A Novel Approach for Target Detection and Classification Using Canonical Correlation Analysis

Abstract

We present a novel detection approach, detection with canonical correlation (DCC), for target detection without prior information on the interference. We use the maximum canonical correlations between the target set and the observation data set as the detection statistic, and the coefficients of the canonical vector are used to determine the indices of components from a given target library, thus enabling both detection and classification of the target components that might be present in the mixture. We derive an approximate distribution of the maximum canonical correlation when targets are present. For applications where the contributions of components are non-negative, non-negativity constraints are incorporated into the canonical correlation analysis framework and a recursive algorithm is derived to obtain the solution. We demonstrate the effectiveness of DCC and its nonnegative variant by applying them on detection of surface-deposited chemical agents in Raman spectroscopy.

Appendix: Distribution of the Multiple Correlation

In this appendix, we evaluate the distribution of the square of multiple correlation, which can be written as [13]:

$$ R^2=\frac{\mathbf{y}^T\mathbf{T}\left(\mathbf{T}^T\mathbf{T}\right)^{-1}\mathbf{T}^T\mathbf{y}}{\mathbf{y}^T\mathbf{y}}\triangleq\alpha $$

(17)

where $\mathbf{y}\sim\mathcal{N}\left(\mathbf{t}_1,\sigma^2\mathbf{I}\right)$ and T is an any given matrix, such that T = [t ₁, ⋯ ,t _L ].

We rotate the coordinates of y so that the last L coordinates are in the subspace spanned by the columns of T. Given the singular value decomposition of $\mathbf{T}^T\mathbf{T}=\mathbf{Q}{\boldsymbol{\Lambda}}\mathbf{Q}^T$, we define $\mathbf{F}=\mathbf{Q}{\boldsymbol\Lambda}^{-\frac{1}{2}}$ so that H ^− 1 = FF ^T. We construct an N × N orthonormal matrix P = [P ₁  P ₂], where P ₂ = TF, and u = P ^T y. We can see that

$$ \begin{array}{rll} E\left\{\mathbf{u}\right\}&=&\mathbf{P}^T E\left\{\mathbf{y}\right\}=\mathbf{P}^T\mathbf{t}_1 \\ &=&\mathbf{P}^T\mathbf{T}\mathbf{e}_1 \\ &=&\left[\begin{array}{l} \mathbf{P}^T_1 \\ \mathbf{P}^T_2 \end{array} \right]\mathbf{P}_2\mathbf{F}^{-1}\mathbf{e}_1 \\ &=&\left(\begin{array}{l} \mathbf{0} \\ \mathbf{f} \end{array} \right) \end{array} $$

where $\mathbf{e}_1=[1, 0, \cdots, 0]^T$, and $\mathbf{f} \triangleq \mathbf{F}^{-1}\mathbf{e}_1$. Since P is orthonormal, the covariance matrix of u is the same with y, i.e., σ ² I, thus, we have

$$ \mathbf{u}=\left(\begin{array}{c} \mathbf{u}^{(1)} \\ \mathbf{u}^{(2)} \end{array} \right), $$

where

$$ \begin{array}{rll} \mathbf{u}^{(1)} &=& \left(\begin{array}{c} u_1 \\ \vdots \\ u_{N-L} \\ \end{array} \right) \sim\mathcal{N}\left(\mathbf{0},\sigma^2\mathbf{I}\right),~{\rm and}\\ \mathbf{u}^{(2)} &=& \left(\begin{array}{c} u_{N-L+1} \\ \vdots \\ u_{N} \\ \end{array} \right) \sim\mathcal{N}\left(\mathbf{f},\sigma^2\mathbf{I}\right). \end{array} $$

Let

$$ \begin{array}{rll} \mathbf{G}&=&\left(\mathbf{T}^T\mathbf{T}\right)^{-1}\mathbf{T}^T\mathbf{y} \\ &=&\mathbf{FF}^T\mathbf{F}^{-T}\mathbf{P}_2^T\mathbf{Pu} \\ &=&\mathbf{F}\left[ \mathbf{0}~ \mathbf{I} \right]\mathbf{u}\\ &=&\mathbf{Fu}^{(2)}, \end{array} $$

then

$$ \begin{array}{rll} \mathbf{y}^T\mathbf{T}\left(\mathbf{T}^T\mathbf{T}\right)^{-1}\mathbf{T}^T\mathbf{y}&=&\mathbf{G}^T\mathbf{T}^T\mathbf{TG} \\ &=&\left(\mathbf{u}^{(2)}\right)^T\mathbf{F}^T\mathbf{T}^T\mathbf{TFu}^{(2)} \\ &=&\left(\mathbf{u}^{(2)}\right)^T\mathbf{u}^{(2)}=\sum_{i=N-L+1}^{N}u_i^2.\end{array} $$

(18)

