## 3 Design of Control Strategy
### Upper Control Strategy
\begin{array}{c}\frac{F_{xrl}d+F_{xfl}(cos\delta\cdot d-sin\delta \cdot 4)}{I_{z}};\ \text{{control}}\ \text{{of}}\ \text{{left}\ \text{{side}}}\\ \frac{-F_{xrr}d-F_{xrf}(cos\delta\cdot d-sin\delta\cdot a)}{I_{z}};\ \text{{control}}\ \text{{of}}\ \text{{right}}\ \text{{side}} \end{array} \right. \tag{26}\]  
The equation of state is formulated into the error form as shown in Equations (27) and (28). \[\dot{e_{1}}=\dot{\beta_{d}}-\dot{\beta}=\dot{\beta_{d}}-(-\omega+s)=-(\omega_{ d}-\omega)+\big{(}\omega_{d}+\dot{\beta_{d}}-s\big{)}=-e_{2}+s^{\prime} \tag{27}\]  
\[\dot{e_{2}}=\dot{\omega_{d}}-\dot{\omega}=\dot{\omega_{d}}-z-U=U^{\prime} \tag{28}\]  
where \(e_{1}\) is the error of the side slip angle; \(e_{2}\) is the error of the yaw rate. Inside, \(s^{\prime}=\big{(}\omega_{d}+\dot{\beta_{d}}-s\big{)}\); \(U^{\prime}=\dot{\omega_{d}}-z-U\). Presented in matrix form as in formula (29). \[\left[ \begin{array}{c}\dot{e_{1}}\\ \dot{e_{2}}\end{array} \right]=A\left[ \begin{array}{c}e_{1}\\ e_{2}\end{array} \right]+BU^{\prime}+\left[ \begin{array}{c}1\\ 0\end{array} \right]s^{\prime} \tag{29}\]  
Inside, \(A=\left[ \begin{array}{cc}0&-1\\ 0&0\end{array} \right]\); \(B=\left[ \begin{array}{c}0\\ 1\end{array} \right]\). To determine the controllability of the system, \(Rank(B,AB)=2\), when the matrix satisfies the full rank. In this paper, the feedback gain matrix for closed-loop control is designed by the LQR control algorithm to define the performance index function as shown in Equation (30). \[J=\frac{1}{2}\int\ e^{T}Qe+U^{\prime T}RU^{\prime} \tag{30}\]  
where \(Q\) is the error weight matrix; \(R\) is the input weighting factor; the Riccati equation can be obtained as shown in Equation (31). \[-PA-A^{T}P+PBR^{-1}B^{T}P-Q=0 \tag{31}\]Solving the algebraic Riccati equation, we can get the \(P\), then the feedback gain matrix can be equal to \(K\), \(K=-R^{-1}B^{T}Px(t)\), and then \(U^{\prime}=\dot{\omega_{d}}-z-\mathrm{U}=-k_{1}e_{1}-k_{2}e_{2}\), thus \(U=k_{1}e_{1}+k_{2}e_{2}+\dot{\omega_{d}}-z\), \(K=\left[ \begin{array}{cc}k_{1}&k_{2}\end{array} \right]\). Therefore, the brake for the left wheel is as in Equation (32):  
\[U=k_{1}e_{1}+k_{2}e_{2}+\dot{\omega_{d}}-z=\frac{F_{xrl}d+F_{xfl}(cos\delta \cdot d-sin\delta\cdot a)}{I_{z}} \tag{32}\]  
Then we get the Equation (33):  
\[F_{xrl}d+F_{xfl}(cos\delta\cdot d-sin\delta\cdot a)=I_{2}\Big{(}k_{1}e_{1}+k_{ 2}e_{2}+\dot{\omega_{d}}\Big{)}-z_{1} \tag{33}\]  
the brake for the right wheel is as in Equation (34):  
\[U=k_{1}e_{1}+k_{2}e_{2}+\dot{\omega_{d}}-z=\frac{-F_{xrl}d-F_{xfr}(cos\delta \cdot d+sin\delta\cdot a)}{I_{z}} \tag{34}\]  
After that, we can get the Equation (35):  
\[F_{xrr}d+F_{xfr}(cos\delta\cdot d+sin\delta\cdot a)=-I_{2}\Big{(}k_{1}e_{1}+k_ {2}e_{2}+\dot{\omega_{d}}\Big{)}+z_{2} \tag{35}\]  
Due to the small longitudinal acceleration, the front and rear axle-load transfer is small. The front and rear longitudinal force is distributed according to the static axle load: \(\frac{F_{xfl}}{F_{xr}}=\frac{b}{a}\). The constraint of maximum longitudinal force is based on the current friction circle to constrain the maximum longitudinal force, which is defined as Equation (36):  
\[F_{xi}\leq\sqrt{\left(F_{zi}t\right)^{2}-F_{yi}^{2}} \tag{36}\]  
In addition, due to the existence of driving force, the calculated target longitudinal force can be a negative value, if its absolute value is not greater than the current driving force of the tire.