﻿ 利用集群内测距和对目标测向的协同定位方法
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Cooperative positioning method using distance measurement within a cluster and direction finding of a target
ZHONG Rijin, CHEN Qifeng
School of Aeronautics and Astronautics, Central South University, Changsha 410083, China
Abstract: When the relative position measurement accuracy between multiple aircrafts is significantly higher than the absolute positioning accuracy of the aircraft, a relative measurement is added between the aircrafts, and the coordinated processing of the positioning measurement of each aircraft can improve the absolute positioning accuracy. This article aims at the situation that the cluster aircraft cannot measure the distance to the non-cooperative target but can only find the direction. By introducing the mutual ranging between the cluster aircrafts, the positioning accuracy of the cluster aircraft and the non-cooperative target can be improved. For this incomplete relative measurement situation, a cooperative positioning solution method that can improve the positioning accuracy is presented, including the nonlinear static optimization estimation and the linearized least squares estimation method. This coordinated positioning scheme does not need to perform a complete relative position measurement, but can reasonably decompose the ranging and direction finding in the group relative measurement and reduce the requirements for the configuration of aircraft measurement equipment. The validity of the method is verified by the simulation.
Keywords: ranging    direction finding    cluster    collaborative positioning    static optimization

1 不完整测量与协同定位原理

1.1 不完整测量模型

 图 1 对Target测角的几何模型 Fig. 1 Geometric model for Target angle measurement

 $\left\{ {\begin{array}{*{20}{l}} {{\varphi _i} = {\rm{arctan}}\frac{{{y_t} - {y_i}}}{{{x_t} - {x_i}}}}\\ {{\theta _i} = {\rm{arctan}}\frac{{{z_t} - {z_i}}}{{\sqrt {{{({x_t} - {x_i})}^2} + {{({y_t} - {y_i})}^2}} }}}\\ {{d_{ij}} = \sqrt {{{({x_i} - {x_j})}^2} + {{({y_i} - {y_j})}^2} + {{({z_i} - {z_j})}^2}} }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} i,j = 1,2, \cdots ,n;j \ne i} \end{array}} \right.$ （1）

 $\left[ {\begin{array}{*{20}{l}} {{{\hat \varphi }_i}}\\ {{{\hat \theta }_i}}\\ {{{\hat d}_{ij}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {{\varphi _i}}\\ {{\theta _i}}\\ {{d_{ij}}} \end{array}} \right] + \left[ {\begin{array}{*{20}{l}} {{v_{{\varphi _i}}}}\\ {{v_{{\theta _i}}}}\\ {{v_{{d_{ij}}}}} \end{array}} \right]\;\:i,j = 1,2, \cdots ,n;j \ne i$ （2）

1.2 协同定位原理

 $\left\{ {\begin{array}{*{20}{l}} {{{\tilde x}_i} = {x_i} + {v_{{x_i}}}}\\ {{{\tilde y}_i} = {y_i} + {v_{{y_i}}}}\\ {{{\tilde z}_i} = {z_i} + {v_{{z_i}}}} \end{array}\;\:i = 1,2, \cdots ,n} \right.$ （3）

 $\left\{ {\begin{array}{*{20}{l}} {{x_i} = {x_{\rm{t}}} + \Delta {x_i}}\\ {{y_i} = {y_{\rm{t}}} + \Delta {y_i}\;\:i = 1,2, \cdots ,n}\\ {{z_i} = {z_{\rm{t}}} + \Delta {z_i}} \end{array}} \right.$ （4）

xyz方向的测量相互独立，则可仅考虑x方向的协同定位，测量模型可表示为

 ${\tilde x_i} - \Delta {x_i} = {x_{\rm{t}}} + {v_x}\;\:i = 1,2, \cdots ,n$ （5）

 ${\hat x_t} = \frac{1}{n}\sum\limits_{i = 1}^n {({{\tilde x}_i} - \Delta {x_i})}$ （6）

 $\begin{array}{l} {\rm{Var}}[{{\hat x}_{\rm{t}}}] = \frac{{ {\rm{Var}} [({{\tilde x}_i} - \Delta {x_i})]}}{n} = \frac{{\sigma _x^2}}{n}\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} i = 1,2, \cdots ,n \end{array}$ （7）

