Multiple Small Subs - Geddes Approach - Page 54

john k... · 2nd February 2009, 02:33 AM

Todd,

I am not advocating listening in an anechoic chamber though I have head sound reproduction in a few. (At one time I was involved in active noise suppression for submarines). Low frequency reproduction and sound reproduction above the Schroeder frequency are very different. Above the Schroeder frequency we are able to discern the difference between the direct and reflected sound in a room and the reverberant field adds “space” to the recording (artificial as it may be, but seemingly necessary for the illusion). Below the Schroeder frequency all we hear is the room. I agree that the energy would have to be increased if the room were free of modal excitation at low frequency. After all, in an enclosed space the SPL is related to 1/V the rate at which the sound decays is dependent on dissipation and transmission through the wall, a simple case of conservation of energy.

As for minimum phase, that again seems to be very dependent on where you are in the room, which modes contribute what to the SPL at the listening point, and how close the listening position is to a source. The only way to be sure the response is MP is to eliminate the modal contributions.

Maybe I'm looking at this in a different way than might be considered conventional but if I just look at it from the same point of view, well why would I bother? But I listen to what everyone says and then look into it and draw my own conclusions and attempt to move forward. I don't mind if I hit a dead end. At least I know I explored all the options.

But, as I said earlier, if you allow modal excitement to enter the picture then the quest for even low frequency response is impossible because the SPL at any listening point depends of the spatial variation of the modes as the listener moves around the room. You simple can not have even low frequency response over a wide area in an acoustically small room if modes are excited.

By the way, the impulse response was a simulation. FWIW, here is the same type of simulation for a dipole with dipole axis aligned with the room axis so that axial and tangential modes normal to the dipoles axis are not excited. Again, green = what the MP response would look like, blue the in room response eq to the same amplitude as the MP response. The comment All Modes means that all modes were included in the simulations, but not necessarily excited due to positioning.

and here is the result for the same dipole position but rotated 45 degrees:

markus76 · 2nd February 2009, 03:09 AM

Todd, I'm not kidding and I've read Floyd's book. I would love to see your data.

Best, Markus

P.S. Ethan Winer did some real world tests: http://www.realtraps.com/eq-traps.htm

cap'n todd · 2nd February 2009, 04:19 AM

John, I want to response to your points. Not much time now, but I would be interested to see what the room reponse in your simulation looked like before eq, to put things in perspective.

cap'n todd · 2nd February 2009, 04:29 AM

Quote:

Originally posted by markus76
Todd, I'm not kidding and I've read Floyd's book. I would love to see your data.

Best, Markus

Hi Markus,

For starters, page 248, fig 13.24 (not a waterfall, but the data is there). It is perhaps a bit harder to see in a waterfall, but I can find it tomorrow. So, do you think that the room has so much excess phase resposne that bringing a sharp peak (easily as narrow as 2 Hz) down in level using a properly matched filter would not reduce the ringin significantly at that frequency? Or did I misunderstand you? To say that there is still some ringing afterwards does not mean you didn't remove most of it.

And BTW, just because there is excess phase, does not mean it is audible. ..and even if so, that says nothing about wehter it is detrimental or not. Whats a little excess phase between friends?

john k... · 2nd February 2009, 12:49 PM

Quote:

Originally posted by cap'n todd
John, I want to response to your points. Not much time now, but I would be interested to see what the room response in your simulation looked like before eq, to put things in perspective.

I did that work last May and I all that remains of it is what I used for my web page. The discussion on room modes starts here where I look at the behavior of monopoles, dipoles and cardioids in a room. In that part I show the Green's function approach that we have been discussing and verify that for a rectangular room the result is in good agreement with a FEM approach. Then in the in the next part. At the bottom of the this section I discuss the minimum phase/non-minimum phase aspects of the response. Additionally, I show that as the listening position moves closer to the source the response does become minimum phase. In the figures at the bottom of the page it may not be clear but the phase from the room simulation is in green and the thin, dark blue line is the minimum phase based on the simulated amplitude response obtained form the Hilbert-Bode transformation.

