Question

我想用matlab进行CHAN算法仿真，求程序

Answer 1

function X = Chan_3BS(MSP,R,Noise)
% Chan 算法，利用3BS对MS进行定位；
% CHAN_3BS:
% 参数说明：
% Noise: 测距误差方差.
% R： 小区半径.
% Also see: Chan_3BS.

% 参数检测:
if nargout ~=1,
error('Too many output arguments!');
end
if nargin ~= 3,
error('input arguments error!');
end

% 算法开始
MS = R*MSP;
BS = R*NetworkTop(3);

% A矩阵:
X21 = BS(1,2) - BS(1,1);
X31 = BS(1,3) - BS(1,1);
Y21 = BS(2,2) - BS(2,1);
Y31 = BS(2,3) - BS(2,1);
A = inv([X21,Y21;X31,Y31]);

% B矩阵:
R1 = sqrt((BS(1,1) - MS(1))^2 + (BS(2,1) - MS(2))^2);
R2 = sqrt((BS(1,2) - MS(1))^2 + (BS(2,2) - MS(2))^2);
R3 = sqrt((BS(1,3) - MS(1))^2 + (BS(2,3) - MS(2))^2);

R21 = R2 - R1 + MeaNoise(Noise); % 需要加噪声
R31 = R3 - R1 + MeaNoise(Noise);
B = [R21;R31];

% C矩阵:
K1 = BS(1,1)^2 + BS(2,1)^2;
K2 = BS(1,2)^2 + BS(2,2)^2;
K3 = BS(1,3)^2 + BS(2,3)^2;
C = 0.5*[R21^2 - K2 + K1; R31^2 - K3 + K1];

% 一元二次方程的系数：
a = B'*A'*A*B - 1;
b = B'*A'*A*C + C'*A'*A*B;
c = C'*A'*A*C;

% 方程的两个根：
root1 = abs((-b + sqrt(b^2 - 4*a*c))/(2*a));
root2 = abs((-b - sqrt(b^2 - 4*a*c))/(2*a));

% 检验方程的根：
if root1 < R,
EMS = -A*(B*root1 + C);
else
EMS = -A*(B*root2 + C);
end

% 输出结果：
if nargout == 1,
X = EMS;
else
disp(EMS);
end