CN104965191A - Two-site time difference positioning method - Google Patents

Abstract

In view of the state of the art that the current plane time difference positioning needs to use three sites, the invention provides a method of only using two sites to realize target positioning. Firstly, through carrying out simplified processing on a one-dimensional two-base positioning equation, a single-base direction finding formula with a measurement basis in the midpoint of the two sites is acquired; then, a least square method is used for correcting accuracy of the single-base direction finding formula, on the basis of features that a compensation function obtained through curve fitting is not sensitive to target distance changes, a method of using a target bearing ideal value, a single-base path difference measurement value and the actually-used base line length value for compensating direction finding deviation is given. On the basis, according to a same difference relationship between the single-base path difference and the two-base path difference proved by a two-base three-site path difference equation, and an actually-measured single-base path difference value and an arrival angle through correction, two path difference values corresponding to a one-dimensional two-base array are obtained recursively, and thus two independent positioning equations meeting requirements of two-dimensional plane positioning can be equivalently acquired.

Description

A kind of dual station time difference positioning method

Technical field

The present invention relates to radio-location technology, be specifically related to a kind of dual station time difference positioning method.

Background technology

Positioning using TDOA is that a kind of time of arrival of being measured the same pulse of same radiation source by multiple receiver is poor, is then obtained the method for radiation source positions according to the position of each platform and the time difference by central processing station.Under modern measure technology condition, due to the raising of time difference measurements precision, time difference position technolot has become a kind of more accurate localization method, and is widely used.

The mistiming being arrived two stations by target emanation signal can determine the hyperboloid that a pair is focus with two stations, the hyperbolic position therefore time difference locating technology is otherwise known as.According to existing hyperbolic navigation positioning principle, two dimensional surface at least must use three stations, obtain two independently path differences thus, in order to construct two independently positioning equations.Three dimensions then needs to arrange four stations, and in order to produce three pairs of hyperboloids, intersect to obtain line by face, face, segment-Facet intersection invocation point realizes the location Calculation to target location.

Summary of the invention

The state of the art of three websites must be used for existing plane positioning using TDOA, the object of the invention is to utilize through single base direction finding value of Curve fitting compensation, and the equal difference relation between single double-basis path difference give a kind of dual station time difference measurement that only uses can the method for realize target location.

The present invention is achieved through the following technical solutions.

First be single base direction finding formula at single baseline point midway place by obtaining a witness mark benchmark to the simplify processes of one dimension double-basis eikonal equation.

Subsequently, least square method is adopted to give the compensation method that effectively can be improved single base direction finding accuracy.First according to the target range that location systems art index is given, or by believable information estimating target distance; The theoretical value of the angle of arrival, the actual base length node-by-node algorithm path difference used is utilized subsequently on the detection interval of whole target azimuth; Further for the consideration to substantial measurement errors, and unusual in order to avoid calculating appearance, artificial correction is carried out to path difference theoretical value; Then pre-estimating the deviation of the target angle of arrival by the path difference value through revising, the correction to single base direction finding value can be realized by least square method on this basis.

Although utilize the method for curve can obtain the direction finding Exact Solutions of restraining, curve-fitting method inherently needs the Exact Solutions using direction finding, and single base direction finding only can provide approximate solution comparatively accurately.The reason finally breaking away from this predicament analyzes to find that the penalty function obtained by curve is mainly relevant to the change of base length and path length difference, and insensitive to the change of target range, especially when base length is less than 100 kilometers.Just based on this characteristic, just can adopt the approximate evaluation value of target radial distance, by the ideal value of target azimuth, base length and actual measurement path difference, calculate the offset of direction finding deviation.

In fact, based on this feature, another kind of compensation way is according to system works distance, and precalculate and store the direction finding bias compensation value corresponding to certain specific base length, the actual measurement approximate value then when real work according to the target angle of arrival carries out real-time calling.

On the basis by compensating the accurate direction finding solution of acquisition, can demonstrate,prove have a kind of equal difference relation between single base path difference and double-basis path difference by double-basis three station eikonal equation, based on this equal difference relation, utilize single base path difference value of actual measurement and the angle of arrival through revising, just recursion can obtain two the path difference values corresponding to one dimension double-basis array, the acquisition that just energy is of equal value thus meets two location-independent equations required for two dimensional surface location.

Specifically comprise the following steps:

The time difference measurement of step 1, the baseline mid point angle of arrival.If form a positioning using TDOA system by two stations, utilize the time difference of measuring and obtaining, be calculated as follows the target angle of arrival θ in two station baseline midpoint:

$\sin \mathrm{\θ}=\frac{\mathrm{\Δ}{R}_{13}}{2D_0}=\frac{v_c\mathrm{\Δt}}{2D_0}---\left(1\right)$

In formula: Δ R ₁₃it is the path difference between two stations; v _cfor the light velocity; Δ t is the time difference, D ₀for base length.

Step 2, on whole object position-sensitive detecting interval, calculate the theoretical value of path difference, and revised.According to the target range that location systems art index is given, or by believable information estimating target distance, utilize the theoretical value of the angle of arrival, the actual base length used, the distance by cosine law node-by-node algorithm baseline two ends:

$r_{1i}=\sqrt{r_2^2+D_0^2+2r_2{D}_0\sin {\mathrm{\θ}}_i}---\left(2\right)$

$r_{3i}=\sqrt{r_2^2+D_0^2-2r_2{D}_0\sin {\mathrm{\θ}}_i}---\left(3\right)$

Obtain path difference thus:

ΔR _i＝r _1i-r _3i(4)

In formula: r ₂it is baseline midpoint target range.Different from formula (1), for succinctly, the subscript of the path difference calculated value when the different angle of arrival no longer comprises site information.

For the consideration to measuring error, and unusual in order to avoid occurring when calculating, artificial correction is carried out to path difference theoretical value;

ΔR _ai＝a ₀ΔR _i(5)

In formula: a ₀< 1, span is about about 0.5.

