583 lines
17 KiB
TeX
583 lines
17 KiB
TeX
\psection{Introduction}
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Blabla
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\newpage
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\section{Modelisation}
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\begin{equation*}
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\dot{x_{1}}, \dot{x_{2}} = f(e_{1},e_{2},q)
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\end{equation*}
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\begin{equation*}
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\left\{ \begin{aligned}
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x_{1}(\zeta,t) &= \frac{\partial^{2}\omega}{\partial\varepsilon^{2}}\\
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x_{2}(\zeta,t) &= \rho(\zeta)\frac{\partial\omega}{dt}
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\end{aligned} \right.
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\end{equation*}
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\begin{equation*}
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\left\{ \begin{aligned}
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e_{1}(\zeta,t) &= EI(\zeta)x_{1}\\
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e_{2}(\zeta,t) &= \frac{1}{\rho(\zeta)}x_{2} = \frac{\partial\omega}{\partial t}
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\end{aligned} \right.
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\end{equation*}
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\subsection{Question 1}
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\begin{equation*}
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\left\{\begin{aligned}
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\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})\\
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\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}
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\end{aligned}\right.
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\end{equation*}
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\begin{equation}
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\Rightarrow\boxed{\left\{\begin{aligned}
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\dot{x_{1}} &= \frac{\partial^2 e_2}{\partial\zeta^2} \\
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\dot{x_{2}} &= -\frac{\partial^2 e_1}{\partial\zeta^2}-q(\zeta,t)
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\end{aligned}\right. }
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\end{equation}
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\subsection{Question 2}
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\begin{align*}
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\left\{\begin{aligned}
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x_1 &\approx \phi^Tx_{1d}(t) \\
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x_2 &\approx \phi^Tx_{2d}(t)
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\end{aligned}\right. & &
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\left\{\begin{aligned}
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e_1 &\approx \phi^Te_{1d}(t) \\
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e_2 &\approx \phi^Te_{2d}(t)
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\end{aligned}\right.
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\end{align*}
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En utilisant la première ligne de l'équation 1, on trouve :
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\begin{equation*}
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\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_{2*d}
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\end{equation*}
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\begin{equation}
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\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}
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\end{equation}
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Intégration par partie (IPP) :
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\begin{equation*}
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\int_{0}^{L}u\cdot \dot{v} d\zeta = [u\cdot v]_{0}^{L} - \int_{0}^{L}\dot{u}\cdot v d\zeta
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\end{equation*}
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Dans notre cas :
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\begin{align*}
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\left\{\begin{aligned}
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u&=\phi(\zeta) \\
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\dot{v} &= \ddot{\phi}(\zeta)^T
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\end{aligned}\right. &
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\Rightarrow\left\{\begin{aligned}
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\dot{u} &= \dot{\phi}(\zeta) \\
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v &= \dot{\phi}(\zeta)^T
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\end{aligned}\right.
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\end{align*}
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On applique plusieurs fois :
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\begin{equation*}
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\begin{aligned}
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\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 \\
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&= \phi(L)\dot{\phi}(L)^T - \underbrace{\phi(0)\dot{\phi}(0)^T }_\text{=0} - \int_{0}^{L}\dot{\phi}(\zeta)\dot{\phi}(\zeta)^T d\zeta \\
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&= \phi(L)\dot{\phi}(L)^T - \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 \\
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&= \phi(L)\dot{\phi}(L)^T - \dot{\phi}(L)\phi(L)^T + \int_{0}^{L} \ddot{\phi}(\zeta)\phi(\zeta)^T d\zeta = D
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\end{aligned}
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\end{equation*}
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On continue avec ça en remplaçant dans l'équation 2 :
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\begin{equation*}
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\Rightarrow \boxed{E\dot{x}_{1d} = De_{2d}}
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\end{equation*}
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Puis on a dans la deuxième ligne de l'équation 1 :
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\begin{equation*}
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\begin{aligned}
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\phi^T(\zeta)\dot{x}_{2d} &= - \frac{\partial^2}{\partial\zeta^2} (\phi(\zeta)^T e_{1d}) - q(\zeta, t) \\
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\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}}
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\end{aligned}
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\end{equation*}
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Puis on trouve D :
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\begin{equation*}
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\begin{aligned}
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D &= \phi(L)\dot{\phi}(L)^T - \phi(L)\dot{\phi}(L)^T + \int_{0}^{L}\ddot{\phi}(\zeta)\phi(\zeta)^Td\zeta \\
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\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 \\
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\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
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\end{aligned}
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\end{equation*}
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Donc :
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\begin{equation*}
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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}
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\end{equation*}
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Or on sait que :
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\begin{equation}
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\Rightarrow\left\{\begin{aligned}
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\frac{de_1}{d\zeta} = u(t) &\Rightarrow \dot{\phi}(L)^T e_{1d} = u(t) \\
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e_1 (L,t) = 0 &\Rightarrow \phi(L)^T e_{1d} = 0
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\end{aligned}\right.
