回顾上节
动力学方程
J ⏟ a 1 x ¨ + b ⏟ a 2 x ˙ ∣ x ˙ ∣ + m g l ⏟ a 3 s i n x = u \underbrace{J}_{a_1}\ddot{x}+\underbrace{b}_{a_2}\dot{x}\left | \dot{x} \right |+\underbrace{mgl}_{a_3}sinx=u
a 1 J x ¨ + a 2 b x ˙ ∣ x ˙ ∣ + a 3 m g l s in x = u
x ~ ( t ) = x ( t ) − x d ( t ) \tilde{x}(t)= x(t)-x_d(t) x ~ ( t ) = x ( t ) − x d ( t ) s = x ~ ˙ + λ x ~ = x ˙ − x ˙ r s=\dot{\tilde{x}}+\lambda \tilde{x}=\dot{x}-\dot{x}_r s = x ~ ˙ + λ x ~ = x ˙ − x ˙ r x ˙ r = x ˙ d − λ x ~ \dot{x}_r= \dot{x}_d-\lambda \tilde{x} x ˙ r = x ˙ d − λ x ~ s = x ( n ) − x r ( n ) s=x^{(n)}-x_r^{(n)} s = x ( n ) − x r ( n )
y = [ x ¨ r x ˙ ∣ x ˙ ∣ s i n x ] \boldsymbol{y}=[\ddot{x}_r ~~\dot{x}\left |\dot{x} \right |~~sinx ] y = [ x ¨ r x ˙ ∣ x ˙ ∣ s in x ]
u = y a ^ − k s u=\boldsymbol{y}\hat{\boldsymbol{a}}-ks u = y a ^ − k s a ^ ˙ = − P y T s \boldsymbol{\dot{\hat{a}}}= -\boldsymbol{P}\boldsymbol{y}^Ts a ^ ˙ = − P y T s
a ~ ( t ) = a ^ ( t ) − a ( t ) \boldsymbol{\tilde{a}}(t)=\boldsymbol{\hat{a}(t)}-\boldsymbol{a}(t) a ~ ( t ) = a ^ ( t ) − a ( t )
V = 1 2 J s 2 + 1 2 a ~ T P − 1 a ~ V=\frac{1}{2}Js^2+\frac{1}{2}\boldsymbol{\tilde{a}}^T\boldsymbol{P}^{-1}\boldsymbol{\tilde{a}} V = 2 1 J s 2 + 2 1 a ~ T P − 1 a ~
V ˙ = s ( u − y a ) = − k s 2 + s y a ~ + a ^ ˙ T P − 1 a ~ ⏟ = 0 = − k s 2 ⩽ 0 \dot{V}=s(u-\boldsymbol{y}\boldsymbol{a}) =-ks^2+\underbrace{s\boldsymbol{y}\tilde{\boldsymbol{a}}+\boldsymbol{\dot{\hat{a}}}^T\boldsymbol{P}^{-1}\boldsymbol{\tilde{a}}}_{=0}=-ks^2 \leqslant 0 V ˙ = s ( u − y a ) = − k s 2 + = 0 s y a ~ + a ^ ˙ T P − 1 a ~ = − k s 2 ⩽ 0
V ˙ → 0 ⇒ s → 0 ⇒ x ~ → 0 , x ~ ˙ → 0 \dot{V} \rightarrow 0 \Rightarrow s \rightarrow 0 \Rightarrow \tilde{x} \rightarrow0,\dot{ \tilde{x}} \rightarrow 0 V ˙ → 0 ⇒ s → 0 ⇒ x ~ → 0 , x ~ ˙ → 0
Barbalat引理
V bounded often to prove V ¨ \ddot{V} V ¨
验证V ¨ \ddot{V} V ¨ V ¨ = d d t V ˙ = − 2 k s ⋅ s ˙ = − 2 k s ( x ¨ − x ¨ r ) \ddot{V}= \frac{\mathrm{d} }{\mathrm{d} t}\dot{V}=-2ks\cdot \dot{s}=-2ks(\ddot{x}-\ddot{x}_r) V ¨ = d t d V ˙ = − 2 k s ⋅ s ˙ = − 2 k s ( x ¨ − x ¨ r ) ⇒ \Rightarrow ⇒ a ^ \boldsymbol{\hat{a}} a ^ ⇒ \Rightarrow ⇒ x ~ \tilde{x} x ~ x ~ ˙ \dot{\tilde{x}} x ~ ˙ ⇒ \Rightarrow ⇒ x x x x ˙ \dot{x} x ˙ V ¨ \ddot{V} V ¨
