Decentralized Fuzzy-Neural Identification and I-Term Adaptive Control of Distributed Parameter Bioprocess Plant

Abstract

The chapter proposed to use of a Recurrent Neural Network Model (RNNM) incorporated in a fuzzy-neural multi model for decentralized identification of an aerobic digestion process, carried out in a fixed bed and a recirculation tank anaerobic wastewater treatment system. The analytical model of the digestion bioprocess represented a distributed parameter system, which is reduced to a lumped system using the orthogonal collocation method, applied in four collocation points. The proposed decentralized RNNM consists of five independently working Recurrent Neural Networks (RNN), so to approximate the process dynamics in four different measurement points plus the recirculation tank. The RNN learning algorithm is the second order Levenberg-Marquardt one. The comparative graphical simulation results of the digestion wastewater treatment system approximation, obtained via decentralized RNN learning, exhibited a good convergence, and precise plant variables tracking. The identification results are used for I-term direct and indirect (sliding mode) control obtaining good results.

Appendices

Appendix 1: Detailed Derivation of the Recursive Levenberg-Marquardt Optimal Learning Algorithm for the RTNN

First of all we shall describe the optimal off-line learning method of Newton, then we shall modify it passing through the Gauss-Newton method and finally we shall simplify it so to obtain the off line Levenberg-Marquardt learning which finally will be transformed to recursive form (see [34] for more details).

The quadratic cost performance index under consideration is denoted by J _k (W), where W is the RTNN vector of weights with dimension N _w subject of iterative learning during the cost minimization. Let us assume that the performance index is an analytic function so all its derivatives exist.

Let us to expand J _k (W) around the optimal point of W(k) which yields:

$$\begin{aligned} J_{k} \left( {\text{W}} \right) \approx & \,J_{k} \left( {{\text{W}}\left( k \right)} \right) + \nabla J_{k}^{T} \left( {{\text{W}}\left( k \right)} \right)\left[ {{\text{W}} - {\text{W}}\left( k \right)} \right] \\ & \, + \frac{1}{2}\left[ {{\text{W}} - {\text{W}}\left( k \right)} \right]^{T} \nabla^{2} J_{k} \left( {{\text{W}}\left( k \right)} \right)\left[ {{\text{W}} - {\text{W}}\left( k \right)} \right] \\ \end{aligned}$$

(A.1)

where ∇J(W) is the gradient of J(W) with respect to the weight vector W:

$$\nabla J\left( {\text{W}} \right) = \left[ {\begin{array}{*{20}c} {\frac{\partial }{{\partial {\text{w}}_{1} }}J\left( {\text{W}} \right)} \\ {\frac{\partial }{{\partial {\text{w}}_{2} }}J\left( {\text{W}} \right)} \\ \vdots \\ {\frac{\partial }{{\partial {\text{w}}_{{N_{W} }} }}J\left( {\text{W}} \right)} \\ \end{array} } \right]$$

(A.2)

and ∇² J(W) is the Hessian matrix defined as:

$$\nabla^{2} J\left( {\text{W}} \right) = \left[ {\begin{array}{*{20}c} {\frac{{\partial^{2} }}{{\partial {\text{w}}_{1}^{2} }}J\left( {\text{W}} \right)} & {\frac{{\partial^{2} }}{{\partial {\text{w}}_{1} \partial {\text{w}}_{2} }}J\left( {\text{W}} \right)} & \cdots & {\frac{{\partial^{2} }}{{\partial {\text{w}}_{1} \partial {\text{w}}_{{N_{w} }} }}J\left( {\text{W}} \right)} \\ {\frac{{\partial^{2} }}{{\partial {\text{w}}_{2} \partial {\text{w}}_{1} }}J\left( {\text{W}} \right)} & {\frac{{\partial^{2} }}{{\partial {\text{w}}_{2}^{2} }}J\left( {\text{W}} \right)} & \cdots & {\frac{{\partial^{2} }}{{\partial {\text{w}}_{2} \partial {\text{w}}_{{N_{w} }} }}J\left( {\text{W}} \right)} \\ \vdots & \vdots & \ddots & \vdots \\ {\frac{{\partial^{2} }}{{\partial {\text{w}}_{{N_{w} }} \partial {\text{w}}_{1} }}J\left( {\text{W}} \right)} & {\frac{{\partial^{2} }}{{\partial {\text{w}}_{{N_{w} }} \partial {\text{w}}_{2} }}J\left( {\text{W}} \right)} & \cdots & {\frac{{\partial^{2} }}{{\partial {\text{w}}_{{N_{w} }}^{2} }}J\left( {\text{W}} \right)} \\ \end{array} } \right]$$

