brucek elaborates on bandwidth:
OK here's my more detailed explanation of relative and absolute bandwidth as I see it.
I'll just talk about cuts here in my examples, but the same holds for gains. Gain is the same, but I didn't want to add confusion.
When you enter a BW (bandwidth) of one octave into the front panel of a BFD, the filter will be one octave above and one octave below the center frequency  but that's the easy answer. Where I measure the bandwidth of one octave is the rub, and takes some explaination. Try not to go to sleep during the discussion.
Here are the simple rules to follow.
Absolute bandwidth = bandwidth at the 3dB point.
Relative bandwidth = absolute bandwidth divided by the center frequency.
Q = inverse of the relative bandwidth = 1/relative.
Front panel entered bandwidth is the absolute bandwidth divided by two.
For a cut, the absolute bandwidth is defined as the width of the filter at its 3dB down points expressed in octaves. Did you notice I said 3dB "down". Big difference from saying  3dB cut. This isn't 3dB like the gain dial readout of  3dB. This is the relative 3dB less than zero dB for a given cut (or boost) value. So the 3dB down point will be different for every cut value. But the absolute bandwidth at that 3dB down point will remain the same for different cuts, but then of course, the bandwidth where the filter flatlines into zero dB will get wider and narrower as the cut is changed.....
For example, that means if the front panel entered cut is 6dB from zero, then you have to use a mathematical expression to calculate the value at which the curve will cross 3dB "down" from zero when we use  6dB as the cut, (given your front panel octave width selection) . This makes sense because obviously if we cut by  2dB on the front panel from zero, we couldn't say the absolute bandwidth was measured at absolute  3dB cut because  2dB cut is less than 3dB cut. We'd have to calculate for the selected cut of 2dB where 3dB "down" point was located.
Lets take an example of three filters. The filters maintain a constant absolute bandwidth of 2/3 octave centered around 50Hz. 1/3 octave is entered into the front panel for BW. 1/3 octave less (40Hz) and 1/3 octave greater (63Hz) than 50 Hz. The only variable is the cut of 3dB and 6dB and 12dB.
I have calculated the 3dB "down" from zero points at which the filter must cross the 40Hz and 63Hz absolute bandwidth of 2/3 octave for a front panel entered cut of 3dB, 6dB and 12dB. Each value will be 3dB down from zero at a different spot, but when they are indeed at 3dB down, they all possess the same absolute bandwidth of 2/3 octave from 40Hz to 63Hz.
Here's my three calculations  ignore if you're not familiar with decibel and antilog calculations.
Let x = 12dB cut calculation for value of cut at 3dB down from zero.
x = [(antilog 3dB/20) x cut value ]  cut value
x = [(antilog 3dB/20) x 12 ]  12
x =  3.50dB
Let x = 6dB cut calculation for value of cut at 3dB down from zero.
x = [(antilog 3dB/20) x cut value ]  cut value
x = [(antilog 3dB/20) x 6 ]  6
x =  1.75dB
Let x = 3dB cut calculation for value of cut at 3dB down from zero.
x = [(antilog 3dB/20) x cut value ]  cut value
x = [(antilog 3dB/20) x 3 ]  3
x =  0.87dB
So what am I saying is that for a 50Hz center frequency filter with an entered value of BW of 20/60, with a cut of 12dB, the at 40Hz and 63Hz the cut will be 3.5dB. These are the absolute bandwidth 3dB down points. The filter will of course extend out further until it reaches zero effect, but most of it's power is in the absolute area.
So for this filter example here's the values from my initial list of definitions.
Center frequency = 50Hz
BW = 2/3 octave from 40Hz to 63Hz = Absolute bandwidth = bandwidth at the 3dB down point.
Relative bandwidth = (63  40) / 50 = 0.46 = absolute bandwidth divided by the center frequency.
Q = 1 / 0.46 = 2.17 = inverse of the relative bandwidth = 1 / relative.
Front panel entered bandwidth is the absolute bandwidth divided by two = 2/3 / 2 = 1/3 or 20/60
Anyway, the math and the design graph seem to support each other, so I think I have this correct  I may have made some math mistakes. The design graph isn't super resolute, so it's hard to get exact readings off it, but its close enough to use to make filter decisions  it's easier than doing antilogs anyway.
If you had "any" cut, then it's always possible to calculate the absolute bandwidth 3dB down from zero point  and it will always be negative less than zero.
For example, if I have a 2dB cut filter at 50Hz that has an absolute bandwidth of 2/3 octave from 40Hz to 63Hz, then the 3dB down point at 40 and 63Hz would be:
Let x = 2dB cut calculation for value of cut at 3dB down from zero.
x = [(antilog 3dB/20) x cut value ]  cut value
x = [(antilog 3dB/20) x 2 ]  2
x =  0.58dB
So at the center frequency there would be 2dB cut and at 40Hz & 63Hz there would still be 0.58dB cut.
brucek
