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2021-04-02 反步法示例

發(fā)布時間:2025/3/15 编程问答 15 豆豆
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一階系統(tǒng)控制

x˙=u+φ1(x)y=x\begin{aligned} \dot{x}&=u+\varphi_1(x) \\ y&=x \end{aligned} x˙y?=u+φ1?(x)=x?

u=?cz1?φ1(x)+y˙du=-c z_{1}-\varphi_1(x)+\dot{y}_ozvdkddzhkzd u=?cz1??φ1?(x)+y˙?d?

z1=y?ydz_{1}=y-y_ozvdkddzhkzd z1?=y?yd?

二階系統(tǒng)控制

x˙1=x2+φ1(x1)x˙2=u+φ2(x1,x2)y=x1\begin{aligned} \dot{x}_{1}&=x_{2}+\varphi_{1}\left(x_{1}\right) \\ \dot{x}_{2}&=u+\varphi_{2}\left(x_{1}, x_{2}\right) \\ y&=x_{1} \end{aligned} x˙1?x˙2?y?=x2?+φ1?(x1?)=u+φ2?(x1?,x2?)=x1??

u=?c2z2?φ2(x1,x2)+?α1?x1x˙1+?α1?ydy˙d+?α1?y˙dy¨d?z1α1=?c1z1?φ1(x1)+y˙d\begin{aligned} u&=-c_{2} z_{2}-\varphi_{2}\left(x_{1}, x_{2}\right)+\frac{\partial \alpha_{1}}{\partial x_{1}} \dot{x}_{1}+\frac{\partial \alpha_{1}}{\partial y_ozvdkddzhkzd} \dot{y}_ozvdkddzhkzd+\frac{\partial \alpha_{1}}{\partial \dot{y}_ozvdkddzhkzd} \ddot{y}_ozvdkddzhkzd-z_{1} \\ \alpha_{1}&=-c_{1} z_{1}-\varphi_{1}\left(x_{1}\right)+\dot{y}_ozvdkddzhkzd \end{aligned} uα1??=?c2?z2??φ2?(x1?,x2?)+?x1??α1??x˙1?+?yd??α1??y˙?d?+?y˙?d??α1??y¨?d??z1?=?c1?z1??φ1?(x1?)+y˙?d??

z1=y?ydz2=x2?α1\begin{aligned} z_{1}&=y-y_ozvdkddzhkzd \\ z_{2}&=x_{2}-\alpha_{1} \end{aligned} z1?z2??=y?yd?=x2??α1??

三階系統(tǒng)控制

x˙1=x2+φ1(x1)x˙2=x3+φ2(x1,x2)x˙3=u+φ3(x1,x2,x3)y=x1\begin{aligned} \dot{x}_{1}&=x_{2}+\varphi_{1}\left(x_{1}\right) \\ \dot{x}_{2}&=x_{3}+\varphi_{2}\left(x_{1}, x_{2}\right) \\ \dot{x}_{3}&=u+\varphi_{3}\left(x_{1}, x_{2}, x_{3}\right) \\ y&=x_{1} \end{aligned} x˙1?x˙2?x˙3?y?=x2?+φ1?(x1?)=x3?+φ2?(x1?,x2?)=u+φ3?(x1?,x2?,x3?)=x1??

