Question

Design a phase lead controller using the Root Locus Method for the system described by G(s). Ensure that there is a reduction of MORE than 40% of the original settling time, additionally, enure that an overshoot of 12% is not exceeded.

G(s)=1/(s³+13s²+32s+20)

Answer 1

matlab:

clc;

clear all;

close all;

s=tf('s');

g=1/(s^3+13*s^2+32*s+20);

rlocus(g)

File Edit View Insert Tools Desktop Window Help 08H3 Way. I E = 0 Root Locus System: g Gain: 34.6 Pole: -1.28 + 1.89 Damping: 0.561 Overshoot (%): 11.9 Frequency (rad/s): 2.29 Imaginary Axis (seconds) Real Axis (seconds)

Salt L {wn 128 fram from the Root locus to 1 615) = (s$+ 138+ 325 +40) 12% overshost point is Spa -1-28 + 195, The settling time te= y 2 4 = 3.125 and to 12% overshoot A JFSZ is a sz 0.56 100 . The new settling time Required is 0.4x3+125= 1.25 i Es= 4 =125 = {wna 3.2 and to Ewn {=0.56 , Wna 5.7143 The new domi non poles saz-son I wrotes are fyz - 3.23 4.73J The open loop poles of Gis) are s=-10.13,-1.43 1 0.955: The Angle doji Ciency of 'S go to the open loop poles of GIS) is -------74.735 Odoy = 180 -0,-ą -03 0,= 180- Tan (679) il po 70.955 = 180- nom (s ) 4 -0.955 = Tañ (4:33) z odej = -76.7 pole - zero g. SA is Angle Birector of IBSO ona - Z -3.2 sa 3.68 -3:2 4 1.77 -10:13 Sa ger -- 4 alatt 56 A7124 LPSE = 7617

- -- - ---- -..- _ 2 4.73 LESA = (Aso = 124 = 620 [PSA = 2asz = 767 - 38-35* In ole esp: Tan (62-38.35)= 4.73 » esa 10:8 BS in The pole is at 10.8 +3:2 = 14 lly in ble szc Ton (62-38.35-34) = Zc 3 Zc= -0.8639 o The zero is 3.2-0.8089 = 2.3361 E The Lead Controller is - Ge(s) = K (s+2.3361) (8 +14) The gain is found Giteia inc using magnitude 1 K (st 2.3361) ! (8+14) (3 + 135+ 325+20) I @sa-3.2 +4.13] K: 517.0558 Gls) = 517.0558 (S+ 2.3361) (5+14)