Two-Dimensional Direction-of-Arrival Estimation in Acoustic Vector Sensor Array via Constrained Tensor Decomposition

Abstract

The canonical polyadic decomposition (CPD) of higher-order tensors, a.k.a. PARAFAC, has shown excellent performance in two-dimensional direction of arrival (DOA) estimation using the acoustic vector sensor array (AVSA). However, most existing studies pay little attention to the manifold matrix structure of the AVSA during the CPD and are designed for uncorrelated sources. This paper presents a constrained CPD-based algorithm for DOA estimation using a uniform linear AVSA, whose manifold matrix is highly structured. Specifically, the manifold matrix equals to the Khatri-Rao product of a Vandermonde matrix and a proportional column-norm matrix. We show that DOA estimation accuracy is further improved by incorporating the prior structured information. Besides, we also extend the Toeplitz decorrelation technique to the AVSA to handle possibly correlated sources. The algorithm does not require iteration or peak searching and thus is computationally effective. Numerical simulations verify the effectiveness and superior performance of the algorithm.

Appendices

Appendix A Proof of (16)

Proof

From the construction, one obtains

$$\begin{aligned} {\mathcal {R}}(:,:,i)= {{\textbf {A}}} \bar{{{\textbf {D}}}}_i {{\textbf {A}}}^H,\quad (i=1,2,3,4), \end{aligned}$$

(A.1)

where $\bar{{{\textbf {D}}}}_i = \sum _{j=1}^4 {{\textbf {D}}}_{i,j}/4$. Let ${{\textbf {d}}}_i \in {\mathbb {C}}^{K\times 1}$ be the vector consisting of the diagonal elements of the diagonal matrix $\bar{{{\textbf {D}}}}_i$, then ${{\textbf {d}}}_i(k) = \xi _k {{\textbf {B}}}(i,k)$ with $\xi _k = \sigma _k^2 (1+u_k+v_k+w_k)/4$. Stacking the ${\mathcal {R}}(:,:,i)$ into a long vector gives

$$\begin{aligned} \text {vec}\{{\mathcal {R}}(:,:,i)\} = ({{\textbf {A}}}^*\odot {{\textbf {A}}}) {{\textbf {d}}}_i. \end{aligned}$$

(A.2)

Then, the mode-3 unfolding of ${\mathcal {R}}$ can be expressed by

$$\begin{aligned} {\textbf {R}}_{3}&=[\text {vec}\{{{\textbf {R}}}(:,:,1)\},\dots ,\text {vec}\{{{\textbf {R}}}(:,:,4)\}] \nonumber \\&= ({{\textbf {A}}}^*\odot {{\textbf {A}}}) [{{\textbf {d}}}_1, \dots , {{\textbf {d}}}_4] \nonumber \\&= ({{\textbf {A}}}^*\odot {{\textbf {A}}}) {\tilde{{\textbf {B}}}}^T. \end{aligned}$$

(A.3)

where ${\tilde{{\textbf {B}}}} = {{\textbf {B}}}\cdot \text {Diag}\{\xi _1,\dots ,\xi _K\}$. Comparing with (5), ${\mathcal {R}}$ can be denoted as ${\mathcal {R}}=[[{{\textbf {A}}}, {{\textbf {A}}}^*, {\tilde{{\textbf {B}}}}]]$. $\square$

Appendix B Proof of (25)

Proof

The (l, m)-th element of ${{\textbf {R}}}_{i,j}$ can be expressed as

$$\begin{aligned} r_{l,m}&= E \left\{ \left( \sum _{k=1}^{K} {{\textbf {B}}}(i,k) z_k^{l} s_k[t] \right) \left( \sum _{k'=1}^{K} {{\textbf {B}}}(j,k') z_{k'}^{m} s_{k'}[t] \right) ^H \right\} \nonumber \\&= \sum _{k=1}^K {{\textbf {B}}}(i,k) \left( \sum _{k'=1}^{K} {{\textbf {B}}}(j,k') E\left\{ s_k[t]s_{k'}^*[t]\right\} z_{k'}^{-m} \right) z_k^l \nonumber \\&= \sum _{k=1}^K d_{i,j,m,k} z_k^l,\quad (l,m=-M,\dots ,0,\dots ,M), \end{aligned}$$

