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Simulink Difference Equation Implementation

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Sezer Memis
Sezer Memis on 15 Dec 2020
Commented: Sezer Memis on 22 Dec 2020
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.


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Sezer Memis
Sezer Memis on 16 Dec 2020
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, this output occurs.
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) ;
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
Walter Roberson on 20 Dec 2020
If you are using a discrete system then use memory block.
If you are using a continuous system, then you would have difficulty implementing something like the trapazoid rule.
Paul on 20 Dec 2020
Hard to say how to make the two equivalent without knowing exactly what's inside the block C(s)1.

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Accepted Answer

Paul on 20 Dec 2020
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;
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;
And here is the scope from the model showing that the implementation works as expected:

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Sezer Memis
Sezer Memis on 22 Dec 2020
Thank you so much. It works well. You are great!

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