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?
Answer & step-by-step solution
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°.
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.
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.
किसी कंट्रोल सिस्टम का अभिलाक्षणिक समीकरण s(s + 4)(s² + 2s + s) + k(s + 1) = 0 है। रूट लोकाई के अनंतस्पर्शियों (asymptotes) के कोण क्या हैं?
उत्तर और चरणबद्ध हल
रूट-लोकस अनंतस्पर्शी कोण केवल ओपन-लूप पोल और ज़ीरो के अंतर पर निर्भर करते हैं: θ_k=(2k+1)180°/(n−m)। यहाँ ओपन-लूप संरचना से तीन अनंतस्पर्शी बनते हैं, अतः कोण 60°, 180° और 300° हैं।
विस्तृत हल (मूल भाषा में)
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.
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.
કોઈ કંટ્રોલ સિસ્ટમનું લાક્ષણિક સમીકરણ s(s + 4)(s² + 2s + s) + k(s + 1) = 0 છે. રૂટ લોકસના અનંતસ્પર્શીઓ (asymptotes) ના ખૂણા કયા છે?
જવાબ અને પગલાવાર ઉકેલ
રૂટ-લોકસ અનંતસ્પર્શી ખૂણા ફક્ત ઓપન-લૂપ પોલ અને ઝીરોના તફાવત પર આધાર રાખે છે: θ_k=(2k+1)180°/(n−m). અહીં ઓપન-લૂપ રચનાથી ત્રણ અનંતસ્પર્શી બને છે, તેથી ખૂણા 60°, 180° અને 300° છે.
વિગતવાર ઉકેલ (મૂળ ભાષામાં)
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.
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.
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