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I encountered error while running program(runga kutta solving using function handle) for solving a set of differential equations and applying a controller 'u'. How to tackle it out?

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SHIJINA P P
SHIJINA P P 2016-7-20
评论： Geoff Hayes 2016-7-21
matrix dimension doesn't match in line 86 while solving the attached file 'abc'
  2 个评论
SHIJINA P P
SHIJINA P P 2016-7-21
I want out(4,i) to be an array, but when that is used in u (via sdot), error occurs, Is it due due the presence of 'u' in function handle?
Geoff Hayes
Geoff Hayes 2016-7-21
u appears to be used on a line following the one that generates the error. If you use the MATLAB debugger, what do you see?

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回答（1 个）

Geoff Hayes
Geoff Hayes 2016-7-21
Shijina - use the MATLAB debugger to step through the code and see what is happening. Consider these lines which initializes a variable named out as
x=[21900;301.4*2;0;0;2*pi/180;0];
out( :,1)=x;
and so out is a 6x1 array. Now, in your for loop, you do some work and then update out as
out(:,i+1)=x;
and so out is now a 6x2 array. The problem is with line 86
sdot=(1/d)*(-out(4,:)-(rho*A*out(2,:)*Cl/(2*m)-(g/out(2,:)-...
out(2,:)/(R+out(1,:)))*cos(out(3,:)))-rho*A*out(2,:)*Cl/(2*m)-(g/out(2,:)-...
out(2,:)/(R+out(1,:)))*cos(out(3,:))-lambda*(rho*(out(2,:))^2*A*D/(2*I)*(-...
Cm+Cmq*D*out(4,:)/out(2,:))))
Look how out is used. On the first iteration, when out is a 6x1 array, out(4,:) is a scalar as are all other instances where we access this variable. But on the subsequent iteration of the loop, out(4,:) (and the others) will be 1x2 arrays. Is this what you want? Or do you just want the scalar from the previous iteration i.e.
out(4,i)
were i is the indexing variable in your for loop.

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