How to solve a system of equations whose unknowns values are exponents with cole-cole methods

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HONG CHENG 2021-3-20

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https://ww2.mathworks.cn/matlabcentral/answers/778227-how-to-solve-a-system-of-equations-whose-unknowns-values-are-exponents-with-cole-cole-methods

评论： Walter Roberson 2021-3-20

采纳的回答： Walter Roberson

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I want to solve this equation,

I want to calculate the value of B1,B2

I use this equation

eh = 45.;
kl = 4.15;
de1 = 6200;
fc1 = 500*10^3;
de2 = 990;
fc2 = 2.70*10^6;
b1 = 0.86;
b2=0.98;
f=10^4;
% ev=8.854187817 * 10^(-12);
ev=8.854 * 10^(-12); %pF/m
e0 = eh+de1/(1+j*f/fc1)^b1+de2/(1+j*f/fc2)^b2+kl/(j*2*pi*f*ev)
e1 = real(e0);
e2 = -imag(e0);
%%
x0 = [b1,b2];
% options=optimset('MaxFunEvals',1e4,'MaxIter',1e4);
options = optimoptions('fsolve','Display','iter','TolFun',1e-30,'TolX',1e-30);
[x,FVAL,EXITFLAG,OUTPUT,JACOB] = fsolve(@(x)fun_cole_cole(x,fc1,fc2,f,de1,de2,eh,kl,ev,e1,e2),x0,options);

The fun_cole_cole function is shown below

function f = fun_cole_cole(x,fc1,fc2,f,de1,de2,eh,kl,ev,e1,e2)  % b c可以是随意的参数
A1 = fc1^x(1)+f^x(1)*cos(pi*x(1)/2);
A2 = fc2^x(2)+f^x(2)*cos(pi*x(2)/2);
B1 = f^x(1)+sin(pi*x(1)/2);
B2 = f^x(2)+sin(pi*x(2)/2);
D1 = de1*fc1^x(1);
D2 = de2*fc2^x(2);
w = 2*pi*f;
f1 = A1*D1/(A1^2+B1^2)+A2*D2/(A2^2+B2^2)+eh-e1;
f2 = B1*D1/(A1^2+B1^2)+B2*D2/(A2^2+B2^2)+kl/(w*ev)-e2;
f = [f1;f2];

But I got this result

No solution found.
fsolve stopped because the problem appears to be locally singular.

When I set options as

options=optimset('MaxFunEvals',1e4,'MaxIter',1e4);

I got this

No solution found.
fsolve stopped because the relative size of the current step is less than the
default value of the step size tolerance squared, but the vector of function values
is not near zero as measured by the default value of the function tolerance.

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Answer 1

Walter Roberson 2021-3-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/778227-how-to-solve-a-system-of-equations-whose-unknowns-values-are-exponents-with-cole-cole-methods#answer_653132

编辑：Walter Roberson 2021-3-20

在 MATLAB Online 中打开

This reacts well to a standard trick of converting a system of equations into a least-squared minimization.

The fsolve() is not needed; I kept it in to show you that the result of the fmincon() checks out with fmincon()

format long g
eh = 45.;
kl = 4.15;
de1 = 6200;
fc1 = 500*10^3;
de2 = 990;
fc2 = 2.70*10^6;
b1 = 0.86;
b2=0.98;
f=10^4;
% ev=8.854187817 * 10^(-12);
ev=8.854 * 10^(-12); %pF/m
e0 = eh+de1/(1+j*f/fc1)^b1+de2/(1+j*f/fc2)^b2+kl/(j*2*pi*f*ev)
e0 = 
           7233.00405024904 -      7459936.41901046i
e1 = real(e0);
e2 = -imag(e0);
%%
x0 = [b1,b2];
% options=optimset('MaxFunEvals',1e4,'MaxIter',1e4);
options = optimoptions('fsolve','Display','iter','TolFun',1e-3,'TolX',1e-3);
f = @(x)fun_cole_cole(x,fc1,fc2,f,de1,de2,eh,kl,ev,e1,e2);
syms x [1 2]
residue = sum(f(x).^2);
R  = matlabFunction(residue,'vars',{x});
[best, fval] = fmincon(R, x0)
Local minimum possible. Constraints satisfied.

