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Butterworth function help request (mathematics)

george2525

Jan 30, 2015
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Hello,

I have been having issues with the butterworth squared magnitude function for a while. I understand it enough to use it but its derivation is causing me a headache and realy bothers me. If anyone understands it mathematically it would be good to get some pointers.

upload_2018-12-23_10-51-0.png
upload_2018-12-23_10-51-45.png
here are the typical Butterworth formulas in mosts texts.

Then comes the step where we should find the poles in the S domain. The following relation is often used:
upload_2018-12-23_10-53-53.png
I understand this

But what causes me confusion is that this is only true if s = jw. IE we have discarded the real parts of any poles (which we need)

Then the following is used to formulate the pole locations (on a circle)
upload_2018-12-23_10-56-47.png
Here is where I get lost. If we set w = S/j then doesnt |H(jw)|^2 become |H(S)|^2 ?

And if not, then why not? it is never written that way.

Also if we have already discarded the real parts of any poles by initially setting S = jw, then how do they magically re-appear in H(s)H(-S) to trace the classic butterworth pole circle?

upload_2018-12-23_11-2-13.png

Maybe I have missed a mathematical rule somewhere?

I hope this question makes sense. If you are willing to help and need further clarification, please ask. Thanks to anyone who can help.
 

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Harald Kapp

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this is only true if s = jw.
This is a typical simplification made (and valid) for steady state signals.
If we set w = S/j then doesnt |H(jw)|^2 become |H(S)|^2 ?
It does. And by the same reasoning as in this equation:
upload_2018-12-23_10-53-53-png.44286

where |H(jω)|2 = H(jω)*H(-jω) it can also be written as |H(s)|2 = H(s)*H(-s) = |H(jω)|2
I see no contradiction here.
And if not, then why not? it is never written that way.
Is it not? I don't know and I do not have an answer to that question.

how do they magically re-appear in H(s)H(-S) to trace the classic butterworth pole circle?
That is maybe due to sloppiness of the writer(s). It's simply easier to write 's' instead of 'jω' (I know because I had to use the windows symbol table to copy a small omega letter for this text).[/sup]
 

george2525

Jan 30, 2015
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It does. And by the same reasoning as in this equation:
upload_2018-12-23_10-53-53-png.44286

where |H(jω)|2 = H(jω)*H(-jω) it can also be written as |H(s)|2 = H(s)*H(-s) = |H(jω)|2
I see no contradiction here.

Thanks for the reply.

OK lets say H(S) = S
where S = A+jB

H(S)*H(-S) = (A+jB)(-A-jB) = B^2 - A^2 - j2AB

but |H(S)|^2 = A^2+B^2

which is not the same!

My point being that they are only the same IF A = 0

and so somewhere in the theory the real part of our poles (the A's) are brought back into the equation so we can properly place our poles. This is what I dont understand.
 

Harald Kapp

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OK lets say H(S) = S
where S = A+jB

H(S)*H(-S) = (A+jB)(-A-jB) = B^2 - A^2 - j2AB

but |H(S)|^2 = A^2+B^2

which is not the same!
It is the same for s=jω as in that case A=0 and your equation reduces to
H(S)*H(-S) = (A+jB)(-A-jB) = B^2 - A^2 - j2AB = B2
compared to
|H(S)|^2 = A^2+B^2 = B2
which is identical. Keep in mind that the condition under which this holds true is s=jω which is valid for steady state signals only. In a strict sense there are no steady state signals as any practical signal has a time where it starts and a time where it ends. On the other hand this is a simplification that holds true when the duration of a signal is longer than the settling time of the system under consideration. A sufficiently long non-steady signal can then not be discerned from a steady signal. From the point of view of the system the signal is steady.

The mathematics as used in engineering obviously does not always follow the strict mathematical regimen - engineers are lazy, I should know ;)
 
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