Question

Note: Please draw the Root Locus plots using Rules and verify your results with Matlab Commands. Enclose both plots. For the unity feedback system, with the following transfer functions (as shown in problems 1 through 4), sketch the Root- Locus plot and find the following: (a) The break-away and break-in points (b) The jw-axis crossing (c) The angle of departures / arrivals at complex poles and zeros. (d) The range of the gain K, to keep the system stable. Problem 1: TOK G (S) = ----------- s (s + 10) (s +20)

Answer 1

Answer-1 Im(s) Re(s) -20 -10 Given XX GC) = lok SC S+ S+20) The poles of G(s) are at s=0, s=-10 and 5=-20. Here, the root locus starts from poles to and & as shown and then meet at A and then moves towards right half of s-plane. In this process, it cuts imaginary ax is at points 'B' and 'C'. From pole b2 root locus moves away on real axis towards left half of s-plane.

(9) Let's find break -alway point now. This point is 'A' here. Chracteristic quaton for a unity feedback system It G(S) = 0 {s(s+ 10) (5 +20) = 0 + 10k} → S3+ 30s2 + 2005+ 10k =D k- - (53 +3052 + 2005) 10 OS Let's finds values for which dk =o. dk - (352 +685 +200) 0 10 → 382 + 605 + 200 = 0 S = -60 02_12.8200) 6 S = -10 + 10 E Si= -15.77 Sq = - 4.226

NS2 = Now, let's check the values of k for S, and S2. klo - - (S23 + 30 S2²+ 20052). 385 10 → kls2 = 38.5 kls = -15,3 +308;+2005) - 38.5 oo k value at Sy is positive. Therefore, it is desired k value at point A. Leto find the poles at k = 38.5; We have = It 6/32 = 0 + $3 + 306+ 200 s + 10=0 → 53 +30s+ 2005 +385 = 0 Finding roots by calculator, we have Os=-4.226 - 4.226,- 21.547

Therefore are same out of above 3 poles, first two poles and this is the boint where the root loci mect at A. Therefore, breakaway boint (A) = 64.226 +jo) by We shall use R-H criterion for this. First, We shall find for what values of k the system is stable. The limit of stability will give us the jw axis crossing. Characteristic equation → 55+ 3052 + 200 S + 10k=0 RH Table * * * $31 2000 5230 lok o s' 6000-10k) o so Ilok 6000 For stability 600D-TOK >O 30 Ci) 10k>0

Therefore k o , and K 2600 axis. Therefore, tock<600 Anso (d) At k= 600 the root locus crosses jw Let's find poles for k=600 33 + 3052 +2005 + 6000 =0 → S = -30, £ 1 14.142 Therefore > The imaginary axis crossings are > s=+ j 14. 142 Anso (6) (CL As you can see from root locus , of departure are > Departure at to = K (radians) ? ! $1 = 0 (radians) → ! ! Þ2 = (radians) Angolc)

MATLAB Code ==>

close all
clear all

s = tf('s'); %% defining laplace variable 's'

%% Continuous time transfer function, G(s) ==>
G = 10 /(s*(s+10)*(s+20)); %% defines loop transfer function
rlocus(G) %% plots the root locus

Result ==>

Root Locus Imaginary Axis (seconds) -70 -60 -50 0 10 20 30 40 30 -20 -10 Real Axis (seconds)

To help you with the confirmation of break-away point, imaginary axis crossing and corresponding gain; I am plotting magnified root locus diagram with data points (cursors) ==>

Root Locus Imaginary Axis (seconds) System: G Gain: 38.5 Pole: 4.27 -9.98-08i Damping: 1 Oversh System: G Freque Gain: 600 Pole: 0.00146 - 14.1i Damping: -0.000103 Overshoot (%): 100 Frequency (rad/s): 14.1 -60 -50 -40 -30 100 10 20 Real Axis (seconds)

NOTE - For any other help or discussion related to the problem, please let me know in the comments. Also, if you feel that the images are not clear, then please let me know. I shall upload them again.