fmincon stopped because the size of the current step is less than
the value of the step size tolerance and constraints are 
satisfied to within the value of the constraint tolerance.
best = 1×2
          1.09674543072127         0.653154719676118
fval = 
      1.68411061989985e-11
[x,FVAL,EXITFLAG,OUTPUT,JACOB] = fsolve(f,best,options)
                                         Norm of      First-order   Trust-region
 Iteration  Func-count     f(x)          step         optimality    radius
     0          3     1.68383e-11                       0.00096               1

Equation solved at initial point.

fsolve completed because the vector of function values at the initial point
is near zero as measured by the value of the function tolerance, and
the problem appears regular as measured by the gradient.
x = 1×2
          1.09674543072127         0.653154719676118
FVAL = 2×1
      2.15581258089514e-06
      3.49152833223343e-06
EXITFLAG = 
     1
OUTPUT = struct with fields:
       iterations: 0
        funcCount: 3
        algorithm: 'trust-region-dogleg'
    firstorderopt: 0.000960191844270147
          message: '↵Equation solved at initial point.↵↵fsolve completed because the vector of function values at the initial point↵is near zero as measured by the value of the function tolerance, and↵the problem appears regular as measured by the gradient.↵↵<stopping criteria details>↵↵Equation solved. The final point is the initial point.↵The sum of squared function values, r = 1.683830e-11, is less than sqrt(options.FunctionTolerance) = 3.162278e-02.↵The relative norm of the gradient of r, 9.601918e-04, is less than options.OptimalityTolerance = 1.000000e-03.↵↵'
JACOB = 2×2
          91.0225376370537            112.5927734375
         -331.207215298002                  -134.375
function f = fun_cole_cole(x,fc1,fc2,f,de1,de2,eh,kl,ev,e1,e2)  % b c可以是随意的参数
A1 = fc1^x(1)+f^x(1)*cos(pi*x(1)/2);
A2 = fc2^x(2)+f^x(2)*cos(pi*x(2)/2);
B1 = f^x(1)+sin(pi*x(1)/2);
B2 = f^x(2)+sin(pi*x(2)/2);
D1 = de1*fc1^x(1);
D2 = de2*fc2^x(2);
w = 2*pi*f;
f1 = A1*D1/(A1^2+B1^2)+A2*D2/(A2^2+B2^2)+eh-e1;
f2 = B1*D1/(A1^2+B1^2)+B2*D2/(A2^2+B2^2)+kl/(w*ev)-e2;
f = [f1;f2];
end

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HONG CHENG 2021-3-20

在 MATLAB Online 中打开

I think I have another question. I am not sure if my understanding is right or not.

we can use

[best, fval] = fmincon(R, x0)

to finds a constrained minimum of a function of several variables.

after we get the values of best, we use them as the intial values for fsolve

options = optimoptions('fsolve','Display','iter','TolFun',1e-3,'TolX',1e-3);
[x,FVAL,EXITFLAG,OUTPUT,JACOB] = fsolve(f,best,options)

But finally, we get the same results and also get the values of EXITFLAG,OUTPUT,JACOB

Walter Roberson 2021-3-20

As I wrote, "The fsolve() is not needed; I kept it in to show you that the result of the fmincon() checks out with fmincon()"

fmincon() is a minimizer; it could potentially have returned a value that was not a root, or it might have gotten stuck just a little away from a root. Any time you switch between techniques (I recoded fsolve as fmincon) then at the beginning you want to reassure yourself that the new technique does as well as the old one would have done. Once you have tried a few cases and seen that it does work, you can get rid of the fsolve call.

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