Nonlinear quality-relevant process monitoring based on maximizing correlation neural network

Abstract

Quality-relevant fault detection aims to reveal whether quality variables are affected when a fault is detected. For current industrial processes, kernel-based methods focus on the nonlinearity within process variables, which is insufficient for obtaining nonlinearities of quality variables. Alternatively, neural network is an option for nonlinear prediction. However, these models are driven by predictive errors on samples. For quality-relevant tasks, the key is to capture the trends of quality variables. Therefore, this study proposes a new model, namely, maximizing correlation neural network (MCNN), to predict the quality-relevant information intuitively. The MCNN is trained to maximize the linear correlation between quality variables and the combinations of nonlinear representations mapped by a multilayer feedforward network. As such, fault detection can be implemented in the quality-relevant and irrelevant subspaces on the basis of the deep most correlated representations of process variables. Considering that different variables have different sensitivities to quality at various locations due to their nonlinear relationship, fault backpropagation is designed in the MCNN to isolate the faulty variables on the basis of real-time faulty information. Finally, numerical example and Tennessee Eastman process are used to evaluate the proposed method, which exhibits a competitive performance.

Appendix 1: Derivation of MCNN gradient

In order to perform the mini-batch back propagation algorithm to train MCNN, gradient of the objective $\frac{\partial \rho }{{\partial {{\varvec{\uptheta}}}_{k} }}$ is needed, where ${{\varvec{\uptheta}}}_{k}$ is one of the trainable parameters of MCNN. Based on the chain rule, we can get that

$$\frac{\partial \rho }{{\partial {{\varvec{\uptheta}}}_{k} }} = \sum\limits_{ij} {\frac{\partial \rho }{{\partial {\mathbf{F}}_{ij} }}\frac{{\partial {\mathbf{F}}_{ij} }}{{\partial {{\varvec{\uptheta}}}_{k} }}}$$

(23)

Since $\rho = \left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)^{ - 1/2} = \left[ {\left( {{\mathbf{S}}_{xx}^{ - 1/2} {\mathbf{S}}_{xy} } \right)^{T} \left( {{\mathbf{S}}_{xx}^{ - 1/2} {\mathbf{S}}_{xy} } \right)} \right]^{ - 1/2},$ we firstly show that

$$\frac{\partial \rho }{{\partial \left( {{\mathbf{S}}_{xy} } \right)_{a} }} = \left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)^{ - 1/2} \left( {{\mathbf{K}}^{T} {\mathbf{S}}_{xx}^{ - 1/2} } \right)_{a}$$

(24)

$$\frac{\partial \rho }{{\partial \left( {{\mathbf{S}}_{xx} } \right)_{ab} }} = - \frac{1}{2}\left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)^{ - 1/2} \left( {{\mathbf{S}}_{xy} {\mathbf{S}}_{xx}^{ - 1} } \right)_{a} \left( {{\mathbf{S}}_{xx}^{ - 1} {\mathbf{S}}_{xy} } \right)_{b}$$

(25)

Proof of (32) and (33) is illustrated as follows.

$$\begin{aligned} \frac{{\partial \rho }}{{\partial \left( {{\mathbf{S}}_{{xy}} } \right)_{a} }} & = \frac{{\partial \rho }}{{\partial \left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)}}\sum\limits_{b} {\frac{{\partial \left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)}}{{\partial \left( {\mathbf{K}} \right)_{b} }}\frac{{\partial \left( {\mathbf{K}} \right)_{b} }}{{\partial \left( {{\mathbf{S}}_{{xy}} } \right)_{a} }}} = \frac{1}{2}\left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)^{{ - 1/2}} \sum\limits_{b} {2\left( {\mathbf{K}} \right)_{b} \left( {{\mathbf{S}}_{{xx}}^{{ - 1/2}} } \right)_{{ba}} } , \\ & = \left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)^{{ - 1/2}} \left( {{\mathbf{K}}^{T} {\mathbf{S}}_{{xx}}^{{ - 1/2}} } \right)_{a} \\ \end{aligned}$$

