Question

PrtSc Insert Delete Num Lock 7 Hon Q13. Using root locus, determine the following for the close loop system below. [6 Marks) 6(6):6+3) G(s) = st 4 s(s + 2) Y(5) Hoo G(s) i. Open loop transfer function poles and zeros [2 Marks] ii. On the above template, neatly sketch the root locus of the system [3 Marks] un iii. Is the system stable for all values of K [1 Mark] YES YES NO Faculty of Engineering Technology and Science Page 9 of 17

Answer 1

Given trems for function. * ** 1494 1 2) Gls) = S +4 s(s+2) 1. Open loop soros, roots of numeratos is eros ily, S+4 =0 - ne., s=-4 is zero. open loop poles, roots of demoninatol is polis , 0 = نے میاء ren, sro & SP-4 polos. 1. 2. Root locus Steps!!- Root locus on real axis, Represent pols of acros on splane - Cie., sum of) Coming from oo to wards, if no of combination R exists. of poles & 30 Root locus otherwise not doesn't exist 1 of poles & geros is odd on particular section,

polis-0 polos 2 30- sum=3 Rols=2! poles / 368052o, 3uros=0 sum=2 I sum=1 el doon't RL exists 3croso RL disn't exist X - pole Cpole represnts with x) 0 - zero (acro represuts with 'o) I RL exist exist -2 -4 * Note! Every RL bromch starts from (pole & înfinily) endsat (aero of infinity see arrows in bronches. Centrioce (a) = sum of real parts of poles no of sumof real parts seros pols- nooot acros . c = (0-2) - (-4) 2y = 2 180 = Angle of assymtotes (zert!) 18 a-1 = 180 (201 +1) = 360° acros hence Angle of arraival / Angle af departure No complex poles / No complex doesn't exist

Break in/ Breakaway point ® At this point RL split into two bronchus It com be find by doko Consider charcuterstic equation, I+ GLS) HCS) = 0 It K(s+4) S S(ST2) 45 s(s+2) + K (5+4) = 0 kls+4) = - S(5+2) k= -s(572) (8+4) 7 (5+2)(s+4) +s(s+4) – s(s+2) = 0 → 5276s+8+5745 -s? 25 = 0 > S +85+ 8 = 0 = Suvz = -1:1716, -6.828 Crossing passong points on jw . No crossing points on jw since No assymtore going imaginary axis side.

Complete Root local quico @ -70/716 K-6.820 -4 'break away point (comis blew two poles) Breakin points (comos blw two acros. Stabiloly By seeing RL, system unstable on if and Right only if RL exists Half of s-plane It does exists in Rit of s-plane but one pole at seos so system is critically stable for k=0 stable for all k o