what is the problem with the second 'IF' loop ? Program terminates after entering into the loop.
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
kintali narendra
el 7 de Jun. de 2016
Respondida: kintali narendra
el 8 de Jun. de 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];
0 comentarios
Respuesta aceptada
Más respuestas (0)
Ver también
Categorías
Más información sobre Simulink Functions en Help Center y File Exchange.
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!