[johnk...]

I ran some simulations my self to see what the response directly behind various open configurations looked like.

Starting at the top are a series to 3 simulations to provide confidence in the last two results. The top is a simulation of an 18" U-frame using SoundEasy's Enclosure design. This is a limped parameter model based on the work of Backman, IIRC. The next plot is an undamped U-frame 18" long by 9.5" wide x 9.5" high, opened at the back simulated using SoundEasy's 3-D finite element code. The plot is the response at the center of the read exit opening. The third figure is the result for the same 18" u frame using the King, Kreskovsky (mostly King) MathCAD worksheet. Note that all three give very similar results. The 4th plot down is for a  SoundEasy 3d finite element simulation of a 48" tall, 24" wide, 12" deep U with only the back opened. This would be close to Martins baffle with folded wings and the top closed. The driver is assumed to be centered on the 48" x 24" front baffle. The response is ",measured" in the center of the rear opening, directly behind the driver. The last figure is the same as the 4th but with the top opened. Frequency range is 50 to 1k Hz.

I think the lower two results are consistent with where I measures the response directly behind one of my NaO Mini baffle with (red) and without (green) "wings".

[MJK]

If I look a little closer at plots, I see a couple of interesting features that should help us understand better.  In the top three plots, the smaller U frame, looking at the MathCad result I believe the impact of the acoustic impedance boundary condition at the open end are easily seen, the successive peaks become less and less pronounced.  I can only conclude that the SoundEasy simulation does not include this type of frequency dependent boundary condition so all the peaks exhibit a much higher Q.  Also, the first peak in the MathCad model is a little lower in frequency due to the contribution of acoustic mass at lower frequencies from this same boundary condition.

Keeping these differences in mind, when we look at the next two plots from SoundEasy I have to wonder if the peaks and dips are overstated.  If there is significant damping applied at the much larger opening for this geometry I would not expect these peaks and dips to exist and the response would become more uniform.  The second plot should show higher damping compared to the first due to the larger opening.  This is typical of FE solutions, the acoustic boundary conditions tend to be limited to fixed, in other words frequency independent, constraints on pressure or velocity or in some cases the ratio of the two.  My conclusion is that the SoundEasy results probably predict the frequencies correctly but are calculated with too little acoustic damping and therefore overstate the response.

Returning to the discussion of deeper bass with the wings folded back or extended.  In simple terms the bass response of a dipole is directly related to the baffle size and shape.  Picturing the size of the baffle as setting the distance between two simple sources, one positive and one negative, then in simple terms if we calculate the distance traveled from the center of the front source to the center of the rear source this might help decide which geometry produces deeper bass. Looking at the two cases :

Wings Folded Back

d = 12 + 24 + 24 = 60 inches (assuming the rear wave is planer in the rear cavity)

Wings Extended

d = 12 + 24 + (24^2 + 12^2)^1/2 + 12 =  74.8 in (same assumption as before)

I would conclude that the wings extended would produce deeper bass.  This somewhat makes sense when you think about the front wave traveling to the edge of the front baffle then wrapping around and picture in your mind the baffle step response one would expect if this was a monopole system.  The wide baffle would have a baffle step loss much lower in frequency.  But whatever logic is offered, sitting and listening to the Lowther OB system I have the impression of deeper bass with the wings extended.

I have also modeled U Frames in my latest MathCad worksheets.  These worksheets include the calculation of group delay.  I don't see negative group delay in my results, I don't understand the concept of negative group delay.  Physically how does one end up with a phase plot that has a slope such that a negative group delay results?  At this point negative group delay does not make physical sense to me.

[AJinFLA]

JohnK,

can Soundeasy simulate a U-baffle with a hole in the bottom leading into an expanding, folded TL like JohninCR has built? http://www.audiocircle.com/index.php?topic=34399.msg313229#msg313229
It would be fascinating to see the predicted response vs JohninCR's (eventual) measurements.
Strong output into the 50's using a driver with a Mms of only 9.7g seems to suggest greater potential than a simple U, stuffed or not.

cheers,

AJ

p.s. MJK, nice to have you back. Hope you had a Happy New Years.

[JohninCR]

JohnK & MJK,

I need to help get you guys on the same page. 

