Effacer les filtres
Effacer les filtres

Cycle counter with reset

6 vues (au cours des 30 derniers jours)
Luccas S.
Luccas S. le 15 Déc 2021
Commenté : Sargondjani le 18 Déc 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

Réponse acceptée

Sargondjani
Sargondjani le 16 Déc 2021
you can use the function mod. For example:
if mod(n,5) == 0
or variants thereof.
  2 commentaires
Luccas S.
Luccas S. le 17 Déc 2021
I think it would be interesting to work with if (mod(n,5)==0) && (PE(n,1))>p_fix)
or is it not necessary?
Sargondjani
Sargondjani le 18 Déc 2021
Yes, think that's what you need. But my advice is to always check if an algoirthm does what you expect.

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