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SCIENTIA SINICA Informationis, Volume 49, Issue 5: 570-584(2019) https://doi.org/10.1360/N112018-00341

Multi-source passive localization via multiple unmanned aerial vehicles

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  • ReceivedDec 30, 2018
  • AcceptedMar 10, 2019
  • PublishedMay 8, 2019

Abstract

Passive source localization via multiple unmanned aerial vehicles (UAVs) is a key technology for the practical application of military reconnaissance. Previous studies primarily focused on the localization of a single source. This study explores the multi-source passive localization problem using time difference of arrival (TDOA) measurements. We formulate the localization problem as a constrained weighted least squares problem. The formulation is an indefinite quadratically constrained quadratic programming problem, which is non-convex and NP-hard. To obtain approximate programming with linear constraints, an iterative constrained weighted least squares (CWLS) algorithm is proposed to perform a linearization procedure on the quadratic equality constraints. Theoretical analysis reveals that the proposed algorithm, if converges, can lead to a global optimal solution of the formulated problem. The results of the Monte Carlo experiment indicate that the proposed algorithm quickly converges in most situations and offers better localization accuracy compared with the previous two-step weighted least squares method.


Funded by

国家自然科学基金(61873217)

西南民族大学中央高校基本科研业务费(2019NQN28)


Supplement

Appendix

定理3.1的证明

将优化问题(22)中的$m$个线性等式约束写成矩阵形式: \begin{equation} {\boldsymbol A}\bar{{\boldsymbol u}}= \bf 0. \tag{32}\end{equation} 根据矩阵广义逆理论 1), 线性方程(32)的通解可以表示为 \begin{equation*} \bar{{\boldsymbol u}} ={{\boldsymbol A}}^\dagger \bf{0}+({\boldsymbol I}-{{\boldsymbol A}}^\dagger{\boldsymbol A}){\boldsymbol \xi} =({\boldsymbol I}-{{\boldsymbol A}}^\dagger{\boldsymbol A}){\boldsymbol \xi},\end{equation*} 其中${\boldsymbol~\xi}~\in~\mathbb{R}^{4m}~$可以为任意向量. 显然${{\boldsymbol~A}}$是一个行满秩的矩阵, 有 \begin{equation*} {{\boldsymbol A}}^\dagger={{\boldsymbol A}}'({\boldsymbol A}{{\boldsymbol A}}')^{-1},\end{equation*} 因此, 线性方程(32)的通解也可表示为 \begin{align} \bar{{\boldsymbol u}} = {\boldsymbol P}{\boldsymbol \xi}. \tag{33} \end{align} 将通解(33)代入优化问题(22)的目标函数, 得到 \begin{equation*}\bar{{\boldsymbol u}}'\bar{{\boldsymbol W}}\bar{{\boldsymbol u}}-2\bar{{\boldsymbol h}}'\bar{{\boldsymbol u}} = {\boldsymbol \xi}'{{\boldsymbol P}}'\bar{{\boldsymbol W}}{\boldsymbol P}{\boldsymbol \xi}-2\bar{{\boldsymbol h}}'{\boldsymbol P}{\boldsymbol \xi},\end{equation*} 其中${\boldsymbol~P}$是一个对称矩阵. 根据矩阵伪逆的基本性质, 有 \begin{equation*}{{\boldsymbol P}} \bar{{\boldsymbol W}}{\boldsymbol P}\big({{\boldsymbol P}} \bar{{\boldsymbol W}}{\boldsymbol P}\big)^\dagger{\boldsymbol P}\bar{{\boldsymbol h}}={\boldsymbol P}\bar{{\boldsymbol h}},\end{equation*} 因此, \begin{eqnarray}\bar{{\boldsymbol u}}'\bar{{\boldsymbol W}}\bar{{\boldsymbol u}}-2\bar{{\boldsymbol h}}'\bar{{\boldsymbol u}} = \big({\boldsymbol \xi}-({\boldsymbol P}\bar{{\boldsymbol W}}{\boldsymbol P})^\dagger{\boldsymbol P}\bar{{\boldsymbol h}}\big)'{{\boldsymbol P}} \bar{{\boldsymbol W}}{\boldsymbol P}\big({\boldsymbol \xi}-({\boldsymbol P}\bar{{\boldsymbol W}}{\boldsymbol P})^\dagger{\boldsymbol P}\bar{{\boldsymbol h}}\big) -\bar{{\boldsymbol h}}'({\boldsymbol P}\bar{{\boldsymbol W}}{\boldsymbol P})^\dagger\bar{{\boldsymbol h}}. \tag{34} \end{eqnarray} 显然, 最小化二次目标函数等价于取${\boldsymbol~\xi}=({\boldsymbol~P}\bar{{\boldsymbol~W}}{\boldsymbol~P})^\dagger{\boldsymbol~P}\bar{{\boldsymbol~h}}.$ 由于${\boldsymbol~P}({\boldsymbol~P}\bar{{\boldsymbol~W}}{\boldsymbol~P})^\dagger=({\boldsymbol~P}\bar{{\boldsymbol~W}}{\boldsymbol~P})^\dagger$且$({\boldsymbol~P}\bar{{\boldsymbol~W}}{\boldsymbol~P})^\dagger$是一个对称矩阵, 有 \begin{equation} {\boldsymbol \xi}=({\boldsymbol P}\bar{{\boldsymbol W}}{\boldsymbol P})^\dagger\bar{{\boldsymbol h}}. \tag{35}\end{equation} 将式(35)代入通解(33)中, 可得式(23)是优化问题(22)的最优解.

Ben-Israel A, Greville T N E. Generalized Inverses: Theory and Applications. 2nd ed. Hoboken: Wiley, 2002.

定理4.1的证明

从方程(28) 和 (29)的条件可知, $\hat{\bar{{\boldsymbol~u}}}$ 一定是优化问题: \begin{equation}\min_{\bar{{\boldsymbol u}}} \bar{{\boldsymbol u}}'\bar{{\boldsymbol W}}\bar{{\boldsymbol u}}-2{\bar{{\boldsymbol h}}}'\bar{{\boldsymbol u}}+\sum_{j=1}^m\lambda_j{\bar{{\boldsymbol u}}}'{\boldsymbol C}_j\bar{{\boldsymbol u}} \tag{36}\end{equation} 的最优解, 那么, 对原问题(19)的任意可行解${\bar{{\boldsymbol~u}}}$都有 \begin{align*}\bar{{\boldsymbol u}}'\bar{{\boldsymbol W}}\bar{{\boldsymbol u}}-2{\bar{{\boldsymbol h}}}'\bar{{\boldsymbol u}} &=\bar{{\boldsymbol u}}'\bar{{\boldsymbol W}}\bar{{\boldsymbol u}}-2{\bar{{\boldsymbol h}}}'\bar{{\boldsymbol u}}+\sum\limits_{j=1}^m\lambda_j{\bar{{\boldsymbol u}}}'{\boldsymbol C}_j\bar{{\boldsymbol u}} \\ & \geq \hat{\bar{{\boldsymbol u}}}'\bar{{\boldsymbol W}}\hat{\bar{{\boldsymbol u}}}-2{\bar{{\boldsymbol h}}}'\hat{\bar{{\boldsymbol u}}}+\sum\limits_{j=1}^m\lambda_j{\hat{\bar{{\boldsymbol u}}}}'{\boldsymbol C}_j\hat{\bar{{\boldsymbol u}}} \\ \nonumber&= \hat{\bar{{\boldsymbol u}}}'\bar{{\boldsymbol W}}\hat{\bar{{\boldsymbol u}}}-2{\bar{{\boldsymbol h}}}'\hat{\bar{{\boldsymbol u}}}, \end{align*} 其中第1个等式成立是由于${\bar{{\boldsymbol~u}}}$是原问题(19)的可行解, 最后一个等式成立是由于方程(30)的条件. 因此, $\hat{\bar{{\boldsymbol~u}}}$一定是优化问题(19)的全局最优解.