 $\left\{ {\begin{array}{*{20}{l}} {{{\hat x}_i} = {{\hat x}_{\rm{t}}} + \Delta {x_i}}\\ { {\rm{Var}} [{{\hat x}_i}] = {\rm{Var}} [{{\hat x}_{\rm{t}}}] = \frac{{\sigma _x^2}}{n}\;\:i = 1,2, \cdots ,n} \end{array}} \right.$ （8）

2 不完整测量下的协同定位算法

2.1 非线性优化求解

1) 非线性优化模型

 \left\{ \begin{align} & {{f}_{d}}=\sum\limits_{i=1}^{n-1}{\sum\limits_{j=i+1}^{n}{\frac{1}{\sigma _{{{d}_{ij}}}^{2}}}}{{[\sqrt{{{({{x}_{i}}-{{x}_{j}})}^{2}}+{{({{y}_{i}}-{{y}_{j}})}^{2}}+{{({{z}_{i}}-{{z}_{j}})}^{2}}}-{{{\hat{d}}}_{ij}}]}^{2}} \\ & {{f}_{\varphi }}=\sum\limits_{i=1}^{n}{\frac{1}{\sigma _{{{\varphi }_{i}}}^{2}}}{{\left( \text{arctan}\frac{{{y}_{\text{t}}}-{{y}_{i}}}{{{x}_{\text{t}}}-{{x}_{i}}}-{{{\hat{\varphi }}}_{i}} \right)}^{2}} \\ & {{f}_{\theta }}=\sum\limits_{i=1}^{n}{\frac{1}{\sigma _{{{\theta }_{i}}}^{2}}}{{\left( \text{arctan}\frac{{{z}_{\text{t}}}-{{z}_{i}}}{\sqrt{{{({{x}_{\text{t}}}-{{x}_{i}})}^{2}}+{{({{y}_{\text{t}}}-{{y}_{i}})}^{2}}}}-{{{\hat{\theta }}}_{i}} \right)}^{2}} \\ & \underset{\begin{smallmatrix} {{x}_{i}},{{y}_{i}},{{z}_{i}},{{x}_{\text{t}}},{{y}_{\text{t}}},{{z}_{\text{t}}} \\ i=1,2,\cdots ,n \end{smallmatrix}}{\mathop{\text{min}}}\,f={{f}_{d}}+{{f}_{\varphi }}+{{f}_{\theta }} \\ \end{align} \right. （9）

2) 初值的选取

 $\left\{ \begin{array}{*{35}{l}} {{y}_{ij\_\text{t0}}}=\frac{-{{y}_{i0}}\text{tan}{{\varphi }_{j}}+{{y}_{j0}}\text{tan}{{\varphi }_{j}}+\text{tan}{{\varphi }_{i}}\text{tan}{{\varphi }_{j}}{{x}_{i0}}-\text{tan}{{\varphi }_{i}}\text{tan}{{\varphi }_{j}}{{x}_{j0}}}{\text{tan}{{\varphi }_{i}}-\text{tan}{{\varphi }_{j}}} \\ {{x}_{ij\_\text{t0}}}=\frac{-{{y}_{i0}}+{{y}_{j0}}+{{x}_{i0}}\text{tan}{{\varphi }_{i}}-{{x}_{j0}}\text{tan}{{\varphi }_{j}}}{\text{tan}{{\varphi }_{i}}-\text{tan}{{\varphi }_{j}}} \\ \end{array} \right.$ （10）

xij_t0yij_t0代入俯仰角θiθj的计算公式便可以计算出非合作目标zij_t0的值:

 $\begin{array}{*{20}{l}} {{z_{i{j_ - }{\rm{t0}}}} = \frac{1}{2}[{z_i} - {\rm{tan}}{\theta _i}\sqrt {{{({x_{{\rm{t0}}}} - {x_i})}^2} + {{({y_{{\rm{t0}}}} - {y_i})}^2}} + }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {z_j} - {\rm{tan}}{\theta _j}\sqrt {{{({x_{{\rm{t0}}}} - {x_j})}^2} + {{({y_{{\rm{t0}}}} - {y_j})}^2}} ]} \end{array}$ （11）

2.2 线性化求解

 $\left[ {\begin{array}{*{20}{l}} {\bar x}\\ {\bar y}\\ {\bar z} \end{array}} \right] = \left[ {\begin{array}{*{20}{l}} {\bar x}\\ {\bar y}\\ {\bar z} \end{array}} \right] + \left[ {\begin{array}{*{20}{l}} {{v_{\bar x}}}\\ {{v_{\bar y}}}\\ {{v_{\bar z}}} \end{array}} \right]$ （12）

