How can I use the iteration to store variables? It gives error Subscript indices must either be real positive integers or logicals.

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Yokuna
Yokuna 2021-6-17
评论： Yokuna 2021-6-20
I wrote the code for 2 agents, but facing problem in storing variables in iteration. As the error shows Subscript indices must either be real positive integers or logicals. I want to rum the loop for 50 iterations.
t0=0;
tf=20;
tspan=[t0:1:tf];
x0=[3;0;0];
u1(1)=3; u2(1)=-3;
E1(1)=2; E2(1)=2;
itt=50;N=4;
y1=zeros
y1(1)=3; y2(1)=-1;
a=[0 -1;-1 0];
% For 1 agent
for i=1:itt
options=odeset('RelTol',1e-3,'AbsTol',1e-6);
[t,x]=ode23(@new,tspan,x0,options)
y1_est(i)=y1(i-1)+E1(i)*(du1(i));
err1(i)=y1(i)-y1_est(i);
if (sign(E1(i))<0)
E1(i)=E1(1);
else
E1(i)=E1(i-1)+(eta*du1(i-1)*err1(i-1))/(mu+du1(i-1)^2)
end
end
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
function xdot = new(t,x)
x1=x(1);
x2=x(2);
x3=x(3);
X1=[x1;x2;x3];
A=[0.5 0.035 0.025;0.0255 0.6 -0.99;0.75 0.03 0.025];
B=[0.2;0.2;0];
C=[1 0 1];
y1=C*X1;
xdot1 =A*X1+B*u1;
xdot =[xdot1];
end

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采纳的回答

Geoff Hayes
Geoff Hayes 2021-6-17
Shivanshu - presumably the error is with
for i=1:itt
options=odeset('RelTol',1e-3,'AbsTol',1e-6);
[t,x]=ode23(@new,tspan,x0,options)
y1_est(i)=y1(i-1)+E1(i)*(du1(i));
because on the first iteration of the list, i is 1 yet the code tries to access y1(i-1)...hence the error. I think that you need to reconsider your how you are accessing data from these arrays. I was going to suggest starting from 2 (i.e. for i=2:itt) but I suspect that will lead to other issues.
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Geoff Hayes
Geoff Hayes 2021-6-18
If I just use dummy values for Y1 and Y2, then I see the same error as you. It appears that the output from new1 is no longer a column vector/array but a matrix
xdot =
2.1000 2.2440
0.6765 0.8205
2.2500 2.2500
-0.5750 -0.7190
-1.5900 -1.7340
0.0250 0.0250
and this may be related to how your u1 and u2 are computed. I recommend using the MATLAB debugger to step through the code to determine why this happens.

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