CN102967870A - Fast iterative method of global position system (GPS) positioning - Google Patents

Abstract

The traditional global position system (GPS) positioning method, such as the method of Mohinder S. and the like, includes two steps: positioning initial position (error-free positioning) solving and inverse-matrix-based position iterative solving. The problem of the method is that error-free positioning obviously has no solution due to the fact that if the error-free positioning has solutions, four balls have to intersect at one point, errors are inevitable and the harsh requirement is hardly (zero probability) to meet. In positioning solving, inverse matrix operation is needed in each iteration, accordingly operation complexity is increased, operation time is prolonged, the final result cannot ensure that the quadratic sum of the distances of all satellites and corresponding pseudo-ranges reaches the minimum, and the minimum quadratic sum is the common standard in noise solving. The invention provides a novel GPS positioning method which includes a new positioning initial position solving method and a position iterative solving method not related to inverse matrix operation. The novel GPS positioning method reduces operation complexity, increases GPS positioning timeliness and enables GPS positioning to have wide application occasions.

Description

The iteratively faster method of a kind of GPS location

Technical field

The present invention relates to the iteratively faster method of a kind of GPS location.

Background technology

The position of four Navsats of note is (x _i, y _i, z _i), apart from anchor point (x ₀, y ₀, z ₀) pseudorange be respectively R _i(i=1,2,3,4), then

$R_i=\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}+{\mathrm{\ξ}}_i(i=\mathrm{1,2,3,4})---\left(1\right)$

ξ wherein _iThe expression pseudorange error.

There is the whole bag of tricks from system of equations (1), to find the solution position location (x ₀, y ₀, z ₀).Take the people's such as Mohinder S. method as example, its method can be summarized as following two steps:

1. find the solution free from error orientation problem

Suppose that pseudorange is error free in the system of equations (1), namely

R _i ²＝(x _i-x ₀) ²+(y _i-y ₀) ²+(z _i-z ₀) ² (i＝1，2，3，4) (2)

Then have

R _i＝x _i ²+y _i ²+z _i ²+x ₀ ²+y ₀ ²+z ₀ ²-2x _ix ₀-2y _iy ₀-2z _iz ₀

Or

R _i-(x _i ²+y _i ²+z _i ²)+r ²＝C _rr-2x _ix ₀-2y _iy ₀-2z _iz ₀ (i＝1，2，3，4)

X wherein ₀ ²+ y ₀ ²+ z ₀ ²=r ²+ C _Rr, r is earth radius, C _RrBe the clock jitter modified value.Obtain thus following system of equations:

AX＝Y

Wherein

$A=\left(\begin{array}{cccc}-2x_1& {-2y}_1& {-2z}_1& 1\\ {-2x}_2& {-2y}_2& {-2z}_2& 1\\ -{2x}_3& {-2y}_3& {-2z}_3& 1\\ {-2x}_4& {-2y}_4& {-2z}_4& 1\end{array}\right),$ $X=\left(\begin{array}{c}{x}_0\\ y_0\\ z_0\\ C_{\mathrm{rr}}\end{array}\right),$ $Y=\left(\begin{array}{c}{R}_1-({x_1}^2+{y_1}^2+{z_1}^2)+r^2\\ R_2-({x_2}^2+{y_2}^2+{z_2}^2)+r^2\\ R_3-({x_3}^2+{y_3}^2+{z_3}^2)+r^2\\ R_4-({x_4}^2+{y_4}^2+{z_4}^2)+r^2\end{array}\right)$

When matrix A is nonsingular, can solve thus

$\left(\begin{array}{c}{x}_0\\ y_0\\ z_0\\ C_{\mathrm{rr}}\end{array}\right)=X=A^{-1}Y$

2. find the solution the aircraft orientation problem

Utilize Taylor exhibition formula to be with system of equations (2) approximate representation of clock jitter