 $\left\{ {\begin{array}{*{20}{l}} {\bar x = \frac{1}{n}({x_{10}} + {x_{20}} + \cdots + {x_{n0}})}\\ {\bar y = \frac{1}{n}({y_{10}} + {y_{20}} + \cdots + {y_{n0}})}\\ {\bar z = \frac{1}{n}({z_{10}} + {z_{20}} + \cdots + {z_{n0}})} \end{array}} \right.$ （13）

 $\left[ {\begin{array}{*{20}{c}} {{{\hat \varphi }_i}}\\ {\hat \theta }\\ {{{\hat d}_{ij}}}\\ {\bar {\hat x}}\\ {\bar {\hat y}}\\ {\bar {\hat z}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {{\varphi _i}}\\ {{\theta _i}}\\ {{d_{ij}}}\\ {\bar x}\\ {\bar y}\\ {\bar z} \end{array}} \right] + \left[ {\begin{array}{*{20}{c}} {{v_{{\varphi _i}}}}\\ {{v_{{\theta _i}}}}\\ {{v_{{d_{ij}}}}}\\ {{v_{\bar x}}}\\ {{v_{\bar y}}}\\ {{v_{\bar z}}} \end{array}} \right]\;\:i,j = 1,2, \cdots ,n;j \ne i$ （14）

 $\begin{array}{l} {\varphi _i} = {a_{1i}}{x_i} + {a_{2i}}{y_i} + {a_{3i}}{x_t} + {a_{4i}}{y_t} + {a_{5i}}\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} i = 1,2, \cdots ,n \end{array}$ （15）

 $\left\{ {\begin{array}{*{20}{l}} {{a_{1i}} = \frac{1}{k}({y_{{\rm{t0}}}} - {y_{i0}})}\\ {{a_{2i}} = - \frac{1}{k}({x_{{\rm{t0}}}} - {x_{i0}})}\\ {{a_{3i}} = - \frac{1}{k}({y_{{\rm{t0}}}} - {y_{i0}})}\\ {{a_{4i}} = \frac{1}{k}({x_{{\rm{t0}}}} - {x_{i0}})}\\ {{a_{5i}} = {\rm{arctan}}\frac{{{x_{{\rm{t0}}}} - {x_{i0}}}}{{{y_{{\rm{t0}}}} - {y_{i0}}}}}\\ {k = {{({x_{{\rm{t0}}}} - {x_{i0}})}^2} + {{({y_{{\rm{t0}}}} - {y_{i0}})}^2}} \end{array}} \right.$ （16）

 $\begin{array}{l} {\theta _i} = {b_{1i}}{x_i} + {b_{2i}}{y_i} + {b_{3i}}{z_i} + {b_{4i}}{x_t} + {b_{5i}}{y_t} + {b_{6i}}{z_t} + {b_{7i}}\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} i = 1,2, \cdots ,n \end{array}$ （17）

 $\left\{ \begin{array}{l} \begin{array}{*{20}{l}} {{b_{1i}} = \frac{1}{g}({x_{{\rm{t0}}}} - {x_{i0}})({z_{{\rm{t0}}}} - {z_{i0}})}\\ {{b_{2i}} = \frac{1}{g}({y_{{\rm{t0}}}} - {y_{i0}})({z_{{\rm{t0}}}} - {z_{i0}})} \end{array}\\ \begin{array}{*{20}{l}} {{b_{3i}} = - \frac{1}{g}{{({x_{{\rm{t0}}}} - {x_{i0}})}^2} + {{({y_{{\rm{t0}}}} - {y_{i0}})}^2}}\\ {{b_{4i}} = - \frac{1}{g}({x_{{\rm{t0}}}} - {x_{i0}})({z_{{\rm{t0}}}} - {z_{i0}})} \end{array}\\ \begin{array}{*{20}{l}} {{b_{5i}} = - \frac{1}{g}({y_{{\rm{t0}}}} - {y_{i0}})({z_{{\rm{t0}}}} - {z_{i0}})}\\ {{b_{6i}} = \frac{1}{g}{{({x_{{\rm{t0}}}} - {x_{i0}})}^2} + {{({y_{{\rm{t0}}}} - {y_{i0}})}^2}} \end{array}\\ \begin{array}{*{20}{l}} {{b_{7i}} = {\rm{arctan}}\frac{{{z_{{\rm{t0}}}} - {z_{i0}}}}{{\sqrt {{{({x_{{\rm{t0}}}} - {x_{i0}})}^2} + {{({y_{{\rm{t0}}}} - {y_{i0}})}^2}} }}}\\ {g = [{{({x_{{\rm{t0}}}} - {x_{i0}})}^2} + {{({y_{{\rm{t0}}}} - {y_{i0}})}^2} + }\\ {{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {{({z_{{\rm{t0}}}} - {z_{i0}})}^2}] \times \sqrt {{{({x_{{\rm{t0}}}} - {x_{i0}})}^2} + {{({y_{{\rm{t0}}}} - {y_{i0}})}^2}} } \end{array} \end{array} \right.$ （18）

