汽车悬架系统的智能采样数据输出反馈控制方法

摘要

一种汽车悬架系统的智能采样数据输出反馈控制方法，其步骤如下：建立起汽车悬架系统动态模型；确定出汽车底盘的弹性质量m_s(t)和车轮部件的非弹性质量m_u(t)的变化范围；构造两个关键物理量来用于评价控制方法的性能优劣；建立起汽车悬架系统的状态空间模型；得到整体性的汽车悬架系统的模糊状态空间模型；设计一种智能采样数据输出反馈控制器；进行汽车悬架系统的在线控制。本发明能够实现汽车悬架系统的高性能控制目标，满足驾驶过程中的高舒适度以及高安全性；降低了控制方法对检测技术的要求和检测元件配置的要求，提高了汽车悬架系统的可靠性和经济性；实现了用户驾乘体验的个性自主化，彰显了以客户为中心的设计理念。

主权项

一种汽车悬架系统的智能采样数据输出反馈控制方法，其特征在于：它包括以下技术步骤：1)应用力学原理建立起如下汽车悬架系统动态模型：$m_s\left(t\right){\stackrel{··}{z}}_s\left(t\right)+c_s[{\stackrel{·}{z}}_s\left(t\right)-{\stackrel{·}{z}}_u\left(t\right)]+k_s[z_s\left(t\right)-z_u\left(t\right)]=u\left(t\right),---\left(1\right)$$\begin{array}{c}{m}_u\left(t\right){\stackrel{··}{z}}_u\left(t\right)+c_s[{\stackrel{·}{z}}_u\left(t\right)-{\stackrel{·}{z}}_s\left(t\right)]+k_s[z_u\left(t\right)-z_s\left(t\right)]+k_t[z_u\left(t\right)-z_r\left(t\right)]+c_t[{\stackrel{·}{z}}_u\left(t\right)-{\stackrel{·}{z}}_r\left(t\right)]=\\ -u\left(t\right)\end{array}---\left(2\right)$其中，m_s(t)为汽车底盘的弹性质量，单位为Kg；m_u(t)为车轮部件的非弹性质量，单位为Kg；u(t)为汽车悬架系统的控制输入量，单位为N；z_s(t)为m_s以水平地面为起始点垂直向上方向上的坐标位置，单位为m；z_u(t)为m_u以水平地面为起始点垂直向上方向上的坐标位置，单位为m；z_r(t)为以水平地面为起始点垂直向上方向上路面接触点坐标位置，单位为m；c_s为汽车悬架系统的阻尼系数，单位为N/(m/s)；k_s为汽车悬架系统的硬度系数，单位为N/m；c_t为汽车轮胎的阻尼系数，单位为N/(m/s)；k_t为汽车轮胎的硬度系数，单位为N/mp；2)基于汽车的机械结构特性和可允许乘客数目及质量的变化情况，确定出m_s(t)和m_u(t)的变化范围为：m_s(t)∈[m_smin,m_smax]和m_u(t)∈[m_umin,m_umax]；3)考虑与高舒适度和高安全性相关的主要影响因素，构造如下两个物理量来用于评价控制方法的性能优劣，$z_1\left(t\right)={\stackrel{··}{z}}_s\left(t\right),---\left(3\right)$$z_2\left(t\right)={\left[\begin{array}{cc}\frac{z_s\left(t\right)-z_u\left(t\right)}{z_{\max }}& \frac{k_t(z_u\left(t\right)-z_r\left(t\right))}{(m_s\left(t\right)+m_u\left(t\right))g}\end{array}\right]}^T,---\left(4\right)$其中，g为重力常数，单位为N/Kg；z_max为汽车悬架系统的最大偏移量，单位为m；且有|z_s(t)‑z_u(t)|≤z_max和k_t(z_u(t)‑z_r(t))<(m_s(t)+m_u(t))g同时成立；4)根据所述步骤1)中给出的汽车悬架系统动态模型，建立起汽车悬架系统的状态空间模型：$\left\{\begin{array}{c}\stackrel{·}{x}\left(t\right)=A\left(t\right)x\left(t\right)+B_w\left(t\right)w\left(t\right)+B\left(t\right)u\left(t\right)\\ z_1\left(t\right)=C_1\left(t\right)x\left(t\right)+D_1\left(t\right)u\left(t\right)\\ z_2\left(t\right)=C_2\left(t\right)x\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}\right.