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\end{equation}
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Donc :
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\begin{equation*}
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\boxed{E\dot{x}_{2d} = - e_{1d}D^T - \phi(L)u(t) - F_{ext}}
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\end{equation*}
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Puis on a :
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\begin{equation*}
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\left\{\begin{aligned}
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y(t) &= -e_2 (L,t) \\
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e_2 (L,t) &\approx \phi(L)^T e_{2d}
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\end{aligned}\right.
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\end{equation*}
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Ce qui nous donne l'approximation :
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\begin{equation*}
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\boxed{y(t) = -\phi(L)^T e_{2d}}
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\end{equation*}
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\subsection{Question 3}
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On a :
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\begin{equation*}
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\left\{\begin{aligned}
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e_1 = EIx_1 = x_1 &\Rightarrow e_{1d} = x_{1d} \\
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e_2 = \frac{1}{\rho(\zeta)}x_2 = x_2 &\Rightarrow e_{2d} = x_{2d}
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\end{aligned}\right.
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\end{equation*}
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Donc :
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\begin{equation*}
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\left\{\begin{aligned}
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E\dot{X}_{1d} &= Dx_{2d} \\
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E\dot{X}_{2d} &= -D^T x_{1d} - \phi(L)u(t) - 0 \\
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y &= -\phi(L)^T x_{2d}
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\end{aligned}\right.
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\end{equation*}
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Ce qui nous donne :
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\begin{equation*}
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A = \begin{pmatrix}
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0 & E^{-1}D \\
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-E^{-1}D^T & 0
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\end{pmatrix}, \ B = \begin{pmatrix}
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0 \\
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-E^{-1}\phi(L)
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\end{pmatrix}, \ C = \begin{pmatrix}
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0 & -\phi(L)^T
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\end{pmatrix}
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\end{equation*}
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Pour :
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\begin{equation*}
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\left\{\begin{aligned}
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\zeta = L = 1 \\
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\phi(1) = \begin{pmatrix}
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0 \\
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1 \\
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0 \\
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0
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\end{pmatrix}
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\end{aligned}\right. \Rightarrow d\acute{e}cla \ A, \ B, \ C \ en \ MATLAB
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\end{equation*}
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\begin{equation*}
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A = \begin{pmatrix}
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0 & 0 & 0 & 0 & 114 & 6 & 12 & -2 \\
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0 & 0 & 0 & 0 & -54 & 6 & -2 & 4 \\
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0 & 0 & 0 & 0 & -1188 & -12 & -114 & 6 \\
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0 & 0 & 0 & 0 & -828 & -12 & -54 & 0 \\
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6 & 54 & 4 & -2 & 0 & 0 & 0 & 0 \\
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-6 & -114 & -2 & 12 & 0 & 0 & 0 & 0 \\
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-12 & -828 & -6 & 54 & 0 & 0 & 0 & 0 \\
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-12 & -1188 & -6 & 114 & 0 & 0 & 0 & 0 \\
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\end{pmatrix}, \ B = \begin{pmatrix}
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0 \\
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0 \\
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0 \\
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0 \\
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4 \\
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-16 \\
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-60 \\
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-120 \\
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\end{pmatrix}, \ C = \begin{pmatrix}
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0 & 0 & 0 & 0 & 0 & -1 & 0 & 0
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\end{pmatrix}
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\end{equation*}
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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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\begin{equation*}
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C_\omega(p) = \begin{bmatrix}
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\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
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\end{bmatrix}
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\end{equation*}
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\begin{equation*}
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\phi(L)^T = \begin{bmatrix}
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2p^3 - 3p^2 + 1 & 3p^2 - 2p^3 & p^3 - 2p^2 + p & p^3 - p^2
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\end{bmatrix}
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\end{equation*}
