Analyzing Sound Through a Rectangular Slot

[Alfred Cann]

I am stumped analyzing the transmission of sound through a rectangular slot in a thin rigid wall. I have found online that a square or round aperture acts as a highpass filter with a 6dB/oct slope and a corner frequency where the wavelength is on the order of twice the aperture diameter.

I have reasoned that a rectangular aperture would act as a highpass filter with a 6dB/oct slope up to a corner frequency where the wavelength is twice the slot length, then a 3 dB/oct slope up to a corner frequency where the wavelength is twice the slot width. I would like to have this confirmed or refuted.

 

[trurle]

Summary:: Sound transmission through a rectangular slot in a thin rigid wall

I am stumped analyzing the transmission of sound through a rectangular slot in a thin rigid wall. I have found online that a square or round aperture acts as a highpass filter with a 6dB/oct slope and a corner frequency where the wavelength is on the order of twice the aperture diameter.

I have reasoned that a rectangular aperture would act as a highpass filter with a 6dB/oct slope up to a corner frequency where the wavelength is twice the slot length, then a 3 dB/oct slope up to a corner frequency where the wavelength is twice the slot width. I would like to have this confirmed or refuted.

The rigid zero-thickness wall is indeed produce a node in air velocity, resulting in high pass filter. In practice, the effect from air resistance in finite depth hole/slot is dominant, resulting in low-pass filter. I doubt you can neglect the thickness of wall.
https://www.sciencedirect.com/science/article/pii/S1877705811060425

 

[Alfred Cann]

To solve a complex problem, I always approach it with simplifications first, in order to develop basic insights. So let's stick to my statement of the problem with negligible wall thickness. In case you're interested, this problem arose in connection with the propagation through the 1x5 ft. pass-through window of the kitchen of a small restaurant. The kitchen crew said they could hear an electric bass better than an acoustic bass. I conjectured that was because the electric bass has a higher harmonic content, but I want to put some theoretical clothes on that.

 

[trurle]

To solve a complex problem, I always approach it with simplifications first, in order to develop basic insights. So let's stick to my statement of the problem with negligible wall thickness. In case you're interested, this problem arose in connection with the propagation through the 1x5 ft. pass-through window of the kitchen of a small restaurant. The kitchen crew said they could hear an electric bass better than an acoustic bass. I conjectured that was because the electric bass has a higher harmonic content, but I want to put some theoretical clothes on that.

In your particular case (restaurant room with size of few wavelengths) i would suspect multipath (or standing wave) effects rather than small corrections due harmonic contents. Are the acoustic bass and electric bass were in exactly same location? Acoustic bass may be simple in location placing a pressure node closer to kitchen window.

 

[Alfred Cann]

There are many confounding factors:
1. Multipath effects in dining room and kitchen
2. Electric bass speaker on the floor and behind the player, thus not in exactly the same location as the center of radiation of the acoustic bass
3. Radiation of the acoustic bass from a much bigger area, different radiation pattern, and different frequency sensitivity of radiation pattern
4. Subjective adjustment of electric bass to same loudness (at the band) as acoustic bass. Probably within 3dB, maybe 6dB
5. Although both play the same fundamental tones, 41 Hz to about 200 Hz, the higher harmonic content of the electric not only gets through the window better, but is also heard better (equal loudness contours).

Nevertheless, my intuition says the window makes a big difference but I would like to verify that. I treat it as a piston and calculate a loss of 15.6-4.4dB for frequencies of 41-200Hz.

 

[trurle]

There are many confounding factors:
1. Multipath effects in dining room and kitchen
2. Electric bass speaker on the floor and behind the player, thus not in exactly the same location as the center of radiation of the acoustic bass
3. Radiation of the acoustic bass from a much bigger area, different radiation pattern, and different frequency sensitivity of radiation pattern
4. Subjective adjustment of electric bass to same loudness (at the band) as acoustic bass. Probably within 3dB, maybe 6dB
5. Although both play the same fundamental tones, 41 Hz to about 200 Hz, the higher harmonic content of the electric not only gets through the window better, but is also heard better (equal loudness contours).

Nevertheless, my intuition says the window makes a big difference but I would like to verify that. I treat it as a piston and calculate a loss of 15.6-4.4dB for frequencies of 41-200Hz.

I think without properly measuring 1 (intensity patterns across rooms) and 4 (sound output of both devices), any calculation would be a speculation. I even not sure your wall can be effectively attenuating at such low frequency. Better to walk around with sound meter before jumping to conclusions.

 

[Alfred Cann]

Thank you. I still would like to know, just for academic interest, if my analysis of a slot in an ideal wall is correct.

 

[trurle]

Thank you. I still would like to know, just for academic interest, if my analysis of a slot in an ideal wall is correct.

https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19760016859.pdf
from page 46 seems the high-pass action for small opening at ~6 dB/octave (convert 3 dB/octave flow slope to 6 dB/octave power slope, and revert 1/omega horizontal axis) is confirmed by experiment. The curve shape seems to be different from simple high-pass electrical filter (not a straight line in log-log coordinates at low omega).

