How to figure FR with bass boost?
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 Multi-Way Conventional loudspeakers with crossovers

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Dave Jones
diyAudio Member

Join Date: Jul 2003
Location: Silicon Valley
Graph:
Attached Images
 pgraph.png (6.4 KB, 81 views)
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Davy Jones

Dave Jones
diyAudio Member

Join Date: Jul 2003
Location: Silicon Valley
Code:
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Davy Jones

Dave Jones
diyAudio Member

Join Date: Jul 2003
Location: Silicon Valley
Here, this is a better picture. I cut the graph off at 20 Hz rather than 10.
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 pgraph2.png (5.8 KB, 73 views)
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Davy Jones

 3rd June 2004, 05:02 PM #14 Ron E   diyAudio Member     Join Date: Jun 2002 Location: USA, MN Oops, I didn't look at the box equations too closely. To modify, you can take the square root of the box response equation So: P(f)= Fn4/sqrt((Fn4-C*Fn2+A)^2+Fn2*(D*Fn2-B)^2) A(f) = ff2/sqrt((1-ff2)^2+(ff2/Q^2)) B(f) = 1/sqrt((1-fc2)^2+(fc2/Q^2)) The response of the woofer + HP filter + LP crossover is then 20*log10(A(f)*B(f)*P(f)) You will want to try to graph with semilog scales to get decent looking output. You notice that you get 5dB at 30 Hz rather than at the peak, that is because the formula I gave you is for dB at te hfilter frequency, not at the peak. Looks Good! __________________ Our species needs, and deserves, a citizenry with minds wide awake and a basic understanding of how the world works. --Carl Sagan Armaments, universal debt, and planned obsolescence--those are the three pillars of Western prosperity. —Aldous Huxley
Dave Jones
diyAudio Member

Join Date: Jul 2003
Location: Silicon Valley
Quote:
 Originally posted by Ron E Clarifying: The boost is a 2nd order highpass filter with the equation: A(f) = ff2/sqrt((1-ff2)^2+(ff2/Q^2)) ...and yes, the Q determines the dB of peaking. 20*log10(1.778)=5dB The shape of the curve is defined by your box design, the peaking filter and the crossover frequency. There are a lot of independent variables now, so the system can get complicated pretty fast. When applying a boost circuit, it is best to apply boost at the Fb of the vented system. Remember, a lowpass passes the lows and cuts the highs, and a highpass passes the highs and cuts the lows. The peaking lowpass filter here cuts the lows below trhe tuning frequency and protects the vented driver below resonance. This is better known as a 6th order alignment. The crossover is a lowpass filter - and a 2nd order lowpass has the equation: B(f) = 1/sqrt((1-fc2)^2+(fc2/Q^2)), where : fc2 = (F/F_crossover)^2 The response of the woofer + HP filter + LP crossover is then 20*log10(A(f)*B(f)*P(f))

I'm confused again. What is F_crossover? That's a new one. And there's a whole new part to the formula. Sorry if I'm being dense.

Are you showing how to model both the bass boost and the infrasonic filter that plate amps usually have? And if so, is F_crossover the frequency of the knee of the infrasonic (high pass) filter? Or what?
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Davy Jones

Dave Jones
diyAudio Member

Join Date: Jul 2003
Location: Silicon Valley
Quote:
 Originally posted by Dave Jones I'm confused again. What is F_crossover? That's a new one. And there's a whole new part to the formula. Sorry if I'm being dense. Are you showing how to model both the bass boost and the infrasonic filter that plate amps usually have? And if so, is F_crossover the frequency of the knee of the infrasonic (high pass) filter? Or what?
Light dawns over Dunderhead. Are you showing how to incorporate the effect of the sub's crossover? That must be it. F_crossover must be where I set the crossover frequency knob on the sub. Duh.
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Davy Jones

Dave Jones
diyAudio Member

Join Date: Jul 2003
Location: Silicon Valley
I got it. Thanks again. Here's the chart on a log scale:
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 log_boost.png (5.6 KB, 50 views)
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Davy Jones

Dave Jones
diyAudio Member

Join Date: Jul 2003
Location: Silicon Valley
The boost graph appears to be right. I don't think I've got the crossover right though.

Quote:
 B(f) = 1/sqrt((1-fc2)^2+(fc2/Q^2)), where : fc2 = (F/F_crossover)^2
1) What is the Q in that equation? It's not the Q from the other equations is it? The crossover and boost are independent aren't they?

2) I presume "f" is "F", right?

I think when I get this figured out, I'm armed for bear.
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Davy Jones

 4th June 2004, 08:11 PM #19 Dave Jones   diyAudio Member   Join Date: Jul 2003 Location: Silicon Valley Hmmmm.... If I use Q = 1/sqrt(2) in the crossover equation, I get what looks like it might be the right curve. Below is the combined boost and crossover. What's that Q supposed to be? __________________ Davy Jones
Dave Jones
diyAudio Member

Join Date: Jul 2003
Location: Silicon Valley
Combined:
Attached Images
 combined.png (5.7 KB, 40 views)
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Davy Jones

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