Question

2. Nise(9.6) For a unity feedback system KG(s) (s 6) G(s) T(s) (s 2)(s3)(s 5) 1 + KG(s) a) Given a K 4.60, .707, on the 135 line, find the operating point on the root locus NOTE: use the fact that 1 + KG(s) 0 at all points on the root locus, so K 1 and G(s)l 12(2k 1)180. Or use geometry using the point knowing that cose LKG(s) 1 = and a wn and b b) Find the steady state error, and the steady state error constant c) What is the settling time & peak time for the current system? d) Design a proportional derivative compensator to reduce the settling time by 2. What is Ge(s), and what is the new Gn(s)= Ge(s) G(s), Where should the new zero be added at? What is the new gain at this new zero value. e) Design a lead (derivative) compensator to reduce the settling time by 3. What is Ge(s), and what is the new Gn (s) Ge(s) G(s), Where should the new zero & pole be added at? f) plot the step response for the old & new systems to show the improvement

Answer 1

Answen Given that For a writy feedback syftem . .. (5+6). (5+2) (5+3)(5+5) G (s): - K24.6 The Characteristic equation IHK G(s) is s²+ 105²+35.657576 The roots of the above equation one 52-5.3661, -2.3169 2. 31643 The Operating pointis – 2.31691 2. 31645 ) K =416 6(s). 46 (5+6) (S+ 2) (s+3) (5+5)

6 since Thin in a type-o system the steady state constant is ke Position constant Where Kp G (0) - 4.676, 0.92 2x375 The steady state equis essa I = 0,5903 1tke c) since the operating point in se 23 52-2,31691 2.31645 one in the from S= - É Wn I wn Vis25 ta 2 ts a Ewn = 1.7264. 4 2.3169

- tp = II, I II = 1,3562 I Wd waffs2 2.3164 ③ d) The new setting time is 1.7264 = 0.8632 2 - ts =ų - 0.8632 B Ewn So Ewn 4.6339 with E= 0.707 Wn- 6.5543 The new dominant pulls are o Sa = -{ wnt while25 Sd = -416339€ 4.635 The Angle dealing of Sa to S= -2, -3, -5, 6 2 180-03-02-037 Og = -60.8359

6+ the PD Compensato be © GC (s)= k (s+Q) Then a must be placed at 60, 8359° famid ---|- 4.63 6 -5 4,63 -3 - a= 4.63 +4.63 1. Tan (60x8359) = 6:7263 Gr(s)= k (5+67263)

The gain 'k' in found using magnitude Créterial 6 k (5+6.7263) x (5+6) (5+2) (5+3) (5+5) 52-4.63+4.63T K=4.947 Sel 4.63 a. -463

step response:

clc;

clear all;

close all;

s=tf('s');

k=4.6;

g=(s+6)/((s+2)*(s+3)*(s+5));

gc=4.947*(s+6.7263);

step(feedback(k*g,1),feedback(g*gc,1))

legend('uncompensated system','compensated system')

File Edit View Insert Tools Desktop Window Help Step Response System: compensaved svetom Setting Time (seconds):0.782 System: uncompensated system Setting time seconds): 1.83 Amplitude Time (seconds)