$R_i={(\frac{{\∂R}_i}{\∂x_0}\mathrm{\δx}+\frac{{\∂R}_i}{{\∂y}_0}\mathrm{\δy}+\frac{{\∂R}_i}{{\∂z}_0}\mathrm{\δz})}_{(x_0,y_0,z_0)=(x_{\mathrm{mon}},y_{\mathrm{mon}},z_{\mathrm{mon}})}+C_{\mathrm{rr}}+{\mathrm{\ξ}}_i$ (i＝1，2，3，4)

Wherein

$\frac{{\∂R}_i}{{\∂x}_0}=\frac{-(x_i-x_0)}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}$

$\frac{{\∂R}_i}{{\∂y}_0}=\frac{-(y_i-y_0)}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}$

$\frac{{\∂R}_i}{{\∂z}_0}=\frac{-(z_i-z_0)}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}$

ξ _iBe Taylor exhibition formula error term, δ x=x-x _Nom,, δ y=y-y _Nom, δ z=z-z _NomObtain thus

${\left(\begin{array}{cccc}\frac{{\∂R}_1}{{\∂x}_0}& \frac{{\∂R}_1}{{\∂y}_0}& \frac{{\∂R}_1}{{\∂z}_0}& 1\\ \frac{{\∂R}_2}{{\∂x}_0}& \frac{{\∂R}_2}{{\∂y}_0}& \frac{{\∂R}_2}{{\∂z}_0}& 1\\ \frac{{\∂R}_3}{{\∂x}_0}& \frac{{\∂R}_3}{{\∂y}_0}& \frac{{\∂R}_3}{{\∂z}_0}& 1\\ \frac{{\∂R}_4}{{\∂x}_0}& \frac{{\∂R}_4}{{\∂y}_0}& \frac{{\∂R}_4}{{\∂z}_0}& 1\end{array}\right)}_{(x_0,y_0,z_0)=(x_{\mathrm{mon}}{,y}_{\mathrm{mon}},z_{\mathrm{mon}})}\left(\begin{array}{c}\mathrm{\δx}\\ \mathrm{\δy}\\ \mathrm{\δy}\\ C_b\end{array}\right)=\left(\begin{array}{c}{R}_1\\ R_2\\ R_3\\ R_4\end{array}\right)$

So

$\left(\begin{array}{c}x\\ y\\ z\\ C_b\end{array}\right)=\left(\begin{array}{c}{x}_{\mathrm{nom}}\\ y_{\mathrm{nom}}\\ y_{\mathrm{nom}}\\ 0\end{array}\right)+{\left(\begin{array}{cccc}\frac{{\∂R}_1}{{\∂x}_0}& \frac{{\∂R}_1}{{\∂y}_0}& \frac{{\∂R}_1}{{\∂z}_0}& 1\\ \frac{{\∂R}_2}{{\∂x}_0}& \frac{{\∂R}_2}{{\∂y}_0}& \frac{{\∂R}_2}{{\∂z}_0}& 1\\ \frac{{\∂R}_3}{{\∂x}_0}& \frac{{\∂R}_3}{{\∂y}_0}& \frac{{\∂R}_3}{{\∂z}_0}& 1\\ \frac{{\∂R}_4}{{\∂x}_0}& \frac{{\∂R}_4}{{\∂y}_0}& \frac{{\∂R}_4}{{\∂z}_0}& 1\end{array}\right)}_{(x_0,y_0,z_0)=(x_{\mathrm{mon}},y_{\mathrm{mon}},z_{\mathrm{mon}})}^{-1}\left(\begin{array}{c}{R}_1\\ R_2\\ R_3\\ R_4\end{array}\right)---\left(3\right)$

Solution (x with free from error orientation problem ₀y ₀z ₀) ^TBe (x _Nomy _Nomz _Nom) ^TInitial vector, utilize (3) can obtain the solution of orientation problem by interative computation.

There are the following problems for the people's such as Mohinder S. method:

Free from error orientation problem (2) obviously be one without the solution problem, because system of equations (2) has solution to mean that four balls must meet at a bit, for the requirement of harshness like this, as long as actual error exists, " hardly " (probability is zero) may have solution.Even add clock jitter modified value C _Rr, situation is still without any change.So C _RrAppearance not only meaningless, also can increase operand.

