新的双基地MIMO雷达角度估计方法

doi:10.3969/j.issn.1000-436x.2012.12.016

摘要/Abstract

摘要：

摘要：为降低双基地MIMO雷达前端数据处理的计算量，构造出了“扩展”信号子空间，并根据此信号子空间的特点，提出了多项式求根-空域滤波的收发角度估计算法，避免了二维谱搜索，实现了目标角度的自动配对，并推导了多目标和单目标下双基地 MIMO 雷达角度估计的克拉默-拉奥下界（CRB）。研究表明空域滤波时，泰勒级数展开的阶数越高，目标的角度估计精度越好；当目标各发射角度相隔较近时，仍能得到较好的估计结果；在低信噪比时，估计精度优于ESPRIT算法，在高信噪比时，2种算法的估计精度均接近于CRB。

关键词: 双基地MIMO雷达, 角度估计, “扩展”信号子空间, 多项式求根, 空域滤波

Abstract:

The extended signal subspace was constructed to reduce the computation load of former data processing of bistatic MIMO radar.According to the characteristic of this signal subspace,the polynomial rooting-spatial filtering algorithm was proposed to estimate the direction of arrivals (DOA) and direction of departure (DOD).It avoided the conventional two-dimensional spectrum searching and the estimated parameters were paired automatically without extra pairing operation.Furthermore,Cramer-Rao bounds (CRB) for DOA and DOD estimation were derived under multi-target and single target in bistatic MIMO radar.The results illustrate that when spatial sector is filt the higher order of the Taylor series expansion,the better angular accuracy of targets can be derived.The proposed algorithm can get good estimation performance even when the DODs of targets are close to each other.And the polynomial rooting-spatial filtering algorithm can perform better than ESPRIT algorithm in resolution in SNR,and both of them are close to the CRB in the high SNR.

Key words: bistatic MIMO radar, angle estimation, extended signal subspace, polynomial rooting, spatial filtering

郑志东,张剑云,屈金佑,林秀清. 新的双基地MIMO雷达角度估计方法[J]. 通信学报, 2012, 33(12): 123-132.

Zhi-dong ZHENG,Jian-yun ZHANG,Jin-you QU,Xiu-qing LIN. Novel method for angle estimating of bistatic MIMO radar[J]. Journal on Communications, 2012, 33(12): 123-132.

图/表 13

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表1

收发角度组合对应的代价函数值"

收发角度组合	Q_ML/dB
${\hat{j}}_{1}$ =?39.929°, $q_{1}$ =?20.471°	14.658
${\hat{j}}_{1}$ =?39.929°, $q_{2}$ =?0.150°	16.635
${\hat{j}}_{1}$ =?39.929°, $q_{3}$ =20.172°	16.684
${\hat{j}}_{2}$ =0.153°, $q_{1}$ =?20.471°	16.241
${\hat{j}}_{2}$ =0.153°, $q_{2}$ =?0.150°	15.333
${\hat{j}}_{2}$ =0.153°, $q_{3}$ =20.172°	16.912
${\hat{j}}_{3}$ =30.875°, $q_{1}$ =?20.471°	16.395
${\hat{j}}_{3}$ =30.875°, $q_{2}$ =?0.150°	16.942
${\hat{j}}_{3}$ =30.875°, $q_{3}$ =20.172°	15.312

表1

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