John, Martin's baffle has a fixed U.  In addition, it has wings that he can fold out to end up with a big wide front plus a narrower U in the back, not just a wide flat baffle.  Martin, John is saying that some damping in the U should result in even deeper bass extension.

Regarding this negative group delay thing.  A negative GD, to me, means sound is getting somewhere before it is supposed to.  The only explanation that seems to hold water is that the air in the cavity is behaving as a "lumped mass", so you don't get the full propagation delay from the back of the driver to the rear edges of the U shape.  Instead the rear wave source is moved closer to the rear edge of an undamped U.  Does this lumped mass behavior have a direct correlation with resonance, or is it coincidence that damping cures both?  Could this have relevance to the argument about damping's affect on the speed of sound in a TL over on the Decware forum?

[johnk...]

Damn it. I had a response type in here but it was lost. I'm not going to retype it but I will summarize. For the U SE and the MathCAD worksheet shows qualitative agreement. The amplitudes may be different, but the effects are similar. Therefore I feel confident that the results for the 48 x 24 x 12 U with and without open top are also qualitatively correct. Amplitudes may vary.

I'm not particularly concerned with Martins exact configuration, just whether or not resonances in a widely open U with/without top can be excited.

Negative GD. It's just a ways of looking at this.

Acoustic wave propagates down a duct at the speed of sound. That is the physics. So the rear radiation should be delayed by L/C compared to the front. But look at this figure

The phase of the rear response of this undamped U, measured at the rear exit is the same (except for noise) as the phase of the front response measured at the dust cover up to about 75 Hz. Given that the acoustic waves from the rear of the driver must travel a distance of L to the rear exit you would expect that they would be delayed by L/C relative to the front. But they are not. This is because of the negative GD of the resonances. If you look at the first figure above the negative GD is constant at low frequency. What than means is that the phase is increasing with frequency in a linear manor. The propagation delay yield a phase which decrease linear with frequency. So we have Phi = aF at low frequency from the resonance canceled by Phi = -aF from the propagation delay.

Martin, you can run your worksheet for an undamped U frame and look at the polar response well below the resonance and you will see a dipole response. How ever you want to reason it out in you mind, that result means any internal propagation delay associated with the length of the U doesn't appear in the rear SPL response. You won't see negative GD because I am using that in the sense that the rear response is just the front response, delayed, with the duct transfer function (the resoannce) superimposed on it. It is supposed to be a simplistic way to look at this. Maybe it's only simplistic to me. Obviously you wont' see this in the net response a the rear of the U. I'm just trying to break it down in simple terms; Driver response x propogation delay x duct transfer function = output. The driver's volume velocity is the same front and rear so the difference in the front and rear response, in simplified terms, is due to the delay and duct TF.

AJ, Not sure. I'm have to try and set it up. But I'm not going to do that. 

[JohninCR]

John,

I understand a resonance causing phase shift, at least I think I do, however, playing content with frequencies only below that first resonant frequency there is no resonance, so what causes this constant negative GD?  I want to understand the cause, so I can figure out alternatives to combat it.

[MJK]

you can run your worksheet for an undamped U frame and look at the polar response well below the resonance and you will see a dipole response. How ever you want to reason it out in you mind, that result means any internal propagation delay associated with the length of the U doesn't appear in the rear SPL response. You won't see negative GD because I am using that in the sense that the rear response is just the front response, delayed, with the duct transfer function (the resoannce) superimposed on it.

OK, I think I understand what you are calculating.  At low frequencies (a relative term) the air in the pipe acts like a lumped mass and moves together as a rigid body with the back of the driver's cone.  So the air motion in front of the driver's cone and at the open end of the pipe have the same magnitude but are exactly 180 degrees out a phase, a classic textbook dipole.  As the frequency increases, the air in the pipe starts to "flex" due to the distributed mass and stiffness of the air column and a magnitude and phase difference starts to appear as rigid body motion transitions to wave motion eventually leading up to the fundamental resonance.  But I think at some point the same phenomenon will occur for a bass reflex, a longer TL, and a back loaded horn but at much low frequencies.  The only unique thing about the U or H frame is the higher frequencies that the rigid body motion of the air slug continues to occur due to the relatively short lengths involved.  My MathCad models include both behaviors and the transition is accounted for in the simulation.  I think this also explains the magnitude and phase curves in your plotted data.

