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多快照无网格压缩波束形成

Multiple snapshot grid free compressive beamforming.

作者信息

Park Yongsung, Choo Youngmin, Seong Woojae

机构信息

Department of Naval Architecture and Ocean Engineering, Seoul National University, Seoul, 151-744, South Korea.

Department of Defense System Engineering, Sejong University, Seoul, 143-747, South Korea.

出版信息

J Acoust Soc Am. 2018 Jun;143(6):3849. doi: 10.1121/1.5042242.

DOI:10.1121/1.5042242
PMID:29960424
Abstract

Compressive sensing (CS) based estimation technique utilizes a sparsity promoting constraint and solves the direction-of-arrival (DOA) estimation problem efficiently with high resolution. In this paper a grid free CS based DOA estimation technique is proposed, which uses sequential multiple snapshot data. Conventional CS technique suffers from a basis mismatch issue, while grid free CS technique is relieved of basis mismatch problem. Moreover, when the DOAs are stationary, multiple snapshot processing provides stable estimates over fluctuating single snapshot processing results. For multiple snapshot processing, the generalized version of total variation norm (group total variation norm) is implemented to impose a common sparsity pattern of multiple snapshot solution vectors in a continuous angular domain. Furthermore, an extended version is proposed using the singular value decomposition technique to mitigate computational complexity resulting from a large number of multiple snapshots. Data from SWellEx-96 are used to examine the proposed method. From the experimental data, it was observed that the present method not only offers high resolution even when the sources are coherent, but also the basis mismatch in the conventional CS method can be avoided.

摘要

基于压缩感知(CS)的估计技术利用稀疏性促进约束,高效地解决了高分辨率到达方向(DOA)估计问题。本文提出了一种基于无网格CS的DOA估计技术,该技术使用连续多快照数据。传统的CS技术存在基失配问题,而无网格CS技术则避免了基失配问题。此外,当DOA静止时,多快照处理相对于波动的单快照处理结果提供了稳定的估计。对于多快照处理,采用总变分范数的广义形式(组总变分范数)在连续角度域中强加多个快照解向量的公共稀疏模式。此外,还提出了一种扩展形式,使用奇异值分解技术来减轻由于大量多快照而导致的计算复杂度。利用SWellEx-96的数据来检验所提出的方法。从实验数据中可以观察到,本方法不仅在源相干时也能提供高分辨率,而且可以避免传统CS方法中的基失配问题。

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