Borrar filtros
Borrar filtros

How do I define an output vector as real and positive

1 visualización (últimos 30 días)
Christos
Christos el 21 de Sept. de 2014
Editada: Roger Wohlwend el 22 de Sept. de 2014
Hi all
I am currently running this script to fit a complex function on my data using lsqcurvefit.
clear all;
hold off E1=importdata('E1.mat'); %E1 and DR1 are the names of my arrays
DR=importdata('DR1.mat');
hold on plot(E1,DR)
%x(1)=C
%x(2)=f
%x(3)=Eg
%x(4)=G
x0= [1 0 1 0.02 ];
F=@(x,E1)x(1)*exp(1i*x(2))./((E1-x(3)-1i*x(4))^2);
[x,resnorm,~,exitflag,output] = lsqcurvefit(F,x0,E1,DR);
plot(E1,F(x,E1),'--r')
hold off
It is working perfectly except for the part where all the fitted parameters are returned as complex numbers when they should be real and positive. Is there some way to fix this?
Thanks in advance
  1 comentario
Star Strider
Star Strider el 21 de Sept. de 2014
I don’t know what you are doing, but (from the Euler identities), exp(1i*x(2)) will be evaluated essentially as cos(x(2)) + i*sin(x(2)), and you are also multiplying x(4)^2 by 1i. If you do not want complex parameters, you will likely have to re-write your model. I would go to the literature to see what others have done in similar situations.

Iniciar sesión para comentar.

Respuestas (1)

Roger Wohlwend
Roger Wohlwend el 22 de Sept. de 2014
Editada: Roger Wohlwend el 22 de Sept. de 2014
The same question occured recently on Matlab answers: Link to the question.

Categorías

Más información sobre Get Started with Curve Fitting Toolbox en Help Center y File Exchange.

Community Treasure Hunt

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

Start Hunting!

Translated by