 $\begin{array}{l} {d_{ij}} = {c_{1i}}{x_i} + {c_{2i}}{y_i} + {c_{3i}}{z_i} + {c_{4i}}{x_j} + {c_{5i}}{y_j} + {c_{6i}}{z_j}\\ {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} i,j = 1,2, \cdots ,n;j \ne i \end{array}$ （19）

 $\left\{ \begin{array}{l} \begin{array}{*{20}{l}} {{c_{1i}} = \frac{1}{l}({x_{i0}} - {x_{j0}})}\\ {{c_{2i}} = \frac{1}{l}({y_{i0}} - {y_{j0}})} \end{array}\\ \begin{array}{*{20}{l}} {{c_{3i}} = \frac{1}{l}({z_{i0}} - {z_{j0}})}\\ {{c_{4i}} = - \frac{1}{l}({x_{i0}} - {x_{j0}})} \end{array}\\ \begin{array}{*{20}{l}} {{c_{5i}} = - \frac{1}{l}({y_{i0}} - {y_{j0}})}\\ {{c_{6i}} = - \frac{1}{l}({z_{i0}} - {z_{j0}})} \end{array}\\ l = \sqrt {{{({x_{i0}} - {x_{j0}})}^2} + {{({y_{i0}} - {y_{j0}})}^2} + {{({z_{i0}} - {z_{j0}})}^2}} \end{array} \right.$ （20）

 $\mathit{\boldsymbol{\hat Z}} = \mathit{\boldsymbol{AX}} + \mathit{\boldsymbol{B}}$ （21）

 $\mathit{\boldsymbol{A}} = \left[ {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&0&0& \cdots &0&0&0&{{a_{31}}}&{{a_{41}}}&0\\ {}&{}&{}&{}& \vdots &{}&{}&{}&{}&{}&{}\\ \cdots &{{a_{1i}}}&{{a_{2i}}}&0& \cdots &0&0&0&{{a_{3i}}}&{{a_{4i}}}&0\\ \cdots &{{b_{1i}}}&{{b_{2i}}}&{{b_{3i}}}& \cdots &0&0&0&{{b_{4i}}}&{{b_{5i}}}&{{b_{6i}}}\\ \cdots &{{c_{1ij}}}&{{c_{2ij}}}&{{c_{3ij}}}& \cdots &{{c_{4ij}}}&{{c_{5ij}}}&{{c_{6ij}}}& \cdots &0&0\\ {}&{}&{}&{}& \vdots &{}&{}&{}&{}&{}&{}\\ \cdots &0&{{C_{1(n - 1)n}}}&{{C_{2(n - 1)n}}}&{{C_{3(n - 1)n}}}&{{C_{4(n - 1)n}}}&{{C_{5(n - 1)n}}}&{{C_{6(n - 1)n}}}&0&0&0\\ {1/n}&0&0&{1/n}& \cdots &{1/n}&0&0&0&0&0\\ 0&{1/n}&0&0&{1/n}& \cdots &{1/n}&0&0&0&0\\ 0&0&{1/n}&0&0&{1/n}& \cdots &{1/n}&0&0&0 \end{array}} \right]$ （22）
 $\mathit{\boldsymbol{B}} = \left[ {\begin{array}{*{20}{l}} {{a_{51}}}&{{b_{71}}}&0& \cdots &0&{{a_{5n}}}&{{b_{7n}}}&0& \cdots &0 \end{array}} \right]$ （23）

A矩阵中存放式(16)、式(18)、式(20)中泰勒展开后X(x1, y1, z1, …xn, yn, zn, xt, yt, zt)各项的系数，B矩阵中存放俯仰角和方位角的泰勒展开后的常数项，即式(16)和式(18)中的常数项。根据最小二乘法由式(21)可以得到

