CN102914764A

CN102914764A - Time difference positioning method for resisting sampling synchronous error of receiver

Info

Publication number: CN102914764A
Application number: CN2012103800688A
Authority: CN
Inventors: 罗来源; 万群; 文飞; 熊瑾煜
Original assignee: FIFTY SEVENTH RESEARCH INSTITUTE OF CHINESE PEOPLE'S LIBERATION ARMY GENERAL STAFF; University of Electronic Science and Technology of China
Current assignee: FIFTY SEVENTH RESEARCH INSTITUTE OF CHINESE PEOPLE'S LIBERATION ARMY GENERAL STAFF; University of Electronic Science and Technology of China
Priority date: 2012-11-24
Filing date: 2012-11-24
Publication date: 2013-02-06

Abstract

本发明专利主要针对在接收机采样时钟存在同步误差的情况下的高精度到达距离差(RDOA)定位方法。多个接收机采样时钟采样的同步，一般利用全球定位系统(GPS)提供时钟基准，实现高精度同步采样。但由于各个接收机采样时钟输出的脉冲信号的前沿存在误差，使得接收机的采样在GPS时钟脉冲激发后存在一个采样周期的随机误差，导致RDOA定位精度恶化。本发明利用采样时钟同步误差的随机变化规律克服采样时钟同步误差对RDOA定位的影响，显著的提高了RDOA定位精度。The patent of the present invention is mainly aimed at a high-precision distance difference of arrival (RDOA) positioning method when there is a synchronization error in the receiver sampling clock. The synchronization of multiple receiver sampling clock sampling generally uses the Global Positioning System (GPS) to provide a clock reference to achieve high-precision synchronous sampling. However, due to the error in the leading edge of the pulse signal output by each receiver sampling clock, there is a random error of one sampling period in the sampling of the receiver after the GPS clock pulse is excited, which leads to the deterioration of the RDOA positioning accuracy. The invention utilizes the random variation law of the sampling clock synchronization error to overcome the influence of the sampling clock synchronization error on the RDOA positioning, and significantly improves the RDOA positioning accuracy.

Description

A time-difference positioning method against receiver sampling synchronization error

技术领域 technical field

本发明属于被动定位方法，尤其涉及多个接收机采样时钟存在同步误差时的高精度、快速到达距离差定位方法。 The invention belongs to a passive positioning method, in particular to a high-precision and fast arrival distance difference positioning method when there are synchronization errors in sampling clocks of multiple receivers. the

背景技术 Background technique

高精度被动定位在雷达、声纳、卫星和通信等系统的中具有重要的应用价值。到达距离差(Range Difference Of Arrival，RDOA)定位是一种重要的高精度定位方法，在声纳、雷达及其它无线电定位导航领域获得了广泛的应用。现有的RDOA定位方法大致可分为几何交叉法和统计判决法两大类。其中，几何交叉法主要利用RDOA测量值所包含的信号源和接收机之间的几何位置关系，直接计算目标的位置；统计判决法可以利用多余的RDOA测量值，根据测量误差的统计特性，优化定位结果。统计判决法有很多，主要包括KF(Kalman Filtering)、TSE(Taylor Series Expansion)、DAC(Divide And Conquer)、SI(Spheral Interpolation)、SX(Spheral Intersection)、PX(Plane intersection)以及一些改进方法。这些经典定位方法在被动定位中已经获得广泛应用。 High-precision passive positioning has important application value in systems such as radar, sonar, satellite and communication. Range Difference Of Arrival (RDOA) positioning is an important high-precision positioning method, which has been widely used in sonar, radar and other radio positioning and navigation fields. The existing RDOA positioning methods can be roughly divided into two categories: geometric intersection method and statistical judgment method. Among them, the geometric intersection method mainly uses the geometric position relationship between the signal source and the receiver contained in the RDOA measurement value to directly calculate the position of the target; the statistical judgment method can use the redundant RDOA measurement value, according to the statistical characteristics of the measurement error, to optimize positioning results. There are many statistical judgment methods, mainly including KF (Kalman Filtering), TSE (Taylor Series Expansion), DAC (Divide And Conquer), SI (Spheral Interpolation), SX (Spheral Intersection), PX (Plane intersection) and some improved methods. These classic positioning methods have been widely used in passive positioning. the

