Matlab transfer function problem

[MechatronO]

I have a transfer function

L(s) = A(s)/B(s)

that I call Ls.

However, when defining

H(s)= L(s)/(1+L(s))

by

%%
Ls =
 
  -70.6s^2 - 2003 s + 3375
  --------------------------
   s^3 + 119.6 s^2 + 2806 s

%%

Hs = Ls / (1+Ls)

things go wild. The highest exponent power doubles or so and I get several extra poles or zeros. What I should get is

H(s) = A(s) / ( B(s)+A(s) )

which is the exact result. This I can achieve in MATLAB by

[As,Bs] = tfdata(Ls,'v');

Hs= tf(As,Bs+As)

and the result from this are exactly the expected.

However, why won't

 Hs = Ls / (1+Ls)

work?

Are matlabs routines that bad? Or am I doing something wrong?

 

[marcusl]

Is s a vector?

 

[MechatronO]

No,

s= tf('s');

 

[marcusl]

%%
Ls =
 
  -70.6s^2 - 2003 s + 3375
  --------------------------
   s^3 + 119.6 s^2 + 2806 s

%%
[/QUOTE]
This doesn't look like a valid Matlab statement to me. I would write it as 
Ls = ( -70.6s^2 - 2003 s + 3375 ) / ( s^3 + 119.6 s^2 + 2806 s );
(It presumes that s has already been defined earlier.)

 

[alphy]

I understand your frustration with this issue. It is important to note that MATLAB is a powerful tool for scientific and mathematical computations, but it is not infallible. It is possible that there may be a bug in the transfer function calculation routine, or that there is a limitation in the way MATLAB handles certain transfer functions.

In order to troubleshoot this issue, I would suggest reaching out to MATLAB technical support or consulting with other experts in the field who have experience with similar transfer function problems. They may be able to provide insight into the underlying issue and offer potential solutions.

In the meantime, it is always a good practice to double-check your calculations and try alternative methods, as you have already done with the [As,Bs] = tfdata(Ls,'v') and tf(As,Bs+As) methods. This approach can help identify any potential errors or limitations in the code.

Overall, it is important to remember that while MATLAB is a powerful tool, it is not perfect. As scientists, it is our responsibility to critically evaluate our results and seek out alternative methods when needed.