﻿ 基于旋转干涉仪的近场源参数估计算法
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 雷达学报  2015, Vol. 4 Issue (3): 287-294  DOI: 10.12000/JR14146 0

### 引用本文 [复制中英文]

[复制中文]
Ma Jing-tao, Tao Hai-hong, Xie Jian, et al..Rotating interferometer-based algorithm for parameter estimation of near-field source[J]. Journal of Radars, 2015, 4(3): 287-294. DOI: 10.12000/JR14146.
[复制英文]

### 文章历史

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Rotating interferometer-based algorithm for parameter estimation of near-field source
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National Laboratory of Radar Signal Processing Xidian University, Xi’an , 710071, China
Abstract: This paper proposes a novel rotating interferometer-based algorithm for parameter estimation of near-field source. The algorithm exploits the rotation of a single long baseline interferometer and the integration operation of the phase to unwrap the phase ambiguity. Thus the contradiction between the maximal unambiguous angle and the direction finding accuracy in the single baseline interferometer can be efficiently eliminated in the near-field source scenario. The algorithm can obtain the closed form solutions of the elevation and azimuth angles as well as range estimations for near-field source with two receiving sensors. It avoids the construction of high-order cumulant matrices and multi-dimension search. Simultaneously, it alleviates the requirment for multi-baseline channel consistency. Compared with the conventional double long baselines interferometer method, the proposed algorithm achieves higher parameter estimation accuracy and better performance in ambiguity resolution. Moreover, the proposed algorithm enjoys a simple structure, which is easy to be implemented in engineering application. Simulation results demonstrate the effectiveness and validity of the proposed algorithm.
Key words: Near-field source    Rotating interferometer    Two-dimensional angle estimation    Range estimation    Unwrap phase ambiguity
1 引言

2 问题模型 2.1 近场源信号模型

 图 1 阵列结构示意图 Fig.1 Structure of array
 ${X_1}(t) = S(t){{\mathop{\rm e}\nolimits} ^{{\mathop{\rm j}\nolimits} {\tau _1}}} + {n_1}(t)$ (1)

 \begin{aligned} {\tau _1} & = 2π f\Delta t = 2π f\frac{{{y_1} - r}}{{\mathop{\rm c}\nolimits} } \hspace{72pt} \\ & = \frac{{2π }}{\lambda }(\sqrt {{r^2} + {d \hspace{1pt} ^2} - 2rd\sin \psi } - r)\\ & = \frac{{2π r}}{\lambda }\Big( \sqrt {1 + \frac{{{d \hspace{1pt} ^2}}}{{{r^2}}} - \frac{{2d\sin \psi }}{r}} - 1 \Big) \end{aligned} (2)

 $\sqrt {1 + x} \approx \left( {1 + \frac{x}{2} - \frac{{{x^2}}}{8} + \frac{{{x^3}}}{{16}} + ···} \right),\,\left| x \right| < 1$ (3)

 \begin{aligned} {\tau _1} & = \frac{{2π r}}{\lambda }\Bigg( {1 + \frac{1}{2}\left( {\frac{{{d \hspace{1pt} ^2}}}{{{r^2}}} - \frac{{2d\sin \psi }}{r}} \right) } \\ & \quad - {\frac{1}{8}{{\left( {\frac{{{d \hspace{1pt} ^2}}}{{{r^2}}} - \frac{{2d\sin \psi }}{r}} \right)}^2} + ··· - 1} \Bigg)\\ & \approx - 2π \frac{d}{\lambda }\sin \psi + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\cos ^2}\psi \end{aligned} (4)

2.2 近场源测向模糊问题

f = angle (X1(t)/X0(t))

 $\begin{array}{l} \phi = - 2π \frac{{d\sin \psi }}{\lambda } + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\cos ^2}\psi = {\phi _1} + {\phi _2} \end{array}$ (5)

n=sin${\psi}$ ，则相位差可表示为f= $- 2π \frac{d}{\lambda }n + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}(1 - {n^2})$ ，进而可得fn的关系图，如图 2所示。由图 2可知，n=-1时，相位差取最大值fmax = 2pd/λ; n=1时，相位差取最小值fmin = -2pd/λ。

 图 2 近场源相位差与n的关系图 Fig.2 Relationship between phase difference and n

3 用于近场源的旋转干涉仪参数估计算法

 图 3 天线旋转时的阵列模型 Fig.3 Array model of rotating interferometer
3.1 解模糊算法

 ${y_1}' = \sqrt {{d \hspace{1pt} ^2} + {{(r\sin \theta )}^2} - 2dr\sin \theta \cos (\varphi - \omega t)}$ (6)