(A.3)

Taking the gradient of the Eq. (A.1) with respect to W and equating it to zero, we obtained:

$$\nabla J_{k} \left( {{\text{W}}\left( k \right)} \right) + \nabla^{2} J_{k} \left( {{\text{W}}\left( k \right)} \right)\left[ {{\text{W}} - {\text{W}}\left( k \right)} \right] = 0$$

(A.4)

Deriving (A.4) for W, we have:

$${\text{W}} = {\text{W}}\left( k \right) - \left( {\nabla^{2} J_{k} \left( {{\text{W}}\left( k \right)} \right)} \right)^{ - 1} \nabla J_{k} \left( {{\text{W}}\left( k \right)} \right)$$

(A.5)

Finally, we obtain the Newton’s learning algorithm as:

$${\text{W}}\left( {k + 1} \right) = {\text{W}}\left( k \right) - \left( {\nabla^{2} J_{k} \left( {{\text{W}}\left( k \right)} \right)} \right)^{ - 1} \nabla J_{k} \left( {{\text{W}}\left( k \right)} \right)$$

(A.6)

where W(k + 1) is the weight vector minimizing J _k (W) in the instant k; i.e. J _k (W)|_W=W(k+1) is min.

Let us suppose that W(k) is the weight vector that minimize J _k − 1(W) in the instant k–1, then:

$$\left. {\nabla J_{k - 1} \left( {\text{W}} \right)} \right|_{{{\text{W}} = {\text{W}}\left( k \right)}} = \nabla J_{k - 1} \left( {{\text{W}}\left( k \right)} \right) = 0$$

(A.7)

The performance index is defined as:

$$J_{k} \left( {\text{W}} \right) = \frac{1}{2}\sum\limits_{q = 1}^{k} {\alpha^{k - q} } {\text{E}}_{q}^{T} \left( {\text{W}} \right){\text{E}}_{q} \left( {\text{W}} \right)$$

(A.8)

$$J_{k} \left( {\text{W}} \right) = \frac{1}{2}\sum\limits_{q = 1}^{k} {\left( {\alpha^{k - q} \sum\limits_{j = 1}^{L} {e_{j,q}^{2} \left( {\text{W}} \right)} } \right)}$$

(A.9)

where: 0 < α ≤ 1 is a forgetting factor, q is the instant of the corresponding error vector, E _q represented the qth error vector, e _j,q is the jth element of E _q , k es is the final instant of the performance index. The ith element of the gradient is:

$$\left[ {\nabla J_{k} \left( {\text{W}} \right)} \right]_{i} \,=\, \frac{{\partial J_{k} \left( {\text{W}} \right)}}{{\partial {\text{w}}_{i} }} = \sum\limits_{q = 1}^{k} {\left( {\alpha^{k - q} \sum\limits_{j = 1}^{L} {e_{j,q} \left( {\text{W}} \right)\frac{{\partial e_{j,q} \left( {\text{W}} \right)}}{{\partial {\text{w}}_{i} }}} } \right)}$$

(A.10)

$$\begin{aligned} \left[ {\nabla J_{k} \left( {\text{W}} \right)} \right]_{i} \,=\, & \sum\limits_{q = 1}^{k} {\left( {\alpha^{k - q} \sum\limits_{j = 1}^{L} {e_{j,q} \left( {\text{W}} \right)\frac{{\partial \left( {r_{j,q} - y_{j,q} \left( {\text{W}} \right)} \right)}}{{\partial {\text{w}}_{i} }}} } \right)} \\ = & \, - \sum\limits_{q = 1}^{k} {\left( {\alpha^{k - q} \sum\limits_{j = 1}^{L} {e_{j,q} \left( {\text{W}} \right)\frac{{\partial y_{j,q} \left( {\text{W}} \right)}}{{\partial {\text{w}}_{i} }}} } \right)} \\ \end{aligned}$$

