Non-Gaussian Process Monitoring and Fault Diagnosis

Abstract

Compared with PLS and PCA, independent component analysis (ICA) uses higher-order statistical information (above third-order) of the signal, so as to extract the non-Gaussian characteristic. In recent years, ICA was commonly used as a fault diagnosis method in the field of non-Gaussian process monitoring.

Appendix

Proposition 1 Assuming the kernel function is a radial basis, such that $k({\varvec{x}}_{i} ,{\varvec{x}}_{j} ) = \exp ( - \left\| {x_{i} - x_{j} } \right\|^{2} /{\text{c}})$, for a given sample ${\varvec{x}}^{b}$, if ${\varvec{x}}^{b} = {\varvec{0}}$, then $\Psi ({\varvec{x}}^{b} )$ is approximately zero.

Proof Assume that ${\varvec{X}}_{R} \in \Re^{{{\text{N}} \times {\text{m}}}}$ is the original training matrix containing ${\text{N}}$ samples, and ${\text{m}}$ is the number of variables contained in each sample. ${\varvec{x}}_{R}^{b} \in \Re^{{1 \times {\text{m}}}}$ is an original test sample. The centralized operations of ${\varvec{X}}_{R}$ and ${\varvec{x}}_{R}^{b}$ are performed as follows:

$$ {\varvec{X}} = {\varvec{X}}_{R} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{X}}_{R} , $$

(9.110)

$$ {\varvec{x}}^{b} = {\varvec{x}}_{R}^{b} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{X}}_{R} . $$

(9.111)

By performing the first-order Taylor series expansion, $k({\varvec{x}}_{i} ,{\varvec{x}}_{j} )$ can be expressed as:

$$ k({\varvec{x}}_{i} ,{\varvec{x}}_{j} ) = \sum\limits_{{{\text{n}} = 0}}^{1} {\frac{{\left[ { - \frac{{\left\| {{\varvec{x}}_{i} - {\varvec{x}}_{j} } \right\|^{2} }}{{\text{c}}}} \right]^{{\text{n}}} }}{{{\text{n}}!}}} + o(\Delta {\varvec{x}}^{2} ) \, = \, 1 - \frac{{\left\| {{\varvec{x}}_{i} - {\varvec{x}}_{j} } \right\|^{2} }}{{\text{c}}} + o(\Delta {\varvec{x}}^{2} ). $$

(9.112)

Let ${\text{dis}}({\varvec{x}}_{i} ,{\varvec{x}}_{j} )$ be the Euclidean distance between ${\varvec{x}}_{i}$ and ${\varvec{x}}_{j}$, that:

$$ {\text{dis}}({\varvec{x}}_{i} ,{\varvec{x}}_{j} ) = \left\| {{\varvec{x}}_{i} - {\varvec{x}}_{j} } \right\|^{2} = {\varvec{x}}_{i} {\varvec{x}}_{i}^{{\text{T}}} + {\varvec{x}}_{j} {\varvec{x}}_{j}^{{\text{T}}} - 2{\varvec{x}}_{i} {\varvec{x}}_{j}^{{\text{T}}} . $$

(9.113)

Then, the Euclidean distance matrix of ${\varvec{X}}_{R}$ can be written as:

$${\varvec{\varTheta}}={\varvec{\varGamma}}\,{\varvec{1}}_{{\text{N}}}^{{\text{T}}} + {\varvec{1}}_{{\text{N}}}{\varvec{\varGamma}}^{{\text{T}}} - 2{\varvec{X}}_{R} {\varvec{X}}_{R}^{{\text{T}}} , $$

(9.114)

where ${\varvec{\varTheta}}_{{{\text{i}},{\text{j}}}} = {\text{dis}}({\varvec{x}}_{i} ,{\varvec{x}}_{j} )$ and ${\varvec{\varGamma}}= [x_{1} x_{1}^{{\text{T}}} ,x_{2} x_{2}^{{\text{T}}} , \ldots ,x_{{\text{N}}} x_{{\text{N}}}^{{\text{T}}} ]^{{\text{T}}} \in \Re^{{{\text{N}} \times 1}}$. Define an operation factor ${\varvec{H}} = {\varvec{I}}_{{\text{N}}} - (1/N){\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}}$, and then we can obtain ${\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{H}}^{{\text{T}}} = ({\varvec{H1}}_{{\text{N}}}^{{}} )^{{\text{T}}} = ({\varvec{1}}_{{\text{N}}}^{{}} - (1/N){\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{1}}_{{\text{N}}} )^{{\text{T}}} = {\varvec{0}}$.

