## Magnetude designed biquad problem !

lereme
KVRer

21 posts since 25 Mar, 2012
Okay so you was true !!

I've got to complex conjugate roots for my numerator !

r1 : (0.9578011987+i*0.1134391591) and r2 : (0.9578011987+i*-0.1134391591)

for the purpose of the exercise admit we got :
r1 : (1.05...+i*0.13...) and r2 : (1.05...-i*0.13...)

oh shit, these solutions are lying outside the unit circle. we reflect them inside th unit circle by taking :

r1 = real(r1)-i*img(r1)
--------------------
real(r1)^2+img(r1)^2

and magically we got :

r1 : (0.9578011987+i*0.1134391591) and r2 : (0.9578011987+i*-0.1134391591)

we can write (I suppose)

b2*(z-r1)*(z-r2) <=> b2*z^2+b1*z+b0

But how does we get back our coefficients ? we develop the expression and hope the imaginary parts will leave (branching) ?
Last edited by lereme on Sun Jan 20, 2013 4:55 pm, edited 1 time in total.
lereme
KVRer

21 posts since 25 Mar, 2012
Ok ! I will have a sleep it's 1:30 am in France !
Thank you very much for your help I really appreciate !

Since 2 years I was convinced that poles and zero are the coefficients themselves, for the first time in 2 years I post a question about DSP in a forum and I understand everything... (my eyes are opening !)

I had a look at your plugins ! I also made a few have a look : http://kingtao.factoryset.org

This is a really good day
Music Engineer
KVRAF

3753 posts since 8 Mar, 2004, from Berlin, Germany
lereme wrote:we can write (I suppose)

b2*(z-r1)*(z-r2) <=> b2*z^2+b1*z+b0

But how does we get back our coefficients ? we develop the expression and hope the imaginary parts will leave (branching) ?

yes, you just multiply out the product-form (left hand side) and due to the fact that the roots always come in complex conjugate pairs (unless they are real), the imaginary parts will cancel. i'm currently not entirely sure about the overall scale-factor b2 in front of the product form, though.
lereme
KVRer

21 posts since 25 Mar, 2012
Okay !

I'm back from siscilia and working hard !

Okay for the roots I understand that

I'm a bit surprised about the first formula of you paper. I don't understant how come the formula of the squared magnetude.

b0^2+b1^2+b2^2 + ...
--------------------
a0^2+a1^2+a2^2 + ...

I've tried to develop the equation (to obtain the form of a sum of geometriq suites) (I'm not sure about the english of that math term, but...)

Maybe you can explain that ?

If your are a bit bored about explaining I can totally understand !

lereme
KVRer

21 posts since 25 Mar, 2012
maybe one question more.

we can imagine modify your design to make a system with one derivative and 4 gains.

have you considered adding poles or zeros to your design, in the idea of giving more constraints ? (2 derivatives, 3 points for example)
Music Engineer
KVRAF

3753 posts since 8 Mar, 2004, from Berlin, Germany
lereme wrote:I'm a bit surprised about the first formula of you paper. I don't understant how come the formula of the squared magnetude.

b0^2+b1^2+b2^2 + ...
--------------------
a0^2+a1^2+a2^2 + ...

I've tried to develop the equation (to obtain the form of a sum of geometriq suites) (I'm not sure about the english of that math term, but...)

Maybe you can explain that ?

it can be derived by replacing z with e^(j*w) in the transfer-function and then considering H * H' (the prime denoting the complex conjugate). i've given the derivation in the appendix of my thesis, beginning at the bottom of page 84:

http://rs-met.com/documents/dsp/Magisterarbeit.pdf

although the text is german, the language of equations is universal, so hopefully it will be legible. i guess, some dsp textbooks have it as well, although i can't name one from the top of my head.

in this other thread, we discuss some more general cases:

http://www.kvraudio.com/forum/viewtopic.php?t=371738

seems like, when deriving the biquad magnitude from the general case, sines will be involved. the formulas must then somehow be equivalent due to some trigonometric identities, i suppose.
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