## 2 Hydraulic Simulation and Theoretical Analysis
### Theoretical Analysis of the Low-Frequency Characteristic
#### 2.2.1 Analysis of Pressure in Different Stages Without Internal Leakage  
In the rodless chamber, when no internal leakage exists, the effective volume \(V_{eff}(P)\) and the net flow \(Q(P)\) are as follows:  
\[V_{eff}(P)=A_{1}x+V_{1} \tag{6}\]  
\[Q(P)=q_{1}-vA_{1}\frac{\rho(P_{1})}{\rho(0)} \tag{7}\]  
Figure 4: The pressure fluctuation amplitude at the main frequency with different internal leakage levels.  
where \(x\) is the displacement of piston rod, \(V_{1}\) is the initial volume of the rodless chamber, \(V_{2}\) is the initial volume of the rod chamber, \(v\) is the movement speed of the piston rod, \(q_{1}\) is the oil inflow into the rodless chamber, \(q_{2}\) is the oil inflow into the rod chamber.  
The effective bulk modulus within the volume \(B_{eff}(P)\) can be calculated as:  
\[B_{wall}(P)=\frac{1+w_{comp}\times P}{w_{comp}} \tag{8}\]  
\[\frac{1}{B_{eff}(P)}=\frac{1}{B_{fluid}(P)}+\frac{1}{B_{wall}(P)} \tag{9}\]  
where \(B_{wall}(P)\) is the fluid bulk modulus, \(w_{comp}\) is the wall compliance, \(B_{fluid}(P)\) is the fluid bulk modulus.  
The time derivative of pressure is:  
\[\frac{dP}{dt}=\frac{B_{eff}(P)\times Q(P)}{V_{eff}(P)} \tag{10}\]  
From the above equation, the state equation of the pressure of a single hydraulic cylinder is described as follows [35]:  
\[\dot{P_{1}}=\frac{K_{B}\Big{(}q_{1}-vA_{1}\frac{\rho(P_{1})}{\rho(0)}\Big{)}}{ A_{1}x+V_{1}} \tag{11}\]  
\[\dot{P_{2}}=\frac{K_{B}\Big{(}-q_{2}+vA_{2}\frac{\rho(P_{2})}{\rho(0)}\Big{)}} {V_{2}-A_{2}x} \tag{12}\]  
where \(K_{B}\) is the effective elastic modulus of oil.  
When the Phase 1 starts, \(v=0\). Driven by \(q_{1}\), \(v\), \(P_{1}\) and \(P_{2}\) gradually increase. Then, in the Phase 2, \(v=v_{s}\), \(P_{1}\) and \(P_{2}\) gradually increase to \(P_{1}^{max}\) and \(P_{2}^{max}\) (\(\dot{P_{1}}=0\), \(\dot{P_{2}}=0\), \(P_{1}^{max}>P_{2}^{max}\)), and the following formula holds:  
\[P_{1}^{max}A_{3}+(P_{1}^{max}-P_{2}^{max})A_{2}=F_{L} \tag{13}\]  
In Phase 3, the \(q_{1}\) rapidly decreases to 0. Correspondingly, according to the Equations (1), (11) and (12), the reduction of \(q_{1}\) can forces the decrease of \(P_{1}\), and then \(v_{s}\) decreases, which force the decrease of \(P_{2}\). As a result, \(P_{1}^{max}\) and \(P_{2}^{max}\) gradually decrease to \(P_{1}^{s}\) and \(P_{2}^{s}\), and \(P_{1}^{s}\geq P_{2}^{s}\) holds, which can be proved as follows:  
Firstly, \(P_{1}^{max}\)\(>P_{1}^{s}\) and \(P_{1}^{max}\)\(>P_{2}^{s}\) holds. In Phase 4, \(v=0\) and \(a=0\) holds. Then, according to the Equation (1), the following formula holds:  
\[P_{1}^{s}A_{1}-P_{2}^{s}A_{2}=F_{L} \tag{14}\]  
\[P_{1}^{s}A_{3}+\left(P_{1}^{s}-P_{2}^{s}\right)A_{2}=F_{L} \tag{15}\]  
There are two cases: \(P_{1}^{s}<P_{2}^{s}\) or \(P_{1}^{s}\geq P_{2}^{s}\). If \(P_{1}^{s}<P_{2}^{s}\), \(P_{1}\geq P_{2}\) and \(P_{1}^{s}A_{3}<P_{1}A_{3}\) hold in Phase 2. To satisfy Equations (4) and (15), \(P_{1}^{s}-P_{2}^{s}>P_{1}^{max}-P_{2}^{max}>0\) conflicts with \(P_{1}^{s}<P_{2}^{s}\). Therefore, \(P_{1}^{s}\geq P_{2}^{s}\) holds, and the proof is over.  
In Phase 4, the net inflow \(q_{1}=0\) and \(q_{2}=0\) hold. According to Equations (11) and (12), \(P_{1}^{s}\) and \(P_{2}^{s}\) remain unchanged, and the hydraulic cylinder is in a stable pressure maintenance state.