Find the solution in the iterative process of aircraft orientation problem, what each step was obtained all is the approximate coordinates of position of aircraft, but net result can not guarantee to reach minimum with the quadratic sum of the difference of each satellite distance and corresponding pseudorange, and the latter is the general standard of separating noise problem.

In addition, all there is the computing of finding the inverse matrix in two steps, both increased the complicacy of computing, also prolonged operation time.

Summary of the invention

Patent of the present invention proposes the iteratively faster method that a kind of new GPS locates, this brand-new GPS localization method does not need to carry out the inverse matrix computing, reduced the complicacy of computing, increased the real-time of GPS location and made it to have more more widely application scenario.

For solving the problems of the technologies described above, the present invention has adopted following technical scheme.

Method of the present invention comprises two steps, namely locates initial position (free from error location) and finds the solution iterative with the position.Location initial position solution procedure be at first obtain centered by four gps satellites, any three intersection point in four spheres that corresponding pseudorange is radius, with this initial position as the location, intersection point provides with the formula form.The method of position iterative is in the system of equations that consists of of three equations of position coordinates to be positioned that the coordinate substitution of initial position is derived in advance, utilizes this system of equations to carry out iterative.The condition that iterative process stops is that the distance of facing mutually between the position that twice iteration obtain is no more than some setting values (deciding on positioning accuracy request).

One of characteristics of the present invention are that the location initial position only needs the substitution Formula For Solving, and the position iterative does not relate to the inverse matrix computing, thereby simple operation, and locator data is real-time.Two of characteristics of the present invention are that the quadratic sum take the difference of each satellite distance and corresponding pseudorange reaches minimum as criterion in the solution procedure, thereby have improved bearing accuracy.

Description of drawings

Fig. 1 be centered by three gps satellites, corresponding pseudorange is the synoptic diagram of three spheres intersect of radius.L among the figure ₁₂And l ₁₃Represent respectively first and second spheres, the public secant of first and the 3rd spherical projection behind plane, three gps satellite places.The intersection point of three satellites is inevitable on through the intersection point of these two secants and the straight line vertical with plane, three gps satellite places.

Embodiment

1. method:

The position of four Navsats of note is (x _i, y _i, z _i), anchor point (x ₀, y ₀, z ₀) pseudorange be respectively R _i(i=1,2,3,4), then

$R_i=\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}+{\mathrm{\ξ}}_i$ (i＝1，2，3，4)

ξ wherein _iThe expression pseudorange error.Because the existence of pseudorange error, above-mentioned solving equations are actually " error sum of squares is a minimum " problem, it is minimum that the quadratic sum of the difference of the coordinate of the anchor point of namely obtaining and each satellite distance and corresponding pseudorange reaches, and represents to be exactly to ask (x with mathematical form ₀, y ₀, z ₀), make

$F=\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{(\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}-R_i)}^2$

Reach minimum value.

1.1 the solution of minimum problems

Order

$\frac{\∂F}{\∂x_0}=-2\underset{i=1}{\overset{4}{\mathrm{\Σ}}}(\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}-R_i)\frac{x_i-x_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}=0$

Namely

$\underset{i=1}{\overset{4}{\mathrm{\Σ}}}(x_i-x_0-\frac{x_i-x_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}{R}_i)=0$

$x_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{x}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{x_i-x_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

In like manner

$y_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{y}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{y_i-y_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

$z_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{z}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{z_i-z_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

Therefore obtain one about the elements of a fix (x ₀, y ₀, z ₀) Nonlinear System of Equations

$x_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{x}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{x_i-x_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

$y_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{y}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{y_i-y_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

$z_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{z}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{z_i-z_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

This Nonlinear System of Equations can not direct solution, but can find the solution with the interative computation method, and problem is that interative computation can not be avoided the local minimum problem.Yet, determine the overall situation minimum (minimum) or the key of local minimum is choosing of initial point.If global minimum " near " choose initial point and carry out interative computation, interative computation just can obtain global minimum; If but chose initial point in the zone away from global minimum, what interative computation obtained just might be local minizing point.And if choose initial point in the enough little neighborhood of global minimum, what interative computation obtained must be global minimum.Therefore as long as in the enough little neighborhood of global minimum, choose an initial point (x ₀, y ₀, z ₀), carry out interative computation in the above-mentioned system of equations of substitution and just can obtain anchor point.

