# Cycle counter with reset

8 visualizaciones (últimos 30 días)
Luccas S. el 15 de Dic. de 2021
Comentada: Sargondjani el 18 de Dic. de 2021
I apologize for not having developed any code for this part, because I have no idea how to do it. So, I'm going to work with the block diagram and what I've created so far.
I can't think of a way to do this:
Basically, when p_fix is ​​found it is to check if PE>p_fix every 5 cycles repeatedly until IC occurs. If PE>p_fix is ​​not respected during these 5 cycles, it is to evaluate 5 cycles again and again....
What has been programmed so far:
for n = 1:size(t,1)
if n>=4
X = [Ia(n-1,1) Ia(n-2,1) ; Ia(n-2,1) Ia(n-3,1)];
future = [Ia(n,1) ; Ia(n-1,1)];
C = X\future;
Ia_future(n,1) = C(1,1)*Ia(n,1)+C(2,1)*Ia(n-1,1);
PE(n,1)=Ia(n,1)+Ia_future(n,1);
p(n,1) = (1+0.2)*max(PE);
if PE(n,1)>p(n,1)
p_fix = p(n,1);
end
end
end ##### 0 comentariosMostrar -1 comentarios más antiguosOcultar -1 comentarios más antiguos

Iniciar sesión para comentar.

Sargondjani el 16 de Dic. de 2021
you can use the function mod. For example:
if mod(n,5) == 0
or variants thereof.
##### 2 comentariosMostrar 1 comentario más antiguoOcultar 1 comentario más antiguo
Sargondjani el 18 de Dic. de 2021
Yes, think that's what you need. But my advice is to always check if an algoirthm does what you expect.

Iniciar sesión para comentar.

### Categorías

Más información sobre Whos 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!