Avancement
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3 changed files with 138 additions and 42 deletions
34
BE_Beam.m
34
BE_Beam.m
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@ -173,20 +173,20 @@ plot(eig_zoh3,'x');
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hold off;
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%%
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sys_CL_Ts = c2d(sys_CL, 10, 'zoh');
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sys_CL_Ts = c2d(sys_CL, 1, 'zoh');
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figure
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step(sys_CL_Ts)
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%% Question 11
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K = lqr(sys, eye(size(A)), 1);
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H_lqr = inv(Cw_L*inv(-A+B*K)*B) ;
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K_lqr = lqr(sys, eye(size(A)), 1);
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H_lqr = inv(Cw_L*inv(-A+B*K_lqr)*B) ;
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%% Question 12
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sys_CL_lqr = ss(A-B*K, B*H_lqr, C, 0);
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sys_CL_lqr = ss(A-B*K_lqr, B*H_lqr, C, 0);
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[y, t, x] = lsim(sys_CL_lqr, cmd, t);
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u = H_lqr*cmd'-(K*x')' ;
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u = H_lqr*cmd'-(K_lqr*x')' ;
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Question12 = figure ;
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plot(t, u, t, y, t, cmd, t, (Cw_L*x')');
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xlabel("Temps (s)");
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@ -201,12 +201,12 @@ end
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%% Question 15
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Bp = [0; 0; 0; 0; 1/2*L^4-L^3+L; L^3-1/2*L^4; 1/4*L^4-2/3*L^3+1/2*L^2; 1/4*L^4-1/3*L^3];
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q = -5 * double (t' >= 4);
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sys_perturbations = ss(A-B*K, [B*H_lqr Bp], C, [0 0]);
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q0 = -5 * double (t' >= 4);
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sys_perturbations = ss(A-B*K_lqr, [B*H_lqr Bp], C, [0 0]);
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%%Question 16
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[y, t, x] = lsim(sys_perturbations, [cmd ; q], t');
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u = H_lqr*cmd'-(K*x')' ;
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[y, t, x] = lsim(sys_perturbations, [cmd ; q0], t');
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u = H_lqr*cmd'-(K_lqr*x')' ;
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Question16 = figure ;
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plot(t, u, t, y, t, cmd, t, (Cw_L*x')');
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xlabel("Temps (s)");
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@ -240,4 +240,18 @@ legend("$u(t)$", "$y(t)$", "$w_c(L,t)$", "$w(L,t)$", 'Interpreter', 'latex');
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title("Simulation du retour d'états avec action intégrale et perturbations") ;
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if SAVE
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exportgraphics(Question19, './latex/Illustrations/Question19.pdf')
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end
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end
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%%
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sys_integral_d = c2d(sys_integral, 0.5, 'zoh');
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t_d = 0:0.5:10; % Define the time vector for discrete simulation
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cmd_d = double (t_d >= 1) ;
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q0_d = -5 * double(t_d >= 4);
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[y_d, t_d, x_d] = lsim(sys_integral_d, [cmd_d ; q0_d], t_d);
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u_d = -(K_aug*x_d')' ;
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Question20 = figure ;
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plot(t_d, u_d, t_d, y_d, t_d, cmd_d, t_d, ([Cw_L 0]*x_d')');
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xlabel("Temps (s)");
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ylabel("Amplitude");
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legend("$u(t)$", "$y(t)$", "$w_c(L,t)$", "$w(L,t)$", 'Interpreter', 'latex');
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title("Simulation du retour d'états avec action intégrale et perturbations") ;
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@ -25,7 +25,7 @@ Blabla
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\end{equation*}
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\subsection{Exercice 1}
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\subsection{Question 1}
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\begin{equation*}
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\left\{\begin{aligned}
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@ -42,7 +42,7 @@ Blabla
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\end{equation}
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\subsection{Exercice 2}
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\subsection{Question 2}
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\begin{align*}
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\left\{\begin{aligned}
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@ -156,7 +156,7 @@ Ce qui nous donne l'approximation :
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\boxed{y(t) = -\phi(L)^T e_{2d}}
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\end{equation*}
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\subsection{Exercice 3}
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\subsection{Question 3}
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On a :
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\begin{equation*}
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@ -229,7 +229,7 @@ Pour :
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\end{equation*}
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\subsection{Exercice 4}
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\subsection{Question 4}
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\begin{equation*}
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\omega(p, t) = C_\omega(p)x_d(t)
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\end{equation*}
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@ -247,7 +247,7 @@ Pour :
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\end{equation*}
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\includegraphics[width=\textwidth]{Illustrations/Question4}
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\subsection{Exercice 5}
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\subsection{Question 5}
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\includegraphics[width=\textwidth]{Illustrations/Question5}\begin{align*}
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y = -\phi(L)^Te_{2d} && y = -e_2(L, t) & & e_2(\zeta, t) \approx \phi(\zeta)^Te_{2d}(t)
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\end{align*}
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@ -267,7 +267,7 @@ Pour :
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\end{align*}
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\subsection{Exercice 6}
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\subsection{Question 6}
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\begin{equation*}
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\begin{bmatrix}
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x_1 \\ x_2
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@ -303,56 +303,138 @@ $C_1$ et $C_2$ ?????????????????
