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Doubt math

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Nuno
Nuno on 6 Apr 2011
Commented: Image Analyst on 15 Oct 2015
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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Accepted Answer

Matt Tearle
Matt Tearle on 7 Apr 2011
OK, to expand on the cyclist's answer:
  1. rewrite your equation in the form f(I) = 0: ICC - IR.*(exp((V1+Rs*I)./(m.*VT))-1) - ((V1+Rs.*I)/Rp) - I = 0
  2. 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;
  3. 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).

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Walter Roberson
Walter Roberson on 11 Apr 2011
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
Nuno on 11 Apr 2011
This expression:
-(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
But, how do you transform the expression in this form?
Walter Roberson
Walter Roberson on 11 Apr 2011
I used a different symbolic package to get that, but it is likely that solve() like Tim showed should be able to handle it.

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More Answers (3)

Matt Fig
Matt Fig on 6 Apr 2011
What do you mean "without math methods?" MATLAB uses only math methods as far as I know...

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Matt Tearle
Matt Tearle on 7 Apr 2011
Yeah, we wouldn't want any awkwardly named functions...
**cough CUMTRAPZ cough***
Image Analyst
Image Analyst on 12 Oct 2012
Or "ASSEMPDE".
Image Analyst
Image Analyst on 15 Oct 2015
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.

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the cyclist
the cyclist on 6 Apr 2011
You could use the function "fzero" to solve this equation.

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Nuno
Nuno on 7 Apr 2011
But how fzero resolve this problem?

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Tim Zaman
Tim Zaman on 6 Apr 2011
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
Walter Roberson on 6 Apr 2011
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
Nuno on 7 Apr 2011
Ups... I don't understand...
Walter Roberson
Walter Roberson on 7 Apr 2011
"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.

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