11 Ranging by Global Navigation Satellite Systems (GNSS)

Transcription

1 Algebraic Geodesy and Geoinformatics PART II APPLICATIONS 11 Ranging by Global Navigation Satellite Systems (GNSS) Overview First the observation equations are developed for implicit and explicit error definition, and simplified by preliminary elimination. These equations of the 4- Point Problem can be solved in symbolic way via different methods: Sturmfels method, Dixon resultant, standard Groebner basis, reduced Groebner basis and Global Symbolic Solver (GSS). All of these methods give the same result, a quadratic monomial. In addition Global Numerical Solver (GNS ) can provide a fast, numerical solution. For N- Point Problem the solution of the implicit and explicit error representation are different. However, one can use the result of the 4- Point Problem with the Gauss- Jacobi combinatorial algorithm for N- points, too, but the weights should be computed on the basis of the equations of the explicit distance error representation. ALESS model for explicit error representation can be also generated and solved by homotopy method as well as with Extended Newton- Raphson method and local minimization method employing a subset solution of the Gauss-Jacobi algorithm as initial condition Problem definition Observation equations Throughout history, position determination has been one of the most important tasks of mountaineers, pilots, sailor, civil engineers etc. In modern times, Global Navigation Satellite Systems (GNSS) provide an ultimate method to accomplish this task. If one has a hand held GNSS receiver, which measures the travel time of the signal transmitted from the satellites, the distance travelled by the signal from the satellites to the receiver can be computed by multiplying the measured time by the speed of light in vacuum. The distance of the receiver from the i-th GNSS satellite, the pseudo- range observations, di is related to the unknown position of the receiver, {x1, x2, x3 } by di = Hx1 - ai L2 + H x2 - bi L2 + Hx3 - ci L2 + x4 where 8ai, bi, ci <, i = 0, 1, 2, 3 are the coordinates of the ith satellite. The distance is influenced also by the satellite and receiver clock biases. The satellite clock bias can be modelled while the receiver clock bias has to be considered as an unknown variable, x4. This means, we have four unknowns, consequently we need four satellite to provide a minimum observation. The general form of the equation for the i-th satellite is fi = Hx1 - ai L2 + Hx2 - bi L2 + Hx3 - ci L2 - Hx4 - di L2 The residual of this type of equation represents the error implicitly. However in geodesy the explicit distance error definition is usual, namely,

2 2 RangingGNSS_11.nb The residual of this type of equation represents the error implicitly. However in geodesy the explicit distance error definition is usual, namely, The relation between the two expressions, which implies that if f i = 0 then g i = 0 and vica versa. Therefore, in case of four observations, determined system, we employ the first expression, which is easy to handle as a polynomial. The observation equations, Clear"Global " e1 x1 a 0 2 x2 b 0 2 x3 c 0 2 x4 d 0 2 ; e2 x1 a 1 2 x2 b 1 2 x3 c 1 2 x4 d 1 2 ; e3 x1 a 2 2 x2 b 2 2 x3 c 2 2 x4 d 2 2 ; e4 x1 a 3 2 x2 b 3 2 x3 c 3 2 x4 d 3 2 ; Let us suppose that the observation data are, data a , a , a , a , b , b , b , b , c , c , c , c , d , d , d , d ; Preliminary elimination First, this system of polynomials, will be transformed into a system of linear equations and a quadratic equation. Let us expand and multiply by minus one, and arranged the original equations, eqsl e1l, e2l, e3l, e4l MapSortExpand &, e1, e2, e3, e4 x1 2 x2 2 x3 2 x4 2 2 x1 a 0 a x2 b 0 b x3 c 0 c x4 d 0 d 0 2, x1 2 x2 2 x3 2 x4 2 2 x1 a 1 a x2 b 1 b x3 c 1 c x4 d 1 d 1 2, x1 2 x2 2 x3 2 x4 2 2 x1 a 2 a x2 b 2 b x3 c 2 c x4 d 2 d 2 2, x1 2 x2 2 x3 2 x4 2 2 x1 a 3 a x2 b 3 b x3 c 3 c x4 d 3 d 3 2 Subtruct the fourth equation from the other three ones,

