Those equations are bad - for comparision see:

Research Využití aproximačních funkcí pro kaskádní syntézu filtrů

http://www.rfcafe.com/references/electrical/butter-poles.htm

Those equations are bad - for comparision see:

Research Využití aproximačních funkcí pro kaskádní syntézu filtrů

Following with interest

Aparna Sathya Murthy - thank you for your response, but unfortunately, I could not find any answer to my question.

The table of poles in the s-domain Prof Lutz von Wangenheim

Dear Lutz,

welcome,

Please Look at the following post. It may satisfy your requirement. Butterworth poles lie along a circle and are spaced at equal angular distances around a circle.

Pole locations pk are calculated as follows, where K=1,2,...,n.   n is the filter order.

pk= - alphak + jbeta = - sin (pi (2k-1)/ 2n) +jcos (pi (2k-1)/ 2n).

The quality factor of the pole pairs according to their similarity with the resonance circuit response Qp= 1/2 alpha,

and consequently the overall Qt = product of Qpk for all k values of the filter.

So one can get an expression for the quality factor Qt in terms of the alpha pf the poles as:

Qpt= product of ( 1/2 alphak) for k =1:n

= Product of ( 1/2 sin(pi (2k-1)/ 2n) for k=1:n

One can calcualte Qpt for even order filters and find it =0.707 and odd order filters and find it .5 as you observed it.

Best wishes

@Aparna Sathya Murthy : Thank you, but these tables are not new for me. They can be found in each relevant textbook...

Dear Abdelhalim.....yes, I think we are approaching the same line....

For a unit circle, we always have a pole frequency wp=1 - and the pole Q depends on the cos (alpha) only.

Hence, Qt involves the product of cos functions only.

Using cos (x)*cos(y)=cos(x+y)+cos(x-y) we can show that, indeed, for second order Butterwirth functions (inserting the known Qp values) the Q-product is Qt=0.7071.

But I am looking for a general prove (without inserting actual Qp values) independent on the order of n. This should possible!!

Final comment/question: For the Q-product you have chosen the term "overall Q". But I think this product has a mathematical meaning only without any practical relevance. Otherwise, a similar term (with a fixed value) should exist for all other lowpass approximations (which is not the case).

Today - I can give some more detailed information about the mentioned (mathematical ) problem:

1) I am sure that - of course - the main reason for the observed property of Butterworth polynominals is that the poles are located at a unity circle - equally spaced!! Therfore, for even orders we can identify a certain symmetry below and above the 45 deg line in the upper left quarter of the s-plane.

2.)Symmetry properties: The real part of the lowest pole (P1 with lowest Q1) is identical to the imag. part of the largest pole (Pn with the largest Qn).

3.) Therefore, using the cos and sin function we can find a kind of "pairing" by expressing Qn with Q1...and so forth.

4.) More than that, applying some mathematical addition theorems, the product of several Q-pairings (expressed as a product of cos-functions) can be reduced to a relatively simple expression.

5.) EXAMPLE: n=12.

For a Butterworth function (n=12) we have 6 poles in the upper left quarter of the s-plane forming 3 symmetrical pole pairs.

With Q=1/2cos(phi) we need the product

Q1*...*Q6=1/64[cos(phi1)*.....*cos(phi6)]=1/64[P1P2P3]

With the products (and using the symmetry properties, see point 2)

P1=cos (phi1)cos (phi6)=0.5*sin(2phi1)

P2=cos (phi2)cos (phi5)=0.5*sin(2phi2)

P3=cos (phi3)cos (phi4)=0.5*sin(2phi3)

We arive at a general formula

P1P2P3=(1/32)[2sin(2phi2)*cos(2phi3-2Phi1)] .

With phi1=7.5° ; phi2=22.5° ; phi3=37.5°

we have

P1P2P3=(2/32)*0.7071*0.5 and finally

Q1*....*Q6=1/64[P1P2P3]=1/(2*0.7071)=0.7071 q.e.d.