c = -1; % define parameter first x = fsolve(@(x) myfun(x,c),[-5;-5]) %x1和x2的初值都定为-5 结果如下. x = 0.5671 0.5671 fval = -1.0640e-08 exitflag = 1 分析如下:x1=0.5671,x2=0.5671,差值在-8个数量级，结果理想。 求解方程： Parameter. Value. Description ... fminunc,fsolve , lsqcurvefit, ... 优化函数 数据库性能优化 非线性数据 线性函数归一化 非SPJ优化 非凸优化 ...

Us=fsolve(@(Us) Uval(Us,Ud,Sbt,Uf,Vg) , init);..... Code above works on matlab. The equation is parametric. And depends on "Ud" "Sbt" "Uf" "Vg" parameters. "Us" is unknown (like x) "Ud" is 1*100 matrix and rest are constants. In matlab after this code "Us" becomes 1*100 matrix and has been solved 100 times according to "Ud" . Your Matlab function will be the complex and fsolve function statement in your earlier code, and replace the variable theta with u (this is the Matlab Fcn block convention); return the output dimension as theta1 and theta2 for each of the complex and fsolve functions respectively. 57591 5 18 0. fsolve stopped because the last step was ineffective. However, the vector of function values is not near zero, as measured Note the message - the result you have obtained isn't valid and matlab yells full of it, just read the message. However if you take another initial guess you can obtain the solution easily.