Low-Order Approximation of System

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
2 replies · 1K views
sandy.bridge
Messages
797
Reaction score
1

Homework Statement


I have a pretty simple question. I was going over an older exam when I encountered something that did not quite make sense to me.

If [tex]\frac{(2s+5)(-s+0.5)}{(s+3)(s^2+0.1s+0.01)}[/tex],

find a low order approximation for the system.

I understand that the pole at s=-3 can be neglected, and that we can drop the terms containing the zeros. I also know that we need to consider the DC gain of these portions when dropping those terms for low-order approximation. What I do not understand, is how the DC gain from (2s+5) term is 5/2, rather than merely 5. Wouldn't you simply plug in a zero for s?
 
on Phys.org
My approach is as follows:
Divide numerator and denominator by "2". Thus, in the numerator we have (s+2.5). Now - as a first approximation, the zero at "-2.5" and the pole at "-3" cancel each other.
This leads to a "low-order approximation" of the given function. Why do you think, that you can "neglect" the pole at "-3" ?
 
The behaviour of the system will primarily be governed by the poles that are close the s-axis (relative to the pole at s=-3). The solution reduces the equation to 5/(12(s^2+0.1s+0.01)), but I was certain that it should be 5/(6(s^2+0.1s+0.01)).

EDIT* I plotted it in MATLAB and determined that their solution is wrong. Thanks!
 
Last edited: