# Delay Analysis and the Control of Electro-Hydrostatic Actuators
## 2 Materials and Methods
### Model of Hydraulic System  
The most basic EHA hydraulic system consists of a hydraulic pump and a piston-cylinder. There are also many nonlinear factors in the hydraulic system. Therefore, the AMESIM was used to build a hydraulic system model, which could better reflect the characteristics of the hydraulic components. As the hydraulic system's power element, the hydraulic pump can convert the mechanical energy into hydraulic energy and transfer energy to the piston cylinder to push the load. The flow of the pump can be calculated as follows:  
\[Q=D_{P}\cdot n-C_{1}(P_{1}-P_{2})-C_{2}P_{1}, \tag{6}\]  
\(n\) is the input speed of the hydraulic pump, \(C_{1}\), \(C_{2}\) is the inner leakage constant and outer leakage constant, \(P_{1}\) and \(P_{2}\) are the outlet and the inlet pressure of the pump. The motor drives the hydraulic pump, and the pump's input torque is equal to the motor's output torque. The output torque of the pump is:  
\[T=Q(P_{1}-P_{2})/w, \tag{7}\]  
where the torque of the friction and the moment of inertia of the pump are ignored. The friction and the moment of inertia can be ignored when the load is large enough. The bidirectional hydraulic lock is a kind of pressure protection element composed of two parallel piloted check valves, and they cross-connect the inlet and the pilot port. The hydraulic lock can overcome the influence caused by the load's disturbance, but it may lead to a delay. Therefore, the hydraulic lock is a critical component that should be analyzed. The mechanical structure and the symbol of the hydraulic lock are shown in Figure 5. The piloted check valve A is located at the inlet of the piston-cylinder, while the piloted check valve B is located at the outlet. Figure 4: The simulation and experimental results of the motor system. (**a**) The simulation results. (**b**) the experimental results. The opening of the valve satisfies the function [34]. \[x_{v}=\frac{P_{in}-P_{out}-P_{crack}+P_{p3}\cdot ratio}{K_{s}}, \tag{8}\]  
The dimensionless valve opening is \(x_{v}\), \(P_{in}\) and \(P_{out}\) are the inlet and the outlet pressure of the valve, \(P_{crack}\) is the check valve cracking pressure, \(P_{p3}\) is the pilot pressure, ratio is the pilot area ratio, and \(K_{s}\) is the equivalent spring rate. When the valve is not fully opened, it is equivalent to a throttle. The flow is associated with differential pressure and dimensionless valve opening [34]:  
\[Q=kx_{v}\sqrt{\frac{2\Delta P}{\rho}}, \tag{9}\]  
The flow coefficient is \(k\), \(\Delta P\) is the pressure difference of the inlet and the outlet of the valve, and \(\rho\) is the fluid density. The valve is fully open when the pilot pressure reaches the following condition [34]. \[P_{p3}\geq\frac{P_{out}-P_{in}+P_{crack}}{ratio}, \tag{10}\]  
According to Equations (8)-(10), the expression of the check valve is:  
\[Q= \begin{cases}0,&P_{p3}\leq\frac{P_{out}-P_{in}+P_{crack}}{ratio}\cap P_{in} \leq P_{out}+P_{crack},\\ Q=kx_{\sigma}\sqrt{\frac{2\Delta P}{\rho}},&P_{p3}>\frac{P_{out}-P_{in}+P_{crack }}{ratio}\cup P_{in}>P_{out}+P_{crack},\end{cases}  \tag{11}\]  
According to the above Equation (9), the hydraulic lock may cause a delay. When the valve is not open, the output of flow is 0. When the valve is open, the flow starts to output. The delay time depends on the valve opening time. The piston cylinder is the primary execution component of the hydraulic system, which outputs force and displacement, according to the flow rate continuity and the force balance equation:  
\[Q=A\frac{dx}{dt}+C_{g}(P_{1}-P_{2})+\frac{V_{t}}{4\beta_{e}}\frac{d(P_{1}-P_{ 2})}{dt}, \tag{12}\]  
\[F=A(P_{1}-P_{2})=m_{t}\frac{d^{2}x}{dt^{2}}+B_{c}\frac{dx}{dt}+kx+F_{t}, \tag{13}\]  
Then, the mathematical model of the piston cylinder can be obtained.