﻿ 双基地声呐探测距离误差估计
 舰船科学技术  2022, Vol. 44 Issue (23): 128-131    DOI: 10.3404/j.issn.1672-7649.2022.23.025 PDF

Estimation on distance error of bistatic sonar detection
SHU Xiang-lan, SUN Rong-guang, XU Jing-feng
Acoustic Center, Naval Submarine Academy, Qingdao 266199, China
Abstract: Based on the principle of bistatic sonar orientation, we build the model of estimation on distance error considering the measured error of time, direction and acoustic velocity, and detailedly analyse the mean square error influenced by eccentricity of ellipse and orientation. The outcome of simulation indicate that the orientation and position of the targets apart from the sender and receiver impact the estimated precision of distance. On the condition of the distance of sender and receiver fixed, the position more apart from the sender and receiver, the distance estimation lower effect influenced by the orientation of targets, and the value of mean square error of distance keeps in a lower scope. According to the model of estimation presented by this paper, deploy the bistatic sonar reasonably can realized high precision and detected targets in a long distance.
Key words: error     mean square error     sonar deployment     orientation precision
0 引　言

1 双基地声呐定位原理

 图 1 双基地声呐配置 Fig. 1 The deployment for bistatic soar

 ${(X - R)^2} = {R^2} + {S^2} - 2RS\cos \gamma，$ (1)

 $R = \dfrac{{{X^2} - {S^2}}}{{2(X - S\cos \gamma )}}。$ (2)

 ${t_X}{\text{ = }}t + {t_S} ，$ (3)

 $R = \dfrac{{V(t_X^2 - t_S^2)}}{{2({t_X} - {t_S}\cos \gamma )}}。$ (4)
2 双基地声呐距离误差模型

 $R = \dfrac{{{{(U{t_X})}^2} - {{(V{t_S})}^2}}}{{2(U{t_X} - V{t_S}\cos \gamma )}}。$ (5)

 $R = \dfrac{{V(\tau _X^2 - t_S^2)}}{{2({\tau _X} - {t_S}\cos \gamma )}}。$ (6)

 $e = \dfrac{S}{X} = \dfrac{{V{t_S}}}{{U{t_X}}} = \dfrac{{{t_S}}}{{{\tau _X}}}，$ (7)

 图 2 时间测量误差示意图 Fig. 2 Timing estimation error

 图 3 方位测量误差示意图 Fig. 3 Bearing estimation error
 $R = \dfrac{{V{t_S}(1 - {e^2})}}{{2e(1 - e\cos \gamma )}} = \dfrac{S}{{2e(1 - e\cos \gamma )}} ，$ (8)

 $\dfrac{{\Delta R}}{R} = \dfrac{1}{R}\dfrac{{\partial R}}{{\partial V}}\Delta V + \dfrac{1}{R}\dfrac{{\partial R}}{{\partial {\tau _X}}}\Delta {\tau _X} + \dfrac{1}{R}\dfrac{{\partial R}}{{\partial {t_S}}}\Delta {t_S} + \dfrac{1}{R}\dfrac{{\partial R}}{{\partial \gamma }}\Delta \gamma。$ (9)

 $\left\{ {\begin{array}{*{20}{l}} {\begin{array}{*{20}{l}} {\dfrac{1}{R}\dfrac{{\partial R}}{{\partial V}} = \dfrac{1}{V}} ，\\ {\dfrac{1}{R}\dfrac{{\partial R}}{{\partial {\tau _X}}} = \dfrac{{2{\tau _X}}}{{\tau _X^2 - t_S^2}} - \dfrac{1}{{{\tau _X} - {t_S}\cos \gamma }}} ，\\ {\dfrac{1}{R}\dfrac{{\partial R}}{{\partial {t_S}}} = \dfrac{{ - 2{\tau _X}}}{{\tau _X^2 - t_S^2}} + \dfrac{{\cos \gamma }}{{{\tau _X} - {t_S}\cos \gamma }}} ，\end{array}} \\ {\dfrac{1}{R}\dfrac{{\partial R}}{{\partial \gamma }}= -\dfrac{{{t_S}\sin \gamma }}{{{\tau _X} - {t_S}\cos \gamma }}} 。\end{array}} \right.$ (10)

 \begin{aligned}[b] & \dfrac{{\Delta R}}{R} = \dfrac{{\Delta V}}{V} + \left( {\dfrac{{2{\tau _X}}}{{\tau _X^2 - \tau _S^2}} - \dfrac{1}{{{\tau _X} - {t_S}\cos \gamma }}} \right)\Delta {\tau _X} - \\ & \begin{array}{*{20}{l}} {}&{ \left( {\dfrac{{2{t_S}}}{{\tau _X^2 - \tau _S^2}} - \dfrac{{\cos \gamma }}{{{\tau _X} - {t_S}\cos \gamma }}} \right)} \end{array}\Delta {t_S} - \dfrac{{{t_S}\sin \gamma }}{{{\tau _X} - {t_S}\cos \gamma }}\Delta \gamma。\end{aligned} (11)

