tr= 1.21 (s)Ge(s) =Y(S)MP/ - 16-3/-ess)%3D9= 0:5Wn= 2 rad Isecondels) Pp(s) - (s)%3D%3DSd = -1t 13jDeteumine) angle of deficiency(ii) Location of pole Cund zeioofCompensat or Compute T and xUse compensaton and apply unit negatwefedback.Anower:(i) - 60°(i) poles at s: -4zeo at s =-1| = 1 x = 0- 25

Given: Specifications are tr=1.21 sec, MP%=16.3 %, ξ=0.5, ωn=2 rad/sec Open loop transfer function…

tr= 1.21(s) MP./ = 16.3%. G=0.5 Wn = 2 rad /second Sd = −1+√³j Determine (i) angle of deficiency (ii) Location of pole and zero of Compensator Compute I and x Gc(S) = Y(s) e(s) e(s)=(s)-(s Use compensation and apply unit negative feedback. Answer:. (1) -60° (ii) Poles at S= -4 Zero at S=-1 T= 1 α = 0.25

tr= 1.21(s) MP./ = 16.3%. G=0.5 Wn = 2 rad /second Sd = −1+√³j Determine (i) angle of deficiency (ii) Location of pole and zero of Compensator Compute I and x Gc(S) = Y(s) e(s) e(s)=(s)-(s Use compensation and apply unit negative feedback. Answer:. (1) -60° (ii) Poles at S= -4 Zero at S=-1 T= 1 α = 0.25

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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