Step 3, utilize least square method to the correction of target angle of arrival measured value.By the theoretical value of the angle of arrival, actual base length and corrected path difference value calculate the deviation data (θ of the angle of arrival _i, Δ ε _i):

Δε _i＝2D ₀sinθ _i-ΔR _ai(6)

And the linear combination function be constructed as follows:

Δε＝c ₁f ₁+c ₂f ₂+c ₃f ₃+c ₄f ₄+c ₅f ₅+c ₆f ₆+c ₇f ₇+c ₈f ₈(7)

Wherein: c _ifor undetermined constant.F _iknown function for following:

$f_1=\frac{\mathrm{\&Delta;}{R}_{\mathrm{ai}}}{2D_0}$

$f_2=\sqrt{1-{\left(\frac{\mathrm{\&Delta;}{R}_{\mathrm{ai}}}{2D_0}\right)}^2}$

f ₃＝f ₁+f ₂

$f_4=f_1{f}_2^2{f}_3$

f ₅＝(f ₁+f ₂) ³

f ₆＝(f ₁+f ₂) ⁷

f ₇＝(f ₁-f ₂) ³

f ₈＝(f ₁-f ₂) ⁸

Following linear equation is set up by least square method:

Ac＝y (8)

Wherein:.

$y=\left[\begin{array}{c}\mathrm{\&Delta;}{\mathrm{\&epsiv;}}_1\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_2\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_i\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_m\end{array}\right]---\left(10\right)$

Solve coefficient thus:

c＝A\y (11)

The deviation delta ε that so far just can try to achieve by curve fitting method revises single base direction finding value:

$\sin \mathrm{\&theta;}=\frac{|\mathrm{\&Delta;}{R}_a+\mathrm{\&Delta;\&epsiv;}|}{2D_0}---\left(12\right)$

In formula: Δ R _a=a ₀Δ R ₁₃for the path difference measured value through revising.Absolute value in formula is to eliminate in the imaginary part phenomenon that likely there will be close to deviation delta ε during baseline axis direction.

Practical application also can precalculate according to operating distance and store the direction finding bias compensation value corresponding to certain specific base length, and the actual measurement approximate value then when real work according to the target angle of arrival carries out real-time calling.

Step 4, demonstrate,prove to obtain by double-basis three station eikonal equation the equal difference relation that has between single base path difference and double-basis path difference, based on this equal difference relation, utilizing single base path difference value of actual measurement and the angle of arrival through revising, just recursion can obtain two the path difference values corresponding to one dimension double-basis array.

From geometric relationship, single base path difference of corresponding array total length is the path difference sum of adjacent two baselines:

ΔR ₁₃＝r ₁-r ₃＝(r ₁-r ₂)+(r ₂-r ₃)＝ΔR ₁+ΔR ₂(13)

And the length sum of adjacent two baselines equals array total length.As by single base path difference by one dimension double-basis path difference direction finding formula:

$\sin {\mathrm{\&theta;}}_2=\frac{(D_0^2-\mathrm{\&Delta;}{R}_1^2)\mathrm{\&Delta;}{R}_2+(D_0^2-\mathrm{\&Delta;}{R}_2^2)\mathrm{\&Delta;}{R}_1}{D_0(2D_0^2-\mathrm{\&Delta;}{R}_1^2-\mathrm{\&Delta;}{R}_2^2)}---\left(14\right)$

In path difference Δ R ₁or Δ R ₂displace, then have:

$\sin \mathrm{\&theta;}=\frac{(D_0^2-\mathrm{\&Delta;}{R}_1^2)(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)+[D_0^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)}^2]\mathrm{\&Delta;}{R}_1}{D_0\left[2D_0^2-\mathrm{\&Delta;}{R}_1^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)}^2\right]}---\left(15\right)$

$\sin \mathrm{\&theta;}=\frac{[D_0^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)}^2]\mathrm{\&Delta;}{R}_2+(D_0^2-\mathrm{\&Delta;}{R}_2^2)(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)}{D_0[2D_0^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)}^2-\mathrm{\&Delta;}{R}_2^2]}---\left(16\right)$

Therefrom can obtain the result being similar to arithmetic series:

$\mathrm{\&Delta;}{R}_1=\frac{\mathrm{\&Delta;}{R}_{13}}{2}+\mathrm{\&Delta;a}---\left(17\right)$

$\mathrm{\&Delta;}{R}_2=\frac{\mathrm{\&Delta;}{R}_{13}}{2}-\mathrm{\&Delta;a}---\left(18\right)$

Wherein:

$\mathrm{\&Delta;a}=0.5\sqrt{\frac{\mathrm{\&Delta;}{R}_{13}^2(\mathrm{\&Delta;}{R}_{13}+2D_0\sin {\mathrm{\&theta;}}_2)-4D_0(\mathrm{\&Delta;}{R}_{13}^2\sin {\mathrm{\&theta;}}_2+D_0\mathrm{\&Delta;}{R}_{13}-2D_0^2\sin {\mathrm{\&theta;}}_2)}{(\mathrm{\&Delta;}{R}_{13}+2D_0\sin {\mathrm{\&theta;}}_2)}}---\left(19\right)$

Step 5, try to achieve the distance and bearing in array end points target by one dimension double-basis positioning equation, at the distance and bearing of baseline right-hand member place target be:

$r_3=\frac{2D_0^2+2\mathrm{\&Delta;}{R}_2^2-\mathrm{\&Delta;}{R}_{13}^2}{2(\mathrm{\&Delta;}{R}_{13}-2\mathrm{\&Delta;}{R}_2)}---\left(20\right)$

$\sin {\mathrm{\&theta;}}_3=\frac{\mathrm{\&Delta;}{R}_2}{D_0}-\frac{D_0^2-\mathrm{\&Delta;}{R}_2^2}{2D_0{r}_3}---\left(21\right)$

Wherein:

ΔR ₁₃＝v _cΔt (22)

$\mathrm{\&Delta;}{R}_2=0.5v_c\mathrm{\&Delta;t}-0.5\sqrt{\mathrm{\&Delta;\&epsiv;}}\sqrt{\frac{4D_0^2-{\left(v_c\mathrm{\&Delta;t}\right)}^2}{(2v_c\mathrm{\&Delta;t}+\mathrm{\&Delta;\&epsiv;})}}---\left(23\right)$

Characteristic of the present invention:

1. decrease the quantity of detection site, not only reduce arrangement and the use cost of positioning system, but also reduce the difficulty of addressing.

