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 $k_N=54.26$

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

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

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

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

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

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

The Euler-Lagrange equations of motion of the system are given as follows

$$\begin{pmatrix} (M+m) & mL\cos{q_1} & 0\\ mL\cos{q_1} & mL^2+\Theta & \dfrac{\Theta}{2}\\ 0 & \dfrac{\Theta}{2} & \Theta \end{pmatrix} \begin{pmatrix} \ddot{x}\\ \ddot{q_1}\\ \ddot{q_2} \end{pmatrix} + \begin{pmatrix} -mL\sin{q_1\dot{q_1}^2}\\ -mLg_g\sin{q_1}\\ 0 \end{pmatrix} = \begin{pmatrix} Q_x\\ Q_1\\ Q_2 \end{pmatrix},$$

where $Q_x (N)$ is the generalized force pushing the cart in the horizontal “$x$” direction, $Q_1$ and $Q_2$ are torques in $(N · m)$ rotating the beam of the crane around a horizontal axis orthogonal to “$x$” and counter-rotating the hamper at the free end of the beam to avoid turning out the worker from the hamper. $L (m)$ denotes the lenght of the crane’s beam, $g_g$ ($m/s^2$) is the gravitational acceleration, $m$ ($kg$) and $\Theta$ $(kg · m^2)$ denote the momentum (with respect to its own center of mass that was supposed to be on the rotational axle) and the mass of the hamper.

Consider a pendulum system with Coulomb friction and external perturbation

$$ \ddot {\theta} = \frac{1}{J}u - \frac{g}{L}\sin \theta - \frac{V_s}{J}\dot{\theta } - \frac{P_s}{J}\mathrm{sgn}(\dot{\theta}) + \upsilon, $$

where parameters have the following values $M=1.1$, $L=0.9$, $J=ML^2=0.891$, $V_s=0.18$, $P_s=0.18$, $P_s=0.45$, $g=9.815$, and $v$ is an uncertain external perturbation $|\upsilon| \leq 1$.

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

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}. $