Approximating Characteristics of Fuzzy Systems With Arbitrary Membership Function

Abstract

While quadrilateral membership function is used as general membership function, Stone-Weierstrass theorem is utilized to prove that fuzzy systems with arbitrary membership functions are capable of approximating any continuous function on a compact set. It is a generalization for theory of basic fuzzy systems approximating any continuous non-linear function. The performance of fuzzy systems to approximate random non-linear function is the theoretical basis of fuzzy system's application in system identification.

Content

《1 引言》

1 引言

模糊系统是模糊数学在自动控制中的应用。1965年Zadeh , 提出了模糊集合理论, 1974年英国学者Mamdani首先将模糊理论用于锅炉和蒸汽机的控制。20世纪80年代以来, 日本采用模糊控制技术的产品大量上市, 模糊技术在地铁、机器人、交通管理、市场预测等领域得到了广泛应用。当前, 模糊理论及其应用正向深度和广度进一步发展 ^[1,2], 成为智能控制理论的重要分支。北京师范大学的李洪兴教授利用变论域自适应模糊控制方法对四级倒立摆进行控制 ^[3], 成功地证明模糊控制是适用的、可靠的、精确的。

模糊系统的一个重要作用就是用于系统辩识, 从输入输出的角度, 模糊系统是一个非线性函数, 那么模糊系统能不能任意的逼近任意一个非线性函数呢?如果答案是肯定的, 那么就为模糊控制的可靠性、精确性提供了理论依据。文献[4]用基本模糊系统逼近任意连续非线性函数, 证明了单点模糊器、高斯隶属函数、乘积推理、加权平均解模糊的模糊系统对任意非线性连续函数的任意逼近性, 文献[5,6,7]证明了对于定义在紧集上的连续可微函数可用隶属函数为三角形的分层模糊系统来逼近, 文献[8]从矩阵的角度, 证明了具有任意形状隶属函数的分层模糊系统对紧集上连续函数的逼近性质。笔者证明, 对于任意形状隶属函数的模糊系统对连续的非线性函数具有任意的逼近性。

《2 预备知识》

2 预备知识

为了证明任意形状隶属函数的模糊系统对连续的非线性函数的逼近性, 首先给出相关的定义和定理。

《2.1 四边形隶属函数》

2.1 四边形隶属函数

定义1 四边形隶属函数令[a, d]⊂R, 模糊集A的四边形隶属函数是R上的一个连续函数, 其定义为

$μ_{A} (x; a, b, c, d, Η) = {\begin{cases} Ι (0), x \in [a, b) \\ Η, x \in [b, c] \\ D (x), x \in [c, d] \\ 0, x \in R - (a, d) \end{cases} (1)$ $μ_{A} (x; a, b, c, d, Η) = {\begin{cases} Ι (0), x \in [a, b) \\ Η, x \in [b, c] \\ D (x), x \in [c, d] \\ 0, x \in R - (a, d) \end{cases} (1)$

式中a≤b≤c≤ d, 0≤H≤1, 0≤I (x) ≤1是[a, b]上的一个非减函数, 0≤D (x) ≤1是 (c, d) 上的一个非增函数, 如图1, 当模糊集A为标准模糊集 (即H =1) 时, 其隶属函数可以简单地记为μ_A (x, a, b, c, d) 。

《图1》

图1 四边形隶属函数

Fig.1 Quadrilateral membership function

对于四边形隶属函数:若I (x) = (x-a) / (b-a) , D (x) = (x-d) / (c-d) , 则为梯形隶属函数;若满足上述条件下, 且b=c, 则为三角隶属函数;若 $a = + \infty, b = c = \bar{x} ‚ d = - \infty$ $a = + \infty, b = c = \bar{x} ‚ d = - \infty$ , 且 $Ι (x) = D (x) = \exp (- ((x - \bar{x}) / σ)^{2})$ $Ι (x) = D (x) = \exp (- ((x - \bar{x}) / σ)^{2})$ , 则为高斯隶属函数;若a=b=c=d, h=1, 则单点隶属函数, 所以, 可以把四边形隶属函数看成任意隶属函数。

《2.2 Stone-Weierstrass定理》

2.2 Stone-Weierstrass定理

定理1 Stone-Weierstrass定理若Z是一个定义在紧集U上的连续实函数的集合, 如果:

1) Z为一个代数, 即Z对加法、乘法和标量乘法是封闭的;

2) Z能离析U上的各点, 即对任意x, y∈U, 若x≠y, 则存在f∈Z, 使f (x) ≠f (y) ;

3) Z在U上任意一点不为零, 即对任意x∈U, 均存在f∈Z, 使f (x) ≠0;

