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I am using multiple digital MEMS microphones in an array which are 8mm apart. Now I recorded a sine wave of 20kHz for calibration (at angle 0°) and want to measure the phase difference over time between all microphones. Sampling frequency is 48 kHz.

My approach would be to perform a hilbert transformation (because the recorded signal is real) and taking the angles of the complex samples.

Is there a better method to do this? I want to know whether the phase difference

  1. stays constant
  2. is reasonably small

Thank you!

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  • $\begingroup$ 20 kHz sine wave seems like the worst possible choice for that type of measurement. What is your specific acoustic environment ? Do are you planning to do this in anechoic chamber ? $\endgroup$
    – Hilmar
    Mar 15, 2022 at 13:35
  • $\begingroup$ I did the measurements in an anechoic chamber - the constant sine was chosen because I want to see the phase drift over time $\endgroup$
    – nik124
    Mar 15, 2022 at 13:57
  • $\begingroup$ the relative phase between the microphones will depend on frequency and what happens at 20 kHz is unlikely to be representative at "normal" audio frequency. But maybe you want the array to work mostly at 20 kHz. $\endgroup$
    – Hilmar
    Mar 15, 2022 at 14:13

1 Answer 1

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The Hilbert Transform idea would definitely work but may be overly complicated. An FFT would also provide phase and amplitude information for each bin, and the phase of the dominant bins where the tone resides can be compared to determine relative phase of each, assuming the data captures are synchronized.

In this case with a reasonably small phase another simple approach would be to multiply each waveform with a reference tone offset by a quarter cycle (90°) and filter. The waveforms which should be synchronously captured should be leveled to the same amplitude prior to taking the product since the phase result is both sensitive to amplitude and actual phase. Given two sinusoids, the average of the product is proportional to the cosine of the phase difference between the two sinusoids, by delaying one of the two by 90 degrees, the result will be proportional to the sine, and for small angles $\sin(\theta) \approx \theta$ when the angle is given in radians. For example an angle as large at 10 degrees which is 0.1745 radians has the result $\sin(0.1745) = 0.1736$ with an error that would most likely be below the noise of the result.

This is seen in the trigonometric relationship of the product of two sinusoids:

$$\cos(2\pi f t) \cos(2\pi f t + \theta) =\frac{1}{2} \cos(4\pi f t + \theta) + \frac{1}{2}\cos(\theta)$$

We see the product is the sum and the difference of the two frequencies and phase offsets. When the frequencies are the same we get a tone at double the frequency (which we filter out in the average, or low pass filter) and a tone at DC which is proportional to the cosine of the angle.

Building on this and the need to level each tone prior to taking the product (to remove amplitude sensitivity), a best approach would be to hard limit each received tone, ensuring the threshold is properly set for a 50% duty cycle in each resulting square wave. The low frequency result of this product (which is a simple XOR function that is averaged) will have a linear result versus phase, providing an unambiguous linear output versus phase over a $\pm 90$ degree phase range. Again, the reference tone should be delayed by 90° to center the result in the usable range.

The product and average solution has the benefit of minimizing noise in estimate versus any approach that would for example compare just one zero crossing, it is important however that the reference waveform be synchronized to the test waveform to eliminate inevitable drift errors between the two.

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  • $\begingroup$ Thanks a lot for your answer! I think synchronization is not a problem if I just compare microphones to each other (they sample synchronously already) right? $\endgroup$
    – nik124
    Mar 15, 2022 at 14:01
  • $\begingroup$ As long as they are indeed synchronously sampled- that is very important. $\endgroup$ Mar 15, 2022 at 16:27
  • $\begingroup$ But yes compare to each other- but remember the 90 degree delay- May be easier to just synthesize a coherent reference: use sine to transmit and cosine to compare $\endgroup$ Mar 15, 2022 at 16:32

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