roots in Simulink
11 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Cosmin Darab
le 9 Jan 2012
Réponse apportée : Muhammet Öztürk
le 4 Mai 2017
hy, I'm having some differences between matlab and simulink when using function roots():
"p=[1 xc b z]; xc= 0.005117782794100; b= 4.267869584093415e-008; z= -1.778279410038923e-019; roots(p);"
in matlab the result is: -0.005109429866868 -0.000008352931399 0.000000000004167
and in simulink(i put the code in a embeded matlab function): -0.005109429866875 -0.000008352927225 -0.000000000000000 I have to use the positive root but simulink does not have one. Am I doing something wrong? Does Simulink use diferent compiler? Thank you
0 commentaires
Réponse acceptée
Titus Edelhofer
le 9 Jan 2012
Hi Cosmin,
I took a look at the implementation of roots for the Embedded MATLAB Function block (<matlabroot>\toolbox\eml\lib\matlab\polyfun\roots.m). It's stated there:
% Limitations:
% Output is always variable size.
% Output is always complex.
% Roots may not be in the same order as MATLAB.
% Roots of poorly conditioned polynomials may not match MATLAB.
The last sentence is what makes you the headache (and yes, your polynomial is badly conditioned!). If you take a look at the plot you will see, that the curve hardly touches the x-axis.
I have a suggestion though: the value -z/b is a (very) good approximation of the root you are looking for ...?
Titus
Plus de réponses (2)
Titus Edelhofer
le 9 Jan 2012
Hi Cosmin,
up to roundoff error both solutions are fine. If you take the second solution and compute the polynomial within MATLAB:
r2=[-0.005109429866875 -0.000008352927225 -0.000000000000000];
p2=conv(conv([1 -r2(1)], [1, -r2(2)]), [1 -r2(3)])
% what's the difference to p?
p-p2
ans =
1.0e-017 *
0 0 -0.168584041049737 0.017782794100389
So: both solutions are equally "good"...
Titus
Muhammet Öztürk
le 4 Mai 2017
I have got a similar question; I have a code polynomial as
Ka=1.8e-5; Kw=1e-14; sigma=0.0120; ksi=0.0482; %sigma=0.0608; %ksi=0.0406;
Poli=[1 Ka+ksi Ka*(ksi-sigma)-Kw -Kw*Ka]; Hcal = roots(Poli)
Matlab gives the results as correct Hcal =
-0.0482
-0.0000
0.0000
But in simulink function block I take the answer Hcal = -0.0482 + 0.0000i -0.0000 + 0.0000i -0.0000 + 0.0000i
İn simulink I can not take the correct answer. How can I solve this problem.
0 commentaires
Voir également
Catégories
En savoir plus sur String dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!