[Scilab-users] Problem with partial fraction decomposition (dfss)

[Tin Nguyen]

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!
>
>
>
>
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