本发明专利主要针对在接收机采样时钟存在同步误差的情况下的高精度RDOA定位方法。在无源定位系统中为实现多个接收机采样时钟的同步，一般利用全球定位系统(GPS)提供时钟基准，实现高精度同步采样。其基本原理是各个接收机采样时钟经GPS的时钟脉冲激发实现各个接收机采样时钟输出的脉冲信号的前沿与GPS时间同步。但由于各个接收机采样时钟输出的脉冲信号的前沿存在误差，使得接收机的采样在GPS时钟脉冲激发后存在一个采样周期的随机误差，导致RDOA定位精度恶化。因此，有必要发展一种抗接收机采样时钟同步误差的高精度定位方法。 The patent of the present invention is mainly aimed at the high-precision RDOA positioning method under the condition that there is a synchronization error in the sampling clock of the receiver. In order to realize the synchronization of sampling clocks of multiple receivers in a passive positioning system, the Global Positioning System (GPS) is generally used to provide a clock reference to realize high-precision synchronous sampling. The basic principle is that the sampling clocks of each receiver are excited by the clock pulse of GPS to realize the synchronization of the leading edge of the pulse signal output by each receiver sampling clock with GPS time. However, due to the error in the leading edge of the pulse signal output by each receiver sampling clock, there is a random error of one sampling period in the sampling of the receiver after the GPS clock pulse is excited, which leads to the deterioration of the RDOA positioning accuracy. Therefore, it is necessary to develop a high-precision positioning method that is resistant to receiver sampling clock synchronization errors. the

发明内容 Contents of the invention

本发明的目的是针对接收机采样时钟存在同步误差的情况提供一种高精度的RDOA定位方法。 The purpose of the present invention is to provide a high-precision RDOA positioning method for the situation that there is a synchronization error in the sampling clock of the receiver. the

根据该方法，利用实际RDOA定位系统中接收机采样时钟同步误差的特点，首先建立了相应的同步误差模型，然后在估计目标位置时同时估计出采样同步误差，最终通过对消掉采样同步误差来得到高精度定位。 According to this method, using the characteristics of the synchronization error of the receiver sampling clock in the actual RDOA positioning system, the corresponding synchronization error model is first established, and then the sampling synchronization error is estimated at the same time when the target position is estimated, and finally the sampling synchronization error is eliminated by canceling the sampling synchronization error. Get high-precision positioning. the

本发明方法包括如下步骤： The inventive method comprises the steps:

1)根据接收机采样同步误差模型和系统参数(采样率)，构造离散同步误差集。 1) Construct a discrete synchronization error set according to the receiver sampling synchronization error model and system parameters (sampling rate). the

2)取离散同步误差集中的一点在测量距离差中进行对消，然后使用最小二乘法估计出对应的位置参数，并计算估计残差。遍历误差集，得到与离散同步误差集相对应的位置参数估计集和残差集。 2) Take a point in the discrete synchronization error set and cancel it in the measurement distance difference, then use the least square method to estimate the corresponding position parameters, and calculate the estimated residual. The error set is traversed to obtain the position parameter estimation set and the residual set corresponding to the discrete synchronization error set. the

3)搜索残差集，以其中最小残差对应的位置参数估计为最优估计，同时也可得到对应的采样同步误差估计。 3) Search the residual set, take the position parameter estimate corresponding to the minimum residual as the optimal estimate, and at the same time obtain the corresponding sampling synchronization error estimate. the

附图说明 Description of drawings

图1显示了在距离测量方差为σ_l＝1.5米时本发明方法和SI方法的定位误差累计概率曲线。 Fig. 1 shows the positioning error cumulative probability curves of the method of the present invention and the SI method when the distance measurement variance is σ _l =1.5 meters.

图2显示了在距离测量方差为σ₁＝3.0米时本发明方法和SI方法的定位误差累计概率曲线。 Fig. 2 shows the positioning error cumulative probability curves of the method of the present invention and the SI method when the distance measurement variance is σ ₁ =3.0 meters.

