CN103716879A - Novel wireless positioning method by adopting distance geometry under NLOS environment - Google Patents

Abstract

The invention relates to a novel wireless positioning method by adopting distance geometry under the NLOS environment. On the basis of TOA measuring values, the method provides two kinds of algorithms based on the distance geometry theory under the honeycomb cell environment on two-dimensional space, and the two algorithms include the filtering algorithm and the constraint algorithm. Firstly, the distance geometry relationships between base stations and between the base stations and a mobile station are utilized for constructing a constraint function, the filtering algorithm conducts position estimation by selecting a set of measuring values with the relatively small error from a plurality of sets of measuring values, and the constraint algorithm conducts nonlinearly constrained optimization on the obtained measuring values to a certain extent and then conducts position estimation. It is guaranteed that under the situation that non-sight-distance measuring values are not identified and removed in advance, the non-sight-distance error can be effectively restrained, and the positioning accuracy can be improved.

Description

Under NLOS environment, adopt the wireless location new method of geometric distance

Technical field

The present invention relates to the Cellular Networks location technology under a kind of nlos environment, particularly relate to the wireless location new method that adopts geometric distance under NLOS environment.

Background technology

In Cellular Networks location technology now, TOA model orientation algorithm is a kind of important location algorithm, a plurality of setting circle intersection points of demand solution, but under nlos environment, position error is very large; The method of conventional elimination non line of sight error is to differentiate null method, it certainly will will carry out a large amount of data samplings and data processing is set up required experience database in early stage, by these prior informations, judge whether the measurement data obtaining comprises non line of sight error, and this can cause great amount of calculation.In view of this consideration, a kind of TOA location algorithm based on geometric distance is proposed, the Blumenthal embedding theorems that it be take in distance geometry are got up as Foundation, do not need to differentiate in advance whether measurement data comprises NLOS error, but be optimized according to geometrical-restriction relation value of adjusting the distance of distance between travelling carriage and base station.Existing relevant research has obtained certain result according to geometric distance theory, but also there is no relevant research in Cellular Networks environment.

Summary of the invention

For above the deficiencies in the prior art, the present invention proposes to adopt under a kind of NLOS environment the wireless location new method of geometric distance, suppresses non line of sight error, improves positioning precision to solve under a kind of nlos environment.

Object of the present invention is achieved through the following technical solutions:

Under NLOS environment, adopt the wireless location new method of geometric distance, this new location method comprises geometric distance filtering method DGF and geometric distance constrained procedure DGC,

The step of described DGF algorithm is as follows:

101. travelling carriages obtain the many groups measured value from base station, and each group measured value is the signal propagation time that comes from different base station that travelling carriage receives at synchronization, is designated as (t ₁, t ₂, t ₃, t ₄), wherein subscript represents different base stations;

102. calculate according to following formula each group measured value,, for each group measured value, all can obtain a calculated value, choose in all measured values and meet the following conditions and make one group of calculating formula value minimum:

For the signal that can receive four base stations:

$\min ({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-t_2^2+t_3^2-t_4^2-{4a}^2);$

For the signal that can receive three base stations:

$\min ({\mathrm{<figure-callout\; id="1"\; label="p"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">p</figure-callout>}}_1^2+p_2^2+p_3^2-v^4({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_2^2+{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_3^2+t_2^2{t}_3^2)+4a^2),$

Wherein ${\mathrm{<figure-callout\; id="1"\; label="p"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">p</figure-callout>}}_1=v^2{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-{2a}^2,p_2=v^2{t}_2^2-{2a}^2,p_3=v^2{t}_3^2-2a^2,$ A is half of adjacent base station distance, and v is transmission of wireless signals speed;

103. solve one group of following formula of measured value substitution choosing in step 102, and the value of (x, the y) of acquisition is the position coordinates of travelling carriage under the coordinate system of setting:

Four base stations: $\left\{\begin{array}{c}x=\frac{\sqrt{3}{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-3t_2^2-t_3^2+3t_4^2)}{24a}+\frac{\sqrt{3}a}{2}\\ y=\frac{v^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2+t_2^2-t_3^2-t_4^2)}{8a}+\frac{3a}{2}\end{array}\right.$

Three base stations: $\left\{\begin{array}{c}x=\frac{\sqrt{3}{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-2t_2^2+t_3^2)}{12a}+\frac{\sqrt{3}}{3}a\\ y=\frac{v^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-t_3^2)}{4a}+a\end{array}\right.;$

The step of described DGC method is as follows:

201. travelling carriages obtain the measured value from base station, and the signal propagation time that comes from different base station for travelling carriage receives at synchronization, is designated as (t _m1, t _m2, t _m3, t _m4), wherein subscript represents different base stations;

202. set unknown parameter (ε ₁, ε ₂..., ε _i), the base station number of i for participating in calculating, calculates following nonlinear optimal problem, is met one group of (ε of this nonlinear optimal problem ₁, ε ₂..., ε _i) numerical value, wherein a is half of adjacent base station distance, v is transmission of wireless signals speed:

For the signal that can receive four base stations:

$\min J=\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}_i^2$

s.t.

