Optimal sensor placement for digital twin based on mutual information and correlation with multi-fidelity data

Abstract

This paper proposes an optimal sensor placement algorithm based on mutual information and correlation, which is particularly suitable for solving sensor placement problems in digital twins. Digital twins roughly comprise two data types: (1) the sensor data collected from a physical entity and (2) the simulation data collected from the physics-based simulation models. The proposed method takes advantage of the two data types to select sensor locations. Specifically, it determines sensor locations by maximizing the mutual information between the selected and uninstrumented candidate locations. In addition, the correlation between the sensor data and simulation data is characterized by a covariance matrix and used in the determination of mutual information. Moreover, the proposed method is also applicable in multiworking conditions by using clustering algorithms. To verify the effectiveness and applicability of the proposed sensor placement method, it was compared with two other sensor placement methods in terms of reconstruction accuracy through a digital twin case of a crane boom. The results show that the proposed method exhibits excellent performance.

Appendices

Appendix 1: Deduction of Eq. (7)

The detailed deduction, $I\left(X;Y\right)=H\left(X\right)-H(X|Y)$:

$$I\left(X;Y\right)=\iint p\left(x,y\right)\mathrm{log}\frac{p\left(x,y\right)}{p\left(y\right)p\left(x\right)}dxdy$$

$$=\iint p(y)p(x|y)\mathrm{log}p\left(x|y\right)dxdy-\iint p(x,y)\mathrm{log}p(x)dxdy$$

$$=\int p(y)\left[\int p(x|y)\mathrm{log}p\left(x|y\right)dx\right]dy-\int \mathrm{log}p(x)\left[\int p(x,y)dy\right]dx$$

(64)

$$=-\int p\left(y\right)H\left(X|Y=y\right)dy-\int p(x)\mathrm{log}p(x)dx$$

$$=-H\left(X|Y\right)+H(X)$$

$$=H\left(X\right)-H(X|Y)$$

Appendix 2: Entropy of the Gaussian

Let $X$ be a random variable that is normally distributed with mean $\mu$ and standard deviation $\sigma$:

$$X\sim \mathcal{N}\left(\mu ,{\sigma }^{2}\right).$$

(65)

Then, its entropy is

$$H\left(X\right)=-\int p\left(x\right)\mathrm{log}p\left(x\right)dx =-{\mathbb{E}}\left[\mathrm{log}p\left(x\right)\right]$$

$$=-{\mathbb{E}}\left[\mathrm{log}\mathcal{N}(\mu ,{\sigma }^{2})\right]$$

$$=-{\mathbb{E}}\left[\mathrm{log}\left[{(2\pi {\sigma }^{2})}^{-1/2}\mathrm{exp}\left(-\frac{1}{2{\sigma }^{2}}{(x-\mu )}^{2}\right)\right]\right]$$

(66)

$$=-{\mathbb{E}}\left[\mathrm{log}\left[{(2\pi {\sigma }^{2})}^{-1/2}\right]-\frac{1}{2{\sigma }^{2}}{(x-\mu )}^{2}\right]$$

$$=\frac{1}{2}log \left(2\pi {\sigma }^{2}\right)+\frac{1}{2{\sigma }^{2}}{\mathbb{E}}\left[{(x-\mu )}^{2}\right]$$

$$=\frac{1}{2}\mathrm{log}(2\pi {\sigma }^{2})+ \frac{1}{2}.$$

From Eq. (66), it can be seen that entropy is a function of the variance ${\sigma }^{2}$. This also supports the statement that entropy is a measure of uncertainty.

Appendix 3: Block matrix inversion

The block matrix inversion works as follows. For a nonsingular matrix

$$\mathbf{A}= \left[\begin{array}{cc}{\mathbf{P}}_{1}& {\mathbf{R}}_{1}\\ {\mathbf{R}}_{1}^{\mathrm{T}}& {\mathbf{Q}}_{1}\end{array}\right]$$

(67)

${\mathbf{A}}^{-1}$ can be expressed as.