则对U上的任意连续实函数g (x) 和任意ε>0, 均存在f∈Z, 使 $\sup_{x \in U} | f (x) - g (x) | < ε$ $\sup_{x \in U} | f (x) - g (x) | < ε$ 成立。

《3 模糊系统的逼近性能》

3 模糊系统的逼近性能

《3.1 模糊系统输出表达式》

3.1 模糊系统输出表达式

对于一个多输入单输出 (MISO) 的模糊系统, 设输入变量为x₁, x₂, …, x_n, 用矢量x=[x₁, x₂, …, x_n]^T表示, x的论域是实空间上的紧密集, 即x∈U∈Rⁿ;设系统的输出变量为y, y的论域是实数域上的紧密集, 即y∈V⊂R, 则模糊规则的一般形式为:

R^j: if x₁ is A $_{1}^{j}$ $_{1}^{j}$ and x₂ is A $_{2}^{j}$ $_{2}^{j}$ and… and x_n is A $_{n}^{j}$ $_{n}^{j}$ , then y is B^j

其中, j=1, 2, …, M, M为规则数;A $_{i}^{j}$ $_{i}^{j}$ 是x_i的模糊集合, 隶属函数为μ $_{A i}^{j}$ $_{A i}^{j}$ (x_i) , i=1, 2, …, n;B^j是y的模糊集合, 隶属函数为μ^j_B (y) 。

采用广义取式推理, t—范数选代数积算子, s—范数选最大算子, 得输出

$μ_{B^{'}} (y) = \max_{j = 1}^{Μ} [\sup_{x \in U} (μ_{A^{'}} (x) \prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}) μ_{B}^{j} (y))] (2)$ $μ_{B^{'}} (y) = \max_{j = 1}^{Μ} [\sup_{x \in U} (μ_{A^{'}} (x) \prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}) μ_{B}^{j} (y))] (2)$

若模糊集A′是模糊单点, 则式 (2) 简化为

$μ_{B^{'}} (y) = \max_{j = 1}^{Μ} [\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}) μ_{B}^{j} (y)] (3)$ $μ_{B^{'}} (y) = \max_{j = 1}^{Μ} [\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}) μ_{B}^{j} (y)] (3)$

若 $\bar{y}$ $\bar{y}$ ^j为第j个模糊集B^j的中心, 采用加权平均解模糊, 得模糊系统输出:

$y = f (x) = \frac{\sum_{j = 1}^{Μ} \bar{y}^{j} (\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}))}{\sum_{j = 1}^{Μ} (\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}))} (4)$ $y = f (x) = \frac{\sum_{j = 1}^{Μ} \bar{y}^{j} (\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}))}{\sum_{j = 1}^{Μ} (\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}))} (4)$

《3.2 定理2》

3.2 定理2

定理2 令Z是所有形如式 (4) 的模糊系统的集合, A $_{i}^{j}$ $_{i}^{j}$ 为四边形隶属函数, 对于紧密集U∈Rⁿ上的连续实函数g (x) , 给定ε>0, 必存在f∈Z, 使 $\sup_{x \in U} | f (x) - g (x) | < ε$ $\sup_{x \in U} | f (x) - g (x) | < ε$ 成立。

1) 证明Z是代数

令 f₁, f₂∈Z, c∈R, 写出f₁, f₂表达式:

$\begin{array}{l} f_{1} (x) = \frac{\sum_{j_{1} = 1}^{Μ_{1}} \bar{y}_{1}^{j} (\prod_{i = 1}^{n} μ_{A_{1} i}^{j} (x_{i}))}{\sum_{j_{1} = 1}^{Μ_{1}} (\prod_{i = 1}^{n} μ_{A_{1} i}^{j} (x_{i}))}, \\ f_{2} (x) = \frac{\sum_{j_{2} = 1}^{Μ_{2}} \bar{y}_{2}^{j} (\prod_{i = 1}^{n} μ_{A_{2} i}^{j} (x_{i}))}{\sum_{j_{2} = 1}^{Μ_{2}} (\prod_{i = 1}^{n} μ_{A_{2} i}^{j} (x_{i}))}, \\ 则 f_{1} (x) f_{2} (x) = \\ \sum_{j_{1} = 1}^{Μ_{1}} \sum_{j_{2} = 1}^{Μ_{2}} (\bar{y}_{1}^{j_{1}} \bar{y}_{2}^{j_{2}}) [\prod_{i = 1}^{n} μ_{A_{1} i}^{j_{1}} (x_{i}) μ_{A_{2} i}^{j_{2}} (x_{i})] / \\ \sum_{j_{1} = 1}^{Μ_{1}} \sum_{j_{2} = 1}^{Μ_{2}} [\prod_{i = 1}^{n} μ_{A_{1} i}^{j_{1}} (x_{i}) μ_{A_{2} i}^{j_{2}} (x_{i})] (5) \\ f_{1} (x) + f_{2} (x) = \sum_{j_{1} = 1}^{Μ_{1}} \sum_{j_{2} = 1}^{Μ_{2}} (\bar{y}_{1}^{j_{1}} + \\ \bar{y}_{2}^{j_{2}}) [\prod_{i = 1}^{n} μ_{A_{1} i}^{j_{1}} (x_{i}) μ_{A_{2} i}^{j_{2}} (x_{i})] / \\ \sum_{j_{1} = 1}^{Μ_{1}} \sum_{j_{2} = 1}^{Μ_{2}} [\prod_{i = 1}^{n} μ_{A_{1} i}^{j_{1}} (x_{i}) μ_{A_{2} i}^{j_{2}} (x_{i})] (6) \end{array}$ $\begin{array}{l} f_{1} (x) = \frac{\sum_{j_{1} = 1}^{Μ_{1}} \bar{y}_{1}^{j} (\prod_{i = 1}^{n} μ_{A_{1} i}^{j} (x_{i}))}{\sum_{j_{1} = 1}^{Μ_{1}} (\prod_{i = 1}^{n} μ_{A_{1} i}^{j} (x_{i}))}, \\ f_{2} (x) = \frac{\sum_{j_{2} = 1}^{Μ_{2}} \bar{y}_{2}^{j} (\prod_{i = 1}^{n} μ_{A_{2} i}^{j} (x_{i}))}{\sum_{j_{2} = 1}^{Μ_{2}} (\prod_{i = 1}^{n} μ_{A_{2} i}^{j} (x_{i}))}, \\ 则 f_{1} (x) f_{2} (x) = \\ \sum_{j_{1} = 1}^{Μ_{1}} \sum_{j_{2} = 1}^{Μ_{2}} (\bar{y}_{1}^{j_{1}} \bar{y}_{2}^{j_{2}}) [\prod_{i = 1}^{n} μ_{A_{1} i}^{j_{1}} (x_{i}) μ_{A_{2} i}^{j_{2}} (x_{i})] / \\ \sum_{j_{1} = 1}^{Μ_{1}} \sum_{j_{2} = 1}^{Μ_{2}} [\prod_{i = 1}^{n} μ_{A_{1} i}^{j_{1}} (x_{i}) μ_{A_{2} i}^{j_{2}} (x_{i})] (5) \\ f_{1} (x) + f_{2} (x) = \sum_{j_{1} = 1}^{Μ_{1}} \sum_{j_{2} = 1}^{Μ_{2}} (\bar{y}_{1}^{j_{1}} + \\ \bar{y}_{2}^{j_{2}}) [\prod_{i = 1}^{n} μ_{A_{1} i}^{j_{1}} (x_{i}) μ_{A_{2} i}^{j_{2}} (x_{i})] / \\ \sum_{j_{1} = 1}^{Μ_{1}} \sum_{j_{2} = 1}^{Μ_{2}} [\prod_{i = 1}^{n} μ_{A_{1} i}^{j_{1}} (x_{i}) μ_{A_{2} i}^{j_{2}} (x_{i})] (6) \end{array}$

$c f_{1} (x) = \frac{\sum_{j = 1}^{Μ_{1}} c \bar{y}_{1}^{j} (\prod_{i = 1}^{n} μ_{A_{1} i}^{j} (x_{i}))}{\sum_{j = 1}^{Μ_{1}} (\prod_{i = 1}^{n} μ_{A_{1} i}^{j} (x_{i}))} (7)$ $c f_{1} (x) = \frac{\sum_{j = 1}^{Μ_{1}} c \bar{y}_{1}^{j} (\prod_{i = 1}^{n} μ_{A_{1} i}^{j} (x_{i}))}{\sum_{j = 1}^{Μ_{1}} (\prod_{i = 1}^{n} μ_{A_{1} i}^{j} (x_{i}))} (7)$

