PID controller and PID tuner for a SIMO system

Hello,

I've been trying to search how to tune my PID using pidTuner and pidtune but I dont know how should I do it since my system has 1 input and 2 outputs. Because of this, I can´t use pidTuner as it seems that only works for SISO plants or 2 DOF PID controllers. Another thing to say is, that, this is my first control script in matlab, so I´m sure that there is a better way to create the system itself or the PID. Any help is highly appreciated!

So far, I managed to create the control scheme, but I need help for tunning the PID.

%% INIT
clear variables;
close all;
clc;

%% Reference input
% time
time=150;

%SINUSOIDAL
t = (0:0.1:time);
A=4;
w=0.8;
u1=A*sin(w*t);
%dr=zeros(1,length(r)-1);
%dr(1,:)=diff(r)./diff(t);
du1=gradient(u1,t);

%STEP
u2=step(tf(1,1,'InputDelay',10),t);
du2=gradient(u2,t);
%% variable initialization
[a,b,Kp,Kd,Ki,deltaT]= initVar();
%define space state system
f2 = @(t,X) [X(2); a*X(1)+b*X(2)];

%% DEFINE SYSTEM
% x1'= x2;
% x2'= a*x1+b*x2+u
As=[0 1;a b];
Bs=[0;1];
Cs=[1 0;0 1];
Ds=[0;0];
sys = ss(As,Bs,Cs,Ds);
systf=tf(sys);

%% PID
FC=pid(Kp,Ki,Kd,deltaT);

%% LOOP
% close the loop with the PID controller
size(systf*FC)
feedin=1;
feedout=1;
csys = feedback(systf*FC,feedin,feedout,1);
%output
ys= lsim(csys,u2,t);
rf=ys(:,1);
drf=ys(:,2);

max(rf)
max(drf)

%% PLOT
% plot position and speed of the system
figure
subplot(2,1,1)
plot(t,rf,'k', 'LineWidth',2)
hold on
plot(t,u2,'r', 'LineWidth',2)
xlabel('time(s)','fontweight','bold','FontSize',12)
ylabel('position','fontweight','bold','FontSize',12)
title('function evolution')
subplot(2,1,2)
plot(t,drf,'k', 'LineWidth',2)
hold on
plot(t,du2,'r', 'LineWidth',2)
xlabel('time(s)','fontweight','bold','FontSize',12)
ylabel('speed','fontweight','bold','FontSize',12)

%plot surface generated by the function
% [x1,x2] = meshgrid(-5:5);
% surff=a*x1+b*x2;
% figure
% surf(x1,x2,surff)
% view(135,45)


%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%% FUNCTION
function [a,b,Kp,Kd,Ki,deltaT]= initVar()
%coefs
a=-0.3;
b=-1.1;

%gains
Kp=3;
Kd=0.8;
Ki=0.9;

deltaT=1e-1;
end

Answer 1

[Sam Chak]

Hi @Mikel

Your system is

.

If you algebraically design u as

,

then the closed-loop becomes

.

Since you didn't specify the performance requirements, here is just a little test on the theoretical calculation. This method assumes that you can measure the velocity directly.

% parameters
a    = -0.3;
b    = -1.1;

% system
As   = [0 1; a b];
Bs   = [0; 1];
Cs   = [1 0; 0 1];
Ds   = [0; 0];
sys  = ss(As, Bs, Cs, Ds);
Gp   = tf(sys)

Gp =
 
  From input to output...
               1
   1:  -----------------
       s^2 + 1.1 s + 0.3
 
               s
   2:  -----------------
       s^2 + 1.1 s + 0.3
 
Continuous-time transfer function.

% gains designed by a simple algebra manipulation
Kp   = 0.7;
Ki   = 0.0;
Kd   = 0.9;
Tf   = 0.1;
Gc   = pid(Kp, Ki, Kd, Tf)

Gc =
 
               s    
  Kp + Kd * --------
             Tf*s+1 

  with Kp = 0.7, Kd = 0.9, Tf = 0.1
 
Continuous-time PDF controller in parallel form.

% closed-loop
Gcl  = feedback(Gc*Gp(1), 1)

Gcl =
 
          9.7 s + 7
  --------------------------
  s^3 + 11.1 s^2 + 21 s + 10
 
Continuous-time transfer function.

% step(Gcl, 10)     % can see the effect of zero that causes the overshoot
% Add a prefilter at the Reference Input to cancel out the zero (numerator of Gcl)
Gf   = tf(10, [9.7 7])

Gf =
 
     10
  ---------
  9.7 s + 7
 
Continuous-time transfer function.

% closed-loop with prefilter
Gclf   = minreal(series(Gf, Gcl))

Gclf =
 
              10
  --------------------------
  s^3 + 11.1 s^2 + 21 s + 10
 
Continuous-time transfer function.

% system response
[y, t] = step(Gclf, 10);
plot(t, y), grid on

% PID output (I think so)
Gu   = feedback(Gc, Gp(1));
Gfu  = minreal(series(Gf, Gu))

Gfu =
 
      10 s^2 + 11 s + 3
  --------------------------
  s^3 + 11.1 s^2 + 21 s + 10
 
Continuous-time transfer function.

step(Gfu, 10), grid on

Note: The following Block diagram in Simulink should produce the same outputs as demonstrated in MATLAB:

[Mikel]

thank you! I have another question, is it possible using the feedback function to acces to the values of the PID output? or for that should I build a PID manually doing all the calculations by myself?

[Sam Chak]

You are welcome @Mikel. I have edited my Answer to show you how to obtain the PID output as well.

If you find the explanations, MATLAB code, and block diagram are helpful, please consider accepting ✔ and voting 👍 the Answer. Thanks!