图3显示了在距离测量方差为σ₁＝6.0米时本发明方法和SI方法的定位误差累计概率曲线。 Fig. 3 shows the positioning error cumulative probability curves of the method of the present invention and the SI method when the distance measurement variance is σ ₁ =6.0 meters.

具体实施方式 Detailed ways

RDOA定位涉及到多个观测站同时观测同一辐射源信号。以二维定位为例，发射源和观测站位于同一平面内，各个观测站接收发射源辐射的同一信号，计算相对于参考站的RDOA。这里不考虑RDOA的测量过程。 RDOA positioning involves simultaneous observation of the same emitter signal by multiple observatories. Taking two-dimensional positioning as an example, the emission source and the observation station are located in the same plane, each observation station receives the same signal radiated by the emission source, and calculates the RDOA relative to the reference station. The measurement process of RDOA is not considered here. the

假设在直角坐标系中，N个观测站的位置为S_i(x_i，y_i)，i＝1，2，...，N，辐射源位于S(x，y)，定义如下参量： Assuming that in the Cartesian coordinate system, the positions of N observation stations are S _i ( _xi , y _i ), i=1, 2, ..., N, and the radiation source is located at S(x, y), the following parameters are defined:

S_i到原点的距离 The distance from S _i to the origin

${r r}_{i i} = = \sqrt{{x x}_{i i}^{22} + + {y the y}_{i i}^{22}},, i i = = 1,2 1,2 . . . . . .,, N N - - - - - - ((11))$

S到原点的距离 The distance from S to the origin

$r r = = \sqrt{{x x}^{22} + + {y the y}^{22}} - - - - - - ((22))$

S到S_i的距离 The distance from S to S _i

${d d}_{i i} = = \sqrt{{((x x - - {x x}_{i i}))}^{22} + + {((y the y - - {y the y}_{i i}))}^{22}},, i i = = 1,2 1,2 . . . . . .,, N N - - - - - - ((33))$

S_i和S_j之间测得的发射源信号传播距离差 The measured source signal propagation distance difference between S _i and S _j

d_ij＝d_i-d_j i，j＝1，2，...，N，i≠j (4) d _ij = d _i -d _j i, j = 1, 2, ..., N, i≠j (4)

N个观测站共有

个不同距离测量组合，即产生个距离差，其中只有N-1个是独立的，其余距离差由这N-1个距离差完全确定。不妨选择S₁为参考站，计算一组独立的距离差集合，记为d＝(d₁，d₂，...，d_N)^T，如果存在测量误差，距离差测量值为d_i1＝d_i-d₁+n_i1，i＝2，3，...，N，其中n_i1表示S_i和S₁之间的距离差测量误差，写成向量形式N＝(n₂₁，n₃₁，...，n_N1)^T。对于距离差测量误差的分布模型，假设n_i1服从零均值的分布，且n_i1和n_j1相互独立，这里i，j＝2，...，N；i≠j。 A total of N observation stations

a combination of different distance measurements, that yields distance differences, of which only N-1 are independent, and the remaining distance differences are completely determined by these N-1 distance differences. It may be advisable to choose S ₁ as the reference station, and calculate a set of independent distance difference sets, recorded as d=(d ₁ , d ₂ ,...,d _N ) ^T , if there is a measurement error, the measured value of the distance difference is d _i1 = d _i -d ₁ +n _i1 , i=2, 3,..., N, where n _i1 represents the distance difference measurement error between S _i and S ₁ , written in vector form N=(n ₂₁ , n ₃₁ , ..., n _N1 ) ^T . For the distribution model of distance difference measurement error, it is assumed that n _i1 obeys the distribution of zero mean, and n _i1 and n _j1 are independent of each other, where i, j=2,...,N; i≠j.

在多个接收机采样的同步存在误差时，距离差测量值模型可表示为d_i1＝d_i-d₁+n_i1+v_i1，i＝2，3，...，N，其中v_i1表示S_i和S₁之间的距离差测量由同步引起的误差，且v_i1在离散集{-θ，0，θ}中均匀分布，v_i1～U{-θ，0，θ}，θ为一个采样周期内的信号传播距离。 When there are errors in the synchronization of multiple receiver samples, the distance difference measurement value model can be expressed as d _i1 =d _i -d ₁ +n _i1 +v _i1 , i=2, 3, ..., N, where v _i1 Indicates that the distance difference between S _i and S ₁ measures the error caused by synchronization, and v _i1 is uniformly distributed in the discrete set {-θ, 0, θ}, v _i1 ~ U{-θ, 0, θ}, θ is the signal propagation distance in one sampling period.

