# Sliding Mode Control of Laterally Interconnected Air Suspensions
## 3 Sliding Mode Controller for Interconnection  
The main aim of the controller is to minimize the roll angle subject to road disturbances, thereby providing better handling. Considering that the interconnection decouples the vertical and roll vibrations and only has an effect on the roll movement of the sprung mass, let \(e\) be the error, defined as  
\[e=\theta_{ref}-\theta, \tag{17}\]  
where \(\theta_{ref}\) is the reference roll angle of sprung mass and equals 0. Let \(\mathbf{C}=[c,1]\), error vector \(\mathbf{E}=\left[e\,\dot{e}\right]^{T}\), then the sliding surface \(s\) is defined as follows  
\[s=\mathbf{C}\mathbf{E}=c\left(\theta_{ref}-\theta\right)+\dot{\theta}_{ref}- \dot{\theta}, \tag{18}\]  
By selecting the Lyapunov function,  
\[V=\frac{1}{2}s^{2}. \tag{19}\]  
To ensure the existence and attractiveness of the sliding mode surface, it is sufficient to choose a control law such that  
\[\frac{dV}{dt}\triangleq\dot{V}=s\dot{s}<0. \tag{20}\]  
Therefore, the sliding mode controller is designed by the constant rate reaching law to meet the above condition:  
\[\dot{s}=-\delta sgn(s), \tag{21}\]  
where \(\delta>0\) is the sliding mode switching control gain. When the generalized sliding mode arrival condition \(\dot{s}s<0\), the approaching law will be satisfied. Combining the nonlinear and coupling characteristics of the interconnected air suspension, the control law can be calculated step by step. According to Equation (17) and (18), and considering that \(\theta_{ref}\) is a constant and equals 0, Equation (21) can be expressed as  
\[c\dot{\theta}+\ddot{\theta}=\delta sgn(s). \tag{22}\]  
Noticing that \(\ddot{\theta}\) can be replaced according to Equation (7)  
\[\frac{B_{f}}{2I_{r}}(F_{1}-F_{2})+\frac{B_{r}}{2I_{r}}(F_{3}-F_{4})=\delta sgn (s)-c\dot{\theta}. \tag{23}\]  
Because the torques for the sprung mass roll motion provided by suspensions of front and rear axes are related to the pressure of each air spring, \(F_{1}\)-\(F_{2}\) and \(F_{3}\)-\(F_{4}\) can be calculated as follows  
\[\left\{ \begin{array}{l}F_{1}-F_{2}=\frac{P_{1}+P_{2}}{P_{1}+P_{2}+P_{3}+P_{4 }}\frac{I_{r}}{B_{f}}\left(\delta sgn(s)-c\dot{\theta}\right)\\ F_{3}-F_{4}=\frac{P_{3}+P_{4}}{P_{1}+P_{2}+P_{3}+P_{4}}\frac{I_{r}}{B_{r}} \left(\delta sgn(s)-c\dot{\theta}\right)\end{array} \right., \tag{24}\]  
According to Equation (8), each suspension force includes spring force and damping force, and the spring force can be calculated with the effect area \(A_{ci}\), volume \(V_{i}\), and air mass \(m_{i}\), and the initial pressure \(p_{i0}\) of the air spring depends on Equation (2). Among them, only air mass has matters with the interconnection control. Thus, Equation (24) can be written as  
\[\left\{ \begin{array}{l}m_{1}{}^{\kappa}S_{p1}-m_{2}{}^{\kappa}S_{p2}=S_{f1} \\ m_{3}{}^{\kappa}S_{p3}-m_{4}{}^{\kappa}S_{p4}=S_{f2}\end{array} \right., \tag{25}\]where  
\[\left\{ \begin{array}{l}S_{pi}=P_{i0}\big{(}\frac{V_{i0}}{m_{i0}V_{i}}\big{)}^{ x}A_{ei}\\ S_{f1}=\frac{P_{1}+P_{2}}{P_{1}+P_{2}+P_{3}+P_{4}}\frac{I_{f}}{B_{f}}\big{(} \delta sgn(s)-c\dot{\theta}\big{)}-c_{1}\dot{f}d_{1}+c_{2}\dot{f}d_{2}-P_{a}(A _{e1}-A_{e2})\\ S_{f2}=\frac{P_{2}+P_{4}}{P_{1}+P_{2}+P_{3}+P_{4}}\frac{I_{f}}{B_{f}}\big{(} \delta sgn(s)-c\dot{\theta}\big{)}-c_{3}\dot{f}d_{3}+c_{4}\dot{f}d_{4}-P_{a}(A _{e3}-A_{e4})\end{array} \right. \tag{26}\]  
where \(A_{ei}\) and \(V_{i}\) can be calculated with Equations (5) and (3). Considering that the interconnection only makes the air mass flow from one air spring to the other, the total air mass among the air springs is a constant, thus an optimization problem can be established to find the expected air mass \(m_{i}\) of each air spring that best meet Equation (25) within the range of \([0,m_{(2|-10)}+m_{(2|0)}]\), \(j=1,2\). Minimizing the square values of the left and right sides' differences of Equation 25 as the optimization target, the problem can be described as follows  
\[