# Equivalent Linearization Methods for a Control System with Clutching Inerter Damper
## 5 ELM3 Based on SLM  
### Statistical Linearization Procedure  
In this section, the statistical linearization method (SLM) of the nonlinear SDOF system with the CID under stationary stochastic excitation is studied. The responses of the CID system are considered to have a Gaussian distribution. Firstly, Equation (9) and Equation (10) can be rewritten as  
\[m\ddot{x}+c\dot{x}+kx+g\big{(}\dot{x},\ddot{x}\big{)}=-m\ddot{x}_{g} \tag{44}\]  
\[g\big{(}\dot{x},\ddot{x}\big{)}=\mu m\ddot{x}\Big{(}\mathrm{sgn}\big{(}\dot{x} \ddot{x}\big{)}+\mathrm{sgn}^{2}\big{(}\dot{x}\ddot{x}\big{)}\Big{)}/2 \tag{45}\]  
where \(g\big{(}\dot{x},\ddot{x}\big{)}\) is the nonlinear part of Equation (9) and \(sgn\big{(}\dot{x}\ddot{x}\big{)}\) is the sign function. The difference between Equation (44) and Equation (11) is  
\[\varepsilon=g\big{(}\dot{x},\ddot{x}\big{)}-b_{eq}\ddot{x}-c_{eq}\dot{x} \tag{46}\]  
Minimizing \(E\big{\{}\varepsilon^{2}\big{\}}\), which is the expectation of \(\varepsilon^{2}\), is suggested here:  
\[E\Big{\{}\varepsilon^{2}\Big{\}}=E\Big{\{}\big{(}g\big{(}\dot{x},\ddot{x} \big{)}-b_{eq}\ddot{x}-c_{eq}\dot{x}\big{)}^{2}\Big{\}} \tag{47}\]  
Figure 5: Square of velocity response. Here \(b_{eq}\) and \(c_{eq}\) are solved by the partial differential equation:  
\[\frac{\partial}{\partial b_{eq}}E\Big{\{}\epsilon^{2}\Big{\}}=\frac{\partial}{ \partial c_{eq}}E\Big{\{}\epsilon^{2}\Big{\}}=0 \tag{48}\]  
For the case that \(\dot{x}\)(t) and \(\ddot{x}\)(t) are Gaussian processes, this procedure finally leads to Reference [14]:  
\[b_{eq}=E\bigg{\{}\frac{\partial g}{\partial\ddot{x}}\bigg{\}} \tag{49}\]  
\[c_{eq}=E\bigg{\{}\frac{\partial g}{\partial\ddot{x}}\bigg{\}} \tag{50}\]  
Substituting Equation (45) into Equation (49), \(b_{eq}\) is obtained:  
\[ \begin{array}{ll}b_{eq}&=E\Big{\{}\frac{\partial g}{\partial\ddot{x}}\Big{\}} \\ &=E\bigg{\{}0.5\mu m\frac{\partial}{\partial\ddot{x}}\big{(}\ddot{x}(sgn\big{(} \dot{x}\ddot{x}\big{)}+sgn^{2}\big{(}\dot{x}\ddot{x}\big{)})\big{)}\Big{\}}\\ &=0.5\mu mE\big{\{}sgn\big{(}\dot{x}\ddot{x}\big{)}+sgn^{2}\big{(}\dot{x} \ddot{x}\big{)}\big{\}}+0.5\mu mE\Big{\{}\frac{\partial}{\partial\ddot{x}} \big{(}\mbox{sgn}\big{(}\dot{x}\ddot{x}\big{)}+sgn^{2}\big{(}\dot{x}\ddot{x} \big{)}\big{)}\Big{\}}\\ &=0.5\mu mE\big{\{}\mbox{sgn}\big{(}\dot{x}\ddot{x}\big{)}+sgn^{2}\big{(}\dot {x}\ddot{x}\big{)}\big{\}}\end{array}  \tag{51}\]  
where \(\delta\big{(}\ddot{x}\big{)}\) is the Dirac function of \(\ddot{x}\). Here \(sgn\big{(}\dot{x}\ddot{x}\big{)}\) is not a standard sign function due to the different time period for \(\dot{x}\ddot{x}>0\) and \(\dot{x}\ddot{x}<0\). This can also be found in Figure 4. To obtain a more accurate result of Equation (51), considering the time-history of \(sgn\big{(}\dot{x}\ddot{x}\big{)}\) in the first circle of free vibration in Figure 4:  
\[sgn\big{(}\dot{x}\ddot{x}\big{)}=\left\{ \begin{array}{c}1,\;0<t\leq t_{1}\\ -1,\;t_{1}<t\leq t_{2}\\ 1,\;t_{2}<t\leq t_{3}\\ -1,\;t_{3}<t\leq t_{4}\end{array} \right. \tag{52}\]  
\[sgn^{2}\big{(}\dot{x}\ddot{x}\big{)}=\left\{ \begin{array}{c}0,t=0,t_{1},t_{ 2},t_{3},t_{4}\\ 1,t=\mbox{ other time}\end{array} \right. \tag{53}\]  
\[E\big{\{}sgn\big{(}\dot{x}\ddot{x}\big{)}\big{\}}=\frac{2t_{1}-t_{2}}{t_{2}}= 1+\frac{2\big{(}1-\sqrt{1+\mu}\big{)}}{\mu} \tag{54}\]  
\[E\Big{\{}sgn^{2}\big{(}\dot{x}\ddot{x}\big{)}\Big{\}}\;=1 \tag{55}\]  
in which \(t_{1}\) and \(t_{2}\) are obtained in Equation (26) and Equation (27), respectively. Substituting Equation (54) and Equation (55) into Equation (51), a more accurate solution of \(b_{eq}\) is obtained:  
\[b_{eq}=\Big{(}1+\mu-\sqrt{1+\mu}\Big{)}m \tag{56}\]  
along with the equivalent inertance-mass ratio:  
\[\mu_{eq}=\frac{b_{eq}}{m}=\Big{(}1+\mu-\sqrt{1+\mu}\Big{)} \tag{57}\]Substituting Equation (45) into Equation (50), \(c_{eq}\) is obtained  
\[