3 RangingGNSS_11.nb q TableeqsLi eqsl4, i, 1, 3 Simplify 2 x1 a 0 a x1 a 3 a x2 b 0 b x2 b 3 b x3 c 0 c x3 c 3 c x4 d 0 d x4 d 3 d 3 2, 2 x1 a 1 a x1 a 3 a x2 b 1 b x2 b 3 b x3 c 1 c x3 c 3 c x4 d 1 d x4 d 3 d 3 2, 2 x1 a 2 a x1 a 3 a x2 b 2 b x2 b 3 b x3 c 2 c x3 c 3 c x4 d 2 d x4 d 3 d 3 2 This is a system of three linear equations, and they can be written as, g 1 a 0,3 x1 b 0,3 x2 c 0,3 x3 d 0,3 x4 e 0,3 ; g 2 g , 0 1 x1 a 1,3 x2 b 1,3 x3 c 1,3 x4 d 1,3 e 1,3 g 3 g , 0 2 x1 a 2,3 x2 b 2,3 x3 c 2,3 x4 d 2,3 e 2,3 The coefficients a i,3, b i,3, c i,3, d i,3, e i,3, i 0,.. 2, can be determined as, coeffs0 TableCoefficientqi 1, x1, Coefficientqi 1, x2, Coefficientqi 1, x3, Coefficientqi 1, x4 Factor, i, 0, 2; which are the coefficients of the variables x 1, x 2, x 3, x 4. The constant part is, coeffs1 Tableqi coeffs0i.x1, x2, x3, x4 Simplify, i, 1, 3; Therefore, all of the coefficients are, coeffs TableUnioncoeffs0i, coeffs1i, i, 1, 3 2 a 0 a 3, 2 b 0 b 3, 2 c 0 c 3, 2 d 0 d 3, a 0 2 a 3 2 b 0 2 b 3 2 c 0 2 c 3 2 d 0 2 d 3 2, 2 a 1 a 3, 2 b 1 b 3, 2 c 1 c 3, 2 d 1 d 3, a 1 2 a 3 2 b 1 2 b 3 2 c 1 2 c 3 2 d 1 2 d 3 2, 2 a 2 a 3, 2 b 2 b 3, 2 c 2 c 3, 2 d 2 d 3, a 2 2 a 3 2 b 2 2 b 3 2 c 2 2 c 3 2 d 2 2 d 3 2 Let us assign these coefficients to the linear system, coeffsn Flatten TableInner1 2 &, a i,3, b i,3, c i,3, d i,3, e i,3, coeffsi 1, List, i, 0, 2 a 0,3 2 a 0 a 3, b 0,3 2 b 0 b 3, c 0,3 2 c 0 c 3, d 0,3 2 d 0 d 3, e 0,3 a 0 2 a 3 2 b 0 2 b 3 2 c 0 2 c 3 2 d 0 2 d 3 2, a 1,3 2 a 1 a 3, b 1,3 2 b 1 b 3, c 1,3 2 c 1 c 3, d 1,3 2 d 1 d 3, e 1,3 a 1 2 a 3 2 b 1 2 b 3 2 c 1 2 c 3 2 d 1 2 d 3 2, a 2,3 2 a 2 a 3, b 2,3 2 b 2 b 3, c 2,3 2 c 2 c 3, d 2,3 2 d 2 d 3, e 2,3 a 2 2 a 3 2 b 2 2 b 3 2 c 2 2 c 3 2 d 2 2 d 3 2 In addition, we take one of the nonlinear equations, let say the fourth one, e4 x1 a 3 2 x2 b 3 2 x3 c 3 2 x4 d 3 2 Now, we shall solve the linear system for the variables x 1, x 2, x 3, with x 4 as parameter. It means, the relations x 1 = g 1 (x 4 ), x 2 = g 2 (x 4 ) and x 3 = g 3 (x 4 ) will be computed. To do that, different elimination methods can be employed.

4 4 RangingGNSS_11.nb 11-2 GPS 4-Point Problem Sturmfels method The Sturmfels approach can be employed to solve the linear system of {g 1, g 2, g 3 }. Depending on which variable one wants, the original system is rewritten such that this particular variable is hidden (i.e. is treated as a constant). If our interest is to solve x 1 = g 1 (x 4 ), the equations are first homogenized using a new variable x 5 and consider the variables x 1 and x 4 as well as the constant part as parameters, f1 a 0,3 x1 d 0,3 x4 e 0,3 x5 b 0,3 x2 c 0,3 x3; f2 a 1,3 x1 d 1,3 x4 e 1,3 x5 b 1,3 x2 c 1,3 x3; f3 a 2,3 x1 d 2,3 x4 e 2,3 x5 b 2,3 x2 c 2,3 x3; The Jacobian determinant then becomes, Jx1 Det x2 f1 x3 f1 x5 f1 x2 f2 x3 f2 x5 f2 x2 f3 x3 f3 x5 f3 ; Then the solution for x 1 as function of x 4 is, solx1 SolveJx1 0, x1 x1 x4 b 2,3 c 1,3 d 0,3 x4 b 1,3 c 2,3 d 0,3 x4 b 2,3 c 0,3 d 1,3 x4 b 0,3 c 2,3 d 1,3 x4 b 1,3 c 0,3 d 2,3 x4 b 0,3 c 1,3 d 2,3 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 Similarly, the homogenized system for x 2 = g 2 (x 4 ) then f4 b 0,3 x2 d 0,3 x4 e 0,3 x5 a 0,3 x1 c 0,3 x3; f5 b 1,3 x2 d 1,3 x4 e 1,3 x5 a 1,3 x1 c 1,3 x3; f6 b 2,3 x2 d 2,3 x4 e 2,3 x5 a 2,3 x1 c 2,3 x3; Jx2 Det and its solution is, x1 f4 x3 f4 x5 f4 x1 f5 x3 f5 x5 f5 x1 f6 x3 f6 x5 f6 ; solx2 SolveJx2 0, x2 x2 x4 a 2,3 c 1,3 d 0,3 x4 a 1,3 c 2,3 d 0,3 x4 a 2,3 c 0,3 d 1,3 x4 a 0,3 c 2,3 d 1,3 x4 a 1,3 c 0,3 d 2,3 x4 a 0,3 c 1,3 d 2,3 a 2,3 c 1,3 e 0,3 a 1,3 c 2,3 e 0,3 a 2,3 c 0,3 e 1,3 a 0,3 c 2,3 e 1,3 a 1,3 c 0,3 e 2,3 a 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 Finally x 3 g 3 x 4 leads to f7 c 0,3 x3 d 0,3 x4 e 0,3 x5 a 0,3 x1 b 0,3 x2; f8 c 1,3 x3 d 1,3 x4 e 1,3 x5 a 1,3 x1 b 1,3 x2; f9 c 2,3 x3 d 2,3 x4 e 2,3 x5 a 2,3 x1 b 2,3 x2;

