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Error : In an assignment A(I) = B, the number of elements in B and I must be the same. Help !!!!

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Ahmad Sheikh
Ahmad Sheikh 2015-6-19
关闭： MATLAB Answer Bot 2021-8-20
I am working on takagi sugeno fuzzy modelling and a write a code like but its giving an error i dont understand why my function is like this.
function derivative_x = project_aft(t,x)
derivative_x=zeros(5,1);
z1 = 1342999.9 .* x(1) .* x(2) - 1342999.9 .* x(1) .* 1.1983 + 7748604.2 .* q .* x(2) + 7748604.2 .* x(3) .*1 - 172.19 .* 1.962.* x(2) + 172.19 .*1.962;
z2 = 684505.6 .* ( 1.962 + 11.1 .* (x(3) ./ (x(1).^2)));
z3 = (3874302.146 ./ x(1)) .* (1.962 .* x(2) +1.962);
a1 = 0.0269;
a2 = -0.0660;
b1 = -7.91;
b2 = -8.73;
c1 = -0.923;
c2 = -2.09;
a= a1-a2;
b= b1-b2;
c= c1-c2;
M11= (a1 - z1) ./ a;
M21= (z1 - a2) ./ a;
M12= (b1 - z2) ./ b;
M22= (z2 - b2) ./ b;
M13= (c1 - z3) ./ c;
M23= (z3 - c2) ./ c;
h1= M11* M12* M13
h2= M11* M12* M23
h3= M11* M22* M13
h4= M11* M22* M23
h5= M21* M12* M13
h6= M21* M12* M23
h7= M21* M22* M13
h8= M21* M22* M23
h=h1+h2+h3+h4+h5+h6+h7+h8
derivative_x(1)= 0.0001.*h1.*x(1)-1.4915.*h1.*x(2)-1.1387.*h1.*x(3)+14.9782.*h1.*x(4)-0.0001.*h1.*x(5)-0.0001.*h2.*x(1)-1.4915.*h2.*x(2)-1.1370.*h2.*x(3)+14.9782.*h2.*x(4)-0.0001.*h2.*x(5)-0.0001.*h3.*x(1)-1.4914.*h3.*x(2)-1.1387.*h3.*x(3)+14.9782.*h3.*x(4)-0.0001.*h3.*x(5)-0.0001.*h4.*x(1)-1.4914.*h4.*x(2)-1.1371.*h4.*x(3)+14.9782.*h4.*x(4)-0.0001.*h4.*x(5)-1.4915.*h5.*x(2)-1.1385.*h5.*x(3)+14.9784.*h5.*x(4)-0.0001.*h5.*x(5)-1.4915.*h6.*x(2)-1.1369.*h6.*x(3)+14.9784.*h6.*x(4)-0.0001.*h6.*x(5)-1.4914.*h7.*x(2)-1.1386.*h7.*x(3)+14.9784.*h7.*x(4)-0.0001.*h7.*x(5)-1.4914.*h8.*x(2)-1.1370.*h8.*x(3)+14.9784.*h8.*x(4)-0.0001.*h8.*x(5);
In simple
8
derivative_x(1)= ∑ h(i) [A(i) * x]
i=1
the other editor file where i called the above function is
Initial_Time=0;
Final_Time=100;
[x1,x2] = meshgrid(-2:0.1:2);
[t,x] = ode45(@project_aft, [Initial_Time Final_Time], [0 10 0 0 0]);
plot(t,x(:,1));
xlabel('time')
ylabel('state x1k')
  2 个评论
Guillaume
Guillaume 2015-6-19
Please, give us the entire error message (everything in red), particularly the part that shows the line where the error occurs.
Note that instead of writing a very long expression which increases the risk of mistakes, you would be much better off storing your h and A in a column vector, that is:
h = [M11 .* M12 .* M13;
M11 .* M12 .* M13;
...
M21 .* M22 .* M23];
A = [0.0001; -1.4915; ....; -0.0001];
Then derivative_x(1) is simply:
derivative_x(1) = sum(h .* A .* x);

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