sufficiently, a ^ \hat{\boldsymbol{a}} a ^ a a a ⇔ \Leftrightarrow ⇔ y d T y d ⩾ 0 \boldsymbol{y}_d^T \boldsymbol{y}_d \geqslant 0 y d T y d ⩾ 0
8.8 Robust Adaptive Control
系统
J ⏟ a 1 x ¨ + b ⏟ a 2 x ˙ ∣ x ˙ ∣ + m g l ⏟ a 3 s i n x = f ( x , x ˙ , t ) + u \underbrace{J}_{a_1}\ddot{x}+\underbrace{b}_{a_2}\dot{x}\left | \dot{x} \right |+\underbrace{mgl}_{a_3}sinx=f(x,\dot{x},t)+u
a 1 J x ¨ + a 2 b x ˙ ∣ x ˙ ∣ + a 3 m g l s in x = f ( x , x ˙ , t ) + u
∣ f ^ ( x , t ) − f ( x , t ) ∣ ⩽ F ( x , t ) \left | \hat{f}(\boldsymbol{x},t)-f(\boldsymbol{x},t) \right |\leqslant F(\boldsymbol{x},t) f ^ ( x , t ) − f ( x , t ) ⩽ F ( x , t ) f ^ ( x , t ) \hat{f}(\boldsymbol{x},t) f ^ ( x , t ) F ( x , t ) F(\boldsymbol{x},t) F ( x , t )
want the system to behave after some transients as if a \boldsymbol{a} a
定义变量s Δ = s − Φ s a t ( s / Φ ) s_{\Delta} = s-\Phi sat(s/\Phi ) s Δ = s − Φ s a t ( s /Φ )
∣ s Δ ∣ ⩽ Φ ⇔ s Δ = 0 \left | s_{\Delta} \right | \leqslant \Phi \Leftrightarrow s_{\Delta} = 0 ∣ s Δ ∣ ⩽ Φ ⇔ s Δ = 0 d d t s Δ 2 = s Δ ⋅ s ˙ \frac{\mathrm{d} }{\mathrm{d} t}s_{\Delta}^2=s_{\Delta}\cdot \dot{s} d t d s Δ 2 = s Δ ⋅ s ˙
李雅普诺夫函数
V = 1 2 J s Δ 2 + 1 2 a ~ T P − 1 a ~ V=\frac{1}{2}Js^2_{\Delta}+\frac{1}{2}\boldsymbol{\tilde{a}}^T\boldsymbol{P}^{-1}\boldsymbol{\tilde{a}}
V = 2 1 J s Δ 2 + 2 1 a ~ T P − 1 a ~
求微分
V ˙ = s Δ ⋅ J s ˙ = s Δ ( u + f ( x , t ) − y a ) (8.6) \dot{V}= s_{\Delta} \cdot J\dot{s}= s_{\Delta} (u+f(\boldsymbol{x},t)-\boldsymbol{y}\boldsymbol{a}) \tag{8.6}
V ˙ = s Δ ⋅ J s ˙ = s Δ ( u + f ( x , t ) − y a ) ( 8.6 )
若选择控制律u = y a ^ − k s a t ( s / Φ ) − f ^ ( x , t ) ⏟ R o b u s t C o n t r o l u= \boldsymbol{y}\boldsymbol{\hat{a}}-\underbrace{k~sat(s/\Phi )-\hat{f}(\boldsymbol{x},t)}_{Robust~~Control} u = y a ^ − R o b u s t C o n t ro l k s a t ( s /Φ ) − f ^ ( x , t ) a ~ ( t ) = a ^ ( t ) − a ( t ) \boldsymbol{\tilde{a}}(t)=\boldsymbol{\hat{a}(t)}-\boldsymbol{a}(t) a ~ ( t ) = a ^ ( t ) − a ( t )