5 RangingGNSS_11.nb Jx3 Det x1 f7 x2 f7 x5 f7 x1 f8 x2 f8 x5 f8 x1 f9 x2 f9 x5 f9 ; solx3 SolveJx3 0, x3 x3 x4 a 2,3 b 1,3 d 0,3 x4 a 1,3 b 2,3 d 0,3 x4 a 2,3 b 0,3 d 1,3 x4 a 0,3 b 2,3 d 1,3 x4 a 1,3 b 0,3 d 2,3 x4 a 0,3 b 1,3 d 2,3 a 2,3 b 1,3 e 0,3 a 1,3 b 2,3 e 0,3 a 2,3 b 0,3 e 1,3 a 0,3 b 2,3 e 1,3 a 1,3 b 0,3 e 2,3 a 0,3 b 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 Substituting the obtained expressions of x 1 g 1 x 4, x 2 g 2 x 4 and x 3 g 3 x 4 in the fourth equation e 4 (x 1, x 2, x 3, x 4 ), G e4. solx11, 1, solx21, 1, solx31, 1; This is a quadratic equation for x 4 ExponentG, x4, List 0, 1, 2 2 The coefficients of this equation, h 2 x 4 + h 1 x 4 + h 0 = 0 are quite long expressions, therefore here we do not display them. h2 CoefficientG, x4 2 Simplify; h1 CoefficientG, x4 Simplify; h0 SimplifyG h2 x4 2 h1 x4; The actual numeric solution of the original system means the evaluation of these coefficients with the numerical data, h2c h2. coeffsn. data h1c h1. coeffsn. data h0c h0. coeffsn. data and then solving the quadratic equation for variable x 4, solx4 Solveh2c x4 ^ 2 h1c x4 h0c 0, x4 x , x The two solutions are x41, x42 x4. solx , Substituting these values in x 1 = g 1 (x 4 ), we get x 1 X1, X2 Mapx1. solx11, 1. coeffsn. data. x4 &, x41, x , Similarly, the values for x 2 are Y1, Y2 Mapx2. solx21, 1. coeffsn. data. x4 &, x41, x ,

6 6 RangingGNSS_11.nb and for x 3 Z1, Z2 Mapx3. solx31, 1. coeffsn. data. x4 &, x41, x , We can select the proper solution according to their norms, NormX1, Y1, Z NormX2, Y2, Z The second position is in space, consequently, the first solution is admissible, SetPrecisionX1, Y1, Z1, , , In order to get the symbolic expression for the coefficients of the quadratic equation, other methods also can be used Dixon Resultant Resultant Dixon Now we can solve the original system, eliminating the variables x 2 and x 3 to get x 1 = g 1 (x 4 ), AbsoluteTimingdrx1 DixonResultantg 1, g 2, g 3, x2, x3, u2, u3; , Null drx1 x1 a 2,3 b 1,3 c 0,3 x1 a 1,3 b 2,3 c 0,3 x1 a 2,3 b 0,3 c 1,3 x1 a 0,3 b 2,3 c 1,3 x1 a 1,3 b 0,3 c 2,3 x1 a 0,3 b 1,3 c 2,3 x4 b 2,3 c 1,3 d 0,3 x4 b 1,3 c 2,3 d 0,3 x4 b 2,3 c 0,3 d 1,3 x4 b 0,3 c 2,3 d 1,3 x4 b 1,3 c 0,3 d 2,3 x4 b 0,3 c 1,3 d 2,3 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 This is a linear expression contains only x 1 and x 4, Exponentdrx1, x1, x2, x3, x4 1, 0, 0, 1 Then the solution for x 1 solx1 Solvedrx1 0, x1 x1 x4 b 2,3 c 1,3 d 0,3 x4 b 1,3 c 2,3 d 0,3 x4 b 2,3 c 0,3 d 1,3 x4 b 0,3 c 2,3 d 1,3 x4 b 1,3 c 0,3 d 2,3 x4 b 0,3 c 1,3 d 2,3 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 Similarly, for the two additional variables, x 2 = g 2 (x 4 ) and x 3 = g 3 (x 4 ), drx2 DixonResultantg 1, g 2, g 3, x1, x3, u1, u3 x2 a 2,3 b 1,3 c 0,3 x2 a 1,3 b 2,3 c 0,3 x2 a 2,3 b 0,3 c 1,3 x2 a 0,3 b 2,3 c 1,3 x2 a 1,3 b 0,3 c 2,3 x2 a 0,3 b 1,3 c 2,3 x4 a 2,3 c 1,3 d 0,3 x4 a 1,3 c 2,3 d 0,3 x4 a 2,3 c 0,3 d 1,3 x4 a 0,3 c 2,3 d 1,3 x4 a 1,3 c 0,3 d 2,3 x4 a 0,3 c 1,3 d 2,3 a 2,3 c 1,3 e 0,3 a 1,3 c 2,3 e 0,3 a 2,3 c 0,3 e 1,3 a 0,3 c 2,3 e 1,3 a 1,3 c 0,3 e 2,3 a 0,3 c 1,3 e 2,3

