1

I've found the closed loop transfer function but I want to know the order of it. So, the idea is to factor the denominator by the highest exponent of s to get an expression like this one: $$ l(s)=g(s)r(s)=k \frac{b_ms^m+b_{m-1}s^{m-1}+\dots+1}{s^N(a_{n-N}s^{n-N}+a_{n-N-1}s^{n-N-1}+\dots+1)} $$

I tried collect, factor terms but none of them seem to work for my purpose. 

Collect[FullSimplify[(3.72 10^9 s - 5.73 10^6 s^2 + 57904 s^3 + 
     2.79 10^11)/((s + 64.4) (270 s + s^2 + 246193) (170 s + s^2 + 
       37812))], s]
Miguel Duran Diaz
  • 355
  • 1
  • 7

1 Answers1

1

It looks like you're looking for the leading term of the series expansion of your polynomial around the expansion point 0. One idea is given in my answer to Get leading series expansion term?:

expr = (3.72 10^9 s-5.73 10^6 s^2+57904 s^3+2.79 10^11)/((s+64.4) (270 s+s^2+246193) (170 s+s^2+37812));
expr /. s -> s + O[s]^2 //TeXForm

$0.465386-0.00362407 s+O\left(s^2\right)$

Here is another example:

expr = (1 + s + s^2)/(s^3(1+s)(1+s^2)(1+s^3));
expr /. s -> s + O[s]^2 //TeXForm

$\frac{1}{s^3}+O\left(\frac{1}{s^2}\right)$

Carl Woll
  • 130,679
  • 6
  • 243
  • 355
  • I'm sorry but I'm not familiar with that notation $O(s^2)$. In your example there, when it appears $O(s^2)$ does it mean the leading coefficient is $s^2$? – Miguel Duran Diaz Oct 14 '17 at 15:13