How to solve 3 TDOA equations with 3 variables x,y,z

syms x y z 
x1 = 0.60; y1 = 0.00; z1 = 0.70;
x2 = 0.20; y2 = 1.30; z2 = 0.70;
x3 = 0.80; y3 = 1.05; z3 = 0.20;
x4 = 0.00; y4 = 0.25; z4 = 0.20;
a = 0.19845; b = 0.08791; c = 0.11492;
eqn1 = a==sqrt((x-x2)^2+(y-y2)^2+(z-z2)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
eqn2 = b==sqrt((x-x3)^2+(y-y3)^2+(z-z3)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
eqn3 = c==sqrt((x-x4)^2+(y-y4)^2+(z-z4)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
S = solve([eqn1,eqn2,eqn3],[x,y,z])
S = struct with fields:
    x: (193873*83486204818584396915878923974835^(1/2))/17900867984687600000000 + 2237507060134434351121/5594021245214875000000
    y: 29092239452758845515869/44752169961719000000000 - (36989*83486204818584396915878923974835^(1/2))/4475216996171900000000
    z: (3723*83486204818584396915878923974835^(1/2))/11188042490429750000000 + 417333164652058423709/716034719387504000000
xnum = double(S.x)
xnum = 0.4989
ynum = double(S.y)
ynum = 0.5746
znum = double(S.z)
znum = 0.5859

Code for fsolve:

X0 = [1,1,1];
x1 = 0.60; y1 = 0.00; z1 = 0.70;
x2 = 0.20; y2 = 1.30; z2 = 0.70;
x3 = 0.80; y3 = 1.05; z3 = 0.20;
x4 = 0.00; y4 = 0.25; z4 = 0.20;
a = 0.19845; b = 0.08791; c = 0.11492;
fun=@(x,y,z)[a-(sqrt((x-x2)^2+(y-y2)^2+(z-z2)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2)), ...
             b-(sqrt((x-x3)^2+(y-y3)^2+(z-z3)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2)), ...
             c-(sqrt((x-x4)^2+(y-y4)^2+(z-z4)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2))];
X = fsolve(@(x)fun(x(1),x(2),x(3)),X0);
Equation solved.

fsolve completed because the vector of function values is near zero
as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
x = X(1)
x = 0.4989
y = X(2)
y = 0.5746
z = X(3)
z = 0.5859

2 个评论
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Venkata Naresh 2023-11-22

Hi, In the code you provided, what does the values a = 0.19845; b = 0.08791; c = 0.11492;

mean

Torsten 2023-11-22

编辑：Torsten 2023-11-22

You are given four points X1, X2, X3 and X4 in three-dimensional space.

You search for a fifth point X for which

the distance difference between (X and X2) and (X and X1) is "a"

the distance difference between (X and X3) and (X and X1) is "b"

the distance difference between (X and X4) and (X and X1) is "c"

Here, a, b and c are given values.

I don't know the application behind this problem, maybe in surveying and mapping.

If you are interested, you should google "TDOA equations" from the title of the message.

请先登录，再进行评论。

Answer 2

Torsten 2022-1-21

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1633255-how-to-solve-3-tdoa-equations-with-3-variables-x-y-z#answer_878895

在 MATLAB Online 中打开

If there are no specialized methods to solve the TDOA equations (did you take a look into the literature ?), I suggest trying

syms x1 x2 x3 x4 y1 y2 y3 y4 z1 z2 z3 z4 x y z a b c
eqn1 = a==sqrt((x-x2)^2+(y-y2)^2+(z-z2)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
eqn2 = b==sqrt((x-x3)^2+(y-y3)^2+(z-z3)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
eqn3 = c==sqrt((x-x4)^2+(y-y4)^2+(z-z4)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
S = solve([eqn1,eqn2,eqn3],[x,y,z])

If this is not successful, use "fsolve".

3 个评论
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Torsten 2022-1-21

Then follow the two ways I gave you.

Ebrahim Abdullah Ahmed Albabakri 2022-1-21

在 MATLAB Online 中打开

This is the output of solve

S = 
  struct with fields:
    x: [1×1 sym]
    y: [1×1 sym]
    z: [1×1 sym]

fsolve shows an error

    Error using fsolve (line 180)
FSOLVE requires the following inputs to be of data type double: 'X0'.

and This is the code

syms x1 x2 x3 x4 y1 y2 y3 y4 z1 z2 z3 z4 x y z a b c
x1 = 0.60; y1 = 0.00; z1 = 0.70;
x2 = 0.20; y2 = 1.30; z2 = 0.70;
x3 = 0.80; y3 = 1.05; z3 = 0.20;
x4 = 0.00; y4 = 0.25; z4 = 0.20;
a = 0.19845; b = 0.08791; c = 0.11492;
eqn1 = a==sqrt((x-x2)^2+(y-y2)^2+(z-z2)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
eqn2 = b==sqrt((x-x3)^2+(y-y3)^2+(z-z3)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
eqn3 = c==sqrt((x-x4)^2+(y-y4)^2+(z-z4)^2) - sqrt((x-x1)^2+(y-y1)^2+(z-z1)^2);
S = solve([eqn1,eqn2,eqn3],[x,y,z])