Problem with pidtune function

For the following code I keep getting just an integral element to my controller even though I am specifying PID. I need PID to get the response I'm looking for but unsure of what I'm doing wrong. tfp4 comes from a state space which I am sure is correct and isn't causing any errors that should effect my response.

[C,info] = pidtune(tfp4, 'PID')

tuned = feedback(C*tfp4, 1);

t = linspace(0, 100, 1000);

u = 3*pi/180*heaviside(t);

[y42, t42] = lsim(tuned, u, t);

y42 = y42*180/pi;

figure()

plot(t42, y42)

[Paul]

will be difficult for anyone to help w/o the tfp4 model, which can be saved in a .mat file and added to the question using the Paperclip icon in the Insert menu.

If I had to guess, I'd say that pidtune is able to meet its internally generated design goals with just the integral control; the call to pidtune isn't specifying any design goals in particular, i.e., the resopnse you're looking for.

[Collin Lightfoot]

Thanks. I’ll try setting some response parameters and see if that gets me anywhere.

Answer 1

[Sam Chak]

Hi @Collin Lightfoot

According to the documentation, if the tuning algorithm can achieve satisfactory performance and robustness using a lower-order controller than what is specified for the desired PID controller type, the pidtune function will return a 'C' controller with fewer actions than specified. In your case, 'C' could be an Integral-only controller, even though the type is 'PID'.

Additionally, if no performance specifications are provided, the algorithm will choose a crossover frequency based on the plant dynamics and automatically design for a target phase margin of 60°

%% 4th-order transfer function plant (tfp4)
num = [-1 4 6 4 1];
den = [ 1 4 6 4 1];
Gp  = tf(num, den)

Gp =
 
  -s^4 + 4 s^3 + 6 s^2 + 4 s + 1
  ------------------------------
  s^4 + 4 s^3 + 6 s^2 + 4 s + 1
 
Continuous-time transfer function.

%% Attempt to implement a PID Controller
% C0 = pidstd(1, 1, 1e-6);      % baseline controller
% [C, info] = pidtune(Gp, C0)
[C, info] = pidtune(Gp, 'PID')

C =
 
        1 
  Ki * ---
        s 

  with Ki = 0.667
 
Continuous-time I-only controller.

info = struct with fields:
                Stable: 1
    CrossoverFrequency: 1
           PhaseMargin: 90.0000

%% Closed-loop system
Gcl = feedback(C*Gp, 1)

Gcl =
 
    -0.6667 s^4 + 2.667 s^3 + 4 s^2 + 2.667 s + 0.6667
  ------------------------------------------------------
  s^5 + 3.333 s^4 + 8.667 s^3 + 8 s^2 + 3.667 s + 0.6667
 
Continuous-time transfer function.

%% Plot results
subplot(211)
step(Gp, 10), grid on, legend('Original Plant')
subplot(212)
step(Gcl), grid on, legend('Compensated Plant with I-only controller', 'location', 'East')

[Sam Chak]

However, you can try using this trick to potentially force the pidtune algorithm to design a controller 'C' of the same type and form as the baseline controller 'C0', although there is no guarantee of success.

In general, it is considered good engineering practice not to restrict the controller to a specific control structure as long as satisfactory performance and robustness are achieved using a simpler solution. There have been instances where 'overkill' control solutions have been published in academic journals to address what appears to be a simple and localized problem.

%% 4th-order transfer function plant (tfp4)
num = [-1 4 6 4 1];
den = [ 1 4 6 4 1];
Gp  = tf(num, den)

Gp =
 
  -s^4 + 4 s^3 + 6 s^2 + 4 s + 1
  ------------------------------
  s^4 + 4 s^3 + 6 s^2 + 4 s + 1
 
Continuous-time transfer function.

%% Attempt to implement a PID Controller
C0 = pidstd(1, 1, 0)    % baseline controller

C0 =
 
             1      1 
  Kp * (1 + ---- * ---)
             Ti     s 

  with Kp = 1, Ti = 1
 
Continuous-time PI controller in standard form

[C, info] = pidtune(Gp, C0)

C =
 
             1      1 
  Kp * (1 + ---- * ---)
             Ti     s 

  with Kp = 6.67e-05, Ti = 0.0001
 
Continuous-time PI controller in standard form

info = struct with fields:
                Stable: 1
    CrossoverFrequency: 1
           PhaseMargin: 90.0057

%% Closed-loop system
Gcl = feedback(C*Gp, 1)

Gcl =
 
  -6.667e-05 s^5 - 0.6664 s^4 + 2.667 s^3 + 4 s^2 + 2.667 s + 0.6667
  ------------------------------------------------------------------
        s^5 + 3.334 s^4 + 8.667 s^3 + 8 s^2 + 3.667 s + 0.6667
 
Continuous-time transfer function.

%% Plot results
subplot(211)
step(Gp, 10), grid on, legend('Original Plant')
subplot(212)
step(Gcl), grid on, legend('Compensated Plant with I-only controller', 'location', 'East')

[Collin Lightfoot]

Thank you!

[Sam Chak]

You're welcome! If you have additional questions on how to improve the performance, feel free to ask.