(B1)

where

$$\begin{aligned} \sigma _{k,k'}&= E\{s_k[t]s_{k'}^*[t]\}, \end{aligned}$$

(B2)

$$\begin{aligned} d_{i,j,m,k}&= {{\textbf {B}}}(i,k) \left( \sum _{k'=1}^{K} {{\textbf {B}}}(j,k') \sigma _{k,k'} z_{k'}^{-m} \right) . \end{aligned}$$

(B3)

From the construction of ${{\textbf {T}}}_{i,j,m}$ in (24), its $(l_1,l_2)$-th element equals to $r_{l_1-l_2,m}$ for $l_1,l_2 = 1,\dots ,M+1$. Note that the first element of ${{\textbf {R}}}_{i,j}$ is indexed by $(-M,-M)$ while that of ${{\textbf {T}}}_{i,j,m}$ is indexed by (1, 1) for a formal convenience.

On the other hand, the $(l_1,l_2)$-th entry of ${{\textbf {V}}} {{\textbf {D}}}_{i,j,m} {{\textbf {V}}}^H$ equals to

$$\begin{aligned} \sum _{k=1}^K d_{i,j,m,k}{{\textbf {v}}}_{k}(l_1) {{\textbf {v}}}_{k}^H(l_2)&= \sum _{k=1}^K d_{i,j,m,k}z_k^{l_1-1} z_{k}^{1-l_2} = r_{l_1-l_2,m}, \end{aligned}$$

(B4)

where ${{\textbf {v}}}_k = [1,z_k,\dots ,z_k^M]^T$ is the k-th column of ${{\textbf {V}}}$. Thus, (25) is proven. $\square$

Appendix C Proof of full column rank of ${\tilde{{\textbf {B}}}}\odot {{\textbf {A}}}$

Proof

The Vandermonde matrix ${{\textbf {A}}} \in {\mathbb {C}}^{N\times K}$ is full column rank if $K<N$ holds. To show ${\tilde{{\textbf {B}}}} \odot {{\textbf {A}}}$ is full column rank, a lemma is introduced as follows: $\square$

Lemma 1

[Full Rank of Khatri-Rao Product [22]] If $k_{{{\textbf {W}}}}+k_{{{\textbf {Z}}}} \ge J+1$, then ${{\textbf {Z}}}\odot {{\textbf {W}}}$ is full column rank J, where ${{\textbf {W}}}$ is $I_1\times J$ and ${{\textbf {Z}}}$ is $I_2 \times J$.

Using Lemma 1 and the fact that $k_{{\textbf {A}}} = \min (N,K)$ and $k_{{\tilde{{\textbf {B}}}}}=\min (3,K)$, one obtains that $k_{{\textbf {A}}} + k_{{\tilde{{\textbf {B}}}}} = \min (N,K) + \min (3,K)$. Therefore, ${\tilde{{\textbf {B}}}} \odot {{\textbf {A}}}$ is full column rank if $1<K<N$ holds.

Appendix D Proof of the PCN constrained solution

Solving (38) seems to be challenging because of the non-positive definiteness of the PCN constraint. Fortunately, it can be solved by the following lemma.

Lemma 2

[Optimal Scaling Lemma [2]] For any column-wise constraint set ${\mathbb {S}}$ it holds that

$$\begin{aligned} \min _{{{\textbf {b}}} \in {\mathbb {S}}} \left\| {{\textbf {X}}} - {{\textbf {a}}}{{\textbf {b}}}^H \right\| _F^2 \Longleftrightarrow \min _{{{\textbf {b}}} \in {\mathbb {S}}} \left\| {{\textbf {b}}}^{LS} - {{\textbf {b}}} \right\| _2^2, \end{aligned}$$

(D.1)

where ${{{\textbf {b}}}}^{LS}= {{\textbf {X}}}^H{{\textbf {a}}}/\Vert {{\textbf {a}}}\Vert _2^2$. In other words, the optimal solution of the CLS problem is simply the projection of the (unconstrained) LS solution onto the constraint set ${\mathbb {S}}$.