$$\left[\begin{array}{cc}{\mathbf{P}}_{1}^{-1}+{\mathbf{P}}_{1}^{-1}{\mathbf{R}}_{1}{({\mathbf{Q}}_{1}-{\mathbf{R}}_{1}^{\mathrm{T}}{\mathbf{P}}_{1}^{-1}{\mathbf{R}}_{1})}^{-1}{\mathbf{R}}_{1}^{\mathrm{T}}{\mathbf{P}}_{1}^{-1}& -{\mathbf{P}}_{1}^{-1}{\mathbf{R}}_{1}{({\mathbf{Q}}_{1}-{\mathbf{R}}_{1}^{\mathrm{T}}{\mathbf{P}}_{1}^{-1}{\mathbf{R}}_{1})}^{-1}\\ -{({\mathbf{Q}}_{1}-{\mathbf{R}}_{1}^{\mathrm{T}}{\mathbf{P}}_{1}^{-1}{\mathbf{R}}_{1})}^{-1}{\mathbf{R}}_{1}^{\mathrm{T}}{\mathbf{P}}_{1}^{-1}& {({\mathbf{Q}}_{1}-{\mathbf{R}}_{1}^{\mathrm{T}}{\mathbf{P}}_{1}^{-1}{\mathbf{R}}_{1})}^{-1}\end{array}\right]$$

(68)

where, ${\mathbf{P}}_{1}$, ${\mathbf{R}}_{1}$, and ${\mathbf{Q}}_{1}$ are submatrices.

Appendix 4: Proof of Eq. (57)

Proposition: If symmetric matrix A is permutated using the same elementary row and column transformations to obtain matrix B, then the same sequence of elementary transformations can be used on ${\mathbf{A}}^{-1}$ to obtain ${\mathbf{B}}^{-1}$.

Proof: Performing the same elementary row and column transformations to matrix $\mathbf{A}$ means that matrix $\mathbf{A}$ right and left multiply by the same matrix $\mathbf{P}$. Let $\mathbf{B}=\mathbf{P}\mathbf{A}\mathbf{P}$, where $\mathbf{A}$ and B are two symmetric matrices, $\mathbf{P}$ is an elementary matrix. Then, ${\mathbf{B}}^{-1}={(\mathbf{P}\mathbf{A}\mathbf{P})}^{-1}={\mathbf{P}}^{-1}{\mathbf{A}}^{-1}{\mathbf{P}}^{-1}=\mathbf{P}{\mathbf{A}}^{-1}\mathbf{P}$.

$\mathbf{B}=\mathbf{P}\mathbf{A}\mathbf{P}$ denotes that matrix $\mathbf{B}$ is obtained by applying an elementary row and an elementary column transformation to $\mathbf{A}$. For instance, matrix $\mathbf{B}$ can be obtained by exchanging the $i$ th and $j$ th rows of $\mathbf{A}$ and then exchanging the $i$ th and $j$ th columns of $\mathbf{A}$. Next, ${\mathbf{B}}^{-1}=$ $\mathbf{P}{\mathbf{A}}^{-1}\mathbf{P}$ denotes that matrix ${\mathbf{B}}^{-1}$ can be obtained by applying the same elementary row and column transformations to ${\mathbf{A}}^{-1}$. That is, matrix ${\mathbf{B}}^{-1}$ can also be obtained by exchanging the $i$ th and $j$ th rows of ${\mathbf{A}}^{-1}$ and then exchanging the $i$ th and $j$ th columns of ${\mathbf{A}}^{-1}$.

In Sect. 3.2, matrix ${\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{(\eta )}$ is obtained from ${\mathbf{C}}_{\mathcal{U},\mathcal{U}}$ by exchanging the $\eta$ th row and $\left|\mathcal{U}\right|$ th row of matrix ${\mathbf{C}}_{\mathcal{U},\mathcal{U}}$ and then exchanging the $\eta$ th and $\left|\mathcal{U}\right|$ th columns of ${\mathbf{C}}_{\mathcal{U},\mathcal{U}}$. Hence, matrix ${\left({\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{(\eta )}\right)}^{-1}$ can also be obtained from $\left({\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{-1}\right)$ by the same manipulations applied to ${\mathbf{C}}_{\mathcal{U},\mathcal{U}}$. Therefore, the $(\left|\mathcal{U}\right|,\left|\mathcal{U}\right|)th$ element of the matrix ${\left({\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{(\eta )}\right)}^{-1}$ equals to the $(\eta ,\eta )th$ element of ${\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{-1}$, i.e., ${\left({\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{(\eta )}\right)}_{\left|\mathcal{U}\right|,\left|\mathcal{U}\right|}^{-1}= {\left( {\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{-1}\right)}_{\eta ,\eta }$, where ${\left(\bullet \right)}_{\left|\mathcal{U}\right|,\left|\mathcal{U}\right|}$ denotes the $(\left|\mathcal{U}\right|,\left|\mathcal{U}\right|)th$ element of the matrix ${\left({\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{(\eta )}\right)}^{-1}$ and ${\left({\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{-1}\right)}_{\eta ,\eta }$ denotes the $(\eta ,\eta )$ th element of ${\mathbf{C}}_{\mathcal{U},\mathcal{U}}^{-1}$.