what is the wrong wtih the “fsolve” function to solve the problem？

the figure is the lightning voltage,if the maxium amplitude of it is Um, the green line

crosses the two points:(t1,90%*Um) and (t2,30%*Um),and it intersect at x-axis and horizontal

line of Um。tf is called wave-front time，and the Tt is the half-tail time。Usuallly,we

suppose the voltage can express as the following equation(also double exponential equation):

u(t)=U*ζ*(exp(-a*t)-exp(-b*t)). Now, we know the tf,Tt and U(the maxium amplitude)，

the ζ，a，and b to be solved. this question can be solved by the matlab function fsolve. i

have defined the following fuction in matlab:

%--------------------------------------------------------------------%

    function Vc = imp_f(x,tf,Tt,U)
    %UNTITLED2 Summary of this function goes here
    %   Detailed explanation goes here
    % X is the [a；β；ζ；t1；t2]
    % a/β/ζ are the parameters in function ：i（t）=U*ζ*(exp(-a*t)-exp(-β*t))；
    % t1/t2/Tt/tf：(t1,90%*Um) and (t2,30%*Um) have the same meaning in the previous figure ，U is voltage crest，kV;

    Vc=[tf-(x(4)-x(5))/0.6;                           % tf=(x(4)-x(5))/0.6

        0.3*U-x(3)*(exp(-x(1)*x(5))-exp(-x(2)*x(5))); % cross the point (t2,30%*Um)

        0.9*U-x(3)*(exp(-x(1)*x(4))-exp(-x(2)*x(4))); % cross the point (t1,90%*Um)

        0.5*U-x(3)*(exp(-x(1)*(Tt+1.5*x(5)-0.5*x(4)))-exp(-x(2)*(Tt+1.5*x(5)-0.5*x(4)))); %cross the point  
                                                                                           (Tt+1.5*t2-0.5*t1,50%*Um)
        U-x(3)*(exp(-x(1)*log(x(2)/x(1))/(x(2)-x(1)))-exp(-x(2)*log(x(2)/x(1))/(x(2)-x(1))))];
                                                                                         %cross the maximu value point  
                                                                                            (log(β/a)/(β-a),Um)

    end

then use the following code to find the answer：

%--------------------------------------------------------------------------
tf=1.2;    %
Tt=50;     %
U=1.0;    %
x0=[1;1;1;1;1]; %初值；
[x,fval]=fsolve(@(x) imp_f(x,tf,Tt,U),x0);  % Call solver
%---------------------------------------------------------------

MATLAB give the following wrong answer：

%---------------------------------------------------------------
Error using trustnleqn (line 28)
Objective function is returning undefined values at initial point. FSOLVE cannot continue.

Error in fsolve (line 376)
    [x,FVAL,JACOB,EXITFLAG,OUTPUT,msgData]=...

Error in l_impluse (line 11)
[x,fval]=fsolve(@(x) imp_f(x,tf,Tt,U),x0);  % Call solver

%--------------------------------------------------------------

and i donot know the reason? and how to solve the problem?

Answer 1

[Walter Roberson]

@(x) imp_f(x,tf,tt,U) references the non-existent variable tt . There is a Tt a few lines earlier in the code but not tt

[cheng sy]

thanks for Walter Roberson. but tt is formal parameters，not actual.

[Walter Roberson]

You edited your code to fix that problem after I responded.

Have a look at your section

        0.5*U-x(3)*(exp(-x(1)*(Tt+1.5*x(5)-0.5*x(4)))-exp(-x(2)*(Tt+1.5*x(5)-0.5*x(4)))); %cross the point  
                                                                                           (Tt+1.5*t2-0.5*t1,50%*Um)

In that code, (Tt+1.5*t2-0.5*t1,50%*Um) is on a separate line, so it is not part of the comment "%cross the point", and so is being executed as code. But it has a comma in the middle of it and does not have a ')', so it is invalid syntax. You need to comment out that line.

[cheng sy]

thanks for Walter Roberson ,i comment out the line,but the matlab still give the wrong message:

%---------------------------------------------------------------
Error using trustnleqn (line 28)
Objective function is returning undefined values at initial point. FSOLVE cannot continue.
Error in fsolve (line 376)
    [x,FVAL,JACOB,EXITFLAG,OUTPUT,msgData]=...
Error in l_impluse (line 11)
[x,fval]=fsolve(@(x) imp_f(x,tf,Tt,U),x0);  % Call solver
%---------------------------------------------------------
and how to solve the problem? waiting for your answer.

[Torsten]

You divide by x(2)-x(1) which is 0 at the start.

Furthermore, be careful that during iteration, x(2)/x(1) remains positive. Otherwise, log(x(2)/x(1)) will produce complex results.

Best wishes

Torsten.

[cheng sy]

thank you for your remind.