assessment of the control loop performance using the Integral of the absolute error (IAE) performance index

I am not getting anything related to P_Measure.IAE = trapz(t, abs(e));

I used the code below:

clear all;

close all;

%% define num and den of tf

%Gp defining

num=[0.8];

den=[10 1];

Gp=tf(num,den,'IODelay',9)

%Gc defining

kc=(0.5*10)/(0.8*9);

Ti=10;

Gc=kc+tf([kc],[Ti]);

%defining Gd

Gd=tf([-4],[1 15], 'IODelay',9)

% close loop transfer function

%change in set point

GCL1=Gc*Gp/(1+Gc*Gp)

GCL1= minreal(GCL1)

%disturbnance

GCL2=Gd/(1+Gc*Gp)

GCL2= minreal(GCL2)

% Simulate closed loop tranfer function

T_Final = 100;

[y1, t1] = step(Gp, T_Final)

[y2,t2]=step(GCL1,T_Final)

[y3,t3]=step(GCL2,T_Final)

[y4,t4]=step(Gd,T_Final)

plot(t2,y2,'k','linewidth',2)

hold on

plot(t3,y3,'y','linewidth',2)

hold on

plot(t1,y1,'m','linewidth',2)

hold on

plot(t4,y4,'c','linewidth',2)

hold off

legend('Setpoint changes','Disturbance changes', 'Gp', 'Gd')

ylabel('output')

xlabel('time(min)')

%

G = 1 / (1 + Gc * Gp);

T_Final=100;

function [P_Measure] = Performance(G, T_Final);

[e, t] = step(G, T_Final);

P_Measure.IAE = trapz(t, abs(e));

end

Answer 1

[Akanksha]

Hey @Amnia Al-Hammadi,

The issue is caused due to usage of P_Measure.IAE = trapz(t, abs(e)); inside a function, but did not define the error signal e ie e = setpoint - output correctly for your closed-loop setpoint change.

Also, in R2018b, you cannot define a function at the end of a script (unless you use a separate function file or a live script).

This could easily be resolved. Please go through the code below that will help you to get the required output.

% Control Loop Performance Assessment using IAE
clear all;
close all;
clc

%% 1. Define Process (Gp) and Controller (Gc)
num = [0.8];
den = [10 1];
Gp = tf(num, den, 'IODelay', 9);

kc = (0.5 * 10) / (0.8 * 9);
Ti = 10;
Gc = kc + tf([kc], [Ti]);

%% 2. Closed-Loop Transfer Function for Setpoint Change
GCL = feedback(Gc * Gp, 1);  % Unity feedback

%% 3. Simulate Step Response (Setpoint Change)
T_Final = 100;
[y, t] = step(GCL, T_Final);  % y is the output, t is the time vector

%% 4. Calculate Error Signal (e = r - y)
r = ones(size(t));  % Unit step setpoint
e = r - y;          % Error at each time point

%% 5. Calculate IAE (Integral of Absolute Error)
IAE = trapz(t, abs(e));

%% 6. Display and Plot Results
disp(['Integral of Absolute Error (IAE): ', num2str(IAE)]);

figure;
plot(t, y, 'b', 'LineWidth', 2); hold on;
plot(t, r, 'r--', 'LineWidth', 1.5);
plot(t, e, 'k:', 'LineWidth', 1.5);
legend('Output y(t)', 'Setpoint r(t)', 'Error e(t)');
xlabel('Time (min)');
ylabel('Value');
title(['Closed-loop Response and Error, IAE = ', num2str(IAE)]);
grid on;

I have attached the screenshots of the solution I got, whilst running the above code in R2018b MATLAB.

Also, for futher references on you can type "doc <keyword you need help with>" to open the documentation for the MATLAB R2018b release as shown below.