$\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_1-{\mathrm{\&epsiv;}}_2+{\mathrm{\&epsiv;}}_3-{\mathrm{\&epsiv;}}_4-{\mathrm{<figure-callout\; id="1"\; label="t\; m"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_m^1+t_{m2}^2-t_{m3}^2+t_{m4}^2+4a^2=0\\ {\mathrm{\&epsiv;}}_i>0,i=\mathrm{0,1,2,3}\end{array}\right.$

For the signal that can receive three base stations:

$\min J=\underset{i=1}{\overset{3}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}^2$

s.t.

$\left\{\begin{array}{c}{\mathrm{\&epsiv;}}^T\mathrm{A\&epsiv;}+{\mathrm{\&epsiv;}}^{T}b+c=0\\ {\mathrm{\&epsiv;}}_i>0,i=\mathrm{0,1,2,3}\end{array}\right.$

ε=(ε wherein ₁, ε ₂, ε ₃), $A=\left(\begin{array}{ccc}{8a}^2& -4a^2& -4a^2\\ -4a^2& 8a^2& -4a^2\\ -4a^2& -{4a}^2& 8a^2\end{array}\right),b=\left(\begin{array}{ccc}-16a^2& 8a^2& {8a}^2\\ 8a^2& -16a^2& 8a^2\\ 8a^2& 8a^2& -16a^2\end{array}\right)\left(\begin{array}{c}{v}^2{\mathrm{<figure-callout\; id="1"\; label="t\; m"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_m^1\\ v^2{t}_{m2}^2\\ v^2{t}_{m3}^2\end{array}\right)+\left(\begin{array}{c}32a^2\\ 32a^2\\ 32a^2\end{array}\right),$

$c=128a^6+8a^2{v}^4({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^4+t_2^4+t_3^4-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_2^2-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_3^2-t_2^2{t}_3^2)-32a^4{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2+t_2^2+t_3^2);$

203. (the ε that step 202 is obtained ₁, ε ₂..., ε _i) numerical value substitution following formula, try to achieve travelling carriage after optimization to the distance parameter value of each base station:

$d_i=v\sqrt{t_{\mathrm{mi}}^2-\mathrm{\&epsiv;}}$

204. least square methods solve: X=(A ^ta) ^-1a ^ty, the vectorial X calculating is a bivector, is the position coordinates of travelling carriage under the coordinate system of setting,

Wherein

$A=-2\left(\begin{array}{cc}(x_1-x_k)& (y_1-y_k)\\ (x_2-x_k)& (y_2-y_k)\\ L& L\\ (x_{k-1}-x_k)& (y_{k-1}-y_k)\end{array}\right),Y=\left(\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">d</figure-callout>}}_1^2-d_k^2-x_1^2+x_k^2-{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">y</figure-callout>}}_1^2+y_k^2\\ d_2^2-d_k^2-x_2^2+x_k^2-y_2^2+y_k^2\\ L\\ d_{k-1}^2-d_k^2-x_{k-1}^2+x_k^2-y_{k-1}^2+y_k^2\end{array}\right)$

(x _i, y _i) be the coordinate figure of base station i, d _ithe travelling carriage obtaining for step 203 is to the distance parameter value of base station i.

The invention has the advantages that:

The present invention can, guaranteeing in the situation that do not need to carry out in advance discriminating and the elimination of non line of sight measured value, effectively to suppress non line of sight error, improve positioning precision.The present invention has deposited at cellular network base stations cloth, has stronger practicality.

Accompanying drawing explanation

Fig. 1: non line of sight error location schematic diagram;

Fig. 2: location, cellular cell schematic diagram;

Fig. 3: four architecture simulation results;

Fig. 4: three architecture simulation results;

Fig. 5: RMSE simulation result under four base station different N LOS errors;

Fig. 6: RMSE simulation result under three base station different N LOS errors;

Fig. 7: three base stations and four base station simulation result comparisons;

Fig. 8: DGF algorithm flow chart;

Fig. 9: DGC algorithm flow chart.

Embodiment

The present invention adopts signal transmission universal model, consider that between travelling carriage (MS) and base station (BS), barrier stops (being non line of sight error) such as caused signal reflex, refractions, the measured value measuring is so compared with free from error accurate measured value and is had a larger deviation, is shown below:

t _m=t+ε (1)

Wherein, t _mfor actual measured value, t is measured value under view distance environment, and ε is non line of sight error, is a larger positive deviate.Traditional targeting scheme is due to the impact of non line of sight error in the parameter of using, cause the reduction of positioning precision, as shown in Figure 2, the measured value that travelling carriage obtains from 3 base stations is due to the impact of non line of sight error, cause positioning result not converge, there is deviation with actual position.This programme utilizes geometric distance correlation theory in the many groups measured value obtaining, and finds out one group of one group of measured value approaching most under view distance environment, and deviation is minimum, and it is positioned as locator data.

The Blumenthal embedding theorems that the present invention be take in distance geometry are got up as Foundation, do not need to differentiate in advance whether measurement data comprises non line of sight error, but be optimized according to geometrical-restriction relation value of adjusting the distance of distance between travelling carriage and base station, wherein geometric distance bounding theory is as follows:

If S is r dimension theorem in Euclid space E _rin orderly point set, S={P ₁, P ₂, LP _n, note d _ijfor P _i, P _jspacing, with d _ijthe matrix forming for element is E _rin the orderly distance matrix on point set S, and have

d _ij=d _ji,d _ij>0(i≠j),d _ii=0 (2)

Should be noted, not all that meet above-mentioned condition is all the fixed E of dimension simultaneously _rin distance matrix.