u=?c3z3?φ3(x1,x2,x3)+?α2?x1x˙1+?α2?x2x˙2+?α2?ydy˙d+?α2?y˙dy¨d+?α2?y¨dyd(3)?z2α2=?c2z2?φ2(x1,x2)+?α1?x1x˙1+?α1?ydy˙d+?α1?y˙dy¨d?z1α1=?c1z1?φ1(x1)+y˙d\begin{aligned} u&=-c_{3} z_{3}-\varphi_{3}\left(x_{1}, x_{2}, x_{3}\right)+\frac{\partial \alpha_{2}}{\partial x_{1}} \dot{x}_{1}+\frac{\partial \alpha_{2}}{\partial x_{2}} \dot{x}_{2} +\frac{\partial \alpha_{2}}{\partial y_ozvdkddzhkzd} \dot{y}_ozvdkddzhkzd+\frac{\partial \alpha_{2}}{\partial \dot{y}_ozvdkddzhkzd} \ddot{y}_ozvdkddzhkzd+\frac{\partial \alpha_{2}}{\partial \ddot{y}_ozvdkddzhkzd} y^{(3)}_ozvdkddzhkzd-z_{2} \\ \alpha_{2}&=-c_{2} z_{2}-\varphi_{2}\left(x_{1}, x_{2}\right)+\frac{\partial \alpha_{1}}{\partial x_{1}} \dot{x}_{1}+\frac{\partial \alpha_{1}}{\partial y_ozvdkddzhkzd} \dot{y}_ozvdkddzhkzd+\frac{\partial \alpha_{1}}{\partial \dot{y}_ozvdkddzhkzd} \ddot{y}_ozvdkddzhkzd-z_{1} \\ \alpha_{1}&=-c_{1} z_{1}-\varphi_{1}\left(x_{1}\right)+\dot{y}_ozvdkddzhkzd \end{aligned} uα2?α1??=?c3?z3??φ3?(x1?,x2?,x3?)+?x1??α2??x˙1?+?x2??α2??x˙2?+?yd??α2??y˙?d?+?y˙?d??α2??y¨?d?+?y¨?d??α2??yd(3)??z2?=?c2?z2??φ2?(x1?,x2?)+?x1??α1??x˙1?+?yd??α1??y˙?d?+?y˙?d??α1??y¨?d??z1?=?c1?z1??φ1?(x1?)+y˙?d??

z1=y?ydz2=x2?α1z3=x3?α2\begin{aligned} z_{1}&=y-y_ozvdkddzhkzd \\ z_{2}&=x_{2}-\alpha_{1} \\ z_{3}&=x_{3}-\alpha_{2} \end{aligned} z1?z2?z3??=y?yd?=x2??α1?=x3??α2??

三階動態(tài)面控制

x˙1=x2+φ1(x1)x˙2=x3+φ2(x1,x2)x˙3=u+φ3(x1,x2,x3)y=x1\begin{aligned} \dot{x}_{1}&=x_{2}+\varphi_{1}\left(x_{1}\right) \\ \dot{x}_{2}&=x_{3}+\varphi_{2}\left(x_{1}, x_{2}\right) \\ \dot{x}_{3}&=u+\varphi_{3}\left(x_{1}, x_{2}, x_{3}\right) \\ y&=x_{1} \end{aligned} x˙1?x˙2?x˙3?y?=x2?+φ1?(x1?)=x3?+φ2?(x1?,x2?)=u+φ3?(x1?,x2?,x3?)=x1??

u=?c3z3?φ3(x1,x2,x3)+ε˙2α2=?c2z2?φ2(x1,x2)+ε˙1α1=?c1z1?φ1(x1)+y˙d\begin{aligned} u&=-c_{3} z_{3}-\varphi_{3}\left(x_{1}, x_{2}, x_{3}\right)+\dot{\varepsilon}_{2} \\ \alpha_{2}&=-c_{2} z_{2}-\varphi_{2}\left(x_{1}, x_{2}\right)+\dot{\varepsilon}_{1} \\ \alpha_{1}&=-c_{1} z_{1}-\varphi_{1}\left(x_{1}\right)+\dot{y}_ozvdkddzhkzd \end{aligned} uα2?α1??=?c3?z3??φ3?(x1?,x2?,x3?)+ε˙2?=?c2?z2??φ2?(x1?,x2?)+ε˙1?=?c1?z1??φ1?(x1?)+y˙?d??

z1=y?ydz2=x2?ε1z3=x3?ε2\begin{aligned} z_{1}&=y-y_ozvdkddzhkzd \\ z_{2}&=x_{2}-\varepsilon_{1} \\ z_{3}&=x_{3}-\varepsilon_{2} \end{aligned} z1?z2?z3??=y?yd?=x2??ε1?=x3??ε2??

τ1ε˙1+ε1=α1τ2ε˙2+ε2=α2\begin{aligned} \tau_{1} \dot{\varepsilon}_{1}+\varepsilon_{1}&=\alpha_{1} \\ \tau_{2} \dot{\varepsilon}_{2}+\varepsilon_{2}&=\alpha_{2} \end{aligned} τ1?ε˙1?+ε1?τ2?ε˙2?+ε2??=α1?=α2??

濾波器誤差: Y1=ε1?α1,Y2=ε2?α2Y_{1}=\varepsilon_{1}-\alpha_{1}, Y_{2}=\varepsilon_{2}-\alpha_{2}Y1?=ε1??α1?,Y2?=ε2??α2?

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