定理4.2的证明

首先证明如果序列$\{\hat{{\boldsymbol~u}}^{k},~k=0,1,\ldots\}$收敛到$\hat{{\boldsymbol~u}}$, 那么算法alg1涉及的变量序列$\{\bar{{\boldsymbol~u}}^{k}\}$, $\{{\bar{{\boldsymbol~W}}}^{k}\}$, $\{{\bar{{\boldsymbol~h}}}^{k}\}$ 和 $\{{{\boldsymbol~P}}^{k}\}$ 也是收敛的. 由算法alg1的第$5$步和第$10$步更新策略可得 $$\bar{{\boldsymbol u}}^{k}=2\hat{{\boldsymbol u}}^{k}-\hat{{\boldsymbol u}}^{k-1},$$ 因此, $\lim_{k\rightarrow~\infty}\bar{{\boldsymbol~u}}^k~=2\lim_{k\rightarrow~\infty}\hat{{\boldsymbol~u}}^k-\lim_{k\rightarrow~\infty}\hat{{\boldsymbol~u}}^{k-1} =2\hat{{\boldsymbol~u}}-\hat{{\boldsymbol~u}}=\hat{{\boldsymbol~u}}.$ 根据式(10), (D), (15)和(18)可知, 权矩阵${\boldsymbol~W}$是关于目标位置${\boldsymbol~u}$的连续函数, 因此如果序列$\{\hat{{\boldsymbol~u}}^{k},~k=0,1,\ldots\}$收敛到$\hat{{\boldsymbol~u}}$, 那么其更新的权矩阵序列$\{{\boldsymbol~W}(\hat{{\boldsymbol~u}}^{k}),~k=0,1,\ldots\}$也将收敛, 记其极限为${\boldsymbol~W}(\hat{{\boldsymbol~u}})$. 由于式(20)和(21)可知$\bar{{\boldsymbol~W}}$和$\bar{{\boldsymbol~h}}$均是关于${\boldsymbol~W}$的连续函数, 所以算法alg1更新的序列$\{{\bar{{\boldsymbol~W}}}^{k}\}$和$\{{\bar{{\boldsymbol~h}}}^{k}\}$均收敛, 记其极限分别为 $\bar{{\boldsymbol~W}}(\hat{{\boldsymbol~u}})$和 $\bar{{\boldsymbol~h}}(\hat{{\boldsymbol~u}})$,即 \begin{equation}\bar{{\boldsymbol W}}(\hat{{\boldsymbol u}}) = \lim_{k\rightarrow \infty} {\bar{{\boldsymbol W}}}^{k}, \bar{{\boldsymbol h}}(\hat{{\boldsymbol u}}) = \lim_{k\rightarrow \infty} {\bar{{\boldsymbol h}}}^{k}. \tag{37}\end{equation} 同理, 由式(24)更新的序列$\{{{{\boldsymbol~P}}}^{k}\}$也是收敛的, 记其极限为${{\boldsymbol~P}}(\hat{{\boldsymbol~u}})$.

根据定理3.1,算法alg1中的$\bar{{\boldsymbol~u}}^{k}$是近似CWLS多目标定位问题: \begin{align} \min_{\bar{{\boldsymbol u}}} \bar{{\boldsymbol u}}'\bar{{\boldsymbol W}}^k\bar{{\boldsymbol u}}-2(\bar{{\boldsymbol h}}^k)'\bar{{\boldsymbol u}} \mbox{s.t.} (\hat{{\boldsymbol u}}^k)'{\boldsymbol C}_j\bar{{\boldsymbol u}}=0 (j=1,\ldots,m) \tag{38} \end{align} 的最优解.由于问题(38)是一个凸优化问题, 因此其最优解$\bar{{\boldsymbol~u}}^{k}$必定满足该问题的Karush-Kuhn-Tucher (KKT)条件 2), 即存在$\bar{\lambda}_j^{k}~\in~\mathbb{R}~(j=1,\ldots,m)$, 使得下面的方程组: \begin{eqnarray}&\displaystyle \bar{{\boldsymbol W}}^{k}\bar{{\boldsymbol u}}^{k}+\sum\limits_{j=1}^m \bar{\lambda}_j^{k}{\boldsymbol C}_j\hat{{\boldsymbol u}}^{k} = \bar{{\boldsymbol h}}^{k}, \tag{39} \\ &\displaystyle(\hat{{\boldsymbol u}}^{k})'{\boldsymbol C}_j\bar{{\boldsymbol u}}^{k} =0 (j=1,\ldots,m) \tag{40} \end{eqnarray} 成立. 根据权矩阵的构造式(10), (D), (15)和(20)可知, 算法alg1中的矩阵$\bar{{\boldsymbol~W}}^{k}$在每次迭代中都是正定矩阵, 因此由(39)式可得 \begin{equation}\bar{{\boldsymbol u}}^{k}= (\bar{{\boldsymbol W}}^{k})^{-1} \left(\bar{{\boldsymbol h}}^{k}- \sum_{j=1}^m \bar{\lambda}_j^{k}{\boldsymbol C}_j\hat{{\boldsymbol u}}^{k}\right). \tag{41}\end{equation} 将(41)式代入(40)中, 有 \begin{equation}{\boldsymbol a}_j^{k}\bar{{\boldsymbol h}}^{k}= \sum_{j=1}^m {\boldsymbol a}_j^{k}{\boldsymbol C}_j' \hat{{\boldsymbol u}}^{k} \bar{\lambda}_j^{k} (j=1,\ldots,m), \tag{42}\end{equation} 其中${\boldsymbol~a}_j^{k}=(\hat{{\boldsymbol~u}}^{k})'{\boldsymbol~C}_j(\bar{{\boldsymbol~W}}^{k})^{-1}~.$ 根据(31)和(37),可得 \begin{equation}{\boldsymbol a}_j \triangleq \lim_{k\rightarrow \infty} {\boldsymbol a}_j^{k}= \hat{{\boldsymbol u}}'{\boldsymbol C}_j(\bar{{\boldsymbol W}}(\hat{{\boldsymbol u}}))^{-1} (j=1,\ldots,m), \tag{43}\end{equation} 因此, 方程组(42)的解$(\bar{\lambda}_1^{k},\ldots,\bar{\lambda}_m^{k})$也是收敛的, 记其极限为$(\hat{\lambda}_1,\ldots,\hat{\lambda}_m).$