The matrix that meets above-mentioned condition is " semi-metric matrix ", if be just in time E _rthe distance matrix of upper one orderly point set, claims it to embed E _r; If can embed E _rbut can not embed E _r-1, claim it to E _rembedding can not be degenerated.

Blumenthal embedding theorems ^[6]to a semi-metric matrix, can not degenerate and embed E _rprovided necessary and sufficient condition.Double metric matrix D and in order point set S={P ₁, P ₂, L, P _n, can on S, define binary function d (P _i, P _j)=di _j(1≤i, j≤n) is P _i, P _jbetween press the semi-metric of D definition.If

and it is constant that in S', element retains precedence, i.e. S'={P _i, P _j, L, P _k, 1≤i≤j≤L≤k≤n, the semi-metric matrix on S' is

$D^{\&prime;}\left(\begin{array}{cccc}{d}_{\mathrm{ii}}& d_{\mathrm{ij}}& L& d_{\mathrm{ik}}\\ d_{\mathrm{ji}}& d_{\mathrm{jj}}& L& d_{\mathrm{ik}}\\ M& M& & M\\ d_{\mathrm{ki}}& d_{\mathrm{kj}}& L& d_{\mathrm{kk}}\end{array}\right){\stackrel{\&OverBar;}{D}}^{\&prime;}=\left(\begin{array}{ccccc}{d}_{\mathrm{ii}}^2& d_{\mathrm{ij}}^2& L& d_{\mathrm{ik}}^2& 1\\ d_{\mathrm{ji}}^2& d_{\mathrm{jj}}^2& L& d_{\mathrm{jk}}^2& 1\\ M& M& & M& M\\ d_{\mathrm{ki}}^2& d_{\mathrm{kj}}^2& L& d_{\mathrm{kk}}^2& 1\\ 1& 1& \mathrm{<figure-callout\; id="1"\; label="L"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">L</figure-callout>}& 1& 0\end{array}\right)---\left(3\right)$

Its square of edged battle array is

and will

determinant is designated as Δ (i, j, L, k).

Theorem 1:Blumenthal theorem: establishing D is one n * n semi-metric matrix, it is not degenerated and embeds E _rnecessary and sufficient condition be, D is done to an order adjustment, adjust later half metric matrix D _ameet following condition:

Condition 1:

(-1) ²·Δ(1,2)>0,(-1) ³·Δ(1,2,3)>0,L,(-1) ^r·Δ(1,2,L,r)>0

Condition 2:

For meeting r+1<u, v≤n positive integer u, v has

Δ(1,2,L,r+1,u)=Δ(1,2,L,r+1,v)=Δ(1,2,L,r+1,u,v)=0

In two dimensional surface theorem in Euclid space, for this theorem special case E ₂situation, has:

Theorem 2: establishing D is one n * n semi-metric matrix, it is not degenerated and embeds E ₂necessary and sufficient condition be: D is done to an order adjustment, adjusts later half metric matrix D _ameet following condition:

Condition one: Δ (1,2) >0, Δ (1,2,3) <0,

Condition two: for meeting r+1<u, v≤n positive integer u, v have Δ (1,2,3, u)=Δ (1,2,3, v)=Δ (1,2,3, u, v)=0.

Wherein: $\mathrm{\&Delta;}\left(\mathrm{1,2}\right)=\left|\begin{array}{ccc}0& d_{12}^2& 1\\ d_{12}^2& 0& 1\\ 1& 1& 0\end{array}\right|=2d_{12}^2>0$ Inevitable, Δ (1,2,3) <0 represents by D _afront 3 conllinear not after embedding; (1,2,3, (1,2,3, v)=Δ (1,2,3, u, v)=0 represents semi-metric matrix square edged battle array to u)=Δ to Δ

order is 4.

In two dimensional surface space, Δ (1,2 in the condition 2 of theorem 2,3, u)=Δ (1,2,3, v)=Δ (1,2,3, u, v)=0, can only draw the distance matrix square edged battle array equation relation of 4 or 5 points, remove travelling carriage, participate in so effective base station that location survey value obtains and only have 3 or 4 most.Therefore,, under the environment of cellular cell, choose 4 adjacent base stations as shown in Figure 2 as position reference base station.

When four reference base station participate in location, choose orderly point set (O, A, B, M, C), Yi Zhiqi meets Blumenthal embedding theorems, can obtain its square of edged battle array

$\stackrel{\&OverBar;}{D}=\left(\begin{array}{cccccc}0& 4a^2& 12a^2& v^2{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2& 4a^2& 1\\ {4a}^2& 0& 4a^2& v^2{t}_2^2& 4a^2& 1\\ 12a^2& 4a^2& 0& v^2{t}_3^2& 4a^2& 1\\ v^1{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2& v^2{t}_2^2& v^2{t}_3^2& 0& v^2{t}_4^2& 1\\ 4a^2& 4a^2& 4a^2& v^2{t}_4^2& 0& 1\\ 1& 1& 1& 1& 1& 0\end{array}\right)---\left(4\right)$

(t wherein ₁, t ₂, t ₃, t ₄) being respectively (O, A, B, C) to the time measured value of a M, the distance of adjacent base station is 2a.

