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In Reply to: RE: MMG MiniDSP and bi-amp Questions posted by neolith on February 23, 2012 at 15:32:28
Hey Neo,
If the phase of the tweeter and woofer is incorrect you will hear a drop in the output around 800-1000 hz.
I had assumed that the 1st order slopes have the tweeter and woofer in phase. When I was messing with the allocator software to mimic the stock slopes I never could get the right graph. Davey showed me that even in this config the tweeter and woofer are wired out of phase so any crossover mod should keep that same orientation.
So "incorrect" in this case is in phase with each other???
For instance, here is the curve with the drivers out of phase:
and here it is with them in phase:
I guess I thought the drivers were supposed to be in phase on 1st order slopes. Where can I read more about that stuff??
Afterwards we discovered faith; it's all you need
Follow Ups:
If you have a symmetric first order crossover than the phase difference of the LP filter and and HP filter is a constant 90 degrees at all frequencies. It doesn't matter if the polarity is reversed since the difference is still 90 degrees.
However this does hold up with an asymmetric crossover and in that case changing the polarity may very well result in a different phase relation and a "dip".
On the violin: "Heaven reward the man who first hit on the idea of sawing the innards of a cat with the tail of horse."
Neo mentions: "If you have a symmetric first order crossover than the phase difference of the LP filter and and HP filter is a constant 90 degrees at all frequencies. It doesn't matter if the polarity is reversed since the difference is still 90 degrees."
Neo, that's precisely what I thought at first. It didn't work quite that way in practice but I'd have to look up the charts to refresh my mind. (It's Friday at work and my mind is having a major dip at the xover point : - ))
The phase relations are determined by the math. Here's two graphs that show the difference between a symmetric 1st order crossing at 800 hz and an asymmetric 1st order also crossing at 800 hz.
With the symmetric the phase difference is a constant 90 and the output of normal polarity and inverted polarity are superimposed on each other.
With the asymmetric, the phase difference varies with frequency and the normal polarity creates a dip while the inverted polarity has a much shallower dip.
On the violin: "Heaven reward the man who first hit on the idea of sawing the innards of a cat with the tail of horse."
Neo, I was wondering if we were talking about 2 different things. By asymmetric, I meant the original MMG xover with the mix of 1st/2nd order slopes. In this case, the current design would be symmetric for having 1st/1st.Perhaps the symmetry you refer to is with regard to another set of parameters? I believe you mean that the points for each driver are at the same frequency. This latter case is when one should expect to see no difference? I guess this is it. Which would explain why, by being asymmetric in this sense, I didn't see what I expected from a current design. Of course, MMGs don't want to use the same points because, hey, that would be boring and conventional. LOL!
So, both MMG xover designs share this kind of approach to frequency point asymmetry. In addition, the original design is further asymmetric in the slopes used.
All of which helps us have so much fun. : - ))
Edits: 02/24/12
It's my understanding that a symmetric crossover is like the typical L-R in that the fc of the LP and HP are the same as are the orders of the two filters. Anything else is asymmetric. Magnepan almost exclusivly uses asymmetric crossovers with the exception of the SMG(a)(series 1st order) and IIa (parallel 3rd order).
On the violin: "Heaven reward the man who first hit on the idea of sawing the innards of a cat with the tail of horse."
Edits: 02/25/12
Hint. You should think polar response here. Horizontal polar response, to be more specific.
Cheers,
Dave.
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