## 3 Nonlinear Modeling of the Semi-Active Energy Regenerative Suspension with an EHA
### Mathematical Model of the Semi-Active Energy Regenerative Suspension with an EHA  
When the mathematical model of the EHA actuator is established, the friction between the piston, the cylinder wall, and the internal leakage of the system is ignored. Thus, the damping force generated by the hydraulic cylinder can be expressed as:  
\[F=\Delta P\cdot A \tag{5}\]  
where \(\Delta P\) is the pressure drop between the upper and lower surface of the piston in the hydraulic cylinder, and \(A\) is the effective piston area. We take into account the hydraulic pipeline's pressure loss, which includes the pressure loss along the path and the local pressure loss [19]. The pressure loss along the path of the hydraulic pipeline in the system is expressed as:  
\[\Delta P_{\chi}=\lambda\frac{l}{d}(\frac{\rho\
u^{2}}{2}) \tag{6}\]  
where \(\lambda\) is the along resistance coefficient, \(l\) is the length of the hydraulic pipe, \(d\) is the diameter of the hydraulic pipe, \(\rho\) is the hydraulic oil's density, and \(\
u\) is the oil flow rate in the pipeline. The local pressure loss of the hydraulic pipeline in the system is expressed as:  
\[\Delta P_{\zeta}=\zeta\frac{\rho\
u^{2}}{2} \tag{7}\]  
where \(\zeta\) is the local resistance coefficient. It can be seen that the total pressure loss of the hydraulic pipeline in the system can be obtained as follows:  
\[\Delta P_{g}=\Delta P_{\lambda}+\Delta P_{\zeta}=\left(\lambda\frac{l}{d}+ \zeta\right)\frac{\rho\
u^{2}}{2} \tag{8}\]  
according to the continuity equation of liquid flow, which is given by:  
\[Q_{d}=A\cdot\
u_{g}=\frac{\pi d^{2}}{4}\
u \tag{9}\]  
where \(Q_{d}\) is the flow of the hydraulic motor, and \(\
u_{g}\) is the speed of the piston rod. According to the working principle of the semi-active energy regenerative suspension with an EHA, it can be seen that under the action of body vibration the oil in the hydraulic cylinder enters the hydraulic motor and drives the hydraulic motor to work, and the output torque of the hydraulic motor drives the generator to generate electricity. At the same time, the angular velocity and output torque of the hydraulic motor meet the following relationships:  
\[\omega_{0}=\frac{2\pi Q_{d}}{q}\eta_{\
u} \tag{10}\]  
\[T_{d}=\frac{\Delta p_{d}q}{2\pi}\eta_{m} \tag{11}\]where \(q\) is the hydraulic motor displacement, \(\eta_{v}\) is the volumetric efficiency of the hydraulic motor, \(\Delta p_{d}\) is the pressure drop between the inlet and the outlet of the hydraulic motor, and \(\eta_{m}\) is the mechanical efficiency of the hydraulic motor. The generator converts the mechanical energy of the hydraulic motor into electrical energy, so the input torque of the generator is equal to the output torque of the hydraulic motor. Therefore, the output voltage \(U\) and the output torque \(T_{g}\) of the generator meet the following relationships:  
\[U=E-I(r+R)=IR_{0} \tag{12}\]  
\[T_{g}=J\ddot{\theta}+k_{t}I \tag{13}\]  
where \(E\) is the induced electromotive force, \(I\) is the loop current, \(r\) is the internal resistance of the generator, \(R\) is the external load resistance of the generator, \(R_{0}\) is the equivalent resistance of the energy regenerative circuit, \(J\) is the moment of inertia, \(\ddot{\theta}\) is the angular acceleration of the generator, and \(k_{t}\) is the torque constant of the generator. Among them, the induction electromotive force is expressed as:  
\[E=k_{e}n=\frac{30}{\pi}k_{e}\omega \tag{14}\]  
where \(k_{e}\) is the back electromotive force constant of the generator, \(n\) is the generator rotor speed, and \(\omega\) is the generator rotor angular velocity. The hydraulic motor is directly connected with the generator through the coupling, so their angular velocity and torque are equal, which is expressed as \(\omega_{0}=\omega\) and \(T_{d}=T_{g}\).