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【Paper】2021_Observer-Based Controllers for Incrementally Quadratic Nonlinear Systems With Disturbanc

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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：

${x˙=Ax+Bu+Ep(q)+Ewwy=Cx+Du+Fwwq=Cqx(1)\left\{\begin{aligned} \dot{{x}} =& A {x} + B u + E p({q}) + E_w w \\ {y} =& C {x} + D u + F_w w \\ {q} =& C_q {x} \end{aligned}\right. \tag{1}$

3 LMI-Based Conditions for robust global stablization of incrementally quadratic nonlinear systems

觀測器設計為：
${x^˙=Ax^+Bu+Ep(q^+L1(y^?y))+L2(y^?y)y^=Cx^+Duq^=Cqx^(11)\left\{\begin{aligned} \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q} + L_1(\hat{y} - y)~) + L_2(\hat{y} - y) \\ \hat{y} =& C \hat{x} + D u \\ \hat{q} =& C_q \hat{x} \end{aligned}\right. \tag{11}$

反饋控制輸入為
$u(t)=k(x^)(12)u(t) = k(\hat{x}) \tag{12}$

基于觀測器控制器（12）的閉環系統可以表示為
${x˙=(A+BK1)x+(E+BK2)p+BΔk+Ewwe˙=(A+L2C)e?Eδp+(Ew+L2Fw)w(15)\left\{\begin{aligned} \dot{{x}} =& (A+BK_1) {x} + (E + BK_2) p + B \Delta k + E_w w \\ \dot{e} =& (A+L_2 C) e - E \delta p + (E_w + L_2 F_w) w \\ \end{aligned}\right. \tag{15}$

定義 $(\begin{matrix}x \\ e \\ \end{matrix})$ ，動力學（15）可以表示為
$z˙=Acz+H1p+H2δp+H3Δp+H4w(17)\dot{z} = A_c z + H_1 p + H_2 \delta p + H_3 \Delta p + H_4 w \tag{17}$

A. Block diagonal parameterization

4 ETC design

A. Configuration I: the controller channel is implemented by ETM

控制輸入為：

$u(t)=K1x^s(t)+K2p(Cqx^s(t))(71)u(t) = K_1 \hat{x}_s(t) + K_2 p(C_q \hat{x}_s(t)) \tag{71}$

B. Configuration II: the controller and observer channels are both implemented by ETMs

$ys(t)=y(tky)?t∈[tky,tk+1y)y_s(t) = y(t_k^y) ~~~~ \forall t \in [t_k^y, t_{k+1}^y)$

這里， $t_0^y = 0$ 并且觸發時刻 $t1y,t2y,?t_1^y, t_2^y, \cdots$ 由以下觸發規則決定：
$tk+1y=inf?{t∣t≥tky+τy,∥ye(t)∥>σy∥y(t)∥+?y}(83)t_{k+1}^y = \inf \{ t | ~~~t \ge t_k^y + \tau_y, ~~~\|y_e(t)\| > \sigma_y \| y(t) \| + \epsilon_y \} \tag{83}$

其中 $y_e(t) = y_s(t) - y(t)$ 并且 $τy,σy,?y\tau_y, \sigma_y, \epsilon_y$ 都是設定好的正數。

結合采樣信息 $y_s(t)$ ，觀測器變為
${x^˙=Ax^+Bu+Ep(q^+L1(y^?ys))+L2(y^?ys)y^=Cx^+Duq^=Cqx^(84)\left\{\begin{aligned} \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q} + L_1(\hat{y} - \red{y_s})~) + L_2(\hat{y} - \red{y_s}) \\ \hat{y} =& C \hat{x} + D u \\ \hat{q} =& C_q \hat{x} \end{aligned}\right. \tag{84}$

其中 $L_1, L_2$ 是設計矩陣。

基于觀測器控制輸入 $u (t)$ 如（71）所示，其中 $x^s(t)\hat{x}_s(t)$ 更新在時刻 $t1u,t2u,?t_1^u, t_2^u, \cdots$
$x^s(t)=x^(tku)?t∈[tku,tk+1u)\hat{x}_s(t) = \hat{x} (t_k^u) ~~~~ \forall t \in [t_k^u, t_{k+1}^u)$

