﻿ 基于双圆阵列的轴频场源定位方法
 舰船科学技术  2020, Vol. 42 Issue (11): 110-115    DOI: 10.3404/j.issn.1672-7649.2020.11.022 PDF

Research on source of the shaft-rate electric field location based on double circle array
WANG Hai-guang, WANG Xiang-jun, ZHANG Jian-chun
Naval University of Engineering, Wuhan 430033, China
Abstract: In order to locate the ship through the shaft-rate electric field signal, a positioning method of the source of shaft-rate electric field based on double circular array is proposed. Using horizontal time harmonic electric dipole to model the shaft-rate electric field generated by the ship in shallow sea. The connecting rod with three-component electric field sensors on both ends of it was installed on the seabed. The connecting rod rotates at a certain speed to simulate the circular detection array to measure the shaft-rate signal. The differencial wave of electric field intensity produced by the source of shaft-rate electric field at both ends of the diameter of the two circular arrays at a certain depth and its variation with the position of the source of field are obtained by simulation calculation. The correctness of this rule is verified by ship model experiments. The simulation results show that, in quasi-near-field environment, the location of the source of field can be achieved by measuring the azimuth angle and the position relationship between the two circular arrays. The positioning accuracy is high.
Key words: horizontal electric dipole     double circle array     shaft-rate electric field     differential wave of electric field strength     quasi-near field
0 引　言

1 系统模型

 图 1 理论模型 Fig. 1 Theoretical mode

 $\begin{split} &\left\{ \begin{array}{l} {x_1} = {{r}} \cdot \cos {\theta _1},\\ {y_1} = {{r}} \cdot \sin {\theta _1},\\ {z_1} = D, \end{array} \right.\;\;\;\;\;\;\;\;\;\left\{ \begin{array}{l} x'_1 = {{r}} \cdot \cos ({\text{π}} + {\theta _1}),\\ y'_1 = {{r}} \cdot \sin ({\text{π}} + {\theta _1}),\\ z'_1 = D, \end{array} \right.\\ &\left\{ \begin{array}{l} {x_{\rm{2}}} = a + {{r}} \cdot \cos {\theta _{\rm{2}}},\\ {y_{\rm{2}}} = {{r}} \cdot \sin {\theta _{\rm{2}}},\\ {z_{\rm{2}}} = D, \end{array} \right.\;\;\left\{ \begin{array}{l} x'_2 = a + {{r}} \cdot \cos ({\text{π}} + {\theta _2}),\\ y'_2 = {{r}} \cdot \sin ({\text{π}} + {\theta _2}),\\ z'_2 = D\text{。} \end{array} \right. \end{split}$ (1)
2 电场强度差值波推导

 ${\nabla ^2}{{A}} + {k^2}{{A}} = - \mu {J_s}\text{，}$ (2)
 $E = - j\omega \left[ {{{A}} + \frac{1}{{{k^2}}}\nabla \left( {\nabla \cdot {{A}}} \right)} \right]\text{。}$ (3)

 $\begin{array}{l} {\left. {\left[ {\dfrac{{\nabla \cdot {{{A}}_i}}}{{k_i^2}} - \dfrac{{\nabla \cdot {{{A}}_j}}}{{k_j^2}}} \right]} \right|_{{L_{ij}}}} = 0,\\ {\left. {\left[ {\dfrac{{{{{A}}_{in}}}}{{{\mu _i}}} - \dfrac{{{{{A}}_{jn}}}}{{{\mu _j}}}} \right]} \right|_{{L_{ij}}}} = 0,\\ {\left. {\left[ {\dfrac{\partial }{{\partial z}}\left( {\dfrac{{{{{A}}_{it}}}}{{{\mu _i}}} - \dfrac{{{{{A}}_{jt}}}}{{{\mu _j}}}} \right)} \right]} \right|_{{L_{ij}}}} = 0\text{。} \end{array}$

 $\left\{ \begin{array}{l} {\nabla ^2}{{{A}}_{1z}} + k_1^2{{{A}}_{1z}} = 0\text{，} \\ {\nabla ^2}{{{A}}_{2z}} + k_2^2{{{A}}_{2z}} = - \mu {J_s}\delta (\left| {R - R'} \right|)\text{，} \\ {\nabla ^2}{{{A}}_{3z}} + k_3^2{{{A}}_{3z}} = 0\text{，} \end{array} \right.$ (4)

