TDOA算法原理

TDOA(Time Difference of Arrival)——时间差到达算法，利用了几何数学中双曲线的特点—— 双曲线上的任意点到达两焦点的距离差是固定值。 这个距离差它天然可以抹去用户设备(UE)和基站的之间时钟误差。

P

1

C

1

=

c

⋅

(

t

11

+

Δ

t

)

P_1C_1 = c·(t_{11}+\Delta t)

P1​C1​=c⋅(t11​+Δt)其中

Δ

t

\Delta t

Δt是UE和基站之间的钟差(在UE与基站不完全同步的情况下)，这个钟差我们没法直接获得。

P

1

C

2

=

c

⋅

(

t

12

+

Δ

t

)

P_1C_2 = c·(t_{12}+\Delta t)

P1​C2​=c⋅(t12​+Δt)

则

∣

P

1

C

1

−

P

1

C

2

∣

=

c

⋅

(

t

11

−

t

12

)

则|P_1C_1-P_1C_2 |=c·(t_{11}-t_{12})

则∣P1​C1​−P1​C2​∣=c⋅(t11​−t12​)可见这里的钟差

Δ

t

\Delta t

Δt被消除了，之后使用数学方法求出两个双曲线的焦点。但这同时也暗示着 基站的时钟需要同步才能被消除。 所以TDOA算法特性：UE和基站无需同步，基站之间需要同步，最少三个基站能测得焦点。

Chan算法介绍

在TDOA的解算方法上，有直接求解析解的Chan算法、Fang算法。也有迭代算法如Taylor算法(它是通过不断计算当前误差来调整参数，这个误差需要真实的位置标签来对比，但我们有真实标签后为什么还需要估计呢？这个是我对Taylor算法的疑惑，欢迎大家一起探讨👏)。基于解析解的方差理论上是没有误差的，它只受限于计算机的计算精度。

Chan算法公式推导

上述算法中，

A

x

=

r

0

C

+

D

Ax=r_0C+D

Ax=r0​C+D，先求

A

x

=

C

Ax=C

Ax=C的解

x

a

x_a

xa​，再求

A

x

=

D

Ax=D

Ax=D的解

x

b

x_b

xb​，再求

r

0

r_0

r0​，最后按照

x

=

r

0

x

a

+

x

b

x=r_0x_a+x_b

x=r0​xa​+xb​组合起来。

Chan算法实现2D

<code>

%%

clc;

clear;

close all;

format long;

figure;

%设置UE位置

for i=1:100

ue_x = randi(100);

ue_y = randi(100);

scatter(ue_x, ue_y, '*');

hold on;

%注意注意！！！基站的y不能全部相同，否则在第57行的矩阵A第二列元素全为0，Ax=C或Ax=Ds时求不出唯一解

stations = [-40 0; 20 0; 40 50; 10 10]; %第四个基站是为了提出伪解

hold on;

r0_real = distance(ue_x, ue_y,stations(1,1), stations(1,2));

r1_real = distance(ue_x, ue_y,stations(2,1), stations(2,2));

r2_real = distance(ue_x, ue_y,stations(3,1), stations(3,2));

r3_real = distance(ue_x, ue_y,stations(4,1), stations(4,2));

%ri_real只是用来计算tds，实际上它会带有时钟误差，而这个误差我们不能直接得到

tds = [r1_real-r0_real r2_real-r0_real, r3_real-r0_real];

position = TDOA(stations, tds);

scatter(position(1), position(2), 'o');

hold on;

end

%%

function [position] = TDOA(stations, tds)

x0 = stations(1,1);y0 = stations(1,2);

x1 = stations(2,1);y1 = stations(2,2);

x2 = stations(3,1);y2 = stations(3,2);

x3 = stations(4,1);y3 = stations(4,2);

r10 = tds(1);

r20 = tds(2);

r30 = tds(3); %ue对3号基站和0号基站的距离差，真实的

scatter(x0,y0,120,'d', 'filled'); text(x0,y0,'Anchor1');

scatter(x1,y1,120,'d', 'filled'); text(x1,y1,'Anchor2');

scatter(x2,y2,120,'d', 'filled'); text(x2,y2,'Anchor3');

hold on;

x10 = x1 - x0;

x20 = x2 - x0;

y10 = y1 - y0;

y20 = y2 - y0;

k0 = x0^2 + y0^2;

k1 = x1^2 + y1^2;

k2 = x2^2 + y2^2;