Considering Eqs. 17, 18 and u ^T u = y ^T PP ^T y = y ^T y, we can write

$$ \begin{array}{rll} \frac{\alpha}{1-\alpha}&=&\frac{\mathbf{y}^T\mathbf{T}\left(\mathbf{T}^T\mathbf{T}\right)^{-1}\mathbf{T}^T\mathbf{y}} {\mathbf{y}^T\mathbf{y}-\mathbf{y}^T\mathbf{T}\left(\mathbf{T}^T\mathbf{T}\right)^{-1}\mathbf{T}^T\mathbf{y}} \\ &=&\frac{\sum_{i=N-L+1}^{N}u_i^2}{\sum_{i=1}^{N}u_i^2-\sum_{i=N-L+1}^{N}u_i^2} \\ &=&\frac{\sum_{i=N-L+1}^{N}u_i^2/\sigma^2}{\sum_{i=1}^{N-L}u_i^2/\sigma^2}. \end{array} $$

(19)

The numerator in Eq. 19 follows a non-central χ ²-distribution with L degrees of freedom, and the denominator follows a χ ²-distribution with N − L degrees of freedom. The numerator and denominator are independent since u _i ’s are independently distributed.

We define

$$ h\triangleq \frac{N-L}{L}\cdot\frac{\alpha}{1-\alpha}=\frac{\frac{1}{L}\sum_{i=N-L+1}^{N}u_i^2/\sigma^2}{\frac{1}{N-L}\sum_{i=1}^{N-L}u_i^2/\sigma^2}, $$

which has a non-central F-distribution with L and N − L degrees of freedom and a non-centrality parameter given by

$$\lambda=\sum\limits_{i=N-L+1}^{N}\frac{E\{u_i\}^2}{\sigma^2}=\frac{\|\mathbf{f}\|^2}{\sigma^2}. $$

(20)

The probability density function of h is

$$ \begin{array}{rll} p(h)&=&e^{-\frac{\lambda}{2}}\sum\limits_{k=0}^{\infty}\frac{\left(\frac{\lambda}{2}\right)^k}{\beta\left(\frac{N-L}{2},\frac{L}{2}+k\right)k!} \left(\frac{L}{N-L}\right)^{\frac{L}{2}+k} \\ && \times \left(\frac{N-L}{N-L+Lh}\right)^{\frac{N}{2}+k}h^{\frac{L}{2}+k-1}, \end{array} $$

where $\beta(x,y)=\int_0^1{t^{x-1}(1-t)^{y-1}dt}$ is the beta function. Since

$$ \frac{dh}{d\alpha}=\frac{N-L}{L}\cdot\frac{1}{\left(1-\alpha\right)^2}, $$

then

$$ \begin{array}{rll} p(\alpha)&=&p(h)\left|\frac{dh}{d\alpha}\right| \\ \!&=&\!e^{-\frac{\lambda}{2}}(1-\alpha)^{-2}\!\sum\limits_{k=0}^{\infty}\!\frac{\left(\frac{\lambda}{2}\right)^k} {\beta\!\left(\frac{N-L}{2},\!\frac{L}{2}\!+\!k\right)k!}\! \left(\!\frac{L}{N-L}\!\right)^{\!\!\frac{L}{2}+k-1}\\ && \times \left(1-\alpha\right)^{\frac{N}{2}+k} \left(\frac{N-L}{L}\frac{\alpha}{1-\alpha}\right)^{\frac{L}{2}+k-1} \\ &=&e^{-\frac{\lambda}{2}}(1-\alpha)^{\frac{N-L}{2}-1}\sum\limits_{k=0}^{\infty} \frac{\left(\frac{\lambda}{2}\right)^k\left(\alpha\right)^{\frac{L}{2}+k-1}} {\beta\left(\frac{N-L}{2},\frac{L}{2}+k\right)k!}. \end{array} $$

The expectation of α can be obtained as:

$$ \begin{array}{rll} E\left\{\alpha\right\}&=&\int_0^1{\alpha p(\alpha)d\alpha}\\ &=&e^{-\frac{\lambda}{2}}\sum\limits_{k=0}^{\infty}\frac{\left(\frac{\lambda}{2}\right)^k} {\beta\left(\frac{N-L}{2},\frac{L}{2}+k\right)k!} \\&&\times\int_0^1{(1-\alpha)^{\frac{N-L}{2}-1}\left(\alpha\right)^{\frac{L}{2}+k}d\alpha} \\ &=&e^{-\frac{\lambda}{2}}\sum\limits_{k=0}^{\infty}\frac{\left(\frac{\lambda}{2}\right)^k}{\beta\left(\frac{N-L}{2},\frac{L}{2}+k\right)k!} \\ &&\times\; {\beta\left(\frac{N-L}{2},\frac{L}{2}+k+1\right)}\\ &=&e^{-\frac{\lambda}{2}}\sum\limits_{k=0}^{\infty}\frac{\left(\frac{\lambda}{2}\right)^k}{k!}\cdot\frac{L+2k}{N+2k}. \end{array} $$

The last step is obtained by using the following equalities:

$$ \beta(x,y)=\frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)}~~{\rm and}~~\Gamma(n+1)=n\Gamma(n), $$

where $\Gamma(x)=\int_0^{\infty}{t^{x-1}e^{-t}dt}$ is the gamma function.