\int_{0}^{L}\phi(\zeta)\ddot{\phi}(\zeta)^T d\zeta - \underbrace{\int_{0}^{L}\phi(\zeta)q(\zeta,t)d\zeta}_{=F_{ext}}
	\end{aligned}
\end{equation*}

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 \\
		\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*}

Or on sait que : 

\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. & &
	\Rightarrow & &
	\boxed{E\dot{x}_{2d} = - e_{1d}D^T - \phi(L)u(t) - F_{ext}}
\end{align*}

Puis on a ceci :

\begin{align*}
	\left\{\begin{aligned}
		y(t) &= -e_2 (L,t) \\
		e_2 (L,t) &\approx	\phi(L)^T e_{2d} 
	\end{aligned}\right. & &
	\Rightarrow & &
	\boxed{y(t) = -\phi(L)^T e_{2d}}
\end{align*}


\subsection{Question 3}
On a :

\begin{equation*}
	\left\{\begin{aligned}
		e_1 = EIx_1 = x_1 &\Rightarrow e_{1d} = x_{1d} \\
		e_2 = \frac{1}{\rho(\zeta)}x_2 = x_2 &\Rightarrow e_{2d} = x_{2d}
	\end{aligned}\right.
\end{equation*}

Donc :

\begin{equation*}
	\left\{\begin{aligned}
		E\dot{X}_{1d} &= Dx_{2d} \\
		E\dot{X}_{2d} &= -D^T x_{1d} - \phi(L)u(t) - 0 \\
		y &= -\phi(L)^T x_{2d}
	\end{aligned}\right.
\end{equation*}

Ce qui nous donne :

\begin{equation*}
	A = \begin{pmatrix}
		0 & E^{-1}D \\
		-E^{-1}D^T & 0
	\end{pmatrix}, \  B = \begin{pmatrix}
		0 \\
		-E^{-1}\phi(L)
	\end{pmatrix}, \ C = \begin{pmatrix}
		0 & -\phi(L)^T
	\end{pmatrix}
\end{equation*}

Pour :

\begin{equation*}
	\left\{\begin{aligned}
		\zeta = L = 1 \\
		\phi(1) = \begin{pmatrix}
			0 \\
			1 \\
			0 \\
			0
		\end{pmatrix}
	\end{aligned}\right. 
\end{equation*}

\begin{equation*}
	A = \begin{pmatrix}
		0 & 0 & 0 & 0 & 114 & 6 & 12 & -2 \\
		0 & 0 & 0 & 0 & -54 & 6 & -2 & 4 \\
		0 & 0 & 0 & 0 & -1188 & -12 & -114 & 6 \\
		0 & 0 & 0 & 0 & -828 & -12 & -54 & 0 \\
		6 & 54 & 4 & -2 & 0 & 0 & 0 & 0 \\
		-6 & -114 & -2 & 12 & 0 & 0 & 0 & 0 \\
		-12 & -828 & -6 & 54 & 0 & 0 & 0 & 0 \\
		-12 & -1188 & -6 & 114 & 0 & 0 & 0 & 0 \\
	\end{pmatrix}, \  B = \begin{pmatrix}
		0 \\
		0 \\
		0 \\
		0 \\
		4 \\
		-16 \\
		-60 \\
		-120 \\
	\end{pmatrix}, \ C = \begin{pmatrix}
		0 & 0 & 0 & 0 & 0 & -1 & 0 & 0
	\end{pmatrix}
\end{equation*}


\subsection{Question 4}
\begin{equation*}
	\omega(p, t) = C_\omega(p)x_d(t)
\end{equation*}
	
\begin{equation*}
	C_\omega(p) = \begin{bmatrix}
		\frac{p^2(2p^3-5p^2+10)}{20} & -\frac{p^4(2p-5)}{20} & \frac{p^3(3p^2-10p+10)}{60} & \frac{p^4(3p-5)}{60} & 0 & 0 & 0 & 0
	\end{bmatrix}
\end{equation*}

\begin{equation*}
	\phi(L)^T = \begin{bmatrix}
		2p^3 - 3p^2 + 1 & 3p^2 - 2p^3 & p^3 - 2p^2 + p & p^3 - p^2
	\end{bmatrix}
\end{equation*}
\includegraphics[width=\textwidth]{Illustrations/Question4}