For three architectures, choose base station O, A, C, for orderly point set (O, A, C, M), Yi Zhiqi meets Blumenthal embedding theorems so, in like manner can obtain its square of edged battle array

$\stackrel{\&OverBar;}{D}=\left(\begin{array}{ccccc}0& {4a}^2& {4a}^2& v^2{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2& 1\\ 4a^2& 0& 4a^2& v^2{t}_2^2& 1\\ 4a^2& 4a^2& 0& v^2{t}_3^2& 1\\ v^2{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2& v^2{t}_2^2& v^2{t}_3^2& 0& 1\\ 1& 1& 1& 1& 0\end{array}\right)---\left(5\right)$

(t wherein ₁, t ₂, t ₃) being respectively (O, A, C) to the time measured value of a M, the distance of adjacent base station is 2a.

1, geometric distance filtering algorithm

If can obtain many group measured values, in these measured values, have so and must have one group of measured value to compare deviate minimum with actual value, approach most the measured value under LOS.Geometric distance filtering algorithm (Distance geometric filtering, calls DGF in the following text) will be attempted finding out this group in these measured values and approach one group of measured value in LOS situation most, and it is positioned as locator data.

During four architectures, according to the condition two of Blumenthal theorem, make in (3) formula

calculate and abbreviation, can obtain

${\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-t_2^2+t_3^2-t_4^2-{4a}^2---\left(6\right)$

(t wherein ₁, t ₂, t ₃) while being respectively error free (O, A, B, C) to the time measured value of some M, a is half of adjacent base station distance.And actual measured value is owing to having comprised NLOS error, cannot obtain as the equation of (5) formula, therefore, can from many groups measured value, choose

one group of value minimum as location survey value.In TOA model, obtain measuring equation group

$\left\{\begin{array}{c}{(x-x_1)}^2+{(y-y_1)}^2=v^2{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2\\ {(x-x_2)}^2+{(y-y_2)}^2=v^2{t}_2^2\\ {(x-x_3)}^2+{(y-y_3)}^2=v_2{t}_3^2\\ {(x-x_4)}^2+{(y-y_4)}^2=v^2{t}_4^2\end{array}\right.---\left(7\right)$

Appoint and to get 3 setting circles, its equation subtracts each other between two three straight lines that obtain and intersects at a point (as shown in Figure 2), and 4 setting circles are appointed respectively and got 3, can obtain four intersection points

according to centroid algorithm, location of mobile station is:

$(\frac{{{\mathrm{<figure-callout\; id="1"\; label="x\; M"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">x</figure-callout>}}_M}_1+{x_M}_2+{x_M}_3+{x_M}_4}{4},\frac{{{\mathrm{<figure-callout\; id="1"\; label="y\; M"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">y</figure-callout>}}_M}_1+{y_M}_2+{y_M}_3+{y_M}_4}{4})$

By base station (O, A, B, C) coordinate substitution (7) formula, solve and can obtain location of mobile station coordinate:

$\left\{\begin{array}{c}x=\frac{\sqrt{3}{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-3t_2^2-t_3^2+3t_4^2)}{24a}+\frac{\sqrt{3}a}{2}\\ y=\frac{v^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2+t_2^2-t_3^2-t_4^2)}{8a}+\frac{3a}{2}\end{array}\right.---\left(8\right)$

Location survey value substitution (8) formula obtaining is resolved.

In like manner, when adopting three architectures, can obtain and retrain accordingly equation and be

${\mathrm{<figure-callout\; id="1"\; label="p"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">p</figure-callout>}}_1^2+p_2^2+p_3^2-v^4({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_2^2+{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_3^2+t_2^2{t}_3^2)+{4a}^2=0---\left(9\right)$

Wherein ${\mathrm{<figure-callout\; id="1"\; label="p"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">p</figure-callout>}}_1=v^2{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-{2a}^2,p_2=v^2{t}_2^2-{2a}^2,p_3=v^2{t}_3^2-2a^2$

Corresponding position coordinates is

$\left\{\begin{array}{c}x=\frac{\sqrt{3}{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-2t_2^2+t_3^2)}{12a}+\frac{\sqrt{3}}{3}a\\ y=\frac{v^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-t_3^2)}{4a}+a\end{array}\right.---\left(10\right)$

2, geometric distance bounding algorithm

Because how NLOS error has compared a larger overgauge with LOS, geometric distance bounding algorithm (Distance geometric constraint, call DGC in the following text) utilize geometrical relationship to make measured value deduct certain deviate, thus make the one group of measured value obtaining can be closer to the measured value under LOS.

During four architectures, known (6) formula:

${\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-t_2^2+t_3^2-t_4^2-4a^2=0$

Due to the existence of NLOS, measured value has a large positive deviation than actual value, therefore can establish measured value

$t_{\mathrm{mi}}^2=t_i^2+{\mathrm{\&epsiv;}}_i---\left(11\right)$

ε wherein _i>=0.