在定位计算中，将源到参考站的距离d₁作为一个分离参量单独估计，而是将其作为和辐射源坐标相互独立的冗余参量直接作最小二乘估计，在得到d₁和辐射源坐标的初始估计后，再利用它们之间存在的约束关系，构造另一个独立的线性最小二乘问题，以得到较为精确的定位估计。在小误差域，如果已知测量误差的统计特性，还可以进行加权处理。 In the positioning calculation, the distance d ₁ from the source to the reference station is estimated separately as a separate parameter, but it is used as a redundant parameter independent of the coordinates of the radiation source for least squares estimation, and after obtaining d ₁ and the radiation source After the initial estimation of the coordinates, another independent linear least squares problem is constructed by using the constraint relationship between them to obtain a more accurate positioning estimation. In the small error domain, if the statistical characteristics of the measurement error are known, weighting can also be performed.

本发明定位方法的具体实施步骤如下： The specific implementation steps of the positioning method of the present invention are as follows:

1)构造离散同步误差集，由误差v_i1的模型，令 1) Construct a discrete synchronization error set, from the model of error v _i1 , let

$v v = = [\begin{matrix} {v v}_{21 twenty one} \\ {v v}_{3131} \\ . . \\ . . \\ . . \\ {v v}_{N N 11} \end{matrix}]$

则v有M＝3^N-1种组合，令v的集合为I。 Then v has M=3 ^N-1 combinations, let the set of v be I.

2)对任意v∈I，计算d_i1＝d_i1-v_i1，i＝2，3，...，N。引入冗余变量d₁，从(1)到(4)式变换得到线性定位方程 2) For any v∈I, calculate d _i1 =d _i1 -v _i1 , i=2, 3, . . . , N. Introduce the redundant variable d ₁ , transform from (1) to (4) to obtain the linear positioning equation

$\frac{11}{22} \cdot &Center Dot; (({d d}_{i i 11}^{22} - - {r r}_{i i}^{22} + + {r r}_{11}^{22})) = = {- - x x}_{i i 11} x x - - {y the y}_{i i 11} y the y - - {d d}_{i i 11} {d d}_{11},, i i = = 22,, . . . . . .,, N N - - - - - - ((55))$

只考虑小误差情况，写成矩阵形式 Only consider the case of small errors, written in matrix form

δ＝Ap+ε (6) δ＝Ap+ε (6)

其中 in

$A = [\begin{matrix} {- x}_{21} & {- y}_{21} & {- d}_{21} \\ . & . & . \\ . & . & . \\ . & . & . \\ {- x}_{N 1} & {- y}_{N 1} & {- d}_{N 1} \end{matrix}],$ $p = [\begin{matrix} x \\ y \\ d_{1} \end{matrix}],$ $A = [\begin{matrix} {- x}_{twenty one} & {- the y}_{twenty one} & {- d}_{twenty one} \\ . & . & . \\ . & . & . \\ . & . & . \\ {- x}_{N 1} & {- the y}_{N 1} & {- d}_{N 1} \end{matrix}],$ $p = [\begin{matrix} x \\ they \\ d \\ _{1} \end{matrix}],$

x_i1＝x_i-x₁，y_i1＝y_i-y₁，i＝2，...，N； x _i1 =xi _-x ₁ , y _i1 =y _i -y ₁ , i=2, . . . , N;

假设Q_ε＝cov{ε}＝T₁·Q₀·T₁，p的加权最小二乘估计 Assuming Q _ε =cov{ε}=T ₁ ·Q ₀ ·T ₁ , weighted least squares estimation of p

${\overset{^^}{p p}}_{WLS WLS} = = {(({A A}^{T T} {Q Q}_{ϵ ϵ}^{- - 11} A A))}^{- - 11} {A A}^{T T} {Q Q}_{ϵ ϵ}^{- - 11} δ δ - - - - - - ((88))$