\subsection{Question 5}
\includegraphics[width=\textwidth]{Illustrations/Question5}\begin{align*}
	y  = -\phi(L)^Te_{2d} && y = -e_2(L, t) & & e_2(\zeta, t) \approx \phi(\zeta)^Te_{2d}(t)
\end{align*}

\begin{align*}
	x_1 = \frac{\partial^2\omega}{\partial\zeta^2}(\zeta,t)\approx\phi(\zeta)^2x_{1d} & & x_2(\zeta,t) = \rho(\zeta)\frac{\partial\omega}{\partial t}(\zeta,t) \approx\phi(\zeta)^Tx_{2d}
\end{align*}

\begin{align*}
	\begin{bmatrix}
		\dot{x}_{1d} \\ \dot{x}_{1d}
	\end{bmatrix} = A \begin{bmatrix}
	x_{1d} \\ x_{2d}
	\end{bmatrix} + Bu & & y = C \begin{bmatrix}
	x_{1d} \\ x_{2d}
	\end{bmatrix}
\end{align*}


\subsection{Question 6}
\begin{equation*}
	\begin{bmatrix}
		x_1 \\ x_2
	\end{bmatrix} = \begin{bmatrix}
		\frac{\partial^2\omega}{\partial\zeta^2}(\zeta,t) \\
		\rho(\zeta)\frac{\partial\omega}{\partial t}(\zeta,t)
	\end{bmatrix} = \begin{bmatrix}
		\phi(\zeta)^T & 0000 \\
		0000 & \phi(\zeta)^T
	\end{bmatrix} \begin{bmatrix}
		x_{1d} \\ x_{2d}
	\end{bmatrix}
\end{equation*}

\begin{equation*}
	\frac{\partial^2\omega}{\partial\zeta^2}(\zeta,t) = \underbrace{\begin{bmatrix}
		\phi(\zeta)^T & 0 & 0 & 0 & 0 
	\end{bmatrix}}_{C.I.=0+0}x_{d}(t)
\end{equation*}

\begin{equation*}
	\begin{aligned}
		\frac{\partial\omega}{\partial\zeta}(\zeta,t) &= \int\frac{\partial^2\omega}{\partial\zeta^2}(\zeta,t)d\zeta = x_d(t)\int\begin{bmatrix}
			\phi(\zeta)^T & 0 & 0 & 0 & 0 
		\end{bmatrix}d\zeta + C_1(t) \\
		\omega(\zeta,t) &= \int\int\frac{\partial^2\omega}{\partial^2\zeta^2}(\zeta,t)d\zeta^2 = x_d(t)\int\int\begin{bmatrix}
			\phi(\zeta)^T & 0 & 0 & 0 & 0 
		\end{bmatrix}d\zeta^2 + \underbrace{\int C_1(t)d\zeta + C_2(t)}_{\zeta C_1(t) + C_2(t)}
	\end{aligned}
\end{equation*}

$C_1$ et $C_2$ ?????????????????

\section{Retour de sortie}

\subsection{Question 7}
La sortie est donnée par $y(t) = C\,x_d(t)$, donc le système en boucle fermée s’écrit :

\begin{equation*}
\begin{cases}
	\dot{x}_d(t) = (A - B C k)\,x_d(t) + B H\,w_c(L,t) \\
	w(L,t) = C_w(L)\,x_d(t)
\end{cases}
\end{equation*}

En régime permanent, $\dot{x}_d(t)=0$, donc :
\begin{equation*}
	0 = (A - B C k)\,x_d + B H\,C_w(L)\,x_d
	\quad \Rightarrow \quad
	x_d = -(A - B C k)^{-1} B H w_c(L,t)
\end{equation*}

En multipliant l’expression par $C_w(L)$, on trouve :
\begin{equation*}
	w(L,t) = C_w(L)\,x_d = -C_w(L)(A - B C k)^{-1} B H w_c(L,t)
\end{equation*}

Pour assurer le suivi $w(L,t) = w_c(L,t)$, il faut :
\begin{equation*}
	- C_w(L)(A - B C k)^{-1} B H = I
\end{equation*}

D’où : 
\begin{equation*}
	\boxed{H = -\left(C_w(L)(A - B C k)^{-1} B\right)^{-1}}
\end{equation*}

Finalement, après calcul sous MATLAB, on obtient $H = -3$.