Obtain

${\mathrm{\&epsiv;}}_1-{\mathrm{\&epsiv;}}_2+{\mathrm{\&epsiv;}}_3-{\mathrm{\&epsiv;}}_4-{\mathrm{<figure-callout\; id="1"\; label="t\; m"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_m^1+t_{m2}^2-t_{m3}^2+t_{m4}^2+4a^2=0---\left(12\right)$

Therefore measured value optimization problem is converted into following nonlinear optimal problem:

$\min J=\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}^2$

s.t. （13）

$\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_1-{\mathrm{\&epsiv;}}_2+{\mathrm{\&epsiv;}}_3-{\mathrm{\&epsiv;}}_4-{\mathrm{<figure-callout\; id="1"\; label="t\; m"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_m^1+t_{m2}^2-t_{m3}^2+t_{m4}^2+4a^2=0\\ {\mathrm{\&epsiv;}}_i>0,i=\mathrm{0,1,2,3}\end{array}\right.$

Obtain ε ₁, ε ₂, ε ₃, ε ₄, show that TOA optimizes measured value

$t_i^{\&prime;}=\sqrt{t_{\mathrm{mi}}^2-{\mathrm{\&epsiv;}}_i}---\left(14\right)$

Optimization distance value d _i=vt ' _i

Substitution LS Algorithm for Solving base station coordinates again,

X=(A ^TA) ^-1A ^TY (15)

Wherein

$A=-2\left(\begin{array}{cc}(x_1-x_k)& (y_1-y_k)\\ (x_2-x_k)& (y_2-y_k)\\ L& L\\ (x_{k-1}-x_k)& (y_{k-1}-y_k)\end{array}\right),Y=\left(\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">d</figure-callout>}}_1^2-d_k^2-x_1^2+x_k^2-{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">y</figure-callout>}}_1^2+y_k^2\\ d_2^2-d_k^2-x_2^2+x_k^2-y_2^2+y_k^2\\ L\\ d_{k-1}^2-d_k^2-x_{k-1}^2+x_k^2-y_{k-1}^2+y_k^2\end{array}\right)$

(x _i, y _i) be BS coordinate, d _ifor the distance measure after optimizing.

During three architectures, can obtain corresponding nonlinear optimization restricted problem and be:

$\min J=\underset{i=1}{\overset{3}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}^2$

s.t. (16)

$\left\{\begin{array}{c}{\mathrm{\&epsiv;}}^T\mathrm{A\&epsiv;}+{\mathrm{\&epsiv;}}^{T}b+c=0\\ {\mathrm{\&epsiv;}}_i>0,i=\mathrm{0,1,2,3}\end{array}\right.$

ε=(ε wherein ₁, ε ₂, ε ₃), $A=-2\left(\begin{array}{cc}(x_1-x_k)& (y_1-y_k)\\ (x_2-x_k)& (y_2-y_k)\\ L& L\\ (x_{k-1}-x_k)& (y_{k-1}-y_k)\end{array}\right),Y=\left(\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">d</figure-callout>}}_1^2-d_k^2-x_1^2+x_k^2-{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">y</figure-callout>}}_1^2+y_k^2\\ d_2^2-d_k^2-x_2^2+x_k^2-y_2^2+y_k^2\\ L\\ d_{k-1}^2-d_k^2-x_{k-1}^2+x_k^2-y_{k-1}^2+y_k^2\end{array}\right)$

$c=128a^6+8a^2{v}^4({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^4+t_2^4+t_3^4-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_2^2-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_3^2-t_2^2{t}_3^2)-32a^4{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2+t_2^2+t_3^2);$

Geometric distance filtering algorithm (Distance geometric filtering, DGF) and geometric distance bounding algorithm (Distance geometric constraint, DGC) be two kinds of different data processing methods in geometric distance localization method, according to algorithm characteristic and simulation result, known DGF algorithm need be measured multi-group data, has greater probability to obtain the situation of LOS measured value while being applicable to stationary node, NLOS mean error compared with Datong District; DGC algorithm can be used for mobile node location, and when NLOS mean error is little, effect is better.Under as the environment of Fig. 2, the specific implementation method of DGF and DGC is as follows, as Figure 8-9.

1:DGF algorithm:

(1) travelling carriage obtains the many groups measured value from base station, and each group measured value is the signal propagation time that comes from different base station that travelling carriage receives at synchronization, is designated as (t ₁, t ₂, t ₃, t ₄), wherein subscript represents different base stations;

(2) each group measured value is calculated according to following formula,, for each group measured value, all can obtain a calculated value, choose in all measured values meet the following conditions (making calculating formula value minimum) one group:

Four base stations (can receive the signal of O in Fig. 2, A, B, tetra-base stations of C):

$\min ({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-t_2^2+t_3^2-t_4^2-{4a}^2);$

Three base stations (can receive the signal of O in Fig. 2, A, tri-base stations of C):

$\min ({\mathrm{<figure-callout\; id="1"\; label="p"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">p</figure-callout>}}_1^2+p_2^2+p_3^2-v^4({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_2^2+{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_3^2+t_2^2{t}_3^2)+4a^2),$

