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FCPN Approach for Uncertain Nonlinear Dynamical System with Unknown Disturbance

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Abstract

In this work, we have used a fuzzy counter-propagation network (FCPN) model to control different discrete-time, uncertain nonlinear dynamic systems with unknown disturbances. Fuzzy competitive learning (FCL) is used to process the weight connection and make adjustments between the instar and the outstar of the network. FCL paradigm adopts the principle of learning, used for calculation of the Best Matched Node (BMN) in the instar–outstar network. FCL provides a control of discrete-time uncertain nonlinear dynamic systems having dead zone and backlash. The errors like mean absolute error (MAE), mean square error (MSE), and best fit rate, etc. of FCPN are compared with networks like dynamic network (DN) and back propagation network (BPN). The FCL foretells that the proposed FCPN method gives better results than DN and BPN. The success and enactments of the proposed FCPN are validated through simulations on different discrete-time uncertain nonlinear dynamic systems and Mackey–Glass univariate time series data with unknown disturbances over BPN and DN.

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Acknowledgments

The author is gratefully acknowledged the financial assistance provided by the All India Council of Technical Education (AICTE) in the form of Research Promotion Scheme (RPS) project in 2012.

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Authors and Affiliations

  1. Madhav Institute of Technology & Science, Gwalior, 474005, India

    Vandana Sakhre, Uday Pratap Singh & Sanjeev Jain

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  1. Vandana Sakhre
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  2. Uday Pratap Singh
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  3. Sanjeev Jain
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Correspondence to Uday Pratap Singh.

Appendices

Appendix A: Dynamic Learning for CPN

Learning stability are fundamental issues for CPN, however there are few studies on the learning issue of FNN. BPN for learning is not always successful because of its sensitivity to learning parameters. Optimal learning rate always change during the training process. Dynamic learning of CPN is carried out using lemma 1, lemma 2 and lemma 3 [37, 38].

Assumption

Let us assume ϕ(x) is a sigmoidal function, if it is a bounded, continuous, and increasing function. Since input to the neural network in this model is bounded, we consider the lemma 1, and lemma 2 as given below:

Lemma 1

Let ϕ(x) be a sigmoid function and Ω be a compact set in \({\mathbb{R}}^{n} , \,{\text{and}} \quad f:{\mathbb{R}}^{n} \to {\mathbb{R}}\) on Ω is a continuous function and for arbitrary ∈ > 0, ∃ a integer N and real constants c i , θ i , w ij , i = 1, 2,…, N, j = 1, 2,…, n. Such that

$$\bar{f}\left( {x_{1} ,x_{2} , \ldots ,x_{n} } \right) = \mathop \sum \limits_{i = 1}^{N} c_{i} \emptyset \left( {\mathop \sum \limits_{j = 1}^{n} w_{ij} x_{i} - \theta_{i} } \right)$$
(72)

Satisfies

$$\mathop {\hbox{max} }\limits_{x \in \varOmega } ||f\left( {x_{1} ,x_{2} , \ldots ,x_{n} } \right) - \bar{f}\left( {x_{1} ,x_{2} , \ldots ,x_{n} } \right)|| < \in$$
(73)

Using lemma 1, dynamic learning for a three-layered CPN can be formulated where the hidden-layer transfer functions are ϕ(x) and the transfer function at the output layer is linear.

Let all the vectors be column vectors, and superscript \(d_{p}^{k}\) refer to specific output vectors component.