\subsection{Question 8}
\includegraphics[width=\textwidth]{Illustrations/Question8}


\subsection{Question 9}


\subsection{Question 10}



\section{Retour d'état}

\subsection{Question 11}
En utilisant la fonction \texttt{lqr()} de MATLAB avec $Q = I_8$ et $R = 1$, on trouve la matrice de gain $K$ suivante :
\begin{equation*}
	K = \begin{pmatrix}
		0.97 & -14.62 & 0.66 & 1.32 & 19.32 & 0.39 & 2.40 & -2.11
	\end{pmatrix}
\end{equation*}

\subsection{Question 12}
\includegraphics[width=\textwidth]{Illustrations/Question12}


\subsection{Question 13}
\includegraphics[width=\textwidth]{Illustrations/Question13}


\subsection{Question 14}
\includegraphics[width=\textwidth]{Illustrations/Question14}


\section{Rejection de perturbation}

\subsection{Question 15}
On sait que :
\begin{equation*}
	F_{ext} = \int_{0}^{L}\phi(\zeta)q(\zeta,t)d\zeta
	\quad \Rightarrow \quad 
	F_{ext} = q_0(t) \cdot \int_{0}^{L}\phi(\zeta)d\zeta \text{ car } q(\zeta,t) = q_0(t)
\end{equation*}
Or on a :
\begin{equation*}
	E\dot{x}_{2d} = - e_{1d}D^T - \phi(L)u(t) - F_{ext}
\end{equation*}
Donc, puisque $\dot{x}_{d}(t) = A x_d(t) + B u(t) + B_p q_0(t)$, on trouve :
\begin{equation*}
B_p =
\begin{pmatrix}
    0_{4 \times 1} \\
    -\int_{0}^{L}\phi(\zeta)d\zeta
\end{pmatrix}
=
\begin{pmatrix}
    0_{4 \times 1} \\
    \frac{1}{2} \\
    \frac{1}{2} \\
    \frac{1}{12} \\
    -\frac{1}{12}
\end{pmatrix}
\end{equation*}

\subsection{Question 16}
\includegraphics[width=\textwidth]{Illustrations/Question16}
Notre système ne permet pas de garantir une erreur nulle en régime permanent face
à une perturbation constante.

\subsection{Question 17}
Nous souhaitons garder les mêmes valeurs propres que celles obtenues lors de la question 11
avec la LQR. La valeur propre supplémentaire doit être plus à droite pour permettre
de rejeter la perturbation. On chosit $\lambda_9 = -2$.
On trouve alors : 
\begin{equation*}
	\lambda = \begin{pmatrix}
		-84.66 & -7.56 \pm 68.77i & -27.97 \pm 17.24i & -15.91 & -4.38 \pm 2.66i & -2
	\end{pmatrix}^T
\end{equation*}

\subsection{Question 18}
En utilisant la fonction \texttt{place()} de MATLAB, on trouve la matrice de gain $K$ suivante :
\begin{equation*}
K_{aug} =
\begin{pmatrix}
    K_1 & K_i
\end{pmatrix}
\quad \text{avec} \quad
K_1 \in \mathbb{R}^{1 \times 8}, \;
K_i \in \mathbb{R}
\end{equation*}

\begin{equation*}
	K_{aug} = \begin{pmatrix}
		-0.89 & -16.59 & 0.28 & 1.68 & 19.39 & -0.84 & 2.36 & -1.94 & 12.00
	\end{pmatrix}
\end{equation*}


\subsection{Question 19}
\includegraphics[width=\textwidth]{Illustrations/Question19}
En utilisant l'action intégrale, on parvient à rejeter la perturbation constante 
et à garantir une erreur nulle en régime permanent.

\subsection{Question 20}