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Problem 4 the closed-loop characteristic equation 1 KG(s) = 0 The open-loop transfer function is G(8) poles/zeros marked imaginary axis crossing and K root locus K→∞0 ← axes correctly labeled correct K values marked 3m(s) K=0 Ko0j2 (8² + 4) (8 + 1)³ √3 K 8 K00 j2 correct real-axis branches arrows on branches j√3 K=8 Plot the root locus for Re(s) correct complete

Problem 4 the closed-loop characteristic equation 1 KG(s) = 0 The open-loop transfer function is G(8) poles/zeros marked imaginary axis crossing and K root locus K→∞0 ← axes correctly labeled correct K values marked 3m(s) K=0 Ko0j2 (8² + 4) (8 + 1)³ √3 K 8 K00 j2 correct real-axis branches arrows on branches j√3 K=8 Plot the root locus for Re(s) correct complete

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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please explain the steps for sketching this root locus by hand 

Problem 4 the closed-loop characteristic equation 1 KG(s) = 0 The open-loop transfer function is G(s) poles/zeros marked imaginary axis crossing and K root locus K→∞0 axes correctly labeled correct K values marked 3m {s} K=0 K→∞0 j2 8₁ K →∞0 j2 1 K +3 8-K K +3 4K 1 j√3 K=8 Characteristic polynomial: (8-1)³ K(s²+4)=8³(3+K)8² +38 +1+4K Routh-Hurwitz: 3 (s² + 4) (s + 1)³ 4K +1 0 j√3 K = 8 0 correct real-axis branches arrows on branches Plot the root locus for Re{s} correct complete closed-loop system stable for 0.25 <K <8 ju-axis poles for K = 8: Auxiliary polynomial: A(s) = (K + 3)² + 4K +1=11s² + 33=11(²+3) Then ju-axis poles at 1j√3
Transcribed Image Text:Problem 4 the closed-loop characteristic equation 1 KG(s) = 0 The open-loop transfer function is G(s) poles/zeros marked imaginary axis crossing and K root locus K→∞0 axes correctly labeled correct K values marked 3m {s} K=0 K→∞0 j2 8₁ K →∞0 j2 1 K +3 8-K K +3 4K 1 j√3 K=8 Characteristic polynomial: (8-1)³ K(s²+4)=8³(3+K)8² +38 +1+4K Routh-Hurwitz: 3 (s² + 4) (s + 1)³ 4K +1 0 j√3 K = 8 0 correct real-axis branches arrows on branches Plot the root locus for Re{s} correct complete closed-loop system stable for 0.25 <K <8 ju-axis poles for K = 8: Auxiliary polynomial: A(s) = (K + 3)² + 4K +1=11s² + 33=11(²+3) Then ju-axis poles at 1j√3
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