Small/easy Challenge

Hi Clay,

>
>Partial fraction decomp allows you to write the factored 3rd order s
equati=
>on as a sum. Plug in your bilinear transform at this point and reduce the
a=
>lgebra.
>

I'm not sure you're interested in this at all but I think that your
proposal of using partial fraction decomposition is not as 'easy' as your
sentence seems to indicate. It does not seem to be easier that just trying
to get to the allpass decomposed form directly from the non-decomposed form
of the transfer function. The issue is that after the polynomial division
there is this constant term (not dependent on z) and there is no 'easy'
way, at far as I can tell, to put this term into the first and second order
terms such that the allpass form comes out.

wolframalpha and the symbolic toolbox in matlab (2004 version) spit out the
partial fraction decomposition in milliseconds but so far I have not been
able to make them spit out the allpass decomposition. I'm not sure if the
reason is due to some assumptions not being made or whether the symbolic
engines don't find the allpass form 'interesting' enough to display. Using
pencil and paper is soooo last millenium ;)

Anyway, getting the expression for the coefficients is indeed very easy as
you said as they can be read directly from the factored form of the
denominator after doing the bilinear transformation.
cheers	 

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Hi again Clay,

>Hi Clay,
>
>>
>>Partial fraction decomp allows you to write the factored 3rd order s
>equati=
>>on as a sum. Plug in your bilinear transform at this point and reduce
the
>a=
>>lgebra.
>>
>

Please disregard my last post. The partial fraction decomposition should of
course be done before the bilinear transformation. 
cheers

	 

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On Friday, September 27, 2013 6:31:46 AM UTC-4, niarn wrote:
> Hi again Clay,
> 
> 
> 
> >Hi Clay,
> 
> >
> 
> >>
> 
> >>Partial fraction decomp allows you to write the factored 3rd order s
> 
> >equati=
> 
> >>on as a sum. Plug in your bilinear transform at this point and reduce
> 
> the
> 
> >a=
> 
> >>lgebra.
> 
> >>
> 
> >
> 
> 
> 
> Please disregard my last post. The partial fraction decomposition should of
> 
> course be done before the bilinear transformation. 
> 
> cheers
> 
> 
> 
> 	 
> 
> 
> 
> _____________________________		
> 
> Posted through www.DSPRelated.com

I'm glad you got it working.

Clay

>
>I'm glad you got it working.
>
>Clay
>

Me 2 :), thanks

How would you then rate this question (easy, medium, not fun, not
interesting, just lame, etc).

Why do both of the below systems of 3 equations give the solution to
problem? What do they correspond to?
(Note that the two systems of equations are not equivalent but they share
the same real solution!)


(u1+v1)*(2*K-1) + (v2+v1*u1)*(2*K+1) + v2*u1*(2*K+3) = -2*K+3 
(v2+v1*u1)*(1 - K^3) + (u1+v1)*(-K^3-1) + v2*u1*(-1-K^3) = K^3-1
(v2+v1*u1)*(1+2*K^2) + (u1+v1)*(1+2*K^2) + v2*u1*(-3+2*K^2) = -2*K^2+3


u1+v1/v2 = 3*(1-K^3)/(1+K^3)
1/v2+u1*v1/v2 = 3
u1/v2 = (1-K^3)/(1+K^3)	 

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