Question

Sketch the root-locus plot of a unity feedback system. Determine the asymptotes of the root loci. Find the points where root loci cross the imaginary axis and the value of at the crossing points. Find the breakaway point. K(s+9) G(s) =- H(S)=1 s(s+2) (s+5)

Answer 1

Que) sketch the root - locus plot of a unity fedback system, determine the asymptotes of the root locie Find the points where root loci cross the imaginary axis and the value of k at the crossing points. Find the breakway point H(s)= 1, G(s) = K (+9) 818+2)(8+5) Points for the plotting_Of Root locus - 0 Origination point - Root locus will originate either from the open loop pole or infinite but the priority will be given to the open loop and the value of gain k at the origination point will be zero. Lokso xt ♡ Termination point : The foot locus will terminate either at the open loop zeron or infinite but the priority will to of gaink and the loop zero open be given the value at the termination point will be infinity. infinity. KE DK Lukao oksa a ② existance of the root locus on the real axin- root locus will exint only on that section of real axin which contains add no. of poles a zeroes towards its right side.

Breakaway points - It is the maximum value of k for which the root locus will exint on the real axin. when k exceeds this maximum value the roul locus will break into two parts in the complex conjugate location procedure to find the breakaway point- 17615) H(s)=0 K= f(s) For the given open unity feedback system - 1+ 6(S)H(s) = 0 1 + K (0+9) $10+2) 1975) (832+20 X 8+5)+k(+9) 0 8 19+2)(8+5) 2) 18+5)=0 - 0 834582+28?t100+ ks+gk = 0 $+ 78?+ 40+K)+gk=0 - K (8+) = 83 + 732+100 -K= 834 78?+ 301 Us +9) dr - do -0 + (+9) (383+140) -(034784+10s).1 (879)2 383+ 140+278+1268 - 83-702-108 = 0 253 +70°+2782 - JOD +1260 =0 20% + 34 82 +206870

207(92+175 +53)= 0_ break away pointesa 1 8=0, -4.11, -12.887 valid break away points [8=0 gain -K = 0 +7/0) + Solo) 0 +9 (K = 0. angle of asymptotes – We will calculate angle of arymptote only for those branches of foot locus which will either terminate or originate from infinity OK=(2K+12180° . p-2 where k=0, 1, 2 - Kmax kmax=P-2-1 where p in the total no. of open doop poles. z in the total no. of open loop zerosas For the given unity fodback system OK = (2K+1) 480" (2K+1)1809 = (2x +1) 90° 3-1 K=0,1 100 = 900, 270 Centooid :-> DC = EP-kz - -2 where, sp in the sum of the location of real part of the open loop poles the sum of the location of real part of the open loop "zern 2

K200 CO 170 1 Kao - mok**** s=9 ses 8=-2 I 270 31 (centroid) eo V centroid = 0-2-5-(-9) = -7+9 = 1 2 3 - 1 For the points where root loci cross the imaginary axina 831 I (10tk) ук so the root loci goes infinity on the Imaginary axine 7 (10tk)-g 9K 7(10+K)-9 = 0 7 70+ 76 -9=0 7K 4-61 Kmar = -61 = -8.714 2 782+ gk = 0 ző + 9x-8.719=0 702-9x8.71% 82-11-20 58=-3-34