# How can I use the matlab function like fcn block

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titor7 on 16 Feb 2023
Answered: Paul on 18 Feb 2023
Hello guys,
Ex. when i write basic spring-damper equation, i got error. In fcn block, it would be written like (-1/m2)*(k1*(u(3)-u(1))+k2*(u(3)-u(5))+c1*(u(4)-u(2))+c2*(u(4)-u(6))). How can I use the matlab function like fcn block?
function y = fcn(u)
m2=40;
c1=1000;
c2=0;
k1=13000;
k2=220000;
y =(-1/m2)*(k1*(u(3)-u(1))+k2*(u(3)-u(5))+c1*(u(4)-u(2))+c2*(u(4)-u(6)));
end
titor7 on 17 Feb 2023
hi @Paul, x1_dotdot is different.
function y = fcn(u)
m1=290;
c1=1000;
k1=13000;
y = (-1/m1)[(k1*(u1-u3))+(c1*(u2-u4)];
end
titor7 on 17 Feb 2023
I'm modelling quarter car suspension system. Step signal is the force to move car to foward.
Normally, if i would with fcn function and scope u1(displacement of x1), it would be like this.

Sulaymon Eshkabilov on 17 Feb 2023
There was one missing block from the step input signal that is a derivative block. Note that using the derivative block is not recommended to use that might lead to unstable results.
Another important point is that the damping coefficient c2 =0, and if c2 =c1, then the system response will become damped otherwise it vibrates for a quite long time. You should adjust stiffness and adamping coefficient values in both MATLAB Fcn blocks.
The relative tolerance was also tightened (1e-4) in the model settings.
See the attached model with two corrections.
All the best.
Here are the simulation results:
Warning: An error occurred while evaluating "loc_createToolchain" in "/MATLAB/toolbox/coder/autosar/rtwTargetInfo.p": Unrecognized function or variable 'autosarroot'.
This custom registration is not loaded.
Warning: Solver is encountering difficulty in simulating model 'M_Fcn_Simulink' at time 1.0000000000000038. Simulink will continue to simulate with warnings. Please check the model for errors.
Warning: Solver was unable to reduce the step size without violating minimum step size of 3.55271E-15 for 1 consecutive times at time 1. Solver will continue simulation with the step size restricted to 3.55271E-15 and using an effective relative error tolerance of 0.0135238, which is greater than the specified relative error tolerance of 0.0001. This usually may be caused by the high stiffness of the system. Please check the system or increase the solver Number of consecutive min steps violation parameter.
plot(S.ScopeData{1}.Values.Time, S.ScopeData{1}.Values.Data, 'r-')

Sam Chak on 17 Feb 2023
I think this one should be .
In this blog, there is an article on the "Quarter Car Modeling" by João Marabisa.
You can refer the blocks used in the modeling.
Also check out this tutorial:
titor7 on 17 Feb 2023
I know how to do this type modelling actually. I just want to do this using matlab function also. Thank you much for anyway.

Paul on 18 Feb 2023
Hi titor7,
The reason for the error in the original model show in this comment is because Simulink didn't have enough information to determine the dimensions of y outputs of the Matlab Function blocks.
This problem can be addressed in one of two ways. The "clean" way is to double click on the Matlab Function block, go to the function tab and click "Edit Data." On the Symbols pane click on 'y' and in the Property Inspector change the Size from - 1, which means inherited, to 1. Do the same for the other Matlab Function block.
The quick way is to add a line of code to the Matlab Function block to force the parser to recognize that y is a scalar. Like this
function y = fcn(u)
y = 0; % define y as a scalar
m2=40;
c1=1000;
c2=0;
k1=13000;
k2=220000;
y =(-1/m2)*(k1*(u(3)-u(1))+k2*(u(3)-u(5))+c1*(u(4)-u(2))+c2*(u(4)-u(6)));
end
function y = fcn(u)
y = 0; % define y as a scalar
m1=290;
c1=1000;
k1=13000;
%y = (-1/m1)[(k1*(u1-u3))+(c1*(u2-u4)];
y = (-1/m1)*(k1*(u(1)-u(3)))+(c1*(u(2)-u(4)));
end
You'll still have to deal how to approximate the derivative of the step function, but at least now the model will update and run.

Sulaymon Eshkabilov on 17 Feb 2023
MATLAB Fcn block can be embedded - see attached Simulink model of your exercises with some modifications. It works perfectly fine. There is one importnat point is that has been overlooked in the MATLAB fcn syntax - see the following syntax of the MATLAB fcn block:
function y = fcn(u1, u2, u3, u4, u5, u6)
m2=40;
c1=1000;
c2=0;
k1=13000;
k2=220000;
y =(-1/m2)*(k1*(u3-u1)+k2*(u3-u5)+c1*(u4-u2)+c2*(u4-u6));
end
titor7 on 17 Feb 2023
It's worked kind of but i want to scope each u1,u2,u3... and it should be like this.

titor7 on 17 Feb 2023
Edited: titor7 on 17 Feb 2023
Sulaymon Eshkabilov on 17 Feb 2023
Edited: Sulaymon Eshkabilov on 17 Feb 2023
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