2D-DOA estimation with spatially separated “long” crossed-dipoles array based on three-way compressive sensing

Abstract

For the angle estimation demand of real-world electromagnetic vector sensor (EMVS) array radar, in this paper we propose a two-dimensional direction of arrival (2D-DOA) estimation method with spatially separated “long” crossed-dipoles array based on three-way compressive sensing. First, the low-dimension random matrix is utilized to compressively sample the received data in tensor style. Second, the compressed factor matrices are achieved through parallel factor decomposition. Third, the cosines’ estimations are derived from the support locations of sparse vectors recovered with orthogonal matching pursuit algorithm. Finally, 2D-DOA estimation is calculated according to the relationship between cosines. Compared with the traditional ESPRIT method, the proposed method can estimate parameters with higher accuracy under the condition of low signal-to-noise ratio and a few number of snapshots, which is conducive to improve the robustness of angle estimation with real-world EMVS array radar.

Appendix

1.1 A. Derivation for signal modeling based on matrix style

The common received data can be expressed as in matrix style (Li et al., 2018)

$$ {\mathbf{Y}} = \sum\limits_{k = 1}^{K} {\left( {{\mathbf{q}}_{y} (v_{k} ) \otimes {\mathbf{q}}_{x} (u_{k} ) \otimes {\mathbf{b}}_{k} } \right){\mathbf{s}}_{k}^{{\text{T}}} } + {\mathbf{N}} = \left( {{\mathbf{Q}}_{y} \oplus {\mathbf{Q}}_{x} \oplus {\mathbf{B}}} \right){\mathbf{S}}^{{\text{T}}} + {\mathbf{N}} \in {\mathbb{C}}^{2MN \times G} $$

(39)

Transposing both sides of (39), we can obtain

$$ {\mathbf{X}} = {\mathbf{Y}}^{{\text{T}}} = {\mathbf{S}}\left( {{\mathbf{Q}}_{y} \oplus {\mathbf{Q}}_{x} \oplus {\mathbf{B}}} \right)^{{\text{T}}} + {\mathbf{N}}^{{\text{T}}} \in {\mathbb{C}}^{G \times 2MN} $$

(40)

According to (5) and (6) in Definition 3, (40) can be transformed into the following tensor form

$$ {\mathbf{\mathcal{X}}} = {\mathbf{\mathcal{I}}}_{4,K} \times_{1} {\mathbf{Q}}_{y} \times_{2} {\mathbf{Q}}_{x} \times_{3} {\mathbf{B}} \times_{4} {\mathbf{S}} + {\mathbf{\mathcal{N}}} \in {\mathbb{C}}^{N \times M \times 2 \times G} $$

(41)

1.2 B. Derivation for CRB

The CRB is only related to the array form, not to the type of estimation algorithm. Therefore, we will still use the traditional matrix method to derive the CRB. Denote ${{\varvec{\uptheta}}} = [\theta_{1} ,\theta_{2} , \ldots ,\theta_{K} ]$ and ${\varvec{\phi}} = [\phi_{1} ,\phi_{2} , \ldots ,\phi_{K} ]$. Then the Fisher information matrix can be given by

$$ {\mathbf{J}} = \left[ {\begin{array}{*{20}c} {{\mathbf{J}}_{{{\mathbf{\theta \theta }}}} } & {{\mathbf{J}}_{{{{\varvec{\uptheta}}}{\varvec{\phi}}}} } \\ {{\mathbf{J}}_{{{\varvec{\phi}}{{\varvec{\uptheta}}}}} } & {{\mathbf{J}}_{{\varvec{\phi \phi }}} } \\ \end{array} } \right] $$

(42)

Since we assume that the source is a random unknown signal, according to Trees (2002), the subarray in Fisher information matrix can be expressed as

$$ {\mathbf{J}}_{{{\mathbf{hk}}}} (i,j) = G \cdot {\text{Tr}}\left( {{\mathbf{R}}^{ - 1} \frac{{\partial {\mathbf{R}}}}{{\partial {\mathbf{h}}_{i} }}{\mathbf{R}}^{ - 1} \frac{{\partial {\mathbf{R}}}}{{\partial {\mathbf{k}}_{j} }}} \right) $$

(43)

where ${\text{Tr}}( \cdot )$ denotes the trace of a matrix, R denotes covariance matrix of the received data (in matrix form)

$$ {\mathbf{R}} = \sum\limits_{k = 1}^{K} {\sigma_{{s_{k} }}^{2} {\mathbf{d}}_{k} {\mathbf{d}}_{k}^{{\text{H}}} } + \sigma_{n}^{2} {\mathbf{I}}_{2MN} $$

