Dual-mode global stabilization of high-order saturated integrator chains: LMI-based MPC combined with a nested saturated feedback

Abstract

This paper considers the problem of high-performance global stabilization of an integrator chain via a bounded control at the presence of input disturbance. While nested saturated feedback (NSF) is known as the most inspiring existing solution in the literature, we shall highlight the inherent shortcomings of this approach which cause a poor performance in terms of convergence rate. Then, a novel dual-mode control scheme combining an improved NSF law with a linear matrix inequality (LMI)-based model predictive controller (MPC) is developed to overcome the weaknesses of pure NSF. By offline calculations, a set of nested robust invariant attraction regions and their attributed feedback gains are computed respecting the known bound for disturbance. In online implementation, an improved NSF law is adopted (as the first mode) to bring states within the largest invariant set. The proposed LMI-based state feedback law is applied thereafter (as the second mode) to bring state vector toward the origin with a fast convergence rate. In both modes, the well-known extended state observer (ESO) is employed to effectively suppress the external disturbance. The superiority of the proposed globally stabilizing scheme in terms of faster convergence and improved disturbance rejection ability is demonstrated comparing performances of foremost existing solutions in case of a triple integrator chain.

Appendix A: Proof of Lemma 1

The induction method is followed. For $i=1$, computing derivative of $z_1={\tilde{x}}_1+\sigma _{0}^{'}(z_{0})$ and applying (13), leads to ${\dot{z}}_1=z_2-\sigma _{1}^{'}(z_{1})$. Also, for $i=2$ one can show

$$\begin{aligned} {\dot{z}}_2= & {} \dot{{\tilde{x}}}_2 + \frac{\partial \sigma _{1}^{'}}{\partial z_{1}}{\dot{z}}_1 =\underbrace{z_3- \sigma _{2}^{'}(z_2)}_{\dot{{\tilde{x}}}_2={\tilde{x}}_3}\nonumber \\&+\frac{\partial \sigma _{1}^{'}}{\partial z_{1}}(z_2-\sigma _1^{'}(z_1)). \end{aligned}$$

(44)

For $i=k$ assume ${\dot{z}}_k$ can be calculated as

$$\begin{aligned} {\dot{z}}_k= & {} z_{k+1}-\sigma _k^{'}(z_k)\nonumber \\&+\sum _{j=1}^{k-1}\prod _{r=1}^{j}\frac{\partial \sigma _{k-r}^{'}}{\partial z_{k-r}}\big (z_{k-j+1}-\sigma _{k-j}^{'}(z_{k-j})\big ), \end{aligned}$$

(45)

for $\forall k\le n-2$. Now, to complete induction procedure, it is required to prove

$$\begin{aligned} {\dot{z}}_{k+1}= & {} z_{k+2}-\sigma _{k+1}^{'}(z_{k+1})\nonumber \\&+\sum _{j=1}^{k}\prod _{r=1}^{j}\frac{\partial \sigma _{k+1-r}^{'}}{\partial z_{k+1-r}}\nonumber \\&\big (z_{k-j+2}-\sigma _{k-j+1}^{'}(z_{k-j+1})\big ). \end{aligned}$$

(46)

To that end, taking derivative from $z_{k+1}$ gives

$$\begin{aligned} {\dot{z}}_{k+1}= & {} \dot{{\tilde{x}}}_{k+1}+\frac{\partial \sigma _{k}^{'}}{\partial z_{k}}{\dot{z}}_k =\underbrace{z_{k+2}- \sigma _{k+1}^{'}(z_{k+1})}_{\dot{{\tilde{x}}}_{k+1}={\tilde{x}}_{k+2}}\nonumber \\&+\frac{\partial \sigma _{k}^{'}}{\partial z_{k}}{\dot{z}}_k, \end{aligned}$$

(47)

where incorporating (45) and (47) with some rearrangements results in (46) which completes the induction procedure showing validity of (16). As that last step, taking derivative from $z_{n}={\tilde{x}}_{n}+\sigma _{n-1}^{'}(z_{n-1})$ gives

$$\begin{aligned} {\dot{z}}_{n}= & {} \dot{{\tilde{x}}}_{n}+\frac{\partial \sigma _{n-1}^{'}}{\partial z_{n-1}}{\dot{z}}_{n-1}\nonumber \\= & {} \underbrace{\sigma _{u_m}(u)+{\tilde{d}}-{\tilde{y}}_d^{(n)}}_{\dot{{\tilde{x}}}_{n}} +\frac{\partial \sigma _{n-1}^{'}}{\partial z_{n-1}}{\dot{z}}_{n-1}. \end{aligned}$$

(48)

Substituting (16) for $i=n-1$ in (48), leads to (15).