$$\begin{align*} \dot{x}_1 &= x_1 = x_1x_4 + \theta_1x_1^2 \theta_2x_3 + x_2u_1 + u_2 + u_3 \\ \dot{x}_2 &= x_2\cos{x_3} + \theta_3x_4 + u_1 + u_4 \\ \dot{x}_3 &= \sin{x_2} - x_3 \\ \dot{x}_4 &= x_3 - x_4 + x_1x_4 + \theta_1x_1^2 + (1 + x_2)u_1 + u_2 + u_3 \\ y_1 &= x_1 \\ y_2 &= x_2, \end{align*}$$

where $\theta_1 = 0.5$, $\theta_2 = 2$, and $\theta_3 = 1$ are unknown constant parameters.

The motor dynamics are mapped by aset of five highly coupled nonlinear differential equations as given by

$$\begin{align*} \dfrac{{\mathrm d}i_{qs}^r}{{\mathrm d}t} &= \dfrac{1}{L_{\Sigma}} \left[-\dfrac{L_{RM}^2r_s+M^2r_r}{L_{RM}}i_{qs}^r + \dfrac{r_rM}{L_{RM}}\Psi_{qr}^r - (L_\Sigma i_{ds}^r + M \Psi_{dr}^r) \omega_r + L_{RM}u_{qs}^r\right] \\ \dfrac{{\mathrm d}i_{ds}^r}{{\mathrm d}t} &= \dfrac{1}{L_{\Sigma}} \left[-\dfrac{L_{RM}^2r_s+M^2r_r}{L_{RM}}i_{ds}^r + \dfrac{r_rM}{L_{RM}}\Psi_{dr}^r + (L_\Sigma i_{qs}^r + M \Psi_{qr}^r) \omega_r + L_{RM}u_{ds}^r\right] \\ \dfrac{{\mathrm d}\Psi_{qr}^r}{{\mathrm d}t} &= \dfrac{r_r M}{L_{RM}}i_{qs}^r - \dfrac{r_r}{L_{RM}}\Psi_{qr}^r \\ \dfrac{{\mathrm d}\Psi_{dr}^r}{{\mathrm d}t} &= \dfrac{r_r M}{L_{RM}}i_{ds}^r - \dfrac{r_r}{L_{RM}}\Psi_{dr}^r \\ \dfrac{{\mathrm d}\omega_r}{{\mathrm d}t} &= -\dfrac{B_m}{J}\omega_r + \dfrac{P}{2J}\left[\dfrac{P}{2}\dfrac{M}{L_{RM}}(i_{qs}^r\Psi_{dr}^r - i_{ds}^r-\Psi_{qr})-T_L\right], \end{align*}$$

where $L_{\Sigma} = L_{SM}L_{RM}-M^2$. For the load torque we assume the expression $T_L=c_2\omega_r^2 + c_3\omega_r^3$.

The state vector is given by

$x(t) = [i_{qs}^r, i_{ds}^r, \Psi_{qr}^r, \Psi_{dr}^r, \omega_r]^{\mathrm T} = [x_1, x_2, x_3, x_4, x_5]^{\mathrm T}.$

A schematic of the CSTR plant is shown in the attached image. The process dynamics are described by

$$\begin{align*} \dot{C}_{a} &=\frac{q}{V}(C_{a0}-c_{a})-a_{0}C_{a}e^{-\frac{E}{RT_{a}}} \\ \dot{T}_{a} &=\frac{q}{V}(T_{f}-T_{a})+a_{1}C_{a}e^{-\frac{E}{RT_{a}}}+a_{3}q_{c}\left(1-e^{\frac{a_{2}}{q_{c}}}\right)(T_{cf}-T_{a}), \end{align*}$$

where the variables $C_a$ and $T_a$ are the concentration and temperature of the tank, respectively; the coolant flow rate $q_c$ is the control input and the parameters of the plant are defined in the attached table. Within the tank reactor, two chemicals are mixed and react to produce a product compound $A$ at a concentration $C_a(t)$ with the temperature of the mixture being $T(t)$. The reaction is both irreversible and exothermic.