A = [x10 y10; x20 y20];

C = -[r10; r20];

D = [(k1-k0-r10^2)/2; (k2-k0-r20^2)/2];

%求解Ax = r0 * C + D

a = A\C;

b = A\D;

%求解r0

A_ = a(1)^2 + a(2)^2-1;

B_ = a(1) * (b(1) - x0) + a(2) * (b(2) - y0);

C_ = (x0 - b(1))^2 + (y0 - b(2))^2;

if B_^2-A_*C_ < 0

position = [Nan, Nan];

else

r0_1 = -(B_+sqrt(B_^2-A_*C_))/A_;

r0_2 = -(B_-sqrt(B_^2-A_*C_))/A_;

X1 = a * r0_1 + b;

X2 = a * r0_2 + b;

%剔除错误解：方法一：UE和基站时钟尽量同步。方法二：增加观测站(本例使用)

if abs(r30-(distance(X1(1),X1(2),x3,y3)-distance(X1(1),X1(2),x0,y0))) < 1e-8

position = X1;

else

position = X2;

end

end

end

%%

function dist = distance(x1,y1,x2,y2)

dist = sqrt((x1-x2)^2 + (y1-y2)^2);

end

上述代码需要注意三个地方！！！

算距离差时，不加上绝对值，这样可以排除掉一半的解(两双曲线相交有2至4个交交点)计算

r

0

r_0

r0​时可能有伪解，需要增加观测站或牺牲一定精度来排除另外一个解。(本文是增加了一个基站)。二维定位时，不允许所有的基站在

y

y

y轴的数值相等。

上述代码得到的结果

上图中*表示UE的真实位置，o表示UE的计算位置，可以看到每个UE的位置都被正确解算了。

Chan算法实现3D

<code>

%%

clc;

clear;

close all;

format long;

% tmp = unifrnd(0,255,4,2);

% x1 = tmp(1,1); y1 = tmp(1,2); % Anchor1

% x2 = tmp(2,1); y2 = tmp(2,2); % Anchor2

% x3 = tmp(3,1); y3 = tmp(3,2); % Anchor3

figure;

%随机生成100个UE位置，并对其进行TDOA计算

correct_sum = 0;

uncorrect_sum = 0;

for i=1:100

ue_x = randi(60)+randn();

ue_y = randi(60)+randn();

ue_z = randi(60)+randn();

scatter3(ue_x, ue_y, ue_z, 120, '*');

hold on;

%注意注意！！！基站的z不能全部相同，否则在第67行的矩阵A第三列元素全为0，Ax=C或Ax=Ds时求不出唯一解

stations = [-40 0 5; 20 0 15; 40 50 5; 0 0 5; 10 10 10];

r0_real = distance(ue_x, ue_y, ue_z, stations(1,1), stations(1,2), stations(1,3));

r1_real = distance(ue_x, ue_y, ue_z, stations(2,1), stations(2,2), stations(2,3));

r2_real = distance(ue_x, ue_y, ue_z, stations(3,1), stations(3,2), stations(3,3));

r3_real = distance(ue_x, ue_y, ue_z, stations(4,1), stations(4,2), stations(4,3));

r4_real = distance(ue_x, ue_y, ue_z, stations(5,1), stations(5,2), stations(5,3));

tds = [r1_real-r0_real r2_real-r0_real r3_real-r0_real r4_real-r0_real];

