How can I get coder to generate c code for a script with the solve function

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Cordell Smith
Cordell Smith 2016-7-29
评论： Walter Roberson 2017-10-31
I am trying to get c code from a Matlab algorithm using Matlab coder but the code uses the solve function which is not supported in coder. I tried solving for the variable that I am trying to obtain but I do not get the same answer. I trying to solve for current in the equation below:
syms current
eqn = log((pi()*exp(1)*(radius(l).^2)*CD)/deltaE_max) == log(current);
soln2 = solve(eqn, current);
I_cath(l) = soln2;
Can anyone tell me how to solve for current without using the solve function?
  2 个评论
Markus Hahnenkamm
Markus Hahnenkamm 2017-10-27
编辑：Markus Hahnenkamm 2017-10-27
I am currently experiencing the same issue. I would like to integrate the solve in a c code. Is there a workaround?
Walter Roberson
Walter Roberson 2017-10-27
Markus Hahnenkamm:
There is no way to include solve() in generated code. You need to attempt to find a closed form solution in MATLAB and then code that solution to get a numeric result. You might need to use fzero() or fsolve(). You will probably find it useful to look at the 'file' option of matlabFunction()

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回答（2 个）

Walter Roberson
Walter Roberson 2017-10-27
Your code is
eqn = log((pi()*exp(1)*(radius(l).^2)*CD)/deltaE_max) == log(current);
Current is only found on the right hand side, in a fairly simple usage. To solve for it, take exp() of both sides:
exp(log((pi()*exp(1)*(radius(l).^2)*CD)/deltaE_max)) == exp(log(current))
which gives
(pi*CD*exp(1)*radius(l)^2)/deltaE_max == current
which gives a direct calculation for current based upon the other values. solve() is not needed.
Cordell Smith
Cordell Smith 2017-10-31
编辑：Walter Roberson 2017-10-31
Sorry I copied the equation incorrectly here is the actually equation:
for l = 1:radius_el
syms current
eqn = (4*pi()*A*B*deltaE_max) / current + log((pi()*exp(1)*(radius(l).^2)*CD)/deltaE_max) == log(current);
soln = solve(eqn, current);
current is actually on both sides of the equation so it was not such a simple solution. I used a work around by setting both sides of the equation equal to each other and got the approximate solutions by finding where they intersected using the code below:
for l = 1:radius_el
current=[.00001:0.00000001:.01];
y2=log(current);
y3=(4*pi()*A*B*deltaE_max) ./ current + log((pi()*exp(1)*(radius(l).^2)*CD)/deltaE_max);
dif=abs(y2-y3);
soln=find(dif==min(dif(:)))+1000;
  2 个评论
Walter Roberson
Walter Roberson 2017-10-31
syms A B deltaE_max radius(l) CD
eqn = (4*sym('pi')*A*B*deltaE_max) / current + log((sym('pi')*exp(sym(1))*(radius(l).^2)*CD)/deltaE_max) == log(current);
soln = solve(eqn, current)
This gives
(4*A*B*deltaE_max*pi)/wrightOmega(- log(1/(4*A*B*deltaE_max*pi)) - log((pi*CD*exp(1)*radius(l)^2)/deltaE_max))
Most of this can have code generated. The part that cannot is the WrightOmega. It is probably easier to find a replacement code if you rewrite in terms of LambertW, which happens to give
4*A*B*deltaE_max*pi / lambertw(4*A*B*deltaE_max^2*exp(-1)/(CD*radius(l)^2))
In that regard, I find https://github.com/DarkoVeberic/LambertW . Or you might be satisfied with using http://lasp.colorado.edu/cism/CISM_DX/code/CISM_DX-0.50/required_packages/octave-forge/main/specfun/lambertw.m
Walter Roberson
Walter Roberson 2017-10-31
In the special case that A, B, deltaE_max, and CD are fixed ahead of time, and radius does not vary much, you could consider doing a taylor expansion. This would involve a number of uses of LambertW(4*A*B*deltaE_max^2*exp(-1)/(nominal_R^2*CD)) but under the special situation I mention that could be calculated ahead of time.

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