這里， $t_0^u = 0$ 并且觸發時刻 $t1u,t2u,?t_1^u, t_2^u, \cdots$ 由以下觸發規則決定：
$tk+1u=inf?{t∣t≥tku+τu,∥x^e(t)∥>σu∥x^(t)∥+?u}(85)t_{k+1}^u = \inf \{ t | ~~~t \ge t_k^u + \tau_u, ~~~\|\hat{x}_e(t)\| > \sigma_u \| \hat{x}(t) \| + \epsilon_u \} \tag{85}$

其中 $x^e(t)=x^(tk)?x^(t)\hat{x}_e(t) = \hat{x}(t_k) - \hat{x}(t)$ 并且 $τu,σu,?u\tau_u, \sigma_u, \epsilon_u$ 都是設定好的正數。注意 $x^\hat{x}$ 和 $x^e\hat{x}_e$ 的信息可以從設計的觀測器中獲得。

5 Simulation example

$x˙1=x2x˙2=?sin?(x1)+u+wy=x1\begin{aligned} \dot{x}_1 =& x_2 \\ \dot{x}_2 =& -\sin(x_1) + u + w \\ y =& x_1 \\ \end{aligned}$

觀測器設計為
${x^˙=Ax^+Bu+Ep(q^+L1(y^?y))+L2(y^?y)y^=Cx^+Duq^=Cqx^(11)\left\{\begin{aligned} \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q} + L_1(\hat{y} - y)~) + L_2(\hat{y} - y) \\ \hat{y} =& C \hat{x} + D u \\ \hat{q} =& C_q \hat{x} \end{aligned}\right. \tag{11}$

仿真時參數假設為 $L1=?1,L2=[?5.1294,?18.0352]T,p(q^)=sin?(q^),Cq=(1,0)L_1 = -1, L_2 = [-5.1294, -18.0352]^\text{T}, p(\hat{q}) = \sin(\hat{q}), C_q = (1, 0)$ .

因為觀測值 $y^\hat{y}$ 常由傳感器獲得，因此這里假設 $y^=y\hat{y} = y$ ，于是有
${x^˙=Ax^+Bu+Ep(q^)q^=Cqx^\left\{\begin{aligned} \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q}~) \\ \hat{q} =& C_q \hat{x} \\ \end{aligned}\right.$

反饋控制輸入為
$u(t)=k(x^)(12)u(t) = k(\hat{x}) \tag{12}$

$u(t)=K1x^s(t)+K2p(Cqx^s(t))(71)u(t) = K_1 \hat{x}_s(t) + K_2 p(C_q \hat{x}_s(t)) \tag{71}$
其中參數假設為 $K_1 = (-7.3936, -3.9937), K_2 = 1$ .
而 $w$ is uniformly generated from $w_0, w_0]$ .

結合上述分析過程，可以通過以下方式求解觀測值 $x^\hat{x}$
${q^=Cqx^u=K1x^(t)+K2p(q^)x^˙=Ax^+Bu+Ep(q^)\left\{\begin{aligned} \hat{q} =& C_q \hat{x} \\ u =& K_1 \hat{x}(t) + K_2 p(~\hat{q}~) \\ \dot{\hat{x}} =& A \hat{x} + B u + E p(~\hat{q}~) \\ \end{aligned}\right.$

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

$ys(t)=y(tky)?t∈[tky,tk+1y)y_s(t) = y(t_k^y) ~~~~ \forall t \in [t_k^y, t_{k+1}^y)$

其中 $y_e(t) = y_s(t) - y(t)$ 并且 $τy,σy,?y\tau_y, \sigma_y, \epsilon_y$ 都是設定好的正數。

$k=0,t1y=inf?{t∣t≥0+1.07?1e?4,∥ye(t)∥>0.0017∥y(t)∥+0.005}k=1,t2y=inf?{t∣t≥t1y+1.07?1e?4,∥ye(t)∥>0.0017∥y(t)∥+0.005}k=2,t3y=k=3,t4y=\begin{aligned} k = 0, t_1^y =& \inf \{ t | ~~~t \ge 0 + 1.07*1e-4, ~~~\|y_e(t)\| > 0.0017 \| y(t) \| + 0.005 \} \\ k = 1, t_2^y =& \inf \{ t | ~~~t \ge t_1^y + 1.07*1e-4, ~~~\|y_e(t)\| > 0.0017 \| y(t) \| + 0.005 \} \\ k = 2, t_3^y =& \\ k = 3, t_4^y =& \\ \end{aligned}$

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