position = TDOA(stations, tds);

if distance(position(1), position(2), position(3), ue_x, ue_y, ue_z) < 1e-8

correct_sum = correct_sum + 1;

else

uncorrect_sum = uncorrect_sum + 1;

end

scatter3(position(1), position(2), position(3), 'o', 'r');

hold on;

end

%%

function [position] = TDOA(stations, tds)

x0 = stations(1,1);y0 = stations(1,2);z0 = stations(1,3);

x1 = stations(2,1);y1 = stations(2,2);z1 = stations(2,3);

x2 = stations(3,1);y2 = stations(3,2);z2 = stations(3,3);

x3 = stations(4,1);y3 = stations(4,2);z3 = stations(4,3);

x4 = stations(5,1);y4 = stations(5,2);z4 = stations(5,3);

r10 = tds(1);

r20 = tds(2);

r30 = tds(3);

r40 = tds(4);

scatter3(x0,y0,z0,120,'d', 'filled'); text(x0,y0,z0,'Station1');hold on;

scatter3(x1,y1,z1,120,'d', 'filled'); text(x1,y1,z1,'Station2');hold on;

scatter3(x2,y2,z2,120,'d', 'filled'); text(x2,y2,z2,'Station3');hold on;

scatter3(x3,y3,z3,120,'d', 'filled'); text(x3,y3,z3,'Station4');hold on;

hold on;

% r21 represents the TDOA between anchor1 and anchor2

% r31 represents the TDOA between anchor1 and anchor3

x10 = x1 - x0;

x20 = x2 - x0;

x30 = x3 - x0;

y10 = y1 - y0;

y20 = y2 - y0;

y30 = y3 - y0;

z10 = z1 - z0;

z20 = z2 - z0;

z30 = z3 - z0;

k0 = x0^2 + y0^2 + z0^2;

k1 = x1^2 + y1^2 + z1^2;

k2 = x2^2 + y2^2 + z2^2;

k3 = x3^2 + y3^2 + z3^2;

A = [x10 y10 z10; x20 y20 z20; x30 y30 z30];

C = -[r10; r20; r30];

D = [(k1-k0-r10^2)/2; (k2-k0-r20^2)/2; (k3-k0-r30^2)/2];

%求解Ax = r0 * C + D

a = A\C;

b = A\D;

%求解r0

A_ = a(1)^2 + a(2)^2 + a(3)^2 -1;

B_ = a(1) * (b(1) - x0) + a(2) * (b(2) - y0) + a(3) * (b(3) - z0);

C_ = (x0 - b(1))^2 + (y0 - b(2))^2 + (z0 - b(3))^2;

r0_1 = -(B_+sqrt(B_^2-A_*C_))/A_;

r0_2 = -(B_-sqrt(B_^2-A_*C_))/A_;

X1 = a * r0_1 + b;

X2 = a * r0_2 + b;

%剔除错误解：方法一：UE和基站时钟尽量同步。方法二：增加观测站(本例使用)

if abs(r40-(distance(X1(1),X1(2),X1(3),x4,y4,z4)-distance(X1(1),X1(2),X1(3),x0,y0,z0))) < 1e-8

position = X1;

else

position = X2;

end

end

%%

function dist = distance(x1,y1,z1,x2,y2,z2)

dist = sqrt((x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2);

end

这份代码是随机生成了100个三维点，然后使用Chan算法解算位置。

上述代码也需要注意三个地方！！！

算距离差时，不加上绝对值，这样可以排除掉一半的解(两双曲线相交有2至4个交交点)计算

r

0

r_0

r0​时可能有伪解，需要增加观测站或牺牲一定精度来排除另外一个解。(本文是增加了一个基站)。二维定位时，不允许所有的基站在

z

z

z轴的数值相等。

上述代码的运行结果

经过测试，所有的解算误差都小于

1

0

−

8

10^{-8}

10−8