因为μ^j₁_A₁i, μ^j₂_A₂i为式 (1) 所示的四边形隶属函数, μ^j₁_A₁i, μ^j₂_A₂i所构成的函数分别由 (0, I₁I₂, H₁I₂, D₁I₂, I₁H₂, H₁H₂, D₁H₂, I₁D₂, H₁D₂, D₁D₂) 组成, 如果把μ^j₁_A₁iμ^j₂_A₂i看成是隶属函数, 则必然也是一个四边形隶属函数, 图2为μ^j₁_A₁i, μ^j₂_A₂i构成的四边形隶属函数的各种形式, 把 $(\bar{y}_{1}^{j_{1}} \bar{y}_{2}^{j_{2}}) ‚ (\bar{y}_{1}^{j_{1}} + \bar{y}_{2}^{j_{2}}) ‚ c \bar{y}_{1}^{j_{1}}$ $(\bar{y}_{1}^{j_{1}} \bar{y}_{2}^{j_{2}}) ‚ (\bar{y}_{1}^{j_{1}} + \bar{y}_{2}^{j_{2}}) ‚ c \bar{y}_{1}^{j_{1}}$ 看作是模糊集合的中心, 则式 (5) 至式 (7) 仍是式 (4) 的形式, 即f₁ (x) , f₂ (x) ∈Z, f₁ (x) +f₂ (x) ∈Z, cf₁ (x) ∈Z, 所以Z是代数。

《图2》

图2μ^j₁_A₁i, μ^j₂_A₂i构成的四边形隶属函数

Fig.2 Quadrilateral membership function composed by μ^j₁_A₁i, μ^j₂_A₂i

2) 证明Z能离析U上的各点

对于式 (4) 的模糊系统, 给定任意的x⁰, z⁰∈U, 则有:

取 $Μ = 2, \bar{y}^{1} = 0, \bar{y}^{2} = 1$ $Μ = 2, \bar{y}^{1} = 0, \bar{y}^{2} = 1$ , 则

$\begin{array}{l} f (x^{0}) = \frac{\sum_{j = 1}^{Μ} \bar{y}^{j} (\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}^{0}))}{\sum_{j = 1}^{Μ} (\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}^{0}))} = \\ \frac{\prod_{i = 1}^{n} μ_{A i}^{2} (x_{i}^{0})}{\prod_{i = 1}^{n} μ_{A i}^{1} (x_{i}^{0}) + \prod_{i = 1}^{n} μ_{A i}^{2} (x_{i}^{0})} = \\ 1 / (1 + \prod_{i = 1}^{n} μ_{A i}^{1} (x_{i}^{0}) / \prod_{i = 1}^{n} μ_{A i}^{2} (x_{i}^{0})), \\ f (z^{0}) = \frac{\sum_{j = 1}^{Μ} \bar{y}^{j} (\prod_{i = 1}^{n} μ_{A i}^{j} (z_{i}^{0}))}{\sum_{j = 1}^{Μ} (\prod_{i = 1}^{n} μ_{A i}^{j} (z_{i}^{0}))} = \\ \frac{\prod_{i = 1}^{n} μ_{A i}^{2} (z_{i}^{0})}{\prod_{i = 1}^{n} μ_{A i}^{1} (z_{i}^{0}) + \prod_{i = 1}^{n} μ_{A i}^{2} (z_{i}^{0})} = \\ 1 / (1 + \prod_{i = 1}^{n} μ_{A i}^{1} (z_{i}^{0}) / \prod_{i = 1}^{n} μ_{A i}^{2} (z_{i}^{0})), \end{array}$ $\begin{array}{l} f (x^{0}) = \frac{\sum_{j = 1}^{Μ} \bar{y}^{j} (\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}^{0}))}{\sum_{j = 1}^{Μ} (\prod_{i = 1}^{n} μ_{A i}^{j} (x_{i}^{0}))} = \\ \frac{\prod_{i = 1}^{n} μ_{A i}^{2} (x_{i}^{0})}{\prod_{i = 1}^{n} μ_{A i}^{1} (x_{i}^{0}) + \prod_{i = 1}^{n} μ_{A i}^{2} (x_{i}^{0})} = \\ 1 / (1 + \prod_{i = 1}^{n} μ_{A i}^{1} (x_{i}^{0}) / \prod_{i = 1}^{n} μ_{A i}^{2} (x_{i}^{0})), \\ f (z^{0}) = \frac{\sum_{j = 1}^{Μ} \bar{y}^{j} (\prod_{i = 1}^{n} μ_{A i}^{j} (z_{i}^{0}))}{\sum_{j = 1}^{Μ} (\prod_{i = 1}^{n} μ_{A i}^{j} (z_{i}^{0}))} = \\ \frac{\prod_{i = 1}^{n} μ_{A i}^{2} (z_{i}^{0})}{\prod_{i = 1}^{n} μ_{A i}^{1} (z_{i}^{0}) + \prod_{i = 1}^{n} μ_{A i}^{2} (z_{i}^{0})} = \\ 1 / (1 + \prod_{i = 1}^{n} μ_{A i}^{1} (z_{i}^{0}) / \prod_{i = 1}^{n} μ_{A i}^{2} (z_{i}^{0})), \end{array}$