Lastly a rather novel woofer setup that I have been experimenting with recently is described. It is only intended for single position listening but it potentially make eq very simple, if required at all.

gedlee · 2nd February 2009, 01:51 PM

I've been disconnected for a few days.

I wanted to get back to something that John posted. Yes, I agree that Morse (more likely Ingard given that statment is new from Morses original text) does make the statement about the seperation of the free field Green's function from X(r), which is claimed as being the solution to the homogeneous wave equation.

Here is why I don't see how that can be true. The free field Greens function g(r) has a finite velocity magnitude everywhere. Thus if placed inside of any boundary it will have a finite velocity magnitude normal to that boundary (not necessarily everywhere, but necessarily somewhere). For X(r) plus g(r) to have a zero normal velocity on that boundary (the boundary condition for a rigid walled room), X(r) would have to have the inverse of g(r) on the boundary and hence X(r) would have to contain sources on the boundary. If it has sources then it is not the solution of the "homogeneous wave equation".

This is a minor point and has no bearing on the discussion, and hence not really worthy of elaboration, but I did want to explain why I do not readily accept that X(r) is the solution to the homogenrous wave eqauation. There certainly does exist some X(r) that works as described above, and it can be found from Green's theorem.

Since the rooms sound field above the Schroeder frequency is never flat or smooth (its statistics are well known), I've never argued that the modal region needs to be any smoother than that. Only that it needs to be smoother than it will be with a single source. So the fact that it can't be perfectly smooth because of modal effects is not important. It just needs to be "smooth enough".

markus76 · 2nd February 2009, 01:52 PM

Quote:

Originally posted by cap'n todd For starters, page 248, fig 13.24 (not a waterfall, but the data is there). It is perhaps a bit harder to see in a waterfall, but I can find it tomorrow.

Why is it harder to see in a waterfall? A high resolution CSD should reveal ringing clearly. Modal ringing can be seen as "ridges".

Quote:

Originally posted by cap'n todd So, do you think that the room has so much excess phase resposne that bringing a sharp peak (easily as narrow as 2 Hz) down in level using a properly matched filter would not reduce the ringin significantly at that frequency? Or did I misunderstand you? To say that there is still some ringing afterwards does not mean you didn't remove most of it.

When you attenuate a modal peak then you attenuate its "ridge" accordingly. That's a good thing but doesn't necessarily mean that the decay rate of the ringing was reduced. If the system is minimum phase and Q and frequency of the EQ matches the peak then the "ridge" should be gone completely. I haven't had the chance to see a waterfall of a real listening room that shows this.

Quote:

Originally posted by cap'n todd And BTW, just because there is excess phase, does not mean it is audible. ..and even if so, that says nothing about wehter it is detrimental or not. Whats a little excess phase between friends?

I guess that's the core problem, to relate physical measures to perceptual metrics. I'm not a physicist and I do not care if a room behaves minimum phase at low frequencies or not. The fundamental questions (to me) are more like: How much deviation in low frequency response would you allow? How long is a mode allowed to ring before it becomes audible? What decay time is best?

john k... · 2nd February 2009, 02:58 PM

Since a few, many or all of you may doubt the results I posted for impulse response based on a simulated in room woofer response I though I would make a measurement of a woofer in my room and use that response as the basis. The woofer is a 12" woofer of a Martin Logan Monolith speaker, slightly modified to have a response defined by a Q = 0.5, 2nd order 20 Hz HP cascaded with a Q = 0.5, 2nd order 120 Hz LP. The first figure shows the measured amplitude response at my listening position overlayed with the target. It's not too bad. The second figure shows the raw phase response with woofer to mic delay removed, roughly 4 M worth, in green. The blue phase response is the phase after equalizing the response to the target amplitude, shown above, using minimum phase equalization. The purple phase response is the minimum phase of the target. So we can see that in this case the phase, even after removal of excess delay, is not minimum phase. The last figure is the simulated impulse response for, Red, the unequalized response, Blue, the response equalized to the target amplitude, and purple, the the impulse of the minimum phase target. The impulse was 1 msec wide.

