Solve an equation and get a solution that depends on a variable

Hi everyone! I need some help with getting a solution to a equation. I am trying to solve a equation using solve. However i dont get any solution to the equation, i only get Empty sym: 0-by-1. I can solve the equation manually, the problem is that i have other equation that i cant solve manually and would like a better way to solve them. Is there any good way to solve a equation and get a solution that has a dependent variable in return. Here is what i have tried so far.

syms Nin(t)
% Vin = (sign(t - 200/11)/2 + 1/2)*((11*t)/100000 - 1/500) - (11*t)/100000 + 1/500
eq1 = Nin == Constant * Vin;
sol = solve(eq1, t);

sol -> Empty sym: 0-by-1

Manually

Nin = Constant * Vin;
-> Nin = Constant * (sign(t - 200/11)/2 + 1/2)*((11*t)/100000 - 1/500) - (11*t)/100000 + 1/500

Any guidance is appreciated.

[Torsten]

You request the equation to be solved for t, not for Nin.

[Jonas Ström Seez]

Forgot to change that. However when i try

sol = solve(eq1, Nin);

the solution still is Empty sym: 0-by-1.

Have i thought wrong. Is there some other way to solve for Nin.

[Torsten]

syms Nin t

instead of

syms Nin(t)

[Jonas Ström Seez]

It worked! Thank you very much!

Answer 1

[Walter Roberson]

syms Nin(t)
% Vin = (sign(t - 200/11)/2 + 1/2)*((11*t)/100000 - 1/500) - (11*t)/100000 + 1/500
eq1 = Nin == Constant * Vin;
sol = solve(eq1, t);

That code requests that an unknown function, Nin(t) be compared to Constant times an expression in t, and that the result be solved for the t that makes the two sides equal. The result would implicitly be of the form

t = some function of Nin(t) and Constant * Vin

except it cannot be, because the right hand side of solving for t cannot contain t, but it must because Nin(t) is an unknown function of t.

Therefore solve() must fail.

Nin = Constant * Vin;
-> Nin = Constant * (sign(t - 200/11)/2 + 1/2)*((11*t)/100000 - 1/500) - (11*t)/100000 + 1/500

That is not a solution for t, it would simply make the two sides of eqn1 the same as each other, in which case solving for t would have to give all t.