则
V ˙ = s Δ ( u + f ( x , t ) − y a ) = s Δ ( − k s a t ( s / Φ ) + y a ~ + f ( x , t ) − f ^ ( x , t ) ) + a ^ ˙ T P − 1 a ~ \begin{align*}
\dot{V}&=s_{\Delta} (u+f(\boldsymbol{x},t)-\boldsymbol{y}\boldsymbol{a}) \\
&=s_{\Delta}(-k~sat(s/\Phi) +\boldsymbol{y}\boldsymbol{\tilde{a}}+f(\boldsymbol{x},t)-\hat{f}(\boldsymbol{x},t))+\boldsymbol{\dot{\hat{a}}}^T\boldsymbol{P}^{-1}\boldsymbol{\tilde{a}}\tag{8.7}
\end{align*}
V ˙ = s Δ ( u + f ( x , t ) − y a ) = s Δ ( − k s a t ( s /Φ ) + y a ~ + f ( x , t ) − f ^ ( x , t )) + a ^ ˙ T P − 1 a ~ ( 8.7 )
因为s Δ s a t ( s / Φ ) = s Δ s g n ( s ) = ∣ s Δ ∣ s_{\Delta}sat(s/\Phi) =s_{\Delta}sgn(s)= \left | s_{\Delta} \right | s Δ s a t ( s /Φ ) = s Δ s g n ( s ) = ∣ s Δ ∣
V ˙ = − k ∣ s Δ ∣ + s Δ y a ~ + ( f ( x , t ) − f ^ ( x , t ) ) s Δ + a ^ ˙ T P − 1 a ~ (8.8) \dot{V}= -k\left | s_{\Delta} \right |+s_{\Delta}\boldsymbol{y}\boldsymbol{\tilde{a}}+(f(\boldsymbol{x},t)-\hat{f}(\boldsymbol{x},t))s_{\Delta}+ \boldsymbol{\dot{\hat{a}}}^T\boldsymbol{P}^{-1}\boldsymbol{\tilde{a}} \tag{8.8}
V ˙ = − k ∣ s Δ ∣ + s Δ y a ~ + ( f ( x , t ) − f ^ ( x , t )) s Δ + a ^ ˙ T P − 1 a ~ ( 8.8 )
根据式(8.5),取a ^ ˙ = − P y T s \boldsymbol{\dot{\hat{a}}}= -\boldsymbol{P}\boldsymbol{y}^Ts a ^ ˙ = − P y T s s y a ~ + a ^ ˙ T P − 1 a ~ = ( s y + a ^ ˙ T P − 1 ) a ~ = 0 s\boldsymbol{y}\tilde{\boldsymbol{a}}+\boldsymbol{\dot{\hat{a}}}^T\boldsymbol{P}^{-1}\boldsymbol{\tilde{a}}= (s\boldsymbol{y}+\boldsymbol{\dot{\hat{a}}}^T\boldsymbol{P}^{-1})\boldsymbol{\tilde{a}}=0 s y a ~ + a ^ ˙ T P − 1 a ~ = ( s y + a ^ ˙ T P − 1 ) a ~ = 0
V ˙ = − k ∣ s Δ ∣ + ( f − f ^ ) s Δ (8.9) \dot{V}= -k\left | s_{\Delta} \right |+(f-\hat{f})s_{\Delta} \tag{8.9}
V ˙ = − k ∣ s Δ ∣ + ( f − f ^ ) s Δ ( 8.9 )
取k = F + η k=F+\eta k = F + η
V ˙ = − k ∣ s Δ ∣ + ( f − f ^ ) s Δ = − ( F + η ) ∣ s Δ ∣ + ( f − f ^ ) s Δ \begin{align*}
