I am building a matlab simulation that applies room compensation to a 3D audio rendering system, and I'm searching for the frequency dependent sound absorption coefficients of various materials in order to extract the reflection coefficients.

However all the tables I found provide absorption coefficients for different octave frequency bands that have center frequency from 125Hz to 4kHz or at most 8kHz, but in my case I need to consider frequencies up to 20kHz.

My question is why are tables for sound absorption coefficients limited to such frequency range and how should I consider the coefficients for higher frequencies(i.e. higher than 10kHz) and also for lower ones(i.e. lower than 80 Hz)?

I found the tables in the annex section of the book "Auralization" by Dr. Michael Vorländer, which uses the ISO 354:2003 standard in order to perform the measurements.

The only resource where I found tables for frequency values higher than 8kHz was this paper: https://www.degruyter.com/downloadpdf/j/aoa.2013.38.issue-2/aoa-2013-0020/aoa-2013-0020.pdf where it is proposed an alternative method w.r.t. the previously mentioned standard in order to take the measurements from 50Hz to 50kHz, however these are provided just for very few materials.

  • $\begingroup$ Hey Luca, I'm not an acoustics expert myself, but I guess future readers, and among them, potential answerers would be very interested in which literature you found these tables – this is something often asked for on DSP.SE, and it might incentivize people to work harder on an answer. Also, citing your sources is pretty much always a good idea :) $\endgroup$ – Marcus Müller Mar 14 '17 at 23:12
  • $\begingroup$ Please don't cross-post, especially not without referring to the other version of a question: physics.stackexchange.com/questions/318807/… $\endgroup$ – Marcus Müller Mar 14 '17 at 23:18
  • $\begingroup$ Hi! I am sorry, I didn't know it wasn't polite to cross and I removed the question from Physics Stack Exchange. I also added some references on my question here, thanks for the tips! $\endgroup$ – Luca Mar 15 '17 at 9:34
  • $\begingroup$ If you are looking for more of coefficients up to 8kHz for various materials, then take a look at this excel sheet. $\endgroup$ – jojek Mar 15 '17 at 10:50

Great question. After about 10kHz, most of the energy is lost due to air absorption depending on your distance to the source. Your room model could approximate this with a lowpass filter whose rolloff is approximated by distance, but don't quote me on that.

| improve this answer | |
  • $\begingroup$ Thanks for the answer!! I tried use a lowpass filter before, but the quality of the rendering degrades too much! Maybe a solution could be to consider the absorption coefficients of the walls as 100% for frequencies higher than 10kHz, in order to simulate the air attenuation? $\endgroup$ – Luca Mar 15 '17 at 10:42
  • $\begingroup$ Sidenote: could the downvoter explain their downvote? No, I didn't say use a single lowpass filter, I said try a dynamic one dependent on distance in your source-receiver model. Walls do not absorb 100% of high frequencies unless they are perhaps made of heavy velour, so using alpha = 1 will not be physically accurate. $\endgroup$ – panthyon Mar 15 '17 at 14:53

The main reason why the tables don't have it, is that it's hard to measure. The measurement technique in ISO 354:2003 relies on measuring the difference in reverberation times in a reverberation rooms with and with/out a material sample. At higher frequencies, the reverb time is dominated by air absorption and and the sound field becomes less and less diffuse which violates the basic assumption behind the measurement. It's not uncommon to measure absorption coefficients larger than 1. That's not good !

I wouldn't mingle air absorption and wall absorption. They are different effects and should be modelled separately. They are basically additive. At 20 kHz air absorption is around 50-ish dB/100m. The early reflections in a room have a travel distance of only a few meters so there is some attenuation, but there is still plenty of energy left.

Keep in mind that wall absorption measurement is NOT done at 8 kHz but it's averaged over the 8 kHz octave, so the data covers frequencies up to 11.5 kHz. If you need to go hire, you can often simply extrapolate to 16 kHz by looking at the difference between the 4 kHz and 8 kHz value.

Easiest is to extrapolate in dB for the reflection coefficient. Example: 4 kHz = 0.6, 8 kHz = 0.8. Reflection coefficients are 0.4 and 0.2 or -4dB and -7dB respectively. So it's 3 dB drop of reflected energy per octave. Extrapolation yields -10 dB for 16 kHz, which is a reflection coefficient of 0.1 or an absorption of 0.9.

Something like this $$\alpha_{16} = 1-10^{0.1 \cdot (2 \cdot 10 \cdot \log_{10}(1-\alpha_8)-10 \cdot \log_{10}(1-\alpha_4))}$$

| improve this answer | |
  • 1
    $\begingroup$ There is ISO 10534 standard for measurements of absorption coefficient using the Impedance Tube. $\endgroup$ – jojek Mar 15 '17 at 19:45
  • $\begingroup$ Yep, but if I understand correctly that only gives you reflection/absorption for frontal incidence and not spatially averaged for a diffuse sound field $\endgroup$ – Hilmar Mar 15 '17 at 21:44
  • $\begingroup$ Indeed, physical absorption coefficient is not the same as one measured in the reverberation chamber. Nonetheless it's still useful as it gives you an idea of what are the properties of sample. $\endgroup$ – jojek Mar 16 '17 at 8:25
  • $\begingroup$ Hi thanks a lot for your detailed answer!! However I was wondering how is it reasonable to extrapolate coefficients for higher frequencies using the way you propose? I understand that this would be just an approximation but I checked and the relation that you suggest does not hold between the other frequency bands $\endgroup$ – Luca Mar 16 '17 at 12:34

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.