If you don't believe the impulse responses I can post a comparison of the measure impulse for the driver and the simulation of that impulse. (Hint: they are identical).

Anyway, I think this illustrates the fact that simple smoothing the amplitude for a woofer isn't enough. While it improves the initial attack, the long time response is even worst than the unequalized response, both of which ring on forever compared to the target.

cap'n todd · 2nd February 2009, 03:21 PM

Quote:

Originally posted by markus76

Why is it harder to see in a waterfall? A high resolution CSD should reveal ringing clearly. Modal ringing can be seen as "ridges".

High resolution in time or frequency? No, I just meant visually, because of hte projection angle. Like if you look at someones nose from 45 degrees its harder to see how long it is compared to from the side.

Quote:

Originally posted by markus76

When you attenuate a modal peak then you attenuate its "ridge" accordingly. That's a good thing but doesn't necessarily mean that the decay rate of the ringing was reduced. If the system is minimum phase and Q and frequency of the EQ matches the peak then the "ridge" should be gone completely. I haven't had the chance to see a waterfall of a real listening room that shows this.

If you use anything close to a matched filter, you will reduce the slope of the resonance mag response, and thus the Q of the resonance. THis reduces the decay time, not the level (at least in the minimum phase sense, which I contend is more true than not).

john k... · 2nd February 2009, 03:30 PM

Quote:

Originally posted by gedlee
I've been disconnected for a few days.

Here is why I don't see how that can be true. The free field Greens function g(r) has a finite velocity magnitude everywhere. Thus if placed inside of any boundary it will have a finite velocity magnitude normal to that boundary (not necessarily everywhere, but necessarily somewhere). For X(r) plus g(r) to have a zero normal velocity on that boundary (the boundary condition for a rigid walled room), X(r) would have to have the inverse of g(r) on the boundary and hence X(r) would have to contain sources on the boundary. If it has sources then it is not the solution of the "homogeneous wave equation".

I guess it depends on how we define homogenious. I would suggest that the equation for X(r) is homogeneous. The solution required is that to the homogeneous boundary value problem with boundary conditions such that the velocity of of the bounding surfaces is that required by the wall admittance, zero for rigid walls.

2nd February 2009, 02:33 AM	#531
john k... diyAudio Member Join Date: Aug 2004 Location: US	Todd, I am not advocating listening in an anechoic chamber though I have head sound reproduction in a few. (At one time I was involved in active noise suppression for submarines). Low frequency reproduction and sound reproduction above the Schroeder frequency are very different. Above the Schroeder frequency we are able to discern the difference between the direct and reflected sound in a room and the reverberant field adds “space” to the recording (artificial as it may be, but seemingly necessary for the illusion). Below the Schroeder frequency all we hear is the room. I agree that the energy would have to be increased if the room were free of modal excitation at low frequency. After all, in an enclosed space the SPL is related to 1/V the rate at which the sound decays is dependent on dissipation and transmission through the wall, a simple case of conservation of energy. As for minimum phase, that again seems to be very dependent on where you are in the room, which modes contribute what to the SPL at the listening point, and how close the listening position is to a source. The only way to be sure the response is MP is to eliminate the modal contributions. Maybe I'm looking at this in a different way than might be considered conventional but if I just look at it from the same point of view, well why would I bother? But I listen to what everyone says and then look into it and draw my own conclusions and attempt to move forward. I don't mind if I hit a dead end. At least I know I explored all the options. But, as I said earlier, if you allow modal excitement to enter the picture then the quest for even low frequency response is impossible because the SPL at any listening point depends of the spatial variation of the modes as the listener moves around the room. You simple can not have even low frequency response over a wide area in an acoustically small room if modes are excited. By the way, the impulse response was a simulation. FWIW, here is the same type of simulation for a dipole with dipole axis aligned with the room axis so that axial and tangential modes normal to the dipoles axis are not excited. Again, green = what the MP response would look like, blue the in room response eq to the same amplitude as the MP response. The comment All Modes means that all modes were included in the simulations, but not necessarily excited due to positioning. and here is the result for the same dipole position but rotated 45 degrees: __________________ John k.... Music and Design NaO Dipole Loudspeakers. "We have no right to assume that any physical laws exist, or if they have existed up to now, that will continue to exist in a similar manner in the future." Max Planck

