## 2 Problem Description and Model Analysis
### Model Analysis
#### 2.2.1 Model Establishment
The volume chamber at the top of the cylinder plunger constantly interacts with the hydraulic system to create different pressures. According to the node volume cavity method [40], the mathematical model of pressure in the hydraulic cavity is established as follows:  
\[p_{oil}(t)=\frac{E_{p}}{\sum V_{hp}}\int_{0}^{t}q_{oil}dt+p_{oil}^{0} \tag{3}\]  
where \(p_{oil}\) is the hydraulic pressure of the plunger of the hydraulic cylinder; \(E_{p}\) is the volume elastic modulus of the hydraulic fluid, which is a function of pressure; \(p_{oil}^{0}\) is the initial pressure of the plunger volume chamber; \(\sum V_{hp}\) is the volume of the upper cavity of the piston; and \(q_{oil}\) is the hydraulic oil flow. Because of this assumption, the delay of the pipeline and solenoid valve is not considered in this paper. Based on the incompressibility of oil, the hydraulic pressure depends on the flow of the hydraulic oil and the movement of the external load. The piston in the hydraulic cylinder overcomes the reciprocating motion of friction (\(f_{hp}\)) and load force (\(F_{l}\)) under the action of hydraulic pressure in the chamber. When combined with the pressure established via Equation (3) in the chamber, the equation of motion of the piston is as follows:  
\[\left\{ \begin{array}{l}\frac{\mathrm{d}^{2}x_{hp}}{\mathrm{d}t^{2}}=\frac{p_ {oil}\times A_{hp}-f_{hp}-F_{l}}{M_{hp}+M_{mu}}\\ \frac{\mathrm{d}\,p_{oil}}{\mathrm{d}\,t}=\frac{\mathrm{d}V_{oil}}{\mathrm{d}t }\times\frac{E_{0}\times(1-e^{-(0.4+p_{oil}\times 2\times 10^{-7})})}{v_{hp}^{0}+ \frac{A_{hp}}{\mathrm{d}t^{2}}\times A_{hp}}\end{array} \right. \tag{4}\]  
where \(\frac{\mathrm{d}V_{oil}}{\mathrm{d}t}\) is the flow of hydraulic oil in the hydraulic cylinder; \(A_{hp}\) is the cross-sectional area of the plunger of the hydraulic cylinder; \(p_{oil}\) is the hydraulic pressure of the plunger of the hydraulic cylinder; \(E_{0}\) is the volume elastic modulus of the hydraulic fluid; \(V_{hp}^{0}\) is the initial volume of the plunger volume chamber; \(f_{hp}\) is the friction of the plunger of the hydraulic cylinder; \(F_{l}\) is the load force of the hydraulic cylinder; \(M_{hp}\) and \(M_{mu}\) are the masses of the plunger and moving part of the unloader, respectively; and \(x_{hp}\) is the displacement of the piston. The load force of the hydraulic cylinder is the same as that of the unloader: \(F_{l}=F_{hy}\). The movement states of the hydraulic cylinder plunger and unloader are the same: \(x_{mu}=x_{hp}\). According to the hydraulic schematic shown in Figure 1, during the ejection process, the hydraulic oil is pumped out of the tank by the hydraulic pump, supplied in one direction to the hydraulic cylinder, and returned to the tank through the one-way intake relief valve. Therefore, the flow rate of hydraulic oil obtained by the hydraulic cylinder comes from the difference between the displacement of the pump and the flow rate of the relief valve. During plunger withdrawal, as the return relief valve is connected in series with the hydraulic circuit (not shown in Figure 1), the outgoing flow of the hydraulic cylinder is mainly dependent on the flow of the return relief valve. According to the principles of hydraulic systems, the flow rate of hydraulic oil in the hydraulic cylinder is related to the pressure settings of the intake and return relief valves. The equation of the flow rate of hydraulic oil in the hydraulic cylinder can be expressed as:  
\[\frac{\mathrm{d}V_{oil}}{\mathrm{d}t}= \begin{cases}\frac{\mathrm{d}V_{pump}}{ \mathrm{d}t}-\frac{\mathrm{d}V_{rr,in}}{\mathrm{d}t}(p_{in})&t_{1}\leq t\leq t_ {2}\\ 0&t_{2}<t<t_{3}\\ \frac{\mathrm{d}V_{rr,in}}{\mathrm{d}t}(p_{{}_{bf}})&t_{3}\leq t\leq t_{4} \end{cases}