7 RangingGNSS_11.nb Exponentdrx2, x1, x2, x3, x4 0, 1, 0, 1 solx2 Solvedrx2 0, x2 x2 x4 a 2,3 c 1,3 d 0,3 x4 a 1,3 c 2,3 d 0,3 x4 a 2,3 c 0,3 d 1,3 x4 a 0,3 c 2,3 d 1,3 x4 a 1,3 c 0,3 d 2,3 x4 a 0,3 c 1,3 d 2,3 a 2,3 c 1,3 e 0,3 a 1,3 c 2,3 e 0,3 a 2,3 c 0,3 e 1,3 a 0,3 c 2,3 e 1,3 a 1,3 c 0,3 e 2,3 a 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 and drx3 DixonResultantg 1, g 2, g 3, x1, x2, u1, u2 x3 a 2,3 b 1,3 c 0,3 x3 a 1,3 b 2,3 c 0,3 x3 a 2,3 b 0,3 c 1,3 x3 a 0,3 b 2,3 c 1,3 x3 a 1,3 b 0,3 c 2,3 x3 a 0,3 b 1,3 c 2,3 x4 a 2,3 b 1,3 d 0,3 x4 a 1,3 b 2,3 d 0,3 x4 a 2,3 b 0,3 d 1,3 x4 a 0,3 b 2,3 d 1,3 x4 a 1,3 b 0,3 d 2,3 x4 a 0,3 b 1,3 d 2,3 a 2,3 b 1,3 e 0,3 a 1,3 b 2,3 e 0,3 a 2,3 b 0,3 e 1,3 a 0,3 b 2,3 e 1,3 a 1,3 b 0,3 e 2,3 a 0,3 b 1,3 e 2,3 Exponentdrx3, x1, x2, x3, x4 0, 0, 1, 1 solx3 Solvedrx3 0, x3 x3 x4 a 2,3 b 1,3 d 0,3 x4 a 1,3 b 2,3 d 0,3 x4 a 2,3 b 0,3 d 1,3 x4 a 0,3 b 2,3 d 1,3 x4 a 1,3 b 0,3 d 2,3 x4 a 0,3 b 1,3 d 2,3 a 2,3 b 1,3 e 0,3 a 1,3 b 2,3 e 0,3 a 2,3 b 0,3 e 1,3 a 0,3 b 2,3 e 1,3 a 1,3 b 0,3 e 2,3 a 0,3 b 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 After substitution, we have again a quadratic equation for x 4, G e4. solx11, 1, solx21, 1, solx31, 1; ExponentG, x4, List 0, 1, 2 The coefficents of the quadratic equation are, h2d CoefficientG, x4 2 ; h1d CoefficientG, x4; h0d SimplifyG h2d x4 2 h1d x4; The coefficients provided by the Sturmfels method and the Dixon resultant are the same, h2 h2d, h1 h1d, h0 h0d Simplify 0, 0, Groebner basis First, again we want x 1 = g(x 4 ), therefore variables x 2 and x 3 should be eliminated from the Groebner basis, AbsoluteTiminggbx1 GroebnerBasisg 1, g 2, g 3, x1, x2, x3, x4, x2, x3; , Null

8 8 RangingGNSS_11.nb gbx1 x1 a 2,3 b 1,3 c 0,3 x1 a 1,3 b 2,3 c 0,3 x1 a 2,3 b 0,3 c 1,3 x1 a 0,3 b 2,3 c 1,3 x1 a 1,3 b 0,3 c 2,3 x1 a 0,3 b 1,3 c 2,3 x4 b 2,3 c 1,3 d 0,3 x4 b 1,3 c 2,3 d 0,3 x4 b 2,3 c 0,3 d 1,3 x4 b 0,3 c 2,3 d 1,3 x4 b 1,3 c 0,3 d 2,3 x4 b 0,3 c 1,3 d 2,3 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 Now, the basis contains only one equation, Lengthgbx1 1 in which only x 1 and x 4 can be found, Exponentgbx1, x1, x2, x3, x4 1, 0, 0, 1 Therefore x 1 g x 4 can be computed directly, solx1 Solvegbx1 0, x1 Simplify x1 b 2,3 c 1,3 x4 d 0,3 e 0,3 c 0,3 x4 d 1,3 e 1,3 b 1,3 c 2,3 x4 d 0,3 e 0,3 c 0,3 x4 d 2,3 e 2,3 b 0,3 c 2,3 x4 d 1,3 e 1,3 c 1,3 x4 d 2,3 e 2,3 a 2,3 b 1,3 c 0,3 b 0,3 c 1,3 a 1,3 b 2,3 c 0,3 b 0,3 c 2,3 a 0,3 b 2,3 c 1,3 b 1,3 c 2,3 Similarly, in the other cases and gbx2 GroebnerBasisg 1, g 2, g 3, x1, x2, x3, x4, x1, x3; Exponentgbx2, x1, x2, x3, x4 0, 1, 0, 1 solx2 Solvegbx2 0, x2 Simplify x2 a 2,3 c 1,3 x4 d 0,3 e 0,3 c 0,3 x4 d 1,3 e 1,3 a 1,3 c 2,3 x4 d 0,3 e 0,3 c 0,3 x4 d 2,3 e 2,3 a 0,3 c 2,3 x4 d 1,3 e 1,3 c 1,3 x4 d 2,3 e 2,3 a 2,3 b 1,3 c 0,3 b 0,3 c 1,3 a 1,3 b 2,3 c 0,3 b 0,3 c 2,3 a 0,3 b 2,3 c 1,3 b 1,3 c 2,3 gbx3 GroebnerBasisg 1, g 2, g 3, x1, x2, x3, x4, x1, x2; Exponentgbx3, x1, x2, x3, x4 0, 0, 1, 1 solx3 Solvegbx3 0, x3 Simplify x3 a 2,3 b 1,3 x4 d 0,3 e 0,3 b 0,3 x4 d 1,3 e 1,3 a 1,3 b 2,3 x4 d 0,3 e 0,3 b 0,3 x4 d 2,3 e 2,3 a 0,3 b 2,3 x4 d 1,3 e 1,3 b 1,3 x4 d 2,3 e 2,3 a 2,3 b 1,3 c 0,3 b 0,3 c 1,3 a 1,3 b 2,3 c 0,3 b 0,3 c 2,3 a 0,3 b 2,3 c 1,3 b 1,3 c 2,3 After substition them, we get G e4. solx11, 1, solx21, 1, solx31, 1; ExponentG, x4, List 0, 1, 2

