Saturation controller design for fractional-order interval type-2 fuzzy system with distributed time-varying delays

Abstract

This paper presents a novel approach for stabilizing fractional-order (FO) nonlinear systems with distributed time-varying delays and input constraints. It utilizes an interval type-2 fuzzy systems (IT2FS) to effectively manage uncertainties and complexities in the system. By combining various theories, a new methodology is proposed for designing a stabilization controller. Sufficient conditions guarantee the stability of the closed-loop system, addressing challenges of varying delays and controller saturation. The results are expressed as linear matrix inequalities (LMIs), facilitating implementation with the Matlab LMIs toolbox. This study contributes to advancing control theory by developing stabilization techniques for fractional nonlinear systems.

Appendix A

Proof of Lemma 5

Assume that the element at position (i, j) of matrix P is $p_{ij}(i,j=1,2,\ldots ,n)$, the ith component of the function $x(\theta )$ is $x_{i}(\theta )(i=1,2,\ldots ,n)$, then there

$$\begin{aligned}{} & {} \int _{t-d(t)}^tx^T(\theta )\textrm{d}\theta \times P \times \int _{t-d(t)}^tx(\theta )\textrm{d}\theta \\{} & {} \quad =\sum _{i,j=1}^{n}p_{ij}\int _{t-d(t)}^{t}x_{i}(\theta )\textrm{d}\theta \int _{t-d(t)}^{t}x_{j}(\theta )\textrm{d}\theta \\{} & {} \quad =\sum _{i,j=1}^{n}p_{ij}\int \int _{[t-d(t),t]\times [t-d(t),t]}x_{i}(\theta _1)x_{j}(\theta _2)\textrm{d}\theta _1\textrm{d}\theta _2\\{} & {} \quad =\int \int _{[t-d(t),t]\times [t-d(t),t]}x^T(\theta _1)Px(\theta _2)\textrm{d}\theta _1\textrm{d}\theta _2\\{} & {} \quad \le \varpi ^2 \sup _{t-\varpi \le \theta _1,\theta _2<t}x^T(\theta _1)Px(\theta _2). \end{aligned}$$

Just prove that at any moment, when $\theta _1=\theta _2=\theta$, $x^T(\theta _1)Px(\theta _2)$ and $x^T(\theta )Px(\theta )$ must have the same supremum.

In fact, since P is a positive definite matrix, we can get

$$\begin{aligned} (x(\theta _1)-x(\theta _2))^TP(x(\theta _1)-x(\theta _2))\ge 0. \end{aligned}$$

i.e.,

$$\begin{aligned} x^T(\theta _1)Px(\theta _2) \le \frac{x^T(\theta _1)Px(\theta _1)+x^T(\theta _2)Px(\theta _2)}{2}. \end{aligned}$$

The matrix P is positive definite symmetric, so

$$\begin{aligned}{} & {} \sup _{t-\varpi \le \theta _1,\theta _2\le t}\frac{x^T(\theta _1)Px(\theta _1)+x^T(\theta _2)Px(\theta _2)}{2} \\{} & {} \quad {=}\frac{\sup _{t-\varpi {\le } \theta _1{\le } t}x^T(\theta _1)Px(\theta _1)+\sup _{t{-}\varpi \le \theta _2{\le } t}x^T(\theta _2)Px(\theta _2)}{2}\\{} & {} \quad =\sup _{t-\varpi \le \theta \le t}x^T(\theta )Px(\theta )\nonumber \\ \end{aligned}$$

therefore

$$\begin{aligned} \sup _{t-\varpi \le \theta _1,\theta _2\le t}x^T(\theta _1)Px(\theta _2)\le \sup _{t-\varpi \le \theta \le t}x^T(\theta )Px(\theta ). \end{aligned}$$

So when $\theta _1=\theta _2=\theta$, The equal sign of the above equation is established. So the supremum of function $x^T(\theta _1)Px(\theta _2)$ must be obtained at $\theta _1=\theta _2$, so

$$\begin{aligned} \sup _{t-\varpi \le \theta _1,\theta _2\le t}x^T(\theta _1)Px(\theta _2)=\sup _{t-\varpi \le \theta \le t}x^T(\theta )Px(\theta ). \end{aligned}$$

The proof is complete. $\square$