 $\begin{array}{l} {y_1} = \sqrt {{y_1}{{^\prime }^2} + {{(r\cos \theta )}^2}} \\ {\rm{ = }}\sqrt {{d \hspace{1pt} ^2} + {{(r\sin \theta )}^2} - 2rd\sin \theta \cos (\varphi - \omega t) + {{(r\cos \theta )}^2}} \\ {\rm{ = }}\sqrt {{d \hspace{1pt} ^2} + {r^2} - 2rd\sin \theta \cos (\varphi - \omega t)} \end{array}$ (7)

 \begin{aligned} {\tau _1} & = \frac{{2π }}{\lambda }\left( {\sqrt {{r^2} + {d \hspace{1pt} ^2} - 2rd\sin \theta \cos (\varphi - \omega t)} - r} \right)\\ & = \frac{{2π r}}{\lambda }\left( {\sqrt {1 + \frac{{{d \hspace{1pt} ^2}}}{{{r^2}}} - \frac{{2d\sin \theta \cos (\varphi - \omega t)}}{r}} - 1} \right) \end{aligned} (8)

 \begin{aligned} {\tau _1} & = \frac{{2π r}}{\lambda }\Bigg( {1 + \frac{1}{2}\left( {\frac{{{d \hspace{1pt} ^2}}}{{{r^2}}} - \frac{{2d\sin \theta \cos (\varphi - \omega t)}}{r}} \right) } \\ & \quad - {\frac{1}{8}{{\left( {\frac{{{d \hspace{1pt} ^2}}}{{{r^2}}} - \frac{{2d\sin \theta \cos (\varphi - \omega t)}}{r}} \right)}^2} + ··· - 1} \Bigg)\\ & \approx - 2π \frac{{d\sin \theta }}{\lambda }\cos (\varphi - \omega t) \\ & \quad + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}} \left( {1 - {{\sin }^2}\theta {{\cos }^2}(\varphi - \omega t)} \right)\\ & = {\phi _1}\cos (\varphi - \omega t) + {\phi _2} + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta {\sin ^2}(\varphi - \omega t) \end{aligned} (9)

q一定时，天线的旋转可以使相位差f(t)按照近似余弦规律变化，如图 4中实线所示。但是通过数字鉴相器后相位差被限定在(-p,p)范围内，故鉴相器得到的相位差变化曲线存在跳变现象，如图 4中虚线所示。所以，鉴相器的输出并非完整的相位差，从而不能直接根据鉴相器的输出得到f1f2，需采用数字积分器[9]对鉴相器的输出进行积分。

 图 4 旋转一周过程中的相位差变化曲线 Fig.4 Phase difference curve in one rotation

3.2 近场源参数估计

 $\phi (t) = - 2π \frac{{d\sin \theta }}{\lambda } + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\cos ^2}\theta = {\phi _1} + {\phi _2} = {\phi _{{\rm{min}}}}$ (10)

 $\phi (t) = 2π \frac{{d\sin \theta }}{\lambda } + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\cos ^2}\theta = - {\phi _1} + {\phi _2} = {\phi _{{\rm{max}}}}$ (11)

 $\left. \begin{array}{l} {\phi _{{\rm{max}}}} = - {\phi _1} + {\phi _2}\\ {\phi _{{\rm{min}}}} = {\phi _1} + {\phi _2} \end{array} \right\}$ (12a)
 $\left. \begin{array}{l} \quad \quad \quad {\phi _1} = - \frac{1}{2}({\phi _{{\rm{max}}}} - {\phi _{{\rm{min}}}})\\ \quad \quad \quad {\phi _2} = \frac{1}{2}({\phi _{{\rm{max}}}} + {\phi _{{\rm{min}}}}) \end{array} \right\}$ (12b)

 $\left. \begin{array}{l} {\phi _1} = \frac{{ - 2πd\sin \theta }}{\lambda }\\ {\phi _2} = \frac{{π {d \hspace{1pt} ^2}{{\cos }^2}\theta }}{{\lambda r}} \end{array} \right\}$ (13)

 $\sin \theta = \frac{{{\phi _1}\lambda }}{{ - 2π d}}$ (14a)
 $r = {{\pi {d^2}{{\cos }^2}\theta } \over {{\phi _2}\lambda }}\\ \;\, = {{\pi {d^2}} \over {{\phi _2}\lambda }}\left( {1 - {{\phi _1^2{\lambda ^2}} \over {4{\pi ^2}{d^2}}}} \right)$ (14b)

