一階系統控制

$$\begin{array}{rl}\dot{x}& =u+{\phi }_1(x)\\ y& =x\end{array}$$

$$u=?c{z}_1?{\phi }_1(x)+{\dot{y}}_d$$

$$z_1=y?y_d$$

二階系統控制

$$\begin{array}{rl}{\dot{x}}_1& =x_2+{\phi }_1\left(x_1\right)\\ {\dot{x}}_2& =u+{\phi }_2\left(x_1,x_2\right)\\ y& =x_1\end{array}$$

$$\begin{array}{rl}u& =?c_2{z}_2?{\phi }_2\left(x_1,x_2\right)+\frac{?{\alpha }_1}{?x_1}{\dot{x}}_1+\frac{?{\alpha }_1}{?y_d}{\dot{y}}_d+\frac{?{\alpha }_1}{?{\dot{y}}_d}{\ddot{y}}_d?z_1\\ {\alpha }_1& =?c_1{z}_1?{\phi }_1\left(x_1\right)+{\dot{y}}_d\end{array}$$

$$\begin{array}{rl}{z}_1& =y?y_d\\ z_2& =x_2?{\alpha }_1\end{array}$$

三階系統控制

$$\begin{array}{rl}{\dot{x}}_1& =x_2+{\phi }_1\left(x_1\right)\\ {\dot{x}}_2& =x_3+{\phi }_2\left(x_1,x_2\right)\\ {\dot{x}}_3& =u+{\phi }_3\left(x_1,x_2,x_3\right)\\ y& =x_1\end{array}$$

$$\begin{array}{rl}u& =?c_3{z}_3?{\phi }_3\left(x_1,x_2,x_3\right)+\frac{?{\alpha }_2}{?x_1}{\dot{x}}_1+\frac{?{\alpha }_2}{?x_2}{\dot{x}}_2+\frac{?{\alpha }_2}{?y_d}{\dot{y}}_d+\frac{?{\alpha }_2}{?{\dot{y}}_d}{\ddot{y}}_d+\frac{?{\alpha }_2}{?{\ddot{y}}_d}{y}_d^{(3)}?z_2\\ {\alpha }_2& =?c_2{z}_2?{\phi }_2\left(x_1,x_2\right)+\frac{?{\alpha }_1}{?x_1}{\dot{x}}_1+\frac{?{\alpha }_1}{?y_d}{\dot{y}}_d+\frac{?{\alpha }_1}{?{\dot{y}}_d}{\ddot{y}}_d?z_1\\ {\alpha }_1& =?c_1{z}_1?{\phi }_1\left(x_1\right)+{\dot{y}}_d\end{array}$$

$$\begin{array}{rl}{z}_1& =y?y_d\\ z_2& =x_2?{\alpha }_1\\ z_3& =x_3?{\alpha }_2\end{array}$$

三階動態面控制

$$\begin{array}{rl}{\dot{x}}_1& =x_2+{\phi }_1\left(x_1\right)\\ {\dot{x}}_2& =x_3+{\phi }_2\left(x_1,x_2\right)\\ {\dot{x}}_3& =u+{\phi }_3\left(x_1,x_2,x_3\right)\\ y& =x_1\end{array}$$

$$\begin{array}{rl}u& =?c_3{z}_3?{\phi }_3\left(x_1,x_2,x_3\right)+{\dot{\epsilon }}_2\\ {\alpha }_2& =?c_2{z}_2?{\phi }_2\left(x_1,x_2\right)+{\dot{\epsilon }}_1\\ {\alpha }_1& =?c_1{z}_1?{\phi }_1\left(x_1\right)+{\dot{y}}_d\end{array}$$

$$\begin{array}{rl}{z}_1& =y?y_d\\ z_2& =x_2?{\epsilon }_1\\ z_3& =x_3?{\epsilon }_2\end{array}$$

$$\begin{array}{rl}{\tau }_1{\dot{\epsilon }}_1+{\epsilon }_1& ={\alpha }_1\\ {\tau }_2{\dot{\epsilon }}_2+{\epsilon }_2& ={\alpha }_2\end{array}$$

濾波器誤差： $Y_1={\epsilon }_1?{\alpha }_1,Y_2={\epsilon }_2?{\alpha }_2$

總結

以上是生活随笔為你收集整理的2021-04-02 反步法示例的全部內容，希望文章能夠幫你解決所遇到的問題。

如果覺得生活随笔網站內容還不錯，歡迎將生活随笔推薦給好友。