issue #54

improved indices

doc/chap_examples.rst

@@ -518,23 +518,32 @@ are shown in green. Observe that in the congested mode, the density
 :math:`x_2` in the congested part decreases slightly, while the density
 :math:`x_1` upstream of the congested part increases. The blue set above
 the guard is not actually reached, because the state evolves according
-to the green region.
-
+to the green region.
+
+
+.. raw:: html
+
+	<h2>References</h2>
+
+.. [SUN2003] L.Muñoz, X.Sun, R.Horowitz, and L.Alvarez. 2003. Traffic Density
+   Estimation with the Cell Transmission Model. In *Proceedings of the
+   American Control Conference*, 3750–3755. Denver, Colorado, USA.
+      
 Inverted Pendulum On a Cart
 ---------------------------
-
+
 .. _invpendfig:
 
 .. figure:: /pic/chapter06_section05_inpendulum.png
    :alt: invpend
-   :width: 50 %
-
-   Inverted pendulum
-
-Consider a mechanical system consisting of a cart (body 1) and a rod (body 2) attached to the cart with a joint. We assume that movement of the bodies is plane. The system represents an inverted pendulum with unstable equilibrium at :math:`\theta = \frac{\pi}{2}.` There is a control :math:`u(t)` that presents a force applied to the cart along the axis OX. 
-
-Therefore we define
-
+   :width: 50 %
+
+   Inverted pendulum
+
+Consider a mechanical system consisting of a cart (body 1) and a rod (body 2) attached to the cart with a joint. We assume that movement of the bodies is plane. The system represents an inverted pendulum with unstable equilibrium at :math:`\theta = \frac{\pi}{2}.` There is a control :math:`u(t)` that presents a force applied to the cart along the axis OX. 
+
+Therefore we define
+
 -  :math:`\theta` - angle between axis OX and the rod;
 
 -  :math:`m_1` and :math:`m_2` - weights of the bodies 1 and 2 respectively;
@@ -545,43 +554,43 @@ Therefore we define
 
 -  :math:`J_c` - inertia moment of the second body relative to its center of mass;
 
--  :math:`u` - the force applied to cart and we assume :math:`|u| \leqslant \alpha`. 
-
-
-Firstly we consider the system without any control. Writing down expressions for kinetic and potential energies of the bodies 1 and 2 we obtain:
-
-.. math::
-   :label: invpend_energy
-
-   \begin{aligned}
-   T_1 &= \frac{m_1v^2}{2}, \\
-   T_2 &= \frac{J_c\omega^2}{2}, \\
-   \Pi_1 &= 0, \\
-   \Pi_2 &= m_2g\frac{L}{2}\sin(\theta). 
-   \end{aligned}
-
-Here :math:`T_1` and :math:`T_2` stand for kinetic energy of cart and rod respectively, :math:`\Pi_1` and :math:`\Pi_2` stand for potential energy, :math:`v` is a speed of a cart and :math:`\omega` is an angular velocity of the rod. Further we will replace :math:`v` with :math:`\dot x`, :math:`\omega` with :math:`\dot \theta`. 
-
-As there is viscosity, we write down generalized forces:
-
-.. math::
-   :label: invpend_forces
-
-   \begin{aligned}
-   Q_1 &= k_1\dot x, \\
-   Q_2 &= k_2L\dot \theta.
-   \end{aligned}
-
-Writing down the Langrange equations in case of potential and nonpotential forces we obtain:
-
-.. math::
-   :label: invpend_lagrange
-
-   \frac{d}{dt}(\frac{\partial T}{\partial \dot q_i}) - \frac{\partial T}{\partial q_i} + \frac{\partial \Pi}{\partial q_i} = Q_i.
-
-Here the Lagrange coordinates :math:`q_1 = x, q_2 = \theta`, :math:`T = T_1 + T_2, \Pi = \Pi_1 + \Pi_2`.
-
-Applying previously obtained values to the equations and adding the control to the first equation, we get:
+-  :math:`u` - the force applied to cart and we assume :math:`|u| \leqslant \alpha`. 
+
+
+Firstly we consider the system without any control. Writing down expressions for kinetic and potential energies of the bodies 1 and 2 we obtain:
+
+.. math::
+   :label: invpend_energy
+
+   \begin{aligned}
+   T_1 &= \frac{m_1v^2}{2}, \\
+   T_2 &= \frac{J_c\omega^2}{2}, \\
+   \Pi_1 &= 0, \\
+   \Pi_2 &= m_2g\frac{L}{2}\sin(\theta). 
+   \end{aligned}
+
+Here :math:`T_1` and :math:`T_2` stand for kinetic energy of cart and rod respectively, :math:`\Pi_1` and :math:`\Pi_2` stand for potential energy, :math:`v` is a speed of a cart and :math:`\omega` is an angular velocity of the rod. Further we will replace :math:`v` with :math:`\dot x`, :math:`\omega` with :math:`\dot \theta`. 
+
+As there is viscosity, we write down generalized forces:
+
+.. math::
+   :label: invpend_forces
+
+   \begin{aligned}
+   Q_1 &= k_1\dot x, \\
+   Q_2 &= k_2L\dot \theta.
+   \end{aligned}
+
+Writing down the Langrange equations in case of potential and nonpotential forces we obtain:
+
+.. math::
+   :label: invpend_lagrange
+
+   \frac{d}{dt}(\frac{\partial T}{\partial \dot q_i}) - \frac{\partial T}{\partial q_i} + \frac{\partial \Pi}{\partial q_i} = Q_i.
+
+Here the Lagrange coordinates :math:`q_1 = x, q_2 = \theta`, :math:`T = T_1 + T_2, \Pi = \Pi_1 + \Pi_2`.
+
+Applying previously obtained values to the equations and adding the control to the first equation, we get:
 
