- rewrite your equation in the form f(I) = 0: ICC - IR.*(exp((V1+Rs*I)./(m.*VT))-1) - ((V1+Rs.*I)/Rp) - I = 0
- having defined all the constants (ICC, m, VT, etc), make a function of I using an anonymous function handle: f = @(I) ICC - IR.*(exp((V1+Rs*I)./(m.*VT))-1) - ((V1+Rs.*I)/Rp) - I;
- apply fsolve to f
Doubt math
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I have the next expression and my unknown is "I".
I = ICC - IR.*(2.718.^((V1+Rs*I)./(m.*VT))-1) - ((V1+Rs.*I)/Rp);
Exist any function im Matlab that resolve this expression without math methods?
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采纳的回答
Matt Tearle
2011-4-7
OK, to expand on the cyclist's answer:
Like Walter, I'm assuming 2.718.^foo really means e^foo, which, in MATLAB, should be implemented as exp(foo).
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Walter Roberson
2011-4-11
I used a different symbolic package to get that, but it is likely that solve() like Tim showed should be able to handle it.
更多回答(4 个)
Matt Fig
2011-4-6
What do you mean "without math methods?" MATLAB uses only math methods as far as I know...
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Image Analyst
2015-10-15
A new funny one is "removecats" (in the Statistics and Machine Learning Toolbox). People have been trying to use it on youtube and Facebook videos.
Tim Zaman
2011-4-6
I guess what you need is just a solver; for example you define
syms ICC IR V1 Rs m VT Rp;
solve('I = ICC - IR.*(2.718.^((V1+Rs*I)./(m.*VT))-1) - ((V1+Rs.*I)/Rp)')
-untested-
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Walter Roberson
2011-4-7
"How this works" is that the symbolic solver does pattern matching and determines that the expression matches a pattern that it knows how to solve. It then substitutes the components from your actual expression in to the general solution to the kind of problem that it has decided your expression is. It so happens that the pattern matched is one whose answer is expressed in terms of the Lambert W function. _Why_ the Lambert W function is the answer for those kinds of problems is a topic for a series of lectures in complex analysis.
Jenna
2023-2-23
编辑:Jenna
2023-2-23
Unfortunately, there is no built-in MATLAB function that can solve an equation like the one you have given without using any mathematical methods. However, MATLAB does have functions that can help you solve equations numerically using methods like Newton-Raphson or the Bisection method.
Here's an example of how you could use the fsolve function in MATLAB to solve your equation:
% Define the function you want to solve
fun = @(I) ICC - IR.*(2.718.^((V1+Rs*I)./(m.*VT))-1) - ((V1+Rs.*I)/Rp);
% Use fsolve to find the value of I that solves the equation
I = fsolve(fun, x0);
% Display the result
disp(['The value of I that solves the equation is ', num2str(I)]);
In this example, x0 is an initial guess for the value of I. You would need to define the values of ICC, IR, V1, Rs, m, VT, and Rp beforehand.
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