Blind Source Separation of Single Channel Mixture Using Tensorization and Tensor Diagonalization

Abstract

This paper deals with estimation of structured signals such as damped sinusoids, exponentials, polynomials, and their products from single channel data. It is shown that building tensors from this kind of data results in tensors with hidden block structure which can be recovered through the tensor diagonalization. The tensor diagonalization means multiplying tensors by several matrices along its modes so that the outcome is approximately diagonal or block-diagonal of 3-rd order tensors. The proposed method can be applied to estimation of parameters of multiple damped sinusoids, and their products with polynomial.

The work of P. Tichavsky was supported by The Czech Science Foundation through Projects No. 14-13713S and 17-00902S.

A Appendix: Low-Rank Representation of the Sequence \(x(t) = t^n\)

Lemma 5

(Three-way folding of $x(t) = t$ ). An order-3 tensor of size $I\times J\times K$, reshaped (folded) from the sequence $1,2,\ldots , IJK$, where $I,J,K>2$, has multilinear rank-(2, 2, 2) and rank-3, and can be represented as

$$\begin{aligned} \varvec{\mathscr {\MakeUppercase {Y}}}= \varvec{\mathscr {\MakeUppercase {G}}}\times _1 \mathbf{U}_1 \times _2 \mathbf{U}_2 \times _3 \mathbf{U}_3 \end{aligned}$$

(14)

where $\varvec{\mathscr {\MakeUppercase {G}}}$ is a tensor of size $2 \times 2 \times 2$

$$\begin{aligned} \varvec{\mathscr {\MakeUppercase {G}}}(:,1,:)= & {} \left[ \begin{array}{cc} -2 &{} -1 \\ 1 &{} 0 \end{array} \right] , \qquad \varvec{\mathscr {\MakeUppercase {G}}}(:,2,:) = \left[ \begin{array}{cc} 1 &{} 0 \\ 0 &{} 1 \end{array} \right] , \end{aligned}$$

and the three factor matrices are given by

$$\begin{aligned} \mathbf{U}_1= & {} \left[ \begin{array}{cc} 0 &{} 1 \\ \vdots &{} \vdots \\ i &{} i+1\\ \vdots &{} \vdots \\ (I-1) &{} I \end{array} \right] , \;\; \mathbf{U}_2 = \left[ \begin{array}{cc} 0 &{} 1 \\ \vdots &{} \vdots \\ jI &{} jI+1\\ \vdots &{} \vdots \\ (J-1)I &{} (J-1)I+1 \end{array} \right] , \;\; \mathbf{U}_3 = \left[ \begin{array}{cc} 0 &{} 1 \\ \vdots &{} \vdots \\ kIJ &{} kIJ+1\\ \vdots &{} \vdots \\ (K-1)IJ &{} (K-1)IJ+1 \end{array} \right] . \end{aligned}$$

Lemma 6

(Toeplitzation of $x(t) = t$ ). An order-3 Toeplitz tensor of size $I\times J\times K$, tensorized from the sequence $1,2,\ldots , L$, where $L = I+J+K-2$, has multilinear rank-(2, 2, 2) and rank-3, and can be represented as

$$\begin{aligned} \varvec{\mathscr {\MakeUppercase {Y}}}= \varvec{\mathscr {\MakeUppercase {G}}}\times _1 \mathbf{U}_1 \times _2 \mathbf{U}_2 \times _3 \mathbf{U}_3 \end{aligned}$$

(15)

where $\varvec{\mathscr {\MakeUppercase {G}}}$ is a tensor of size $2 \times 2 \times 2$

$$\begin{aligned} \varvec{\mathscr {\MakeUppercase {G}}}(:,:,1)= & {} \left[ \begin{array}{cc} I+4 &{} -(I+3) \\ -(I+3) &{} I+2 \end{array} \right] , \qquad \varvec{\mathscr {\MakeUppercase {G}}}(:,:,2) = \left[ \begin{array}{cc} -(I+3) &{} I+2 \\ I+2 &{} -(I+1) \end{array} \right] , \end{aligned}$$

and the three factor matrices are given by

$$\begin{aligned} \mathbf{U}_1= & {} \left[ \begin{array}{cc} 1 &{} 2 \\ \vdots &{} \vdots \\ i &{} i+1\\ \vdots &{} \vdots \\ I &{} I+1 \end{array} \right] , \;\; \mathbf{U}_2 = \left[ \begin{array}{cc} I &{} I+1 \\ \vdots &{} \vdots \\ j &{} j+1\\ \vdots &{} \vdots \\ I+J-1 &{} I+J \end{array} \right] , \;\; \mathbf{U}_3 = \left[ \begin{array}{cc} I+J-1 &{} I+J-2 \\ \vdots &{} \vdots \\ k &{} k-1\\ \vdots &{} \vdots \\ L &{} L-1 \end{array} \right] \, . \end{aligned}$$

Lemma 7

(Three-way folding of $x(t) = t^n$ ). An order-3 tensor of size $I\times J\times K$, reshaped (folded) from the sequence $x(t) = t^n$, where $n = 1, 2, \ldots$ and $I,J,K>2$, has multilinear rank-($n+1,n+1,n+1$).

Proof

By exploiting the closed-form expression of $x(t) = t$ in Lemma 5, and the property of the Hadamard product stated in Lemma 4 or in (13), we can prove that the tensor reshaped from $x(t) = t^n$ can be fully explained by three factor matrices which have $n+1$ columns, and are defined as $\mathbf{U}_1 = \mathbf{F}(I,n)$, $\mathbf{U}_2 = \mathbf{F}(IJ,n)$ and $\mathbf{U}_3 = \mathbf{F}(IJK,n)$, where

$$\begin{aligned} \mathbf{F}(I,n)= & {} \left[ \begin{array}{ccccc} 0 &{} \cdots &{} 0 &{} \cdots &{} 1 \\ \vdots &{} &{} \vdots \\ i^n &{} \cdots &{} i^k(i+1)^{n-k} &{} \cdots &{} (i+1)^n\\ \vdots &{} &{} \vdots &{} &{} \vdots \\ (I-1) &{} \cdots &{} (I-1)^k I^{n-k} &{} \cdots &{} I^n \end{array} \right] \, . \end{aligned}$$

   $\square$

Lemma 8

(Toeplitzation of $x(t) = t^n$ ). An order-3 Toeplitz tensor of size $I\times J\times K$ of the sequence $x(t) = t^n$, has multilinear rank-($n+1,n+1,n+1$).

Proof

Skipped for lack of space.

   $\square$