Let \(X = \left[ {x_{1} , x_{2} , \ldots , x_{p} } \right] \in {\mathbb{R}}^{L \times P}\), and \(Y = \left[ {y_{1} , y_{2} , \ldots , y_{p} } \right] \in {\mathbb{R}}^{H \times P} O = \left[ {o_{1} , o_{2} , \ldots , o_{p} } \right] \in {\mathbb{R}}^{K \times P}\) be the input-hidden, the output vectors, and \(D = \left[ {d_{1} , d_{2} , \ldots , d_{p} } \right] \in {\mathbb{R}}^{K \times P}\) be the desired vector, respectively, where L, H, and K denote the numbers of input, hidden, and output layer neurons. Let V and W represent the input-hidden and hidden-output layer weight matrices, respectively. The objectives of the network training to minimize an error function J where J is given by

$$J = \frac{1}{2PK}\mathop \sum \limits_{p = 1}^{P} \mathop \sum \limits_{k = 1}^{K} \left( {o_{p}^{k} - d_{p}^{k} } \right)^{2}$$
(74)

Appendix B: Optimal Learning of Dynamic System

While considering a three-layered CPN, the network error matrix is defined as the error between the differences of the desired and the FCPN outputs at any iteration, and is given as [37, 38]

$$E_{t} = O_{t} - D = W_{t} Y_{t} - D = W_{t} V_{t} X - D$$
(75)

The objective of the network to realize the minimization of error given in Eq. (31) is defined as follows:

$${\text{s}}.{\text{t}}.\,J = \frac{1}{2PK}T_{r} \left( {E_{t} E_{t}^{T} } \right)$$
(76)

where \(T_{r}\) represents the trace of matrix. Using gradient-descent method, the updated weight is given by

$$W_{t + 1} = W_{t} - \beta_{t} \frac{\partial J}{{\partial W_{t} }}\,{\text{and}}\, V_{t + 1} = V_{t} - \beta_{t} \frac{\partial J}{{\partial V_{t} }}$$

or

$$W_{t + 1} = W_{t} - \frac{{\beta_{t} }}{PK}E_{t} Y_{t}^{T} {\text{and}}\,V_{t + 1} = V_{t} - \frac{{\beta_{t} }}{PK}W_{t}^{T} E_{t} X^{T}$$
(77)

Using Eq. (22)–(24), we have

$$E_{t + 1} E_{t + 1}^{T} = \left( {W_{t + 1} Y_{t + 1} - D} \right)\left( {W_{t + 1} Y_{t + 1} - D} \right)^{T}$$
(78)

To obtain the minimum error for multilayer network, by simplification of above equation, we have

$$J_{t + 1} - J_{t} = \frac{1}{2PK}g(\beta )$$
(79)
$$J = \frac{1}{2PK}\mathop \sum \limits_{p = 1}^{P} \mathop \sum \limits_{k = 1}^{K} \left( {o_{p}^{k} - d_{p}^{k} } \right)^{2}$$
(80)

Using Eq. (31), we have

$$E_{t + 1} E_{t + 1}^{T} = \left( {W_{t + 1} V_{t + 1} X - D} \right)\left( {W_{t + 1} V_{t + 1} X - D} \right)^{T}$$

