PI control algorithm manually implemented in matlab - pid

Question

I'm trying to implement a simple script performing a PI control for a cruise control application, but I'm founding some problems with the integral part. Here is my code:
function [] = PI_cruisecontrol()
clc; close all;
t0 = 0; tfinal = 50; dt = 0.001; % time parameters
r = 10; % reference of 10 m/s
m = 1000; % mass
b = 50; % friction coeff. (depends on v)
yp = zeros(tfinal/dt,1); t = yp; % initialize speed and time array
Ki = 40; % integrarl constant
Kp = 800; % proportional constant
int = 0; % itinialize int error
% CONTROL LOOP (Forward-Euler integrator is used to solve the ODE)
for i=t0+2:tfinal/dt
err = r-yp(i-1); % update error
int = int+err; % integral term
u = (Kp*err)+(Ki*int*dt); % action of control
yp(i) = yp(i-1)+((-b*yp(i)/m) + (u/m))*dt; % solve ode for speed
t(i) = t(i)+dt*i; % log the time
end
% Results
figure(1)
plot(t,yp)
title ('Step Response')
xlabel('Time (seconds)')
ylabel('Amplitud')
axis([0 20 0 12])
hold on
reference = ones(tfinal/dt,1)*10;
plot(t,reference,':')
end
And this is how it should be, using predefinided matlab functions:
function [] = PI_cruisecontrol2()
m = 1000;
b = 50;
r = 10;
s = tf('s');
P_cruise = 1/(m*s + b);
Kp = 800;
Ki = 40;
C = pid(Kp,Ki);
T = feedback(C*P_cruise,1);
t = 0:0.1:20;
step(r*T,t)
axis([0 20 0 12])
end
What am I doing wrong in my code?
Thanks!

Answer 1

I managed to fix the problem, working with float variables instead of arrays. Moreover, I added the derivative term (although for this first order problem was not necessary)
Here I left the code:
function [] = aFortran_PI()
clc; close all;
r = 10; % reference of 10 m/s
m = 1000; % mass
b = 50; % friction coeff. (depends on v)
yp = 0; % init response
Kp = 800; % proportional constant
Ki = 40; % proportional constant
Kd = 0; % derivative term is not necessary in this problem
previous_error = 0;
integral = 0;
dt = 0.001;
% CONTROL LOOP
for i=1:20000
error = r-yp; % update error
integral = integral + error*dt; % integral term
derivative = (error-previous_error)/dt; % derivative term
u = Kp*error+Ki*integral+Kd*derivative; % action of control
yp = yp+(-b*yp/m + u/m)*dt % solve ode for velocity
end
end