(A.11)

The matricial form of the performance index gradient is:

$$\nabla J_{k} \left( {\text{W}} \right) = - \sum\limits_{q = 1}^{k} {\alpha^{k - q} {\text{J}}_{{{\text{Y}}_{q} }}^{T} \left( {\text{W}} \right){\text{E}}_{q} \left( {\text{W}} \right)}$$

(A.12)

where the Jacobean matrix of Y _q in the instant q with dimension L × N _w is:

$${\text{J}}_{{{\text{Y}}_{q} }} \left( {\text{W}} \right) = \left[ {\begin{array}{*{20}c} {\frac{{\partial y_{1,q} }}{{\partial w_{1} }}} & {\frac{{\partial y_{1,q} }}{{\partial w_{2} }}} & \cdots & {\frac{{\partial y_{1,q} }}{{\partial w_{{N_{w} }} }}} \\ {\frac{{\partial y_{2,q} }}{{\partial w_{1} }}} & {\frac{{\partial y_{2,q} }}{{\partial w_{2} }}} & \cdots & {\frac{{\partial y_{2,q} }}{{\partial w_{{N_{w} }} }}} \\ \vdots & \vdots & \ddots & \vdots \\ {\frac{{\partial y_{L,q} }}{{\partial w_{1} }}} & {\frac{{\partial y_{L,q} }}{{\partial w_{2} }}} & \cdots & {\frac{{\partial y_{L,q} }}{{\partial w_{Nw} }}} \\ \end{array} } \right]$$

(A.13)

The gradient could be written in the following form:

$$\nabla J_{k} \left( {\text{W}} \right) = \alpha \nabla J_{k - 1} \left( {\text{W}} \right) - {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {\text{W}} \right){\text{E}}_{k} \left( {\text{W}} \right)$$

(A.14)

Then, the h, ith element of the Hessian matrix could be written as:

$$\begin{aligned} \left[ {\nabla^{2} J_{k} \left( {\text{W}} \right)} \right]_{h,i}\, =\, & \frac{{\partial^{2} J_{k} \left( {\text{W}} \right)}}{{\partial w_{h} \partial w_{i} }} \\ = & \sum\limits_{q = 1}^{k} {\alpha^{k - q} \sum\limits_{j = 1}^{L} {\left( {\frac{{\partial e_{j,q} \left( {\text{W}} \right)}}{{\partial w_{h} }}\frac{{\partial e_{j,q} \left( {\text{W}} \right)}}{{\partial w_{i} }} + e_{j,q} \left( {\text{W}} \right)\frac{{\partial^{2} e_{j,q} \left( {\text{W}} \right)}}{{\partial w_{h} \partial w_{i} }}} \right)} } \\ \end{aligned}$$

(A.15)

$$\left[ {\nabla^{2} J_{k} \left( {\text{W}} \right)} \right]_{h,i} \,=\, \sum\limits_{q = 1}^{k} {\alpha^{k - q} \sum\limits_{j = 1}^{L} {\left( {\frac{{\partial e_{j,q} \left( {\text{W}} \right)}}{{\partial w_{h} }}\frac{{\partial e_{j,q} \left( {\text{W}} \right)}}{{\partial w_{i} }} + e_{j,q} \left( {\text{W}} \right)\frac{{\partial^{2} e_{j,q} \left( {\text{W}} \right)}}{{\partial w_{h} \partial w_{i} }}} \right)} }$$

(A.16)

$$\nabla^{2} J_{k} \left( {\text{W}} \right) = \sum\limits_{q = 1}^{k} {\alpha^{k - q} \left( {{\text{J}}_{{{\text{Y}}_{q} }}^{T} \left( {\text{W}} \right){\text{J}}_{{{\text{Y}}_{q} }} \left( {\text{W}} \right) + \sum\limits_{j = 1}^{L} {e_{j,q} \left( {\text{W}} \right)\nabla^{2} e_{j,q} \left( {\text{W}} \right)} } \right)}$$

(A.17)

Equating:

$$\sum\limits_{j = 1}^{L} {e_{j,q} \left( {\text{W}} \right)\nabla^{2} e_{j,q} \left( {\text{W}} \right)} \approx 0$$

(A.18)

we could obtain directly the Gauss-Newton method of optimal learning.

