Question

Problem 3 (30 points) Given the following unity feedback system we wish to sketch the root locus of KG() = -16+-10) for 0 < K<0. (a) Indicate the following on the above s-plane (show all your works): 1) (2 points) Finite poles and zeros of G(3) ii) (2 points) real axis section of root locus i.e. real axis roots) m) (4 points) departure angles and amival angles if any

Answer 1

solution Grisani 6446) MAY (541) Guirer KG H (6) OLK SO s(5+2) (5710) Part ② 0 Poles pzz: 550, 5=2, S5-10 Keros 750, 1p2=3] -- -- 0 @ Real anis section of root locus: Alal ascis dection is part of Rool Locus j sum of poles and tends to the rightes section o an odd number Retua righ * since we do not have any complex polesteros Angle of Departure Arrival o not applicable for this particular Tolosje Function ® Break away and Breakcin poort general concepti A break away points enests betcullen tewo adjacent poles is the segment between them is past of root locus.

From Figli, we can conclude that these exists at least ome breakaway pent between 1 s=-2 and são Now to obtain breakaway point we merke at so Writing the characteristic Equilton it KG HIS) =0 7 17:16:42] (5710). To Ks - 5(5+2)(5+10) <- 15152+125+20)] k=-{53 + 125 + 20s Now 4-0 * 4-[5971252-42057 20 → 352 +245 +20=0 or s4 185 700 € • S= -247)(2452-4137120) - 6 * S= -0.95, -7.06 Went we check the validity of Drealcaway pent From Pig), we conclude that since s2-7.00 is not on the root loters → it cannot be a valid break away

point Hence 1s=-0.951 is a valid break away point Nort: Break-in penila : VA o Angle of assumptates: Ov=Qx+1 180° K= 0,1,2 -- D- we have (P-2)=3 > There ase 3 asymptotes Assymptotes are guidelines fos RL Banches to go to infinity .: Angles are Ør) 1. Oo= 180° = 60° 04 = (3) 890° - 180° Oz= 5(150), 3000 Angle of Asseymptotes : 60,1800, 300"] Controido Centroid is point of intersection of Assimptotes on real axis. GAZ { real part of Poles - Real Post of teros P-Z = 0-2-10 -[0] = -12 2-4

centroid T&A=-4) © jw- crossings: Yo check whether our Root locus crosses the jo anis or not we make ure of Routh Criteria we have the chasacteristic Feguation HKGH6) 20 * H KS 27 (5470) 5° = s(stz) (5+10)+x=0 O 5371252+206 +K=0 Constructing Routh Antay 33 / 20 s2 12 K o gl 240-x 12 so k Now Por Root Locuy to cross jw ares, entene how should be equal to her s'yow can be made zero ý 240-k_0 or TK = 240 |

Thus for K= 240 Aurillary equation is 125 +IC=0 1252724070 » s2 = -240 = -20 Oz/s2 j4.471 Thees Root lecus creme jw avis at s= 1j4:47 Part Sketching Root locus: Paz General pointe? * Root locus is symmetrical about heal anis Root Lecus: Red Color Assomptetes: Green و infinita بالا Towards. -10 Towards 6E-K -0.95 Coreak Mray point)

Just For Reference:

Matlab Code:

clc;clear all;close all;
syms s
K=1;% Gain Value
num=1; % Defining the Numerator Coefficients
den=sym2poly(s*(s+2)*(s+10)); % Defining the Denominator Polynomial
GH=tf(num,den); % Generating the transfer function
figure,
rlocus(K*GH); % Plotting the Root Locus

Plots:

Normal Plot:

Root Locus Imaginary Axis (seconds) -20 -30 35 - -30 -25 -20 -15 -10 Real Axis (seconds)

Plot with showing approximate jw crossings and breakaway point

Root Locus System: untitled1 Gain: 238 System: until Pole: -0.00421 +4.461 Gain: 9.04 Damping: 0.000943 Pole: -0.945 Overshoot (%): 99.7 Damping: 0. Frequency (rad/s): 4.46 Overshoot (%). U Frequency (rad/s): 0.945 Imaginary Axis (seconds) Real Axis (seconds)