(44)

where ${\mathbf{d}}_{k} { = }{\mathbf{q}}_{x} (u_{k} ) \otimes {\mathbf{q}}_{y} (v_{k} ) \otimes {\mathbf{b}}_{k}$ denotes the joint steer vector, ${\mathbf{I}}_{2MN}$ denotes an $2MN \times 2MN$ identity matrix, $\sigma_{{s_{k} }}^{2}$ and $\sigma_{n}^{2}$ represent the power of the signal and noise, respectively. The first-order partial derivatives of the covariance matrix ${\mathbf{R}}$ with respect to $\theta_{k}$ and $\phi_{k}$ are given by

$$ \frac{{\partial {\mathbf{R}}}}{{\partial \theta_{k} }} = \frac{{\sigma_{{s_{k} }}^{2} \partial {\mathbf{d}}_{k} {\mathbf{d}}_{k}^{{\text{H}}} }}{{\partial \theta_{k} }} = \sigma_{{s_{k} }}^{2} \frac{{\partial {\mathbf{d}}_{k} }}{{\partial \theta_{k} }}{\mathbf{d}}_{k}^{{\text{H}}} + \sigma_{{s_{k} }}^{2} {\mathbf{d}}_{k} \frac{{\partial {\mathbf{d}}_{k}^{{\text{H}}} }}{{\partial \theta_{k} }} $$

(45)

$$ \frac{{\partial {\mathbf{R}}}}{{\partial \phi_{k} }} = \frac{{\sigma_{{s_{k} }}^{2} \partial {\mathbf{d}}_{k} {\mathbf{d}}_{k}^{{\text{H}}} }}{{\partial \phi_{k} }} = \sigma_{{s_{k} }}^{2} \frac{{\partial {\mathbf{d}}_{k} }}{{\partial \phi_{k} }}{\mathbf{d}}_{k}^{{\text{H}}} + \sigma_{{s_{k} }}^{2} {\mathbf{d}}_{k} \frac{{\partial {\mathbf{d}}_{k}^{{\text{H}}} }}{{\partial \phi_{k} }} $$

(46)

The detailed derivation process of $\frac{{\partial {\mathbf{d}}_{k} }}{{\partial \theta_{k} }}$ and $\frac{{\partial {\mathbf{d}}_{k} }}{{\partial \phi_{k} }}$ are given as follows

$$ \begin{aligned} \frac{{\partial {\mathbf{d}}_{k} }}{{\partial \theta_{k} }} & = \left[ {{\mathbf{c}}_{1} \odot {\mathbf{q}}_{x} (u)} \right] \otimes {\mathbf{q}}_{y} (v) \otimes {\mathbf{b}}_{k} + {\mathbf{q}}_{x} (u) \otimes \left[ {{\mathbf{c}}_{2} \odot {\mathbf{q}}_{y} (v)} \right] \otimes {\mathbf{b}}_{k} \\ & \quad + {\mathbf{q}}_{x} (u) \otimes {\mathbf{q}}_{y} (v) \otimes [ - {\mathbf{a}}_{{\theta_{k} }}^{ \circ } \odot {\overline{\mathbf{e}}}_{k} \odot {\mathbf{l}} \odot {\mathbf{t}} - {\mathbf{a}}_{k} \odot {\overline{\mathbf{e}}}_{{\theta_{k} }}^{ \circ } \odot {\overline{\mathbf{e}}}_{k} \odot {\mathbf{l}} \odot {\mathbf{t}} \\ & \quad - {\mathbf{a}}_{k} \odot {\overline{\mathbf{e}}}_{k} \odot {\mathbf{l}}_{{\theta_{k} }}^{ \circ } \odot {\mathbf{t}} - {\mathbf{a}}_{k} \odot {\overline{\mathbf{e}}}_{k} \odot {\mathbf{l}} \odot {\mathbf{t}}_{{\theta_{k} }}^{ \circ } ] \\ \end{aligned} $$

(47)

$$ {\mathbf{c}}_{1} = - {\text{j}}2\pi \cos \theta_{k} \cos \phi_{k} \left[ {0,1, \ldots ,(M - 1)} \right]^{{\text{T}}} \qquad\qquad\qquad\qquad\qquad\qquad$$

