Simulink PID Controller not behaving Properly

Hi All, having some issues with a PID controller project I'm working on using the simulink PID block.

I've been trying to tune the controller manually, but the behavior I'm seeing from the simulink output does not match both the basic step response graph as well as the PID Tuner app.

J = 0.01;
b = 0.1;
K = 0.01;
R = 1;
L = 0.5;
voltage = 10;
stoptime = 10;
timestep= 0.05;

simIn = Simulink.SimulationInput("proj1sim");
simIn = setModelParameter(simIn,'StopTime',stoptime);
simIn = setModelParameter(simIn,'TimeStep',timestep);

%Setup Transfer Function
num = [K];
denom = [J*L R*J+b*L b*R+K^2];

G = tf(num,denom);
%step(G);

%1st Controller 
%PID Parameters
Kp = 60;
Ki = 180;
Kd = 4.5;

out = sim("proj1sim.slx");

output = out.simout;
controller = out.controller;

figure(1);
plot(output);
title('Plant Output');
ylabel('Voltage');
figure(2);
plot(controller);
title('Controller Output');
ylabel('Voltage'); 

The main problem I'm seeing is that when I put any value other than zero for the derivative gain, the system completely breaks down.

As shown in the images, when I use the values given by PID Tuner, the graph makes no sense. But when I sed Kd to zero and change nothing else, it sort of works suddenly.

PID tuner:

Output with PID tuner Values:

Output with Kd=0:

Simulink Model:

Any help would be grealty appreciated.

Answer 1

[Sam Chak]

Hi @Nick

Update: The reason behind the instablity is caused by the Fixed-step size solver.

If you want to let (by default), then readjust the following

timestep = 0.05/10;

so that the new timestep sec is smaller than the time constant of the derivative filter sec. In short, make sure that

.

3 Mar 2023: Not sure what went wrong, but your manually tuned PID gains in Simulink produced the same (stable) results as shown in MATLAB. I have also tested with my tuned PID gains. The closed-loop is also stable.

I suspect that some values in the transfer function block of the Plant are probably incorrectly entered. Cannot confirm without seeing how the values are entered in the Block Parameters.

J = 0.01;
b = 0.1;
K = 0.01;
R = 1;
L = 0.5;
voltage = 10;
stoptime = 10;
timestep= 0.05;

% Setup Transfer Function
num = [K];
den = [J*L R*J+b*L b*R+K^2];
Gp  = tf(num, den)

Gp =
 
             0.01
  ---------------------------
  0.005 s^2 + 0.06 s + 0.1001
 
Continuous-time transfer function.

dcgain(Gp)

ans = 0.0999

step(voltage*Gp, 4)

% My Controller
kp1  = 15.5;
ki1  = 30;
kd1  = 0.21;
Tf1  = 1/12;
Gc1  = pid(kp1, ki1, kd1, Tf1)

Gc1 =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = 15.5, Ki = 30, Kd = 0.21, Tf = 0.0833
 
Continuous-time PIDF controller in parallel form.

Gcl1 = minreal(feedback(Gc1*Gp, 1))

Gcl1 =
 
          36.04 s^2 + 432 s + 720
  ----------------------------------------
  s^4 + 24 s^3 + 200.1 s^2 + 672.2 s + 720
 
Continuous-time transfer function.

step(voltage*Gcl1, 4)

S1 = stepinfo(Gcl1)

S1 = struct with fields:
         RiseTime: 0.5604
    TransientTime: 0.9754
     SettlingTime: 0.9754
      SettlingMin: 0.9009
      SettlingMax: 1.0000
        Overshoot: 0
       Undershoot: 0
             Peak: 1.0000
         PeakTime: 2.3733

% Your Controller
kp2  = 60;
ki2  = 180;
kd2  = 4.5;
N    = 100;     % assuming the default value in the PID block
Tf2  = 1/N;
Gc2  = pid(kp2, ki2, kd2, Tf2)

Gc2 =
 
             1            s    
  Kp + Ki * --- + Kd * --------
             s          Tf*s+1 

  with Kp = 60, Ki = 180, Kd = 4.5, Tf = 0.01
 
Continuous-time PIDF controller in parallel form.

Gcl2 = minreal(feedback(Gc2*Gp, 1))

Gcl2 =
 
            1020 s^2 + 12360 s + 36000
  ----------------------------------------------
  s^4 + 112 s^3 + 2240 s^2 + 1.436e04 s + 3.6e04
 
Continuous-time transfer function.

step(voltage*Gcl2, 4)

S2 = stepinfo(Gcl2)

S2 = struct with fields:
         RiseTime: 0.1637
    TransientTime: 0.8502
     SettlingTime: 0.8502
      SettlingMin: 0.9002
      SettlingMax: 1.0682
        Overshoot: 6.8183
       Undershoot: 0
             Peak: 1.0682
         PeakTime: 0.4425

[Nick]

Thanks for the response, I was able to replicate your results using the pid, minreal, and step functions inside of matlab. Unfortunately the project stipulates that I have to pass the data back from simulink using the to workspace block and create the plots that way, which still seems to be broken. I've updated the transfer funtion block incase that was the issue, replicating what you've shown, but to no avail. Plotting only the no PID response shows the same as your plot, but the PID response with your values shows the same results I was getting before, with the solution rapidly approaching infinity after about 17s according to matlab.

#UPDATE: Just set N to 1 in the PID block and the response seems to be fixed. Does that make sense?

NO PID Only:

The PID Block:

TF Block:

Very frustrating, not sure what I could be wrong here. Could there be a toolbox or something that I'm missing?

[Sam Chak]

@Nick, This is mind-boggling. I'd suggest you to attach the Simulink file (clicking the Paperclip icon) for further investigation.

[Nick]

Right!? Here are both the simulink model and my code files.

Worth noting that lowering the value of N seemd to offer an improvement, but another project requirement is that the controller output shouldnt exceed 25V, but im seeting 200-300v overshoot. any ideas there?

[Sam Chak]

Hi @Nick

Found out that the reason behind the instablity is caused by the Fixed-step size solver.

Because you let N = 100 (untuned), which is equivalent to sec, the time constant of the 1st-order derivative filter is smaller than the selected 0.05 sec time-step in the solver, it cannot accurately produce the output of the derivative action, and thus causing the instability.

In my original PID controller above, the time constant of the 1st-order derivative filter is sec. If you set , and because the timestep , then it will produce a stable response.

Nevertheless, if you want to let (by default value), then readjust the following

timestep = 0.05/10;

so that the new timestep sec is smaller than sec.

If the proposed solution works, please consider accepting ✔ and voting 👍 the Answer. Thanks a bunch! 🙏