Question

A unity feedback system with the forward transfer function G)2)(s +5) is operating with a closed-loop step response that has 15% overshoot. Do the following: a) Evaluate the settling time for a unit step input; b) Design a PD control to yield a 15% overshoot but with a threefold reduction in settling time; c) Evaluate the settling time, overshoot, and steady-state error with the PD control.

Answer 1

I have used MATLAB to find out the gain K where the overshoot is 15%.

MATLAB code to plot the rlocus is given below.

clc;
close all;
clear all;

s = tf('s');
k = 0:0.1:20;
G = 1/(s*(s+2)*(s+5));
figure;rlocus(G,k);

Root Locus 1.5 System: G Gain: 10.9 Pole: -0.7241.2i Dapng 0.517 Overshoot(%): 15 E 0.5 Frequency (radis) 1.4 -0.5 -6 -4 Real Axis (seconds)

From the above figure, it is observed that at gain 10.9, the overshoot is 15%.

Therefore k = 10.9.

settling time for unit step is given by Ts = 4 time constants = 4/ |real part of dominant pole = 4/ 0.724 = 5.52 sec

Now it is required to design a PD controller to meet 15% overshoot but settling time = 5.52/3 = 1.84 sec.

I have used root locus method to determine the PD controller.

OIS closcd loop Polus 1so at s--2-17 35713 s> 2-2 174)- tan 3s1) 17ฯ tan 54 2-2n こうえ = 2.3 . G (5)丨 @S= "-2 . DY + 3.รา』 -|

Therefore the controller is given by C(s) = 19.0245(s+2.3961)

I have used MATLAB to plot he response with the controller.

code:

C = 19.0245*(s+2.3961);
figure;step(feedback(C*G,1));

Response is plotted below.

Step Response System: untitled1 Final Value: 1 System: untitled Peak amplitude: 1.17 Overshoot(%): 17.2 At time (seconds) 0.856I 1-.-:ー· .. ..- : - System: untitled1 Settling Time (seconds): 1.85I 0.2 0.5 Time (seconds)

From the above response it is observed that the settling time is 1.84 sec. steady state error is zero and peak overshoot is 17%