With Lemma 2, the problem in (38) becomes

$$\begin{aligned} \min _{{{\textbf {b}}} \in {\mathbb {P}}^{4\times 1}} \left\| {\tilde{{\textbf {b}}}}_k^{LS} - {{\textbf {b}}} \right\| _2^2, \end{aligned}$$

(D.2)

where ${\tilde{{\textbf {b}}}}_k^{LS}= {\tilde{{\textbf {U}}}}_k^T {{\textbf {a}}}_k^* /\Vert {{\textbf {a}}}_k\Vert _2^2$ is the LS solution in (37). The next step is to project ${\tilde{{\textbf {b}}}}_k^{LS}$ onto the PCN space. Since ${{\textbf {b}}}$ is in the PCN space, it can be parameterized as

$$\begin{aligned} {{\textbf {b}}} = {\tilde{{\textbf {b}}}}_k^{LS}(1) \zeta [1,\cos \alpha \cos \beta ,\cos \alpha \sin \beta ,\sin \alpha ]^T, \end{aligned}$$

(D.3)

where $\zeta$ is used to adjust the length of ${{\textbf {b}}}$ and ${\tilde{{\textbf {b}}}}_k^{LS}(1)$ is used to simplify the subsequent objective function. Defining

$$\begin{aligned} c_i \triangleq {\tilde{{\textbf {b}}}}_k^{LS}(i)/{\tilde{{\textbf {b}}}}_k^{LS}(1), \quad (i= 1,2,3,4), \end{aligned}$$

(D.4)

and substituting (D.3) and (D.4) into the objective function in (D.2), one obtains

$$\begin{aligned} \left\| {\tilde{{\textbf {b}}}}_k^{LS} - {{\textbf {b}}} \right\| _2^2= & {} \left[ {\tilde{{\textbf {b}}}}_k^{LS}(1)\right] ^2 \left[ (1-\zeta )^2 + (c_2-\zeta \cos \alpha \cos \beta )^2 + (c_3-\zeta \cos \alpha \sin \beta )^2 \nonumber \right. \\{} & {} \left. + (c_4-\zeta \sin \alpha )^2 \right] \nonumber \\= & {} \left[ {\tilde{{\textbf {b}}}}_k^{LS}(1)\right] ^2 \!\left[ - 2\zeta (c_1 \!+\! c_2\cos \alpha \cos \beta \!+\! c_3 \!\cos \alpha \sin \beta \!+\! c_4 \sin \alpha ) \!+\! 2\zeta ^2 \right] \nonumber \\{} & {} + \text {const}, \end{aligned}$$

(D.5)

where $\text {const} = \left[ {\tilde{{\textbf {b}}}}_k^{LS}(1)\right] ^2 (1+c_2^2+c_3^2+c_4^2 ) = \left\| {\tilde{{\textbf {b}}}}_k^{LS}\right\| _2^2$. After dropping the constant values and the scale factor in (D.5), the CLS problem in (D.2) is equivalent to find $(\alpha ,\beta ,\zeta )$ that minimizes

$$\begin{aligned} f(\alpha ,\beta ,\zeta ) =&- 2\zeta (c_1 + c_2\cos \alpha \cos \beta + c_3 \cos \alpha \sin \beta + c_4 \sin \alpha ) + 2\zeta ^2. \end{aligned}$$

(D.6)

The minimal of $f(\alpha ,\beta ,\zeta )$ is obtained when ${\partial f}/{\partial \zeta } = {\partial f}/{\partial \alpha } ={\partial f}/{\partial \beta } = 0$. Solving the value of $(\alpha , \beta , \zeta )$ and substituting them into (D.3), the optimal estimates of ${\tilde{{\textbf {b}}}}_k$ simplifies to

$$\begin{aligned} {\tilde{{\textbf {b}}}}_k^{CLS} = \frac{ (1+L){\tilde{{\textbf {b}}}}_k^{LS}(1)}{2} [1,c_2/L,c_3/L,c_4/L]^T, \end{aligned}$$

(D.7)

where $L = \sqrt{c_2^2+c_3^2+c_4^2}$.