Simulate closed loop system with saturated controller output in Matlab

It is entirely possible to simulate a closed-loop system in MATLAB using the ode45 solver. Since ode45 solves problems in the time domain, you will need to obtain the equivalent continuous-time model representation of the Plant transfer function and the PID controller transfer function. I believe all undergrad control theory textbooks cover this procedure, so I will not elaborate on the details.

A   = [0  1
      -4 -1];
B   = [0
       4];
C   = [1  0];
D   =  0;
Sp  = ss(A, B, C, D)
Sp =
 
  A = 
       x1  x2
   x1   0   1
   x2  -4  -1
 
  B = 
       u1
   x1   0
   x2   4
 
  C = 
       x1  x2
   y1   1   0
 
  D = 
       u1
   y1   0
 
Continuous-time state-space model.
Gp  = tf(Sp)    % verify if Gp has the same Plant TF
Gp =
 
       4
  -----------
  s^2 + s + 4
 
Continuous-time transfer function.

Regarding saturation, which refers to a linear input–output mapping with lower and upper bounds, math-oriented control theorists typically use mathematical formulas (as they need to publish in journals), while non-math, tool-dependent, or AI-assisted control practitioners commonly use If-Else conditional statements.

x = linspace(-3, 3, 6001);

m = 1; % Slope

y = m*x; % Linear input–output mapping (straight line)

Lb = -1; % Lower bound

Ub = 1; % Upper bound

for i = 1:length(y)

if y(i) < Lb

sat(i) = Lb;

elseif y(i) > Ub

sat(i) = Ub;

else

sat(i) = y(i);

end

plot(x, sat), grid on, grid minor, ylim([-1.5, 1.5])

xlabel('x'), ylabel('sat(x)'), title('Saturation function')

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Answer 2

Zinea 2024-9-5，7:01

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编辑：Zinea 2024-9-5，8:44

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Hi Cedric Kotitschke,

Simulating a closed-loop system with saturation on the controller output directly in MATLAB (without Simulink) requires a bit of manual intervention since MATLAB's built-in functions like feedback do not support saturation directly. However, you can achieve this by implementing a custom loop with saturation logic. Here's a simple way to do it:

Define the plant and controller.
Implement a custom simulation loop.
Apply saturation to the controller output.

Here are a few points to be noted:

The continuous-time plant and controller are converted to discrete-time using ‘c2d’ to facilitate the simulation loop.
The controller output is saturated using simple ‘min’ and ‘max’ functions.
The loop iteratively calculates the error, computes the controller output, applies saturation, and updates the plant output.

Here's a basic example of how you can achieve the saturation for closed-loop:

% Define the plant and controller 
sysPlant = tf(4, [1 1 4]); 
sysCtrl = pid(10, 3, 5, 0.01); 
% Define simulation parameters 
tFinal = 10; % Final time 
dt = 0.01; % Time step 
time = 0:dt:tFinal; 
% Define saturation limits 
uMin = -10; 
uMax = 10; 
% Initialize variables 
y = zeros(size(time)); % Output 
u = zeros(size(time)); % Control input 
e = zeros(size(time)); % Error 
r = ones(size(time)); % Reference (step input) 
% Convert the systems to discrete-time for simulation 
sysPlantD = c2d(sysPlant, dt); 
sysCtrlD = c2d(sysCtrl, dt); 
% Get the discrete transfer function coefficients 
[numPlant, denPlant] = tfdata(sysPlantD, 'v'); 
[numCtrl, denCtrl] = tfdata(sysCtrlD, 'v'); 
% Initialize states 
xPlant = zeros(length(denPlant) - 1, 1); 
xCtrl = zeros(length(denCtrl) - 1, 1); 
% Simulation loop 
for k = 1:length(time)-1 
    % Calculate the error 
    e(k) = r(k) - y(k); 
    
    % Controller output (before saturation) 
    [uCtrl, xCtrl] = filter(numCtrl, denCtrl, e(k), xCtrl); 
    
    % Apply saturation 
    u(k) = max(min(uCtrl, uMax), uMin); 
    
    % Plant output 
    [y(k+1), xPlant] = filter(numPlant, denPlant, u(k), xPlant); 
end 
% Plot the results 
figure; 
subplot(2,1,1); 
plot(time, y); 
title('System Output'); 
xlabel('Time (s)'); 
ylabel('Output'); 
subplot(2,1,2); 
plot(time, u); 
title('Control Input with Saturation'); 
xlabel('Time (s)'); 
ylabel('Control Input'); 