Solving for the above term, we get

$$\begin{aligned} & = \left[ {\left( {W_{t} - \frac{{\beta_{t} }}{PK}E_{t} Y_{t}^{T} } \right)\left( {V_{t} - \frac{{\beta_{t} }}{PK}W_{t}^{T} E_{t} X^{T} } \right)X - D} \right]\left[ {\left( {W_{t} - \frac{{\beta_{t} }}{PK}E_{t} Y_{t}^{T} } \right)\left( {V_{t} - \frac{{\beta_{t} }}{PK}W_{t}^{T} E_{t} X^{T} } \right)X - D} \right]^{\varvec{T}} \hfill \\ & = \left[ {E_{t} - \frac{{\beta_{t} }}{PK}\left( {W_{t} W_{t}^{T} E_{t} X^{T} X + E_{t} Y_{t}^{T} V_{t} X} \right) + \frac{{\beta_{t}^{2} }}{{\left( {PK} \right)^{2} }}E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X} \right] \hfill \\ & \left[ {E_{t} - \frac{{\beta_{t} }}{PK}\left( {W_{t} W_{t}^{T} E_{t} X^{T} X + E_{t} Y_{t}^{T} V_{t} X} \right) + \frac{{\beta_{t}^{2} }}{{\left( {PK} \right)^{2} }}\left( {E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X} \right)} \right]^{T} \hfill \\ & = E_{t} E_{t}^{T} - \frac{{\beta_{t} }}{PK}\left[ {E_{t} \left( {W_{t} W_{t}^{T} E_{t} X^{T} X} \right)^{T} + E_{t} \left( {E_{t} Y_{t}^{T} V_{t} X} \right)^{T} + \left( {W_{t} W_{t}^{T} E_{t} X^{T} XE_{t}^{T} } \right) + \left( {E_{t} Y_{t}^{T} V_{t} XE_{t}^{T} } \right)} \right] \hfill \\ & \quad + \frac{{\beta_{t}^{2} }}{{\left( {PK} \right)^{2} }}\left[ {E_{t} \left( {E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X} \right)^{T} + E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} XE_{t}^{T} + E_{t} Y_{t}^{T} V_{t} X\left( {W_{t} W_{t}^{T} E_{t} X^{T} X} \right)^{T} + W_{t} W_{t}^{T} E_{t} X^{T} X\left( {E_{t} Y_{t}^{T} V_{t} X} \right)^{T} + E_{t} Y_{t}^{T} V_{t} X\left( {E_{t} Y_{t}^{T} V_{t} X} \right)^{T} + W_{t} W_{t}^{T} E_{t} X^{T} X\left( {W_{t} W_{t}^{T} E_{t} X^{T} X} \right)^{T} } \right] \hfill \\ & \quad - \frac{{\beta_{t}^{3} }}{{\left( {PK} \right)^{3} }}\left[ {W_{t} W_{t}^{T} E_{t} X^{T} X\left( {E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X} \right) + E_{t} Y_{t}^{T} V_{t} X\left( {E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X} \right)^{T} + E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X\left( {W_{t} W_{f}^{T} E_{t} X^{T} X} \right)^{T} + E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X\left( {E_{t} Y_{t}^{T} V_{t} X} \right)^{T} } \right] \hfill \\ & \quad + \frac{{\beta_{t}^{4} }}{{\left( {PK} \right)^{4} }}\left[ {E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X\left( {E_{t} Y_{t}^{T} W_{t}^{T} E_{t} X^{T} X} \right)^{T} } \right] \hfill \\ \end{aligned}$$

For simplification, omit subscript t; then, we have, i.e.,

$$J_{t + 1} - J_{t} = \frac{1}{2PK}\left( {A\beta^{4} + B\beta^{4} + C\beta^{4} + M\beta } \right)$$
(81)

where

$$\begin{aligned} A = \frac{1}{{\left( {PK} \right)^{4} }}T_{r} \left[ {EY^{T} WEX^{T} X\left( {EY^{T} W^{T} EX^{T} X} \right)^{T} } \right] \hfill \\ B = \frac{1}{{\left( {PK} \right)^{3} }}T_{r} \left[ {WW^{T} EX^{T} X\left( {EY^{T} W^{T} EX^{T} X} \right)^{T} + EY^{T} VX\left( {EY^{T} W^{T} EX^{T} X} \right)^{T} + EY^{T} W^{T} EX^{T} X\left( {WW^{T} EX^{T} X} \right)^{T} + EY^{T} W^{T} EX^{T} X\left( {EY^{T} VX} \right)} \right] \hfill \\ C = \frac{1}{{\left( {PK} \right)^{2} }}T_{r} \left[ {\left( {EY^{T} W^{T} EX^{T} X} \right)^{T} + EY^{T} WEX^{T} XE^{T} + EY^{T} VX\left( {WW^{T} EX^{T} X} \right)^{T} + WW^{T} EX^{T} X + \left( {EY^{T} VX} \right)^{T} + EY^{T} VX\left( {EY^{T} VX} \right)^{T} + WW^{T} EX^{T} X\left( {WW^{T} EX^{T} X} \right)^{T} } \right] \hfill \\ M = \frac{1}{PK}T_{r} \left[ {E\left( {WW^{T} EX^{T} X} \right)^{T} + E\left( {EY^{T} VX} \right)^{T} + WW^{T} EX^{T} X + EY^{T} VXE^{T} } \right] \hfill \\ \end{aligned}$$
(82)

where

$$g\left( \beta \right) = \left( {A\beta^{4} + B\beta^{3} + C\beta^{2} + M\beta } \right)$$
(83)

