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Old 2nd February 2009, 10:21 PM   #561
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Quote:
Originally posted by cap'n todd



Hi John, so are you trying to remove all ringing then? Then add some back in to make it sound more natural? I'm dont really think its necessary, but sounds interesting.
I actually think this is a done deal. I have to modify my simulation code before I can verify it theoretically though. And right now I really don't feel like writing code.
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Old 2nd February 2009, 10:25 PM   #562
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youngho, no, I was talking about the rate after the initial drop. Looking at the picture Todd posted, the line on the left hand side decays faster than that on the right side. In the first few ms there's a huge drop but we still do not know if this is enough to render the ringing inaudible.

Todd offers an explanation:

Quote:
So you end up with a corrected response that has two slopes, from the Q of the resonance and the correction filter not cancelling correctly. Should it be surprising that the decay in time has two slopes then?
But doesn't this bring us back to my initial comment that I've never seen a CSD that clearly shows that an EQ completely eliminated ringing?
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Old 2nd February 2009, 10:38 PM   #563
youngho is offline youngho  United States
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Quote:
Originally posted by markus76
youngho, no, I was talking about the rate after the initial drop. Looking at the picture Todd posted, the line on the left hand side decays faster than that on the right side. In the first few ms there's a huge drop but we still do not know if this is enough to render the ringing inaudible.
Hmm, it goes to zero on the right earlier, no? It decays faster on the right initially, then it seems to decay more slowly after that, but it reaches zero earlier. And shouldn't inaudible should be at the level of the noise floor?

Oops, sorry, not a physicist or an engineer. I'l shut up now.
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Old 2nd February 2009, 10:50 PM   #564
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Originally posted by youngho
And shouldn't inaudible should be at the level of the noise floor?
It should be inaudible if the noise floor is inaudible I think the data shown isn't very clear, but it was not me that asserted that "This [data] is not hard to find."
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Old 2nd February 2009, 11:13 PM   #565
youngho is offline youngho  United States
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Quote:
Originally posted by markus76


It should be inaudible if the noise floor is inaudible I think the data shown isn't very clear, but it was not me that asserted that "This [data] is not hard to find."
Just out of curiosity, how would this "modal ringing" be audible above the noise floor, i.e. the "room ringing"? And the graphs were in the book! I was about to refer you to page two-hundred-forty-whatever, myself.
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Old 3rd February 2009, 12:23 AM   #566
gedlee is offline gedlee  United States
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Quote:
Originally posted by john k...



Earl you missinterpreted what I posted, due to my poor gramar.



How about if I rewrote that as, "In simple terms, X must satisfy the wave equation with no sources in the interior. Boundary conditions as required are applied so that when added to g, g + X = G, the solution for the bounded region."

I would agree that solving the wave equation, with appropriate boundary conditions, for X might not be an easy task, but someone must have done it to come up with the modal expansion for G. So, X = G-g must at least be a reasonable representation of X, at least as good as G is a representation of the bounded region Green's function.

John

Agreed! With one caveat. The function X(r) will be different for every source position in the room. There is no single X(r) that will satisfy the boundary conditions for an arbitray source location.

When Morse subtracts off the near field from the Green's function G(r) he does this by only fitting the singularity at the source point. This approach is then only approximate at the walls, but more accurate at the source than the series, because it converges faster. This leaves X(r) independent of the source location, BUT it is only an approximate solution.
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Old 3rd February 2009, 12:39 AM   #567
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Quote:
Originally posted by markus76


It should be inaudible if the noise floor is inaudible I think the data shown isn't very clear, but it was not me that asserted that "This [data] is not hard to find."

I could definately find more, but it is time consuming. If yo have access to the AES journal, you should find more. And in any case I agree that the waterfalls are often hard to read/interpret.

Here's one I measured myself, a resonance at 28 Hz, no eq and eq.
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Old 3rd February 2009, 01:25 AM   #568
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That looks much clearer thanks Todd.
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Old 3rd February 2009, 11:16 AM   #569
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Quote:
Originally posted by gedlee



John

Agreed! With one caveat. The function X(r) will be different for every source position in the room. There is no single X(r) that will satisfy the boundary conditions for an arbitrary source location.

When Morse subtracts off the near field from the Green's function G(r) he does this by only fitting the singularity at the source point. This approach is then only approximate at the walls, but more accurate at the source than the series, because it converges faster. This leaves X(r) independent of the source location, BUT it is only an approximate solution.

I think the problem is the way Morse write the relationship

G(R,Ro) = g(R,Ro) + X(R)

This says X(R) = G(R,Ro) -g(R,Ro) .

Obviously X is a function of Ro. However, X(R) is a solution to (here we go again) the homogeneous wave equation with inhomogeneous boundary conditions (finally got that right). This is were the dependents of X(R) on Ro comes in because the boundary conditions are dependent upon the value of g(R,Ro) evaluated on the bounding surfaces. So while the dependence of X on Ro is not explicit, there is implicit dependence of X on Ro from the BC's. Perhaps it would have been better if Morse had written X(R) is a solution to the homogeneous wave equation with inhomogeneous boundary conditions, BC = F(g(Rb,Ro)) where Rb is the position vector defining the bounding surface. Change the source position and the X changes as well.

I guess I saying that I'm not particularly worried about what Morse says he does or doesn't do. If I have G(R,Ro) and I know g(R,Ro) I can drop in the values of R and Ro and find the value of X (R) for any values of R and Ro in the bounded region. But it's all academic because there is no need to worry about X if we know G(R,Ro).
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Old 3rd February 2009, 12:12 PM   #570
breez is offline breez  Finland
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Default On the effect of EQ on decay rate

By courtesy of the Anti-Mode 8033 subwoofer EQ developer I'm posting a before/after comparison of what the aforementioned little box can do with regards to the decay rate of LF modes.

The time slices are 20 ms each, so from 0-160 ms. The comparison clearly shows faster rate of decay, at least up to 160 ms.
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