Premiere partie finie, bisous à tous

latex/contents.tex

@@ -110,19 +110,172 @@ Puis on a dans la deuxième ligne de l'équation 1 :
 	\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
+	\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}
+	\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*}
+	\boxed{E\dot{x}_{2d} = - e_{1d}D^T - \phi(L)u(t) - F_{ext}}
+\end{equation*}
+
+Puis on a :
+
+\begin{equation*}
+	\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*}
+	\boxed{y(t) = -\phi(L)^T e_{2d}}
+\end{equation*}
 
 \subsection{Exercice 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. \Rightarrow d\acute{e}cla \ A, \ B, \ C \ en \ MATLAB
+\end{equation*}
 
 
 \subsection{Exercice 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*}
 
 
 \subsection{Exercice 5}
+\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{Exercice 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}