2., because only comprising a time difference measurement, so for same base length, the overall measurement error of the time difference locating technology system being more than or equal to three than website quantity is little by the overall measurement error of bistatic location.This is because total observational error will be the component sum of measuring error of multiple different independently metering system, if independently observed quantity is more, then overall measurement error will more be difficult to be lowered.

3., under positioning using TDOA system, the locating and tracking precision of target is main relevant to the relative geometrical relation between target and observer when measuring error etc. remains unchanged.Owing to only there being two stations, the impact of arrangement manner on the locating and tracking precision of target will effectively be reduced, and thereby simplify the analysis of positioning system.

4. not only can be used for early detection, more can be used for in-plant navigator fix.

Accompanying drawing explanation

Fig. 1: one dimension Bistatic

Fig. 2: compensate curve during different radial distances

Fig. 3: relative range error during different stop spacing

Fig. 4: angle measurement error

Embodiment

Further illustrate the present invention below in conjunction with accompanying drawing 1-Fig. 4 how to realize.

Embodiment

A kind of dual station time difference positioning method.Accompanying drawing 1 is one dimension Bistatic; Figure 2 shows compensate curve during different radial distances; Relative range error when accompanying drawing 3 is different stop spacings; Accompanying drawing 4 is angle measurement error.

This patent proposes a kind of dual station time difference positioning method, this cardinal principle of an independently path difference value realize target location that only utilizes is: obtaining on the basis of accurate direction finding value compensating based on single baseline time difference measurement and matching, the equal difference relation had between single base path difference and double-basis path difference is utilized to solve path difference value corresponding to double-basis line, two independent equations be met required for plane positioning that thus just can be of equal value.

This method solving double-basis path difference by the single base path difference of actual measurement must rely on direction finding solution accurately, the main cause being finally able to engineer applied is the impact of change on compensate value of target range is insensitive, just can improve the accuracy of single base direction finding thus by least square method.

One, three path difference location, station

1, fundamental equation

For the one dimension double-basis array shown in accompanying drawing 1, following two geometry subsidiary equations can be listed by the cosine law:

$r_1^2=r_2^2+D_1^2-2D_1{r}_2\cos (90+{\mathrm{\&theta;}}_2)=r_2^2+D_1^2+2D_1{r}_2\sin {\mathrm{\&theta;}}_2---\left(1\right)$

$r_3^2=r_2^2+D_2^2-2D_2{r}_2\cos (90-{\mathrm{\&theta;}}_2)=r_2^2+D_2^2-2D_2{r}_2\sin {\mathrm{\&theta;}}_2---\left(2\right)$

In formula: θ ₂it is the target angle of arrival of central site; D _ithe length of baseline; r _ifor radial distance.

Because having:

x＝r ₂sinθ ₂(3)

In formula: x is the horizontal ordinate in rectangular coordinate system.

Therefore geometry subsidiary equation can be rewritten as:

$r_1^2=r_2^2+D_1^2+2D_{1}x---\left(4\right)$

$r_3^2=r_2^2+D_2^2-2D_{2}x---\left(5\right)$

2, interior point location

According to corresponding to baseline D ₁and D ₂path difference solve positioning equation, then the middle end points 2 of receiving array will become the public connecting end point of adjacent two baselines, claim this kind of locator meams be based on put in baseline path difference location.

Corresponding to baseline D ₁and D ₂eikonal equation respectively:

ΔR ₁＝r ₁-r ₂(6)

ΔR ₂＝r ₂-r ₃(7)

In formula: Δ R _iit is path difference.

Above-mentioned eikonal equation is substituted into geometry assist type (4) and (5), after transposition arranges, has following binary once linear system of equations:

$2D_{1}x-2\mathrm{\&Delta;}{R}_1{r}_2=\mathrm{\&Delta;}{R}_1^2-D_1^2---\left(8\right)$

$2D_{2}x-2\mathrm{\&Delta;}{R}_2{r}_2=D_2^2-\mathrm{\&Delta;}{R}_2^2---\left(9\right)$

Therefrom directly can solve the target range measured based on path difference:

$r_2=\frac{(D_1^2-\mathrm{\&Delta;}{R}_1^2)D_2+(D_2^2-\mathrm{\&Delta;}{R}_2^2)D_1}{2(\mathrm{\&Delta;}{R}_1{D}_2-\mathrm{\&Delta;}{R}_2{D}_1)}---\left(10\right)$

And the target angle of arrival:

$\sin {\mathrm{\&theta;}}_2=\frac{x}{r_2}=\frac{(D_1^2-\mathrm{\&Delta;}{R}_1^2)\mathrm{\&Delta;}{R}_2+(D_2^2-\mathrm{\&Delta;}{R}_2^2)\mathrm{\&Delta;}{R}_1}{(D_1^2-\mathrm{\&Delta;}{R}_1^2)D_2+(D_2^2-\mathrm{\&Delta;}{R}_2^2)D_1}---\left(11\right)$

As equal in adjacent two baselines: D ₁=D ₂=D ₀, then range finding and direction finding expression are:

$r_2=\frac{2D_0^2-\mathrm{\&Delta;}{R}_1^2-\mathrm{\&Delta;}{R}_2^2}{2(\mathrm{\&Delta;}{R}_1-\mathrm{\&Delta;}{R}_2)}---\left(12\right)$

$\sin {\mathrm{\&theta;}}_2=\frac{(D_0^2-\mathrm{\&Delta;}{R}_1^2)\mathrm{\&Delta;}{R}_2+(D_0^2-\mathrm{\&Delta;}{R}_2^2)\mathrm{\&Delta;}{R}_1}{D_0({2D}_0^2-\mathrm{\&Delta;}{R}_1^2-\mathrm{\&Delta;}{R}_2^2)}---\left(13\right)$

3, outer point location

If one of selected path difference is the total length of array, then some outer point will become the points of common connection of two baselines, claim this kind of locator meams to be locate based on the path difference of exterior point.

