79 lines
No EOL
1.4 KiB
TeX
79 lines
No EOL
1.4 KiB
TeX
\textbf{Question 3}
|
|
|
|
Le système devient avec $\rho = EI = 1$ et $q(\zeta,t)=0$ :
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
E\dot{x}_{1d}(t) &= D x_{2d}(t),\\
|
|
E\dot{x}_{2d}(t) &= -D^T x_{1d}(t)-\phi(L)u(t),\\
|
|
y(t) &= -\phi(L)^T x_{2d}(t).
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
En multipliant les deux premières équations par $E^{-1}$, on obtient :
|
|
|
|
\begin{equation}
|
|
\begin{aligned}
|
|
\dot{x}_{1d}(t) &= E^{-1}D x_{2d}(t),\\
|
|
\dot{x}_{2d}(t) &= -E^{-1}D^T x_{1d}(t)-E^{-1}\phi(L)u(t).
|
|
\end{aligned}
|
|
\end{equation}
|
|
|
|
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 \\
|
|
0 & 0 & 0 & 0 & -828 & -12 & -54 & 6 \\
|
|
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{bmatrix}
|
|
\end{equation}
|
|
|
|
\begin{equation}
|
|
B=
|
|
\begin{bmatrix}
|
|
0 \\ -E^{-1}\Phi(L)
|
|
\end{bmatrix}
|
|
=
|
|
\begin{bmatrix}
|
|
0 &
|
|
0 &
|
|
0 &
|
|
0 &
|
|
4 &
|
|
-16 &
|
|
-60 &
|
|
-120
|
|
\end{bmatrix}^{T}
|
|
\end{equation}
|
|
\begin{equation}
|
|
\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}
|
|
Finalement,
|
|
|
|
\begin{equation}
|
|
\boxed{
|
|
\begin{aligned}
|
|
\dot{x}_d(t) &= Ax_d(t)+Bu(t),\\
|
|
y(t) &= Cx_d(t).
|
|
\end{aligned}
|
|
}
|
|
\end{equation} |