## 2 Linear LPM of the HEM  
To better understand the dynamic performance of an HEM with an inertia track, the nonlinear characteristics are initially ignored and the linear time-invariant (LTI) lumped parameter model (LPM) [2; 5; 7; 12] is used, as shown in Figure 2a. This model has been validated and is frequently used to express the performance of an HEM, which supports an engine with mass M. The continuum equations in the upper and lower chambers and fluid dynamic equations about the inertia track are as follows:  
\[C_{1}\dot{p}_{1}(t)+C_{2}\dot{p}_{2}(t)=A_{\rm p}\dot{x}_{\rm e}(t), \tag{1}\]  
\[C_{2}\dot{p}_{2}(t)=Q_{\rm i}(t), \tag{2}\]  
\[p_{1}(t)-p_{2}(t)=l_{\rm i}\dot{Q}_{\rm i}(t)+R_{\rm i}Q_{\rm i}(t), \tag{3}\]  
where \(p_{1}(t)\) and \(p_{2}(t)\) are the fluid pressure of the upper and lower chambers, respectively; \(C_{1}\) and \(C_{2}\) are the compliance in the upper and lower chambers, respectively; \(A_{\rm p}\) is the equivalent piston area of the main rubber; \(x_{\rm e}(t)\) is the displacement of the engine side; \(Q_{\rm i}(t)\) is the flow rate in the inertia track; and \(l_{\rm i}\) and \(R_{\rm i}\) are the inertia of the inertia track and the resistance in the inertia track, respectively. The \(I_{\rm i}\) and \(R_{\rm i}\) can be computed as follows:  
\[I_{\rm i}=\frac{\rho L_{\rm i}}{A_{\rm i}}, \tag{4}\]  
where \(\rho\) is the fluid density; \(L_{\rm i}\) is the length of the inertia track; and \(A_{\rm i}\) is the cross-section area of inertia track. \[R_{\rm i}=\frac{128\mu L_{\rm i}}{\pi D_{\rm h}\,^{4}} \tag{5}\]  
where \(\mu\) is the fluid kinematic viscosity, and \(D_{\rm h}\) is the hydraulic diameter of the inertia track and can be computed as follows:  
\[D_{\rm h}=\frac{4A_{\rm i}}{P}, \tag{6}\]  
where \(P\) is the wetted perimeter of the inertia track. The driving point force at the engine side can be obtained as follows:  
\[F_{\rm D}(t)=k_{\rm r}x_{\rm e}(t)+b_{\rm r}\dot{x}_{\rm e}(t)+A_{\rm p}p_{1}( t), \tag{7}\]  
where \(k_{\rm r}\) and \(b_{\rm r}\) are the stiffness and damping of the main rubber, respectively. \(F_{\rm T}(t)\) is the force transferred at the chassis side. \[F_{\rm T}(t)=k_{\rm r}x_{\rm e}(t)+b_{\rm r}\dot{x}_{\rm e}(t)+A_{\rm p}p_{1}( t) \tag{8}\]  
Transferring Equations (1)-(3) into the Laplace domain, the driving point force on the engine side can be obtained as follows:  
\[F_{\rm D}(s)=k_{\rm r}X_{\rm e}(s)+b_{\rm r}sX_{\rm e}(s)+\frac{A_{\rm p}X_{ \rm e}(s)}{C_{1}}-\frac{A_{\rm p}X_{\rm e}(s)}{C_{1}^{2}(I_{\rm i}s^{2}+R_{\rm i }s+K_{\rm u}+K_{\rm l})}, \tag{9}\]  
where \(K_{\rm u}=1/C_{1}\) is the bulk stiffness of the upper chamber, and \(K_{\rm l}=1/C_{2}\) is the bulk stiffness of the lower chamber. Similarly, the Laplace domain solution for the force transferred to the body side \(F_{\rm T}(s)\) can be obtained as follows:  
\[F_{\rm T}(s)=k_{\rm r}X_{\rm e}(s)+b_{\rm r}sX_{\rm e}(s)+\frac{A_{\rm p}X_{ \rm e}(s)}{C_{1}}-\frac{A_{\rm p}X_{\rm e}(s)}{C_{1}^{2}(I_{\rm i}s^{2}+R_{\rm i }s+K_{\rm u}+K_{\rm l})}. \tag{10}\]  
\[C_{2}\dot{p}_{2}(t)=Q_{\rm i}(t), \tag{2}\]0 Comparison of two lumped parameter models. (**a**) Lumped parameter model of HEM; and (**b**) TMD system consisting of a rubber isolator and a tuned mass damper. The transfer dynamic stiffness \(k_{\rm T}(s)\) and drive point dynamic stiffness \(k_{\rm D}(s)\) are defined as follows:  
\[C_{2}\dot{p}_{2}(t)=Q_{\rm i}(t), \tag{2}\]1  
\[C_{2}\dot{p}_{2}(t)=Q_{\rm i}(t), \tag{2}\]2  
Consider a TMD system supporting an engine (mass M), as shown in Figure 2b, where the TMD system consists of a rubber isolator with stiffness \(k_{\rm r}\) and damping \(b_{\rm r}\) and a TMD with equivalent mass \(M_{\rm ie}\), equivalent stiffness \(k_{\rm ie}\), and equivalent damping \(b_{\rm ie}\).