To say simple, your analysis is correct, but only to accuracy factor of two.

 

[tech99]

Summary:: Sound transmission through a rectangular slot in a thin rigid wall

I am stumped analyzing the transmission of sound through a rectangular slot in a thin rigid wall. I have found online that a square or round aperture acts as a highpass filter with a 6dB/oct slope and a corner frequency where the wavelength is on the order of twice the aperture diameter.

I have reasoned that a rectangular aperture would act as a highpass filter with a 6dB/oct slope up to a corner frequency where the wavelength is twice the slot length, then a 3 dB/oct slope up to a corner frequency where the wavelength is twice the slot width. I would like to have this confirmed or refuted.

For a transverse (EM) wave we see a sudden cut off. For a longitudinal wave (sound) I had presumed that it would pass through the smallest hole. After all, an organ pipe is only a tiny fraction of a wavelength in diameter yet supports a long sound wave.
It looks as if the 6dB/octave slope arises from frictional losses.

 

[Alfred Cann]

trurle:
The NASA paper you cite is completely irrelevant; it deals with fluid flow (as used for sound absorption) not with sound transmission or radiation. Note that their apertures are much smaller than a wavelength; mine are of the order of the wavelength. Look at the literature for radiation from a loudspeaker in an infinite baffle.

tech99:
Sound passes thru a small hole but with a big loss. From the loudspeaker literature it is clear that the 6dB/oct loss is from the decrease of radiation resistance, not friction loss. I envision the following physical mechanism:
When the frequency gets so low that a half wavelength is bigger than the aperture, the air in front of the aperture can slide out of the way sideways more easily than it can push forward, where it would force a longitudinal propagation. For a round or square aperture this is manifested as a 6 dB/oct loss. My conjecture is that in the case of a rectangular or elliptical aperture, at frequencies where only the width is smaller than a half wave, the air could slide away crosswise but not lengthwise. Therefore, the slope would be 3dB/oct. When the frequency gets low enough so that the air can escape in both directions, the slope becomes 6dB/oct. Thus, the curve exhibits 2 corner frequencies with a slope of 3 dB/oct between them. I would like this verified or refuted.

 

[trurle]

When the frequency gets so low that a half wavelength is bigger than the aperture, the air in front of the aperture can slide out of the way sideways more easily than it can push forward

I cannot see why air would "slide sideways". Rigid wall is air velocity node.

Overall, i found one paper on topic which is not behind paywall.
It is in Chinese, but from figures the response of rectangular slot only very vaguely resemble high-pass filter. You cannot see anything like octaves-spanning slopes.
https://www.researchgate.net/publication/334576771_Modal_radiation_impedance_calculation_and_analysis_of_flanged_rectangular_aperture_orifice

 

[Alfred Cann]

1. Remember, a rigid wall is a node only for velocity perpendicular to the wall, not along the wall.
2. Forget the Chinese paper; it's probably not relevant. We already know from the loudspeaker and round aperture literature that the response is a highpass filter, as I have described. And a square can't be that much different. I'm only looking for the modification of the filter curve when the aperture is rectangular or elliptical.

 

[trurle]

1. Remember, a rigid wall is a node only for velocity perpendicular to the wall, not along the wall.
2. Forget the Chinese paper; it's probably not relevant. We already know from the loudspeaker and round aperture literature that the response is a highpass filter, as I have described. And a square can't be that much different. I'm only looking for the modification of the filter curve when the aperture is rectangular or elliptical.

1. Velocity part along wall do not transfer sound through aperture.
2. The slot/opening still transmit power at zero frequency (in case of static pressure difference). Therefore, the "high pass" approximation is invalid for arbitrarily low frequency.

I think the confusion happens because you mix 2 situations:
1) situation with impringing (highly directional) sound wave (your case)
2) Situation with sound emitter at opening, producing omnidirectional wave (speaker simulations you tend to rely on).

These situations do not produce identical results.

 

[Alfred Cann]

trurle said:
"1. Velocity part along wall do not transfer sound through aperture. "
My point exactly -- that energy is lost and not propagated forward. It's like trying to pump up a tire when the air hose has a big leak -- the tire will not get pumped up.
"2. The slot/opening still transmit power at zero frequency (in case of static pressure difference). Therefore, the "high pass" approximation is invalid for arbitrarily low frequency. "
That "power" at zero (or very low) frequency spreads in all directions, hence does not cause any significant sound propagation forward, hence the very high loss.
"I think the confusion happens..."
I disagree with you completely. I have seen analyses that specifically modeled the effect of an impinging sound field as if there were a piston in the aperture.
Aside: why does your name have an e on the end? In Stanislav Lem's Cyberiad the constructor was named Trurl, if I remember right.

