[Scilab-users] Problem with partial fraction decomposition (dfss)
Serge Steer
Serge.Steer at inria.fr
Wed Oct 29 19:25:38 CET 2014
Le 29/10/2014 17:01, Tin Nguyen a écrit :
> Thank you so much, Serge Steer!
> Instead of finding a function like 'residuez' in Matlab, now I need to learn
> many other things. It's not comfortable.
> Hope Scilab will become more convenient for general users like me! :D
Take care that the Matlab function residuez does not produce exactly the
same thing
first the residuez help pages states
"/Numerically, the partial fraction expansion of a ratio of polynomials
is an ill-posed problem. If the denominator polynomial is near a
polynomial with multiple roots, then small changes in the data,
including roundoff errors, can cause arbitrarily large changes in the
resulting poles and residues. You should use state-space (or pole-zero
representations instead. /"
Moreover this function gives you the residual and you have to decide if
some roots are equal or not, to be able to compute the fraction
decomposition and this is not a simple thing to do.
If you do something like
P=poly(ones(1,6),'x','r') //(x-1)^6
r=roots(P);
clf;plot(real(r),imag(r),'*')
You can observe that the roots are not exactly equal to one but located
on a circle of radius approx %eps^(1/6)~=0.002
As noticed in the residuez help page *the problem is ill-posed*
to be convinced just compute the fraction decomposition of 1/(z-1)^6
using residuez and computing back the fraction
>> [r,p,k]=residuez(1,poly(ones(1,6)));
>> [b,a]=residuez(r,p,k)
b =
1.0e+07 *
Columns 1 through 4
-0.5038 + 0.0008i 2.5335 - 0.0041i -5.0956 + 0.0081i 5.1243 -
0.0081i
Columns 5 through 6
-2.5765 + 0.0040i 0.5182 - 0.0008i
a =
1.0000 -6.0000 15.0000 -20.0000 15.0000 -6.0000 1.0000
*The problem is ill-posed,* *so it should be better to avoid its use* .
What is the problem you want to solve?
Serge Steer
>
> -----Original Message-----
> From: users [mailto:users-bounces at lists.scilab.org] On Behalf Of Serge Steer
> Sent: Thursday, October 23, 2014 2:22 AM
> To: users at lists.scilab.org
> Subject: Re: [Scilab-users] Problem with partial fraction decomposition
> (dfss)
>
> Now you are using a too big value for rmax
> -->pf = pfss(h)
> pf =
>
>
> pf(1)
>
> 0.75 - 0.25s
> ------------
> 2
> 1 - 2s + s
>
> pf(2)
>
> 0.25
> ----
> 1 + s
> Some explanation
> pfss converts the transfer function to state space representation and then
> trie to diagonalize the state matrix using the bdiag function
>
> -->S=tf2ss(h);
> -->bdiag(S.A)
> ans =
>
> 1.0000000 - 1.2247449 0.
> 0. 1. 0.
> 0. 0. - 1.
> You can see with result above that there is a 2 by 2 jordan block (for the
> eigenvalue 1) which exhibits a coupling between the two states Now with
> -->bdiag(S.A,1/%eps)
> ans =
>
> 1.0000000 0. 0.
> 0. 1. 0.
> 0. 0. - 1.
> The Jordan block has been broken
>
> [Ab,X]=bdiag(S.A,1/%eps);cond(X)
> ans =
>
> 5.074D+14
> bdiag has broken the jordan block using a singular "base change" so the
> decomposition is a non-sense.
> Serge Steer
>
> Le 21/10/2014 12:07, Tin Nguyen a écrit :
>> Dear Serge Steer,
>> Thank you very much for your response. Your code works fine.
>> However, I still have some problem with this pfss function. For
>> example, I try to decompose a fractional polynomial that has multi-order
> poles.
>> H(z) = 1 / { (1-s)^2 * (1+s) }
>>
>> You can find following from my console:
>> -->s = poly(0,'s');
>>
>> -->h = 1 / ( (1 + s) * (1 - s)^2)
>> h =
>>
>> 1
>> -------------
>> 2 3
>> 1 - s - s + s
>>
>> -->pf = pfss(h, 1/%eps)
>> pf =
>>
>>
>> pf(1)
>>
>> - 9420237.9
>> -------------
>> - 1.0000000 + s
>>
>> pf(2)
>>
>> 9420237.6
>> ---------
>> - 1 + s
>>
>> pf(3)
>>
>> 0.25
>> ----
>> 1 + s
>>
>> I also checked:
>> -->pf(1)+pf(2)
>> ans =
>>
>> 0.75 - 0.25s
>> ------------
>> 2
>> 1 - 2s + s
>>
>> I have no idea about this situation.
>> Can you help me out! Thank you in advance!
>>
>> --Tin Nguyen
>>
>> -----Original Message-----
>> From: users [mailto:users-bounces at lists.scilab.org] On Behalf Of Serge
>> Steer
>> Sent: Tuesday, October 14, 2014 6:36 PM
>> To: users at lists.scilab.org
>> Subject: Re: [Scilab-users] Problem with partial fraction
>> decomposition
>> (dfss)
>>
>> You just need to tune the rmax optionnal parameter pf =
>> pfss(tf2ss(hz),5);length(pf)
>>
>> Serge Steer
>> Le 13/10/2014 10:12, tinnguyen a écrit :
>>> Hi guys,
>>> I'm having trouble using scilab's function pfss to decompose a
>>> polynomial in order to do inverse Z transform. Following is my code:
>>> --> z = poly(0,'z')
>>> --> hz = z^2 / (1 - 6*z + 11*z^2 - 6*z^3) pf = pfss(tf2ss(hz))
>>> --> length(pf)
>>> The fourth command returns '1'. However, I believe it should be 3. I
>>> also did the decomposition manually and got the result as: hz =
>>> 0.5/(1-z) + -1 /
>>> (1-2*z) + 0.5/(1-3*z).
>>> Nevertheless, if I change the numerator of hz from z^2 to z or z^3,
>>> pfss works correctly. I definitely have no idea!!!? :D Please take a
>>> look and correct me if i'm wrong.
>>> Thank you so much!
>>>
>>>
>>>
>>>
>>> --
>>> View this message in context:
>>> http://mailinglists.scilab.org/Problem-with-partial-fraction-decompos
>>> i tion-dfss-tp4031331.html Sent from the Scilab users - Mailing Lists
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