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2nd June 2004, 09:54 PM  #1 
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How to figure FR with bass boost?
The nifty webpage www.diysubwoofers.org has the formulas for calculating frequency response curves for sealed and ported systems. Cool. After about four minutes playing with Python I was making lovely FR plots.
Now then, I want to add in the effect of bass boost. Say for example I have a plate amp that has a 6 dB boost at 30 Hz with a Q of 1.7. What now? How do I adjust the FR curve? Anyone know the formula? Or know where I can find it?
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Davy Jones 
2nd June 2004, 10:04 PM  #2 
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Location: USA, MN

So you have something like:
Vb = net box volume (litres) Fs = driver resonance frequency (Hz) Qts = driver Q at system resonance Fb = box tuning frequency (Hz) Ql = box losses (Ql=7 can be assumed for most cases) then, Fn2 = (F/Fs)^2 Fn4 = Fn2^2 A = (Fb/Fs)^2 B = A/Qts+Fb/(Fs*Ql) C = 1+A+(Vas/Vb)+Fb/(Fs*Qts*Ql) D = 1/Qts+Fb/(Fs*Ql) dBmag = 10*LOG(Fn4^2/((Fn4C*Fn2+A)^2+Fn2*(D*Fn2B)^2)) Multiply the result inside the log expression by: ff2/sqrt((1ff2)^2+(ff2/Q^2)) then take the log........ where ff = F/Filter freq ff2 = ff^2 Q is self explanatory BTW, a sealed box can be modeled with the above equations by setting Fb=0. Have a great day.....
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2nd June 2004, 10:27 PM  #3 
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Beauty! I think I've got it, and the first test graphs look reasonable. Let me just rephrase what you said to be sure I've got it.
First express the unequalized response curve in the form, dbMag(f) = 10*log10(P(f)) where the form of the function P depends on the alignment type (sealed box, or whatever). Let Q = filter_Q ff(f) = f/Filter_freq ff2(f) = ff(f)^2 M(f) = ff2(f)/sqrt((1ff2(f))^2+(ff2(f)/Q^2)) Then the equalized curve is defined by eqdbMag(f) = 10*log10(M(f)*P(f)) Do I got it?
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Davy Jones 
3rd June 2004, 12:11 AM  #4 
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Yup.
Post a screenshot of your results, and the parameters you entered, I should be able to model it and compare. BTW, You can model a 2nd order lowpass with: 1/sqrt((1ff2)^2+(ff2/Q^2))
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3rd June 2004, 03:08 AM  #5  
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Quote:
Will do. Quote:
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3rd June 2004, 03:35 AM  #6 
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Ooops!
Ooops!
The formula does not have a term for the amount of boost. The number I gave, 6dB, was just an example. Back to the drawing board.
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Davy Jones 
3rd June 2004, 04:35 AM  #7 
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boost = 20 * log(Q) ;)
Lowpass filters cut the highs, like the crossover built into most plate amps....
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3rd June 2004, 03:33 PM  #8  
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Quote:
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3rd June 2004, 04:41 PM  #9 
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Clarifying:
The boost is a 2nd order highpass filter with the equation: A(f) = ff2/sqrt((1ff2)^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((1fc2)^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))
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3rd June 2004, 05:03 PM  #10  
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Quote:
eqdbMag(f) = 20*log10(M(f)*P(f)) not eqdbMag(f) = 10*log10(M(f)*P(f)) Here's some Python code and the resulting graph: [I'm having trouble with this. Hold on.]
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