估计的方差为 The estimated variance is

${Q Q}_{p p} = = cov cov {{{\overset{^^}{p p}}_{WLS WLS}}} = = {(({A A}^{T T} {T T}_{11}^{- - 11} {Q Q}_{00}^{- - 11} {T T}_{11}^{- - 11} A A))}^{- - 11} - - - - - - ((99))$

由于 $d_{1}^{2} = {(x - x_{1})}^{2} + {(y - y_{1})}^{2},$ 构造线性方程 because $d_{1}^{2} = {(x - x_{1})}^{2} + {(the y - {the y}_{1})}^{2},$ construct linear equation

V＝Hu+ε₁ (10) V＝Hu+ε ₁ (10)

$V V = = [\begin{matrix} {((p p ((11)) - - {x x}_{11}))}^{22} \\ {((p p ((22)) - - {x x}_{22}))}^{22} \\ {p p}^{22} ((33)) \end{matrix}],,$ $H h = = [\begin{matrix} 11 & 00 \\ 00 & 11 \\ 11 & 11 \end{matrix}],,$ $u u = = [\begin{matrix} {((x x - - {x x}_{11}))}^{22} \\ {((y the y - - {y the y}_{11}))}^{22} \end{matrix}],,$ ${ϵ ϵ}_{11} \approx \approx {T T}_{22} \cdot \cdot δp δp = = [\begin{matrix} 22 ((x x - - {x x}_{11})) & 00 & 00 \\ 00 & 22 ((y the y - - {y the y}_{11})) & 00 \\ 00 & 00 & 22 {d d}_{11} \end{matrix}] \cdot &Center Dot; [\begin{matrix} Δ Δ {p p}_{11} \\ Δ Δ {p p}_{22} \\ Δ Δ {p p}_{33} \end{matrix}]$

假设则u的加权最小二乘估计为 suppose Then the weighted least squares estimate of u is

${\overset{^^}{u u}}_{WLS WLS} = = {(({H h}^{T T} {Q Q}_{{ϵ ϵ}_{11}}^{- - 11} H h))}^{- - 11} {H h}^{T T} {Q Q}_{{ϵ ϵ}_{11}}^{- - 11} V V - - - - - - ((1111))$

估计的方差为 The estimated variance is

${Q Q}_{u u} = = cov cov {{Δ Δ {\overset{^^}{u u}}_{WLS WLS}}} = = {(({H h}^{T T} {Q Q}_{{ϵ ϵ}_{11}}^{- - 11} H h))}^{- - 11} = = {(({H h}^{T T} {T T}_{22}^{- - 11} {Q Q}_{p p}^{- - 11} {T T}_{22}^{- - 11} H h))}^{- - 11} - - - - - - ((1212))$

然后利用u和x＝[x_s y_s]^T之间的关系，计算发射源的定位估计及其定位估计误差协方差。 Then, using the relationship between u and x=[x _s y _s ] ^T , the location estimate of the transmitting source and its location estimation error covariance are calculated.

计算e＝d-v-b，其中b＝[b₂₁...b_M1]^T， $b_{m 1} = \sqrt{{(x_{m} - {\hat{x}}_{s})}^{2} + {(y_{m} - {\hat{y}}_{s})}^{2}} - \sqrt{{(x_{1} - {\hat{x}}_{s})}^{2} + {(y_{1} - {\hat{y}}_{s})}^{2}},$

和

由

得到。让v遍历集合I得到M个u的估计组成的集合U和M个e的估计组成的集合E。 Compute e=dvb, where b=[b ₂₁ ...b _M1 ] ^T ,

b_{m 1} = \sqrt{{(x_{m} - {\hat{x}}_{the s})}^{2} + {({the y}_{m} - {\hat{the y}}_{the s})}^{2}} - \sqrt{{(x_{1} - {\hat{x}}_{the s})}^{2} + {({the y}_{1} - {\hat{the y}}_{the s})}^{2}},

and

Depend on

get. Let v traverse the set I to get the set U of M estimates of u and the set E of M estimates of e.

