Question

Problem 2 Wis) R(s) U(s) Gol (s) D a (s) E(s) H(s) Given a system as in the diagram above, use MATLAB to solve the problems: Assume we want the closed-loop system rise time to be t, 0.18 sec S + Z H(s) 1 Gpl)s(s+)et s(s 1) s + p a) Assume W(s)-0. Draw the root locus of the system assuming compensator consists only of the adjustable gain parameter K, i.e. Dct (s) Determine the approximate range of values of K for which this is true (using the textbook approximation for rise-time. What is the minimum overshoot (maximum damping factor) you can achieve with the P-type control? Choose K to satisfy the rise-time requirement and plot the time-domain response of the closed- loop system to the unit step-input reference signal. What is the settling-time equal to? K b) c) d) A lead-compensator Dci (s) is added to the system, so that the magnitude of zero z 10 Draw Root-locus for various values of Ipl (say, p 20, p-30, etc) and observe the behavior of the closed-loop poles. Pick a suitable value of p, so that the rise-time requirement for the closed-loop system is satisfied, and so that the damping factor is at least 0.5. what is the settling time for this case? Plot the time domain response of the system to a unit-step input. Compare to the unit-step response in part c)

Answer 1

18 2 So So 4 So朮minimum onex Shat

1)matlab code and result

clear;clc;
%% part (a)
Gpl=tf([1],[1 1 0]);
Ls1=Gpl*1;
figure(1);
rlocus(Ls1);
title('Rootlocus plot for proportional controller');
%% part (c)
k1=32.14;
M1=feedback(k1*Ls1,1);
figure(2);
step(M1);
title('step response for proportional controller at required tr');
%% part (d)
Dcl=tf([1 10],[1 71.36]);
Ls2=Gpl*Dcl;
figure(3);
rlocus(Ls2);
title('Rootlocus plot for lead compensator');
k2=767.05;
M2=feedback(k2*Ls2,1);
figure(4);
step(M2);
title('step response for lead compensator at required tr');

Rootlocus plot for proportional controller 0.6 0.4 กั 0.2 -0.2 0.6 1.2 0.4 -0.8 0.6 -0.2 0 0.2 -1 Real Axis (seconds)

step response for proportional controller at required tr 1.8 1.6 1.4 1.2 0.8 0.6 0.4 0.2 0 6 10 12 Time (seconds)

Rootlocus plot for lead compensator 100 80 60 40 F 20 20 40 -60 -80 100 80 -70 -60 -50 -40 -30 20 10 -10 Real Axis (seconds)

step response for lead compensator at required tr 1.4 1.2 0.8 0.6 0.4 0.2 0 0.3 0.1 0.5 0.6 0.9 0.2 0.4 0.7 0.8 Time (seconds)

if you compare both step response the lead compensated response have small rise time and small overshoot