Consider industrial automatic controllers whose control actions are proportional
ID: 2085736 • Letter: C
Question
Consider industrial automatic controllers whose control actions are proportional, integral, proportional-plus-integral, proportional-plus-derivative and proportional-plus-integral-plus-derivative. The transfer functions of these controllers can be given, respectively, by:
Where U(s) is the Laplace transform of u(t), the controller output, and E(s) is the Laplace transform of e(t,), the actuating error signal. USE MATLAB to Sketch the outpout u(t) versus t for each of the five types of controllers when the actuating error signal is: (a) e(t) = unit step function (b) e(t) unit ramp function In sketching curves assume that the numerical values of Kp, Ki, Ti, and Td are given as:
Kp = proportional gain = 4
Ki= integral gain =2
Ti = integral time =2 sec
Td = derivative time 0.8 sec
i need the codes please help.
E(s) U(s) K E(s) U(s) E(s) U(s) TsExplanation / Answer
clc
clear all
close all
t=0:0.1:10;
Kp = 4;
Ki = 2;
Ti = 2;
Td = 0.8;
step_in= (t>0); % step input signal
alpha=2;
ramp_in= alpha*t; % ramp input signal
model=tf(Kp,1); % Kp
[y,t]=lsim(model,step_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Step Response with Kp');
xlim([0 11]);
ylim([0 5]);
[y,t]=lsim(model,ramp_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Ramp Response with Kp');
xlim([0 11]);
ylim([0 85]);
model=tf(Ki,[1 0]); % Ki/s
[y,t]=lsim(model,step_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Step Response with Ki');
%xlim([0 11]);
%ylim([0 5]);
[y,t]=lsim(model,ramp_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Ramp Response with Ki');
%xlim([0 11]);
%ylim([0 85]);
model=tf([Kp*Ti Kp],[Ti 0]); % with Kp and Ti
[y,t]=lsim(model,step_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Step Response with Kp and Ti');
%xlim([0 11]);
%ylim([0 5]);
[y,t]=lsim(model,ramp_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Ramp Response with Kp and Ti');
%xlim([0 11]);
%ylim([0 85]);
model=tf([Kp*Td Kp],1); % with Kp and Td
[y,t]=lsim(model,step_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Step Response with Kp and Td');
%xlim([0 11]);
%ylim([0 5]);
[y,t]=lsim(model,ramp_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Ramp Response with Kp and Td');
%xlim([0 11]);
%ylim([0 85]);
model=tf([Ti*Td Kp*Ti 1],[Ti 0]); % Kp, Ti and Td
[y,t]=lsim(model,step_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Step Response with Kp,Ti and Td');
%xlim([0 11]);
%ylim([0 5]);
[y,t]=lsim(model,ramp_in,t);
figure,
plot(t,y)
xlabel('Time -> t');
ylabel('Ramp Response with Kp,Ti and Td');
%xlim([0 11]);
%ylim([0 85]);
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.