Question

Question 4 a) A simplified form of the open loop transfer function of a chemical process as shown in Figure 4 is 1 G(s) = s(s + 3)(s2 + 2s + 2) where H(s) = 1. Construct the root locus. (5 marks) ii. From the root locus in (a), show the range of gain K at the root loci so that the system will stable. Discuss your answer. (2 marks) i. b) State two important parameters that you got from Bode plot method and explain how you got it theoretically. Next, state the differences between Bode plot and Root Locus method. (3 marks) [10 MARKS] U(s) + Y(s) KG(s) H(s) Figure 4

Answer 1

4 1 G1S) = HIS) = 1 (unity feedback) 5(8+3) (s2+25+2) K open loo = KGIS) = SIS+3) (52+25+2) 0 Root locus is always symmetrical about the real axis. the branch of root locus always start from open loop bole with K=0 and ends either at open loop zero infinite as k=00. 3 4 poles poles :- S=0,-3, -151 zeros ! - No zeros P7Z root locus is equal to the the number of branches of number of 6 Angle of Asymptotes: - Number of Asymptotes = P-z pales. - 4-0 54 ФАЗ (22+1)x180 p-z 4-0,1,2,3 q=0 dis 180 q=1 (2x1+1)x180 q-2 43 = 2*2 +1)x180 q=3 $=(2x3+1)x180 0,45 02= 135 &3 = 225 *4 = 315

Centroid: - centroid is the point of intersection of asymptotes on the real axis. E sum of pole I sum of zeros GA P - Z 0-3-1-1-1-1-o -5 =-1:25 6A= 4 4 intersection of root locus with imaginary axis! characteristic equin 1 + GIS) HIS). -0 K 1+ s (+3) (5?+25+2) s(s+3)(82+25+2) +K=0 s($+25+25+352 +65+6) +K=0 s( 53+552 +85 +6 )+K=0 st+553+852 +65+k=0 R-H criteria '- K se 5x8-6 53 5 =6.8 6 s² K s² 618 6.886-sk 6.8 s' 40.8-5k 6.8

(1) for stability in Kyo 15-8 ob 70 8.9 5K < 40.8 K L 8.16 sange of stability in 0 < K<8:16 Cross the imaginany ar inte 9-1+S 8.9 9-918+458.9 a-feil tus) 8.9 21-5 [s. iw Szt11.095 twati 1.085 W=1095 and see Root locus tw Centopid it X 315