Appendix A: EXACT FORMULATION OF THE BLOCK HANKEL MATRIX

The block Hankel matrix $\mathbf{M}^{(4)}$ can be expressed as

$\displaystyle \mathbf{M}^{(4)}=\mathbf{S}+\mathbf{N},$ (18)

where $\mathbf{S}$ and $\mathbf{N}$ denote signal component and the random noise component that are both in $\mathbf{M}^{(4)}$, respectively. The singular value decomposition (SVD) of $\mathbf{S}$ can be represented as:

$\displaystyle \mathbf{S} = [\mathbf{U}_1^{S}\quad \mathbf{U}_2^{S}]\left[\begin...
...egin{array}{c}
(\mathbf{V}_1^{S})^H\\
(\mathbf{V}_2^{S})^H
\end{array}\right].$ (19)

Because of the deficient rank, the matrix $\mathbf{S}$ can be written as

$\displaystyle \mathbf{S}=\mathbf{U}_1^S\Sigma_1^S(\mathbf{V}_1^S)^H.$ (20)

Because equation A-2 is a SVD of the signal matrix $\mathbf{S}$, the left matrix in equation A-2 is a unitary matrix:

$\displaystyle \mathbf{I}=\mathbf{U}^S(\mathbf{U}^S)^H=[\mathbf{U}_1^S\quad \mat...
...[\begin{array}{c}
(\mathbf{U}_1^S)^H \\
(\mathbf{U}_2^S)^H
\end{array}\right].$ (21)

Combining equations A-1, A-3, and A-4, we can derive:

\begin{displaymath}\begin{split}
\mathbf{M}^{(4)}&=\mathbf{S}+[\mathbf{U}_1^S\qu...
...
(\mathbf{N}^H\mathbf{U}_2^S)^H
\end{array}\right],
\end{split}\end{displaymath} (22)

we can further derive equation A-5 and factorize $\mathbf{M}^{(4)}$ as follows:

$\displaystyle \mathbf{M}^{(4)} = [\mathbf{U}_1^S \quad \mathbf{U}_2^S]\left[\be...
...a_1^S)^H\\
(\Sigma_{2})^{-1}(\mathbf{N}^H\mathbf{U}_2^S)^H
\end{array}\right],$ (23)

where $\Sigma_1$ and $\Sigma_2$ denote diagonal and positive definite matrices. Please note that $\mathbf{M}^{(4)}$ is constructed such that it is close to a square matrix (Oropeza and Sacchi, 2011), and thus the $\Sigma_1$ and $\Sigma_2$ are assumed to be square matrices for derivation convenience. We observe that the left matrix has orthonormal columns and the middle matrix is diagonal. It can be shown that the right matrix also has orthonormal columns. Thus, equation A-6 is an SVD of $\mathbf{M}^{(4)}$. According to the TSVD method, we let $\Sigma_2$ be $\mathbf{0}$ and then the following equation holds:

\begin{displaymath}\begin{split}
\tilde{\mathbf{M}}^{(4)} &= \mathbf{U}_1^S(\mat...
...bf{S} + \mathbf{U}_1^S(\mathbf{U}_1^S)^H\mathbf{N}.
\end{split}\end{displaymath} (24)


2020-12-05