## 3 An Isolator against Impulsive Base Excitation  
Consider a half-cycle sine pulse as the base displacement excitation. \[x_{b}(t)=\left\{ \begin{array}{cc}h\sin\frac{\pi t}{t_{p}},&0\leq t\leq t_{p }\\ 0,&t_{p}<t\end{array} \right., \tag{22}\]  
where \(h\) is the maximum amplitude of the half-cycle sine pulse, and \(t_{p}\) is the duration of the base displacement excitation. Introducing \(\mu=T/2t_{p}\) where \(T=2\pi/\sqrt{k_{p}/m}\), Equation(22) can be rewritten in terms of the non-dimensional time \(\tau\) as follows. \[X_{b}(\tau)=\left\{ \begin{array}{cc}\sin\mu\tau,&0\leq\tau\leq\tau_{p}\\ 0,&\tau_{p}<\tau\end{array} \right., \tag{23}\]where \(\tau_{p}=2\pi t_{p}/T=\omega_{0}t_{p}\). Taking the second-order derivative of \(X_{b}(\tau)\) with respect to non-dimensional time \(\tau\) yields \(X_{b}^{\prime\prime}(\tau)=-\mu^{2}\sin\mu\tau\). Substituting this expression of \(X_{b}^{\prime\prime}(\tau)\) into Equation (6), one has  
\[\left\{ \begin{array}{ll}Y^{\prime\prime}+2\eta\Omega_{n}Y^{\prime}+\Omega_{n }^{2}Y=\frac{\mu^{2}}{(1+\alpha)}\sin\mu\tau,&0\leq\tau\leq\tau_{p}\\ Y^{\prime\prime}+2\eta\Omega_{n}Y^{\prime}+\Omega_{n}^{2}Y=0,&\tau_{p}<\tau \end{array} \right.. \tag{24}\]  
Using the Laplace transform method, the time responses of linear Equation (24) can be determined analytically as in [26]. It is more convenient to assume, for the following closed-form solutions, that the initial displacement \(Y(0)\) and velocity \(Y^{\prime}(0)\) in Equation (24) are both zero. In the under-damped case (\(\eta<1\)), the time responses of this isolator to the half-cycle sine pulse are derived from the results in [26] as  
\[ \begin{array}{l}Y(\tau)=\frac{\mu^{2}\Omega_{n}^{2}}{(1-\beta)} \frac{1}{\left(\Omega_{n}^{2}-\mu^{2}\right)^{2}+\left(2\eta\mu\Omega_{n} \right)^{2}}\Big{[}\left(\Omega_{n}^{2}-\mu^{2}\right)\sin\mu\tau-2\eta\mu \Omega_{n}(\cos\mu\tau-\\ e^{-\eta\Omega_{n}\tau}\cos\left(\sqrt{1-\eta^{2}}\Omega_{n}\tau\right))- \frac{\mu\left(\Omega_{n}^{2}-2\eta^{2}\Omega_{n}^{2}-\mu^{2}\right)e^{-\eta \Omega_{n}\tau}}{\Omega_{n}\sqrt{1-\eta^{2}}}\sin\left(\sqrt{1-\eta^{2}} \Omega_{n}\tau\right)\Big{]},\\ 0\leq\tau\leq\tau_{p}.\end{array}  \tag{25a}\] \[ \begin{array}{l}Y(\tau)=Y\Big{(}\tau_{p}\Big{)}e^{-\eta\Omega_{n}\tau} \cos\left(\sqrt{1-\eta^{2}}\Omega_{n}\tau\right)+\\ \left(Y^{\prime}\Big{(}\tau_{p}\Big{)}+\eta\Omega_{n}Y\Big{(}\tau_{p}\Big{)} \right)e^{-\eta\Omega_{n}\tau}\sin\left(\sqrt{1-\eta^{2}}\Omega_{n}\tau\right) \ \tau_{p}<\tau.\end{array}  \tag{25b}\]  
Equation (25) can be rewritten in terms of the absolute displacement and dimensional time as follows. \[ \begin{array}{l}x(t)=\frac{\mu^{2}\Omega_{n}^{2}}{(1-\beta)} \frac{1}{\left(\Omega_{n}^{2}-\mu^{2}\right)^{2}+\left(2\eta\mu\Omega_{n} \right)^{2}}\Big{[}\left(\Omega_{n}^{2}-\mu^{2}\right)\sin\mu\ \omega_{0}t-2\eta\mu\Omega_{n}(\cos\mu\ \omega_{0}t-\\ e^{-\eta\Omega_{n}\ \omega_{0}t}\cos\left(\sqrt{1-\eta^{2}}\Omega_{n}\ \omega_{0}t \right))-\\ \frac{\mu\Big{(}\Omega_{n}^{2}-2\eta^{2}\Omega_{n}^{2}-\mu^{2}\Big{)}e^{- \eta\Omega_{n}\ \omega_{0}t}}{\Omega_{n}\sqrt{1-\eta^{2}}}\sin\left(\sqrt{1-\eta^{2}} \Omega_{n}\ \omega_{0}t\right)\Big{]}+h\sin\frac{\pi t}{t_{p}},\ 0\leq t\leq t_{p}. \end{array}  \tag{26a}\] \[ \begin{array}{l}x(t)=x\Big{(}t_{p}\Big{)}e^{-\eta\Omega_{n}\omega_{0}t} \cos\left(\sqrt{1-\eta^{2}}\Omega_{n}\omega_{0}t\right)+\left(\frac{x\Big{(}t _{p}\Big{)}}{\omega_{0}}+\right.\\ \left.\eta\Omega_{n}x\Big{(}t_{p}\Big{)}\right)e^{-\eta\Omega_{n}\omega_{0}t} \sin\left(\sqrt{1-\eta^{2}}\Omega_{n}\omega_{0}t\right),\ t_{p}<t.\end{array}  \tag{26b}\]  
To validate the analytical solutions (Equations (25) and (26)), the numerical results of Equation (24) by the fourth-order Runge-Kutta integration are compared with those obtained using the analytical solution (Equation (25a)) for the isolator, with the pulse and isolator parameters set as \(\alpha=1.0\), \(\beta=0.95\), \(\zeta=0.2\), \(\mu=4.0\), and the derived data \(\Omega_{n}\approx 0.16\), \(\eta\approx 0.63\), and \(\tau_{p}\approx 0.78\).