因为μ $_{A i}^{1}$ $_{A i}^{1}$ , μ $_{A i}^{2}$ $_{A i}^{2}$ 均为四边形隶属函数, 由0, I, H, D组成, 隶属函数各边没有重合, 所以当x⁰≠z⁰时,

$\frac{\prod_{i = 1}^{n} μ_{A i}^{1} (x_{i}^{0})}{\prod_{i = 1}^{n} μ_{A i}^{2} (x_{i}^{0})} \neq \frac{\prod_{i = 1}^{n} μ_{A i}^{1} (z_{i}^{0})}{\prod_{i = 1}^{n} μ_{A i}^{2} (z_{i}^{0})},$ $\frac{\prod_{i = 1}^{n} μ_{A i}^{1} (x_{i}^{0})}{\prod_{i = 1}^{n} μ_{A i}^{2} (x_{i}^{0})} \neq \frac{\prod_{i = 1}^{n} μ_{A i}^{1} (z_{i}^{0})}{\prod_{i = 1}^{n} μ_{A i}^{2} (z_{i}^{0})},$

则f (x⁰) ≠f (z⁰) , 因此Z能离析U上的点。

3) 证明Z在U上任意一点不为零

对于任意x⁰∈U, 因为有 $\bar{y}^{j} > 0 (j = 1, 2, \dots, n)$ $\bar{y}^{j} > 0 (j = 1, 2, \dots, n)$ 由式 (4) 得f (x⁰) ≠0。

归纳以上证明, 由Stone-Weierstrass定理可知:f (x) 对于定义在紧集U上的任意连续实函数g (x) 和任意ε>0, 均存在 $\sup_{x \in U} | f (x) - g (x) | < ε$ $\sup_{x \in U} | f (x) - g (x) | < ε$ , 即模糊系统能够任意逼近紧密集U上的任意连续实函数。

《4 实例》

4 实例

对于模糊系统f (x) , 使之逼近定义在U=[-3, 3]上的连续函数g (x) =sin (x) , 精度ε=0.2, 即 $\sup_{x \in U} | f (x) - g (x) | < ε$ $\sup_{x \in U} | f (x) - g (x) | < ε$ 。因为

$\begin{array}{l} \sup_{x \in U} | f (x) - g (x) | \leq \sup_{x \in u} | \frac{\partial g}{\partial x} | Δ x (8) \\ \sup_{x \in u} | \frac{\partial g}{\partial x} | = \sup_{x \in U} | \cos x | = 1, \end{array}$ $\begin{array}{l} \sup_{x \in U} | f (x) - g (x) | \leq \sup_{x \in u} | \frac{\partial g}{\partial x} | Δ x (8) \\ \sup_{x \in u} | \frac{\partial g}{\partial x} | = \sup_{x \in U} | \cos x | = 1, \end{array}$

所以Δx=0.2的模糊系统是满足要求的。在U=[-3, 3]上定义如图3示的隶属度函数的模糊集Aⁱ, 则模糊系统

$f (x) = \frac{\sum_{i = 0}^{30} s i n (- 3 + 0.2 i) μ_{A^{i}} (x)}{\sum_{i = 0}^{30} μ_{A^{i}} (x)} 。$ $f (x) = \frac{\sum_{i = 0}^{30} s i n (- 3 + 0.2 i) μ_{A^{i}} (x)}{\sum_{i = 0}^{30} μ_{A^{i}} (x)} 。$

由图4可见, f (x) 与g (x) =sin (x) 基本一致。

《图3》

图3 实例中的隶属函数

Fig.3 Membership function in the instance

《图4》

图4 模糊系统f (x) 及g (x) =sin x

Fig.4 Fuzzy system f (x) and g (x) =sin x

《5 结论》

5 结论

在基本模糊系统对任意非线性连续函数可任意逼近的基础上, 将模糊系统隶属函数一般化, 用四边形隶属函数代替一般的隶属函数, 对模糊系统逼近性能进行进一步探讨, 是对文献[4]理论的推广, 任意隶属函数模糊系统对任意连续实函数逼近性能的证明, 为一般模糊系统用于非线性系统建模奠定了理论基础, 模糊系统是除多项式函数、神经网络逼近器之外的一个新的函数逼近器, 模糊系统较其他逼近器的优势在于它能够系统而有效地利用语言的能力, 因而被越来越多地应用于非线性系统的辩识和控制之中。

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