FabFilter

Hi,

We have indeed mapped the default Butterworth Q of 0.7071 = sqrt(0.5) to a Q value of 1 in Pro-Q 3. The reason behind this is because if feels more logical to have a default value of 1 instead of a default value of 0.7071. We're not the only plug-in that has done this, lots of other eq plug-ins have applied this as well.

There is quite an easy formula to calculate this:

Pro-Q 3 Q = Butterworth Q / sqrt(0.5)

So...

Bw in Octaves / √(0.5) = 1.4142 ÷ √(0.5 ) = 1.99998081991 = 1 Octave

∴

FabFilter equivalent for 1 Octave - according to your equation.

1 Octave is: 1.99998081991 = 2 (FabFilter Q = 2)

And Two Octaves = 0.948 (FabFilter equivalent Q)

Is this the correct way to apply the equation?

Thanks

F

And is the scale/range of parameters linear or logarithmic?

Can you please give me / us an answer Ralph

The Q Factor
Conversion chart or table
'bandwidth in octaves' N to quality factor Q
www.sengpielaudio.com/calculator-bandwidth.htm

2 Octaves = 0.67 = (please put FabFilter Equivalent after = sign)
1 1/3 Octaves = 1.0444 = ?
2 1/3) Octaves = 0.56 = ?
3 1/3 Octaves = 0.35 = ?
1 Octave = 1.4142 = ?

1/2 Octave = 2.871 = ?
1/3 Octave = 4.3189 = ?
1/4 Octave = 5.7635 = ?
1/12 Octave = 17.3 = ?

3 Octaves = 0.4040 = ?
4 octaves = 0.2667 = ?
5 Octaves = 0.182 = ?
6 Octaves = 0.127 = ?
7 Octaves = 0.089 = ?

1/1/2 Octaves = 0.92 = ?

What is the FabFilter Equivalent???
What are the answers?
We would like the answers!
It shouldn't take you that long - maybe a couple of minutes.

Tell us for God's sake!

Thank you

F

Hi F_W,

That's indeed the correct way to calculate the Q factors. You can just divide the Q factors you mentioned above by sqrt(0.5) and you will get the Pro-Q 3 Q factor. I'm sure you can do the math yourself. It shouldn't take you that long - maybe a couple of minutes :-)

Finally :)

Thank you so much!

For example:
If a centre cut was placed @ 1k and it resulted in affecting the outermost frequencies of 891Hz and 1123Hz.

This yields a result of 4.31 (Q = 4.31) - this is approximately a 1/3 of an Octave.

Would I then have to convert the FabFilter Q with the "given formula" to get the reading for a quarter of an Octave in FabFilter terms?

So... Using the Formula: Q of 4.31 = 6.09 in FabFilter Pro Q3?

i.e.: A third of an Octave = 6.09 in Pro Q3?

I look forward to hearing from you shortly.

F

** Would I then have to convert the *Bandwidth* (Q) with the "given formula" to get the reading for a quarter of an Octave in FabFilter terms?

Made a slight 'typo' there.

regards

F

** Would I then have to convert the *Bandwidth* (Q) with the "given formula" to get the reading for a 1/3 of an Octave in FabFilter terms?

Made another slight 'typo' there.

regards

F

The saying:
"Talk about double Dutch" comes to mind - for this whole scenario!

The phrase “double Dutch” is a common idiom used in English to describe something that is difficult to understand or confusing.

:)

hahahahahaha

I take it that:

Pro R and Pro R2;
Simplon;
Micro;
Timeless 3;
Volcano 3;
Pro C2;
(Did I miss anything?)

Have the same factors attributed to its Q values - and conversion is necessary to obtain the conventional bandwidth in Octaves?

F

***Bump***************************************************

And is the scale/range of parameters linear or logarithmic?

***Bump***************************************************

For example:
If a centre cut was placed @ 1k and it resulted in affecting the outermost frequencies of 891Hz and 1123Hz.

This yields a result of 4.31 (Q = 4.31) - this is approximately a 1/3 of an Octave.

Would I then have to convert the *Bandwidth* (Q) with the "given formula" to get the reading for a 1/3 of an Octave in FabFilter terms?

So... Using the Formula: Q of 4.31 = 6.09 in FabFilter Pro Q3?

i.e.: A third of an Octave = 6.09 in Pro Q3?

***Bump***************************************************

I take it that:

Pro R and Pro R2;
Simplon;
Micro;
Timeless 3;
Volcano 3;
Pro C2;
(Did I miss anything?)

Have the same factors attributed to its Q values - and conversion is necessary to obtain the conventional bandwidth in Octaves?

I would like to know the answers to these three queries as well.

LJ