Corresponding to baseline total length D ₁₃eikonal equation be:

ΔR ₁₃＝r ₁-r ₃(14)

By above formula with corresponding to baseline D ₂path difference formula (7) substitute into geometry assist type (4) and (5), transposition arrangement after have following binary once linear system of equations:

$-2D_{1}x+2(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)r_3=D_1^2+\mathrm{\&Delta;}{R}_2^2-\mathrm{\&Delta;}{R}_{13}^2---\left(15\right)$

$2D_{2}x-2\mathrm{\&Delta;}{R}_2{r}_3=D_2^2+\mathrm{\&Delta;}{R}_2^2---\left(16\right)$

Therefrom directly can solve the target range at website 3 place:

$r_3=\frac{(D_1^2+R_2^2-\mathrm{\&Delta;}{R}_{13}^2)D_2+(D_2^2+\mathrm{\&Delta;}{R}_2^2)D_1}{2[(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)D_2-\mathrm{\&Delta;}{R}_2{D}_1]}---\left(17\right)$

Can solve again:

$\begin{array}{c}x=\frac{D_2^2+\mathrm{\&Delta;}{R}_2^2+2\mathrm{\&Delta;}{R}_2{r}_3}{2D_2}\\ =\frac{D_2^2+\mathrm{\&Delta;}{R}_2^2}{2D_2}+\frac{\mathrm{\&Delta;}{R}_2}{D_2}\frac{[(D_1^2+\mathrm{\&Delta;}{R}_2^2-\mathrm{\&Delta;}{R}_{13}^2)D_2+(D_2^2+\mathrm{\&Delta;}{R}_2^2)D_1]}{2[(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)D_2-\mathrm{\&Delta;}{R}_2{D}_1]}\end{array}---\left(18\right)$

The target angle of arrival of trying to achieve thus on the right side of one dimension double-basis array is:

$\sin {\mathrm{\&theta;}}_3=\frac{x-D_2}{r_3}=\frac{D_2^2+\mathrm{\&Delta;}{R}_2^2}{2D_2{r}_3}+\frac{\mathrm{\&Delta;}{R}_2}{D_2}-\frac{D_2}{r_3}---\left(19\right)$

As equal in adjacent two baselines: D ₁=D ₂=D ₀, then range finding and direction finding solution are:

$r_3=\frac{{2D}_0^2+2\mathrm{\&Delta;}{R}_2^2-\mathrm{\&Delta;}{R}_{13}^2}{2(\mathrm{\&Delta;}{R}_{13}-2\mathrm{\&Delta;}{R}_2)}---\left(20\right)$

$\sin {\mathrm{\&theta;}}_3=\frac{\mathrm{\&Delta;}{R}_2}{D_0}-\frac{D_0^2-\mathrm{\&Delta;}{R}_2^2}{2D_0{r}_3}---\left(21\right)$

Two, the path difference relation between single double-basis

From geometric relationship, the path difference of corresponding array total length equals the path difference sum of adjacent two baselines:

ΔR ₁₃＝r ₁-r ₃＝(r ₁-r ₂)+(r ₂-r ₃)＝ΔR ₁+ΔR ₂(22)

And the length sum of adjacent two baselines equals array total length.As by this relational expression respectively by the path difference Δ R in one dimension double-basis path difference direction finding formula (13) ₁or Δ R ₂displace, then have:

$\sin \mathrm{\&theta;}=\frac{(D_0^2-\mathrm{\&Delta;}{R}_1^2)(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)+[D_0^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)}^2]\mathrm{\&Delta;}{R}_1}{D_0\left[2D_0^2-\mathrm{\&Delta;}{R}_1^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)}^2\right]}---\left(23\right)$

Therefrom can solve two path difference values corresponding to one dimension Bistatic respectively:

$\mathrm{\&Delta;}{R}_1=\frac{\mathrm{\&Delta;}{R}_{13}}{2}+\mathrm{\&Delta;a}---\left(25\right)$

$\mathrm{\&Delta;}{R}_2=\frac{\mathrm{\&Delta;}{R}_{13}}{2}-\mathrm{\&Delta;a}---\left(26\right)$

Wherein:

$\mathrm{\&Delta;a}=0.5\sqrt{\frac{\mathrm{\&Delta;}{R}_{13}^2(\mathrm{\&Delta;}{R}_{13}+2D_0\sin {\mathrm{\&theta;}}_2)-4D_0(\mathrm{\&Delta;}{R}_{13}^2\sin {\mathrm{\&theta;}}_2+D_0\mathrm{\&Delta;}{R}_{13}-2D_0^2\sin {\mathrm{\&theta;}}_2)}{(\mathrm{\&Delta;}{R}_{13}+2D_0\sin {\mathrm{\&theta;}}_2)}}---\left(27\right)$

Demonstrate,prove thus, in one dimension Bistatic, the path difference value corresponding to each single baseline is 1/2nd of the path difference value corresponding to total base length just, then increases and decreases an identical departure.Analog computation shows, small departure Δ a only could be solved by correct when the angle of arrival itself accurately solves, and as adopted approximate direction finding value, then the result obtained will be disabled.

