I've read that wiki as well as about 10 other internet pages. It's quite easy to find the Q for 2nd order but not 4th. How did you know those Q values?
the product is without doubt the controlling criteria. Take for example a 3pole Butterworth. It is simply a passive RC with Q=1/sqrt(2) cascaded with an active 2pole filter with Q=1.0. These two filters do not even need to follow, or be adjacent to, each other in the circuit.I thought that as long as the product of them resulted in 0.7071 it was OK.
Are the Q values for the two 2nd-order stages that critical? I thought that as long as the product of them resulted in 0.7071 it was OK.
I wonder what changing the 4pole L-R to a pair of 2pole stages with Q=1.0 and Q=1/sqrt(2) giving a product of 1/sqrt(2). Would the output be identical to the correct 4pole Butterworth?The two Sallen-Key filter section Qs are 0.54 and 1.31,
The 4pole L-R is not an example of a 4pole Butterworth.
It has a Q=0.5 instead of 1/sqrt(2).
It is not intended to mimic a Butterworth.
I wonder what changing the 4pole L-R to a pair of 2pole stages with Q=1.0 and Q=1/sqrt(2) giving a product of 1/sqrt(2). Would the output be identical to the correct 4pole Butterworth?
Speaker Dave doesn't quite seem to grasp what we are saying and is confusing the issue by talking about other different target Qs and crossover points, but I'm not sure how we can say it any other way.
first order = single pole filter. The Q of an RC single pole filter is 1/sqrt(2)~0.71, not Q=0.5Returning to the Wiki page and looking at the Qs required for Butterworth filters from 1 to 5 you'll find:
1st order Butterworth: Q of .5 (real pole)
Why are you swapping back and forth between Butterworth and Linkwitz Reilly? It seems to be confusing you and the issue.An LR or "Butterworth squared" 4th order will be two cascaded 2nd orders. Since both the Butterworth second orders will have a Q of .7 they can not give the same response as the Q of .541 plus Q of 1.31 of the true 4th order Butterworth. That is why the response is different. As verification, since both of the 2nd orders would be -3dB at the crossover, the LR combination would be -6dB at the crossover point, the Butterworth is always -3dB at the crossover point.
first order = single pole filter. The Q of an RC single pole filter is 1/sqrt(2)~0.71, not Q=0.5
Why are you swapping back and forth between Butterworth and Linkwitz Reilly? It seems to be confusing you and the issue.
That was me. Post #5.My question in post7 was referring only to Butterworth. One respondent shows some of the science explaining why moving the poles around the "Butterworth" circle would change the output.
Clearly. Can anybody else out there back me up on this?You are offering a mixed up set of quotes that seem to be getting nowhere clearly. Your first response (post5) was excellent, but since then I cannot follow your argument.
The dashes mean Q does not have any meaning for first order poles. Biquads have pole pairs, a biquad with a Q of 0.5 has a pair of real poles rather than a pair of complex poles, but it is still two poles and it still ends up at 12dB/octave as your plot shows, playing with the Q alters the shape around the corner frequency, not the ultimate roll-off. To realise odd orders with a biquad-based parametric EQ the EQ would need to offer first order filters as an option (not strictly a biquad) or allow the relevant biquad coefficients to be zeroed.What do the dashes on the 1st and 3rd order mean? Can Butterworth odd orders be realized with parametric filters?