size mismatch in matlab function block

i dont know why there are some errors about size mismatch in my code, because as you can see, variable s has dimension is [1,2],Cl is [2,2] .Therefore i think the result will be [1,2] ,but it doesnt work.Please help me.
My code:
function [Cl,gl,ds] = fcn(q1,q2,dq2,dq1, Tl,Th,de,e,s,t)
%Matrices of dynamic equation.
a = 1.024;b = 0.24;Ro = 0.16;
g = 9.8;
Ml = [a+2*b*cos(q2) Ro+b*cos(q2); Ro+b*cos(q2) Ro];
Cl = [-2*b*sin(q2)*dq2 -b*sin(q2)*dq2; b*sin(q2)*dq1 0];
gl = [2.5*g*Ro*cos(q1+q2)+1.67*0.864*cos(q1) 2.5*g*Ro*cos(q1+q2)];
lamda = 0.6; ep = 3;d =1.2;
theta = [0.99; 0.12];
wl = 0.3+0.2*sin(2*t)+0.3*sin(20*t) - 0.2*sin(10*t)+ 0.3*sin(21*t);
w0 = 0.2 + 0.3*sin(15*t);
% dieu kien cua w estimation
if norm(s) > ep
Row = d*s./norm(s);
else
Row = d*s./ep;
end
we = w0 + Row;
% el(t) = -e(t);
Yl =( - de*Ml*lamda - e*Cl*lamda +gl);
ds =(-de*Ml*lamda -s*Cl - Yl*theta -(2*we-wl)+Tl+Th).*inv(Ml);

 Respuesta aceptada

Fangjun Jiang
Fangjun Jiang el 28 de Oct. de 2020
Editada: Fangjun Jiang el 28 de Oct. de 2020

0 votos

[1 2]*[2 2; 2 2] is correct.
[1 2].*[2 2; 2 2] is wrong (without implicit expansion)

3 comentarios

chuyen hoangcao
chuyen hoangcao el 28 de Oct. de 2020
please take a glance at my code. i set variable ds with [1,2] , after i deleted my '.' , it has this error. But i think '.*' is the multiplication of array?
Fangjun Jiang
Fangjun Jiang el 28 de Oct. de 2020
take a particular set of values for all the input arguments, run the code line by line, stop at the line where the error occured, run indivudual section of that line of code if necessary, you will find where and what caused the error.
chuyen hoangcao
chuyen hoangcao el 29 de Oct. de 2020
it ran when i delete '.' in front of inv 'Ml'.Thank you

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el 28 de Oct. de 2020

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