[Too many input arguments] Hello, I have problem with my s-function,I guess a classical error. When I run the following code

function [sys,x0,str,ts]=rbf_sensorless_test(t,x,u,flag)
switch flag
case 0
[sys,x0,str,ts]=mdlInitializeSizes;
case 1
[sys,x0,str,ts]=mdlDerivatives(t,x,u);
case 3
[sys,x0,str,ts]=mdlOutputs(t,x,u);
case {2, 4, 9 }
sys = [];
otherwise
error(['Unhandled flag = ',num2str(flag)]);
end
function [sys,x0,str,ts]=mdlInitializeSizes
global  Y  L c b
sizes = simsizes;
sizes.NumContStates = 6;
sizes.NumDiscStates = 0;
sizes.NumOutputs = 1;
sizes.NumInputs = 2;
sizes.DirFeedthrough = 0;
sizes.NumSampleTimes = 0;
sys=simsizes(sizes);
x0 = [0*ones(6,1)];
str=[];
ts=[];
c=[-0.5 -0.25 0 0.25 0.5;
    -0.5 -0.25 0 0.25 0.5]; 
b=[0.2 0.2 0.2 0.2 0.2]'; 
L=8.5e-3;
Y=[2 0 0 0 0;0 2 0 0 0;0 0 2 0 0;0 0 0 2 0;0 0 0 0 2];
function sys=mdlDerivatives(u)
global  Y  L c b
x1=u(1); 
x2=u(2);
x=[x1 x2]';
h=zeros(5,1); 
for j=1:1:5
h(j)=exp(-norm(x-c(:,j))^2/(2*b(j)*b(j))); 
end
dw=(1/L)*Y*x2*h;
de=(1/L)*x2;
for i=1:1:5
sys(i)=dw(i);
end
sys(6)=de;
function sys=mdlOutputs(~)
w=[x(1) x(2) x(3) x(4) x(5)]';
e=w'*h+x(6);
sys=e;

Answer 1

[Walter Roberson]

function sys=mdlOutputs(~)
w=[x(1) x(2) x(3) x(4) x(5)]';
e=w'*h+x(6);
sys=e;

You call this function with three arguments but your code says that it expects one argument only and that it should ignore the one argument.

Then your function proceeds to try to use variables x and h that are not defined in that context. You could have gotten x from a parameter if you had not ignored the parameters. But h is defined in another function only, and that other function might not have been called or not called for the current state.

You can use an extra function that calculates h and call that function from both places.

[thinh le duc]

Thanks For the help, I have identified a part of the error, but still have the same error, please help me to continue.

function sys=mdlOutputs(t,x,u)

global c b

x1=u(1);

x2=u(2);

x=[x1 x2]';

h=zeros(5,1);

for j=1:1:5

h(j)=exp(-norm(x-c(:,j))^2/(2*b(j)*b(j)));

end

w=[x(1) x(2) x(3) x(4) x(5)]';

e=w'*h+x(6);

sys=e;

• while executing MATLAB S-function 'rbf_sensorless_test', flag = 3 (output), at time 0.0.
• Too many output arguments.

[Walter Roberson]

[sys,x0,str,ts]=mdlOutputs(t,x,u);

That needs four outputs but the function is only defined with one output.

[thinh le duc]

function [sys,x0,str,ts]=mdlOutputs(t,x,u)
global   c b
x1=u(1); 
x2=u(2);
x=[x1 x2]';
h=zeros(5,1); 
for j=1:1:5
h(j)=exp(-norm(x-c(:,j))^2/(2*b(j)*b(j))); 
end
w=[x(1) x(2) x(3) x(4) x(5)]';
e=w'*h+x(6);
sys=e;

i tried but it did not work!! :((

• while executing MATLAB S-function 'rbf_sensorless_test', flag = 3 (output), at time 0.0.
• Index exceeds array bounds.

[Walter Roberson]

x1=u(1); 
x2=u(2);
x=[x1 x2]';

So your x will be a 2 x 1 vector, overwriting the x that was passed in to the function.

w=[x(1) x(2) x(3) x(4) x(5)]';

There you need the 5th element of the vector that is now only 2 elements.

e=w'*h+x(6);

and there you need the 6th element of the vector that is now only 2 elements.

