# How to find location of a source in TDOA Method

We have 3d system with a source is sending signal and four receivers and we know the coordinate location of these four receivers. We have four Time difference of arrival. How to calculate the location of the source.

Calling $x, y, z$ the coordinates of the source, and $x_i, y_i, z_i$ ($i=1...4$) those of the receivers, the distance $D_i$ between the source and the receiver $i$ is given by $$\sqrt{(x-x_i)^2+(y-y_i)^2+(z-z_i)^2}$$ Considering two receivers $i$ and $j$, the difference $D_i-D_j=D_{ij}$ in the distance from the source is given by

$$D_{ij}=\sqrt{(x-x_i)^2+(y-y_i)^2+(z-z_i)^2}-\sqrt{(x-x_j)^2+(y-y_j)^2+(z-z_j)^2}$$

Now let us call $T_{ij}$ the time difference of signal arrival between receiver $i$ and receiver $j$. Assuming that the signal propagation time is linearly related to the distance according to a proportionality factor $k$ (representing signal velocity), and taking into account that we know the value of $T_{ij}$ for 4 couples of receivers $i,j$, we can write 4 different equations of the form $$D_{ij}=kT_{ij}$$

We therefore obtain a system with four equations and four unknown variables ($x, y, z, k$), that can be solved to get the coordinates of the source.

• @Anatoly- This is exactly how we should proceed. the four equations are not linear. Is it possible for you to spread some more light on solving these non linear equations or converting these non linear equations to linear. – Razack Jul 23 '14 at 6:45
• Trying to solve the system algebrically leads to high-degree equations, which are very complex to elaborate. Probably, the simplest approach could be to use numerical methods to find the solutions of the system. – Anatoly Jul 23 '14 at 10:29
• Any reference of numerical solution could you provide – Razack Jul 23 '14 at 10:53
• I would suggest you the following: onlinelibrary.wiley.com/doi/10.1002/9781118673515.app8/pdf – Anatoly Jul 29 '14 at 21:00
• You don't need numerical methods. Look at $D_{ij}^2$, instead of $D_{ij}$. If you subtract one equation from another, the square terms disappear, and you're left with a linear equation. – bubba Aug 22 '17 at 11:56