How to solve 3 TDOA equations with 3 variables x,y,z
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Ebrahim Abdullah Ahmed Albabakri
on 21 Jan 2022
I have a set of equations which looks like that, how to find x,y,z by coding while all other variables are known?
2 Comments
Accepted Answer
Torsten
on 21 Jan 2022
Code for symbolic solution:
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])
xnum = double(S.x)
ynum = double(S.y)
znum = double(S.z)
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);
x = X(1)
y = X(2)
z = X(3)
2 Comments
Venkata Naresh
on 22 Nov 2023
Hi, In the code you provided, what does the values a = 0.19845; b = 0.08791; c = 0.11492;
mean
Torsten
on 22 Nov 2023
Edited: Torsten
on 22 Nov 2023
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.
More Answers (1)
Torsten
on 21 Jan 2022
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".
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