﻿ 方位和多普勒频移联合的目标要素估计
 舰船科学技术  2016, Vol. 38 Issue (3): 105-110 PDF

Estimation of target elements joining bearings with Doppler frequency shift of line spectrum
ZHAO Jian-xin, XU Guo-jun, GUO Wu-hong
Navy Submarine Academy, Qingdao 266071, China
Abstract: A method of applying target bearings and Doppler frequency shift of line spectrum to estimate the motion parameters of a constant velocity target is presented in this paper. The method is also valid to target motion analysis without observer maneuver. First, two kinds measurement information of target is linearized and two linear models are constructed. Then using least squared method, two pseudo linear filters are obtained, the arithmetic of target elements estimation is obtained by fusion method. The main advantages of this method comparing with the former results are as follows:1) Two linear estimators are two-dimensional systems, so the observability of them is enhanced. 2) It doesn't necessary to estimate the original signal of line spectrum in Doppler frequency shift estimator. Numerical simulations are made to show the performance of arithmetic under different measurement errors, and the effectiveness of the proposed method is verified by experiment results.
Key words: bearings    Doppler frequency shift    target motion analysis
0 引言

 图 1 观测站和目标相对态势 Fig. 1 Relative scenario of observer and target
1 利用多普勒频移信息估计目标运动速度和目标航向 1.1 线谱频率观测模型

fs 为目标的辐射频率，frtt 时刻的接受频率，c 为声速，VwVm 分别为观测器和目标的速度，V 为目标相对速度，CwCm分别为观测器和目标的航向，$\Delta {{F}_{0t}}\overset{\Delta }{\mathop{=}}\,{{F}_{t}}-{{F}_{0}}$ 为 t 时刻与 t0 时刻的方位差，X0Xt 分别为初始时刻 t0t 时刻的目标舷角，D0为目标初始距离。结合图 1，有接收频率

 ${f_{rt}} = {f_s}\left( {1 + \frac{V}{c}\cos {X_t}} \right)$ (1)

 ${f_{rt}} \! = \!{f_s}\left( {1 \!+ \!\frac{{V\cos {X_0}}}{c}\cos \Delta {F_{0t}} - \frac{{V\sin {X_0}}}{c}\sin \Delta {F_{0t}}} \right)$ (2)

 \begin{align} & {{f}_{ri}}={{f}_{s}}\left( 1+\frac{\cos \Delta {{F}_{0i}}}{c}{{\theta }_{1}}-\frac{\sin \Delta {{F}_{0i}}}{c}{{\theta }_{2}} \right)\text{,} \ & i=0\text{,}1\text{,}2\text{,}\cdots n \ \end{align} (3)

 $\frac{{{f}_{ri}}}{{{f}_{r0}}}=\frac{1+\frac{\cos \Delta {{F}_{0i}}}{c}{{\theta }_{1}}-\frac{\sin \Delta {{F}_{0i}}}{c}{{\theta }_{2}}}{1+\frac{1}{c}{{\theta }_{1}}}\text{,}i=1\text{,}2\text{,}\cdots \text{,}n$ (4)

 \begin{align} & {{y}_{i0}}-1=\frac{\cos \Delta {{F}_{0i}}-{{y}_{i0}}}{c}{{\theta }_{1}}-\frac{\sin \Delta {{F}_{0i}}}{c}{{\theta }_{2}}\text{,} \ & i=1\text{,}2\text{,}\cdots \text{,}n \ \end{align} (5)