Three, single base direction finding and curve compensation

1, single base direction finding

According to existing analysis, two adjacent high-order path differences can be thought equal by approximate, so equidistant direction finding formula (13) can be reduced to:

$\sin {\mathrm{\&theta;}}_2\&ap;\frac{(D_0^2-\mathrm{\&Delta;}{R}_2^2)(\mathrm{\&Delta;}{R}_1+\mathrm{\&Delta;}{R}_2)}{{2D}_0(D_0^2-\mathrm{\&Delta;}{R}_1^2)}=\frac{(\mathrm{\&Delta;}{R}_1+\mathrm{\&Delta;}{R}_2)}{2D_0}---\left(28\right)$

Cause:

ΔR ₁₃＝r ₁-r ₃＝(r ₁-r ₂)+(r ₂-r ₃)＝ΔR ₁+ΔR ₂

There is single base direction finding formula:

$\sin {\mathrm{\&theta;}}_2=\frac{\mathrm{\&Delta;}{R}_{13}}{2D_0}---\left(29\right)$

As can be seen here: the witness mark reference point of single baseline direction finding is the point midway place at single baseline, instead of at the end points place, left and right of single baseline.And analog computation result shows, be not only correct to Short baseline to the single base direction finding formula (29) after reference mark shift-corrected, but also can be applicable to comparatively Long baselines.

2, the compensation of deviation

Analog computation shows, simplifies because of approximate, has a small deviation, single base direction finding formula (29) can be rewritten as this reason between the calculated value and theoretical exact value of single base direction finding formula:

$\sin \mathrm{\&theta;}=\frac{\mathrm{\&Delta;}{R}_{13}+\mathrm{\&Delta;\&epsiv;}}{2D_0}---\left(30\right)$

In formula: Δ ε is departure.And in order to simplify subsequent analysis, start the subscript omitting the angle of arrival from here.

By transposition process, formula (30) can be represented as:

Δε＝2D ₀sinθ-ΔR ₁₃(31)

Deviation can be regarded as the difference between approximate path difference and actual measurement path difference thus.On the basis of formula (31), following linear combination function can be constructed:

Δε＝c ₁f ₁+c ₂f ₂+c ₃f ₃+c ₄f ₄+c ₅f ₅+c ₆f ₆+c ₇f ₇+c ₈f ₈(32)

In base: c _ifor undetermined constant; f _ifor known function.

By sunykatuib analysis, utilize only known parameter D ₀with Δ R ₁₃, each function item can be expressed as:

$f_1=\frac{\mathrm{\&Delta;}{R}_{13}}{2D_0}$

$f_2=\sqrt{1-{\left(\frac{\mathrm{\&Delta;}{R}_{13}}{2D_0}\right)}^2}$

f ₃＝f ₁+f ₂

$f_4=f_1{f}_2^2{f}_3$

f ₅＝(f ₁+f ₂) ³

f ₆＝(f ₁+f ₂) ⁷

f ₇＝(f ₁-f ₂) ³

f ₈＝(f ₁-f ₂) ⁸

Some function items are wherein that the mode calculated by hand simulation is obtained, the most basic function f ₁and f ₂in fact be exactly the sine and cosine functions of the approximate target angle of arrival.

By direction finding value and the path difference value acquisition deviation data (θ of whole search coverage _i, Δ ε _i), and set up following linear equation:

Ac＝y (33)

Wherein:

$A=\left[\begin{array}{cccc}{f}_1\left({\mathrm{\&theta;}}_1\right)& f_2\left({\mathrm{\&theta;}}_1\right)& ...& f_8\left({\mathrm{\&theta;}}_1\right)\\ f_1\left({\mathrm{\&theta;}}_2\right)& f_2\left({\mathrm{\&theta;}}_2\right)& & f_8\left({\mathrm{\&theta;}}_2\right)\\ f_1\left({\mathrm{\&theta;}}_i\right)& f_2\left({\mathrm{\&theta;}}_i\right)& & f_8\left({\mathrm{\&theta;}}_i\right)\\ f_1\left({\mathrm{\&theta;}}_m\right)& f_2\left({\mathrm{\&theta;}}_m\right)& ...& f_8\left({\mathrm{\&theta;}}_m\right)\end{array}\right]---\left(34\right)$

$y=\left[\begin{array}{c}\mathrm{\&Delta;}{\mathrm{\&epsiv;}}_1\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_2\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_i\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_m\end{array}\right]---\left(35\right)$

Least square solution be c=A y.After trying to achieve deviation delta ε by matching, when in the end calculating the exact value of the angle of arrival, attention must add absolute value, namely has:

$\sin \mathrm{\&theta;}=\frac{|\mathrm{\&Delta;}{R}_{13}+\mathrm{\&Delta;\&epsiv;}|}{2D_0}---\left(36\right)$

This is to eliminate in curve fitting process, and when close to baseline axis direction, departure Δ ε likely there will be imaginary number phenomenon.

3, to the consideration of path difference measuring error

In fact, when actual measurement, path difference value itself, inevitably with measuring error, to this, can adopt a coefficient to carry out approximate representation path difference measured value departing from exact value:

ΔR _13w＝(1±w)ΔR ₁₃(37)

In formula: ± w simulates measured value to the deviation of exact value.

Simulation subsequently shows, when supposing that measured value is less than ideal value, utilize curve fitting method still can obtain original or even better Compensate approximate result, but when measured value is greater than exact value, by appearance because path difference value is greater than single base length, thus cause the function containing extraction of square root:

$f_2=\sqrt{1-{\left(\frac{\mathrm{\&Delta;}{R}_{13}}{2D_0}\right)}^2}---\left(37\right)$

Abnormal situation is there is when calculating.Be artificial reduction path difference value to this solution, such as, be multiplied by the coefficient (such as getting 0.5) that is less than 1 on the right of expression:

ΔR _13w＝0.5(1±w)ΔR ₁₃

After process like this, still can ensure the convergence of compensation calculation when w gets+0.5, and the calculating of direction finding accuracy is not affected, and still can keep original level.But should be noted that the path difference now in single base direction finding formula when calculating, and when carrying out least square method and calculating, all should adopt the path difference value through artificially simulating amendment.Finally from the needs of actual computation, and in order to be concise in expression, the measuring error coefficient of reflection path difference and artificial correction factor are combined, given path difference amendment type is:

ΔR _13a＝a ₀ΔR ₁₃(38)

In formula: a ₀be one and considered the measuring error of path difference and the coefficient of artificial modifying factor.

