Cycle counter with reset

3 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.

Connectez-vous pour commenter.

Plus de réponses (0)

Catégories

En savoir plus sur Loops and Conditional Statements dans Help Center et File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by