\dot{V}&= -k\left | s_{\Delta} \right |+(f-\hat{f})s_{\Delta} \\
&= -(F+\eta)\left | s_{\Delta} \right |+(f-\hat{f})s_{\Delta} \tag{8.10}
\end{align*}
V ˙ = − k ∣ s Δ ∣ + ( f − f ^ ) s Δ = − ( F + η ) ∣ s Δ ∣ + ( f − f ^ ) s Δ ( 8.10 )
故V ˙ ⩽ − η ∣ s Δ ∣ ⩽ 0 \dot{V}\leqslant -\eta \left | s_{\Delta} \right | \leqslant 0 V ˙ ⩽ − η ∣ s Δ ∣ ⩽ 0
根据Barbalat引理 V ˙ → 0 ⇒ s Δ → 0 \dot{V} \rightarrow 0 \Rightarrow s_{\Delta} \rightarrow 0 V ˙ → 0 ⇒ s Δ → 0
拓展
系统
J ⏟ a 1 x ¨ + b ⏟ a 2 x ˙ ∣ x ˙ ∣ + m g l ⏟ a 3 s i n x = φ ( x ˙ ) + f ( x , x ˙ , t ) + u \underbrace{J}_{a_1}\ddot{x}+\underbrace{b}_{a_2}\dot{x}\left | \dot{x} \right |+\underbrace{mgl}_{a_3}sinx=\varphi (\dot{x})+f(x,\dot{x},t)+u
a 1 J x ¨ + a 2 b x ˙ ∣ x ˙ ∣ + a 3 m g l s in x = φ ( x ˙ ) + f ( x , x ˙ , t ) + u
其中∣ g ^ ( x , t ) − g ( x , t ) ∣ ⩽ F g \left | \hat{g}(\boldsymbol{x},t)-g(\boldsymbol{x},t) \right |\leqslant F_g ∣ g ^ ( x , t ) − g ( x , t ) ∣ ⩽ F g φ ( x ˙ ) = ∑ i n f i n i t y α i ρ i ( x ˙ ) \varphi (\dot{x})=\sum_{infinity}^{} \alpha_i \rho _i(\dot{x}) φ ( x ˙ ) = ∑ in f ini t y α i ρ i ( x ˙ )
Desirable properties of the expansions
(i) At ant x ˙ \dot{x} x ˙ ∑ i n f i n i t y = ∑ f i n i t e N + r e s i d u a l \sum_{infinity}^{}=\sum_{finite~N}^{}+ residual ∑ in f ini t y = ∑ f ini t e N + res i d u a l ↘ \searrow ↘ ↗ \nearrow ↗ ρ i ( x ˙ ) \rho _i(\dot{x}) ρ i ( x ˙ )
∫ e i e j d x ˙ = δ i j = { 1 i f e i = e j 0 i f e i ≠ e j \int e_i~e_jd\dot{x}=\delta_{ij}=\left\{\begin{matrix}
1 &if ~~e_i=e_j \\
0& if~~e_i \neq e_j
\end{matrix}\right.
∫ e i e j d x ˙ = δ ij = { 1 0 i f e i = e j i f e i = e j
∥ ∑ i n f i n i t y − ∑ f i n i t e N ∥ 2 = ∥ ∑ r e s i d u a l α i 2 ∥ \left \| \sum_{infinity}^{}-\sum_{finite~N}^{} \right \|^2=\left \|\sum_{residual}^{}\alpha_i^2 \right \| ∑ in f ini t y − ∑ f ini t e N 2 = ∑ res i d u a l α i 2
参考
【1】Applied Nonlinear Control,Slotine and Li,Prentice-Hall 1991:Exercise 8.8 R6,R7https://www.bilibili.com/video/BV1Tk4y167kR?p=9&vd_source=b1a52fdb6c7481b4e8c78d03e9a9acb0
