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Old 15th April 2006, 11:02 PM   #1
Khron is offline Khron  Romania
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Default Baffle Step Compensation

Ok, i've seen what a line-level BSC circuit can do (as per Rod Elliott - sound.westhost.com ), but i'm real curious.. Does a passive BSC do the same thing? i.e. attenuate sound above a certain frequency in a shelf manner.
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Old 16th April 2006, 12:54 PM   #2
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yes, absolutely.
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Old 16th April 2006, 02:51 PM   #3
Khron is offline Khron  Romania
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http://zaphaudio.com/audio-speaker12.html

In this design, Mr Krutke says he has about 4.5dB of BSC on the woofer. But all he has in the woofer crossover is the inductor (1st order low-pass) and the LC notch filter (for the 8KHz breakup node).

I'm getting a tad confused
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Old 16th April 2006, 03:29 PM   #4
Davey is offline Davey  United States
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Khron,

The woofer driver is a complex impedance and judicious design with a minimal number of components can still yield the desired BSC curve. Take a look at the graph labeled "woofer filter effect." This shows the actual drive to the woofer and the shelving effect between 200-2000Hz that John was targeting.

You'll see this kind of thing a lot with well-designed crossover networks......the actual topology might not look "generic" (or what you'd expect) but it still accomplishes the goal.

Cheers,

Davey.
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Old 16th April 2006, 09:23 PM   #5
Khron is offline Khron  Romania
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Thanks Davey, that does make sense
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