[JohninCR]

Martin,
Thank you very much.  That even helps me understand how that fundamental resonance is
created, since as the flex starts, the change in phase causes a buildup kind of like a sonic
boom as a jet passes through the speed of sound.  Is that at least semi-accurate, except
this is more like the speed of sound slowing down through the speed of the jet?

[MJK]

I don't know anything about jets and passing through the speed of sound.

I like to think about a short slinky that you hang from your hand and then move your hand up and down.  If you start out slow the slinky moves with your hand and does not stretch, this is rigid body motion of the slinky.  As you hand starts to move faster the slinky starts to stretch a little, now the dangling end is not 100% in phase with your hand.  Keep moving your hand faster and the stretch gets more pronounced, the phase difference increases.  At some rate of moving you hand up and down, the free end of the slinky will move a lot and it is the maximum response.  Move you hand up and down a little slower or a little faster and the motion of the free end of the slinky decreases.  The maximum motion occurs at the fundamental resonance and the rate at which you hand moves up and down is the resonant frequency.  You can learn a lot about transmission lines playing with your kid's slinky.

[JohninCR]

Martin,
Perfect analogy thanks.   Short pipe, so short slinky and it happens at higher frequencies.  The slinky thing may even cover the higher harmonics.  Now I have to figure out how to stiffen the same length slinky.

[MJK]

The length of the pipe will determine the resonant frequency, assuming S0 = SL then the equation

L = c / (4 x f)

will size the length.  If the pipe is expanding or tapered this rule of thumb equation does not work and you really need a computer solution to set the length.  The stiffness of the pipe is controlled by the cross-sectional areas S0 and SL.  Bigger areas softer spring, smaller areas stiffer spring.  No rocket science required.

[johnk...]

OK, I think I understand what you are calculating.  At low frequencies (a relative term) the air in the pipe acts like a lumped mass and moves together as a rigid body with the back of the driver's cone.  So the air motion in front of the driver's cone and at the open end of the pipe have the same magnitude but are exactly 180 degrees out a phase, a classic textbook dipole.  As the frequency increases, the air in the pipe starts to "flex" due to the distributed mass and stiffness of the air column and a magnitude and phase difference starts to appear as rigid body motion transitions to wave motion eventually leading up to the fundamental resonance.  But I think at some point the same phenomenon will occur for a bass reflex, a longer TL, and a back loaded horn but at much low frequencies.  The only unique thing about the U or H frame is the higher frequencies that the rigid body motion of the air slug continues to occur due to the relatively short lengths involved.  My MathCad models include both behaviors and the transition is accounted for in the simulation.  I think this also explains the magnitude and phase curves in your plotted data.

Well that is the conventional way to look at it. At low frequency the air acts like a piston and you can ignore the compressibility effects, i.e rigid body motion. At higher frequencies you need to start considering them. You know, lumped parameter, distributed parameter, 1-D wave equ, graduating finally to the full blown 3-d wave eq. The assumptions are what approximations are applied and where are they valid. But they are approximations, not the true physics. After all, if the air really moves like a piston at low frequency (wave length much greater the U length), then why would damping the resonance change the phase a low frequency? We start with an undamped case where the phase of the front at the driver cone surface is basically identical (+180 degrees) to the phase at the rear exit plane, regardless of the length of the U, even though it takes an acoustic disturbance initiated at the rear of the driver L/c seconds to reach the exit plane. Now we damp the resonance and all of a sudden the phase at the rear exit plane shows that L/c delay. Is the air no longer moving like a piston at low frequency? W certainly can't have it both ways. Either it moves like a piston at low frequency all the time or it doesn't.

The convectional lumped parameter analysis of a U frame (Backman, 1999 AES presentation), for example, shows that the lumped parameter model says that at LOW frequency the volume velocity at the exit plane is Uexit = Udriver x 1/(1 +jwRCb) where Cb is the capacitance of the air in the U and R is the resistive damping. That is, Uexit is just a low pass filtered version of Udrive and the delay that comes about with the resistive damping is just the delay of the LP filter. Unfortunately, while this looks good on paper, it is only part of the story. First it contradicts the assumption of pistonic motion of the air in the duct which would require the Uexit always equal Udriver. Second, as I discussed on my U-frame web page, adding the resistive damping doesn't just LP filter Udriver. It also damps the resonance compounding the change in the delay.

