ModerateAssistant Engineer (Electrical), Class-2, Road and Building Department2025✓ Editorially verified

The characteristic equation of a control system is given by s(s + 4)(s² + 2s + s) + k(s + 1) = 0. What are the angles of the asymptotes for the root loci?

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Answer & Explanation

Correct AnswerC. 60°, 180°, 300°

Quick Explanation

Root-locus asymptote angles depend only on the excess number of open-loop poles over zeros: θ_k=(2k+1)180°/(n−m). For the characteristic equation, the open-loop structure gives three asymptotes, so the angles are 60°, 180° and 300°.

Formula / Key Relation

θ_k = (2k+1)180°/(n−m)

For n−m = 3:

k=0 → 60°

k=1 → 180°

k=2 → 300°

Detailed Explanation

Concept & Reasoning

Rewrite the characteristic equation in 1+KG(s)H(s)=0 form and count open-loop poles and finite zeros. If n−m=3, the three asymptotes are equally spaced by 120° and begin at 60°. Their centroid needs pole/zero locations, but the question asks only angles.

How to Solve It in the Exam

Three asymptotes → 60°, 180°, 300°.

Important Exam Point

Angles depend on n−m, not on the numerical value of K.

Related Revision Path

This question sits in the revision path Control Systems → Root Locus → Asymptotes .

Root-Locus Asymptote Angles explanatory diagram

Quick Trick

Three asymptotes → 60°, 180°, 300°.

Common Mistake

Using 0°,120°,240°, which corresponds to even-multiple angles rather than odd-multiple root-locus asymptotes.

Exam Tip

Angles depend on n−m, not on the numerical value of K.

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