Wherein ${\mathrm{<figure-callout\; id="1"\; label="p"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">p</figure-callout>}}_1=v^2{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-{2a}^2,p_2=v^2{t}_2^2-{2a}^2,p_3=v^2{t}_3^2-2a^2,$ A is half of adjacent base station distance, and v is transmission of wireless signals speed;

(3) one group of following formula of measured value substitution choosing in step (2) is solved, the value of (x, the y) of acquisition is the position coordinates of travelling carriage under the coordinate system of setting:

Four base stations: $\left\{\begin{array}{c}x=\frac{\sqrt{3}{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-3t_2^2-t_3^2+3t_4^2)}{24a}+\frac{\sqrt{3}a}{2}\\ y=\frac{v^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2+t_2^2-t_3^2-t_4^2)}{8a}+\frac{3a}{2}\end{array}\right.$

Three base stations: $\left\{\begin{array}{c}x=\frac{\sqrt{3}{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-2t_2^2+t_3^2)}{12a}+\frac{\sqrt{3}}{3}a\\ y=\frac{v^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2-t_3^2)}{4a}+a\end{array}\right.;$

2, DGC algorithm:

(1) travelling carriage obtains the measured value from base station, and the signal propagation time that comes from different base station for travelling carriage receives at synchronization, is designated as (t _m1, t _m2, t _m3, t _m4), wherein subscript represents different base stations;

(2) set unknown parameter (ε ₁, ε ₂..., ε _i), the base station number of i for participating in calculating, calculates following nonlinear optimal problem, is met one group of (ε of this nonlinear optimal problem ₁, ε ₂..., ε _i) numerical value, wherein a is half of adjacent base station distance, v is transmission of wireless signals speed:

Four base stations (can receive the signal of O in Fig. 2, A, B, tetra-base stations of C):

$\min J=\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}_i^2$

s.t.

$\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_1-{\mathrm{\&epsiv;}}_2+{\mathrm{\&epsiv;}}_3-{\mathrm{\&epsiv;}}_4-{\mathrm{<figure-callout\; id="1"\; label="t\; m"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_m^1+t_{m2}^2-t_{m3}^2+t_{m4}^2+4a^2=0\\ {\mathrm{\&epsiv;}}_i>0,i=\mathrm{0,1,2,3}\end{array}\right.$

Three base stations (can receive the signal of O in Fig. 2, A, tri-base stations of C):

$\min J=\underset{i=1}{\overset{3}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}^2$

s.t.

$\left\{\begin{array}{c}{\mathrm{\&epsiv;}}^T\mathrm{A\&epsiv;}+{\mathrm{\&epsiv;}}^{T}b+c=0\\ {\mathrm{\&epsiv;}}_i>0,i=\mathrm{0,1,2,3}\end{array}\right.$

ε=(ε wherein ₁, ε ₂, ε ₃), $A=\left(\begin{array}{ccc}{8a}^2& -4a^2& -4a^2\\ -4a^2& 8a^2& -4a^2\\ -4a^2& -{4a}^2& 8a^2\end{array}\right),b=\left(\begin{array}{ccc}-16a^2& 8a^2& {8a}^2\\ 8a^2& -16a^2& 8a^2\\ 8a^2& 8a^2& -16a^2\end{array}\right)\left(\begin{array}{c}{v}^2{\mathrm{<figure-callout\; id="1"\; label="t\; m"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_m^1\\ v^2{t}_{m2}^2\\ v^2{t}_{m3}^2\end{array}\right)+\left(\begin{array}{c}32a^2\\ 32a^2\\ 32a^2\end{array}\right),$

$c=128a^6+8a^2{v}^4({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^4+t_2^4+t_3^4-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_2^2-{\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2{t}_3^2-t_2^2{t}_3^2)-32a^4{v}^2({\mathrm{<figure-callout\; id="1"\; label="t"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">t</figure-callout>}}_1^2+t_2^2+t_3^2);$

(3) (ε step (2) being obtained ₁, ε ₂..., ε _i) numerical value substitution following formula, try to achieve travelling carriage after optimization to the distance parameter value of each base station:

$d_i=v\sqrt{t_{\mathrm{mi}}^2-\mathrm{\&epsiv;}}$

(4) least square method solves: X=(A ^ta) ^-1a ^ty, the vectorial X calculating is a bivector, is the position coordinates of travelling carriage under the coordinate system of setting,

Wherein

$A=-2\left(\begin{array}{cc}(x_1-x_k)& (y_1-y_k)\\ (x_2-x_k)& (y_2-y_k)\\ L& L\\ (x_{k-1}-x_k)& (y_{k-1}-y_k)\end{array}\right),Y=\left(\begin{array}{c}{\mathrm{<figure-callout\; id="1"\; label="d"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">d</figure-callout>}}_1^2-d_k^2-x_1^2+x_k^2-{\mathrm{<figure-callout\; id="1"\; label="y"\; filenames="HDA0000447287830000023.png"\; state="\{\{state\}\}">y</figure-callout>}}_1^2+y_k^2\\ d_2^2-d_k^2-x_2^2+x_k^2-y_2^2+y_k^2\\ L\\ d_{k-1}^2-d_k^2-x_{k-1}^2+x_k^2-y_{k-1}^2+y_k^2\end{array}\right)$

(x _i, y _i) be the coordinate figure of base station i, d _ithe travelling carriage obtaining for step (3) is to the distance parameter value of base station i.