That the lack of the delay in the rear response when the U is undamped is a result of the resonance can also be demonstrated in a fairly straight forward way. For an undamped U and measure the response at the front. Next, measure that at the rear. It's like the picture in my previous post. But now, instead of acoustically damping the response, electronically equalize the rear response so that its amplitude matches the front response measured without eq applies. You will now find that the equalized rear response has amplitude as the unequalized front response and the phase of the rear will be the phase of the front (inverted) plus a delay of L/C.

Another interesting way to show that the delay, L/c, is still present in the response, but countered by the GD associated with the duct resonance  is by minimum phase analysis. Any causal response can be decomposed into the minimum phase response associated with the amplitude, and some form of all pass response. Measure the rear response and compute the minimum phase. You will find that it will be necessary to add a delay of L/C to the minimum phase to match the measured phase at the exit plane.

This is the problem with these simplified analysis. They may give reasonable engineedring results, but they often hide the true physical nature of the problem.

[johnk...]

John,

I understand a resonance causing phase shift, at least I think I do, however, playing content with frequencies only below that first resonant frequency there is no resonance, so what causes this constant negative GD?  I want to understand the cause, so I can figure out alternatives to combat it.

Take another look at this picture.

The delay is just a result of the phase shift. Phase shifts and time dealys are really the same thing. Just like any filter has a phase shift, you don't have to excite the resonances at the resonant frequency for the phase shift to occur at lower frequency. In the region where the GD is constant it just implies that the phase is varying linear with frequency.

[MJK]

I am sorry John K, I have no idea what you are talking about.  I need things broken down into simple analogies for me to understand.  That is my limitation.  I thought what I wrote was fairly clear and an accurate way of modeling what you had shown in your plots.

[johnk...]

I am sorry John K, I have no idea what you are talking about. 

Don't feel bad Martin. You're not the first to say that.

I'm just trying to make the point that the idea air in a duct moves pistonicly at low frequency (or rather when the wave length is large compared to the  duct length) is an approximation that can lead to reasonable engineering results, but is never really the case. Let me try a couple of figures.



Starting at the lower left I show the front and rear SPL and phase of an undamped U frame. At low frequency the phase is basically identical. (I have inverted the rear phase for comparison). Thus we would expect to see dipole behavior at low frequency. We agree on that. However, both of these response curves were obtained from the impulse responses shown above at the upper left. The upper impulse is that obtained at the rear opening of the U frame; the lower at the front, driver surface.  We see that the impulse for the rear is delayed 1.177 msec = (L/C) relative to the front. But in both cases the SPL data was obtained using an FFt window for the impulse that started at t = 0.0.  Thus the phase of the rear response includes excess phase associated with the 1.177 msec propagation time. Yet the phase at low frequency is the same. So the question is why? How is that excess phase at low frequency removed from the rear response? To the right I show the front response and the front response with a 1.177 msec delay added. And here I show the front with delay added overlaid with the rear response.


What I hope this shows is that below the resonance peak it is apparent that the rear phase shift is less (negative) than what the front is if it were delayed by the 1.177msec propagation time for the duct. At the same time, above the resonance peak it is greater than would be attributed to the propagation delay. At the resonance peak it is the same. Now if you look at the phase response of the Q boost filter in my previous post you will see that if you add the phase of the Q boost filter to the phase of the delayed front response  it starts to look very much like that of the rear response. The Q boost resonant response turns the phase up below the resonance peak which is equivalent to removing delay. Conversely, if you remove the resonance from the rear response at low frequency the phase will turn back down and the excess phase associated with the propagation delay will become apparent again.

So it is not because the air moves pistonicly at low frequency that the front and rear phase of an undamped U  look the same. It is because of the establishment of the resonance which adds and removed enegry from the system  resulting in pistonic like motion at low frequency. Remove the ressonance and you removed the apperance of pistonic motion.

[MJK]

I modeled the U frame you described in my MathCad TL worksheet and got a similar pair of time responses with the relative delay you have shown.  I think my model and your measurements are showing essentially equivalent results.  How one visualizes it and explains it simply so others understand appears to be causing the confusion, I think we are starting to beat nits to death.  I need to move on to some other work.

[JohninCR]

JohnK,

First, I want to thank both of you and Martin for your perseverance on this topic, even if I'm the only one taking so long to get it.

