$$\dot{x}=A_Fx+b_F\tau+e_F(x)$$

with

$A_S = \begin{bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & -K \dfrac{m_n}{\beta_F} & -C \dfrac{m_n}{\beta_F} \\ 0 & 0 & 0 & 1 \\ 0 & 0 & -K \dfrac{m_{bnt}}{\beta_F} & -C \dfrac{m_{bnt}}{\beta_F} \\ \end{bmatrix},\\ x = \begin{bmatrix} z\\ \dot{z}\\ p\\ \dot{p} \end{bmatrix}, b_F = \dfrac{\eta}{\beta_{F}r} \begin{bmatrix} 0\\ m_n\\ 0\\ m_{bn} \end{bmatrix},\\ e_F(x)=\dfrac{1}{\beta_F} \begin{bmatrix} 0\\ \alpha(m_{bnt}g + f_{fF} + m_n(m_ng - k(s_0 - l_0) - f_p))\\ 0\\ -m_nf_{fF} - m_{bnt}(k(s_0-l_0) + f_a)\\ \end{bmatrix}. $

$$\dot{x}=A_Sx+b_S\tau+e_S(x)$$

with

$A_S = \begin{bmatrix} 0 & 1 & 0 & 0 \\ -K\dfrac{J}{\beta_Sr^2} & -C\dfrac{J}{\beta_Sr^2} & K\dfrac{J}{\beta_Sr^2} & C\dfrac{J}{\beta_Sr^2} \\ 0 & 0 & 0 & 1 \\ K\dfrac{m_b}{\beta_s} & C\dfrac{m_b}{\beta_s} & -K\dfrac{m_b}{\beta_s} & -C\dfrac{m_b}{\beta_s} \\ \end{bmatrix},\\ x = \begin{bmatrix} z\\ \dot{z}\\ p\\ \dot{p} \end{bmatrix}, b_S = \dfrac{\eta}{\beta_{S}r} \begin{bmatrix} 0\\ m_n\\ 0\\ m_{bn} \end{bmatrix},\\ e_S(x)=\dfrac{1}{\beta_s} \begin{bmatrix} 0\\ \alpha(ks_0 - m_{bn}g- f_{fS} + m_n(m_ng - ks_0 - f_a)\\ 0\\ 0\\ \end{bmatrix}. $

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

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 $a_3$.