I dont understand, i try to use solve to find x in Matlab Function Block

Question

huy duc 2022-1-18

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https://ww2.mathworks.cn/matlabcentral/answers/1630925-i-dont-understand-i-try-to-use-solve-to-find-x-in-matlab-function-block

评论： Walter Roberson 2022-1-19

function x = fcn(y)
sym a;
r1=7.3e-5;
r2=-1.1e-7;
s1=1.6e-1;
s2=1.38e-3;
s3=-1.6e-5;
t1=1.60e-2;
t2=-1.3;
t3=4.12e2;
A=0.25;
Tel=25;
sol =double(solve( 1.297 + (r1+r2)*Tel*a/A + (s1 + s2*Tel + s3*(Tel^2))*log(((t1 + t2/Tel + t3/(Tel^2))*a/A)+1)-y,a));
x = sol;

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Answer 1

Walter Roberson 2022-1-18

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在 MATLAB Online 中打开

@Benjamin

has shown you how to solve the equation using syms

However, syms requires the MATLAB Interpreter, and so is only available in a MATLAB Function block if you have rapid acceleration turned off completely (or sometimes on the next setting if you are careful); it cannot be used if you need code generation to C or C++ code or deployment.

If you need to generate code or you need deployment, then you need a different approach: you need to use fzero() or fsolve() to find the root of the expression, or you need to work with the formula to find the general solution.

MATLAB is not able to find the general formula, but using Maple, and letting the 1.297 be named C, then

r1=7.3e-5;
r2=-1.1e-7;
s1=1.6e-1;
s2=1.38e-3;
s3=-1.6e-5;
t1=1.60e-2;
t2=-1.3;
t3=4.12e2;
A=0.25;
Tel=25;
C = 1.297;
a = ((Tel^2*s3 + Tel*s2 + s1) * (Tel^2*t1 + Tel*t2 + t3) * ...
     lambertw(Tel^3 * (r1+r2) * ...
       exp((Tel^3 * (r1+r2) - t1 * (C-y) * Tel^2 - t2 * (C-y) * Tel - t3 * (C-y)) / ...
        (Tel^2 * s3 + Tel * s2 + s1) / ...
        (Tel^2 * t1 + Tel * t2 +t3)) / ...
        (Tel^2 * s3 + Tel * s2 + s1) / ...
        (Tel^2 * t1 + Tel * t2 + t3)) ...
     - Tel^3 * (r1+r2)) ...
   * A / Tel / (r1+r2) / (Tel^2 * t1 + Tel * t2 + t3)

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Walter Roberson 2022-1-19

在 MATLAB Online 中打开

opt = optimoptions('Algorithm', 'Levenberg-Marquardt');
iel = fsolve(f, uel, opt);

Walter Roberson 2022-1-19

在 MATLAB Online 中打开

syms x uel

f = uel - (3 + 3 * x + 5 * log(3 * x + 1))

f =

solve(f, x)

ans =

string(ans)

ans = "(5*wrightOmega(uel/5 - log(5) - 2/5))/3 - 1/3"

This tells you that there is a direct solution without using fsolve, using the wrightOmega function

However, wrightOmega is part of the Symbolic Toolbox, and so probably cannot have code generated for it.

You might, however, be able to take advantage of https://www.mathworks.com/matlabcentral/fileexchange/56407-wrightomegaq

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Answer 2

Voss 2022-1-18

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编辑：Voss 2022-1-18

在 MATLAB Online 中打开

Looks like you should use syms rather than sym

fcn(0)
ans = -0.4008
function x = fcn(y)
syms a;
r1=7.3e-5;
r2=-1.1e-7;
s1=1.6e-1;
s2=1.38e-3;
s3=-1.6e-5;
t1=1.60e-2;
t2=-1.3;
t3=4.12e2;
A=0.25;
Tel=25;
sol = double(solve( 1.297 + (r1+r2)*Tel*a/A + (s1 + s2*Tel + s3*(Tel^2))*log(((t1 + t2/Tel + t3/(Tel^2))*a/A)+1)-y,a));
x = sol;
end