## 2 The Vehicle-Pavement-Foundation System: Modeling and Solution
### Vehicle Model and Tire Force  
The objective of this subsection is to formulate the equation representing the force applied by the tire to the pavement. As shown in Figure 1, \(m_{1}\), \(K_{1}\), \(C_{1}\) represent the mass, stiffness coefficient, and damping coefficient of the tire, respectively. \(m_{2}\), \(K_{2}\), \(C_{2}\) represent the mass, stiffness coefficient, and damping coefficient of the suspension system, respectively. The velocity of the vehicle is denoted by \(v\). \(y_{1}\) and \(y_{2}\) are the vertical displacements of the masses \(m_{1}\) and \(m_{2}\), respectively. \(y_{0}\) is the displacement excitation of the pavement to the tire, which can be expressed as  
\[y_{0}=A_{0}\sin(\omega\,t),\quad\omega=\frac{2\pi v}{L_{0}}, \tag{1}\]  
Figure 1: The vehicle–pavement–foundation system. where \(A_{0}\) is the amplitude and \(L_{0}\) is the wavelength of harmonic road roughness [43]. According to [PERSON] principle, the dynamic equations of the vehicle are established as follows:  
\[ \begin{bmatrix}m_{1}&0\\ 0&m_{2}\end{bmatrix}\begin{bmatrix}\hat{y}_{1}\\ \hat{y}_{2}\end{bmatrix} + \begin{bmatrix}C_{1}+C_{2}&-C_{2}\\ -C_{2}&C_{2}\end{bmatrix}\begin{bmatrix}\hat{y}_{1}\\ \hat{y}_{2}\end{bmatrix} + \begin{bmatrix}K_{1}+K_{2}&-K_{2}\\ -K_{2}&K_{2}\end{bmatrix}\begin{bmatrix}y_{1}\\ y_{2}\end{bmatrix} = \begin{bmatrix}K_{1}y_{0}+C_{1}\hat{y}_{0}\\ 0\end{bmatrix}  \tag{2}\]  
The tire exerts a force on the pavement as follows:  
\[F(t)=K_{1}(y_{1}-y_{0})+C_{1}(\hat{y}_{1}-\hat{y}_{0})-(m_{1}+m_{2})g \tag{3}\]  
where \(g\) is the acceleration due to gravity. Equation (2) represents a system of second-order non-homogeneous linear ordinary differential equations. To solve Equation (2), the following Fourier transform pair is defined:  
\[ \begin{split}\hat{y}_{i}(\sigma)=\int_{-\infty}^{\infty}y_{i}(t) e^{-i\sigma t}dt,\\ y_{i}(t)=\frac{1}{2\pi}\int_{-\infty}^{\infty}\hat{y}_{i}(\sigma)e^{i \sigma t}d\sigma,\quad i=0,1,2\end{split}  \tag{4}\]  
Applying Fourier transform to both sides of Equation (2), one has  
\[ \begin{bmatrix}K_{1}+K_{2}-\sigma^{2}m_{1}+i\sigma(C_{1}+C_{2})&-K_{2}-i \sigma C_{2}\\ -K_{2}-i\sigma C_{2}&K_{2}-\sigma^{2}m_{2}+i\sigma C_{2}\end{bmatrix} \begin{bmatrix}\hat{y}_{1}\\ \hat{y}_{2}\end{bmatrix} = \begin{bmatrix}(K_{1}+i\sigma C_{1})\hat{y}_{0}\\ 0\end{bmatrix} , \tag{5}\]  
where \(\hat{y}_{1,2}\) are the Fourier transform of \(y_{1,2}\), respectively, and  
\[\hat{y}_{0}=iA_{0}\pi[\delta(\sigma+\omega)-\delta(\sigma-\omega)], \tag{6}\]  
in which \(\delta(\cdot)\) is the Dirac function. According to [PERSON]'s rule, the solution to Equation (5) can be written as  
\[\hat{y}_{1}=\frac{D_{1}(\sigma)}{D(\sigma)},\quad\hat{y}_{2}=\frac{D_{2}( \sigma)}{D(\sigma)}, \tag{7}\]  
where  
\[D_{1}(\sigma)= \begin{vmatrix}(K_{1}+i\sigma C_{1})\hat{y}_{0}&-K_{2}-i\sigma C _{2}\\ 0&K_{2}-\sigma^{2}m_{2}+i\sigma C_{2}\end{vmatrix} ,\]  
\[D_{2}(\sigma)= \begin{vmatrix}K_{1}+K_{2}-\sigma^{2}m_{1}+i\sigma(C_{1}+C_{2}) &(K_{1}+i\sigma C_{1})\hat{y}_{0}\\ -K_{2}-i\sigma C_{2}&0\end{vmatrix} ,\]  
\[D(\sigma)= \begin{vmatrix}K_{1}+K_{2}-\sigma^{2}m_{1}+i\sigma(C_{1}+C_{2})&-K_ {2}-i\sigma C_{2}\\ -K_{2}-i\sigma C_{2}&K_{2}-\sigma^{2}m_{2}+i\sigma C_{2}\end{vmatrix}