3)对e∈E，搜索使残差代价函数

最小的e对应的

为最终的估计，并得到相应的位置估计

和

3) For e∈E, the search makes the residual cost function

The smallest e corresponds to

as the final estimate and get the corresponding position estimate

and

实验通过对比本发明方法和Chan’s SI方法的定位误差累计概率曲线来评估两种方法的定位性能。实验中使用4个基站，基站位置参数分别为S₁：(-35.9，1.4)km，S₂：(-20.0，0.0)km，S₃：(5.9，2.3)km，S₄：(32.0，7.0)km。定位区域为基站1正前方200～400km、左右-100～100km的矩形区域。把定位区域划分为10km×10km栅格，目标放置在栅格的顶点上(共201×201个目标点)进行实验。采样频率为40MHz/s，采样周期为25ns。测量误差n_l的标准差σ_i分别取1.5米、3.0米和6.0米进行实验。 The experiment evaluates the positioning performance of the two methods by comparing the positioning error cumulative probability curves of the method of the present invention and the Chan's SI method. Four base stations are used in the experiment, and the location parameters of the base stations are S ₁ : (-35.9, 1.4) km, S ₂ : (-20.0, 0.0) km, S ₃ : (5.9, 2.3) km, S ₄ : (32.0, 7.0) km. The positioning area is a rectangular area of 200-400 km directly in front of the base station 1 and -100-100 km left and right. The positioning area is divided into 10km×10km grid, and the target is placed on the apex of the grid (a total of 201×201 target points) for the experiment. The sampling frequency is 40MHz/s, and the sampling period is 25ns. The standard deviation σ _i of the measurement error n _l was taken as 1.5 meters, 3.0 meters and 6.0 meters respectively for experiments.

图1～图3分别示出本发明专利的方法与Chan’s SI方法在σ_i分别取1.5米、3.0米和6.0米情况下的定位误差累计概率曲线。横轴为相对定位误差 Figures 1 to 3 respectively show the positioning error cumulative probability curves of the method of the patent of the present invention and the Chan's SI method when σ _i is 1.5 meters, 3.0 meters and 6.0 meters respectively. The horizontal axis is the relative positioning error

$\frac{\sqrt{{((\overset{^^}{p p} ((11)) - - x x))}^{22} + + {((\overset{^^}{p p} ((22)) - - y the y))}^{22}}}{{d d}_{11}} \times \times 100100 % %$

从图1～图3可以看出，在存在采样同步误差时，本发明专利的方法能明显的提高定位精度。例如，在距离测量的标准差为1.5米时，本发明专利的方法的相对定位误差小于0.5％、1％和1.8％的概率分别为70％、96％和100％，而Chan’s SI方法的相对定位误差小于0.5％、1％和1.8％的概率分别为43％、73％和92％。在时差估计的标准差较大时，本发明专利的方法与Chan’s SI方法相比同样具有明显优势(参见图2和图3)。 It can be seen from FIG. 1 to FIG. 3 that when there is a sampling synchronization error, the method of the patent of the present invention can obviously improve the positioning accuracy. For example, when the standard deviation of the distance measurement is 1.5 meters, the probability that the relative positioning error of the method of the present invention is less than 0.5%, 1% and 1.8% is 70%, 96% and 100%, respectively, while the relative positioning error of the Chan's SI method is The probabilities of positioning errors less than 0.5%, 1% and 1.8% are 43%, 73% and 92%, respectively. When the standard deviation of the time difference estimate is large, the method of the patent of the present invention has obvious advantages compared with the Chan's SI method (see Fig. 2 and Fig. 3). the

虽然已经参考附图对本专利发明的方法以举例方式进行了描述，但是本发明不限于上述这些细节，并且本发明含盖权利要求范围之内的各种变型或改变。 Although the method of the patented invention has been described by way of example with reference to the accompanying drawings, the invention is not limited to such details and the invention encompasses various modifications or changes within the scope of the claims. the

工业应用性 Industrial applicability

可以将本发明提出的定位方法应用于到达距离差定位系统，满足接收机采样时钟存在同步误差的情况下的高精度定位要求。 The locating method proposed by the invention can be applied to a distance-of-arrival locating system to meet the high-precision locating requirement under the condition that there is a synchronization error in the sampling clock of the receiver. the

Claims

1. A time difference location method for anti-receiver sampling synchronization error, characterized in that: utilize the characteristics of receiver sampling clock synchronization error in actual distance difference of arrival (RDOA) positioning system, set up corresponding synchronization error model, estimate target position At the same time, the synchronization error of the sampling clock is estimated, and finally the high-precision positioning is obtained by canceling the synchronization error of the sampling clock. Its specific steps are:

According to the receiver sampling clock synchronization error model and system parameters (sampling rate), a discrete synchronization error set is constructed.

