I'm going to have to give a bit of context for this question to make sense.

I am working on a project which includes audio source localisation in 3-D space through TDoA (Time Difference of Arrival - from here called $\tau_{ij}$, between sensor $i$ and $j$) measurements. In other words I have $m$ sensors that each have a slightly different recording of the audio event. Via cross-correlation methods I can estimate the TDoA (or delay) between each pair of sensors.

With each pair of sensors, $i$ and $j$, we can point to the source position through the following optimization problem:

$$\delta_{ij} = v \tau_{ij}$$ $$h(P_{E},P_{i},P_{j}) = \|P_i - P_E\|_2 - \|P_j - P_E\|_2$$ $$P_E^* = {arg\,min}_{P_E}\sum\limits_{i \neq j}\left(\delta_{ij}-h(P_{E},P_{i},P_{j})\right)^2$$ where $v$ is the speed of sound in the selected medium (consider it ideally constant), $P_E^* = (x_E,y_E,z_E)$ is the source position estimate (after optimization), while $P_i = (x_i,y_i,z_i)$ and $P_j = (x_j,y_j,z_j)$ are the sensor positions.

Note the first two equations. The first one is the distance between the two sensors, calculated from the TDoA estimate. The second one is a hyperbolic surface (see image below, in a 2-D example) that also represents the distance between the sensors. One way to see this is noting (suppose perfect measurements):

$$\|P_i - P_E\|_2 - \|P_j - P_E\| = v \tau_{iE} - v \tau_{jE} = v (\tau_{iE} - \tau_{jE}) = \delta_{ij}$$

In other words, each pair of sensors generate a different hyperbolic surface. We then try find the position by minimizing the function (third equation), by varying $P_E$. For sure there will always be a $P_E$ where $\delta_{ij} = h(P_E,P_i,P_j)$, but it won't necessarily nullify the same equation for different $i$ and $j$. Therefore optimization is necessary.

In a simpler 2-D example, the hyperbolic curves above represent points where the TDoA between two sensors is constant.

Finally, my question is: What is the best geometric arrangement of the $m$ sensors in 3-D space, so that measurements are as precise as possible? For example, a line of sensors is probably not a good idea as the the hyperbolic curves will be parallel to each other.

  • $\begingroup$ Yeah, for 3 sensors don't put them all in a line, or you won't have 2D information, and for 4 sensors don't put them all in a plane, or you won't have 3D information. "that each have a slightly different recording of the audio event" So there's a constraint that the sensors have to be nearer each other than they are to the source? You can't place them around the corners of a room so that the source is in the middle of the sensors, for instance? Because otherwise I suspect making them as far apart as possible is the best solution for resolution (but not SNR) $\endgroup$ – endolith Jul 1 '13 at 23:27
  • $\begingroup$ Well technically its an underwater source localisation problem. There are no corners. :P I am trying to devise an algorithm for the movement of an array of sensors to get as close to the source as possible. $\endgroup$ – jhc Jul 1 '13 at 23:29
  • $\begingroup$ so the sensors are fixed relative to each other and have to move as one unit? they can't be independent? $\endgroup$ – endolith Jul 1 '13 at 23:31
  • $\begingroup$ They are not fixed. I'm just asking because I'm sure there's an ideal arrangement. I imagine it would be one that minimizes the sum of the dot products of all the sensor pairs. In other words a simplex: en.wikipedia.org/wiki/Simplex $\endgroup$ – jhc Jul 1 '13 at 23:36
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    $\begingroup$ @jhc: Easiest to see in 1D, but the idea generalizes. Assume you have a linear array with microphones at positions 0, 1 meter & 2 meter, and a periodic signal with wavelength 0.50 meters coming in from the left side. You wouldn't be able to tell left&right apart because all three signals would be in phase. If you put the third microphone at sqrt(2), the 3 inputs can´t be in phase regardless of wavelength. $\endgroup$ – MSalters Jul 4 '13 at 8:03

This will somewhat depend on the spectrum and properties of the signals and what your dominant source of noise, distortion, jitter, non-linearietes etc. are. Ideally you can resolve every point in space with 3 non-coplanar microphones. A very simple solution is one with 4 microphones: one in the origin and three one unit step displaced in x, y, and z dimension. The arrival difference between X and origin pair gives you the x-coordinate, etc.

You can improve this method by a simple least square approach that takes into account all microphone pairings (6 total). The fit error is metric of how much noise you have in the system.

Microphone spacing depends on spectral content of the signals you are tracking. If it's fairly narrow band you probably want them to be about half a wavelength apart so you maximize the phase difference between adjacent signals. A good robustness metric is something like dx/dt, i.e. you wiggle the arrival time a little and see how much the calculated position moves. For almost coplanar microphones that is a large number which indicates that this setup is very sensitive to noise.

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  • $\begingroup$ I've been thinking of making each 'microphone' a tiny array of microphones so that I can be flexible with my spectrum analysis. $\endgroup$ – jhc Jul 2 '13 at 16:54
  • $\begingroup$ About your first comment (about setting up the sensor positions in a unit step from the origin, and one in the origin.)... Well there really isn't an origin in my problem, as the application of this is to track an underwater target. So the space is vast. Plus, my sensors are supposed to be mobile (installed on an RoV for example). $\endgroup$ – jhc Jul 2 '13 at 16:59
  • $\begingroup$ Well, you need to express the result in some sort of a co-ordinate system, so you'll have to pick some origin. You might as well make that the center of your microphone array. Then, given the actual position of the ROV, you can translate that result into any co-ordinate system you want. I used this primarily to describe the geometry of the mic array. $\endgroup$ – Hilmar Jul 2 '13 at 17:22
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    $\begingroup$ "3 non-coplanar microphones" is impossible. 3 points are always coplanar. $\endgroup$ – endolith Jul 3 '13 at 14:06

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