 \begin{aligned}[b] & \dfrac{{\Delta R}}{R} = \dfrac{{\Delta V}}{V} + \left( {\dfrac{2}{{1 - {e^2}}} - \dfrac{1}{{1 - e\cos \gamma }}} \right)\dfrac{{\Delta {t_X}}}{{{t_X}}}+ \\ & \begin{array}{*{20}{c}} {}&{ \left( {\dfrac{{ - 2{e^2}}}{{1 - {e^2}}} - \dfrac{{e\cos \gamma }}{{1 - e\cos \gamma }}} \right)} \end{array}\dfrac{{\Delta {t_S}}}{{{t_S}}} - \dfrac{{e\sin \gamma }}{{1 - e\cos \gamma }}\Delta \gamma。\end{aligned} (12)

 $\left\{ {\begin{array}{*{20}{l}} {{A_X} = \dfrac{2}{{1 - {e^2}}} - \dfrac{1}{{1 - e\cos \gamma }}} ，\\ {{A_S} = \dfrac{{ - 2{e^2}}}{{1 - {e^2}}} + \dfrac{{e\cos \gamma }}{{1 - e\cos \gamma }}} ，\\ {{A_G} = \dfrac{{ - e\sin \gamma }}{{1 - e\cos \gamma }}} 。\end{array}} \right.$ (13)

 \begin{aligned}[b] & \dfrac{1}{R}(\Delta R)rms = \\ & \sqrt {{{\left( {\dfrac{{\Delta V}}{V}} \right)}^2} + A_X^2{{\left( {\dfrac{{\Delta {t_X}}}{{{t_X}}}} \right)}^2} + A_S^2{{\left( {\dfrac{{\Delta {t_S}}}{{{t_S}}}} \right)}^2} + A_G^2{{(\Delta \gamma )}^2}} \end{aligned} 。 (14)

 图 4 方位角和离心率对误差系数的影响 Fig. 4 The error coefficient affected by bearing and eccentricity

1）离心率越小，即椭圆越扁平， $\left| {{A_X}} \right|$ $\left| {{A_S}} \right|$ $\left| {{A_G}} \right|$ $\gamma$ 的变化越平稳，且 ${A_S}$ $\left| {{A_G}} \right|$ 越接近，而 $\left| {{A_X}} \right|$ 的值相对要大一些。

2）离心率越大，即椭圆越圆满， $\left| {{A_X}} \right|$ $\left| {{A_S}} \right|$ $\left| {{A_G}} \right|$ $\gamma$ 的变化越敏感， $\left| {{A_X}} \right|$ $\left| {{A_S}} \right|$ $\gamma$ 的变化先增大后逐渐平缓最终趋于定值，而 $\left| {{A_G}} \right|$ $\gamma$ 的变化先增大后减小至0。

3）不论离心率取何值，在 $\gamma$ 的绝大部分取值内， $\left| {{A_X}} \right|$ 的值要大于 $\left| {{A_S}} \right|$ $\left| {{A_G}} \right|$ 的值。

3 仿真实例分析

 \begin{aligned}[b] & \dfrac{1}{R}(\Delta R)rms = \\ & \sqrt {{{\left( {\dfrac{{\Delta V}}{V}} \right)}^2} + A_X^2{{\left( {\dfrac{{\Delta {t_X}}}{{{t_X}}}} \right)}^2} + A_S^2{{\left( {\dfrac{{\Delta {t_S}}}{{{t_S}}}} \right)}^2} + A_G^2{{(\Delta \gamma )}^2}}=0.005 \end{aligned}。 (15)

 图 5 目标态势 Fig. 5 The state of target and receiver

 图 6 均方根误差随离心率和方位角的变化 Fig. 6 Mean square error vary with the eccentricity and bearing

4 结　语

 [1] WANG C, YAO Y, WANG Y M, etal. Bistatic soanrt data fusion based on extend kalman filter and its application[C]//Proc of the 2nd International Conference on Computer Engineering and Technology, 2010: 615–617. [2] LIU J B, LLIU W X, ZHOU Q D, et al. Optimal parametric design of bulkhead vibration control for underwater structure[J]. Chinese Journal of Ship Research, 2010, 5(6): 21-25. [3] 阎宜生, 丁玮等 双基地声呐作用距离估计[J]. 西北工业大学学报, 1996, (4): 545–560. [4] 杨丽, 蔡志明 双基地声呐探测范围分析[J]. 兵工学报 2007, 28 (7): 839–843. [5] 张小凤, 赵俊渭, 等 双基地声呐定位精度和算法研究[J]. 系统仿真学报, 2003, 315(10): 1471–1473.