Question

The system in Figure 3 comprises a motor and a contoller. The performance requirements entail a steady state error for ramp input r(t) Ct, smaller than 0.01C. Here, C is a constant. The overshoot for step input must be such that P.0.S 5% and the settling time with a 2% error should be T. 2 seconds. (a) Based on rlocus function, write a piece of MATLAB code which establishes the controller. (b) Create the graph of obtained unit step response and calculate the overshoot and settling time Label the graph properly. (c) Find the influence of disturbance T, (s)-WIs with constant W for the Laplace transform of output given by Y(s) Tfs) Motor and antenna 10 sis +5%s+10) Controller Ys) Position Figure 3: A unity feedback closed loop system with disturbance T (s)

Write as MATLAB code with comments thank you.

Answer 1

MATLAB code is given below in bold letters.

clc;
close all;
clear all;

% define the laplace vaiable s
s = tf('s');

% define the plant
Gp = 10/(s*(s+5)*(s+10));

% define the gain range
k = 0:0.01:20;

% plot the root locus
figure;
rlocus(Gp,k);

root locus plot is shown below

Gain: 8.44 Pole: -1.891.98i Damping: 0.69 Overshoot(%): 4.99 Frequency (rad/s) 2.74 -2 -3 -4 -14 12 -8 -4 Real Axis (seconds)

from the above plot it is observed that for gain K = 8.44 the peak overshoot is <5%.

The settling time corresponding to this is given by 4 / real part of the pole at gain = 8.44

Therefore Ts = 4 / 1.89 ~ 2 sec

the unit step response is plotted below with a gain of K = 8.44.

MATLAB code:

% plot the unit step response
figure;
step(feedback(8.44*Gp,1));grid on;

the response is plotted below.

Step Response 0.8 0.6 0.4 02F 0.5 2.5 3.5 Time (seconds)

Question C

The MATLAB code is given below in bold letters.

% assume W = 1 i.e. unit step disturbance
% then y(s) is plotted below
figure;
step(feedback(8.44*Gp,1)+feedback(Gp,8.44));grid on;
title('y(t) for unit step input and unit step disturbance');

The response is plotted below.

y(t) for unit step input and unit step disturbance 0.8 0.6 0.4 0.2F... 0.5 2.5 3.5 Time (seconds)

From the above response, it is observed that the steady-state error due to unit step disturbance is 0.12. i.e. 12 %