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

1 次查看（过去 30 天）
HONG CHENG
HONG CHENG 2021-3-20
评论： Walter Roberson 2021-3-20
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.

请先登录，再回答此问题。

采纳的回答

Walter Roberson
Walter Roberson 2021-3-20
编辑：Walter Roberson 2021-3-20
Ran in:
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
  5 个评论
显示 3更早的评论
HONG CHENG
HONG CHENG 2021-3-20
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
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.

请先登录，再进行评论。

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Optimization 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by