# Finding x for y = f(x) in simulink

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Phan Van Long 2022 年 1 月 18 日

I have an equation which is uel = 3 + 3 * iel + 3 * log (3 * iel +1) a.k.a uel = f(iel) and I want to find iel ( uel is already known), so I have used fsolve to solve the problem in matlab and it works. Here is the code
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function iel = test(uel)
f = @(x) uel-(3 + 3 * x + 5 * log(3 * x + 1));
iel = fsolve(f,uel);
end
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But when i try to add this code into matlab function block in simulink, the model just don't run. I want my input to be uel and output to be iel. Can anyone please help me on this problem. Thank you so so much
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Phan Van Long 2022 年 1 月 19 日
This is the error message sir @Max Heimann

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### 回答 (1 件)

Paul 2022 年 1 月 19 日

According to the Symbolic Math Toolbox, the solution can be expressed as
%iel = wrightOmega(uel/3 - log(3) - 2/3) - 1/3
assuming that iel and uel are real. If they can be complex, the expression for iel is the same, but there are some other conditions that have to be satisfied.
Check
%doc wrightOmega
Check a couple of test cases (note that I changed the 5 to a 3 in the function test() to match the equation in the question)
uel = 0.3
uel = 0.3000
iel = test(uel)
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
iel = -0.1723
iel = wrightOmega(uel/3 - log(3) - 2/3) - 1/3
iel = -0.1723
uel = 5.4
uel = 5.4000
iel = test(uel)
Equation solved. fsolve completed because the vector of function values is near zero as measured by the value of the function tolerance, and the problem appears regular as measured by the gradient.
iel = 0.2465
iel = wrightOmega(uel/3 - log(3) - 2/3) - 1/3
iel = 0.2465
function iel = test(uel)
% f = @(x) uel-(3 + 3 * x + 5 * log(3 * x + 1)); % doesn't match equation in Question
f = @(x) uel-(3 + 3 * x + 3 * log(3 * x + 1));
iel = fsolve(f,uel);
end

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R2021b

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