4, the variation characteristic of deviation

Based on curve fitting method, need to know direction finding value accurately to the acquisition of deviation data, but measuring the obtainable angle of arrival with the path difference between dual station is only an approximate value, and as do not obtained direction finding solution accurately, then aforesaid whole analysis seems to be only a theoretic proof of pure mathematics.If still can obtain deviation data when the angle of arrival is not also determined accurately, then the penalty method based on curve could be used for engineering reality.

Further sunykatuib analysis proves, as done larger depreciation correction to path difference artificially in advance, the calculated value such as getting path difference is the half of actual measured value, upper trifle has illustrated to do so can't affect the correct of corrected Calculation and accuracy, then when base length is determined, the impact of change on compensate value of target range is very little.If base length is only in the scope below 100 kilometers, then the compensation of deviation will present substantially not with the characteristic of radial distance change.

Figure 2 shows when base length is 50 kilometers, to the deviation compensation curve of different radial distances target.Based on compensating substantially not with the characteristic of radial distance change, we just for specific base length, can obtain by precalculating and preserve deviation data.Single base direction finding formula (29) can be first adopted to know the angle of arrival of target roughly comparatively accurately during practical engineering application, then from the deviation data of storage, select corresponding deviation data in order to revise to direction finding formula according to measured angle value, thus just can obtain direction finding value more accurately.

Analog computation shows, when baseline is larger, such as, is greater than hundreds of kilometer, and the polymerization property of bias compensation value will be deteriorated, and now according to different radial distances, can adopt the mode of segmented compensation.

Four, only based on the positioning error of single base path difference

1, basic skills

Use the geometric relationship shown in accompanying drawing 1, be located at position 1 and 3 place and be provided with two detection sites, and adopt time difference measurement to obtain corresponding to single Long baselines D ₁₃path difference Δ R ₁₃, compensated direction finding deviation by curve subsequently, make to be positioned at Long baselines mid point, the direction finding accuracy at position 2 place effectively improves, and can be used for solving the deviation delta a corresponding to double-basis path difference.Now, recycle relational expression (25) or (26) that channel syndrome obtains, just can accurately obtain each path difference value corresponding to one dimension double-basis array.

On this basis, just the present entity site location of the fructufy obtained by exterior point positioning analysis can be utilized the range finding of target and direction finding:

$r_3=\frac{2D_0^2+2\mathrm{\&Delta;}{R}_2^2-\mathrm{\&Delta;}{R}_{13}^2}{2(\mathrm{\&Delta;}{R}_{13}-2\mathrm{\&Delta;}{R}_2)}---\left(20\right)$

$\sin {\mathrm{\&theta;}}_3=\frac{\mathrm{\&Delta;}{R}_2}{D_0}-\frac{D_0^2-\mathrm{\&Delta;}{R}_2^2}{2D_0{r}_3}---\left(21\right)$

Wherein:

ΔR ₁₃＝v _cΔt

$\mathrm{\&Delta;}{R}_2=0.5v_c\mathrm{\&Delta;t}-0.5\sqrt{\mathrm{\&Delta;\&epsiv;}}\sqrt{\frac{4D_0^2-{\left(v_c\mathrm{\&Delta;t}\right)}^2}{(2v_c\mathrm{\&Delta;t}+\mathrm{\&Delta;\&epsiv;})}}$

2, relative range error

For ease of error analysis, if transition function:

$p=2D_0^2+2\mathrm{\&Delta;}{R}_2^2-\mathrm{\&Delta;}{R}_{13}^2---\left(39\right)$

q＝2(ΔR ₁₃-2ΔR ₂) (40)

Namely have:

$r_3=\frac{p}{q}---\left(41\right)$

When not considering the measuring error of path difference, the differential of distance to time difference Δ t is:

$r_3=\frac{1}{q^2}(q\frac{\&PartialD;p}{\&PartialD;\mathrm{\&Delta;t}}-p\frac{\&PartialD;q}{\&PartialD;\mathrm{\&Delta;t}})---\left(42\right)$

Wherein:

$\frac{\&PartialD;p}{\&PartialD;\mathrm{\&Delta;t}}=4\mathrm{\&Delta;}{R}_2\frac{\&PartialD;{\mathrm{\&Delta;R}}_2}{\&PartialD;\mathrm{\&Delta;t}}-2\mathrm{\&Delta;}{R}_{13}\frac{\&PartialD;\mathrm{\&Delta;}{R}_{13}}{\&PartialD;\mathrm{\&Delta;t}}$

$\frac{\&PartialD;q}{\&PartialD;\mathrm{\&Delta;t}}=2(\frac{\&PartialD;\mathrm{\&Delta;}{R}_{13}}{\&PartialD;\mathrm{\&Delta;t}}-2\frac{\&PartialD;\mathrm{\&Delta;}{R}_2}{\&PartialD;\mathrm{\&Delta;t}})$

And have: $\frac{\&PartialD;\mathrm{\&Delta;}{R}_3}{\&PartialD;\mathrm{\&Delta;t}}=v_c$

For path difference Δ R ₂, in current analysis, temporarily wherein the comprised deviation delta ε being used for compensating path difference is regarded as a small constant, partial differential obtained thus is:

$\frac{\&PartialD;\mathrm{\&Delta;}{R}_2}{\&PartialD;\mathrm{\&Delta;t}}=0.5v_c+0.5\frac{v_c\sqrt{\mathrm{\&Delta;\&epsiv;}}}{(2v_c\mathrm{\&Delta;t}+\mathrm{\&Delta;\&epsiv;})}[v_c\mathrm{\&Delta;t}\sqrt{\frac{2v_c\mathrm{\&Delta;t}+\mathrm{\&Delta;\&epsiv;}}{4D_0^2-{\left(v_c\mathrm{\&Delta;t}\right)}^2}}+\sqrt{\frac{4D_0^2-{\left(v_c\mathrm{\&Delta;t}\right)}^2}{2v_c\mathrm{\&Delta;t}+\mathrm{\&Delta;\&epsiv;}}}]---\left(43\right)$

According to error theory, ignore the error effect of spacing between station, the relative range error only produced by time difference measurement is:

In formula: σ _{Δ t}for the root-mean-square error of time difference measurement.