 ${{A}^{\text{T}}}A\theta ={{A}^{\text{T}}}y$ (6)

\begin{array}{*{35}{l}} \begin{align} & A=\left( \begin{matrix} \frac{\cos \Delta {{F}_{01}}-{{y}_{10}}}{c} & -\frac{\sin \Delta {{F}_{01}}}{c} \\ [8pt]\frac{\cos \Delta {{F}_{02}}-{{y}_{20}}}{c} & -\frac{\sin \Delta {{F}_{02}}}{c} \\ [8pt]\cdots & \cdots \\ [8pt]\frac{\cos \Delta {{F}_{0n}}-{{y}_{n0}}}{c} & -\frac{\sin \Delta {{F}_{0n}}}{c} \\ \end{matrix} \right)\text{;} \\ & \theta =\left( \begin{matrix} {{\theta }_{1}} \\ {{\theta }_{2}} \\ \end{matrix} \right) \\ \end{align} \\ y=\left( \begin{matrix} {{y}_{10}}-1 \\ {{y}_{20}}-1 \\ \vdots \\ {{y}_{n0}}-1 \\ \end{matrix} \right) \\ \end{array}

 $\hat{\theta }={{\left( {{A}^{\text{T}}}A \right)}^{-1}}{{A}^{\text{T}}}y$ (7)

 $\tan {{X}_{0}}=\frac{{{\theta }_{2}}}{{{\theta }_{1}}}$ (8)

 $V = \sqrt {\theta _1^2 + \theta _2^2}$ (9)

 ${V_m} = \sqrt {{V^2} + V_w^2 - 2V{V_w}\cos ({X_0} + {C_w})} \text{,}$ (10)

 $\sin \left( {{C_m} - {C_w}} \right) = \frac{V}{{{V_m}}}\sin \left( {{C_w} + {X_0}} \right)$ (11)
1.2 可观测性分析

$\begin{array}{*{35}{l}} \left( \cos \Delta {{F}_{01}}-{{y}_{10}} \right)\sin \Delta {{F}_{02}}-\left( \cos \Delta {{F}_{02}}-{{y}_{20}} \right)\sin \Delta {{F}_{01}}= \\ \left( \cos \Delta {{F}_{01}}-\frac{{{f}_{r1}}}{{{f}_{r0}}} \right)\sin \Delta {{F}_{02}}-\left( \cos \Delta {{F}_{02}}-\frac{{{f}_{r2}}}{{{f}_{r0}}} \right)\sin \Delta {{F}_{01}}= \\ \sin \left( \Delta {{F}_{02}}-\Delta {{F}_{01}} \right)+\frac{{{f}_{r2}}}{{{f}_{r0}}}\sin \Delta {{F}_{01}}-\frac{{{f}_{r1}}}{{{f}_{r0}}}\sin \Delta {{F}_{02}}= \\ \sin \left( \Delta {{F}_{02}}-\Delta {{F}_{01}} \right)+ \\ \frac{c\left( \sin \Delta {{F}_{01}}-\sin \Delta {{F}_{02}} \right)+V\cos {{X}_{0}}\sin \left( \Delta {{F}_{01}}-\Delta {{F}_{02}} \right)}{c+V\cos {{X}_{0}}}= \\ \frac{c\left[ \sin \left( \Delta {{F}_{02}}-\Delta {{F}_{01}} \right)+\sin \Delta {{F}_{01}}-\sin \Delta {{F}_{02}} \right]}{c+V\cos {{X}_{0}}}= \\ \frac{4c\sin \frac{\Delta {{F}_{02}}-\Delta {{F}_{01}}}{2}\sin \Delta {{F}_{02}}\sin \Delta {{F}_{01}}}{c+V\cos {{X}_{0}}} \\ \end{array}$

2 利用目标方位信息估计目标相对速度距离比

 $\frac{\sin {{F}_{k}}}{\cos {{F}_{k}}}=\text{tan}{{F}_{k}}=\frac{{{D}_{0}}\sin {{F}_{0}}+{{V}_{x}}({{t}_{k}}-{{t}_{0}})}{{{D}_{0}}\cos {{F}_{0}}+{{V}_{y}}({{t}_{k}}-{{t}_{0}})}\text{,}$ (12)

$\Leftrightarrow \sin ({{F}_{k}}-{{F}_{0}})=({{t}_{k}}-{{t}_{0}})\cos {{F}_{k}}\cdot \frac{{{V}_{x}}}{{{D}_{0}}}-({{t}_{k}}-{{t}_{0}})\sin {{F}_{k}}\cdot \frac{{{V}_{y}}}{{{D}_{0}}}$