1.2 finding the solution of initial point

Remember that any three position in four Navsats is (x _i, y _i, z _i), the unit normal vector on this plane, three satellite places is

$r=\frac{(x_2-x_1,y_2-y_1,z_2-z_1)\×(x_3-x_1,y_3-y_1,z_3-z_1)}{\sqrt{{(x_2-x_1)}^2{+(y_2-y_1)}^2+{(z_2-z_1)}^2}\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2+{(z_3-z_1)}^2}}$

(x on this plane ₂, y ₂, z ₂), (x ₃, y ₃, z ₃) and (x ₁, y ₁, z ₁) the line length that connects with the heart be designated as respectively R ₁₂And R ₁₃, public secant is designated as respectively l ₁₂And l ₁₃, they are designated as respectively A with the intersection point of corresponding line of centres ₁₂And A ₁₃, the coordinate that can obtain them is respectively

$A_{12}=(x_1,y_1,z_1)+\frac{{R_{12}}^2+{R_1}^2-{R_2}^2}{{{2R}_{12}}^2}(x_2-x_1,y_2-y_1,y_2-z_1)$

With

$A_{13}=(x_1,y_1,z_1)+\frac{{R_{13}}^2+{R_1}^2-{R_3}^2}{{{2R}_{13}}^2}(x_3-x_1,y_3-y_1,y_3-z_1)$

l ₁₂And l ₁₃Direction vector be respectively

r ₁₂＝r×(x ₂-x ₁，y ₂-y ₁，y ₂-z ₁)

With

r ₁₃＝r×(x ₃-x ₁，y ₃-y ₁，y ₃-z ₁)

So l ₁₂And l ₁₃Parametric equation be respectively A ₁₂+ tr ₁₂And A ₁₃+ hr ₁₃(t, h ∈ (∞ ,+∞)), thereby intersection point A satisfies following system of equations:

$\left\{\begin{array}{c}{\mathrm{tr}}_{12}\left(1\right)-h{r}_{13}\left(1\right)=-A_{12}\left(1\right){+A}_{13}\left(1\right)\\ {\mathrm{tr}}_{12}\left(2\right)-{\mathrm{hr}}_{13}\left(2\right)=-A_{12}\left(2\right){+A}_{13}\left(2\right)\end{array}\right.$

Separate this system of equations and get final product

$t=\frac{\left[A_{12}\left(1\right){-A}_{13}\left(1\right)\right]r_{13}\left(2\right)+[A_{13}\left(2\right)-A_{12}\left(2\right)]r_{13}\left(1\right)}{r_{12}\left(2\right)r_{13}\left(1\right)-r_{12}\left(1\right)r_{13}\left(2\right)}$

R wherein ₁₂(i), r ₁₃(i), A ₁₂(i), A ₁₃(i) be i element of corresponding vector.Therefore the coordinate of A is

$A_{12}+{\mathrm{tr}}_{12}=A_{12}+\frac{\left[A_{12}\left(1\right){-A}_{13}\left(1\right)\right]r_{13}\left(2\right)+[A_{13}\left(2\right)-A_{12}\left(2\right)]r_{13}\left(1\right)}{r_{12}\left(2\right)r_{13}\left(1\right)-r_{12}\left(1\right)r_{13}\left(2\right)}{r}_{12}$

2. calculation procedure

1) calculates

$r=\frac{(x_2-x_1,y_2-y_1,z_2-z_1)\×(x_3-x_1,y_3-y_1,z_3-z_1)}{\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2+{(z_2-z_1)}^2}\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2+{(z_3-z_1)}^2}}$

$A_{12}=(x_1,y_1,z_1)+\frac{{R_{12}}^2+{R_1}^2-{R_2}^2}{{{2R}_{12}}^2}(x_2-x_1,y_2-y_1,y_2-z_1)$