In order to assess the validity of geometric distance bounding algorithm, with MATLAB, algorithm has been carried out to emulation.For convenient simulation, at this, utilized least-squares algorithm to position DGC algorithm, and compared with positioning performance and the DGF algorithm performance of common least-squares algorithm.The value of adjusting the distance of the present invention simultaneously optimization method improves and compares.Simulation result shows, in cellular network base stations, divides and plants, and supposes that base station spacing is 1000m, when non line of sight error to standard deviation at 50m between 500m time, location mean square error preferably all can reach 25m left and right.

The mean square error of having studied algorithm positioning solution under different NLOS errors in emulation, expression formula is:

$\mathrm{RMSE}=\sqrt{{(x-x_0)}^2+{(y-y_0)}^2}---\left(17\right)$

Wherein, (x ₀, y ₀) be the physical location of travelling carriage, the estimated position that (x, y) is travelling carriage.

Under the environment of cellular cell, adopt arrangement of base stations as shown in Figure 1, with four base stations of diamond shape or three base stations of equilateral triangle shape, inner travelling carriage is positioned.

In emulation experiment, suppose that two base station spacings are 1000m, NLOS error mean difference is 250m, double counting 1000 times is also averaged, and obtains four base stations and three kinds of three base stations location simulation result as Fig. 3,4 and Fig. 5 shown in.

From positioning result intuitively, DGF algorithm has suppressed NLOS error preferably, and positioning precision has been compared very big raising with LS algorithm, and its position location extremely approaches travelling carriage actual position; Although the precision of DGC algorithm also improves, be not so good as DGF algorithm obvious.

For different NLOS errors (mean difference be 50m to 500m not etc.), carried out 10 location simulations simulations, each emulation double counting is also averaged for 1000 times, in four base stations and three base station situations, obtains simulation result respectively as shown in Fig. 6, Fig. 7, Fig. 8.

From simulation result, increase along with NLOS error, the positioning precision of LS algorithm obviously declines, DGC algorithm is owing to being to adopt LS algorithm to position after measured value is optimized again, therefore similar with LS algorithm, along with NLOS error increases, positioning precision declines also obvious, but general location precision is better than LS algorithm and learns from this point analysis, if adopt other algorithm (as Taylor algorithm, Chan algorithm etc.) also to have similar result.And DGF algorithm fluctuation and NLOS error size do not have obvious relation, positioning precision is obviously better than other two kinds of algorithms.DGC algorithm is better than three base station situations in the situation that of four base stations, because more measured value can better improve positioning precision under NLOS environment, but for DGF algorithm, base station number is on the impact of positioning precision not obvious, if be because can obtain NLOS error unconspicuous one group of measured value, use minimum base station number also can obtain comparatively desirable positioning result.

Through the analysis to emulated data, why DGC method does not obtain particularly preferred effect, because according to the principle of nonlinear optimization process, in calculating, to the optimization of deviate, can not offset real NLOS error completely, can only obtain micro-optimum results, the optimization measured value finally obtaining is compared with corresponding LOS measured value or is had certain error, and precision is lower when NLOS error is larger.But DGC theoretical method only needs an a small amount of measured value just can be optimized location, thus in the less demanding situation of computation complexity and NLOS error compared with hour considering; DGF method needs many group measured values, after algorithm, getting after filtering a class value positions, with NLOS mean error size without obvious relation, but its precision depends on that group of error minimum in obtained measured value, so the suitable NLOS of being applied in error is large and measured value has very large probability acquisition to approach in the situation of measured value under LOS.Compare with three architectures, four architectures can make DGC algorithm obtain higher positioning precision, and its algorithmic formula is more succinct, and while carrying out nonlinear optimization, computation complexity is lower.

Should be appreciated that the above detailed description of technical scheme of the present invention being carried out by preferred embodiment is illustrative and not restrictive.Those of ordinary skill in the art modifies reading the technical scheme that can record each embodiment on the basis of specification of the present invention, or part technical characterictic is wherein equal to replacement; And these modifications or replacement do not make the essence of appropriate technical solution depart from the spirit and scope of various embodiments of the present invention technical scheme.

Claims (1)

1. The wireless location new method that adopts geometric distance under 1.NLOS environment, is characterized in that, this new location method comprises geometric distance filtering method DGF and geometric distance constrained procedure DGC,

   The step of described DGF algorithm is as follows:

   101. travelling carriages obtain the many groups measured value from base station, and each group measured value is the signal propagation time that comes from different base station that travelling carriage receives at synchronization, is designated as (t ₁, t ₂, t ₃, t ₄), wherein subscript represents different base stations;

   102. calculate according to following formula each group measured value,, for each group measured value, all can obtain a calculated value, choose in all measured values and meet the following conditions and make one group of calculating formula value minimum:

   For the signal that can receive four base stations:

   $\min (t_1^2-t_2^2+t_3^2-t_4^2-{4a}^2);$

   For the signal that can receive three base stations:

   $\min (p_1^2+p_2^2+p_3^2-v^4(t_1^2{t}_2^2+t_1^2{t}_3^2+t_2^2{t}_3^2)+4a^2),$

   Wherein $p_1=v^2{t}_1^2-{2a}^2,p_2=v^2{t}_2^2-{2a}^2,p_3=v^2{t}_3^2-2a^2,$ A is half of adjacent base station distance, and v is transmission of wireless signals speed;

   103. solve one group of following formula of measured value substitution choosing in step 102, and the value of (x, the y) of acquisition is the position coordinates of travelling carriage under the coordinate system of setting:

   Four base stations: $\left\{\begin{array}{c}x=\frac{\sqrt{3}{v}^2(t_1^2-3t_2^2-t_3^2+3t_4^2)}{24a}+\frac{\sqrt{3}a}{2}\\ y=\frac{v^2(t_1^2+t_2^2-t_3^2-t_4^2)}{8a}+\frac{3a}{2}\end{array}\right.$

   Three base stations: $\left\{\begin{array}{c}x=\frac{\sqrt{3}{v}^2(t_1^2-2t_2^2+t_3^2)}{12a}+\frac{\sqrt{3}}{3}a\\ y=\frac{v^2(t_1^2-t_3^2)}{4a}+a\end{array}\right.;$

   The step of described DGC method is as follows:

   201. travelling carriages obtain the measured value from base station, and the signal propagation time that comes from different base station for travelling carriage receives at synchronization, is designated as (t _m1, t _m2, t _m3, t _m4), wherein subscript represents different base stations;

   202. set unknown parameter (ε ₁, ε ₂..., ε _i), the base station number of i for participating in calculating, calculates following nonlinear optimal problem, is met one group of (ε of this nonlinear optimal problem ₁, ε ₂..., ε _i) numerical value, wherein a is half of adjacent base station distance, v is transmission of wireless signals speed:

   For the signal that can receive four base stations:

   $\min J=\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}_i^2$

   s.t.

   $\left\{\begin{array}{c}{\mathrm{\&epsiv;}}_1-{\mathrm{\&epsiv;}}_2+{\mathrm{\&epsiv;}}_3-{\mathrm{\&epsiv;}}_4-t_{m1}^2+t_{m2}^2-t_{m3}^2+t_{m4}^2+4a^2=0\\ {\mathrm{\&epsiv;}}_i>0,i=\mathrm{0,1,2,3}\end{array}\right.$

   For the signal that can receive three base stations:

   $\min J=\underset{i=1}{\overset{3}{\mathrm{\&Sigma;}}}{\mathrm{\&epsiv;}}^2$

   s.t.

   $\left\{\begin{array}{c}{\mathrm{\&epsiv;}}^T\mathrm{A\&epsiv;}+{\mathrm{\&epsiv;}}^{T}b+c=0\\ {\mathrm{\&epsiv;}}_i>0,i=\mathrm{0,1,2,3}\end{array}\right.$

   ε=(ε wherein ₁, ε ₂, ε ₃), $A=\left(\begin{array}{ccc}{8a}^2& -4a^2& -4a^2\\ -4a^2& 8a^2& -4a^2\\ -4a^2& -{4a}^2& 8a^2\end{array}\right),b=\left(\begin{array}{ccc}-16a^2& 8a^2& {8a}^2\\ 8a^2& -16a^2& 8a^2\\ 8a^2& 8a^2& -16a^2\end{array}\right)\left(\begin{array}{c}{v}^2{t}_{m1}^2\\ v^2{t}_{m2}^2\\ v^2{t}_{m3}^2\end{array}\right)+\left(\begin{array}{c}32a^2\\ 32a^2\\ 32a^2\end{array}\right),$

   $c=128a^6+8a^2{v}^4(t_1^4+t_2^4+t_3^4-t_1^2{t}_2^2-t_1^2{t}_3^2-t_2^2{t}_3^2)-32a^4{v}^2(t_1^2+t_2^2+t_3^2);$

   203. (the ε that step 202 is obtained ₁, ε ₂..., ε _i) numerical value substitution following formula, try to achieve travelling carriage after optimization to the distance parameter value of each base station:

   $d_i=v\sqrt{t_{\mathrm{mi}}^2-\mathrm{\&epsiv;}}$

   204. least square methods solve: X=(A ^ta) ^-1a ^ty, the vectorial X calculating is a bivector, is the position coordinates of travelling carriage under the coordinate system of setting,

   Wherein

   $A=-2\left(\begin{array}{cc}(x_1-x_k)& (y_1-y_k)\\ (x_2-x_k)& (y_2-y_k)\\ L& L\\ (x_{k-1}-x_k)& (y_{k-1}-y_k)\end{array}\right),Y=\left(\begin{array}{c}{d}_1^2-d_k^2-x_1^2+x_k^2-y_1^2+y_k^2\\ d_2^2-d_k^2-x_2^2+x_k^2-y_2^2+y_k^2\\ L\\ d_{k-1}^2-d_k^2-x_{k-1}^2+x_k^2-y_{k-1}^2+y_k^2\end{array}\right)$

   (x _i, y _i) be the coordinate figure of base station i, d _ithe travelling carriage obtaining for step 203 is to the distance parameter value of base station i.

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