Recovering a missing transfer function from options: Assume the process open-loop transfer function is G(s) = K / ((s + 1)(s + 4)). In the corresponding root locus, the open-loop poles are located at which points on the real axis?

Difficulty: Easy

Correct Answer: -1, -4

Explanation:


Introduction / Context:
The original stem omitted the explicit transfer function. Applying the Recovery-First Policy, we supply a standard second-order real-pole form G(s) = K / ((s + 1)(s + 4)) that is consistent with the provided multiple-choice options. Root-locus analysis begins by marking the open-loop poles and zeros in the s-plane; their locations define the starting points of the locus branches.



Given Data / Assumptions:

  • Recovered open-loop transfer: G(s) = K / ((s + 1)(s + 4)).
  • No finite zeros.
  • Standard negative feedback configuration.


Concept / Approach:
Open-loop poles are the roots of the denominator of G(s). For G(s) = K / ((s + 1)(s + 4)), the denominator is (s + 1)(s + 4). Setting it to zero gives s = −1 and s = −4. These are the starting points for the root-locus branches as K increases from 0 to ∞.



Step-by-Step Solution:

Denominator: (s + 1)(s + 4) = 0.Solve: s = −1 and s = −4.Therefore, open-loop poles at −1 and −4 on the real axis.


Verification / Alternative check:
Sketching the root locus shows two branches starting at −1 and −4, with breakaway on the real axis between them; asymptotes approach angles ±90° as K → ∞ since there are two poles and zero zeros.



Why Other Options Are Wrong:

  • Positive real locations (1, 4) would indicate unstable open-loop poles.
  • Other negative pairs (−2, −3 or −0.5, −8) do not match the denominator factors (s + 1)(s + 4).


Common Pitfalls:
Confusing closed-loop pole locations (which move with K) with fixed open-loop pole locations; the latter come directly from G(s) at K = 0.



Final Answer:
-1, -4

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