【Paper】2021_Observer-Based Controllers for Incrementally Quadratic Nonlinear Systems With Disturbanc
X. Xu, B. A??kme?e and M. J. Corless, “Observer-Based Controllers for Incrementally Quadratic Nonlinear Systems With Disturbances,” in IEEE Transactions on Automatic Control, vol. 66, no. 3, pp. 1129-1143, March 2021, doi: 10.1109/TAC.2020.2996985.
文章目錄
- 1 Introduction
- 2 Preliminaries
- 3 LMI-Based Conditions for robust global stablization of incrementally quadratic nonlinear systems
- A. Block diagonal parameterization
- 4 ETC design
- A. Configuration I: the controller channel is implemented by ETM
- B. Configuration II: the controller and observer channels are both implemented by ETMs
- 5 Simulation example
1 Introduction
2 Preliminaries
非線形系統描述為:
{x˙=Ax+Bu+Ep(q)+Ewwy=Cx+Du+Fwwq=Cqx(1)\left\{\begin{aligned} \dot{{x}} =& A {x} + B u + E p({q}) + E_w w \\ {y} =& C {x} + D u + F_w w \\ {q} =& C_q {x} \end{aligned}\right. \tag{1}??????x˙=y=q=?Ax+Bu+Ep(q)+Ew?wCx+Du+Fw?wCq?x?(1)
3 LMI-Based Conditions for robust global stablization of incrementally quadratic nonlinear systems
觀測器設計為:
{x^˙=Ax^+Bu+Ep(q^+L1(y^?y))+L2(y^?y)y^=Cx^+Duq^=Cqx^(11)\left\{\begin{aligned} \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q} + L_1(\hat{y} - y)~) + L_2(\hat{y} - y) \\ \hat{y} =& C \hat{x} + D u \\ \hat{q} =& C_q \hat{x} \end{aligned}\right. \tag{11}??????x^˙=y^?=q^?=?Ax^+Bu+Ep(?q^?+L1?(y^??y)?)+L2?(y^??y)Cx^+DuCq?x^?(11)
反饋控制輸入為
u(t)=k(x^)(12)u(t) = k(\hat{x}) \tag{12}u(t)=k(x^)(12)
基于觀測器控制器(12)的閉環系統可以表示為
{x˙=(A+BK1)x+(E+BK2)p+BΔk+Ewwe˙=(A+L2C)e?Eδp+(Ew+L2Fw)w(15)\left\{\begin{aligned} \dot{{x}} =& (A+BK_1) {x} + (E + BK_2) p + B \Delta k + E_w w \\ \dot{e} =& (A+L_2 C) e - E \delta p + (E_w + L_2 F_w) w \\ \end{aligned}\right. \tag{15}{x˙=e˙=?(A+BK1?)x+(E+BK2?)p+BΔk+Ew?w(A+L2?C)e?Eδp+(Ew?+L2?Fw?)w?(15)
定義 z=(xe)z = (\begin{matrix}x \\ e \\ \end{matrix})z=(xe?),動力學(15)可以表示為
z˙=Acz+H1p+H2δp+H3Δp+H4w(17)\dot{z} = A_c z + H_1 p + H_2 \delta p + H_3 \Delta p + H_4 w \tag{17}z˙=Ac?z+H1?p+H2?δp+H3?Δp+H4?w(17)
A. Block diagonal parameterization
4 ETC design
A. Configuration I: the controller channel is implemented by ETM
控制輸入為:
u(t)=K1x^s(t)+K2p(Cqx^s(t))(71)u(t) = K_1 \hat{x}_s(t) + K_2 p(C_q \hat{x}_s(t)) \tag{71}u(t)=K1?x^s?(t)+K2?p(Cq?x^s?(t))(71)
B. Configuration II: the controller and observer channels are both implemented by ETMs
ys(t)=y(tky)?t∈[tky,tk+1y)y_s(t) = y(t_k^y) ~~~~ \forall t \in [t_k^y, t_{k+1}^y)ys?(t)=y(tky?)?????t∈[tky?,tk+1y?)
這里,t0y=0t_0^y = 0t0y?=0 并且觸發時刻 t1y,t2y,?t_1^y, t_2^y, \cdotst1y?,t2y?,? 由以下觸發規則決定:
tk+1y=inf?{t∣t≥tky+τy,∥ye(t)∥>σy∥y(t)∥+?y}(83)t_{k+1}^y = \inf \{ t | ~~~t \ge t_k^y + \tau_y, ~~~\|y_e(t)\| > \sigma_y \| y(t) \| + \epsilon_y \} \tag{83}tk+1y?=inf{t∣???t≥tky?+τy?,???∥ye?(t)∥>σy?∥y(t)∥+?y?}(83)
其中 ye(t)=ys(t)?y(t)y_e(t) = y_s(t) - y(t)ye?(t)=ys?(t)?y(t) 并且 τy,σy,?y\tau_y, \sigma_y, \epsilon_yτy?,σy?,?y? 都是設定好的正數。
結合采樣信息 ys(t)y_s(t)ys?(t),觀測器變為
{x^˙=Ax^+Bu+Ep(q^+L1(y^?ys))+L2(y^?ys)y^=Cx^+Duq^=Cqx^(84)\left\{\begin{aligned} \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q} + L_1(\hat{y} - \red{y_s})~) + L_2(\hat{y} - \red{y_s}) \\ \hat{y} =& C \hat{x} + D u \\ \hat{q} =& C_q \hat{x} \end{aligned}\right. \tag{84}??????x^˙=y^?=q^?=?Ax^+Bu+Ep(?q^?+L1?(y^??ys?)?)+L2?(y^??ys?)Cx^+DuCq?x^?(84)
其中 L1,L2L_1, L_2L1?,L2? 是設計矩陣。
基于觀測器控制輸入 u(t)u(t)u(t) 如(71)所示,其中 x^s(t)\hat{x}_s(t)x^s?(t) 更新在時刻 t1u,t2u,?t_1^u, t_2^u, \cdotst1u?,t2u?,?
x^s(t)=x^(tku)?t∈[tku,tk+1u)\hat{x}_s(t) = \hat{x} (t_k^u) ~~~~ \forall t \in [t_k^u, t_{k+1}^u)x^s?(t)=x^(tku?)?????t∈[tku?,tk+1u?)
這里,t0u=0t_0^u = 0t0u?=0 并且觸發時刻 t1u,t2u,?t_1^u, t_2^u, \cdotst1u?,t2u?,? 由以下觸發規則決定:
tk+1u=inf?{t∣t≥tku+τu,∥x^e(t)∥>σu∥x^(t)∥+?u}(85)t_{k+1}^u = \inf \{ t | ~~~t \ge t_k^u + \tau_u, ~~~\|\hat{x}_e(t)\| > \sigma_u \| \hat{x}(t) \| + \epsilon_u \} \tag{85}tk+1u?=inf{t∣???t≥tku?+τu?,???∥x^e?(t)∥>σu?∥x^(t)∥+?u?}(85)
其中 x^e(t)=x^(tk)?x^(t)\hat{x}_e(t) = \hat{x}(t_k) - \hat{x}(t)x^e?(t)=x^(tk?)?x^(t) 并且 τu,σu,?u\tau_u, \sigma_u, \epsilon_uτu?,σu?,?u? 都是設定好的正數。注意 x^\hat{x}x^ 和 x^e\hat{x}_ex^e? 的信息可以從設計的觀測器中獲得。
5 Simulation example
x˙1=x2x˙2=?sin?(x1)+u+wy=x1\begin{aligned} \dot{x}_1 =& x_2 \\ \dot{x}_2 =& -\sin(x_1) + u + w \\ y =& x_1 \\ \end{aligned}x˙1?=x˙2?=y=?x2??sin(x1?)+u+wx1??
觀測器設計為
{x^˙=Ax^+Bu+Ep(q^+L1(y^?y))+L2(y^?y)y^=Cx^+Duq^=Cqx^(11)\left\{\begin{aligned} \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q} + L_1(\hat{y} - y)~) + L_2(\hat{y} - y) \\ \hat{y} =& C \hat{x} + D u \\ \hat{q} =& C_q \hat{x} \end{aligned}\right. \tag{11}??????x^˙=y^?=q^?=?Ax^+Bu+Ep(?q^?+L1?(y^??y)?)+L2?(y^??y)Cx^+DuCq?x^?(11)
仿真時參數假設為 L1=?1,L2=[?5.1294,?18.0352]T,p(q^)=sin?(q^),Cq=(1,0)L_1 = -1, L_2 = [-5.1294, -18.0352]^\text{T}, p(\hat{q}) = \sin(\hat{q}), C_q = (1, 0)L1?=?1,L2?=[?5.1294,?18.0352]T,p(q^?)=sin(q^?),Cq?=(1,0).