9 RangingGNSS_11.nb Then the coefficients of the quadratic equation are, h2gr CoefficientG, x4 2 ; h1gr CoefficientG, x4; h0gr SimplifyG h2gr x4 2 h1gr x4; We have again the same result, h2 h2gr, h1 h1gr, h0 h0gr Simplify 0, 0, Reduced Groebner basis First, again we want to determine x 1 = g(x 4 ), therefore variables x 2 and x 3 should be eliminated from the Groebner basis, AbsoluteTiminggbx1 GroebnerBasisg 1, g 2, g 3, x1, x2, x3, x4, x2, x3; , Null gbx1 x1 a 2,3 b 1,3 c 0,3 x1 a 1,3 b 2,3 c 0,3 x1 a 2,3 b 0,3 c 1,3 x1 a 0,3 b 2,3 c 1,3 x1 a 1,3 b 0,3 c 2,3 x1 a 0,3 b 1,3 c 2,3 x4 b 2,3 c 1,3 d 0,3 x4 b 1,3 c 2,3 d 0,3 x4 b 2,3 c 0,3 d 1,3 x4 b 0,3 c 2,3 d 1,3 x4 b 1,3 c 0,3 d 2,3 x4 b 0,3 c 1,3 d 2,3 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 Now, the basis contains only one polynomial, Lengthgbx1 1 in which only x 1 and x 4 can be found, Exponentgbx1, x1, x2, x3, x4 1, 0, 0, 1 Therefore x 1 g x 4 can be computed directly, solx1 Solvegbx1 0, x1 Simplify x1 b 2,3 c 1,3 x4 d 0,3 e 0,3 c 0,3 x4 d 1,3 e 1,3 b 1,3 c 2,3 x4 d 0,3 e 0,3 c 0,3 x4 d 2,3 e 2,3 b 0,3 c 2,3 x4 d 1,3 e 1,3 c 1,3 x4 d 2,3 e 2,3 a 2,3 b 1,3 c 0,3 b 0,3 c 1,3 a 1,3 b 2,3 c 0,3 b 0,3 c 2,3 a 0,3 b 2,3 c 1,3 b 1,3 c 2,3 Similarly, in the other cases gbx2 GroebnerBasisg 1, g 2, g 3, x1, x2, x3, x4, x1, x3; Exponentgbx2, x1, x2, x3, x4 0, 1, 0, 1 solx2 Solvegbx2 0, x2 Simplify x2 a 2,3 c 1,3 x4 d 0,3 e 0,3 c 0,3 x4 d 1,3 e 1,3 a 1,3 c 2,3 x4 d 0,3 e 0,3 c 0,3 x4 d 2,3 e 2,3 a 0,3 c 2,3 x4 d 1,3 e 1,3 c 1,3 x4 d 2,3 e 2,3 a 2,3 b 1,3 c 0,3 b 0,3 c 1,3 a 1,3 b 2,3 c 0,3 b 0,3 c 2,3 a 0,3 b 2,3 c 1,3 b 1,3 c 2,3

