PID controller in matlab code

I'm having trouble using a PID controller in a MATLAB script. I don't have the transfer function typically used with PID in MATLAB, but I do have the output error signal. I would like to connect this error signal to the PID code to improve the error correction.

For example, if I have this data (from the black system). I want to reduce the (maximum overshoot , rise time , settling time , ...) by using a PID controller .

clear all
clc
load("Vo_vn.mat")
load("x.mat")
Volt    = Vo_vn;
Time    = x;
OUTS    = stepinfo(Volt,Time);
Desire  = 23.5;
Error   = Vo_vn-Desire;
RiseT   = OUTS.RiseTime

RiseT = 1.3248e-04

SettT   = OUTS.SettlingTime

SettT = 0.0037

OVERSH  = (OUTS.Peak)-Desire

OVERSH = 14.3081

plot(Time,Volt)

[Sam Chak]

Hi @mohammed hussein

If you don't have the system transfer function in the traditional sense, where does the output error signal come from in the MATLAB script? The output signal must be generated from some system, either a transparent mathematical model or a data-driven black box. The PID controller, pid() function, itself does not require a transfer function to operate, unless you are referring to tuning the PID controller, which I understand you implied is intended to “improve the error correction.”

[mohammed hussein]

Dear Sam Chak, thank you very much for your response. My output is derived from a data-driven black box. I have improved my equestion based on your feedback.

[Sam Chak]

Hi @mohammed hussein

You first need to estimate a transfer function that fits the step response data. This is a topic in system identification, and you can use the tfest() function to address this type of problem. Otherwise, you can also intelligently guess the coefficients if you are familiar with the underlying system.

clear all
clc
load("Vo_vn.mat")
load("x.mat")
Volt    = Vo_vn;
Time    = x;
OUTS    = stepinfo(Volt,Time);
Desire  = 23.5;
Error   = Vo_vn-Desire;
RiseT   = OUTS.RiseTime

RiseT = 1.3248e-04

SettT   = OUTS.SettlingTime

SettT = 0.0037

OVERSH  = (OUTS.Peak)-Desire

OVERSH = 14.3081

% Estimated model
s   = tf('s');
w   = 8e3;                      % natural frequency (guess)
K   = w^2;                      % stiffness coeff
zet = 0.16;                     % damping ratio (guess)
D   = 2*zet*w;                  % damping coeff
Gp  = K/(s^2 + D*s + K);        % 2nd-order transfer function

figure
step(Gp, Time(end))             % step response of model
hold on
plot(Time, Volt/Desire, 'r-')   % normalized data
legend('estimated model', 'black box data')
hold off
ylim([0 1.6])

[mohammed hussein]

Thank you for your suggestion

Answer 1

[Sam Chak]

Hi @mohammed hussein

I revisited the problem, and you can perhaps use this as a mini guide. The manually estimated model is not 100% accurate, but it is relatively close. This model is then used by pidtune() to automatically tune the PID gains using a user-defined phase margin, an approach based on a concept you should already be familiar with in electrical engineering.

load("Vo_vn.mat")
load("x.mat")
Volt    = Vo_vn;
Time    = x;
OUTS    = stepinfo(Volt,Time);
Desire  = 23.5;
Error   = Vo_vn-Desire;
RiseT   = OUTS.RiseTime

RiseT = 1.3248e-04

SettT   = OUTS.SettlingTime

SettT = 0.0037

OVERSH  = (OUTS.Peak)-Desire

OVERSH = 14.3081

% Estimated model
s   = tf('s');
w   = 8725;                 % natural frequency (guess)
K   = w^2;                  % stiffness coeff
zet = 0.158;                % damping ratio (guess)
D   = 2*zet*w;              % damping coeff
Gp  = K/(s^2 + D*s + K);    % 2nd-order transfer function

figure
step(Desire*Gp, Time(end))  % step response of model
hold on
plot(Time, Volt, 'r-')      % black box data
title('System Identification stage: Step Response')
legend('estimated model', 'black box data')
hold off
ylim([0 40])

% Auto-tuned PID Controller
opt = pidtuneOptions('PhaseMargin', 75, 'DesignFocus', 'reference-tracking');
Gc  = pidtune(Gp, 'PIDF', opt)

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

  with Kp = 7.02, Ki = 2.35e+04, Kd = 0.000514, Tf = 2.08e-07
 
Continuous-time PIDF controller in parallel form.

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

Gcl =
 
            1.884e11 s^2 + 2.568e15 s + 8.604e18
  ---------------------------------------------------------
  s^4 + 4.804e06 s^3 + 2.017e11 s^2 + 2.934e15 s + 8.604e18
 
Continuous-time transfer function.

% Plot results 
figure
step(Desire*Gcl, 5e-3), grid on     % simulated compensated system (not actual black box system)
hold on
plot(Time, Volt, 'r-')              % black box data
title('Controller Design stage: Step Response')
legend('simulated compensated system', 'uncompensated black box output')
hold off
ylim([0 40])