 ${{B}^{\text{T}}}B\Theta ={{B}^{\text{T}}}Y$ (13)

{{B}^{\text{T}}}B=\left( \begin{matrix} \begin{align} & \sum\limits_{i=1}^{n}{{{\left( {{t}_{i}}-{{t}_{0}} \right)}^{2}}{{\cos }^{2}}{{F}_{i}}}-\sum\limits_{i=1}^{n}{{{\left( {{t}_{i}}-{{t}_{0}} \right)}^{2}}\cos {{F}_{i}}}\sin {{F}_{i}} \\ & -\sum\limits_{i=1}^{n}{{{\left( {{t}_{i}}-{{t}_{0}} \right)}^{2}}\cos {{F}_{i}}}\sin {{F}_{i}}\sum\limits_{i=1}^{n}{{{\left( {{t}_{i}}-{{t}_{0}} \right)}^{2}}{{\sin }^{2}}{{F}_{i}}} \\ \end{align} \\ \end{matrix} \right)

$\begin{gathered} \Theta = \left( {\begin{array}{*{20}{c}} {\frac{{{V_x}}}{{{D_0}}}\frac{{{V_y}}}{{{D_0}}}} \end{array}} \right){\text{,}} \hfill \\ {B^{\text{T}}}Y = \left( {\begin{array}{*{20}{c}} \begin{gathered} \sum\limits_{i = 1}^n {\left( {{t_i} - {t_0}} \right)\cos {F_i}\sin ({F_i} - {F_0})} - \hfill \\ \sum\limits_{i = 1}^n {\left( {{t_i} - {t_0}} \right)\sin {F_i}\sin ({F_i} - {F_0})} \hfill \\ \end{gathered} \end{array}} \right) \hfill \\ \end{gathered}$

 $\hat{\Theta }={{\left( {{B}^{\text{T}}}B \right)}^{-1}}{{B}^{\text{T}}}Y$ (14)

3 算法步骤

1)利用式(14)，估计第 n 时刻目标相对速度距离比；

2)利用式(7)，估计第 n 时刻参数 θ1θ2

3)利用式(10)和式(11)，估计目标速度和航向；

4)利用步骤 1 和步骤 3 得到的参数，估计和参数间的关系，估计目标初始距离。

4 仿真与试验数据验证

4.1 仿真验证

 图 2 方位误差为0，多普勒频移误差为0.1 Hz的解算结果 Fig. 2 Results with zero error of bearing and 0.1 Hz error of Doppler frequency shift

 图 3 方位误差为0，多普勒频移误差为0.3 Hz的解算结果 Fig. 3 Results with zero error of bearing and 0.3 Hz error of Doppler frequency shift

 图 4 方位误差为0.1°，0.3°，0.5°，多普勒频移误差为0 Hz的解算结果 Fig. 4 Results with 0.1°，0.3°，0.5° error of bearing and 0 Hz error of Doppler frequency shift

 图 5 不同距离下方位误差为0.1°，0.3°，0.5°，多普勒频移误差为0.1 Hz的解算结果 Fig. 5 Results with 0.1°，0.3°，0.5° error of bearing and 0.1 Hz error of Doppler frequency shif
4.2 试验数据验证

 图 6 观测器航向、速度和多普勒线谱频移 Fig. 6 Time series of observer course, velocity and Doppler measurement frequency shift

 图 7 目标航向、速度和初始距离估计 Fig. 7 The estimated results of target course, velocity and original distance
5 结语

1)具有声吶设备的观测平台，可以不需要平台机动，能够得到一定精度的目标要素估计；

2)近距离目标与远距离目标相比，多普勒频移变化大，估计精度高；

3)方位误差和多普勒频移误差越大，估计收敛时间越长，目标航向的估计收敛时间短；

4)线谱频率的测量精度对参数估计结果影响显著，建立更先进的频率估计方法是进一步研究的方向之一。

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