# 4th order Butterworth filter Q

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#### 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

#### 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

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

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

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?

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#### 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.

#### speaker dave

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.

Not exactly. An LR 4th order is the cascade of two Butterworth 2nd orders. Being second order (and Butterworth) they would both have a Q of .707. Just because the combination is -6dB down doesn't mean it was a Q of 0.5.

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?

Again, the product of the Qs is not significant. The particular individual Qs are required to define the shape.

David

#### AndrewT

Just because the combination is -6dB down doesn't mean it was a Q of 0.5.
this I do not believe.
John K I want to believe, but I don't understand AC theory sufficiently to let me investigate what is included in his concise response.

#### richie00boy

I too would like more explanation Andrew. I have made 4th-order filters in the past and used different Q's to multiply together to achieve the target Q, and whichever combination of Qs I used always gave the same result in a simulator.

In fact Andrew I have tried the Q=1 and Q=0.7071 you use as an example. Thinking about the individual phase and amplitude responses and how they combine it seems to me that all that is relevant is the Q product. 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.

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#### speaker dave

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.

I fully grasp what you are saying, it just happens to be wrong.

Returning 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)
2nd order: Q of .707 (2nd order)
3rd order: Q of .5 and 2nd order with Q of 1 (real pole and 2nd order)
4th order: Q of .541 and Q of 1.31 (both 2nd order)
5th order: Q of .5, Q of .618 and Q of 1.618 (first is a real pole and other two are 2nd order)

Shifting any of the Qs would make it non-Butterworth. Using different Qs that happen to have the same product would make it non-Butterworth. Sorry but that is the way it is.

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.

Textbook stuff.

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#### AndrewT

Returning 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)
first order = single pole filter. The Q of an RC single pole filter is 1/sqrt(2)~0.71, not Q=0.5
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.
Why are you swapping back and forth between Butterworth and Linkwitz Reilly? It seems to be confusing you and the issue.

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. 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.

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#### speaker dave

first order = single pole filter. The Q of an RC single pole filter is 1/sqrt(2)~0.71, not Q=0.5

I'll backtrack a little from this. If you have a second order with 2 real poles of the same frequency, then that is most definitely a Q of 0.5. Here is a good reference for that. Quality Factor (Q)

With a single real pole I'm not sure that the term Q applies.

I am guessing that you are so used to the notion that "3dB down at the corner = Q of .707" that you can't get past it. The Q of .707 only applies to the 2nd order filter (and some higher order filters that will have one pole pair at that location as well as pole pairs at all the other locations).

Why are you swapping back and forth between Butterworth and Linkwitz Reilly? It seems to be confusing you and the issue.

You made a statement that the Qs of the 2nd order sections that made the Butterworth 4th didn't matter as long as the products were constant. I gave an example of the commonly used LR filter (Butterworth squared) to show that a 4th order with the poles in a non-Butterworth locations would have a different filter shape. Its not about the product of the Q (and John K has confirmed that as well) but the exact locations of the numerous poles for a given filter type. Butterworth requires an equally spaced distribution of poles around the unit circle. Moving some poles up and others down (to maintain a constant product) will force the filter to be non Butterworth. It may look similar for small changes but it is no longer Butterworth.

Wouldn't it stand to reason that the Butterworth Qs must be crucial otherwise textbooks wouldn't give them so specifically?

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.
That was me. Post #5.

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.
Clearly. Can anybody else out there back me up on this?

David

#### sreten

Hi,

I reckon the original question has issues. It most probably meant L/R (cascaded 2nd order Butterworths) for speakers or most probably meant actual 4th order Butterworth for what purpose I don't know.

For the latter the two 2nd order Q's have to be correct as a pair. Those values, and only those values give the maximally extended amplitude response that never exceeds the passband gain.* Allowing a small amount of ripple (Chebyshev family) will give a "better" filter, but its not exactly Butterworth, the same applies to similar filters with a less sharp knee and thus better transient response, they may be near but they are not exactly Butterworth.

rgds, sreten.

* I cannot remember the defining characteristic of Butterworth polynomials that means it must be the most extended flat response, something to do with Vin/Vout =1/square_root(1 + (F/Fc)power_2n) being monotonic.

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#### sregor

Butterworth is just one type of filter, with very specific design constraints. It is not that the products of the poles are .707, but the equation which generates them. For a 4th order system there are infinite number of pairs of Qs whose product is .707, but only one pair which would make it a butterworth filter, which is .54 and 1.31 . For further reading try: http://www.national.com/an/AN/AN-779.pdf I think people are too wrapped up in the Q concept - it is simply a short hand which helps describe the characteristics of a second order system. All higher order systems can be broken down to sets of first and second order systems, and our problem is to make it the one that works best.

#### boris81

Hi,

I'm looking to simulate a Le Cleac'h crossover with 3rd order Butterworth filters as described in here:
Expanded Soundstaging and 3D-Imaging

I don't have the DCX2496 but I'm using a computer based DSP. I can cascade parametric filters but I don't think I'm getting the correct slopes. I was using the Q-factors mentioned earlier.

1st Butterworth = 0.5Q
2nd Butterworth = 0.71Q
3rd Butterworth = 0.5Q + 1Q

and this is what I get (1st=green, 2nd=red, 3rd=blue)
An externally hosted image should be here but it was not working when we last tested it.

but I need this

I researched this issue further and came across this Butterwoth parameters table
An externally hosted image should be here but it was not working when we last tested it.

What do the dashes on the 1st and 3rd order mean? Can Butterworth odd orders be realized with parametric filters?

#### JohnPM

What do the dashes on the 1st and 3rd order mean? Can Butterworth odd orders be realized with parametric filters?
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.

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