10 10 RangingGNSS_11.nb and gbx3 GroebnerBasisg 1, g 2, g 3, x1, x2, x3, x4, x1, x2; Exponentgbx3, x1, x2, x3, x4 0, 0, 1, 1 solx3 Solvegbx3 0, x3 Simplify x3 a 2,3 b 1,3 x4 d 0,3 e 0,3 b 0,3 x4 d 1,3 e 1,3 a 1,3 b 2,3 x4 d 0,3 e 0,3 b 0,3 x4 d 2,3 e 2,3 a 0,3 b 2,3 x4 d 1,3 e 1,3 b 1,3 x4 d 2,3 e 2,3 a 2,3 b 1,3 c 0,3 b 0,3 c 1,3 a 1,3 b 2,3 c 0,3 b 0,3 c 2,3 a 0,3 b 2,3 c 1,3 b 1,3 c 2,3 After substition them, we get G e4. solx11, 1, solx21, 1, solx31, 1; ExponentG, x4, List 0, 1, 2 Then the coefficients of the quadratic equation are, h2gr CoefficientG, x4 2 ; h1gr CoefficientG, x4; h0gr SimplifyG h2gr x4 2 h1gr x4; We have again the same result, h2 h2gr, h1 h1gr, h0 h0gr Simplify 0, 0, Global Symbolic Solver The solution of the system of 4 equations simultaneously in symbolic form, leads to a very large, impractical expression. However the solution of the linear system of g 1, g 2, g 3 with x 4 as parameter is easy, AbsoluteTimingsolGSS3 Solveg 1 0, g 2 0, g 3 0, x1, x2, x3; , Null solgss3 x1 x4 b 2,3 c 1,3 d 0,3 x4 b 1,3 c 2,3 d 0,3 x4 b 2,3 c 0,3 d 1,3 x4 b 0,3 c 2,3 d 1,3 x4 b 1,3 c 0,3 d 2,3 x4 b 0,3 c 1,3 d 2,3 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3, x2 x4 a 2,3 c 1,3 d 0,3 x4 a 1,3 c 2,3 d 0,3 x4 a 2,3 c 0,3 d 1,3 x4 a 0,3 c 2,3 d 1,3 x4 a 1,3 c 0,3 d 2,3 x4 a 0,3 c 1,3 d 2,3 a 2,3 c 1,3 e 0,3 a 1,3 c 2,3 e 0,3 a 2,3 c 0,3 e 1,3 a 0,3 c 2,3 e 1,3 a 1,3 c 0,3 e 2,3 a 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3, x3 a 2,3 b 0,3 a 0,3 b 2,3 a 1,3 x4 d 0,3 e 0,3 a 0,3 x4 d 1,3 e 1,3 a 1,3 b 0,3 a 0,3 b 1,3 a 2,3 x4 d 0,3 e 0,3 a 0,3 x4 d 2,3 e 2,3 a 2,3 b 0,3 a 0,3 b 2,3 a 1,3 c 0,3 a 0,3 c 1,3 a 1,3 b 0,3 a 0,3 b 1,3 a 2,3 c 0,3 a 0,3 c 2,3 The second order equation for solving x 4 is,

11 RangingGNSS_11.nb solgss34 e4. solgss3 x4 d 3 2 b 3 x4 a 2,3 c 1,3 d 0,3 x4 a 1,3 c 2,3 d 0,3 x4 a 2,3 c 0,3 d 1,3 x4 a 0,3 c 2,3 d 1,3 x4 a 1,3 c 0,3 d 2,3 x4 a 0,3 c 1,3 d 2,3 a 2,3 c 1,3 e 0,3 a 1,3 c 2,3 e 0,3 a 2,3 c 0,3 e 1,3 a 0,3 c 2,3 e 1,3 a 1,3 c 0,3 e 2,3 a 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 2 a 3 x4 b 2,3 c 1,3 d 0,3 x4 b 1,3 c 2,3 d 0,3 x4 b 2,3 c 0,3 d 1,3 x4 b 0,3 c 2,3 d 1,3 x4 b 1,3 c 0,3 d 2,3 x4 b 0,3 c 1,3 d 2,3 b 2,3 c 1,3 e 0,3 b 1,3 c 2,3 e 0,3 b 2,3 c 0,3 e 1,3 b 0,3 c 2,3 e 1,3 b 1,3 c 0,3 e 2,3 b 0,3 c 1,3 e 2,3 a 2,3 b 1,3 c 0,3 a 1,3 b 2,3 c 0,3 a 2,3 b 0,3 c 1,3 a 0,3 b 2,3 c 1,3 a 1,3 b 0,3 c 2,3 a 0,3 b 1,3 c 2,3 2 c 3 a 2,3 b 0,3 a 0,3 b 2,3 a 1,3 x4 d 0,3 e 0,3 a 0,3 x4 d 1,3 e 1,3 a 1,3 b 0,3 a 0,3 b 1,3 a 2,3 x4 d 0,3 e 0,3 a 0,3 x4 d 2,3 e 2,3 a 2,3 b 0,3 a 0,3 b 2,3 a 1,3 c 0,3 a 0,3 c 1,3 a 1,3 b 0,3 a 0,3 b 1,3 a 2,3 c 0,3 a 0,3 c 2,3 2 ExponentsolGSS34, x4 2 This is the same equation what we have got by the different elimination techniques, for example with the Sturmfels method, Simplifyh2 x4 2 h1 x4 h0 solgss Global Numeric Solver NSolveg 1 0, g 2 0, g 3 0, e4 0. coeffsn. data, x1, x2, x3, x4 AbsoluteTiming , x , x , x , x , x , x , x , x NumberForm, , x , x , x , x , x , x , x , x Linear Homotopy We have already solved this problem in Section GPS N-point Problem Observation equations In case of m > 4 satellites, the two representations, and

12 12 RangingGNSS_11.nb will be not equivalent in least square sense, namely Let us consider six satellites with the following numerical values, datan a , a , a , a , a , a , b , b , b , b , b , b , c , c , c , c , c , c , d , d , d , d , d , d ; The number of the equations, m 6; Let us see the result for the two different representations. In the first case the general form of the equation for the i-th satellite is, e x1 a i 2 x2 b i 2 x3 c i 2 x4 d i 2 ; The objective to be minimized is the sum of the residium of the equations, f ApplyPlus, Tablee 2. datan, i, 0, m 1 Simplify x x x x x x x x x x x x x x x x x x x x x x x x4 2 2 The global minimum can be found by using the built-in function NMinimize, soln NMinimizef, x1, x2, x3, x4 AbsoluteTiming , , x , x , x , x

