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-boun...@lists.scilab.org] On Behalf Of Serge Steer
Sent: Tuesday, October 14, 2014 6:36 PM
To: users@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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