0
$\begingroup$

Having photo-diodes with a very narrow field of view (+-10 degrees for half power) what should I consider when designing a beam-forming algorithm?

I know that I could use many photo-diodes mounted in an arc to cover 180 degrees for example but the uncertainty of 10 degrees is too big. As an alternative to classical beam-forming, what would be the methods to use to isolate a signal coming from one direction with a high precision (under 1 degree)?

My goal is to analyze the flickering of lamps, knowing their position from analyzing camera data, I want to isolate the signal coming from their direction.

Improve this question
$\endgroup$
0
$\begingroup$

Given the horizontal and vertical component of the distance vector from the source to the sensor array be X and Y, then the angle between the sensor array and the source is given by: $$ \theta = \tan^{-1} (Y/X) $$

If the linear distance between each sensor is $d$, each sensor should experience a delay: $$ \delta t = d \sin(\theta)/c $$

Now if you can measure the delays (high precision measurement will lead to higher accuracy), and have the knowledge of $d$, you can find the angle $\theta$.

Warning: Practical errors and considerations are not taken into account in this approach.

Improve this answer
$\endgroup$
3
  • $\begingroup$ This method would require a sampling in the GHz domain to achieve an accuracy under 1 degree, my poor Arduino won't be able to handle it. Even the impulse response of common photo-diodes doesn't go far below 5 nanoseconds. $\endgroup$ – Mehdi Jul 2 '15 at 12:54
  • $\begingroup$ Then the best I can think of is to play with amplitude. As you know the spatial response of your sensors. However, again in practice noise would make your life very very difficult to get accurate results. I hope someone here comes with a better idea. $\endgroup$ – learner Jul 2 '15 at 16:29
  • $\begingroup$ I searched around more and I actually realized there is some cheap photo-diodes with a field of view of +-75°, this eliminates the problem $\endgroup$ – Mehdi Jul 3 '15 at 8:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.