13 RangingGNSS_11.nb or NumberForm, , , x , x , x , x However, if we employ the norm of the distance instead of the residium of the equations, namely en d i x1 a i 2 x2 b i 2 x3 c i 2 x4; and then the objective is, fn ApplyPlus, Tableen 2. datan, i, 0, m x x x3 2 x x x x3 2 2 x x x x3 2 x x x x3 2 2 x x x x3 2 x x x x3 2 x4 The optimum will be somewhat different, solnn NMinimizefn, x1, x2, x3, x4 AbsoluteTiming , , x , x , x , x or NumberForm, , , x , x , x , x

14 14 RangingGNSS_11.nb Gauss - Jacobi solution Because in case of 4 satellites the two representation are the same, we can use the result of the GPS- 4 Point problem, but we should compute the weights of the Gauss-Jacobi algorithm on the basis of the second representation! First, the subsets should be determined. In our case n 4; m 6; The number of the subsets mn Binomialm, n 15 These subsets are, qs PartitionMap 1 &, FlattenSubsetsRangem, n, n 0, 1, 2, 3, 0, 1, 2, 4, 0, 1, 2, 5, 0, 1, 3, 4, 0, 1, 3, 5, 0, 1, 4, 5, 0, 2, 3, 4, 0, 2, 3, 5, 0, 2, 4, 5, 0, 3, 4, 5, 1, 2, 3, 4, 1, 2, 3, 5, 1, 2, 4, 5, 1, 3, 4, 5, 2, 3, 4, 5 The value of the indices start from zero in correspondence of the indices of the coefficients of the equations. Now, we shall utilize the symbolic solution of the GPS 4-points problem, namely the expressions of the coefficients of the quadratic equation (h 2,h 1,h 0 ). Therefore, we construct a new data list, datap similar to datan,which assignes the proper values to the coefficients of the equations of the subsets. This is the same technique that we have already used. datap TableMapSelectdatan, MemberQqsi, 1, 2 &. 1 0, 2 1, 3 2, 4 3 &, qsi, i, 1, mn; For example, the fourth subset is indexed as qs4 0, 1, 3, 4 and it has the proper data assignments, datap4 a , a , a , a , b , b , b , b , c , c , c , c , d , d , d , d Now, we can employ the symbolic expressions of the coefficients of the quadratic equation for x 4, (h 2, h 1, h 0 ), which were developed for the GPS 4-points problem. Let us consider the result of the Sturmfels approach. These coefficients can be evaluated for all of the 15 combinatorial subsets, (H 2, H 1, H 0 ), H2 Maph2. coeffsn. Flatten &, datap; H1 Maph1. coeffsn. Flatten &, datap; H0 Maph0. coeffsn. Flatten &, datap; It is useful to display these coeffients, H210 TransposeH2, H1, H0;

15 RangingGNSS_11.nb H210c Map. datan &, H210; TableFormNumberFormH210c, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , This table indicates that the 10 th combination has a poor geometry (see first column and Fig.11.1), which fact can be also detected by computing its PDOP (Position Dilution of Precision), see the text book. Then the 15 quadratic equations can be solved for x 4, ListPlotH2, PlotRange All, Joined True Fig Computed values of H 2 coefficients AbsoluteTiming X4 Mapx4. Solve1 x4 ^ 2 2 x4 3 0, x41, 1 &, H210c; , Null These values of x 4 can be substituted into the symbolic relations x 1 = g(x 4 ), x 2 = g(x 4 ) and x 3 = g(x 4 ) developed for GPS 4 - points problem. X1 MapThreadx1. solx11, 1. coeffsn. Flatten1. x4 2 &, datap, X4; X2 MapThreadx2. solx21, 1. coeffsn. Flatten1. x4 2 &, datap, X4; X3 MapThreadx3. solx31, 1. coeffsn. Flatten1. x4 2 &, datap, X4; Let us display these solutions for (x 1, x 2, x 3, x 4 ),

16 16 RangingGNSS_11.nb X TransposeX1, X2, X3, X4; TableFormNumberFormX, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , The poor geometry of the 10 th combination can be also realized in Fig.11.2 as well as in the last column of X. ListPlotX4, PlotRange All, Joined True 500 Computing the arithmetic average, MeanX Fig Computed values of variable x , , , NumberForm, , , , But it is far from the acceptable solution. In order to compute the weights of these solutions, one has to compute the square of the 15 Jacobi determinants. Each has size of 4 4, because of four equations and four variables. Starting with the general form of the i-th equation, en x4 x1 a i 2 x2 b i 2 x3 c i 2 d i The partial derivatives are,

