Question

La viendo la pantalla de Franklin Kuonqui Ver Opciones at View Insert Format Page Shape Tools Window Help Untitled 1 - LibreOffice Draw ARAS Gs= 10 (5+10) (5+5)(S-1) His=1 E.C.:1+kG(s)H(s)=0

solve using the root path

Answer 1

Given G(3) 10 (5+10) (5+5)(51) solve using the root path on boot locus - Step 1:- Identify loci No. of poles p=3 position of poles = -10 -5, 1 No. of Zeros z=0 Loci Max (PZ) = Max (310) 3 > Total Asymptotes = 3-0 3 3 P- Z Zeroes » Center of Asymptotes I Real part of Poles -{ Real part of Oca Poz -10-5+1 -0 - 12 3 - 4.66 =

Angle of Asymptotes 29+1 poz X 180° 9 = 0,1,2 9=0 2xot! 3 X180° = 60° 4-1 0 = 0 = 2xit! X 180° 180° 9=2 O = 2X3+! X180° ? 300° Break away. and break in point 10 dk G(s) H(3) = 1 , ds=0 (stio) (5+5) (5-1) we characteristics equation know Itk a(5) H(S) = 0 ко lt M (5+19) (5+5) (31) 지 33+ 145 ²+ 355-50 + lok 33 + 145² +355-50

-> (33 +1952 +3556-50) +10k = 0 =) K -t (+1452 +355-50) = -to (352+ 285 +35) => dk ds -to (352 + 285 +35) 7-1352 + 285+35) = 0 bt √b²-4ac aa => 35²+285+35 20 -28 (28)2- 4x38 35 2x3 1. S= -1.98, -7.845 2 - 28t 19.07 6 - 28€ 19.07 - 28-19.07 6 2 6 2-1.48, -7.845 Break away point is =-1.48, Because there have 3 branches, but exists only 2 branch " from + 1 to 5 2) -10 to a 5 to 10 doesn't exist cause the righ side have Numben poles. No Break-in point cause there have no aeros. even

Intersecting point to Imaginary axis The characteristics equation is 1+ kacs) H(S) = 0 33+ 1452 +355H50 +10K) using Routh stability criteria 235 s'- 490-450 +10k) 14 14. (-58 +10k) o 52 50 (510-10%) somos . /ک 490+ 50-10k 14 (540-lok 14 500 + lok = (-50MLOK) To determine the value of k 490 + 50-10k 14 23 540 - 10K = 0 549 = 54

Applying Auxillany equation =0 145? + (-50+ lok) K = 54 => 1452 + (-50+590 ) = 0 => 145² 490 => s²= 490 14 g²= - 35 > + j 5.91 = + $ 5:9), - $ 5.91. =) 3 = Angle of departure:- 10 Here GAS) H() = TS Ho) (3+5) (3-1) so In Imaginaniy axis thein are no pole and no zero, so we don't have required to calculate Angle of departure Stability In root locus, their one pole is the right side of the real axis so the system is unstable and In Routh stability criteria, te their 1st column of the table, No. of sign chang 1. so that means system is unstable

Ing / A260° Root Locus DIAGRAM, It Jio +35.91 624.66 Re A2= 180° Re D=1 PP-10 - 591 i) P = 1, 5-10 I is bc =- slo = -1.48 4.66 Break away point i in Break away point =-1-48 I irs Intersecting 1 = (+35:91, -J 5.31) vs Exists Branch is to 5 isto to od vis system is unstable A3=300° Ing

Also I have attached root locus plot using MATLAB by the given transfer function

sys = tf([10],[1 14 35 -50]);
rlocus(sys)

Root Locus 30 0.76 0.64 0.5 0:34 0.16 0.86 20 0.94 10 0.985 35 30 25 20 15 10 Imaginary Axis (seconds") 0.985 - 10 0.94 -20 0.86 0.76 0.64 0.5 0.34 0.16 -30 -40 -30 -20 -10 0 10 20 Real Axis (seconds)