How to solve an implicit handle function with two variables?

조회 수: 2 (최근 30일)
Dor Gotleyb
Dor Gotleyb 2022년 3월 14일
댓글: Dor Gotleyb 2022년 3월 14일
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

답변 (1개)

Torsten
Torsten 2022년 3월 14일
I = @(V) -I0 + lambertw(I0*R*exp(I0*R+V))/R;
  댓글 수: 3
Torsten
Torsten 2022년 3월 14일
편집: Torsten 2022년 3월 14일
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 2022년 3월 14일
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

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Optimization Toolbox에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by