17 RangingGNSS_11.nb de Den, x1, Den, x2, Den, x3, Den, x4 Simplify x1 a i x1 a i 2 x2 b i 2 x3 c i 2, x2 b i x1 a i 2 x2 b i 2 x3 c i 2, x3 c i x1 a i 2 x2 b i 2 x3 c i 2, 1 The numerical values of these partial derivatives will be computed at the corresponding combinatorial solutions. Therefore, the weights, Π j, the square of the 15 Jacobi determinants are, Πs TableMapDetde. i 1, de. i 2, de. i 3, de. i 4. datan 2 &, qsj. x1 Xj, 1, x2 Xj, 2, x3 Xj, 3, x4 Xj, 4, j, 1, mn , , , , , , , , , , , , , , The sum of these weights are sπ, sπs ApplyPlus, Πs Then the weighted solution of the variable x i is X1s, X2s, X3s, X4s MapΠs. &, X1, X2, X3, X4 sπs , , , NumberForm, , , , The result is correct ALESS Equations The equations of the determined model, The prototype of g i, en since x4 x1 a i 2 x2 b i 2 x3 c i 2 d i

18 18 RangingGNSS_11.nb MapDen 2, &, x1, x2, x3, x4 2 x1 a i x4 x1 a i 2 x2 b i 2 x3 c i 2 d i x1 a i 2 x2 b i 2 x3 c i 2, 2 x2 b i x4 x1 a i 2 x2 b i 2 x3 c i 2 d i x1 a i 2 x2 b i 2 x3 c i 2, 2 x3 c i x4 x1 a i 2 x2 b i 2 x3 c i 2 d i, x1 a i 2 x2 b i 2 x3 c i 2 2 x4 x1 a i 2 x2 b i 2 x3 c i 2 d i then in general form, m1 v Map &, i 0 1m i 0 2 x1 a i x4 x1 a i 2 x2 b i 2 x3 c i 2 d i x1 a i 2 x2 b i 2 x3 c i 2, 1m i 0 2 x2 b i x4 x1 a i 2 x2 b i 2 x3 c i 2 d i x1 a i 2 x2 b i 2 x3 c i 2, 1m i 0 1m 2 x3 c i x4 x1 a i 2 x2 b i 2 x3 c i 2 d i x1 a i 2 x2 b i 2 x3 c i 2, 2 x4 x1 a i 2 x2 b i 2 x3 c i 2 d i i 0 In our case m = 6, therefore the numeric form of the equations, vn v. m m. datan Expand; For example the first equation vn1

19 RangingGNSS_11.nb

20 20 RangingGNSS_11.nb

21 RangingGNSS_11.nb x x x x3 2 2 x1 x x x x x x x x3 2 2 x1 x x x x ALESS Numeric Now, to solve the ALESS equations, let us employ the homotopy method. In this case the system is not a polynomial one. We can not generate complex start system and its solution. However, we can use for example, fixed point homotopy. In order to illustrate the robustness of the method, we employ the result of the worst geometry of Gauss-Jacobi solution, that of the combination 10 th. This will be the solution of the start system, X0 X , , , The variables, V x1, x2, x3, x4; The start system itself, gf V FirstX x1, x2, x3, x4 To avoid singularity of the homotopy function, let Γ 1, 1, 1, 1; GeoAlgebra LinearHomotopy Now employing path tracing by integration with high precision, AbsoluteTimingsolH LinearHomotopyFR vn, gf, V, X0, Γ, 10, Λ; , Null solh , , , NumberForm, , , , Displaying the homotopy paths,

22 22 RangingGNSS_11.nb GraphicsArray TableParametricPlotReViΛ, ImViΛ. solh2, , Λ, 0, 1, PlotRange All, BaseStyle FontSize 10, FontFamily "Times", Axes None, FrameLabel "Re", "Im", Frame True, AspectRatio 0.6, PlotLabel StringJoinToStringVi, "Λ 10 6 ", Epilog PointSize0.02, Blue, PointReX01, i 10 6, ImX01, i 10 6, PointSize0.02, Red, PointReViΛ. solh2, 1. Λ , ImViΛ. solh2, 1. Λ , i, 1, LengthV Im

23 RangingGNSS_11.nb Im Fig Homotopy paths Extended Newton- Raphson solution Now, we solve the original overdetermined system employing one of the solutions of the Gauss-Jacobi subset solution. Let us use again the worst combination, X0 X , , , GeoAlgebra NewtonExtended The equations of the overdetermined system, F Tableen, i, 0, m 1. datan x x x3 2 x4, x x x3 2 x4, x x x3 2 x4, x x x3 2 x4, x x x3 2 x4, x x x3 2 x4 AbsoluteTimingsolNE NewtonExtended F, V, X0; , Null The solution is, solne Last , , , NumberForm, , , ,

24 24 RangingGNSS_11.nb The convergence is fast, TakesolNE, 6 NumberForm, 11 & , , , , , , , , , , , , , , , , , , , , , , , Direct Least Square via Local Minimization We have solved the problem with global minimization, here we solve with a local method. Again we use the worst Gauss- Jacobi subset solution as initial guess, OffFindMinimum::"precw" FindMinimumSetPrecisionfn, 30, x1, X10, 1, x2, X10, 2, x3, X10, 3, x4, X10, 4, WorkingPrecision 30, Method "ConjugateGradient " AbsoluteTiming , , x , x , x , x NumberForm, , , x , x , x , x We needed high precision computation to reach acceptable solution, but even doing that the computation time is short. Conclusions For solving GPS ranging problem all of the methods introduced here are efficient. Concerning the implementation out of Mathematica perhaps the results of the Dixon and Groebner methods for 4- point problem, and Gauss- Jacobi as well as Extended Newton- Raphson method for N- point problem are mostly recommended.