Fig. 3 gives relative range error curve during different stop spacing, as seen from the figure, be greater than after 3 kilometers at baseline, in larger angle of arrival region, substantially can meet the technical requirement of 0.1%R.

3, angle measurement error

Have direction finding formula (21) direct differentiation:

$\frac{\&PartialD;{\mathrm{\&theta;}}_3}{\&PartialD;\mathrm{\&Delta;t}}==\frac{1}{D_0\cos {\mathrm{\&theta;}}_3}\frac{\&PartialD;\mathrm{\&Delta;}{R}_2}{\&PartialD;\mathrm{\&Delta;t}}+\frac{1}{2D_0{r}_3\cos {\mathrm{\&theta;}}_3}(2\mathrm{\&Delta;}{R}_2\frac{\&PartialD;\mathrm{\&Delta;}{R}_2}{\&PartialD;\mathrm{\&Delta;t}}+\frac{D_0^2-\mathrm{\&Delta;}{R}_2^2}{r_3}\frac{\&PartialD;r_3}{\&PartialD;\mathrm{\&Delta;t}})---\left(45\right)$

The angle measurement error only produced by time difference measurement is:

${\mathrm{\&sigma;}}_{\mathrm{\&theta;}}=\frac{180}{\mathrm{\&pi;}}\left|\frac{\&PartialD;{\mathrm{\&theta;}}_3}{\&PartialD;\mathrm{\&Delta;t}}\right|{\mathrm{\&sigma;}}_{\mathrm{\&Delta;t}}---\left(46\right)$

Fig. 4 shows angle measurement error curve during different stop spacing.

4, conclude the speech

Analysis result shows, the distance accuracy only obtained by single base time difference measurement than double-basis time difference measurement result as well, this mathematical form should be rational, because, for double-basis positioning using TDOA, general orientation error will be two error component sums that are different, independently time difference measurement, and for single base positioning using TDOA, then only comprises the error component of a time difference measurement.

Claims (6)

1. a dual station time difference positioning method, is characterized in that utilizing through single base direction finding value of Curve fitting compensation, and the equal difference relation between single double-basis path difference obtains and meets required for two dimensional surface location two independently positioning equations.

2. method according to claim 1, is characterized in that by obtaining a witness mark benchmark to the simplify processes of one dimension double-basis positioning equation be single base direction finding formula at dual station mid point:

$\sin \mathrm{\&theta;}=\frac{\mathrm{\&Delta;}{R}_{13}}{2D_0}=\frac{v_c\mathrm{\&Delta;t}}{2D_0}---\left(1\right)$ In formula: Δ R ₁₃it is the path difference between two stations; v _cfor the light velocity; Δ t is the time difference; D ₀for base length.

3. method according to claim 1, is characterized in that calculating path difference theoretical value on whole object position-sensitive detecting interval, and is revised.According to the target range that location systems art index is given, or by believable information estimating target distance, utilize the theoretical value of the angle of arrival, the actual base length used, the distance by cosine law node-by-node algorithm baseline two ends:

$r_{1i}=\sqrt{r_2^2+D_0^2+2r_2{D}_0\sin {\mathrm{\&theta;}}_i}---\left(2\right)$

$r_{3i}=\sqrt{r_2^2+D_0^2-2r_2{D}_0\sin {\mathrm{\&theta;}}_i}---\left(3\right)$

Obtain path difference thus:

ΔR _i＝r _1i-r _3i(4)

In formula: v ₂it is the target range of baseline midpoint.Different from formula (1), for succinctly, the subscript of the path difference calculated value when the different angle of arrival no longer comprises site information.

Further for the consideration to substantial measurement errors, and unusual in order to avoid calculating appearance, artificial correction is carried out to path difference theoretical value;

ΔR _ai＝a ₀ΔR _i(5)

In formula: a ₀< 1, span is about about 0.5.

4. the method according to claim 1,2 and 3, is characterized in that utilizing the measured value of least square method to the target angle of arrival to revise.By the theoretical value of the angle of arrival, actual base length and corrected path difference value calculate the deviation data (θ of the angle of arrival _i, Δ ε _i):

Δε _i＝2D ₀sinθ _i-ΔR _ai(6)

And the linear combination function be constructed as follows:

Δε＝c ₁f ₁+c ₂f ₂+c ₃f ₃+c ₄f ₄+c ₅f ₅+c ₆f ₆+c ₇f ₇+c ₈f ₈(7)

Wherein: c _ifor undetermined constant.F _iknown function for following:

$f_1=\frac{\mathrm{\&Delta;}{R}_{\mathrm{ai}}}{2D_{\stackrel{\&CenterDot;}{0}}}$

$f_2=\sqrt{1-{\left(\frac{\mathrm{\&Delta;}{R}_{\mathrm{ai}}}{2D_0}\right)}^2}$

f ₃＝f ₁+f ₂

$f_4=f_1{f}_2^2{f}_3$

f ₅＝(f ₁+f ₂) ³

f ₆＝(f ₁+f ₂) ⁷

f ₇＝(f ₁-f ₂) ³

f ₈＝(f ₁-f ₂) ⁸

Following linear equation is set up by least square method:

Ac＝y (8)

Wherein:

