what is the problem with the second 'IF' loop ? Program terminates after entering into the loop.
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kintali narendra
on 7 Jun 2016
Answered: kintali narendra
on 8 Jun 2016
I am able to enter into the second "IF" but the simulink program terminates after executing the lines in the second "IF" condition. Simulink runs for the required time without the second "IF" condition.
function [Out] = LMAImplementation_Motor(In)
persistent eWithNoChange eWithDeltaL eWithDeltaK
e = In(1);
L = In(2);
K = In(3);
Clock = In(4);
if (0.0000 <= Clock) && (Clock <= 0.9999)
deltaK = 0; deltaL = 0;
eWithNoChange(end+1) = e;
[m,n] = size(eWithNoChange);
end
if n == 30001 %THIS LINE WAS BOLD
eWithNoChange(:,1) = [];
sizeofeWithNoChange = size(eWithNoChange)
end
if (1.0000 <= Clock) && (Clock <= 1.9999)
deltaL = L*0.1; deltaK = 0;
eWithDeltaL(end+1) = e;
end
Out = [deltaL,deltaK,Clock];
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