A Controller Based on a Class of Affine T–S Fuzzy Models Using Piece-Wise Lyapunov Functions

Abstract

This paper investigates the problem of feedback control for a class of affine T–S fuzzy models using piece-wise Lyapunov functions. Although a large number of works on the issue have been published, several crucial problems still remain open. First, the paper shows what problems arise when using the affine T–S fuzzy model to design a controller, and in turn by employing the S-procedure, what kind of quadratic inequalities are required to help solve the resulting LMIs. It turns out that by partitioning the state space into certain cells based on the information of the antecedents of fuzzy rules, the required quadratic inequalities can be formularised. Taking advantage of the cell partition, a fuzzy controller is proposed using piece-wise Lyapunov functions, in which ensuing problems such as continuity functions used in the piece-wise Lyapunov functions and control input chattering also are addressed. Finally, examples are provided to illustrate the effectiveness of the proposed approach.

Appendix

1.1 Continuity Matrix

The way to construct the continuity matrices is based on the existing works works [33, 34]. Without loss of generality, let us consider there are two antecedent variables $x_1$ and $x_2$, and cell ${{{\mathcal {S}}}}_k$ is corresponding to the i-th partition on $x_1$ and j-th partition on $x_2$.

$$\begin{aligned} {{\bar{F}}}_k=\begin{bmatrix} \bar{\textbf{F}}_iC_1\\ \bar{\textbf{F}}_jC_2\\ I~~0 \end{bmatrix}\in R^{p\times 3}, \end{aligned}$$

(69)

where $k=(i-1)\times n_{x_2}+j$, $p=\sum _{i=1}^2\left( n_{x_i}+1\right) +n$,

$$\begin{aligned} C_1=\begin{bmatrix} 1&{} 0&{}0\\ 0 &{} 0 &{}1 \end{bmatrix},\qquad C_2=\begin{bmatrix} 0&{} 1&{}0\\ 0 &{} 0 &{}1 \end{bmatrix}, \end{aligned}$$

(70)

and the last row may be removed if the resulting ${{\bar{F}}}_k$ are of full column rank, and subsequently $p=\sum _{i=1}^2\left( n_{x_i}+1\right)$ in this case. In the following, constructing $\bar{\textbf{F}}_i=[ \textbf{F}_i~~ \textbf{f}_i]$ is given, while $\bar{\textbf{F}}_j$ can be obtained in the same manner.

Let $v=[v_1, \ldots , v_{n_{x_1}+1}]$ be corner points on $x_1$, which means there are $n_{x_1}$ partitions on $x_i$, and ${{{\mathcal {S}}}}_i=[v_i, v_{i+1}]$ ($i=1, \ldots , {n_{x_1}}$).

• Step 1: Let $\bar{\textbf{F}}_i$ be a $(n_{x_1}+1)$-by-2 zero matrix, and

  $$\begin{aligned} {{{\mathcal {E}}}}=\begin{bmatrix} v_i&{}v_{i+1}\\ 1&{}1 \end{bmatrix}; \end{aligned}$$

• Step 2: replace i-th and $(i+1)$-th rows of $\bar{\textbf{F}}_i$ by ${{{\mathcal {E}}}}^{-1}$.

However, $\textbf{f}_i$ in $\bar{\textbf{F}}_i$ cannot be guaranteed to be zero for ${{{\mathcal {S}}}}_i$ for $i\in {{{\mathcal {I}}}}_0$. Therefore, we modify $\bar{\textbf{F}}_i$ for $i\in {{{\mathcal {I}}}}_0$, and subsequently others related to the modification. Let $\bar{\textbf{F}}_i(j)$, and $\bar{\textbf{F}}_i(j,k)$ be the j-th row, and the element in row j, column k of $\bar{\textbf{F}}_i$, respectively.

• Step 1: Calculate:

  $$\begin{aligned} r&=\left( \bar{\textbf{F}}_{i-1}(i,1)\cdot v_i+\bar{\textbf{F}}_{i-1}(i, 2)\right) /v_i=1/v_i,\\ l&=\left( \bar{\textbf{F}}_{i+1}(i+1,1)\cdot v_{i+1}+\bar{\textbf{F}}_{i+1}(i+1, 2)\right) /v_{i+1}\\&=1/v_{i+1}; \end{aligned}$$

• Step 2: Update $\bar{\textbf{F}}_i$:

  $$\begin{aligned} \textbf{F}_i(i)=[r~~ 0], \quad \textbf{F}_i(i+1)=[l~~0]; \end{aligned}$$

• Step 3: Update $\bar{\textbf{F}}_1\sim \bar{\textbf{F}}_{i-1}$:

  $$\begin{aligned} \bar{\textbf{F}}_{j}(i+1)=[l~~0], \quad \text {for} j=1\sim i-1; \end{aligned}$$

• Step 4: Update $\bar{\textbf{F}}_{i+1}\sim \bar{\textbf{F}}_{n_{x_1}}$:

  $$\begin{aligned} \bar{\textbf{F}}_{j}(i)=[r~~ 0], \text { for} j=i+1\sim n_{x_1}. \end{aligned}$$