---\left(5\right)$其中，x(t)＝[x₁(t) x₂(t) x₃(t) x₄(t)]^T,x₁(t)＝z_s(t)‑z_u(t),x₂(t)＝z_u(t)‑z_r(t),$x_3\left(t\right)={\stackrel{·}{z}}_s\left(t\right),x_4\left(t\right)={\stackrel{·}{z}}_u\left(t\right),w\left(t\right)={\stackrel{·}{z}}_r\left(t\right),y\left(t\right)$y(t)为系统可测输出变量，控制器对于系统的内部信息的获取是通过测量y(t)来实现；$A\left(t\right)=\left[\begin{array}{cccc}0& 0& 1& -1\\ 0& 0& 0& 1\\ -\frac{k_s}{m_s\left(t\right)}& 0& -\frac{c_s}{m_s\left(t\right)}& \frac{c_s}{m_s\left(t\right)}\\ \frac{k_s}{m_u\left(t\right)}& -\frac{k_t}{m_u\left(t\right)}& \frac{c_s}{m_u\left(t\right)}& -\frac{c_s+c_t}{m_u\left(t\right)}\end{array}\right],$$B\left(t\right)=\left[\begin{array}{c}0\\ 0\\ \frac{1}{m_s\left(t\right)}\\ -\frac{1}{m_u\left(t\right)}\end{array}\right],B_w\left(t\right)=\left[\begin{array}{c}0\\ -1\\ 0\\ -\frac{c_t}{m_u\left(t\right)}\end{array}\right],$$C_1\left(t\right)=\left[\begin{array}{cccc}-\frac{k_s}{m_s\left(t\right)}& 0& -\frac{c_s}{m_s\left(t\right)}& \frac{c_s}{m_s\left(t\right)}\end{array}\right],C=\left[\begin{array}{cccc}1& 1& 1& 0\end{array}\right],$$C_2\left(t\right)=\left[\begin{array}{cccc}\frac{1}{z_{\max }}& 0& 0& 0\\ 0& \frac{k_t}{\left(m_s\right(t)+m_u(t\left)\right)g}& 0& 0\end{array}\right],D_1\left(t\right)=\frac{1}{m_s\left(t\right)};$5)根据m_s(t)和m_u(t)的可变特性，两个模糊前件变量选定为和利用Takagi‑Sugeno模糊模型建模规则建立汽车悬架系统的模糊状态空间模型：规则1：如果ξ₁(t)为M₁(ξ₁(t))且ξ₂(t)为N₁(ξ₂(t))，那么$\left\{\begin{array}{c}\stackrel{·}{x}\left(t\right)=A_1\left(t\right)x\left(t\right)+B_{w1}\left(t\right)w\left(t\right)+B_1\left(t\right)u\left(t\right)\\ z_1\left(t\right)=C_{11}\left(t\right)x\left(t\right)+D_{11}\left(t\right)u\left(t\right)\\ z_2\left(t\right)=C_{21}\left(t\right)x\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}\right.;---\left(6\right)$规则2：如果ξ₁(t)为M₁(ξ₁(t))且ξ₂(t)为N₂(ξ₂(t))，那么$\left\{\begin{array}{c}\stackrel{·}{x}\left(t\right)=A_2\left(t\right)x\left(t\right)+B_{w2}\left(t\right)w\left(t\right)+B_2\left(t\right)u\left(t\right)\\ z_1\left(t\right)=C_{12}\left(t\right)x\left(t\right)+D_{12}\left(t\right)u\left(t\right)\\ z_2\left(t\right)=C_{22}\left(t\right)x\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}\right.;---\left(7\right)$规则3：如果ξ₁(t)为M₂(ξ₁(t))且ξ₂(t)为N₁(ξ₂(t))，那么$\left\{\begin{array}{c}\stackrel{·}{x}\left(t\right)=A_3\left(t\right)x\left(t\right)+B_{w3}\left(t\right)w\left(t\right)+B_3\left(t\right)u\left(t\right)\\ z_1\left(t\right)=C_{13}\left(t\right)x\left(t\right)+D_{13}\left(t\right)u\left(t\right)\\ z_2\left(t\right)=C_{23}\left(t\right)x\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}\right.;---\left(8\right)$规则4：如果ξ₁(t)为M₂(ξ₁(t))且ξ₂(t)为N₂(ξ₂(t))，那么$\left\{\begin{array}{c}\stackrel{·}{x}\left(t\right)=A_4\left(t\right)x\left(t\right)+B_{w4}\left(t\right)w\left(t\right)+B_4\left(t\right)u\left(t\right)\\ z_1\left(t\right)=C_{14}\left(t\right)x\left(t\right)+D_{14}\left(t\right)u\left(t\right)\\ z_2\left(t\right)=C_{24}\left(t\right)x\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}\right.;---\left(9\right)$其中，$M_1\left({\xi }_1\right(t\left)\right)=\frac{\frac{1}{m_s\left(t\right)}-\frac{1}{m_{smax}}}{\frac{1}{m_{s\min }}-\frac{1}{m_{smax}}},M_2\left({\xi }_1\right(t\left)\right)=\frac{\frac{1}{m_{smin}}\frac{1}{m_s\left(t\right)}}{\frac{1}{m_{smin}}-\frac{1}{m_{smax}}},$$N_1\left({\xi }_2\right(t\left)\right)=\frac{\frac{1}{m_u\left(t\right)}-\frac{1}{m_{u\max }}}{\frac{1}{m_{u\min }}-\frac{1}{m_{u\max }}},N_2\left({\xi }_2\right(t\left)\right)=\frac{\frac{1}{m_{u\min }}-\frac{1}{m_u\left(t\right)}}{\frac{1}{m_{u\min }}-\frac{1}{m_{u\max }}},$$A_1=\left[\begin{array}{cccc}0& 0& 1& -1\\ 0& 0& 0& 1\\ -\frac{k_s}{m_{s\min }}& 0& -\frac{c_s}{m_{s\min }}& \frac{c_s}{m_{s\min }}\\ \frac{k_s}{m_{u\min }}& -\frac{k_t}{m_{u\min }}& \frac{c_s}{m_{u\min }}& -\frac{c_s+c_t}{m_{u\min }}\end{array}\right],$$B_1=\left[\begin{array}{c}0\\ 0\\ \frac{1}{m_{s\min }}\\ -\frac{1}{m_{u\min }}\end{array}\right],B_{w1}=\left[\begin{array}{c}0\\ -1\\ 0\\ -\frac{c_t}{m_{u\min }}\end{array}\right],$$C_{11}=\left[\begin{array}{cccc}-\frac{k_s}{m_{smin}}& 0& -\frac{c_s}{m_{smin}}& \frac{c_s}{m_{smin}}\end{array}\right],$$C_{21}=\left[\begin{array}{cccc}\frac{1}{z_{max}}& 0& 0& 0\\ 0& \frac{k_t}{(m_{smin}+m_{umin})g}& 0& 0\end{array}\right],D_{11}=\frac{1}{m_{smin}},$$A_2=\left[\begin{array}{cccc}0& 0& 1& -1\\ 0& 0& 0& 1\\ -\frac{k_s}{m_{s\min }}& 0& -\frac{c_s}{m_{s\min }}& \frac{c_s}{m_{s\min }}\\ \frac{k_s}{m_{u\max }}& -\frac{k_t}{m_{u\max }}& \frac{c_s}{m_{u\max }}& -\frac{c_s+c_t}{m_{u\max }}\end{array}\right],$$B_2=\left[\begin{array}{c}0\\ 0\\ \frac{1}{m_{s\min }}\\ -\frac{1}{m_{u\max }}\end{array}\right],B_{w2}=\left[\begin{array}{c}0\\ -1\\ 0\\ -\frac{c_t}{m_{u\max }}\end{array}\right],$$C_{12}=\left[\begin{array}{cccc}-\frac{k_s}{m_{smin}}& 0& -\frac{c_s}{m_{smin}}& \frac{c_s}{m_{smin}}\end{array}\right],$$C_{22}=\left[\begin{array}{cccc}\frac{1}{z_{max}}& 0& 0& 0\\ 0& \frac{k_t}{(m_{smin}+m_{umax})g}& 0& 0\end{array}\right],D_{12}=\frac{1}{m_{s\min }},$$A_3=\left[\begin{array}{cccc}0& 0& 1& -1\\ 0& 0& 0& 1\\ -\frac{k_s}{m_{s\max }}& 0& -\frac{c_s}{m_{s\max }}& \frac{c_s}{m_{s\max }}\\ \frac{k_s}{m_{u\min }}& -\frac{k_t}{m_{u\min }}& \frac{c_s}{m_{u\min }}& -\frac{c_s+c_t}{m_{u\min }}\end{array}\right],$$B_3=\left[\begin{array}{c}0\\ 0\\ \frac{1}{m_{s\max }}\\ -\frac{1}{m_{u\min }}\end{array}\right],B_{w3}=\left[\begin{array}{c}0\\ -1\\ 0\\ -\frac{c_t}{m_{u\min }}\end{array}\right],$$C_{13}=\left[\begin{array}{cccc}-\frac{k_s}{m_{smax}}& 0& -\frac{c_s}{m_{smax}}& \frac{c_s}{m_{smax}}\end{array}\right],$$C_{23}=\left[\begin{array}{cccc}\frac{1}{z_{max}}& 0& 0& 0\\ 0& \frac{k_t}{(m_{smax}+m_{umin})g}& 0& 0\end{array}\right],D_{13}=\frac{1}{m_{smax}},$$A_4=\left[\begin{array}{cccc}0& 0& 1& -1\\ 0& 0& 0& 1\\ -\frac{k_s}{m_{s\max }}& 0& -\frac{c_s}{m_{s\max }}& \frac{c_s}{m_{s\max }}\\ \frac{k_s}{m_{u\max }}& -\frac{k_t}{m_{u\max }}& \frac{c_s}{m_{u\max }}& -\frac{c_s+c_t}{m_{u\max }}\end{array}\right],$$B_4=\left[\begin{array}{c}0\\ 0\\ \frac{1}{m_{s\max }}\\ -\frac{1}{m_{u\max }}\end{array}\right],B_{w4}=\left[\begin{array}{c}0\\ -1\\ 0\\ -\frac{c_t}{m_{u\max }}\end{array}\right],$$C_{14}=\left[\begin{array}{cccc}-\frac{k_s}{m_{smax}}& 0& -\frac{c_s}{m_{smax}}& \frac{c_s}{m_{smax}}\end{array}\right],$$C_{24}=\left[\begin{array}{cccc}\frac{1}{z_{max}}& 0& 0& 0\\ 0& \frac{k_t}{(m_{smax}+m_{umax})g}& 0& 0\end{array}\right],D_{14}=\frac{1}{m_{smax}};$6)根据模糊建模方法，得到整体性的汽车悬架系统的模糊状态空间模型：$\left\{\begin{array}{c}\stackrel{·}{x}\left(t\right)={\Sigma }_{i=1}^4{h}_i\left(\xi \right(t\left)\right)[A_i\left(t\right)x\left(t\right)+B_{wi}\left(t\right)w\left(t\right)+B_i\left(t\right)u\left(t\right)]\\ z_1\left(t\right)={\Sigma }_{i=1}^4{h}_i\left(\xi \right(t\left)\right)[C_{1i}\left(t\right)x\left(t\right)+D_{1i}\left(t\right)u\left(t\right)]\\ z_2\left(t\right)={\Sigma }_{i=1}^4{h}_i\left(\xi \right(t\left)\right)C_{2i}\left(t\right)x\left(t\right)\\ y\left(t\right)=Cx\left(t\right)\end{array}\right.,---\left(10\right)$其中，h₁(ξ(t))＝M₁(ξ₁(t))×N₁(ξ₂(t)),h₂(ξ(t))＝M₁(ξ₁(t))×N₂(ξ₂(t))h₃(ξ(t))＝M₂(ξ₁(t))×N₁(ξ₂(t)),h₄(ξ(t))＝M₂(ξ₁(t))×N₂(ξ₂(t))；7)针对步骤6)所述整体性的汽车悬架系统的模糊状态空间模型，设计一种智能采样数据输出反馈控制器：$\left\{\begin{array}{c}\stackrel{·}{\hat{x}}\left(t\right)={\Sigma }_{i=1}^4{h}_i\left(\xi \right(t\left)\right)\left(A_{ci}\hat{x}\right(t)+A_{cdi}\hat{x}(t_k)+B_{ci}y(t_k\left)\right)\\ u\left(t\right)={\Sigma }_{i=1}^4{h}_i\left(\xi \right(t\left)\right)C_{ci}\hat{x}\left(y\right)\end{array}\right.,t_k\le t<t_{k+1},k=0,1,2,...;---\left(11\right)$其中,A_ci、A_cdi、B_ci、C_ci为对应相应模糊规则的适当维数维控制增益矩阵，它们的计算公式为：这里，矩阵和为非奇异矩阵且满足且以及对称矩阵Q可以通过求解满足如果线性矩阵不等式条件来获得：$\left[\begin{array}{ccc}{\Xi }_{1ii}& {\Xi }_{2ii}& {\Xi }_{3ii}\\ *& -I& 0\\ *& *& {\Xi }_{4ii}\end{array}\right]<0,i=1,2,3,4;---\left(15\right)$$\left[\begin{array}{ccccc}{\Psi }_{1ij}& {\Psi }_{2ij}& {\Psi }_{3ij}& {\Psi }_{4ij}& {\Psi }_{5ij}\\ *& -2I& 0& 0& 0\\ *& *& {\Psi }_{6ij}& {\mathrm{h\mu }}_{6ij}& 0\\ *& *& *& {\Psi }_{7ij}& 0\\ *& *& *& *& {\Psi }_{7ij}\end{array}\right]<0,1\le i<j\le 4;---\left(16\right)$这里，${\Xi }_{1ii}=\left[\begin{array}{cccc}{\lambda }_{1ii}-Q& {\lambda }_{2ii}+Q& 0& {\lambda }_{3ii}\\ *& -2Q& Q& 0\\ *& *& -Q& 0\\ *& *& *& -{\gamma }^{2}I\end{array}\right],$${\Xi }_{2\mathrm{ii}}=\left[\begin{array}{c}{\lambda }_{4\mathrm{ii}}\\ 0\\ 0\\ 0\end{array}\right],{\Xi }_{3\mathrm{ii}}=\left[\begin{array}{c}{\mathrm{h\lambda }}_{5\mathrm{ii}}\\ {\mathrm{h\lambda }}_{6\mathrm{ii}}\\ 0\\ {\mathrm{h\lambda }}_{7\mathrm{ii}}\end{array}\right],$${\lambda }_{3ii}=\left[\begin{array}{c}{B}_{1i}\\ {\mathrm{SB}}_{1i}\end{array}\right],$${\Psi }_{1ij}=\left[\begin{array}{cccc}{\mu }_{1ij}+{\mu }_{1ij}^T-2Q& {\mu }_{2ij}+2Q& 0& {\mu }_{3ij}\\ *& -4Q& Q& 0\\ *& *& -2Q& 0\\ *& *& *& -2{\gamma }^{2}I\end{array}\right],{\Psi }_{2ij}=\left[\begin{array}{c}{\mu }_{4ij}\\ 0\\ 0\\ 0\end{array}\right],{\Psi }_{3ij}=\left[\begin{array}{c}{\mathrm{h\mu }}_{5ij}\\ {\mathrm{h\mu }}_{8ij}\\ 0\\ {\mathrm{h\mu }}_{9ij}\end{array}\right],$${\Psi }_{4\mathrm{ij}}=\left[\begin{array}{c}{\mu }_{6\mathrm{ij}}\\ 0\\ 0\\ 0\end{array}\right]{\Psi }_{5\mathrm{ij}}=\left[\begin{array}{c}{\mu }_{7\mathrm{ij}}\\ 0\\ 0\\ 0\end{array}\right],$${\Psi }_{7ij}=\left[\begin{array}{cc}-I& 0\\ 0& -I\end{array}\right],$${\mu }_{3ij}=\left[\begin{array}{c}{B}_{1i}+B_{1j}\\ {\mathrm{SB}}_{1i}+{\mathrm{SB}}_{1j}\end{array}\right],$${\mu }_{9ij}=\left[\begin{array}{cc}{B}_{1i}^T+B_{1j}^T& B_{1i}^{T}S+B_{1j}^{T}S\end{array}\right],$为矩阵C_2i的第一行的转置向量，为矩阵C_2i的第二行的转置向量,h为控制系统相邻采样时刻的最大间隔，即t_k+1‑t_k≤h，单位为s；θ和ρ为设计者根据汽车悬架系统的高性能控制目标来给定的值，可由经验值给出；γ为采用所述模糊采样控制器得到的控制系统对于外部干扰的抑制指标参考值，其为无量纲数，其值大小可由汽车的使用者在容许范围内按照自身需求来调节设定；8)使用所述步骤7)中给出的智能采样数据输出反馈控制器进行汽车悬架系统的在线控制，使得闭环系统渐近稳定且满足对于外部干扰的抑制指标小于参考值γ。