In the paper, authors assumed that plant parameters $q, C_{a0}, T_f$ and $V$ are at the nominal values given in the attached table. The activation energy $E/R = 1 \times 10^4K$ is assumed to be known. The state variables the input and the output are defined as $x=[x_1,x_2]^{\mathrm T}=[C_a,T_a]^{\mathrm T},u=q_c,y=C_a$. Using this notation, the CSTR plant can be re-expressed as

$$\begin{align*} \dot{x}_{1} &=1-x_{1}-a_{0}x_{1}e^{-\frac{10^4}{{\rm a}_2}} \\ \dot{x}_{2} &=T_{f}-x_{2}+a_{1}x_{1}e^{-\frac{10^4}{{\rm a}_2}}+a_{3}u\left(1-e^{-\frac{a_2}{u}}\right)(T_{cf}-x_{2}) \\ y &= x_{1}, \end{align*}$$

where the unknown constant parameters are $a_0, a_1, a_2$ and $a_3$.

The attached image depicts the kinematic car in the horizontal plane. Let us suppose that the Ackermann steering assumptions hold true, hence all wheels turn around the same point (denoted by P) which lies on the line of the rear axle. It follows that the kinematics of the car can be fully described by the kinematics of a bicycle fitted in the middle of the car (see attached image. The coordinates of the rear axle midpoint are given by $x$ and $y$. The orientation of the car with respect to the axis of $x$ is denoted by 9. The angle of the front wheel of the bicycle with respect to the longitudinal symmetry axis of the car is denoted by $φ$ . One may consider $φ$ or its time derivative $u_2=\dot{φ}$ as input. The longitudinal velocity of the rear axle midpoint is denoted by $u_1$ if it is a control input (two input case) and by $v_{car}$ if not (one input case). All lengths involved in the kinematic calculations, and in particular $l$, equal to one.

$$\begin{align*} \dot{x} &= u_1 \cos{\theta},\\ \dot{y} &= u_1 \sin{\theta},\\ \dot{\theta} &= u_1 \tan{\varphi},\\ \dot{\varphi} &= u_2. \end{align*}$$

The overall model consists of hydraulics (cylinders, valves, pump, etc.) and two degree of freedom linkage. The equations can be combined to form a MIMO state space system. The state vector, $x$, is defined in the table below. The input is the current in two valve solenoids as follows: $u = [i_1, i_2]^T$ . The output is given as, $y = [θ_1, θ_{21}]$. The states of the model are summarized in the table below. The dynamic equations for the linkage and electrohydraulic system in terms of state variables can be written as follows:

$$\begin{align*} \dot{x}_1 &= x_3\\ \dot{x}_2 &= x_4\\ \begin{bmatrix} \dot{x}_3 \\ \dot{x}_4 \end{bmatrix} &= M^{-1}(\tau(x_9, x_{10}, x_{11}, x_{12}, x_{13}) - h(x_1, x_2, x_3, x_4))\\ \dot{x}_5 &= \dfrac{\beta}{V_p}(\omega_px_6/2\pi - x_5K_{Lp}-(Q_{P A,1}(x_5, x_8, x_9) + Q_{P B, 1}(x_8, x_{10} + Q_{P A,2}(x_5, x_{11}, x_{12}) + Q_{P B,2}(x_5, x_{11}, x_{13})))\\ \dot{x}_6 &= [x_7 - x_5 + P_{margin}] G_p\\ \dot{x}_7 &= (\max(x_9,x_{10},x_{12},x_{13})-x_7)1/\tau_p\\ \dot{x}_8 &= (-x_8 + G_vu_1)1/\tau_v\\ \dot{x}_9 &= \dfrac{\beta}{V_{A,1}(x_1)}(Q_{PA,1(x_5,x_8,x_9}) + Q_{TA,1}(x_8,x_9-\dot{V}_{A,1}(x_3))\\ \dot{x}_{10} &= \dfrac{\beta}{V_{B,1}(x_1)}(Q_{PB,1(x_8,x_{10}}) + Q_{TB,1}(x_8,x_{10}-\dot{V}_{B,1}(x_3))\\ \dot{x}_{11} &= (-x_{11} + G_vu_2)1/\tau_v \dot{x}_{12} &= \dfrac{\beta}{V_{A,12}(x_2)}(Q_{PA,2(x_5,x_{11},x_{12}}) + Q_{TA,2}(x_{11},x_{12}-\dot{V}_{A,2}(x_4))\\ \dot{x}_{13} &= \dfrac{\beta}{V_{B,2}(x_2)}(Q_{PB,2(x_5,x_{11},x_{13}}) + Q_{TB,2}(x_{11},x_{13}-\dot{V}_{B,2}(x_4))\\ \end{align*}$$