现在对方程(39)两边取极限, 可得 \begin{equation}\bar{{\boldsymbol W}}(\hat{{\boldsymbol u}}) \hat{{\boldsymbol u}}+\sum_{j=1}^m \hat{\lambda}_j{\boldsymbol C}_j\hat{{\boldsymbol u}} = \bar{{\boldsymbol h}}(\hat{{\boldsymbol u}}). \tag{44}\end{equation} 此外, 对(40)式取极限, 有对$\forall~j=1,\ldots,m$, \begin{equation}\lim_{k\rightarrow \infty} (\hat{{\boldsymbol u}}^{k})'{\boldsymbol C}_j\bar{{\boldsymbol u}}^{k}=\hat{{\boldsymbol u}}'{\boldsymbol C}_j\hat{{\boldsymbol u}}=0. \tag{45}\end{equation}

对比原问题(19)全局最优解的充分性条件(28)$\sim$(30), 还需验证$\bar{{\boldsymbol~W}}(\hat{{\boldsymbol~u}})+\sum_{j=1}^m~\hat{\lambda}_j{\boldsymbol~C}_j~\succeq~\bf~0.$ 根据定理3.1, 算法alg1中的$\bar{{\boldsymbol~u}}^{k}$需要满足 \begin{equation}({\boldsymbol P}^k\bar{{\boldsymbol W}}^k{\boldsymbol P}^k) \bar{{\boldsymbol u}}^{k}= \bar{{\boldsymbol h}}^k, \tag{46}\end{equation} 其中${\boldsymbol~P}^k={\boldsymbol~I}-{{\boldsymbol~A}^k}^\dagger{\boldsymbol~A}^k$, ${\boldsymbol~A}^k=[(\hat{{\boldsymbol~u}}^k)'{\boldsymbol~C}_1;~\ldots;~(\hat{{\boldsymbol~u}}^k)'{\boldsymbol~C}_m].$ 显然矩阵${\boldsymbol~A}^k$是行满秩的, 因此 \begin{equation}\lim_{k\rightarrow \infty} {{\boldsymbol A}^k}^\dagger = {{\boldsymbol A}}(\hat{{\boldsymbol u}})^\dagger, \tag{47}\end{equation} 其中${\boldsymbol~A}(\hat{{\boldsymbol~u}})=[\hat{{\boldsymbol~u}}'{\boldsymbol~C}_1;~\ldots;~\hat{{\boldsymbol~u}}'{\boldsymbol~C}_m].$ 对方程(46)两边取极限, 可得 \begin{equation}\big({\boldsymbol P} (\hat{{\boldsymbol u}}) \bar{{\boldsymbol W}}(\hat{{\boldsymbol u}}){\boldsymbol P}(\hat{{\boldsymbol u}})\big) \hat{{\boldsymbol u}}= \bar{{\boldsymbol h}}(\hat{{\boldsymbol u}}), \tag{48}\end{equation} 其中${\boldsymbol~P}~(\hat{{\boldsymbol~u}})={\boldsymbol~I}-{{\boldsymbol~A}(\hat{{\boldsymbol~u}})}^\dagger{\boldsymbol~A}(\hat{{\boldsymbol~u}}).$ 对比(44)和(48)式, 最终可得 \begin{equation}\bar{{\boldsymbol W}}(\hat{{\boldsymbol u}})+\sum_{j=1}^m \hat{\lambda}_j{\boldsymbol C}_j={\boldsymbol P} (\hat{{\boldsymbol u}}) \bar{{\boldsymbol W}}(\hat{{\boldsymbol u}}){\boldsymbol P}(\hat{{\boldsymbol u}}) \succeq \bf 0. \tag{49}\end{equation} 最终, 从式(44), (45)和(49)可以看出, $(\hat{{\boldsymbol~u}},\hat{\lambda}_1,\ldots,\hat{\lambda}_m~)$ 满足优化问题(19)全局最优解的充分性条件(28)$\sim$(30). 因此, 由算法alg1得到的近似解序列的极限一定是原问题(19)的全局最优解.

