Baffle Step Loss - Why 6dB?

[bmwman91]

This is a technical curiosity of mine. Almost everything I have read pertaining to baffle step compensation (and baffle diffraction simulators) indicates a 6dB loss across the frequency response for a baffled loudspeaker in free space. It is common knowledge that it happens because higher frequencies see a large reflecting surface relative to their wavelength and radiate into 2*pi space. Lower frequencies pass right around it since its size is relatively small compared to their wavelength, and radiate into 4*pi space. OK, so nothing new here.

Why is the loss not 3dB though? When a constant sound power source starts radiating into full space, which is twice the volume of half-space, the energy density is halved. This would indicate that sound intensity is also halved, and intensity is proportional to the square of pressure. So, wouldn't this be a 3dB drop?

The equation for calculating the sound pressure level in the room includes a directivity factor (1 for free-field, 2 for single-plane baffle). A 10*log(2) is 3dB. A 6dB drop would indicate a halving of sound pressure, rather than intensity (which is quartered). An acoustics consultant at the seminar I am at also says that the drop should be 3dB, based upon the same equation. So, where am I getting mixed up (since it seems unlikely that a simple mistake like this would be so widespread)? Thanks.

Aforementioned SPL equation on page 5.
http://www.eurovent-certification.co...-Bill_Cory.pdf

[bmwman91]

Thinking about it in terms of distance from the source also seems to indicate a drop of 3dB.

A doubling of distance halves sound pressure and quarters intensity (-6dB). Doubling the radius of a sphere gives 8*volume and 4*surface area. Since intensity is power/area, it makes perfect sense. Increasing the radius by sqrt(2) gives 2*surface area (and 4*volume), with a halving of intensity and sqrt(half)'ing sound pressure (-3dB).

Now, this differs from the 2*pi-to-4*pi transition since, in that case, both volume and surface area of the virtual sphere double (whereas with a sphere of increasing radius has volume proportional to the square of surface area). Still, the SPL equation, taking directivity factor into account, also indicates -3dB. I wish I had my Acoustics (L.L. Beranek) textbook with me...I bet the answer is in there.

[speaker dave]

Going from 4pi to 2pi there are always 2 things going on. First, the energy that was radiating behind the driver is now constrained to the front. Secondly the resistive air load on the driver is doubling. This is the difference between the airload of a "piston on the end of a pipe" and a piston on an infinite plane.

The second factor gives double the efficiency or +3dB in level. The first item gives a little less, typically 1.5 to 2 dB as the woofer would have some directivity, even at low frequncies.

The end result is 4.5 to 5 dB and a slight uphill slope for the 4 pi case.

David S.

p.s. I designed speakers professionally for over 30 years and never once added baffle step compensation. It was always incorporated into the woofer low pass network.

[bmwman91]

Thank you, David. I am not worrying about the topic for speaker design purposes at the moment, but rather out of intellectual curiosity. I get bothered when my grasp on a technical subject is fuzzy!

Perhaps some of my confusion lies in how you break the effect into two parts. Isn't the resistive load doubling BECAUSE the energy is being constrained to radiation through half the amount of space? I have seen a number of calculations/diagrams showing a frequency plot for a spherical enclosure baffle and a 6dB transition from low to high frequencies. That indicates a doubling (or halving) of sound pressure level and quartering of intensity...but it would seem that doubling the radiation space should double the intensity level, which is a 3dB change.

Sorry if I am being dense. Hopefully there is just some silly oversight that is confusing me here.

[Pano]

Originally Posted by speaker dave

It was always incorporated into the woofer low pass network.

Right were it should be.

[speaker dave]

Originally Posted by bmwman91

Perhaps some of my confusion lies in how you break the effect into two parts. Isn't the resistive load doubling BECAUSE the energy is being constrained to radiation through half the amount of space? I have seen a number of calculations/diagrams showing a frequency plot for a spherical enclosure baffle and a 6dB transition from low to high frequencies. That indicates a doubling (or halving) of sound pressure level and quartering of intensity...but it would seem that doubling the radiation space should double the intensity level, which is a 3dB change.

Sorry if I am being dense. Hopefully there is just some silly oversight that is confusing me here.

Now you've got me thiking about it and I am also confused.

I'm certain that the radiation resistance doubles in 2pi, as I've seen the curves many times. That guarantees double the power radiated, double the efficiency. but wouldn't that be +6, since it is double the power and in half the solid angle?

I don't know if the resistive load is doubling because of the space halving, just that it is.

Let me think about it some and see if I can come up with a better explanation.

David S.

[bmwman91]

Yeah, it is sort of a brain teaser. The acoustics consultant, who has been doing this since well before I was born, certainly sees no reason for a 6dB change...under ideal conditions. Certainly, real-world effects change things, although reflections and such will make the loss smaller rather than larger. Does John K (Zaph) post in here? This seems like something that he could probably elucidate quickly.

[sreten]

Hi,

Its the old chestnut, pressure doubles (as it must), but SPL quadruples, SPL is not pressure,
its pressure squared. Would better if SPL was called Sound Power Level, but it unfortunately isn't.
20log(P1/P2) = [10log(P1/P2)]squared is what you use for SPL. 10log(P1/P2) for pressure.

rgds, sreten.

[speaker dave]

We know that there is a 6dB, (or "approaching" 6dB) difference for all the many curves, such as the orriginal Olson diffraction curves, that we have seen.

Sreten, PWL is the standard term for power level. I think that "pressure doubles (as it must)" isn't the total answer. Although I use that for two woofers in parallel, the woofer in 2pi vs 4pi is a little different case.

Maybe the best way to state it is, when going from 2pi to 4pi the acoustic load cuts in half for a 3dB loss across the board and another 3dB as frontal sound wraps to the back. At upper frequencies where the sound doesn't "wrap" as well you lose an intermediate amount for the observed 4 to 5dB.

Any takers?

David S.

[john k...]

Again, it boils down to simple physics. When the cone moved it displaces a given volume of air giving rise to a wave propagating into space. When the wave is constrained to expand into the front hemisphere the mass displaced by the wave will pass through an area equal to the surface of a hemisphere or some arbitrary radius, R. The mass displaced will be density times surface area times velocity. When the wave expands into free space the area the wave passes through is a sphere which will have twice the surface area of a hemisphere of the same R. Since the mass displaced by the driver is the same and the surface area doubles the velocity at that surface will be 1/2 that of the hemispherical case. The specific acoustic resistance of air in the free field is a constant thus, P = Z x U and since U is 1/2 in full space pressure is also 1/2 that of the hemispherical case. SPL goes like 10 x Log(P^2) or 20 Log(P). Since Pfullspace / P halfspace = 1/2 the full space SPL is -6dB compared to that of 1/2 space (20xLog(1/2) = -6)