4th order Butterworth filter Q

[gainphile]

Quick question,

What is the Q of 4th order butterworth filter ?

And if I want to implement using two cascaded sellen-filter topology (op-amp), what is the Q of each stage?

Thanks

[DSP_Geek]

Google is your friend, and so is Wikipedia:

Butterworth filter - Wikipedia, the free encyclopedia

The two Sallen-Key filter section Qs are 0.54 and 1.31, incidentally.

[gainphile]

Thank you.

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?

[SY]

Q is a parameter for 2nd order filters; the polynomial in the transfer function of a 4th order filter is (naturally) different, as are the means of arriving at the desired coefficients. 4th order filters are generally constructed by cascading two 2nds. DSP-Geek mentioned the required 2nd order Qs for two cascaded filters to get a 4th order Butterworth.

[speaker dave]

Originally Posted by gainphile

Thank you.

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?

About half way down is the table of "Factors of polynomials..." The 4 or 5 place numbers are 1/Q so their inverse is the number you are looking for.

For others: We talk about "Q of .707" so much that some assume it is applicable for all filter orders. In fact it only applies for the second order filter (such as a sealed box simulation). For all other filter orders different coefficients are required to hit a Butterworth shape. In general any higher order filter shape can be split into a series of second order sections + one first order if an odd number is required. Each section will have its own Q and higher order filters wil have a progression from low Q to high Q.

If you understand S domain plots (pole/zero plots) you will see that the Butterworth case positions poles equally around a circle. The ones nearer the jw axis have higher Q. By the way, Chebyshev or Elliptical filters place the poles on an ellipse (hence the name). Since a circle is an ellipse with both axies equal, a Butterworth is a limiting case for the Ellipticals.

David S.

[richie00boy]

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.

[AndrewT]

Originally Posted by richie00boy

I thought that as long as the product of them resulted in 0.7071 it was OK.

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 liked that S domain explanation. It the first time I have noted this and it made me realise where the funny numbers come from.

I look forward to hearing explanations to what effect moving the poles around that circle have on the sound and theoretical correctness of the resulting filters.

[speaker dave]

Originally Posted by richie00boy

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.

Yes, they are critical. If you want a Butterworth shape you need the distribution of low Q to high Q, especially for higher order filters. Having the same Q product isn't equivalent.

One example would be the Linkwitz Riley. It cascades two 2nd order Butterworth. Rather than giving you a 4th order Butterworth you get something with -6dB at the crossover (a softer corner).

David

[AndrewT]

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.

The two Sallen-Key filter section Qs are 0.54 and 1.31,

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?

[john k...]

If you look at the denominator of a 4th order filter transfer function is is

1 + a1 S + a2 S^2 + a3 S^3 + a4 S^4

Expressed as the product of two 2nd order filters with Q1 and Q2 being the Q of each,

1 + s(1/Q1 + 1/Q2) + s^2 (2+ 1/Q1 x 1/Q2)) + s^3 (1/Q1 + 1/Q2) + s^4

You can see that it is the product of the reciprocal of the Qs and the sum of of the reciprocals that determine the shape of the response. Both the sum and the product must be correct. This means that Q1 and Q2 are unique for each type of filter.