$A_{13}=(x_1,y_1,z_1)+\frac{{R_{13}}^2+{R_1}^2-{R_3}^2}{{{2R}_{13}}^2}(x_3-x_1,y_3-y_1,y_3-z_1)$

r ₁₂＝r×(x ₂-x ₁，y ₂-y ₁，y ₂-z ₁)

r ₁₃＝r×(x ₃-x ₁，y ₃-y ₁，y ₃-z ₁)

$t=\frac{\left[A_{12}\left(1\right){-A}_{13}\left(1\right)\right]r_{13}\left(2\right)+[A_{13}\left(2\right)-A_{12}\left(2\right)]r_{13}\left(1\right)}{r_{12}\left(2\right)r_{13}\left(1\right)-r_{12}\left(1\right)r_{13}\left(2\right)}$

$A=A_2+{\mathrm{tr}}_{12}=A_{12}+\frac{\left[A_{12}\left(1\right){-A}_{13}\left(1\right)\right]r_{13}\left(2\right)+[A_{13}\left(2\right)-A_{12}\left(2\right)]r_{13}\left(1\right)}{r_{12}\left(2\right)r_{13}\left(1\right)-r_{12}\left(1\right)r_{13}\left(2\right)}{r}_{12}$

2) with the elements of a fix (x ₀, y ₀, z ₀) Nonlinear System of Equations be rewritten as following iterative equation group

$x_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{x}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{x_i-x_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{\left(k\right)})}^2+{(z_i-z_0^{\left(k\right)})}^2}}\frac{R_i}{4}$

$y_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{y}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{y_i-y_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{0\left(k\right)})}^2+{(z_i-z_0^k)}^2}}\frac{R_i}{4}$

$z_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{z}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{z_i-z_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{\left(k\right)})}^2+{(z_i-z_0^{\left(k\right)})}^2}}\frac{R_i}{4}$

For iterative process is set permissible error δ, and carry out following iteration:

1 °) the iteration initial point is set

Make iteration count variable k=1;

2 °) calculate

$x_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{x}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{x_i-x_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{\left(k\right)})}^2+{(z_i-z_0^{\left(k\right)})}^2}}\frac{R_i}{4}$

$y_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{y}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{y_i-y_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{0\left(k\right)})}^2+{(z_i-z_0^k)}^2}}\frac{R_i}{4}$

$z_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\Σ}}}{z}_i-\underset{i=1}{\overset{4}{\mathrm{\Σ}}}\frac{z_i-z_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{\left(k\right)})}^2+{(z_i-z_0^{\left(k\right)})}^2}}\frac{R_i}{4}$

3 °) if $|(x_0^{(k+1)},y_0^{(k+1)},z_0^{(k+1)})-\left(x_0^{(k+1)}{,y}_0^{(k+1)}{,z}_0^{(k+1)}\right)|<\mathrm{\&delta;},$ Stop iteration;

4 °) value of k adds 1, returns 2 °).

Stop iteration and just obtain the position location

The present invention not technology contents of detailed description is known technology.

Claims (4)

1. the iteratively faster method of GPS location, it is characterized in that: method comprises two steps, namely locates initial position and finds the solution iterative with the position; Location initial position solution procedure is: any three intersection point in four spheres that at first obtain centered by four gps satellites, corresponding pseudorange are radius, with this initial position as the location; The iterative process of position is: with the location the initial position substitution by the positioning equation group

$x_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{x}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{x_i-x_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

$y_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{y}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{y_i-y_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

$z_0=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{z}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{z_i-z_0}{\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}}\frac{R_i}{4}$

In the iterative equation group that obtains, and carry out iteration according to iterative program; The condition that iterative process stops is that the distance of facing mutually between the position that twice iteration obtain is no more than some setting values (deciding on positioning accuracy request).