2nd February 2009, 01:51 PM	#536
gedlee diyAudio Member Join Date: Dec 2004 Location: Novi, Michigan	I've been disconnected for a few days. I wanted to get back to something that John posted. Yes, I agree that Morse (more likely Ingard given that statment is new from Morses original text) does make the statement about the seperation of the free field Green's function from X(r), which is claimed as being the solution to the homogeneous wave equation. Here is why I don't see how that can be true. The free field Greens function g(r) has a finite velocity magnitude everywhere. Thus if placed inside of any boundary it will have a finite velocity magnitude normal to that boundary (not necessarily everywhere, but necessarily somewhere). For X(r) plus g(r) to have a zero normal velocity on that boundary (the boundary condition for a rigid walled room), X(r) would have to have the inverse of g(r) on the boundary and hence X(r) would have to contain sources on the boundary. If it has sources then it is not the solution of the "homogeneous wave equation". This is a minor point and has no bearing on the discussion, and hence not really worthy of elaboration, but I did want to explain why I do not readily accept that X(r) is the solution to the homogenrous wave eqauation. There certainly does exist some X(r) that works as described above, and it can be found from Green's theorem. Since the rooms sound field above the Schroeder frequency is never flat or smooth (its statistics are well known), I've never argued that the modal region needs to be any smoother than that. Only that it needs to be smoother than it will be with a single source. So the fact that it can't be perfectly smooth because of modal effects is not important. It just needs to be "smooth enough". __________________ Earl Geddes Gedlee Website Read about the highly acclaimed Geddes loudspeakers

2nd February 2009, 02:58 PM	#538
john k... diyAudio Member Join Date: Aug 2004 Location: US	Since a few, many or all of you may doubt the results I posted for impulse response based on a simulated in room woofer response I though I would make a measurement of a woofer in my room and use that response as the basis. The woofer is a 12" woofer of a Martin Logan Monolith speaker, slightly modified to have a response defined by a Q = 0.5, 2nd order 20 Hz HP cascaded with a Q = 0.5, 2nd order 120 Hz LP. The first figure shows the measured amplitude response at my listening position overlayed with the target. It's not too bad. The second figure shows the raw phase response with woofer to mic delay removed, roughly 4 M worth, in green. The blue phase response is the phase after equalizing the response to the target amplitude, shown above, using minimum phase equalization. The purple phase response is the minimum phase of the target. So we can see that in this case the phase, even after removal of excess delay, is not minimum phase. The last figure is the simulated impulse response for, Red, the unequalized response, Blue, the response equalized to the target amplitude, and purple, the the impulse of the minimum phase target. The impulse was 1 msec wide. If you don't believe the impulse responses I can post a comparison of the measure impulse for the driver and the simulation of that impulse. (Hint: they are identical). Anyway, I think this illustrates the fact that simple smoothing the amplitude for a woofer isn't enough. While it improves the initial attack, the long time response is even worst than the unequalized response, both of which ring on forever compared to the target. __________________ John k.... Music and Design NaO Dipole Loudspeakers. "We have no right to assume that any physical laws exist, or if they have existed up to now, that will continue to exist in a similar manner in the future." Max Planck

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2nd February 2009, 03:09 AM	#532
markus76 diyAudio Member Join Date: May 2008 Location: Switzerland	Todd, I'm not kidding and I've read Floyd's book. I would love to see your data. Best, Markus P.S. Ethan Winer did some real world tests: http://www.realtraps.com/eq-traps.htm

2nd February 2009, 04:19 AM	#533
cap'n todd diyAudio Member Join Date: Jan 2009	John, I want to response to your points. Not much time now, but I would be interested to see what the room reponse in your simulation looked like before eq, to put things in perspective.