chan演算法-ag真人国际官网

① 我想用matlab進行chan演算法模擬，求程序

function x = chan_3bs(msp,r,noise)
% chan 演算法，利用3bs對ms進行定位；
% chan_3bs:
% 參數說明：
% noise: 測距誤差方差.
% r： 小區半徑.
% also see: chan_3bs.

% 參數檢測:
if nargout ~=1,
error('too many output arguments!');
end
if nargin ~= 3,
error('input arguments error!');
end

% 演算法開始
ms = r*msp;
bs = r*networktop(3);

% a矩陣:
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矩陣:
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); % 需要加雜訊
r31 = r3 - r1 meanoise(noise);
b = [r21;r31];

% c矩陣:
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];

% 一元二次方程的系數：
a = b'*a'*a*b - 1;
b = b'*a'*a*c c'*a'*a*b;
c = c'*a'*a*c;

% 方程的兩個根：
root1 = abs((-b sqrt(b^2 - 4*a*c))/(2*a));
root2 = abs((-b - sqrt(b^2 - 4*a*c))/(2*a));

% 檢驗方程的根：
if root1 < r,
ems = -a*(b*root1 c);
else
ems = -a*(b*root2 c);
end

% 輸出結果：
if nargout == 1,
x = ems;
else
disp(ems);
end

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chan綆楁硶鐨勫師鐞嗘槸浠涔堬紵

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