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

X. Xu, B. Açıkmeşe 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.

1 Introduction

2 Preliminaries

The non-linear system is described as ：

$$\{\begin{array}{rl}\dot{x}=& Ax+Bu+Ep(q)+E_{w}w\\ y=& Cx+Du+F_{w}w\\ q=& C_{q}x\end{array}\text{\hspace{1em}(1)}$$

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

The observer is designed as ：
$$\{\begin{array}{rl}\dot{\hat{x}}=& A\hat{x}+Bu+Ep(\hat{q}+L_1(\hat{y}-y))+L_2(\hat{y}-y)\\ \hat{y}=& C\hat{x}+Du\\ \hat{q}=& C_q\hat{x}\end{array}\text{\hspace{1em}(11)}$$

The feedback control input is
$$u(t)=k(\hat{x})\text{\hspace{1em}(12)}$$

Observer based controller （12） The closed-loop system can be expressed as
$$\{\begin{array}{rl}\dot{x}=& (A+B{K}_1)x+(E+B{K}_2)p+B\Delta k+E_{w}w\\ \dot{e}=& (A+L_{2}C)e-E\delta p+(E_w+L_2{F}_w)w\end{array}\text{\hspace{1em}(15)}$$

Definition $z=(\begin{array}{c}x\\ e\end{array})$, Dynamics （15） It can be expressed as
$$\dot{z}=A_{c}z+H_{1}p+H_2\delta p+H_3\Delta p+H_{4}w\text{\hspace{1em}(17)}$$

A. Block diagonal parameterization

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4 ETC design

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

The control input is ：

$$u(t)=K_1{\hat{x}}_s(t)+K_{2}p(C_q{\hat{x}}_s(t))\text{\hspace{1em}(71)}$$

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

$$y_s(t)=y(t_k^y)\forall t\in [t_k^y,t_{k+1}^y)$$

here ,$t_0^y=0$ And trigger the moment $t_1^y,t_2^y,\cdots$ Determined by the following trigger rules ：
$$t_{k+1}^y=\inf \{t|t\ge t_k^y+{\tau }_y,\Vert y_e(t)\Vert >{\sigma }_y\Vert y(t)\Vert +{ϵ}_y\}\text{\hspace{1em}(83)}$$

among $y_e(t)=y_s(t)-y(t)$ also ${\tau }_y,{\sigma }_y,{ϵ}_y$ They are all positive numbers .

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Combined with sampling information $y_s(t)$, The observer becomes
$$\{\begin{array}{rl}\dot{\hat{x}}=& A\hat{x}+Bu+Ep(\hat{q}+L_1(\hat{y}-y_s))+L_2(\hat{y}-y_s)\\ \hat{y}=& C\hat{x}+Du\\ \hat{q}=& C_q\hat{x}\end{array}\text{\hspace{1em}(84)}$$

among $L_1,L_2$ Design matrix .

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Observer based control input $u(t)$ Such as （71） Shown , among ${\hat{x}}_s(t)$ Update at the moment $t_1^u,t_2^u,\cdots$
$${\hat{x}}_s(t)=\hat{x}(t_k^u)\forall t\in [t_k^u,t_{k+1}^u)$$

here ,$t_0^u=0$ And trigger the moment $t_1^u,t_2^u,\cdots$ Determined by the following trigger rules ：
$$t_{k+1}^u=\inf \{t|t\ge t_k^u+{\tau }_u,\Vert {\hat{x}}_e(t)\Vert >{\sigma }_u\Vert \hat{x}(t)\Vert +{ϵ}_u\}\text{\hspace{1em}(85)}$$

among ${\hat{x}}_e(t)=\hat{x}(t_k)-\hat{x}(t)$ also ${\tau }_u,{\sigma }_u,{ϵ}_u$ They are all positive numbers . Be careful $\hat{x}$ and ${\hat{x}}_e$ The information can be obtained from the designed observer .

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5 Simulation example

$$\begin{array}{rl}{\dot{x}}_1=& x_2\\ {\dot{x}}_2=& -\sin (x_1)+u+w\\ y=& x_1\end{array}$$

The observer is designed as
$$\{\begin{array}{rl}\dot{\hat{x}}=& A\hat{x}+Bu+Ep(\hat{q}+L_1(\hat{y}-y))+L_2(\hat{y}-y)\\ \hat{y}=& C\hat{x}+Du\\ \hat{q}=& C_q\hat{x}\end{array}\text{\hspace{1em}(11)}$$

During simulation, the parameters are assumed to be $L_1=-1,L_2=[-5.1294,-18.0352{]}^{\text{T}},p(\hat{q})=\sin (\hat{q}),C_q=(1,0)$.

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Because the observations $\hat{y}$ Often obtained by sensors , So the assumption here is $\hat{y}=y$, So there is
$$\{\begin{array}{rl}\dot{\hat{x}}=& A\hat{x}+Bu+Ep(\hat{q})\\ \hat{q}=& C_q\hat{x}\end{array}$$

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The feedback control input is
$$u(t)=k(\hat{x})\text{\hspace{1em}(12)}$$

$$u(t)=K_1{\hat{x}}_s(t)+K_{2}p(C_q{\hat{x}}_s(t))\text{\hspace{1em}(71)}$$
The parameter is assumed to be $K_1=(-7.3936,-3.9937),K_2=1$.
and $w$ is uniformly generated from $[-w_0,w_0]$.

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Combined with the above analysis process , Observations can be solved in the following way $\hat{x}$
$$\{\begin{array}{rl}\hat{q}=& C_q\hat{x}\\ u=& K_1\hat{x}(t)+K_{2}p(\hat{q})\\ \dot{\hat{x}}=& A\hat{x}+Bu+Ep(\hat{q})\end{array}$$

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after , In the same way , Combine the formula （1） Find out the system state .

The error diagram is as follows , Used to compare... In the original text Fig.4.

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The next analysis ETC

$$y_s(t)=y(t_k^y)\forall t\in [t_k^y,t_{k+1}^y)$$

here ,$t_0^y=0$ And trigger the moment $t_1^y,t_2^y,\cdots$ Determined by the following trigger rules ：
$$t_{k+1}^y=\inf \{t|t\ge t_k^y+{\tau }_y,\Vert y_e(t)\Vert >{\sigma }_y\Vert y(t)\Vert +{ϵ}_y\}\text{\hspace{1em}(83)}$$

among $y_e(t)=y_s(t)-y(t)$ also ${\tau }_y,{\sigma }_y,{ϵ}_y$ They are all positive numbers .

$$\begin{array}{rl}k=0,t_1^y=& \inf \{t|t\ge 0+1.07\ast 1e-4,\Vert y_e(t)\Vert >0.0017\Vert y(t)\Vert +0.005\}\\ k=1,t_2^y=& \inf \{t|t\ge t_1^y+1.07\ast 1e-4,\Vert y_e(t)\Vert >0.0017\Vert y(t)\Vert +0.005\}\\ k=2,t_3^y=& \\ k=3,t_4^y=& \end{array}$$