(48)

$$ {\mathbf{c}}_{2} = - {\text{j}}2\pi \cos \theta_{k} \sin \phi_{k} \left[ {0,1, \ldots ,(N - 1)} \right]^{{\text{T}}} \qquad\qquad\qquad\qquad\qquad\qquad$$

(49)

$$ {\mathbf{a}}_{{\theta_{k} }}^{ \circ } = \left[ {\begin{array}{*{20}c} { - \sin \theta_{k} \cos \phi_{k} \sin \gamma_{k} {\text{e}}^{{{\text{j}}\eta_{k} }} } \\ { - \sin \theta_{k} \sin \phi_{k} \sin \gamma_{k} {\text{e}}^{{{\text{j}}\eta_{k} }} } \\ \end{array} } \right] \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$

(50)

$$ {\overline{\mathbf{e}}}_{{\theta_{k} }}^{ \circ } = \left[ {0, - {\text{j}}\pi \cos \theta_{k} \sin \phi_{k} } \right]^{{\text{T}}} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad$$

(51)

$$ {\mathbf{l}}_{{\theta_{k} }}^{ \circ } = \left[ {l_{\theta 1} ,l_{\theta 2} } \right]^{{\text{T}}} $$

(52)

$$ l_{\theta 1} = \frac{{ - \lambda \cos \theta_{k} \cos \phi_{k} f_{\theta 1} }}{{\pi \sin^{2} \theta_{xk} \sin (\pi L/\lambda )\sqrt {1 - \sin^{2} \theta_{k} \cos^{2} \phi_{k} } }} $$

(53)

$$ f_{\theta 1} \!=\! \pi L\sin^{2} \theta_{xk} \sin (\pi L\cos \theta_{xk} /\lambda )/\lambda \!-\! \cos \theta_{xk} \left[ {\cos (\pi L\cos \theta_{xk} /\lambda )\! -\! \cos (\pi L/\lambda )} \right] $$

(54)

$$ \theta_{xk} = \arccos \left( {\sin \theta_{k} \cos \phi_{k} } \right) $$

(55)

$$ l_{\theta 2} = \frac{{ - \lambda \cos \theta_{k} \sin \phi_{k} f_{\theta 2} }}{{\pi \sin^{2} \theta_{yk} \sin (\pi L/\lambda )\sqrt {1 - \sin^{2} \theta_{k} \sin^{2} \phi_{k} } }} $$

(56)

$$ f_{\theta 2}\! =\! \pi L\sin^{2} \theta_{yk} \sin (\pi L\cos \theta_{yk} /\lambda )/\lambda\! -\! \cos \theta_{yk} \left[ {\cos (\pi L\cos \theta_{yk} /\lambda )\! -\! \cos (\pi L/\lambda )} \right] $$

(57)

$$ \theta_{yk} = \arccos \left( {\sin \theta_{k} \sin \phi_{k} } \right) $$

(58)

$$ {\mathbf{t}}_{{\theta_{k} }}^{ \circ } = \left[ {\begin{array}{*{20}c} {\frac{{\cos \theta_{xk} \cos \theta_{k} \cos \phi_{k} }}{{\sin^{2} \theta_{xk} \sqrt {1 - \sin^{2} \theta_{k} \cos^{2} \phi_{k} } }}} \\ {\frac{{\cos \theta_{yk} \cos \theta_{k} \sin \phi_{k} }}{{\sin^{2} \theta_{yk} \sqrt {1 - \sin^{2} \theta_{k} \sin^{2} \phi_{k} } }}} \\ \end{array} } \right] $$

(59)

$$ \begin{aligned} \frac{{\partial {\mathbf{d}}_{k} }}{{\partial \phi_{k} }} & = \left[ {{\mathbf{c}}_{3} \odot {\mathbf{q}}_{x} (u)} \right] \otimes {\mathbf{q}}_{y} (v) \otimes {\mathbf{b}}_{k} + {\mathbf{q}}_{x} (u) \otimes \left[ {{\mathbf{c}}_{4} \odot {\mathbf{q}}_{y} (v)} \right] \otimes {\mathbf{b}}_{k} \\ & \quad + {\mathbf{q}}_{x} (u) \otimes {\mathbf{q}}_{y} (v) \otimes [ - {\mathbf{a}}_{{\phi_{k} }}^{ \circ } \odot {\overline{\mathbf{e}}}_{k} \odot {\mathbf{l}} \odot {\mathbf{t}} \\ & \quad - {\mathbf{a}}_{k} \odot {\overline{\mathbf{e}}}_{{\phi_{k} }}^{ \circ } \odot {\overline{\mathbf{e}}}_{k} \odot {\mathbf{l}} \odot {\mathbf{t}} - {\mathbf{a}}_{k} \odot {\overline{\mathbf{e}}}_{k} \odot {\mathbf{l}}_{{\phi_{k} }}^{ \circ } \odot {\mathbf{t}} \\ & \quad - {\mathbf{a}}_{k} \odot {\overline{\mathbf{e}}}_{k} \odot {\mathbf{l}} \odot {\mathbf{t}}_{{\phi_{k} }}^{ \circ } ] \\ \end{aligned} $$

