Generalization of the GPS technology

Question

This might not be a mathematical question, but now we want to generalize the GPS technology.

<--Setting--> There are $n$ radiation source points on the plane $\mathbb{R}^2$.

<--Problem--> Consider the problem to detect the position of the $n$ radiation sources and each radiation power.

<--Example--> For example, we will use drones at the $N$ points with the constant height. In the case of $n = 1$, we can make use of the GPS mechanism and the attenuation rate of the radiation transmitted in the air (this is monotonically decreasing with distance, so it has the inverse function), and the problem is solved in $N=3$.

<--Question--> Considering this problem in the case of general $n\geq 2$, is there an answer?

Thank you for your help.

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EDIT 1 : Of course, to detect the tuple of $n$ pairs of point and radiation power, we have to measure at (more than or equals to) $3n$ points (because the tuple is in $\mathbb{R}^{3n}$). For instance, in the case of $n=1$, the $3$ measuring points must be arranged so as to link a triangle. The question is : how do we have to arrange the $3n$ measuring points to solve the problem?

EDIT 2 : More precisely, this problem can be rewritten to make an algorithm as follows :

Input 1: coordinates of "detectors", $N$($=3n$? more?) points $(x_1,y_1,z_1)$, $\ldots$, $(x_N,y_N,z_N)$ $\in\mathbb{R}^3$.

Input 2: detected radiation powers, $N$ real numbers $\rho_1$, $\ldots$, $\rho_N$ $\in\mathbb{R}$.

(Note : We have the attenuation rate of the radiation transmitted in the air $\rho(r) = c \exp(-\lambda r)$ where $r$ is the distance, and c is the original radiation power, and the lambda is the constant.)

Output : coordinates of the radiation points $(X_1,Y_1)$, $\ldots$, $(X_n,Y_n)$ $\in \mathbb{R}^2$ and the original radiation powers $C_1$, $\ldots$, $C_n\in\mathbb{R}$.

Answer 1

The system of equations is

$$\sum_{j=1}^n C_je^{-\lambda\sqrt{(x_i-X_j)^2+(y_i-Y_j)^2+h^2}}=\rho_i,$$ for $j=1,\cdots N=3n$. (By a proper rescaling of the coordinates, you can make $\lambda=1$.)

This is a Radial Basis Function interpolation problem, where the center point coordinates are also unknown. Just use Newton. I don't think that particular arrangements of the probing points make it simpler to solve. Probably safer to spread them uniformly (equilateral grid).

I suspect that because of symmetries, the solutions could be multiple, and $3n$ points might not be enough; more would help disambiguate.

For better accuracy you may increase the number of probing points and find a Least-Squares optimum using the Levenberg-Marquardt algorithm.

Answer 2

So let me rephrase your question (correct me if I'm wrong). I will also use more standard notation, i.e. $n$ will be the dimension, not the amount of radiation points/detectors (btw why you consider $\mathbb{R}^{3n}$?).

First of all note that the amount of radiation points doesn't matter. Whatever works for one, it works for any number.

Assume that I have "detectors" (i.e. vectors) $A=\{a_1,\ldots, a_k\}\subset\mathbb{R}^n$. What minimal properties $A$ has to satisfy so that any system of equations

$$\begin{cases} d(a_1, x)=b_1 \\ \cdots \\ d(a_k, x)=b_k \end{cases}$$

(where $d$ is some metric, i.e. "the attentuation rate") for given $b_i\in\mathbb{R}$ has at most one solution (so $x$ satisfying the system of equations would be what you call a "radiation point").

It can be shown that in case where $d$ is the Euclidean metric (it probably works for the attentuation rate as well) it is enough if $k=n+1$ and $A$ is affine independent, i.e. the set $\{a_2-a_1, a_3-a_1,\ldots, a_k-a_1\}$ is linearly independent. In case $n=2$ this means that $A$ doesn't lie on a line, in case $n=3$ this means that $A$ doesn't lie on a plane, etc.