Take a point in the discrete synchronization error set to cancel in the distance difference measurement, then use the least square method to estimate the corresponding position parameters, and calculate the estimated residual. The discrete synchronization error set is traversed to obtain the position parameter estimation set and residual error set corresponding to the discrete synchronization error set.

The residual set is searched, and the position parameter estimate corresponding to the smallest residual is the optimal estimate, and the corresponding sampling clock synchronization error estimate can also be obtained at the same time.

2. the time difference positioning method of anti-receiver sampling synchronization error as claimed in claim 1, is characterized in that: determine N-1 dimension vector v=[v ₂₁ ... v _N1 ] by sampling clock synchronization error v _i1 , wherein N represents the number of receivers, and then from the discrete synchronization error set {-θ, 0, θ} of the sampling clock synchronization error v _i1 , determine the set of all possible N-1-dimensional vector v as the error set I:

I = {v_{1}, v_{2}, . . ., v_{3^{v - 1}}}

v ₁ =[-θ, 0, ..., 0], v ₂ = [0, 0, ..., 0], v ₃ = [θ, 0, ..., 0], v ₄ = [ 0, -θ, ..., 0], ...,

Where θ is the signal propagation distance within a sampling period.

3. the time difference positioning method of anti-receiver sampling synchronization error as claimed in claim 1, is characterized in that: get a vector in the error set I, cancel in the distance difference of arrival measurement, then use least squares method to estimate The corresponding position parameters, and calculate the corresponding residuals; traverse the error set I, and obtain the position parameter estimation set and residual set corresponding to the error set I one-to-one.

4. the time difference positioning method of anti-receiver sampling synchronization error as claimed in claim 1, is characterized in that: determine least squares solution is

p=(A ^T A) ^-1 A ^T δ

where () ^T represents the transpose of a matrix or vector,

A A = = [\begin{matrix} {- - x x}_{21 twenty one} & {- - y the y}_{21 twenty one} & {- - \overset{^^}{d d}}_{21 twenty one} \\ . . & . . & . . \\ . . & . . & . . \\ . . & . . & . . \\ {- - x x}_{N N 11} & {- - y the y}_{N N 11} & - - {\overset{^^}{d d}}_{N N 11} \end{matrix}],,

δ δ = = \frac{11}{22} [\begin{matrix} {\overset{^^}{d d}}_{21 twenty one}^{22} - - {r r}_{22}^{22} + + {r r}_{11}^{22} \\ {\overset{^^}{d d}}_{1313}^{22} - - {r r}_{33}^{22} + + {r r}_{11}^{22} \\ . . \\ . . \\ . . \\ {\overset{^^}{d d}}_{N N 11}^{22} - - {r r}_{N N}^{22} + + {r r}_{11}^{22} \end{matrix}]

x _i1 =xi _-x ₁ , y _i1 =y _i -y ₁ ,

v _i1 is the sampling clock synchronization error, that is, the i-1th element of a vector v in the error set I, d _i1 is the measurement of the distance difference between arrivals, i=2,...,N;

(x _i , y _i ) are the position coordinates of N receivers, i=1,...,N;

Make sure the positional arguments are the first two elements of p

Determine positional parameters

The corresponding residual is ε=(δ-Ap) ^T (δ-Ap), where () ^T represents the transposition of the matrix or vector;

Traverse each vector v in the error set I to determine the corresponding position parameters

and residual ε, all corresponding positional parameters

Form the position parameter set, and all the corresponding residuals ε form the residual set.

5. The time-difference positioning method for resisting synchronization error during receiver sampling as claimed in claim 1, characterized in that: search the residual set, and use the position parameter estimate corresponding to the smallest residual as the final signal source position estimate.