 $\left\{ \!\!\!\!\!\begin{array}{l} {{{A}}_{1z}} \!=\!\!\! \displaystyle\sum\limits_{m = - \infty }^\infty \!\!{\cos m\varphi \!\!\int_0^\infty \!\!\!\!{{J_m}} } (\rho \xi )\left[ {{a_1}(\xi ,m){e^{z{\upsilon _1}}} \!\!+\!\! {b_1}(\xi ,m){e^{ - z{\upsilon _1}}}} \right]{\rm{d}}\xi \text{，}\\ {{{A}}_{2z}} \!=\!\!\! \displaystyle\sum\limits_{m = - \infty }^\infty\!\! {\cos m\varphi \!\!\int_0^\infty \!\!\!\!{{J_m}} } (\rho \xi )\left[ {{a_2}(\xi ,m){e^{z{\upsilon _1}}} \!\!+\!\! {b_2}(\xi ,m){e^{ - z{\upsilon _1}}}} \right]{\rm{d}}\xi \!+ \\ \qquad\dfrac{{{\mu _2}Il}}{{4{\text{π}} }}\displaystyle\int_0^\infty {\dfrac{\xi }{{{\upsilon _2}}}{J_0}(\rho \xi ){e^{ - {\upsilon _1}\left| {z - {z_0}} \right|}}d\xi } \text{，}\\ {{{A}}_{3z}} \!=\!\!\! \displaystyle\sum\limits_{m = - \infty }^\infty \!\!{\cos m\varphi \!\!\int_0^\infty \!\!\!\!{{J_m}} } (\rho \xi )\left[ {{a_3}(\xi ,m){e^{z{\upsilon _1}}} \!\!+\!\! {b_3}(\xi ,m){e^{ - z{\upsilon _1}}}} \right]{\rm{d}}\xi \text{。}\\ \end{array} \right.$ (5)

 $\begin{split} {{{A}}_{2z}} =& \int_0^\infty {\left[ {{a_2}(\xi ,0){e^{z{\upsilon _1}}} + {b_2}(\xi ,0){e^{ - z{\upsilon _1}}}} \right]{J_0}(\rho \xi ){\rm{d}}\xi } \cdot {e_z} + \\ &\dfrac{{{\mu _2}Il}}{{4{\text{π}} }}\int_0^\infty {\frac{\xi }{{{\upsilon _2}}}{J_0}(\rho \xi ){e^{ - {\upsilon _1}\left| {z - {z_0}} \right|}}{\rm{d}}\xi } \cdot {e_z} \text{。}\\ \end{split}$ (6)

 $\begin{split}& {a_2}(\xi ,0) = \frac{{\mu Il}}{{4{\text{π}} }}\xi \left( {\frac{{k_1^2{\upsilon _2} + k_2^2{\upsilon _1}}}{{k_2^2{\upsilon _2}}}K - \frac{1}{{{\upsilon _2}}}{e^{ - {\upsilon _2}{z_0}}}} \right),\\ &{b_2}(\xi ,0) = \frac{{\mu Il}}{{4{\text{π}} }}\frac{{\xi (k_1^2{\upsilon _2} - k_2^2{\upsilon _1})}}{{k_2^2{\upsilon _2}}}K,\\ &K = \frac{{k_2^2\left[ {H{e^{{\upsilon _1}(D - {z_0})}} - H'{e^{ - {\upsilon _1}(D - {z_0})}}} \right]}}{{(k_1^2{\upsilon _2} + k_2^2{\upsilon _1})H{e^{D{\upsilon _1}}} + (k_1^2{\upsilon _2} - k_2^2{\upsilon _1})H'{e^{ - D{\upsilon _1}}}}},\\ &H = k_2^2{\upsilon _3} + k_3^2{\upsilon _2},H' = k_2^2{\upsilon _3} - k_3^2{\upsilon _2}\text{。} \end{split}$

3 仿真计算 3.1 圆阵列上电场强度差值分布规律

 图 2 圆阵列上的差值波分布规律 Fig. 2 Distribution of differenceial waves on circular arrays

 图 3 幅值最大处位置角变化规律 Fig. 3 Variation of position angle at maximum amplitude

3.2 极限距离产生的原因

 图 4 仿真电场分量信号 Fig. 4 Component signal of simulatedelectric field

 图 5 仿真电场模值信号 Fig. 5 Module signal of simulated electric field

3.3 基于电场强度差值定位的计算

 图 6 系统投影平面图 Fig. 6 Systematic projection plane

 ${x_0} = a\tan {\theta _2}/(\tan {\theta _2} - \tan {\theta _1})\text{，}$ (8)
 ${y_0} = a\tan {\theta _1}\tan {\theta _2}/(\tan {\theta _2} - \tan {\theta _1})\text{。}$ (9)

4 实验验证

 图 7 实测电场模值信号 Fig. 7 The measured signal of simulated electric field

5 结　语

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