1
$\begingroup$

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]
$\endgroup$

1 Answer 1

1
$\begingroup$

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

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

$\endgroup$
1
  • $\begingroup$ 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 $s^2$? $\endgroup$ Oct 14, 2017 at 15:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.