 .. math:: 
    :label: invpend1
@@ -591,10 +600,10 @@ Applying previously obtained values to the equations and adding the control to t
 .. math::
    :label: invpend2
    
-   (J_c + \frac{m_2L^2}{4})\ddot{\theta}+\frac{m_2gL\cos(\theta)}{2} + k_2\dot{\theta}L = 0 . 
-
-
-After linerarization in the neighbourhood of :math:`\frac{\pi}{2}` we have :math:`\cos(\theta) \approx \frac{\pi}{2} - \theta`.
+   (J_c + \frac{m_2L^2}{4})\ddot{\theta}+\frac{m_2gL\cos(\theta)}{2} + k_2\dot{\theta}L = 0 . 
+
+
+After linerarization in the neighbourhood of :math:`\frac{\pi}{2}` we have :math:`\cos(\theta) \approx \frac{\pi}{2} - \theta`.
 Defining :math:`x_1 = x, x_2=\dot{x}_1, x_3 = \theta` and :math:`x_4=\dot{x}_3`, we can rewrite :eq:`invpend1`-:eq:`invpend2` as a linear system in standard form:
 
 .. math::
@@ -617,12 +626,12 @@ Defining :math:`x_1 = x, x_2=\dot{x}_1, x_3 = \theta` and :math:`x_4=\dot{x}_3`,
    0 \\
    \frac{1}{m_1} \\
    0 \\
-   0 \end{array}\right]u + \left[\begin{array}{cc}
-   0 \\
-   0 \\
-   0 \\
+   0 \end{array}\right]u + \left[\begin{array}{cc}
+   0 \\
+   0 \\
+   0 \\
    -\frac{m_2Lg\pi}{4J_c+m_2L^2} \end{array}\right].   
-
+
 Consider some moment of time :math:`t_1` and final position :math:`x_1(t_1) = x_1, x_2(t_1) = 0, x_3(t_1) = \frac{\pi}{2}, x_4(t_1) = 0`. It is required to calculate the backward reachability sets (tube) for the linearized system :eq:`invpendls` emanating from the given final position and project it onto :math:`(x_1, x_3)` subspace. It is also required to identify whether it’s possible to reach the final position from a given initial position :math:`x_1(t_1) = x_1, x_2(t_1) = v_1, x_3(t_1) = \theta_1, x_4(t_1) = \omega_1` using some admissible control function.
 
 .. raw:: html
@@ -1004,6 +1013,8 @@ Solving collision problem:
    robust longitudinal guidance. Proceedings of the 17th World Congress
    The International Federation of Automatic Control
    Seoul, Korea, July 6-11, 2008
+=======
+Consider some moment of time :math:`t_1` and final position :math:`x_1(t_1) = x^1_1, x_2(t_1) = 0, x_3(t_1) = \frac{\pi}{2}, x_4(t_1) = 0`. It is required to calculate the backward reachability sets (tube) for the linearized system :eq:`invpendls` emanating from the given final position and project it onto :math:`(x_1, x_3)` subspace. It is also required to identify whether it’s possible to reach the final position from a given initial position :math:`x_1(t_0) = x^0_1, x_2(t_0) = v_1, x_3(t_0) = \theta_1, x_4(t_0) = \omega_1` using some admissible control function.
 
 Blending tank with delay
 ------------------------