(26)

$$\begin{gathered} \frac{\partial \rho }{{\partial \left( {{\mathbf{S}}_{xx} } \right)_{ab} }} = \frac{\partial \rho }{{\partial \left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)}}\frac{{\partial \left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)}}{{\partial \left( {{\mathbf{S}}_{xx} } \right)_{ab} }} = \frac{1}{2}\left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)^{ - 1/2} \sum\limits_{df} {\frac{{\partial \left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)}}{{\partial \left( {{\mathbf{S}}_{xx}^{ - 1} } \right)_{df} }}} \frac{{\partial \left( {{\mathbf{S}}_{xx}^{ - 1} } \right)_{df} }}{{\partial \left( {{\mathbf{S}}_{xx} } \right)_{ab} }} \hfill \\ \, = - \frac{1}{2}\left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)^{ - 1/2} \sum\limits_{df} {\left( {{\mathbf{S}}_{xy} {\mathbf{S}}_{xy}^{T} } \right)_{df} \left( {{\mathbf{S}}_{xx}^{ - 1} } \right)_{da} \left( {{\mathbf{S}}_{xx}^{ - 1} } \right)_{bf} } = - \frac{1}{2}\left( {{\mathbf{K}}^{T} {\mathbf{K}}} \right)^{ - 1/2} \left( {{\mathbf{S}}_{xy} {\mathbf{S}}_{xx}^{ - 1} } \right)_{a} \left( {{\mathbf{S}}_{xx}^{ - 1} {\mathbf{S}}_{xy} } \right)_{b} \hfill \\ \end{gathered}$$

(27)

Next, $\frac{{\partial \left( {S_{xx} } \right)_{ab} }}{{\partial {\mathbf{F}}_{ij} }}$ and $\frac{{\partial \left( {S_{xy} } \right)_{a} }}{{\partial {\mathbf{F}}_{ij} }}$ are computed as follows.

$$\frac{{\partial \left( {{\mathbf{S}}_{xx} } \right)_{ab} }}{{\partial {\mathbf{F}}_{ij} }} = \frac{{\partial \left( {{\mathbf{F}}^{T} {\mathbf{AF}}} \right)_{ab} }}{{\partial {\mathbf{F}}_{ij} }} = \left( {{\mathbf{F}}^{T} {\mathbf{AJ}}^{ij} + {\mathbf{J}}^{ji} {\mathbf{AF}}} \right)_{ab}$$

(28)

$$\frac{{\partial \left( {{\mathbf{S}}_{xy} } \right)_{ab} }}{{\partial {\mathbf{F}}_{ij} }} = \frac{{\partial \left( {{\mathbf{F}}^{T} {\mathbf{B}}} \right)_{ab} }}{{\partial {\mathbf{F}}_{ij} }} = \left( {{\mathbf{J}}^{ij} {\mathbf{B}}} \right)_{ab}$$

(29)

where ${\mathbf{J}}^{ij}$ is the single-entry matrix, 1 at $\left( {i,j} \right)$ and zero elsewhere, and ${\mathbf{A}} = \left[ {{\mathbf{I}} - \left( {1/n} \right){\mathbf{1}}_{n} {\mathbf{1}}_{n}^{T} } \right]^{T} \left[ {{\mathbf{I}} - \left( {1/n} \right){\mathbf{1}}_{n} {\mathbf{1}}_{n}^{T} } \right],\;{\mathbf{B}} = \left[ {{\mathbf{I}} - \left( {1/n} \right){\mathbf{1}}_{n} {\mathbf{1}}_{n}^{T} } \right]^{T} {\mathbf{y}}$

Since $\frac{{\partial {\mathbf{F}}_{ij} }}{{\partial {{\varvec{\uptheta}}}_{i} }}$ can be derived in a traditional way in traditional NNs, it is not repeated here and thus the gradient of MCNN is complete for the backpropagation training.