flying
Posts:
10
Registered:
4/13/12


Re: How to deal with complex nonlinear least square problem?
Posted:
Nov 28, 2012 9:06 PM


"Torsten" wrote in message <k958kk$1mm$1@newscl01ah.mathworks.com>... > "flying" wrote in message <k9567n$lea$1@newscl01ah.mathworks.com>... > > Thank you very much! > > > > I've encountered another problem.I met the error "optimization terminated norm of the current step is less than options tolx" when i use 'lsqnonlin' function.This means that the outcome is the same as the initial values i set,so how to deal with it? > > > > Sincerely > > frank > > There seems to be an error in your problem setup, but nothing useful can be said > unless we see the relevant part of your MATLAB code. > > Best wishes > Torsten.
My code is to deal with a multivariables problem.There are many parameters to be optimized.The vectors: 'sigAmpEst' 'alphaEst' 'rEst' have the same dimension. My problem is that i only get the initial parameters when i use 'lsqnonlin',maybe the model to be optimized is too complicated or the parameters to be estimated is too many.I don' t konw why,any good idea?
options = optimset('Display','iter','MaxFunEvals',5e3,'MaxIter',1000,'TolFun',1e7,'TolX',1e11); [x,renorm,residual,exitflag,output] = lsqnonlin(@lsqnonlinCE,[real(sigAmpEst) imag(sigAmpEst) alphaEst rEst],[],[],options);
function y=lsqnonlinCE(z) % Function used in 'lsqnonlin.m'
global pi2 c lj yk1 yk2 fk1 fk2 sigAmpEst
N = length(sigAmpEst); NL = length(yk1); NH = length(yk2); ykce1 =[]; ykes1 =[]; for ii = 1:NL for jj = 1:N ykce1 = [ykce1 (z(jj) + lj*z(jj+N))*exp(z(jj+2*N)*fk1(ii))*exp(lj*2*pi2*z(jj+3*N)*fk1(ii)/c)]; end ykce1 = sum(ykce1); ykes1 = [ykes1 abs(yk1(ii) ykce1)]; end ykce2 =[]; ykes2 =[]; for ii =1:NH for jj = 1:N ykce2 = [ykce2 (z(jj) + lj*z(jj+N))*exp(z(jj+2*N)*fk2(ii))*exp(lj*2*pi2*z(jj+3*N)*fk2(ii)/c)]; end ykce2 = sum(ykce2); ykes2 = [ykes2 abs(yk2(ii) ykce2)]; end y = [ykes1 ykes2]; Sincerely Frank

