Question

B3. Sketch the root locus of the control system shown in the figure below K s(52 +65 +25)

Answer 1

Solution: K 5616586 (openloop transfer function i.e, G (s) k scs&+65 +25) 9Gcs)H(5) = k sC65 +25 HCS) = step!- find the poles & Zeroes GCS) k SC65+25) for poles, SCS +65 +25) = 0 Sco 65 +25 - 0 + 1 6758 S= -6 + 36 - Hx25 2 S=-6 + 36-100 2 S=-3t jy The poles are, S= 0; -3+j4, -3-34 Number of poles = @ 3an Numbed of Zeroes = o = m

stepa- zenges GCSDHS) Centooid - Edealpart of {real past of Files GCS) HG) number of Poles number of zegoes U stepsi- 0-3-3-0 -6 -2 3-0 3 recentooid = -2 Angle of asymptotes 22+1180° number of poles where, q=0,123..6-m-1) -number of Zeroes q=0,1,2, so 9=0 =>QxO+1) 180° 68 ie, =iso 3 3 3 31 q=1 =) (@xl +) 180° -) 3x180 = 180 Gora 9-2 => 2+1 180 5x tso = 300° 3 stephi Any doot locus starts from poles and ends åt zexoes. * poot locus is found in a section of the real axis only if total number of Poles and zegoe's to the right side of section is odd if kso.

-3454 X- lju Conlyimaginary potes) 1 ја 180 160 -ja -3 300 Xes -3-34 -jy in tersect at the Points woite Call asymptotes centroid) steps for breakaway characteristic equation. characteristic I + GCS)H(s) =0 equation . 1+ k SC 65+25) 3 658+a5stk=0 NOW, 3 -s 65"-255 oto (33 + las +25] 35+125 +25 =0

S= -lat PHH-300 * Breakaway point always between 6 = -124 j 12.49 6 =) -24 j2.081 * Breakas Consecutive poles Steps To find out imaginary COSS Ovedi assay Characteristics equation :- 37652+255 +K=0 axis have we to form south's 1 ñ ir 25 ऽ२ K mm dh in an 216 15x 6-15 Kusab solk 50 ik make s' co-efficient equal to zed Alvidad geguration. 25x6-k O 6 K= 150 ₂ kmax.

with steppi-doots of auxiliady Polynomial give inteo Section of soot locus imaginady axis. - Auxillary equation. 68&tk=0 5=ts =+j - +;5 150 6 steps Angle of depasture = 180°+ 246,)H(3) Conly applicable for poles) <GCS), HC5) = { Angle of 5, from zedes - Angle ofs, foom -3+j4 Poles. for -3+j4 Pole, Ist 9=900 & -3, da -jy , = 180°- tancy 3-14 0;= 126.869 180 - 53-13 126.869 da 90

CGCS), HC5,) O - [90+126869] -216.869 Angle of departure = 0o = 180°+ Casjaş) =) 1800-216.869 Do=-36869 Similarly for another conjugate pole, do -36.8690. ie-3-j4

Root locus diagram (15-150) iw -3+j4 -jy 713 -2-j2.081 . Hja 180 ال۔ 68 - 3 2 H-j1 3009 |-ja -ja.com 81 -j3 Xe -3-j4 -j4 35 fig :- doot locus.