According to Eqs. (9.113), and (9.114), the relationship between the ${\varvec{K}}$ and ${\varvec{\varTheta}}$ can be expressed as:

$$ {\varvec{K}} \approx {\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} - \frac{1}{{\text{c}}}{\varvec{\varTheta}}$$

(9.115)

Then, ${\varvec{K}}$ can be mean centered as:

$$ \begin{aligned} {\varvec{K}}^{*} & = ({\varvec{I}}_{{\text{N}}} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} ){\varvec{K}}({\varvec{I}}_{{\text{N}}} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} )^{{\text{T}}} \approx {\varvec{H}}({\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} - \frac{1}{{\text{c}}}{\varvec{\varTheta}}){\varvec{H}}^{{\text{T}}} \\ & = {\varvec{H1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{H}}^{{\text{T}}} - \frac{1}{{\text{c}}}\user2{H\Theta H}^{{\text{T}}} = - \frac{1}{{\text{c}}}\user2{H\Theta H}^{{\text{T}}} \\ & = - \frac{1}{{\text{c}}}{\varvec{H}}(\user2{\Gamma 1}_{{\text{N}}}^{{\text{T}}} + {\varvec{1}}_{{\text{N}}}{\varvec{\varGamma}}^{{\text{T}}} - 2{\varvec{X}}_{R} {\varvec{X}}_{R}^{{\text{T}}} ){\varvec{H}}^{{\text{T}}} = \frac{2}{{\text{c}}}{\varvec{HX}}_{R} {\varvec{X}}_{R}^{{\text{T}}} {\varvec{H}}^{{\text{T}}} \\ & = \frac{2}{{\text{c}}}({\varvec{I}}_{N} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} ){\varvec{X}}_{R} {\varvec{X}}_{R}^{{\text{T}}} ({\varvec{I}}_{{\text{N}}} - \frac{1}{{\varvec{N}}}{\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} )^{{\text{T}}} \\ & = \frac{2}{{\text{c}}}{\varvec{XX}}_{{}}^{{\text{T}}} . \\ \end{aligned} $$

(9.116)

For a certain test sample ${\varvec{x}}_{{}}^{b}$, similar to Eqs. (9.114) and (9.115), its distance vector ${\varvec{\varTheta}}_{b}$ and kernel vector ${\varvec{k}}({\varvec{x}}^{b} )$ can be calculated as:

$${\varvec{\varTheta}}_{b} = {\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{x}}_{R}^{b} {\varvec{x}}_{R}^{b} \,^{{\text{T}}} +{\varvec{\varGamma}}^{{\text{T}}} - 2{\varvec{x}}_{R}^{b} {\varvec{X}}_{R}^{{\text{T}}} , $$

(9.117)

$$ {\varvec{k}}({\varvec{x}}_{{}}^{b} ) \approx {\varvec{1}}_{{\text{N}}}^{{\text{T}}} - \frac{1}{{\text{c}}}{\varvec{\varTheta}}_{b} . $$

(9.118)

Perform the centralization operation on ${\varvec{k}}({\varvec{x}}^{b} )$ as:

$$ \begin{aligned} {\varvec{k}}({\varvec{x}}_{{}}^{b} )^{*} & = ({\varvec{k}}({\varvec{x}}_{{}}^{b} ) - \frac{1}{N}{\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{K}}){\varvec{H}}^{{\text{T}}} \approx ({\varvec{1}}_{{\text{N}}}^{{\text{T}}} - \frac{1}{{\text{c}}}{\varvec{\varTheta}}_{b} - \frac{{1}}{{\text{N}}}{\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{K}}){\varvec{H}}^{{\text{T}}} \\ & \approx - \frac{1}{{\text{c}}}{\varvec{\varTheta}}_{b} {\varvec{H}}^{{\text{T}}} - \frac{{1}}{{\text{N}}}{\varvec{1}}_{{\text{N}}}^{{\text{T}}} ({\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} - \frac{1}{{\text{c}}}{\varvec{\varTheta}}){\varvec{H}}^{{\text{T}}} \\ & { = } - \frac{1}{{\text{c}}}\left[ {{\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{x}}_{R}^{b} {\varvec{x}}_{R}^{b} \,^{{\text{T}}} +{\varvec{\varGamma}}^{{\text{T}}} - 2{\varvec{x}}_{R}^{b} {\varvec{X}}_{R}^{{\text{T}}} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}}^{{\text{T}}} ({\varvec{\varGamma}}\,{\varvec{1}}_{{\text{N}}}^{{\text{T}}} + {\varvec{1}}_{{\text{N}}}{\varvec{\varGamma}}^{{\text{T}}} - 2{\varvec{X}}_{R} {\varvec{X}}_{R}^{{\text{T}}} )} \right]{\varvec{H}}^{{\text{T}}} \\ & { = }\,\frac{2}{{\text{c}}}\left( {x_{R}^{b} {\varvec{X}}_{R}^{{\text{T}}} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{X}}_{R} {\varvec{X}}_{R}^{{\text{T}}} } \right)\left( {{\varvec{I}}_{{\text{N}}} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} } \right)^{{\text{T}}} \\ & { = }\,\frac{2}{{\text{c}}}\left( {x_{R}^{b} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{X}}_{R} } \right)\left( {{\varvec{X}}_{R}^{{}} - \frac{1}{{\text{N}}}{\varvec{1}}_{{\text{N}}} {\varvec{1}}_{{\text{N}}}^{{\text{T}}} {\varvec{X}}_{R}^{{}} } \right)^{{\text{T}}} = \frac{2}{{\text{c}}}x_{{}}^{b} {\varvec{X}}_{{}}^{{\text{T}}} . \\ \end{aligned} $$

(9.119)

If ${\varvec{x}}_{{}}^{b} = {\varvec{0}}$, then ${\varvec{k}}({\varvec{x}}_{{}}^{b} )^{*} = {\varvec{0}}$. After scaling, i.e., $\overline{\user2{k}}({\varvec{x}}_{{}}^{b} ) = {\varvec{k}}({\varvec{x}}_{{}}^{b} )^{*} /[{\text{trace}}({\varvec{K}}^{*} )/{\text{N}}]$, it is obvious that $\overline{\user2{k}}({\varvec{x}}_{{}}^{b} ) = {\varvec{0}}$. From Eq. (9.99), the statistic of ${\varvec{x}}_{{}}^{b}$ can be calculated as:

$$ \Psi ({\varvec{x}}_{{}}^{b} ) = \overline{\user2{k}}({\varvec{x}}_{{}}^{b} ){\varvec{\varSigma}}\,\overline{\user2{k}}({\varvec{x}}_{{}}^{b} )^{{\text{T}}} . $$

(9.120)

Substituting $\overline{\user2{k}}({\varvec{x}}_{{}}^{b} ) = {\varvec{0}}$ into Eq. (50) gives $\Psi ({\varvec{0}})$ = 0.

Proposition 2 For a rank deficiency matrix ${\varvec{R}} \in \Re^{{{\text{d}} \times {\text{l}}}}$, if perform SVD on ${\varvec{RR}}^{T}$ as

$$ {\varvec{RR}}^{{\text{T}}} = \left[ {{\varvec{P}}_{r} \quad {\varvec{P}}_{u} } \right]\left[ \begin{gathered}{\varvec{\var**}}\quad {\varvec{0}} \hfill \\ {\varvec{0}}\quad \;{\varvec{0}} \hfill \\ \end{gathered} \right]\left[ \begin{gathered} {\varvec{P}}_{{\text{r}}}^{{\text{T}}} \hfill \\ {\varvec{P}}_{{\text{u}}}^{{\text{T}}} \hfill \\ \end{gathered} \right], $$

(9.121)

then, it holds the property ${\varvec{P}}_{u} \,^{{\text{T}}} {\varvec{R}} = {\varvec{0}}$.

Proof: According to the properties of SVD, for a given matrix ${\varvec{R}} \in \Re^{{{\text{d}} \times {\text{l}}}}$, if $\gamma = {\text{rank}}({\varvec{R}}) < \min ({\text{d}},{\text{l}})$, then there exist two unitary matrices ${\varvec{U}} \in \Re^{{{\text{d}} \times {\text{d}}}}$ and ${\varvec{V}} \in \Re^{{{\text{l}} \times {\text{l}}}}$ that make.

$$ {\varvec{R}} = {\varvec{U}}\left[ \begin{gathered} {\varvec{Q}}\quad {\varvec{0}} \hfill \\ {\varvec{0}}\quad {\varvec{0}} \hfill \\ \end{gathered} \right]{\varvec{V}}^{{\text{T}}} , $$

(9.122)

where ${\varvec{Q}} = {\text{diag}}(\sigma_{1} ,\sigma_{2} , \ldots ,\sigma_{\gamma } )$, and its diagonal elements are arranged in order $\sigma_{1} \ge \sigma_{2} \ge \ldots \ge \sigma_{\gamma } > 0$.