 $\mathit{\boldsymbol{X}} = {(\mathit{\boldsymbol{A}}{\mathit{\boldsymbol{A}}^{\rm{T}}})^{ - 1}}{\mathit{\boldsymbol{A}}^{\rm{T}}}(\mathit{\boldsymbol{\hat Z}} - \mathit{\boldsymbol{B}})$ （24）

3 协同定位仿真结果分析 3.1 精度评估方法

 $\left\{ {\begin{array}{*{20}{l}} { RMSE{ _x} = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {\frac{1}{m}} \sum\limits_{t = 1}^m ( {x_{it}} - {x_{it,{\rm{ true}}}}{\rm{ }}} {)^2}}\\ { RMSE{ _y} = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {\frac{1}{m}} \sum\limits_{t = 1}^m {{{({y_{it}} - {y_{it,{\rm{ true}}}})}^2}} } }\\ { RMSE{ _z} = \sqrt {\frac{1}{n}\sum\limits_{i = 1}^n {\frac{1}{m}} \sum\limits_{t = 1}^t ( {z_{it}} - {z_{it,{\rm{ true}}}}} {)^2}}\\ {i,j = 1,2, \cdots ,n;j \ne i} \end{array}} \right.$ （25）

3.2 仿真测试初始条件

 飞行器 北向/m 东向/m 高度/m 北向速度/(m·s-1) 东向速度/(m·s-1) 天向速度/(m·s-1) Agent 1 1 150 1 150 1 150 -200 -200 -50 Agent 2 1 300 1 200 1 200 -200 -200 -50 Agent 3 1 300 1 300 1 150 -200 -200 -50 Target 1 000 1 000 1 000 -199 -199 -49

 测角精度/(°) 测距精度/m 自身定位精度/m 0.1 0.2 10
3.3 仿真结果与分析

 图 2 非线性优化求解的均方根误差 Fig. 2 Root mean square error of nonlinear optimization solution

 飞行器 均方根误差/m x轴方向 y轴方向 z轴方向 Agent 1 4.186 9 4.400 6 4.226 8 Agent 2 4.335 3 4.400 8 4.216 5 Agent 3 4.401 0 4.412 2 4.242 5 Target 4.340 0 4.246 6 4.412 2

 飞行器 均方根误差/m x轴方向 y轴方向 z轴方向 Agent 1 5.524 0 5.690 6 5.676 9 Agent 2 5.521 1 5.721 1 5.679 3 Agent 3 5.507 5 5.704 6 5.663 3 Target 5.501 7 5.663 7 5.704 6

 图 3 仅测向时求解的Target均方根误差 Fig. 3 Root mean square error of Target when measuring angle only

 飞行器 均方根误差/m x轴方向 y轴方向 z轴方向 Target 17.467 1 17.804 8 17.263 7

1) 仅测角求取非合作目标位置时，直接采用多飞行器低精度位置值进行计算，加上测角精度的影响，非合作目标定位误差大约在17~18 m，并且不能提高飞行器自身定位。

2) 在测向的同时加入飞行器间测距的信息后，采用非线性静态优化求解方法提高精度效果最好，能够降低飞行器自身的定位误差到4.2~4.4 m，并且对于非合作目标的定位精度大概为4.2~4.4 m。

3) 线性化求解方法效果略差于非线性化直接求解，但远高于仅测向时的定位精度，可降低多飞行器的定位误差精度到5.5~5.7 m，同时对于非合作目标的定位精度大概为5.5~5.7 m。但线性优化求解方法在运算速度上更优，计算量小，可以减少飞行器组的载荷压力。

4 结论

1) 本文研究了集群飞行器在仅能对非合作目标测角的情况下，增加飞行器间测距来进行协同定位的融合处理方法。采用非线性静态优化估计和线性化后最小二乘估计2种求解方法，通过仿真表明，该方法可以提高集群飞行器和非合作目标的定位精度。

2) 该方法可用于在编队飞行的多无人机和利用相对测量提高无人机定位精度；也可用于导弹对非合作目标协同攻击中，在导引头仅能对目标测向时，提高对目标的定位精度。

http://dx.doi.org/10.7527/S1000-6893.2019.23768

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#### 文章信息

ZHONG Rijin, CHEN Qifeng

Cooperative positioning method using distance measurement within a cluster and direction finding of a target

Acta Aeronautica et Astronautica Sinica, 2020, 41(S1): 723768.
http://dx.doi.org/10.7527/S1000-6893.2019.23768