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\section{Retour de sortie}
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\subsection{Exercice 7}
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\subsection{Question 7}
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La sortie est donnée par $y(t) = C\,x_d(t)$, donc le système en boucle fermée s’écrit :
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\begin{equation*}
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\begin{cases}
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\dot{x}_d(t) = (A - B C k)\,x_d(t) + B H\,w_c(L,t) \\
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w(L,t) = C_w(L)\,x_d(t)
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\end{cases}
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\end{equation*}
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En régime permanent, $\dot{x}_d(t)=0$, donc :
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\begin{equation*}
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0 = (A - B C k)\,x_d + B H\,C_w(L)\,x_d
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\quad \Rightarrow \quad
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x_d = -(A - B C k)^{-1} B H w_c(L,t)
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\end{equation*}
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En multipliant l’expression par $C_w(L)$, on trouve :
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\begin{equation*}
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w(L,t) = C_w(L)\,x_d = -C_w(L)(A - B C k)^{-1} B H w_c(L,t)
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\end{equation*}
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Pour assurer le suivi $w(L,t) = w_c(L,t)$, il faut :
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\begin{equation*}
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- C_w(L)(A - B C k)^{-1} B H = I
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\end{equation*}
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D’où :
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\begin{equation*}
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\boxed{H = -\left(C_w(L)(A - B C k)^{-1} B\right)^{-1}}
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\end{equation*}
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Finalement, après calcul sous MATLAB, on obtient $H = -3$.
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\subsection{Exercice 8}
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\subsection{Question 8}
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\includegraphics[width=\textwidth]{Illustrations/Question8}
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\subsection{Exercice 9}
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\subsection{Question 9}
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\subsection{Exercice 10}
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\subsection{Question 10}
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\section{Retour d'état}
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\subsection{Exercice 11}
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\subsection{Question 11}
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En utilisant la fonction \texttt{lqr()} de MATLAB avec $Q = I_8$ et $R = 1$, on trouve la matrice de gain $K$ suivante :
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\begin{equation*}
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K = \begin{pmatrix}
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0.97 & -14.62 & 0.66 & 1.32 & 19.32 & 0.39 & 2.40 & -2.11
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\end{pmatrix}
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\end{equation*}
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\subsection{Question 12}
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\includegraphics[width=\textwidth]{Illustrations/Question12}
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\subsection{Exercice 12}
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\includegraphics[width=\textwidth]{Illustrations/Question12}\begin{align*}
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\subsection{Question 13}
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\subsection{Exercice 13}
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\subsection{Exercice 14}
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\subsection{Question 14}
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\section{Rejection de perturbation}
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\subsection{Exercice 15}
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\subsection{Question 15}
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On sait que :
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\begin{equation*}
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F_{ext} = \int_{0}^{L}\phi(\zeta)q(\zeta,t)d\zeta
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\quad \Rightarrow \quad
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F_{ext} = q_0(t) \cdot \int_{0}^{L}\phi(\zeta)d\zeta \text{ car } q(\zeta,t) = q_0(t)
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\end{equation*}
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Or on a :
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\begin{equation*}
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E\dot{x}_{2d} = - e_{1d}D^T - \phi(L)u(t) - F_{ext}
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\end{equation*}
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Donc, puisque $\dot{x}_{d}(t) = A x_d(t) + B u(t) + B_p q_0(t)$, on trouve :
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\begin{equation*}
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B_p = \begin{pmatrix}
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0_{4 \times 1} \\
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-\int_{0}^{L}\phi(\zeta)d\zeta
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\end{pmatrix}
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= \begin{pmatrix}
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0_{4 \times 1} \\
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1/2 \\
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1/2 \\
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1/12 \\
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-1/12
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\end{pmatrix}
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\end{equation*}
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\subsection{Question 16}
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\includegraphics[width=\textwidth]{Illustrations/Question16}
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Notre système ne permet pas de garantir une erreur nulle en régime permanent face
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à une perturbation constante.
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\subsection{Question 17}
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Nous souhaitons garder les mêmes valeurs propres que celles obtenues lors de la question 11
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avec la LQR. La valeur propre supplémentaire doit être plus à droite pour permettre
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de rejeter la perturbation. On chosit $\lambda_9 = -2$.
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On trouve alors :
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\begin{equation*}
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\lambda = \begin{pmatrix}
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-84.66 & -7.56 \pm 68.77i & -27.97 \pm 17.24i & -15.91 & -4.38 \pm 2.66i & -2
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\end{pmatrix}^T
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\end{equation*}
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\subsection{Question 18}
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En utilisant la fonction \texttt{place()} de MATLAB, on trouve la matrice de gain $K$ suivante :
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\begin{equation*}
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K_{aug} = \begin{pmatrix}
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K_1 & K_i
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\end{pmatrix} \text{ avec } K_1 \in \mathbb{R}^{1 \times 8} \text{ et } K_i \in \mathbb{R}
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K_{aug} = \begin{pmatrix}
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-0.89 & -16.59 & 0.28 & 1.68 & 19.39 & -0.84 & 2.36 & -1.94 & 12.00
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\end{pmatrix}
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\end{equation*}
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\subsection{Exercice 16}
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\includegraphics[width=\textwidth]{Illustrations/Question16}\begin{align*}
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\subsection{Question 19}
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\includegraphics[width=\textwidth]{Illustrations/Question19}
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En utilisant l'action intégrale, on parvient à rejeter la perturbation constante
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et à garantir une erreur nulle en régime permanent.
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\subsection{Exercice 17}
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\subsection{Exercice 18}
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\subsection{Exercice 19}
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\includegraphics[width=\textwidth]{Illustrations/Question19}\begin{align*}
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\subsection{Exercice 20}
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\subsection{Question 20}
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latex/main.pdf
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