[thinh le duc]

Sorry I just started learning this, so there are many mistakes. I tried again but it still didn't work. Thank you so much, please help me!!

function [sys,x0,str,ts]=rbf_sensorless_test(t,x,u,flag)

switch flag
case 0
[sys,x0,str,ts]=mdlInitializeSizes;
case 1
[sys,x0,str,ts]=mdlDerivatives(t,x,u);
case 3
[sys,x0,str,ts]=mdlOutputs(t,x,u);
case {2, 4, 9 }
sys = [];
otherwise
error(['Unhandled flag = ',num2str(flag)]);
end

function [sys,x0,str,ts]=mdlInitializeSizes

global  Y  L c b
sizes = simsizes;
sizes.NumContStates = 6;
sizes.NumDiscStates = 0;
sizes.NumOutputs = 1;
sizes.NumInputs = 2;
sizes.DirFeedthrough = 1;
sizes.NumSampleTimes = 0;
sys=simsizes(sizes);
x0 = [0 0 0 0 0 0]';
str=[];
ts=[];
c=[-0.5 -0.25 0 0.25 0.5;-0.5 -0.25 0 0.25 0.5]; 
b=0.2; 
L=8.5e-3;
Y=[2 0 0 0 0;0 2 0 0 0;0 0 2 0 0;0 0 0 2 0;0 0 0 0 2];

function sys=mdlDerivatives(~,~,u)

global  Y  L c b
m=u(1); 
n=u(2);
m_n=[m n]';
h=zeros(5,1); 

for j=1:1:5
h(j)=exp(-norm(m_n-c(:,j))^2/(2*b^2)); 
end

dw=(1/L)*n*Y*h;
de=(1/L)*n;
sys(1)=dw(1);
sys(2)=dw(2);
sys(3)=dw(3);
sys(4)=dw(4);
sys(5)=dw(5);
sys(6)=de;

function sys=mdlOutputs(~,~,u)

global   c b
m=u(1); 
n=u(2);
w=[x(1) x(2) x(3) x(4) x(5)]';
m_n=[m n]';
h=zeros(5,1); 

for j=1:1:5
h(j)=exp(-norm(m_n-c(:,j))^2/(2*b^2)); 
end

e=w'*h+x(6);

sys=e;

This is my problem:

[Walter Roberson]

You are not building m=sin(t), n=sin(t-pi/3) ?

Your e output uses ϵ but you have not defined that; you have defined . But you are not calculating from n ?

At the moment it is not obvious that you need any output other than e ?

[thinh le duc]

I have provided missing information. I used simulink

The ultimate goal is to calculate "e".

I already calculated the ε

function sys=mdlDerivatives(~,~,u)

global  Y  L c b
m=u(1); 
n=u(2);
m_n=[m n]';
h=zeros(5,1); 

for j=1:1:5
h(j)=exp(-norm(m_n-c(:,j))^2/(2*b^2)); 
end

dw=(1/L)*n*Y*h;  %.............
de=(1/L)*n;      %.............

sys(1)=dw(1);
sys(2)=dw(2);
sys(3)=dw(3);
sys(4)=dw(4);
sys(5)=dw(5);
sys(6)=de;

function sys=mdlOutputs(~,~,u)

global   c b
m=u(1); 
n=u(2);
w=[x(1) x(2) x(3) x(4) x(5)]';
m_n=[m n]';
h=zeros(5,1); 

for j=1:1:5
h(j)=exp(-norm(m_n-c(:,j))^2/(2*b^2)); 
end

ex=x(6);
e=w'*h+ex;

sys=e;

Hope you continue to help me :D

[Walter Roberson]

w=[x(1) x(2) x(3) x(4) x(5)]';

Where is that x vector coming from? You are inside

function sys=mdlOutputs(~,~,u)

which has no x parameter. You do have

global   c b

but there is no x in that.

Why do you have so many state variables? If you were doing an ode then you would have at most states for w, , ϵ and .

Your mdlDerivatives should not be the same as your mdlOutputs . https://www.mathworks.com/matlabcentral/answers/366520-how-can-explain-the-relation-between-mdlderivatives-output-and-mdloutputs-in-s-function-simulink#answer_291095

Your mdlDerivatives exists to calculate derivatives of your model at each time step. If you do not use the derivative anywhere (and MATLAB does not want to calculate them to resolve a transfer function or perhaps algebraic loop) then you do not need to supply that function. MATLAB can also estimate derivatives; mdlDerivatives permits more exact derivatives... when you want derivatives at all.

mdlOutputs is the output of your function, which is not the same as the derivatives.

[thinh le duc]

Thanks so much, I will learn more through what you have instructed. :D