Equation (35) is a polynomial of degree 4, and to obtain optimum value of \(\beta ,\) we have to solve

$$\frac{\partial g}{\partial \beta } = 0,\, {\text{i}} . {\text{e}} .\,\,4A\left( {\beta^{3} + a\beta^{2} + b\beta + c} \right) = 0$$
(84)

where \(a = \frac{3B}{4A} , b = \frac{2C}{4A} , c = \frac{M}{4A}\)

Lemma 2

For solution of general real cubic equation, we use the following Lemma:

$$f\left( x \right) = x^{3} + ax^{2} + bx + c, {\text{Let }}D = - 27c^{2} + 18c ab + a^{2} b^{2} - 4a^{3} b^{3} - 4b^{3}$$
(85)

where D is the discriminant of f(x).

Then

  1. 1.

    If D < 0, f(x) has one real root.

  2. 2.

    If D ≥ 0, f(x) has three real roots.

    1. (a)

      D > 0, f(x) has three different real roots.

    2. (b)

      D = 0, 6b – 2a 2 # 0, f(x) has one single root and one multiple root.

    3. (c)

      D = 0, 6b – 2a 2 = 0, f(x) has one root of three multiplicity.

Lemma 3

For a given polynomial g(β) given Eq. (40) if optimum \(\beta = \left\{ {\beta_{i} |g\left( {\beta_{i} } \right) = \hbox{min} \left( {g\left( {\beta_{1} } \right), g\left( {\beta_{2} } \right), g\left( {\beta_{3} } \right)} \right), i \in \left\{ {1, 2, 3} \right\}} \right\}\), where β i is the real root of \(\frac{\partial g}{\partial \beta },\) then the optimum (β) is the optimal learning rate, and this learning process is stable.

Proof

To find stable learning range of β, consider Lyapunov function

$$V_{t} = Jt^{2} \,{\text{and}} \;\Delta V_{t} = J_{t + 1}^{2} - J_{t}^{2} {\text{if}}\quad \Delta V_{t} < 0$$

and the dynamic system is guaranteed to be stable if \(\Delta V_{t} < 0 \;{\text{i}} . {\text{e}} .\,\;J_{t + 1} - J_{t} < 0\).

Since in the training process, input matrices remain the same during the whole training process, we have to find the range of β which satisfies \(\left( {A\beta^{4} + B\beta^{3} + C\beta^{2} + M\beta } \right) < 0\). Since \(\frac{\partial g}{\partial \beta }\) has at least one real root, one of these roots gives optimum β. Obviously, minimum value of g(β) gives the largest reduction in J t at each step of learning process. Equation (39) shows that g(β) has two or four real roots, one including β = 0. Thus, the minimum value of β shows largest reduction in error at two successive times, minimum value is obtained by differentiating Eq. (30) w.r.t. β, we have from theorem (1).

$$\frac{\partial g}{\partial \beta } = 4A\left( {\beta^{3} + a\beta^{2} + b\beta + c} \right)$$

where \(a = \frac{3B}{4A} , \,b = \frac{2C}{4A} , \,c = \frac{M}{4A}\)

Solving \(\frac{\partial g}{\partial \beta } = 0\), we obtain optimum β, which gives minimum error.

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Sakhre, V., Singh, U.P. & Jain, S. FCPN Approach for Uncertain Nonlinear Dynamical System with Unknown Disturbance. Int. J. Fuzzy Syst. 19, 452–469 (2017). https://doi.org/10.1007/s40815-016-0145-5

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