Simulink Difference Equation Implementation

I have a homework(a). My professor ask me for implement a discrete pid controller in difference equation form to matlab function. This function block should run as a discrete pid controller block in simulink model(b). When simulation began, the Matlab function block is executed once. Because of for loop an array occurs. Then that array signal goes the plant. And this is not usefull. During the simulation what i want is every iteration of for loop gives a value and that value must go to plant. But i could not do this. The same issue happened previous homework(c). I don't even know whether this way is proper or not.

[Alvery]

Think of your function as only solving a single step - it takes the current input, applies the difference function just once to it, and then produces the next result. This leaves you with a challenge - in order to make the difference function work, you also need to know the previous value that was computed.

How do you do memory? you can either store it externally (i.e. not in Matlab code, but in a separate Simulink block) in a 1/Z block, or you can store it internally in the Matlab. Internal storage is easy if you cheat - using 'persistent' variables, but this hides the memory from the solver which potentially messes up backtracking behaviour if your solver setting uses it. The correct alternative is to use stored state.

[Sezer Memis]

Thank you for answer. But is your advice proper for my situation? I don't get it. Here is some figures from previous homework but same issue:

When function run as matlab script, above output occurs.

Same function in Matlab Function block in Simulink, above output occurs. Function code is given as follows:

function y2 = serialProgramming(e1)

T = 0.1;
t_final = 20;
t_initial = 0;
lenght_of_loop = (t_final-t_initial)/T;

y1 = zeros(lenght_of_loop,1);
y2 = zeros(lenght_of_loop,1);
x1 = zeros(lenght_of_loop,1);
x2 = zeros(lenght_of_loop,1);
%initial conditions
x0 = 0; 
x1(1,1) = x0;
x2(1,1) = x0;

for k = 2:lenght_of_loop
   
    x1(k,1) = e1 + 0.9 * x1(k-1,1) ;
    y1(k) = 0.004617 * x1(k-1,1) ;
    
    x1(k-1,1) = x1(k,1) ;
    
    x2(k,1) = y1(k,1) + 0.874 * x2(k-1,1) ;
    y2(k,1) = x2(k,1) + 0.9231 * x2(k-1,1) ;
    
    x2(k-1,1) = x2(k,1) ;
end

Simulink structure is simple like this:

I want the output which is in Simuink like script's output. But i could not do that. If my method is completely wrong, how can i replace a PID(z) block with a difference equation Matlab function in Simulink. The summary :

[Walter Roberson]

If you are using a discrete system then use https://www.mathworks.com/help/simulink/slref/memory.html memory block.

If you are using a continuous system, then you would have difficulty implementing something like the trapazoid rule.

[Paul]

Hard to say how to make the two equivalent without knowing exactly what's inside the block C(s)1.

Answer 1

[Paul]

Maybe this example will help, which essentially illustrates the comment made by Alvery.

Consider a plant P(s) = 1/(s + 5) that is controlled via PI compensation with Ki = Kp = 1. The sample time of the discrete control is Ts = 0.01, with Ts defined as such in the base workspace. This system is modeled in the top half of this diagram, where the sample time parameter for the Zero Order Hold and Discrete Time Integrator blocks is set to Ts.

Now we wish to implement the PI control in a Matlab Function block as in the boltom half of the diagram. We have the following equation for the control input:

u[k] = x[k] + Kp*e[k]  % Kp = 1

The difference equation for x[k] is determined from the z-transform of the integrator:

X(z)/E(z) = Ki*Ts/(z-1)  % Ki = 1
(z-1)*X(z) = Ki*Ts*E(z)
x[k+1] - x[k] = Ki*Ts*e[k]
x[k+1] = Ki*Ts*e[k] + x[k]

so the PI function needs to impement the equations for u[k] and x[k]. Here's the function

function u = PI(e)

persistent x

% initialize the integrator output
if isempty(x)
    x = 0;
end

Kp = 1;
Ki = 1;
Ts = 0.01; % needs to be same value as Ts defined in the workspace

% control at the current time step
u = Kp*e + x;

% update the integrator for the next time step
x = Ki*Ts*e + x;


end

And here is the scope from the model showing that the implementation works as expected:

[Sezer Memis]

Thank you so much. It works well. You are great!