The Eq. (A.17) is reduced to:

$$\nabla^{2} J_{k} \left( {\text{W}} \right) = \sum\limits_{q = 1}^{k} {\alpha^{k - q} {\text{J}}_{{{\text{Y}}_{q} }}^{T} \left( {\text{W}} \right){\text{J}}_{{{\text{Y}}_{q} }} \left( {\text{W}} \right)}$$

(A.19)

and it could also be written as:

$$\nabla^{2} J_{k} \left( {\text{W}} \right) = \alpha \nabla^{2} J_{k - 1} \left( {\text{W}} \right) + {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {\text{W}} \right){\text{J}}_{{{\text{Y}}_{k} }} \left( {\text{W}} \right)$$

(A.20)

Then (A.14) and (A.20) solved for W = W (k) are transformed to:

$$\nabla J_{k} \left( {{\text{W}}\left( k \right)} \right) = \alpha \nabla J_{k - 1} \left( {{\text{W}}\left( k \right)} \right) - {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {{\text{W}}\left( k \right)} \right){\text{E}}_{k} \left( {{\text{W}}\left( k \right)} \right)$$

(A.21)

$$\nabla^{2} J_{k} \left( {{\text{W}}\left( k \right)} \right) = \alpha \nabla^{2} J_{k - 1} \left( {{\text{W}}\left( k \right)} \right) + {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {{\text{W}}\left( k \right)} \right){\text{J}}_{{{\text{Y}}_{k} }} \left( {{\text{W}}\left( k \right)} \right)$$

(A.22)

According to (A.7), the Eq. (A.21) is reduced to:

$$\nabla J_{k} \left( {{\text{W}}\left( k \right)} \right) = - {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {{\text{W}}\left( k \right)} \right){\text{E}}_{k} \left( {{\text{W}}\left( k \right)} \right)$$

(A.23)

Let us define:

$${\text{H}}\left( k \right) = \nabla^{2} J_{k} \left( {{\text{W}}\left( k \right)} \right)$$

(A.24)

Then we could write the Eq. (A.22) in the following form:

$${\text{H}}\left( k \right) = \alpha {\text{H}}\left( {k - 1} \right) + {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {{\text{W}}\left( k \right)} \right){\text{J}}_{{{\text{Y}}_{k} }} \left( {{\text{W}}\left( k \right)} \right)$$

(A.25)

Finally for the learning algorithm (A.6), we could obtain:

$${\text{W}}\left( {k + 1} \right) = {\text{W}}\left( k \right) + \left( {\nabla^{2} J_{k} \left( {{\text{W}}\left( k \right)} \right)} \right)^{ - 1} {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {{\text{W}}\left( k \right)} \right){\text{E}}_{k} \left( {{\text{W}}\left( k \right)} \right)$$

(A.26)

$${\text{W}}\left( {k + 1} \right) = {\text{W}}\left( k \right) + {\text{H}}^{ - 1} \left( k \right){\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {{\text{W}}\left( k \right)} \right){\text{E}}_{k} \left( {{\text{W}}\left( k \right)} \right)$$

(A.27)

The Eq. (A.27) corresponds to the Gauss-Newton learning method where the considered Hessian matrix is an approximation to the real one.

