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I have a microphone array of 4 channels taken as channels [6,10,22,26] from Eigenmike spherical microphone array. I wish to do 3D beamforming and create $8\times 4$ beams, 4-elevations with 8 wavefronts per elevation equally distributed.

I can't seem to work out the math as to the delays between the microphones. I also can't find an "off the shelf" code for that. This is my first beamformer so I am not even sure I am not missing anything. Is delay and sum process all I need?

What are the equations for determining the delays between the channels per beam? Do I need any other equations to create a beamformer?

Specifically, I am trying to create beams in direction from all possible combinations of $\varphi_s \in \{-\frac{3\pi}{8},-\frac{\pi}{8},\frac{\pi}{8},\frac{3\pi}{8}\}$ and $\theta_s\in\{-\frac{7\pi}{8},-\frac{5\pi}{8},-\frac{3\pi}{8},-\frac{\pi}{8},\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8}\}$.

My microphone array contains 4 microphones $m_i=(r_i,\theta_i,\varphi_i)$, at the following positions: \begin{pmatrix} 0.042 & \frac{\pi}{4} & \frac{7\pi}{36} \\ 0.042 & -\frac{\pi}{4} & -\frac{7\pi}{36} \\ 0.042 & \frac{3\pi}{4} & -\frac{7\pi}{36} \\ 0.042 & -\frac{3\pi}{4} & \frac{7\pi}{36} \end{pmatrix}

$r_i$ is in meters. $\varphi_i, \theta_i$ are azimuth and elevation respectively, according to the microphone's documentation, given herein radians.

A quick sketch of the configuration from two views. Note that the distance of all mics is 42mm from the origin:

enter image description here enter image description here

$(x,y,z)$ values in this system is extracted in the following manner to my understanding: $$x=r\cdot \cos(\varphi)\cdot \cos(\theta)$$ $$y=r\cdot \cos(\varphi)\cdot \sin(\theta)$$ $$z=r\cdot \sin(\varphi)$$

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    $\begingroup$ To determine the delays that maximise the response of the combined mic towards a specific direction, you need to know the positions of the individual mics and the "specific direction". See this for a primer. What do you mean with "...8 wavefronts per elevation..." (?) $\endgroup$
    – A_A
    Apr 18, 2019 at 9:23
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    $\begingroup$ You could borrow math from a spherical head model. Here the wave propagates also along the surface of the sphere, so not always in a straight line like in @endolith's answer. $\endgroup$
    – Olli Niemitalo
    Apr 24, 2019 at 4:16
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    $\begingroup$ GeoGebra could be used for sketching spherical coordinates. $\endgroup$
    – Olli Niemitalo
    Apr 25, 2019 at 8:42
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    $\begingroup$ Well, if my quick sketch in octave is correct, you have 4 mics on the same plane. This means that you cannot resolve front-back because a point can be at "two" equal distances ($+d,-d$) from a plane. My initial understanding was that you might be selecting a subset of the 32 mics...not isolating a fixed 4 and doing beamforming just with those. Can you clarify on that point? (And maybe the configuration?) ( @OlliNiemitalo ) $\endgroup$
    – A_A
    Apr 25, 2019 at 9:10
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    $\begingroup$ Might be useful to show how to get $x,y,z$ coordinates from the spherical coordinates, to remove all ambiguity. There seem to be incompatible spherical coordinate conventions out there. $\endgroup$
    – Olli Niemitalo
    Apr 25, 2019 at 12:36

2 Answers 2

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I don't know what the fuss is about. This is a simple vector problem.

"Well, if it is that easy please elaborate. I have 4 microphones on the following locations (0.042,45,35),(0.042,-45,-35),(0.042,135,-35),(0.042,-1355,35). These are coordinates for the microphones with m1=(ri,θi,φi), where −π≤θ≤π and −π2≤φ≤π2. create a beam towards a sourse at (θs,φs)=(22.5,67.5). – havakok"

You need to define five vectors in Cartesian coordinates, one for each mic and one for the beam. The beam vector should be normalized to a length of one.

If the sound source is sufficiently far away, the assumption that the wavefront is coming in as a planar wave is a fairly accurate approximation. Therefore, you simply have to find the projection of each mic location onto the line defined by the beam vector. The projection for each mic is found by dotting the mic vector with the normalized beam vector. The value you get is the distance along the line (where one is the beam vector length). Use the speed of sound to convert this distance to a time shift. Use your sampling rate to convert the time shift to a sample shift. Shift and add. The last mic to receive the signal is not delayed, and the previous mics are delayed relative to their projection's distance to the last mic's projection.

