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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) Ts

Explanation / 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]);

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