因為觀測值 y^\hat{y}y^? 常由傳感器獲得,因此這里假設 y^=y\hat{y} = yy^?=y,于是有
{x^˙=Ax^+Bu+Ep(q^)q^=Cqx^\left\{\begin{aligned} \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q}~) \\ \hat{q} =& C_q \hat{x} \\ \end{aligned}\right.{x^˙=q^?=?Ax^+Bu+Ep(?q^??)Cq?x^?
反饋控制輸入為
u(t)=k(x^)(12)u(t) = k(\hat{x}) \tag{12}u(t)=k(x^)(12)
u(t)=K1x^s(t)+K2p(Cqx^s(t))(71)u(t) = K_1 \hat{x}_s(t) + K_2 p(C_q \hat{x}_s(t)) \tag{71}u(t)=K1?x^s?(t)+K2?p(Cq?x^s?(t))(71)
其中參數假設為 K1=(?7.3936,?3.9937),K2=1K_1 = (-7.3936, -3.9937), K_2 = 1K1?=(?7.3936,?3.9937),K2?=1.
而 www is uniformly generated from [?w0,w0][-w_0, w_0][?w0?,w0?].
結合上述分析過程,可以通過以下方式求解觀測值 x^\hat{x}x^
{q^=Cqx^u=K1x^(t)+K2p(q^)x^˙=Ax^+Bu+Ep(q^)\left\{\begin{aligned} \hat{q} =& C_q \hat{x} \\ u =& K_1 \hat{x}(t) + K_2 p(~\hat{q}~) \\ \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q}~) \\ \end{aligned}\right.??????q^?=u=x^˙=?Cq?x^K1?x^(t)+K2?p(?q^??)Ax^+Bu+Ep(?q^??)?
之后,也可以通過同樣方式,結合公式(1)求出系統狀態。
誤差圖如下,用來對比原文中的 Fig.4.
接下來分析 ETC
ys(t)=y(tky)?t∈[tky,tk+1y)y_s(t) = y(t_k^y) ~~~~ \forall t \in [t_k^y, t_{k+1}^y)ys?(t)=y(tky?)?????t∈[tky?,tk+1y?)
這里,t0y=0t_0^y = 0t0y?=0 并且觸發時刻 t1y,t2y,?t_1^y, t_2^y, \cdotst1y?,t2y?,? 由以下觸發規則決定:
tk+1y=inf?{t∣t≥tky+τy,∥ye(t)∥>σy∥y(t)∥+?y}(83)t_{k+1}^y = \inf \{ t | ~~~t \ge t_k^y + \tau_y, ~~~\|y_e(t)\| > \sigma_y \| y(t) \| + \epsilon_y \} \tag{83}tk+1y?=inf{t∣???t≥tky?+τy?,???∥ye?(t)∥>σy?∥y(t)∥+?y?}(83)
其中 ye(t)=ys(t)?y(t)y_e(t) = y_s(t) - y(t)ye?(t)=ys?(t)?y(t) 并且 τy,σy,?y\tau_y, \sigma_y, \epsilon_yτy?,σy?,?y? 都是設定好的正數。
k=0,t1y=inf?{t∣t≥0+1.07?1e?4,∥ye(t)∥>0.0017∥y(t)∥+0.005}k=1,t2y=inf?{t∣t≥t1y+1.07?1e?4,∥ye(t)∥>0.0017∥y(t)∥+0.005}k=2,t3y=k=3,t4y=\begin{aligned} k = 0, t_1^y =& \inf \{ t | ~~~t \ge 0 + 1.07*1e-4, ~~~\|y_e(t)\| > 0.0017 \| y(t) \| + 0.005 \} \\ k = 1, t_2^y =& \inf \{ t | ~~~t \ge t_1^y + 1.07*1e-4, ~~~\|y_e(t)\| > 0.0017 \| y(t) \| + 0.005 \} \\ k = 2, t_3^y =& \\ k = 3, t_4^y =& \\ \end{aligned}k=0,t1y?=k=1,t2y?=k=2,t3y?=k=3,t4y?=?inf{t∣???t≥0+1.07?1e?4,???∥ye?(t)∥>0.0017∥y(t)∥+0.005}inf{t∣???t≥t1y?+1.07?1e?4,???∥ye?(t)∥>0.0017∥y(t)∥+0.005}?
總結
以上是生活随笔為你收集整理的【Paper】2021_Observer-Based Controllers for Incrementally Quadratic Nonlinear Systems With Disturbanc的全部內容,希望文章能夠幫你解決所遇到的問題。
- 上一篇: 【Paper】2020_异构无人机编队防
- 下一篇: 【Paper】2017_水下潜航器编队海