Electrical Train Optimal Control The electrical motor steady-state equation is $V_a = RI_a+\omega (M_{af}I_f)$
$V_a$ : drum motor voltage $\omega$ : angular velocity $M_{af}$ : mutual induce stator-rotor $I_f$ : rotor current
Motor torque $T = (M_{af}I_f)I_a$
From the equations above, with the addition of a square friction-air drag term, the dynamics of the train can be described from the equations below
$$
\begin{aligned}
& \dot{x}_1=x_2 \\
& \dot{x}_2=-k_1 x_2-k_2 x_2^2+k_3 u
\end{aligned}
$$
where, $x_1$ : the position of the train $x_2$ : the velocity of the train $u$ : the rotor current of the motor
Cost function $J=c_1\left(x_1(T)-x_{1 f}\right)^2+c_2 x_2^2(T)+\int_0^T k_4 x_2 u+R u^2 d t$
The theoretical optimal solution is
$$\dot{\mathbf{x}}=\left[\begin{array}{l}x_1 \ x_2\end{array}\right]=\left[\begin{array}{c}x_2 \ -k_1 x_2-k_2 x_2^2+k_3 u\end{array}\right], \mathbf{x}(0)=\mathbf{0}$$
$$H=L+\mathbf{p} \cdot(A \mathbf{x}+B u)=k_4 x_2 u+R u^2+p_1 x_2+p_2\left(-k_1 x_2-k_2 x_2^2+k_3 u\right)$$
$$\begin{aligned}
& \dot{\mathbf{p}}=-\frac{\partial H}{\partial \mathbf{x}} \Rightarrow\left\{\begin{array}{l}
\dot{p}_1=0 \\\
\dot{p}_2=-k_4 u+k_1 x_2-p_1+2 k_2 x_2 p_2
\end{array}\right. \\\
& \frac{\partial \varphi}{\partial \mathbf{x}}-\mathbf{p}\left(t_f\right)=\mathbf{0} \Rightarrow\left\{\begin{array}{l}
p_1\left(t_f\right)=2 c_1\left(x_1\left(t_f\right)-x_{1 f}\right) \\\
p_2\left(t_f\right)=2 c_2 x_2\left(t_f\right)
\end{array}\right.
\end{aligned}$$
$u^*=\frac{-\left(k_3 p_2+k_4 x_2\right)}{2 R}$ , $u \in\left[I_{\min }, I_{\max }\right]$
$x_{1f} = fixed$
Computationally, the open-loop solution is
For $k_1=0,5, k_2=0,1, k_3=1, c_1=c_2=1000, x_{1 f}=10, R=0,3, k_4=10, I_{\min }=-2, I_{\max }=2, T=10$ , the solution $u$ to the optimal control problem is given the graph
If the initial state was to mildly change, for example x(0) = [0.4, 0.6]
The system develops a steady-state error. In order to mitigate that error, we will introduce a feedback correction term (closed-loop solution)
$$A(t)=\left.\frac{\partial f}{\partial x}\right|_{(x(t), u(t))} \quad B(t)=\left.\frac{\partial f}{\partial u}\right|_{(x(t), u(t))}$$
$$A(t) = \left[\begin{array}{cc}0 & 1 \\ 0 & -k_1-2 k_2 x_2(t)\end{array}\right]$$ $
$$B(t) = \left[\begin{array}{cc}0 \\ k_3\end{array}\right]$$
$$\begin{array}{r}J=\mathbf{y}^T\left(t_f\right) S_f \mathbf{y}\left(t_f\right)+\int_0^{t_f}\left(\mathbf{y}^T(t) Q \mathbf{y}(t)+v^2(t)\right) d t \\ S_f=\left[\begin{array}{cc}20 & 0 \\ 0 & 20\end{array}\right], Q=\left[\begin{array}{cc}2 & 0 \\ 0 & 2\end{array}\right]\end{array}$$
After solving the Riccati equation
$$\begin{aligned}
& \dot{P}+P A+A^T P-P B B^T P+Q=0 \\\
& \Rightarrow\left\{\begin{array}{l}
\dot{P}_{11}=k_3^2 P_{12}^2-2 \\\
\dot{P}_{12}=k_3^2 P_{12} P_{22}-P_{11}-a(t) P_{12} \\\
\dot{P}_{22}=-k_3^2 P_{22}^2-2-2\left(P_{12}+a(t) P_{22}\right)
\end{array}\right. \\\
& v(y)=-k_3\left(P_{12} y_1(t)+P_{22} y_2(t)\right)
\end{aligned}$$
By substituting $u \rightarrow u+v$ , the position and speed steady-state errors vanish