 $\left. \begin{array}{l} \theta = {\rm{arc}}{\rm{sin}}\Big(\frac{{{\phi _1}\lambda }}{{ - 2π d}}\Big)\\ r = \frac{{π {d \hspace{1pt} ^2}}}{{{\phi _2}\lambda }}\left( {1 - \frac{{\phi _1^2{\lambda ^2}}}{{4{π ^2}{d \hspace{1pt} ^2}}}} \right) \end{array} \right\}$ (15)

 $\phi (0) = {\phi _1}\cos \varphi + {\phi _2} + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta (1 - {\cos ^2}\varphi )$ (16)

 $X(t) + \Delta = \phi (t)$ (17)

 ${\phi _1} = - \frac{1}{2}({X_{{\rm{max}}}} - {X_{{\rm{min}}}})$ (18)

 $X''(t) = \phi ''(t)$ (19)

f(t)求2阶导数，得

 \begin{aligned} \phi ''(t) & = - {\phi _1}{\omega ^2}\cos (\varphi - \omega t) \\ & \quad \,+ 2{\omega ^2}π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta \cos [2(\varphi - \omega t)] \\ & = \frac{{2π d\sin \theta }}{\lambda }{\omega ^2}\cos (\varphi - \omega t) \\ & \quad \,+ 2{\omega ^2}π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta \cdot [2{\cos ^2}(\varphi - \omega t) - 1] \end{aligned} (20)

 图 5 相位差变化曲线的2阶导数 Fig.5 Second derivative of phase difference curve

· ω2) + (2ω2p · (d2 / (λr)) · sin2q)；对于sinq < 0,

cos(j-wt)=-1时，相位差2阶导数取最大值$\phi ''{(t)_{{\rm{max}}}} = - \frac{{2π d\sin \theta }}{\lambda }{\omega ^2} + 2{\omega ^2}π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta$ 。

 $\phi ''{(t)_{{\rm{max}}}} = \frac{{2π d\left| {\sin \theta } \right|}}{\lambda }{\omega ^2} + 2{\omega ^2}π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta$ (21)

 \begin{aligned} X''{(t)_{{\rm{max}}}} & = \phi ''{(t)_{{\rm{max}}}} \\ & = \frac{{2π d\left| {\sin \theta } \right|}}{\lambda }{\omega ^2} + 2{\omega ^2}π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta \end{aligned} (22)

 $r = \frac{{2π {\omega ^2}{d \hspace{1pt} ^2}{{\sin }^2}\theta }}{{\lambda X''{{(t)}_{{\rm{max}}}} - 2π d\left| {\sin \theta } \right|{\omega ^2}}}$ (23)

 ${\phi _2} = \frac{1}{2}({\phi _{{\rm{max}}}} + {\phi _{{\rm{min}}}}) = \frac{{π {d \hspace{1pt} ^2}{{\cos }^2}\theta }}{{\lambda r}}$ (24)

 ${\phi _2}_{{\rm{mod}}} = \frac{1}{2}({X_{{\rm{max}}}} + {X_{{\rm{min}}}})$ (25)

 $\Delta = \phi (t) - X(t) = {\phi _2} - {\phi _{{\rm{2mod}}}}$ (26)

 ${X_{{\rm{up}}}}(t) = X(t) + \Delta = \hat \phi (t)$ (27)

Xup(t)为相位差变化曲线f(t)的估计。进而由f(t)的初值以及f1f2来估计方位角j

 $\phi (0) = {\phi _1}\cos \varphi + {\phi _2} + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta {\sin ^2}\varphi$ (28)

 $\phi (0) = {\phi _1}\cos \varphi + {\phi _2} + π \frac{{{d \hspace{1pt} ^2}}}{{\lambda r}}{\sin ^2}\theta (1 - {\cos ^2}\varphi )$ (29)

4 仿真实验和性能分析

 图 6 双长基线干涉仪阵列模型 Fig.6 Array model of DLBI

 图 7 相位差误差随SNR变化关系图 Fig.7 Relationship of phase error versus SNR

 图 8 俯仰、方位测向以及测距均方根误差随信噪比变化关系图 Fig.8 Elevation,azimuth and distance measurements RMSE versus SNR

 图 9 俯仰、方位测向以及测距均方根误差随快拍数变化 Fig.9 Elevation,azimuth and distance measurements RMSE versus snapshots

 $P = \frac{{{\rm{CUA}}}}{{{\rm{Mt}}}} \times {\rm{100}}\%$ (30)

 图 10 正确解模糊概率随信噪比变化 Fig.10 Correctly solving ambiguity probability versus SNR
5 结束语

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