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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Matt Tearle
2011-4-7
0 个投票
OK, to expand on the cyclist's answer:
- 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
Like Walter, I'm assuming 2.718.^foo really means e^foo, which, in MATLAB, should be implemented as exp(foo).
13 个评论
Nuno
2011-4-7
Walter Roberson
2011-4-7
That "rewrite your equation" is intended in a mathematical sense. The MATLAB implementation is as shown by Matt in step 2.
Nuno
2011-4-7
Matt Tearle
2011-4-7
No, use the exact code I gave in step 2 to define f:
f = @(I) ICC - IR.*(exp((V1+Rs*I)./(m.*VT))-1) - ((V1+Rs.*I)/Rp) - I;
This creates f as an anonymous function handle, so f is a function of the dummy variable I. Everything else in that expression is a variable that has already been defined.
Nuno
2011-4-8
Walter Roberson
2011-4-8
No, fsolve(f)
Nuno
2011-4-8
Walter Roberson
2011-4-8
Do you have the optimization toolbox that defines fsolve() ? If not, then
fzero(f,-10)
The answer does not vary with initial starting point.
Matt Tearle
2011-4-8
which fsolve
If you have it, then I = fsolve(f,I0) where I0 is an initial guess (such as -10, like Walter showed).
If not, then I = fzero(f,I0) (again, as Walter showed).
Nuno
2011-4-11
Walter Roberson
2011-4-11
No, you can only solve for a single value of V1 if you are using fzero() . You could solve over multiple V1 if you had the optimization toolkit and fsolve() but the setup would change.
If you go back to the symbolic LambertW expression that I showed, then you should be able to vectorize that.
Nuno
2011-4-11
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
2 个投票
What do you mean "without math methods?" MATLAB uses only math methods as far as I know...
14 个评论
Tim Zaman
2011-4-6
he just wants the "I" out of the middle and last part of the equation singled out.
Matt Fig
2011-4-6
Perhaps...
Nuno
2011-4-6
Matt Tearle
2011-4-6
Matt, it's a dirty little secret that MATLAB was actually written by astrologers with ouija boards. fzero works by summoning demons from the Fourth Circle.
OK, seriously, Nuno, do you mean you want a symbolic answer (ie "rearrange the equation to get I = expression in terms of V1, etc")? Or perhaps a numeric answer ("for a given set of values for V1, IR, etc, I = 0.626782")?
Sean de Wolski
2011-4-6
@Matt T, if fzero summons demons; does bsxfun summon oompa-lumpas to perform the operations?
Matt Fig
2011-4-6
@Sean de, BSXFUN simply is the Fourth Circle.
Matt Tearle
2011-4-7
Oompa Loompa doompety doo
We have a binary expansion for you
Oompa Loompa doompety dee
It has squat to do with binary
Matt Fig
2011-4-7
They had to come up with something, because SXFUN looks too much like sucks-fun - hardly a good name for a built-in function!
the cyclist
2011-4-7
That's not the first thing that sprang to mind when I saw SXFUN.
Matt Fig
2011-4-7
That remark made me laugh out loud, cyclist! Too funny.
Nuno
2011-4-7
Matt Tearle
2011-4-7
Yeah, we wouldn't want any awkwardly named functions...
**cough CUMTRAPZ cough***
Image Analyst
2012-10-12
Or "ASSEMPDE".
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
0 个投票
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-
4 个评论
Nuno
2011-4-6
Walter Roberson
2011-4-6
Symbolically, assuming 2.718 represents exp(1),
-(V1-(-LambertW(-Rs*IR*Rp*exp(Rp*(Rs*ICC+Rs*IR+V1)/(m*VT*(Rp+Rs)))/(-Rs*m*VT-Rp*m*VT))+Rp*(Rs*ICC+Rs*IR+V1)/(m*VT*(Rp+Rs)))*m*VT)/Rs
Ugly. But it is the standard form to solve such equations, which is a fact you would have to know through mathematical experience as the Lambert W function is not one of the obvious ones.
Nuno
2011-4-7
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.
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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