【 ML 】Newton – Raphson Iteration Procedure of TOA

由博文可知：ML ALGORITHMS of TOA - Based Positioning

Newton – Raphson Iteration：

(6)

(7)

(8)

(9)

(10)

对应的 MATLAB 函数：

function H = hessian_ml(X,x,r,sigma2)
% ML Hessian matrix computation
% --------------------------------
% H = hessian_ml(X,x,r,sigma2)
% H = Hessian matrix
% X = matrix for receiver positions
% x = 2D position estimate
% r = TOA measurement vector
% sigma2 = noise variance vector
%
L = size(X,2); % number of receiverst1 = 0;
t2 = 0;
t3 =0;
ds = sum((x*ones(1,L)-X).^2,1);
ds = ds';
for i=1:Lt1 = t1 + (1/sigma2(i))*((x(1)-X(1,i))^2/ds(i)-(r(i)-ds(i)^(0.5))*(x(2)-X(2,i))^2/ds(i)^(1.5));t2 = t2 + (1/sigma2(i))*((x(2)-X(2,i))^2/ds(i)-(r(i)-ds(i)^(0.5))*(x(1)-X(1,i))^2/ds(i)^(1.5));t3 = t3 + (1/sigma2(i))*(r(i)*(x(1)-X(1,i))*(x(2)-X(2,i))/ds(i)^(1.5));
end
H=2.*[t1 t3;t3 t2];

function g = grad_ml(X,x,r,sigma2)
% ML gradient computation
% --------------------------------
% g = grad_ml(X,x,r);
% g = gradient vector
% X = matrix for receiver positions
% x = 2D position estimate
% r = TOA measurement vector
% sigma2 = noise variance vector
%
L = size(X,2); % number of receivers
t1 = 0;
t2 = 0;
ds = sum((x*ones(1,L)-X).^2,1);
ds = ds';
for i=1:Lt1 = t1 + (1/sigma2(i))*(r(i)-ds(i)^(0.5))*(x(1)-X(1,i))/ds(i)^(0.5);t2 = t2 + (1/sigma2(i))*(r(i)-ds(i)^(0.5))*(x(2)-X(2,i))/ds(i)^(0.5);
end
g=-2.*[t1; t2];

先给出一次定位示意图：

定位条件：

Illustration of ML Approach for TOA - Based Positioning in a Single Trial

Consider a 2 - D geometry of L = 4 receivers with known coordinates at (0, 0), (0,10), (10, 0), and (10, 10), while the unknown source position is ( x , y ) = (2, 3).

Note that the source is located inside the square bounded by the four receivers. For presentation simplicity, the range error variance， $\sigma^2_{TOA,l}$ ，is assigned proportional to $d^2_{l}$ .

and we define the signal - to - noise ratio ( SNR ) as $d^2_l/\sigma^2_{TOA,l}$ .

可见，目标真实位置和估计出来的位置重合，可以定位，只不过有点误差而已。

下面分析误差：

做这个测量均方根误差的实验，定位了100次。

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