The dynamic engine model, with parameters for a 1.6 liter, 4-cylinder fuel injected engine, is a two-input/two-output system, given by the following differential equations: $\dot{P}=k_P(\dot{m}_{ai}-\dot{m}_{ao}),$ where $k_p=42.40$

$\dot{N}=k_N(T_i-T_L),$ where $k_N=54.26$

$\dot{m}_{ai}=(1+0.907\theta+0.0998\theta^2)g(P)$

$g{P}= \begin{cases} 1, & P<50.6625 \\ 0.0197(101.325P - P^2)^{\frac{1}{2}}, &P \geq 50.6625 \end{cases}$

$\dot{m}_{ao} = -0.0005968N-0.1336P+0.0005341NP+0.000001757NP^2$

$m_{ao} = \dot{m}_{ao}(t-\tau)/(120N), \ \tau=45/N$

$T_i = -39.22+325024m_{ao}-0.0112\delta^2+0.635\delta+(0.0216+0.000675\delta)N(2\pi/60)-0.000102N^2(2\pi/60)^2$

$T_L = (N/263.17)^2+T_d$.

Coupled nonlinear differential equations describing a process involving a continuous flow stirred tank reactor are given by

$$\begin{align*} \dot{C}_1 &= -C_1u + C_1(1-C_2)e^{C_2/\Gamma} \\ \dot{C}_2 &= -C_2u + C_1(1-C_2)e^{C_2/\Gamma}\dfrac{1+\beta}{1+\beta-C_2}. \end{align*}$$

In these equations, the state variables $C_1$ and $C_2$ represent dimensionless forms of cell mass and amount of nutrients in a constant volume tank, bounded between zero and unity. The control $u$ is the flow rate of nutrients into the tank, and is the same rate at which contents are removed from the tank. The constant parameters $\Gamma$ and $\beta$ determine the rates of cell formation and nutrient consumption; these parameters are set to $\Gamma$= 0.48 and $\beta$ = 0.02 for the nominal benchmark specification.

$$\begin{align*} x_1(k+1) &= 0.9x_1(k)\sin{[x_2(k)]} + \left(2 + 1.5 \dfrac{x_1(k)u_1(k)}{1+x_1^2(k)u_1^2(k)}\right)u_1(k) + \left(x_1(k) + \dfrac{2x_1(k)}{1+x_1^2(k)}\right)u_1(k)\\ x_2(k+1) &= x_3(k)(1+\sin{[4x_3(k)]}+ \dfrac{x_3(k)}{1+x_3^2(k)}\\ x_3(k+1) &= (3 + \sin{[2x_1(k)]})u_2(k)\\ y_1(k)&=x_1(k)\\ y_2(k)&=x_2(k) \end{align*}$$

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

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}. $