Let us now to come back to Eq. (A.19).

$${\text{H}}\left( k \right) = \nabla^{2} J_{k} \left( {\text{W}} \right) = \sum\limits_{q = 1}^{k} {\alpha^{k - q} {\text{J}}_{{{\text{Y}}_{q} }}^{T} \left( {\text{W}} \right){\text{J}}_{{{\text{Y}}_{q} }} \left( {\text{W}} \right)}$$

(A.28)

Here we observe that the product J^T J could be nonsingular which require to perform the following modification of the Hessian matrix:

$${\text{H}}\left( k \right) = {\mathbf{\nabla }}^{2} J_{k} \left( {\text{W}} \right) = \sum\limits_{q = 1}^{k} {\alpha^{k - q} \left( {{\text{J}}_{{{\text{Y}}_{q} }}^{T} \left( {\text{W}} \right){\text{J}}_{{{\text{Y}}_{q} }} \left( {\text{W}} \right) + \rho {\text{I}}} \right)}$$

(A.29)

The Hessian matrix could be written also in the form:

$${\text{H}}\left( k \right) = \alpha {\text{H}}\left( {k - 1} \right) + {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {\text{W}} \right){\text{J}}_{{{\text{Y}}_{k} }} \left( {\text{W}} \right) + \rho {\text{I}}$$

(A.30)

where ρ is a small constant (generally ρ is chosen between 10⁻² and 10⁻⁴).

This modification of the Hessian matrix is essential for the optimal learning method of Levenberg-Marquardt. The computation of the Hessian matrix inverse could be done using the matrix inversion lemma which requires the following modification:

$${\text{H}}\left( k \right) = \alpha {\text{H}}\left( {k - 1} \right) + {\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {\text{W}} \right){\text{J}}_{{{\text{Y}}_{k} }} \left( {\text{W}} \right) + \rho {\text{I}}_{{N_{w} }}$$

(A.31)

where ${\text{I}}_{{N_{w} }}$ is a N _w  × N _w zero matrix except one element (with value 1) corresponding to position i = (k mod N _w ) +1. It could be seen that after N _w iterations, the Eq. (A.31) become equal to Eq. (A.30), i.e.:

$$\sum\limits_{n = k + 1}^{{k + N_{w} }} {\rho {\text{I}}_{{N_{w} }} } \left( n \right) = \rho {\text{I}}$$

(A.32)

Then the Eq. (A.31) is transformed to:

$${\text{H}}\left( k \right) = \alpha {\text{H}}\left( {k - 1} \right) +\Omega ^{T} \left( k \right)\Lambda ^{ - 1} \left( k \right)\Omega \left( k \right)$$

(A.33)

where:

$$\Omega \left( k \right) = \left[ {\begin{array}{*{20}c} {{\text{J}}_{{{\text{Y}}_{k} }} \left( {\text{W}} \right)} \\ {\begin{array}{*{20}c} 0 & \cdots & 0 & 1 & 0 & \cdots & 0 \\ \end{array} } \\ \end{array} } \right]$$

(A.34)

$$\Lambda \left( k \right)^{ - 1} = \left[ {\begin{array}{*{20}c} {\text{I}} & 0 \\ 0 & \rho \\ \end{array} } \right]$$

(A.35)

Then it is easy to apply the matrix inversion lemma, which constitute to the following equation (where the matrices A, B, C and D have compatible dimensions and the product BCD, and the sum A + BCD exists):

$$\left[ {{\text{A}} + {\text{BCD}}} \right]^{ - 1} = {\text{A}}^{ - 1} - {\text{A}}^{ - 1} {\text{B}}\left[ {{\text{DA}}^{ - 1} {\text{B}} + {\text{C}}^{ - 1} } \right]^{ - 1} {\text{DA}}^{ - 1}$$

(A.36)

Let us to apply the following substitutions:

$${\text{A}} = \alpha {\text{H}}\left( {k - 1} \right);{\text{B}} =\Omega ^{T} \left( k \right);{\text{C}} =\Lambda ^{ - 1} \left( k \right);{\text{D}} =\Omega \left( k \right)$$

The inverse of the Hessian matrix H(k) could be computed using the expression:

$$\begin{aligned} {\text{H}}^{ - 1} \left( k \right) = & \,\left[ {\alpha {\text{H}}\left( {k - 1} \right) +\Omega ^{T} \left( k \right)\Lambda ^{ - 1} \left( k \right)\Omega \left( k \right)} \right]^{ - 1} = \alpha^{ - 1} {\text{H}}^{ - 1} \left( {k - 1} \right) \\ & \, - \alpha^{ - 1} {\text{H}}^{ - 1} \left( {k - 1} \right)\Omega ^{T} \left( k \right)\left[ {\Omega \left( k \right)\alpha^{ - 1} {\text{H}}^{ - 1} \left( {k - 1} \right)\Omega ^{T} \left( k \right) +\Lambda \left( k \right)} \right]^{ - 1} \\ & \,\Omega \left( k \right)\alpha^{ - 1} {\text{H}}^{ - 1} \left( {k - 1} \right) \\ \end{aligned}$$