According to the properties of SVD, it holds that

$$ {\varvec{RR}}^{{\text{T}}} = {\varvec{U}}\left[ \begin{gathered} {\varvec{Q}}\quad \;{\varvec{0}} \hfill \\ {\varvec{0}}\quad \;\;{\varvec{0}} \hfill \\ \end{gathered} \right]{\varvec{V}}^{{\text{T}}} {\varvec{V}}\left[ \begin{gathered} {\varvec{Q}}\quad \;{\varvec{0}} \hfill \\ {\varvec{0}}\quad \;\;{\varvec{0}} \hfill \\ \end{gathered} \right]{\varvec{U}}^{{\text{T}}} = {\varvec{U}}\left[ \begin{gathered} {\varvec{Q}}^{2} \quad {\varvec{0}} \hfill \\ {\varvec{0}}\quad \;\;{\varvec{0}} \hfill \\ \end{gathered} \right]{\varvec{U}}^{{\text{T}}} . $$

(9.123)

That is, matrices ${\varvec{R}}$ and ${\varvec{RR}}^{{\text{T}}}$ have the same left singular matrix. Block ${\varvec{U}}$ into ${\varvec{U}} = [{\varvec{U}}_{1} \quad {\varvec{U}}_{2} ],$ ${\varvec{U}}_{1} \in \Re^{{{\text{d}} \times \gamma }} ,{\varvec{U}}_{2} \in \Re^{{{\text{d}} \times ({\text{d}} - \gamma )}}$, then Eqs. (9.122) and (9.123) can be rewritten as

$$ {\varvec{R}} = [{\varvec{U}}_{1} \quad {\varvec{U}}_{2} ]\left[ \begin{gathered} {\varvec{Q}}\quad \;{\varvec{0}} \hfill \\ {\varvec{0}}\quad \;\;{\varvec{0}} \hfill \\ \end{gathered} \right]\left[ \begin{gathered} {\varvec{V}}_{1}^{{\text{T}}} \hfill \\ {\varvec{V}}_{2}^{{\text{T}}} \hfill \\ \end{gathered} \right], $$

(9.124)

$$ {\varvec{RR}}^{T} = \left[ {\begin{array}{*{20}c} {{\varvec{U}}_{1} } & {{\varvec{U}}_{2} } \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {{\varvec{Q}}^{2} } & 0 \\ 0 & 0 \\ \end{array} } \right]\left[ {\begin{array}{*{20}c} {{\varvec{U}}_{1}^{T} } \\ {{\varvec{U}}_{2}^{T} } \\ \end{array} } \right]. $$

(9.125)

Perform left multiplication by matrix $\left[ \begin{gathered} {\varvec{U}}_{1}^{{\text{T}}} \hfill \\ {\varvec{U}}_{2}^{{\text{T}}} \hfill \\ \end{gathered} \right]$ on Eq. (9.124) as follows:

$$ \left[ \begin{gathered} {\varvec{U}}_{1}^{{\text{T}}} \hfill \\ {\varvec{U}}_{2}^{{\text{T}}} \hfill \\ \end{gathered} \right]{\varvec{R}} = \left[ \begin{gathered} {\varvec{Q}}\quad \;{\varvec{0}} \hfill \\ {\varvec{0}}\quad \;\;{\varvec{0}} \hfill \\ \end{gathered} \right]\left[ \begin{gathered} V_{1}^{{\text{T}}} \hfill \\ V_{2}^{{\text{T}}} \hfill \\ \end{gathered} \right]. $$

(9.126)

It is obvious that ${\varvec{U}}_{2}^{{\text{T}}} {\varvec{R}} = {\varvec{0}}$. Comparing Eqs. (9.121) and (9.125), then ${\varvec{U}}_{2} = {\varvec{P}}_{u}$. Thus ${\varvec{P}}_{u}^{{\text{T}}} {\varvec{R}} = {\varvec{0}}$.