(60)

$$ {\mathbf{c}}_{3} = {\text{j}}2\pi \sin \theta_{k} \sin \phi_{k} \left[ {0,1, \ldots ,(M - 1)} \right]^{{\text{T}}}\qquad\qquad\qquad\qquad\qquad\quad\quad$$

(61)

$$ {\mathbf{c}}_{4} = - {\text{j}}2\pi \sin \theta_{k} \cos \phi_{k} \left[ {0,1, \ldots ,(N - 1)} \right]^{{\text{T}}} \qquad\qquad\qquad\qquad\quad\qquad\quad$$

(62)

$$ {\mathbf{a}}_{{\phi_{k} }}^{ \circ } = \left[ {\begin{array}{*{20}c} { - \cos \theta_{k} \sin \phi_{k} \sin \gamma_{k} {\text{e}}^{{{\text{j}}\eta_{k} }} - \cos \phi_{k} \cos \gamma_{k} } \\ {\cos \theta_{k} \cos \phi_{k} \sin \gamma_{k} {\text{e}}^{{{\text{j}}\eta_{k} }} - \sin \phi_{k} \cos \gamma_{k} } \\ \end{array} } \right] \qquad\qquad\quad\quad\qquad\qquad$$

(63)

$$ {\overline{\mathbf{e}}}_{{\phi_{k} }}^{ \circ } = \left[ {\begin{array}{*{20}c} 0 & { - {\text{j}}\pi \sin \theta_{k} \cos \phi_{k} } \\ \end{array} } \right]^{{\text{T}}}\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\quad\qquad $$

(64)

$$ {\mathbf{l}}_{{\varphi_{k} }}^{ \circ } = \left[ {l_{\varphi 1} ,l_{\varphi 2} } \right]^{{\text{T}}} \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\quad\quad$$

(65)

$$ l_{\varphi 1} = \frac{{\lambda \sin \theta_{k} \sin \phi_{k} f_{\theta 1} }}{{\pi \sin^{2} \theta_{xk} \sin (\pi L/\lambda )\sqrt {1 - \sin^{2} \theta_{k} \cos^{2} \phi_{k} } }} \qquad\qquad\qquad\qquad\quad\qquad$$

(66)

$$ l_{\phi 2} = \frac{{ - \lambda \sin \theta_{k} \cos \phi_{k} f_{\theta 2} }}{{\pi \sin^{2} \theta_{yk} \sin (\pi L/\lambda )\sqrt {1 - \sin^{2} \theta_{k} \sin^{2} \phi_{k} } }} \qquad\qquad\qquad\qquad\quad\qquad$$

(67)

$$ {\mathbf{t}}_{{\phi_{k} }}^{ \circ } = \left[ {\begin{array}{*{20}c} {\frac{{\cos \theta_{xk} \sin \theta_{k} \sin \phi_{k} }}{{\sin^{2} \theta_{xk} \sqrt {1 - \sin^{2} \theta_{k} \cos^{2} \phi_{k} } }}} \\ {\frac{{ - \cos \theta_{yk} \sin \theta_{k} \cos \phi_{k} }}{{\sin^{2} \theta_{yk} \sqrt {1 - \sin^{2} \theta_{k} \sin^{2} \phi_{k} } }}} \\ \end{array} } \right] $$

(68)

Substituting (45)–(47) and (60) into (43), we can calculate the Fisher information matrix ${\mathbf{J}}$. Then the CRB of the two parameters can be given by

$$ \left\{ \begin{gathered} {\text{CRB}}\left( {\theta_{k} } \right) = \left[ {{\mathbf{J}}^{ - 1} } \right]_{k,k} \hfill \\ {\text{CRB}}\left( {\phi_{k} } \right) = \left[ {{\mathbf{J}}^{ - 1} } \right]_{K + k,K + k} \hfill \\ \end{gathered}\qquad\qquad \right.\qquad $$

(69)