It should be obvious that you only get one beam direction at a time. Don't be oversold on the "beam" notion either.

Ced

Two Answers:

1) Rounding to the nearest sample should be negligible.

Some ball park figures to consider: Suppose you have a 1000Hz tone. Sound travels at roughly 1130 feet per second. Thus one cycle of the tone is roughly 1.130 feet long, or about 13 and 1/2 inches. At CD quality sampling, about 44 samples. So each sample is about 3/10 of an inch (this is frequency independent). Compared to the length of the cycle, this isn't a lot. However, if you are concerned about frequencies above 10k, this may matter. In that case, a simple linear interpolation in your delay should do the trick.

2) The beam effect is formed by "coordinated interference effects", i.e. signals coming from other directions will be added together in an out of phase manner (only the signals in the desired direction are in phase), thus you are relying on interference to destructively interfere. This is frequency dependent. Also, signals coming in nearly at the beam angle won't be affected much, so perhaps "cone forming" is more accurate.

You didn't mention whether these were omnidirectional mics or not. No bearing on the math, but may influence your results.

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  • $\begingroup$ Thanks. Two questions though. What do you do if the conversion from time to samples needs to be rounded? Is that neglectable? What did you mean by oversold on the "beam"? you meant that I should look at the problem as a cartesian system? $\endgroup$
    – havakok
    Apr 25, 2019 at 14:04
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    $\begingroup$ @havakok, Thanks for the quick acceptance. I've added a followup to address your additional questions. $\endgroup$
    – Cedron Dawg
    Apr 25, 2019 at 14:23
  • $\begingroup$ This doesn't take the reflective/occlusive sphere into account, though $\endgroup$
    – endolith
    Apr 25, 2019 at 14:28
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    $\begingroup$ @endolith, That is true. I just looked at his diagrams and thought they were open space. The link doesn't seem to reflect his description either, i.e. mic count. I don't think it is a significant concern. For the mics facing the source, the sphere is immaterial, the plane wave will reach it before the "surface wave". For the the mics facing away, receiving a slightly delayed signal will not be as impactful since the source signal will be relatively attenuated compared to the forward facing mics signal. $\endgroup$
    – Cedron Dawg
    Apr 25, 2019 at 14:38
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    $\begingroup$ @havakok You can delay by fractions of a sample if necessary. It's an approximation, but it's possible. $\endgroup$
    – endolith
    Apr 25, 2019 at 14:39
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What are the equations for determining the delays between the channels per beam?

It's pretty basic trigonometry: If A and B are microphones, and C is the object you're trying to record, then C's wavefront will hit A before it hits B. So delay A's signal by the same amount of time, to make the recorded waves line up.

enter image description here

So in this case, f is $\sqrt{5^2+1^2} \approx 5.1 $ m long, and g is $\sqrt{6^2+1^2} \approx 6.1$ m long, so B is 0.98 m farther away than A, so the signal in B will be delayed by (0.98 m / c) = 2.89 ms, so delay B by the same amount to make them line up.

If C were infinitely far away, on the same y coordinate, then the delay would be exactly 1 m:

enter image description here

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    $\begingroup$ @havakok endolith is right. First you find the phase center of your array, and then you just look at the angle under which you see the emitter. $\endgroup$
    – Marcus Müller
    Apr 24, 2019 at 17:02
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    $\begingroup$ Straight line calculations won't cut it for elements that are behind the sphere. $\endgroup$
    – Olli Niemitalo
    Apr 25, 2019 at 4:00
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    $\begingroup$ @OlliNiemitalo You're right, I didn't realize there was a sphere $\endgroup$
    – endolith
    Apr 25, 2019 at 4:56
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    $\begingroup$ @A_A there are only four microphone elements so turning off up to three would destroy directivity. Sound does reach the "behind" elements, too. They are pressure mic elements so omni. $\endgroup$
    – Olli Niemitalo
    Apr 25, 2019 at 7:21
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    $\begingroup$ @OlliNiemitalo ...my interpretation of the original post is that there are 4 mics per "elevation ring" (?). See comment of OP to the original question. Will ask for clarification. $\endgroup$
    – A_A
    Apr 25, 2019 at 7:32

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