Question

Due Date: April 20, 2 Problem 2: Consider a unity-feedback control system with the following open-loop transfer function: K G(s)H(s) = s(s2 + 4s + 8) 1. Sketch the root-locus plot. 2. IfK 2, where are the closed-loop poles located? 3. If x = 0.5, where are the closed-loop poles located?

Answer 1

o G(9H(s) - s( 37+45+8) open loop poles P= 3 (0, -2+2? , 2-22) open loop zeros ZEO thê ging! Cem tooid - Ere CP - E Re(z) Rel!) P-z poze 2-0)-o - 2 = centroid * Angle of asymptote & = (22+1)180 ..(p-2-1) p-2 99.0,1,2 Op 18 - 0=60, 180, 360° * Angle of departure : - (+p-92) 006-2+2)) = 180 - (125790) = -45° 0p(-7-3i) 180 - (-135-90) - Yos" = +45 -(45-45) = 180 + 18. - Als -2-2

me point of interseeting at sw-exis is charruetezéstic equation = s(3+45+8) +* = +45+2stko Routh anty: 1 K s 32-K o s' Y K at K:32 50 system is marginally stable Anxilang equation at K = 32 és : 4²+kao 4$+32=0 s- jw = 1.842.828 so, point of intersection at jw-axis are & i2.828 Root locus Jaun -2+2h • Rels) s:-12828

9 2 G(5)H(s)- $(3+45+8) Characteristics equation = $(3+48+8) +222* $} +45°+8520 close loop poles = mots of chemacteristics equaction 3- closed top poles 25: -0.2886 I'ds 50+ 1.866 -1.8550-8.866 3 sp=0.5 :- GLS) (s) (P245) Now: 2 2 W, = 4 2x0.5x0,4 - Ws = 4 Home 16 Charcuteri's tics 662M (3) s($+45+10) equation: 8/5² + 4s +16) +16 = $?+45? +165 +16 close loop poles = roots of characteristics equation of -1.2777 -1,361 + 3.266 À -1.361-3.266 Ans