 

[trurle]

trurle said:
"1. Velocity part along wall do not transfer sound through aperture. "
My point exactly -- that energy is lost and not propagated forward. It's like trying to pump up a tire when the air hose has a big leak -- the tire will not get pumped up.
"2. The slot/opening still transmit power at zero frequency (in case of static pressure difference). Therefore, the "high pass" approximation is invalid for arbitrarily low frequency. "
That "power" at zero (or very low) frequency spreads in all directions, hence does not cause any significant sound propagation forward, hence the very high loss.
"I think the confusion happens..."
I disagree with you completely. I have seen analyses that specifically modeled the effect of an impinging sound field as if there were a piston in the aperture.
Aside: why does your name have an e on the end? In Stanislav Lem's Cyberiad the constructor was named Trurl, if I remember right.

1. Tangential velocity is not a loss. You need more complex calculation with viscosity and friction to come to loss - in simple approximations without viscosity we have only transmission and reflection. If you consider viscosity contribution, it will be most significant in the opening anyway, where only normal velocities are present. As such, your argument is "picking up a random part of problem".
2. Well, i agree some theoretical references you cite are considering hole in the wall being purely high-pass. Such results are inconsistient with either FEM or experimental data i cited before. In such case, i prefer to rely on experiment.

Regarding my net name, it should be Trurl indeed. Decades ago on another forum what name was already taken, so i used an altered form and it stuck.

 

[Alfred Cann]

Sorry, what is FEM?

 

[trurle]

Sorry, what is FEM?

Finite element method.

 

[Alfred Cann]

Just found what I was looking for:
Wright, Julian, Radiation Impedance Calculations for a Rectangular Piston, Journal of the Audio Engineering Society,
V. 38, No. 5, May 1990, pp. 350-354, Fig. 2 and others.
Confirms my conjecture that for rectangular apertures the slope of the radiation resistance is 3dB/oct. between corner frequencies.

 

[trurle]

Just found what I was looking for:
Wright, Julian, Radiation Impedance Calculations for a Rectangular Piston, Journal of the Audio Engineering Society,
V. 38, No. 5, May 1990, pp. 350-354, Fig. 2 and others.
Confirms my conjecture that for rectangular apertures the slope of the radiation resistance is 3dB/oct. between corner frequencies.

I have got the requested paper with experimental result. The 3 dB/octave slope for the sound propagation through slot is can be called correct only in very rough approximation. Actual response of slot more resemble bandstop filter.
See attached file!

 

[Alfred Cann]

I have looked at the 2009 paper you sent. It is almost totally irrelevant. The slits are mostly much narrower than the wavelengths of interest. Furthermore, in many cases they are really ducts of considerable length (what they call depth). This paper comes from the community of noise suppression technology; their interests are quite different from the kind of problem I stated. By the way, I find their terminology confusing, as if they sometimes mix up slit length and depth. In Table V, I must assume the 500mm is the slit length. In Fig. 8, almost all the widths are less than the depths, making them ducts.
You have not said what is your reaction to the 1990 paper I cited. Please do.

 

[trurle]

You have not said what is your reaction to the 1990 paper I cited. Please do.

I think the calculations in "Radiation Impedance Calculations for a Rectangular Piston" is a fine solution for non-existing problem. You cannot make real wall both thin and rigid, therefore experimental community is more concerned with thick walls - mostly to make transmission through wall itself negligible.

 

[Alfred Cann]

I don't think you have made your case at all with that really irrelevant citation, and I think that I've made mine pretty well with the paper I cited. And I'm surprised that no one has yet joined this thread (except for one early post by tech99).
In the case I was analyzing, the wall was about 4 in. thick, thin compared with the narrow dimension of the hole and VERY thin compared with the shortest wavelength of interest. It was definitely much thinner than the walls in that 2009 paper you cited. It was covered with plasterboard on one side and 3/4 in wood on the other side, quite rigid compared with the air in the hole, I think quite adequately thin and rigid to approximate the conditions of the 1990 paper I cited.

 

[trurle]

I don't think you have made your case at all with that really irrelevant citation, and I think that I've made mine pretty well with the paper I cited. And I'm surprised that no one has yet joined this thread (except for one early post by tech99).
In the case I was analyzing, the wall was about 4 in. thick, thin compared with the narrow dimension of the hole and VERY thin compared with the shortest wavelength of interest. It was definitely much thinner than the walls in that 2009 paper you cited. It was covered with plasterboard on one side and 3/4 in wood on the other side, quite rigid compared with the air in the hole, I think quite adequately thin and rigid to approximate the conditions of the 1990 paper I cited.

Well, it is possible to calculate 2nd momentum of area for your wall to estimate rigidity and compare to measured attenuation of solid glass sheet of same rigidity.
As order of magnitude estimate, your wall will have 25~30dB attenuation at 100 Hz falling 3 dB/octave to as low-pass filter. Slot attenuation will be 5-10dB at 100Hz to from range of experimental papers i cited. Therefore, rigid wall attenuation approximation woulld be useful if slot area larger than ~1% (-20dB) of wall area at 100 Hz. or 2.5% at 41 Hz.

 

[Alfred Cann]

glass?? There is no glass in the window!