$A=\left[\begin{array}{cccc}{f}_1\left({\mathrm{\&theta;}}_1\right)& f_2\left({\mathrm{\&theta;}}_1\right)& ..& f_8\left({\mathrm{\&theta;}}_1\right)\\ f_1\left({\mathrm{\&theta;}}_2\right)& f_2\left({\mathrm{\&theta;}}_2\right)& & f_8\left({\mathrm{\&theta;}}_2\right)\\ f_1\left({\mathrm{\&theta;}}_i\right)& f_2\left({\mathrm{\&theta;}}_i\right)& & f_8\left({\mathrm{\&theta;}}_i\right)\\ f_1\left({\mathrm{\&theta;}}_m\right)& f_2\left({\mathrm{\&theta;}}_m\right)& ...& f_8\left({\mathrm{\&theta;}}_m\right)\end{array}\right]---\left(9\right)$

$y=\left[\begin{array}{c}\mathrm{\&Delta;}{\mathrm{\&epsiv;}}_1\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_2\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_i\\ \mathrm{\&Delta;}{\mathrm{\&epsiv;}}_m\end{array}\right]---\left(10\right)$

Solve coefficient thus:

c＝A\y (11)

The deviation delta ε that so far just can try to achieve by curve fitting method revises single base direction finding value:

$\sin \mathrm{\&theta;}=\frac{|\mathrm{\&Delta;}{R}_a+\mathrm{\&Delta;\&epsiv;}|}{2D_0}---\left(12\right)$

In formula: Δ R _a=a ₀Δ R ₁₃for the path difference measured value through revising.Absolute value in formula is to eliminate in the imaginary part phenomenon that likely there will be close to deviation delta ε during baseline axis direction.

Practical application also can according to system works distance, and precalculate and store the direction finding bias compensation value corresponding to certain specific base length, the actual measurement approximate value then when real work according to the target angle of arrival carries out real-time calling.

5. the method according to claim 1 and 4, it is characterized in that demonstrate,proving to obtain by double-basis three station eikonal equation the equal difference relation that has between single base path difference and double-basis path difference, based on this equal difference relation, utilizing single base path difference value of actual measurement and the angle of arrival through revising, just recursion can obtain two the path difference values corresponding to one dimension double-basis array.

From geometric relationship, single base path difference of corresponding array total length is the path difference sum of adjacent two baselines:

ΔR ₁₃＝r ₁-r ₃＝(r ₁-r ₂)+(r ₂-r ₃)＝ΔR ₁+ΔR ₂(13)

And the length sum of adjacent two baselines equals array total length.As by single base path difference by one dimension double-basis path difference direction finding formula:

$\sin {\mathrm{\&theta;}}_2=\frac{(D_0^2-\mathrm{\&Delta;}{R}_1^2)\mathrm{\&Delta;}{R}_2+(D_0^2-\mathrm{\&Delta;}{R}_2^2)\mathrm{\&Delta;}{R}_1}{D_0(2D_0^2-\mathrm{\&Delta;}{R}_1^2-\mathrm{\&Delta;}{R}_2^2)}---\left(14\right)$

In path difference Δ R ₁or Δ R ₂displace, then have:

$\sin \mathrm{\&theta;}=\frac{(D_0^2-\mathrm{\&Delta;}{R}_1^2)(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)+[D_0^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)}^2]\mathrm{\&Delta;}{R}_1}{D_0[2D_0^2-\mathrm{\&Delta;}{R}_1^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_1)}^2]}---\left(15\right)$

$\sin \mathrm{\&theta;}=\frac{[D_0^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)}^2]\mathrm{\&Delta;}{R}_2+(D_0^2-\mathrm{\&Delta;}{R}_2^2)(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)}{D_0[2D_0^2-{(\mathrm{\&Delta;}{R}_{13}-\mathrm{\&Delta;}{R}_2)}^2-\mathrm{\&Delta;}{R}_2^2]}---\left(16\right)$

Therefrom can obtain the result being similar to arithmetic series:

$\mathrm{\&Delta;}{R}_1=\frac{\mathrm{\&Delta;}{R}_{13}}{2}+\mathrm{\&Delta;a}---\left(17\right)$

$\mathrm{\&Delta;}{R}_2=\frac{\mathrm{\&Delta;}{R}_{13}}{2}-\mathrm{\&Delta;a}---\left(18\right)$

Wherein:

$\mathrm{\&Delta;a}=0.5\sqrt{\frac{\mathrm{\&Delta;}{R}_{13}^2(\mathrm{\&Delta;}{R}_{13}+2D_0\sin {\mathrm{\&theta;}}_2)-4D_0(\mathrm{\&Delta;}{R}_{13}^2\sin {\mathrm{\&theta;}}_2+D_0\mathrm{\&Delta;}{R}_{13}-2D_0^2\sin {\mathrm{\&theta;}}_2)}{(\mathrm{\&Delta;}{R}_{13}+2D_0\sin {\mathrm{\&theta;}}_2)}}---\left(19\right)$

6. the method according to claim 1 and 5, is characterized in that trying to achieve distance and bearing in array end points target by one dimension double-basis positioning equation:

$r_3=\frac{2D_0^2+2\mathrm{\&Delta;}{R}_2^2-\mathrm{\&Delta;}{R}_{13}^2}{2(\mathrm{\&Delta;}{R}_{13}-2\mathrm{\&Delta;}{R}_2)}---\left(20\right)$

$\sin {\mathrm{\&theta;}}_3=\frac{\mathrm{\&Delta;}{R}_2}{D_0}-\frac{D_0^2-\mathrm{\&Delta;}{R}_2^2}{2D_0{r}_3}---\left(21\right)$

Wherein:

ΔR ₁₃＝v _cΔt (22)

$\mathrm{\&Delta;}{R}_2=0.5v_c\mathrm{\&Delta;t}-0.5\sqrt{\mathrm{\&Delta;\&epsiv;}}\sqrt{\frac{4D_0^2-{\left(v_c\mathrm{\&Delta;t}\right)}^2}{(2v_c\mathrm{\&Delta;t}+\mathrm{\&Delta;\&epsiv;})}}---\left(23\right)$