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\includegraphics[width=\textwidth]{Illustrations/Question4}
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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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\begin{align*}
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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}
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\end{align*}
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\begin{align*}
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\begin{bmatrix}
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\dot{x}_{1d} \\ \dot{x}_{1d}
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\end{bmatrix} = A \begin{bmatrix}
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x_{1d} \\ x_{2d}
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\end{bmatrix} + Bu & & y = C \begin{bmatrix}
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x_{1d} \\ x_{2d}
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\end{bmatrix}
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\end{align*}
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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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\end{bmatrix} = \begin{bmatrix}
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\frac{\partial^2\omega}{\partial\zeta^2}(\zeta,t) \\
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\rho(\zeta)\frac{\partial\omega}{\partial t}(\zeta,t)
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\end{bmatrix} = \begin{bmatrix}
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\phi(\zeta)^T & 0000 \\
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0000 & \phi(\zeta)^T
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\end{bmatrix} \begin{bmatrix}
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x_{1d} \\ x_{2d}
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\end{bmatrix}
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\end{equation*}
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\begin{equation*}
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\frac{\partial^2\omega}{\partial\zeta^2}(\zeta,t) = \underbrace{\begin{bmatrix}
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\phi(\zeta)^T & 0 & 0 & 0 & 0
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\end{bmatrix}}_{C.I.=0+0}x_{d}(t)
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\end{equation*}
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\begin{equation*}
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\begin{aligned}
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\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}
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\phi(\zeta)^T & 0 & 0 & 0 & 0
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\end{bmatrix}d\zeta + C_1(t) \\
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\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}
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\phi(\zeta)^T & 0 & 0 & 0 & 0
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\end{bmatrix}d\zeta^2 + \underbrace{\int C_1(t)d\zeta + C_2(t)}_{\zeta C_1(t) + C_2(t)}
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\end{aligned}
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\end{equation*}
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$C_1$ et $C_2$ ?????????????????
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\section{Retour de sortie}
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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{Question 8}
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\includegraphics[width=\textwidth]{Illustrations/Question8}
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\subsection{Question 9}
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\subsection{Question 10}
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\section{Retour d'état}
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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{Question 13}
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\subsection{Question 14}
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\section{Rejection de perturbation}
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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{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{Question 20}
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\newpage
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% VIEUX PROJET À RETIRER MAIS ON LE GARDE TEMPORAIREMENT POUR COPIER-COLLER
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\section{VIEUX PROJET À RETIRER MAIS ON LE GARDE TEMPORAIREMENT POUR COPIER-COLLER}
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\vspace{0.25cm}
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Le bille sur rail est une manipulation où le but est de stabiliser une bille sur un rail. Le rail est commandé par une tension, et les données lues sont l'angle du rail et la position de la bille. La position est achevé à l'aide d'un lecture d'impedance.
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\vspace{0.5cm}
|
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\\ \textbf{Le schèma de forces de la bille sur rail:}
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|
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\includegraphics{./Illustrations/Schema_Forces.png}
|
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|
||
\newpage
|
||
|
||
\subsection{Analyse du schèma bloc et setup}
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||
Nous avons remarqué que l'identification du système se fait en bouclé fermé. Voici le schèma bloc désignant le système que nous pouvons manipuler: \{Sett inn bilde av schèma bloc, système rail\}
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|
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\subsection{Mise en oeuvre de N4SID}
|
||
On a utilisé la fonction n4sid() du GIT de Mr. Poussot. Nous avons fait une experiènce temporel, frequentiel et avec Loewner.
|
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Voici le comportement des differents modèles obtenu: \{Sett inn bilde av n4sid\}
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|
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Cela nous avait mené à résumer le systeme du rail à la fonction de transfert suivante :
|
||
$$G(p) = \frac{NUM}{DEN}$$
|
||
|
||
\subsection{Fonction transfert du système: Rail}
|
||
Après avoir trouvé un modèle qui nous va, nous avons ensuite retrouvé la vraie fonction transferte du rail. Avec la relation qui suit:
|
||
\\
|
||
|
||
|
||
%%Lånt av disse her, smarte folk!
