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

1 Ansicht (letzte 30 Tage)
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 Kommentare
Guillaume
Guillaume am 19 Jun. 2015
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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