Basis change for gradient approximation

Question

This is specifically a linear algebra question, but I kind of need to explain context. I suspect that this is simply a basis change problem, but I'm not entirely sure. Also, please feel free to correct any glaring mistakes I make.

The image below should help clarifying some details. Anyways, I have a source emitting a signal. I do not know the position of this source. I do have three sensors that are capable of detecting the signal.

I would like to get an approximation of the gradient of the power, $P$, of this signal as it expands (and loses power density), via measuring differences in the power detected by the sensors.

In other words, I want to find, at a certain $x_o$ and $y_o$: $$ \nabla P(x_o,y_o) $$

If I have sensors ${s_1,s_2,s_3}$, and I call $s_1$ my reference sensor (circled in the image below).

I can then certainly get the power difference, between the pair of sensors:
$$ \Delta P_u = P_1 - P_2\\ \Delta P_v = P_1 - P_3 $$

I can then approximate the gradient, in the {$u,v$}, basis with: $$ \nabla P_{\{u,v\}} = \left[\frac{\Delta P_u}{||s_1-s_2||} \frac{\Delta P_v}{||s_1-s_3||}\right]^T $$

But that is not what I want. I would like the gradient in the normal basis, in other words:
$$ \nabla P_{\{i,j\}} $$

I suspect this can be found with, $M$, a basis change matrix, where the columns are the difference vectors between the sensor positions: $$ M =\begin{bmatrix} |& |\\s_1-s_2 & s_1-s_3\\ |&|\end{bmatrix} $$

Then after normalisation, $M'$: $$ \nabla P_{\{i,j\}} = M' \nabla P_{\{u,v\}} $$

I still get odd answers with this technique, leading me to think that I might be screwing up the math.

Answer 1

Recall the integral definition of the (2d) gradient:

$$\nabla P = \lim_{A \to 0} \frac{1}{A} \oint_{\partial A} P \hat n \, d\ell$$

The curve $\partial A$ bounds an area $A$, and $\hat n$ is the outward normal to the curve.

Now, imagine a parallelogram formed by the vectors $u, v$. Put the sensor $P_1$, the central sensor, on say the right side of the parallelogram, and put the sensor $P_2$ on the left side. In the limit as $A \to 0$, we can approximate the terms of the integral by saying that $P$ takes constant values on these sides. Let $V$ be the length of this side, and $U$ be the length of the other side, so we get terms of the form

$$\lim_{UV \to 0} \frac{V \hat u^* (P_2 - P_1)}{UV} = \lim_{U \to 0} \hat u^* \frac{P_2 - P_1}{U}$$

It's crucial to note that the vector $\hat u^*$ is not $\hat u$ or $\hat v$. Rather, it is a vector orthogonal to $\hat v$.

You should be able to write the gradient, then, as

$$\nabla P \approx \hat u^* \frac{P_2 - P_1}{|s_2 - s_1|} + \hat v^* \frac{P_3 - P_1}{|s_3 - s_1|}$$

Then, write the vectors $\hat u^*, \hat v^*$ in terms of the basis vectors $i, j$ to get the standard basis components. You can use a change of basis matrix for such a purpose, but what's critical is that you realize the gradient is written in terms of the normal vectors $\hat u^*, \hat v^*$, not the tangent vectors $\hat u, \hat v$, and as such, the change of basis matrix is generally different.