(A.37)

$$\begin{aligned} {\text{H}}^{ - 1} \left( k \right) = & \,\alpha^{ - 1} \left\{ {{\text{H}}^{ - 1} \left( {k - 1} \right)} \right. \\ & \, - {\text{H}}^{ - 1} \left( {k - 1} \right)\Omega ^{T} \left( k \right)\left[ {\Omega \left( k \right){\text{H}}^{ - 1} \left( {k - 1} \right)\Omega ^{T} \left( k \right) + \alpha\Lambda \left( k \right)} \right]^{ - 1} \\ & \,\Omega \left( k \right)\left. {{\text{H}}^{ - 1} \left( {k - 1} \right)} \right\} \\ \end{aligned}$$

(A.38)

Let us denote

$${\text{P}}\left( k \right) = {\text{H}}^{ - 1} \left( k \right)$$

and substitute it in the Eq. (A.38), we obtained:

$${\text{P}}\left( k \right) = \alpha^{ - 1} \left\{ {{\text{P}}\left( {k - 1} \right)} \right. - {\text{P}}\left( {k - 1} \right)\Omega ^{T} \left( k \right){\text{S}}^{ - 1} \left( k \right)\Omega \left( k \right)\left. {{\text{P}}\left( {k - 1} \right)} \right\}$$

(A.39)

where:

$${\text{S}}\left( k \right) = \alpha\Lambda \left( k \right) +\Omega \left( k \right){\text{P}}\left( {k - 1} \right)\Omega ^{T} \left( k \right)$$

(A.40)

Finally, the learning algorithm for w is obtained as:

$${\text{W}}\left( {k + 1} \right) = {\text{W}}\left( k \right) + {\text{P}}\left( k \right){\text{J}}_{{{\text{Y}}_{k} }}^{T} \left( {{\text{W}}\left( k \right)} \right){\text{E}}_{k} \left( {{\text{W}}\left( k \right)} \right)$$

(A.41)

where W is a N _w  × 1 vector formed of all RTNN weights (N _w  = L × N + N + N × M).

Using the RTNN topology the weight vector has the following form:

$${\text{W}}\left( k \right) = \left[ {\begin{array}{*{20}c} {c_{1,1} } & \cdots & {c_{L,N} } & {a_{1,1} } & {a_{2,2} } & \cdots & {a_{N,N} } & {b_{1,1} } & \cdots & {b_{N,M} } \\ \end{array} } \right]^{T}$$

(A.42)

and the Jacobean matrix with dimension L × N _w . is formed as:

$${\text{J}}_{{{\text{Y}}_{k} }} \left( {{\text{W}}\left( k \right)} \right) = \left[ {\begin{array}{*{20}c} {{\text{J}}_{{{\text{Y}}_{k} }} \left( {{\text{C}}\left( k \right)} \right)} & {{\text{J}}_{{{\text{Y}}_{k} }} \left( {{\text{A}}\left( k \right)} \right)} & {{\text{J}}_{{{\text{Y}}_{k} }} \left( {{\text{B}}\left( k \right)} \right)} \\ \end{array} } \right]$$

(A.43)

The components of the Jacobean matrix could be obtained applying the diagrammatic method [31]. Using the notation of part 2.2 for (A.43), we could write:

$${\text{DY}}\left[ {{\text{W}}\left( {\text{k}} \right)} \right] = \left[ {{\text{DY}}\left( {{\text{C}}_{\text{ij}} \left( {\text{k}} \right)} \right),{\text{DY}}\left( {{\text{A}}_{\text{ij}} \left( {\text{k}} \right)} \right),{\text{DY}}\left( {{\text{B}}_{\text{ij}} \left( {\text{k}} \right)} \right)} \right].$$

Appendix 2

Table A1 Abbreviations used in the chapter