roots in Simulink

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

Answer 1

[Titus Edelhofer]

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

[Cosmin Darab]

Thank you Titus

I can use the aprox -z/b, that was a lifesavier :)

Answer 2

[Titus Edelhofer]

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

[Cosmin Darab]

I don't get it should I use conv function in simulink?

because it's not working

[Cosmin Darab]

i get what you are saying but I need a positive root because I use it afterthat in a logarithmic expresion

Answer 3

[Muhammet Öztürk]

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.