|
||
% Source - https://tex.stackexchange.com/q/175969
|
||
% Posted by student1, modified by community. See post 'Timeline' for change history
|
||
% Retrieved 2026-04-02, License - CC BY-SA 3.0
|
||
|
||
|
||
% Source - https://tex.stackexchange.com/a/175970
|
||
% Posted by Peter Grill, modified by community. See post 'Timeline' for change history
|
||
% Retrieved 2026-04-02, License - CC BY-SA 3.0
|
||
|
||
|
||
|
||
\tikzstyle{block} = [draw, fill=white, rectangle,
|
||
minimum height=3em, minimum width=6em]
|
||
\tikzstyle{sum} = [draw, fill=white, circle, node distance=1cm]
|
||
\tikzstyle{input} = [coordinate]
|
||
\tikzstyle{output} = [coordinate]
|
||
\tikzstyle{pinstyle} = [pin edge={to-,thin,black}]
|
||
|
||
\begin{tikzpicture}[auto, node distance=2cm,>=latex]
|
||
|
||
\node [input, name=input] {};
|
||
\node [sum, right of=input] (sum) {};
|
||
%%\node [block, right of=sum] (controller) {};
|
||
\node [block, right of=sum,
|
||
node distance=3cm] (system) {$G_{Rail}(s)$};
|
||
|
||
\draw [->] (sum) -- node[name=u] {$u$} (system);
|
||
\node [output, right of=system] (output) {};
|
||
%\node [block, below of=u] (measurements) {Measurements};
|
||
\coordinate [below of=u] (measurements) {};
|
||
|
||
\draw [draw,->] (input) -- node {$r$} (sum);
|
||
%\draw [->] (sum) -- node {$e$} (system);
|
||
\draw [->] (system) -- node [name=y] {$y$}(output);
|
||
%\draw [->] (y) |- (measurements);
|
||
\draw [-] (y) |- (measurements);
|
||
%\draw [->] (measurements) -| node[pos=0.99] {$-$}
|
||
\draw [->] (measurements) -| %node[pos=1.00] {$-$}
|
||
node [near end] {$y_m$} (sum);
|
||
|
||
\coordinate [below=1.7cm of sum] (u1) {};
|
||
\coordinate [below=1.88cm of y] (u2) {};
|
||
\draw[
|
||
decorate,
|
||
decoration={brace, mirror, amplitude=8pt}
|
||
]
|
||
(u1.south west) -- (u2.south east)
|
||
node[midway, below=10pt] {$H(s)$};
|
||
|
||
|
||
%\draw [->]
|
||
\end{tikzpicture}
|
||
\bigskip
|
||
|
||
\begin{equation}
|
||
H(s)=\frac{G(s)}{1+G(s)}\Rightarrow G(s)=\frac{H(s)}{1+H(s)}
|
||
\end{equation}
|
||
|
||
|
||
|
||
\subsection{Calcul du correcteur du système: P}
|
||
On a testé plusieurs valeurs, et conclue que juste un correcteur proportionnel du gain 1 fonctionne très bien.
|
||
\\
|
||
|
||
|
||
\begin{tikzpicture}[auto, node distance=2cm,>=latex]
|
||
|
||
\node [input, name=input] {};
|
||
\node [sum, right of=input] (sum) {};
|
||
\node [block, right of=sum] (controller) {Controleur: P};
|
||
\node [block, right of=controller,
|
||
node distance=3cm] (system) {$G_{Rail}(s)$};
|
||
|
||
\draw [->] (controller) -- node[name=u] {$u$} (system);
|
||
\node [output, right of=system] (output) {};
|
||
%\node [block, below of=u] (measurements) {Measurements};
|
||
\coordinate [below of=u] (measurements) {};
|
||
|
||
\draw [draw,->] (input) -- node {$r$} (sum);
|
||
\draw [->] (sum) -- node {$e$} (controller);
|
||
\draw [->] (system) -- node [name=y] {$y$}(output);
|
||
%\draw [->] (y) |- (measurements);
|
||
\draw [-] (y) |- (measurements);
|
||
%\draw [->] (measurements) -| node[pos=0.99] {$-$}
|
||
\draw [->] (measurements) -| %node[pos=1.00] {$-$}
|
||
node [near end] {$y_m$} (sum);
|
||
|
||
\coordinate [below=1.7cm of sum] (u1) {};
|
||
\coordinate [below=1.88cm of y] (u2) {};
|
||
\draw[
|
||
decorate,
|
||
decoration={brace, mirror, amplitude=8pt}
|
||
]
|
||
(u1.south west) -- (u2.south east)
|
||
node[midway, below=10pt] {$H_C(s)$};
|
||
|
||
|
||
%\draw [->]
|
||
\end{tikzpicture}
|
||
|
||
\subsection{Theorie de la loi de commande}
|
||
Faut mettre des choses ici. Je push!
|
||
|
||
\subsection{Une sous section}
|
||
\subsubsection{Une sous sous section}
|
||
Un mot compliqué\footnote{Une note de bas de page}
|
||
|
||
|
||
\newpage
|
||
|
||
\psection{Conclusion}
|
||
|
||
Une conclusion
|