2. the iteratively faster method of GPS according to claim 1 location, it is characterized in that locating initial position only needs related data substitution formula

$A{=A}_{12}+\frac{\left[A_{12}\left(1\right){-A}_{13}\left(1\right)\right]r_{13}\left(2\right)+[A_{13}\left(2\right)-A_{12}\left(2\right)]r_{13}\left(1\right)}{r_{12}\left(2\right)r_{13}\left(1\right)-r_{12}\left(1\right)r_{13}\left(2\right)}{r}_{12}$

Find the solution, wherein (x _i, y _i, z _i) and R _i(i=1,2,3) are respectively three gps satellites and pseudorange, R ₁₂And R ₁₃Be respectively (x ₂, y ₂, z ₂), (x ₃, y ₃, z ₃) and (x ₁, y ₁, z ₁) distance,

$A_{12}=(x_1,y_1,z_1)+\frac{{R_{12}}^2+{R_1}^2-{R_2}^2}{{{2R}_{12}}^2}(x_2-x_1,y_2-y_1,y_2-z_1)$

$A_{13}=(x_1,y_1,z_1)+\frac{{R_{13}}^2+{R_1}^2-{R_3}^2}{{{2R}_{13}}^2}(x_3-x_1,y_3-y_1,y_3-z_1)$

$r=\frac{(x_2-x_1,y_2-y_1,z_2-z_1)\&times;(x_3-x_1,y_3-y_1,z_3-z_1)}{\sqrt{{(x_2-x_1)}^2{+(y_2-y_1)}^2+{(z_2-z_1)}^2}\sqrt{{(x_3-x_1)}^2+{(y_3-y_1)}^2+{(z_3-z_1)}^2}}$

r ₁₂＝r×(x ₂-x ₁，y ₂-y ₁，y ₂-z ₁)

r ₁₃＝r×(x ₃-x ₁，y ₃-y ₁，y ₃-z ₁)

It all is intermediate vector.

3. the iteratively faster method of a kind of GPS according to claim 1 location, the method for position iterative is:

The position equation group is rewritten as following iterative equation group

$x_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{x}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{x_i-x_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{\left(k\right)})}^2+{(z_i-z_0^{\left(k\right)})}^2}}\frac{R_i}{4}$

$y_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{y}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{y_i-y_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{0\left(k\right)})}^2+{(z_i-z_0^k)}^2}}\frac{R_i}{4}$

$z_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{z}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{z_i-z_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{\left(k\right)})}^2+{(z_i-z_0^{\left(k\right)})}^2}}\frac{R_i}{4}$

For iterative process is set permissible error δ, and carry out following iteration:

1 °) the iteration initial point is set

Put iteration count variable k=1;

2 °) calculate

$x_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{x}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{x_i-x_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{\left(k\right)})}^2+{(z_i-z_0^{\left(k\right)})}^2}}\frac{R_i}{4}$

$y_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{y}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{y_i-y_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{0\left(k\right)})}^2+{(z_i-z_0^k)}^2}}\frac{R_i}{4}$

$z_0^{(k+1)}=\frac{1}{4}\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{z}_i-\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}\frac{z_i-z_0^{\left(k\right)}}{\sqrt{{(x_i-x_0^{\left(k\right)})}^2+{(y_i-y_0^{\left(k\right)})}^2+{(z_i-z_0^{\left(k\right)})}^2}}\frac{R_i}{4}$

3 °) if $|(x_0^{(k+1)},y_0^{(k+1)},z_0^{(k+1)})-\left(x_0^{(k+1)}{,y}_0^{(k+1)}{,z}_0^{(k+1)}\right)|<\mathrm{\&delta;},$ Stop iteration;

4 °) value of k adds 1, returns 2 °).

Stop just to obtain the position location after the iteration

4. the iteratively faster method of a kind of GPS according to claim 1 location is characterized in that the quadratic sum of the distance take each satellite to position to be positioned and the difference of corresponding pseudorange reaches minimum in the iterative process of position and is criterion, namely with function

$F=\underset{i=1}{\overset{4}{\mathrm{\&Sigma;}}}{(\sqrt{{(x_i-x_0)}^2+{(y_i-y_0)}^2+{(z_i-z_0)}^2}-R_i)}^2$

Reaching minimum value is criterion.

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