How to solve an implicit handle function with two variables?

2 vues (au cours des 30 derniers jours)
Dor Gotleyb
Dor Gotleyb le 14 Mar 2022
Commenté : Dor Gotleyb le 14 Mar 2022
Hi,
I have the following handle function:
Vd = @(V,I) V-I*R;
I = @(V,I) I0*(exp(Vd(V,I))-1);
How can I find I(V)=?
I0,R are constants.
Thanks

Réponses (1)

Torsten
Torsten le 14 Mar 2022
I = @(V) -I0 + lambertw(I0*R*exp(I0*R+V))/R;
  3 commentaires
Torsten
Torsten le 14 Mar 2022
Modifié(e) : Torsten le 14 Mar 2022
I don't know what you mean by "In reality my functin (I) is more complex then the the Lambert W function".
I = -I0 + lambertw(I0*R*exp(I0*R+V))/R
solves the equation
I = I0*(exp(V-I*R)-1)
for I.
If your equation is more complex, use "fzero" or "fsolve".
Dor Gotleyb
Dor Gotleyb le 14 Mar 2022
I meant that my function is more complax then just I0*(exp(V-I*R)-1).
its a sum of several functions, but I didn't wont the quastion to be very long, just to understand the consept.
I = @(V,I) I0*(exp(Vd(V,I))-1) + a2*(exp(a3 * Vd(V,I))-1) + a1*sqrt(Vd(V,I)) + ...;
Ok, So I used 'fzero' as you suggested to solve for V=0:
F = @(0,I) -I + I0*(exp(Vd(0,I))-1);
Sol = fzero(F, 0);
And for other voltages I used the previous solution as a guess.
Thank you very mach

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