What I like about Martin's explanation is that it also explains the cause of the resonance to me.  Before I just drop the matter, I have a question.  At what frequency is the impulse response measured? 

It seems to me that if you oscillate a piston in the end of a relatively short pipe very slowly (say once per minute), that the air in the pipe will obviously behave as a lumped mass and move out of the other end at the same time you push it at the other.  Obviously, at high frequency oscillation this won't happen because the air is too compliant (is that the accurate term).  At some point a transition between the two behaviors must occur.  If I understand correctly this transition occurs around the fundamental 1/4 wave resonance of the pipe.  I don't think this contradicts anything either of you guys have said, as long as below resonance it's not really an acoustic wave inside the pipe, just an oscillating lumped mass.

[MJK]

An impulse is an electrical pulse in the time domain.  In theory it is infinitely short with a magnitude of infinity.  In practice it is short enough and produces an audible click.  It excites all frequencies with approximately the same magnitude.

The transition from the lumped mass behavior to the flexible column of air starts at very low frequencies and becomes more pronounced as frequency increases culminating with a very high amplitude response at resonance.  In reality there is always some very small flexibility, and some small amount of phase difference, even at the lowest frequencies.  There is no such thing as a completely rigid lumped mass response for the short column of air.  But at the lowest frequencies, it is not a bad analogy that is easily understood and a reasonably accurate assumption.

[johnk...]

JohnK,

First, I want to thank both of you and Martin for your perseverance on this topic, even if I'm the only one taking so long to get it.

What I like about Martin's explanation is that it also explains the cause of the resonance to me.  Before I just drop the matter, I have a question.  At what frequency is the impulse response measured? 

It seems to me that if you oscillate a piston in the end of a relatively short pipe very slowly (say once per minute), that the air in the pipe will obviously behave as a lumped mass and move out of the other end at the same time you push it at the other.  Obviously, at high frequency oscillation this won't happen because the air is too compliant (is that the accurate term).  At some point a transition between the two behaviors must occur.  If I understand correctly this transition occurs around the fundamental 1/4 wave resonance of the pipe.  I don't think this contradicts anything either of you guys have said, as long as below resonance it's not really an acoustic wave inside the pipe, just an oscillating lumped mass.

Well I don't want to beat this to death any more that I have so I will make one last post and more on as well. I thin Martin answered you impulse question. So all I'll add is that any system of the type we are looking at here has a 1 to 1 correspondence between the impulse and frequency response thorough a mathematical transformation which is the Fourier Transformation. If you don't know what that is that's ok. But many acoustic measurements are made today but some means of generatining an impulse and then transforming the impulse response to a frequency response.

The lumped parameter analysis is fine for the basic understanding of the resonance. A TL is not unlike a vented box in that there is a capacitive element and an inductive element which lead to a LC resonance.

Your observation about low frequency is also fine. As you say, for a relatively short pipe if you push the air at one end very slowly you can expect the air to move as a solid body. That seems reasonable, IF there is no friction or resistance. But when there is friction things are different. Consider the slinky again. At low frequency it moves as one when there is no resistance to motion. But add resistance distributed in sections along the length. Before each section can move the resistance to motion must be cover come. Each section must start to compress a little, one after the other, before the next section can move. So it no longer moves as a solid body and the far end of the slinky doesn't move in phase with the end you are pushing on. This is the effect of damping restoring the delay at low frequency in a U - frame.

[JohninCR]

JohnK,

It seems like we're all in sync now, as long the impulse response doesn't truly reflect the U's rear radiation below the fundamental resonance.

I want to apologize for understanding only small portions of your scientific explanations.  I also want to thank you for sharing your discovery of U-baffle behavior, because I see it as the only way open baffle speakers can break into the consumer market.  It's the only way to minimize size and maintain reasonable bass response without skyrocketing driver costs.  While us OB lovers can appreciate the sonic difference of OB sound, that's probably only marketable to a minor segment of the market.  What is VERY marketable to the consumer market is the same thing that we fight with in our designs, off axis bass cancellation.  Keeping bass in-room is a tremendous and real advantage for anyone with children or close neighbors.  Just look at what Bose does with only smoke and mirrors.  Throw in better clarity for dialogue playback and it's an easy sale.