Boyd S, Vandenberghe L. Convex Optimization. Cambridge: Cambridge University Press, 2004.


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  • Figure 1

    (Color online) Comparison of the localization accuracy in scenario 1 when fixing $\sigma_s$ and varying $\sigma_j$

  • Figure 2

    (Color online) Comparison of the localization accuracy in scenario 1 when fixing $\sigma_j$ and varying $\sigma_s$

  • Figure 3

    (Color online) Comparison of the localization accuracy of source 1 in scenario 2

  • Figure 4

    (Color online) Comparison of the localization accuracy of source 2 in scenario 2

  • Table 1   True positions of UAVs in scenario 1 (m)
    UAV number $i$ $x_i^0$ $y_i^0$ $z_i^0$
    1 300 100 150
    2 400 150 100
    3 300 500 200
    4 350 200 100
    5 $-$100 $-$100 $-$100
    6 200 $-$300 $-$200
  •   

    Algorithm 1 Recursive CWLS multi-source localization algorithm

    Input${\boldsymbol~s}$, $r_{j,i1}~(j=1,\ldots,m,~i=1,\ldots,n)$, ${\boldsymbol~Q}_s$, ${\boldsymbol~Q}_t$;

    Algorithm process:

    Initialize: $k=0$, $\hat{{ u}}^k=({ G}'{G})^{-1}{G}'({ h}-{G}\bar{{ s}}_1)$;

    Update: update $\bar{{\boldsymbol~W}}^k=\bar{{\boldsymbol~W}}(\hat{{\boldsymbol~u}}^k)$ and $\bar{{\boldsymbol~h}}^k=\bar{{\boldsymbol~h}}(\hat{{\boldsymbol~u}}^k)$ following (20) and (21), and then update ${\boldsymbol~P}^k={\boldsymbol~P}(\hat{{\boldsymbol~u}}^k)$ following (24) and (25);

    Calculate: $\bar{{\boldsymbol~u}}^k=({\boldsymbol~P}^k\bar{{\boldsymbol~W}}^k{\boldsymbol~P}^k)^\dagger\bar{{\boldsymbol~h}}^k$;

    $k=k+1$;

    Update: $\hat{{\boldsymbol~u}}^k=(\hat{{\boldsymbol~u}}^{k-1}+\bar{{\boldsymbol~u}}^{k-1})/2$;

    while $\frac{\|\hat{{\boldsymbol~u}}^{k}-\hat{{\boldsymbol~u}}^{k-1}\|}{\|\hat{{\boldsymbol~u}}^{k}\|}~\leq~\delta$ do

    Update: $\bar{{\boldsymbol~W}}^k=\bar{{\boldsymbol~W}}(\hat{{\boldsymbol~u}}^k)$, $\bar{{\boldsymbol~h}}^k=\bar{{\boldsymbol~h}}(\hat{{\boldsymbol~u}}^k)$, ${\boldsymbol~P}^k={\boldsymbol~P}(\hat{{\boldsymbol~u}}^k)$;

    Calculate: $\bar{{\boldsymbol~u}}^k=({\boldsymbol~P}^k\bar{{\boldsymbol~W}}^k{\boldsymbol~P}^k)^\dagger\bar{{\boldsymbol~h}}^k$;

    $k=k+1$;

    Update: $\hat{{\boldsymbol~u}}^k=(\hat{{\boldsymbol~u}}^{k-1}+\bar{{\boldsymbol~u}}^{k-1})/2$;

    end while

    Output:$\hat{{ u}}^k$, $k.$
  • Table 2   The true positions of UAVs in scenario 2 (m)
    UAV number $i$ $x_i^0$ $y_i^0$ $z_i^0$
    1 510 $-$480 30
    2 510 520 30
    3 $-$490 $-$480 30
    4 $-$490 520 30
    5 10 20 $30+500\sqrt{2}$

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