lsqnonlin with jacobian problem

3 次查看（过去 30 天）
Mus Bohr
Mus Bohr 2012-7-19
TE=2:5:5*20;
S(1,:)=104*exp(-TE/10);
options=optimset('Algorithm','levenberg-marquardt','Display','off','Jacobian ','on','Tolfun',1e-6 );
P0=[59 30];
P=lsqnonlin(@test,P0,[],[],options,S,TE)
function [F,J]=test(P,S,TE)
Ft=P(1)*exp(-TE/P(2));
F=S-Ft;
if nargout >1
J(:,1)=exp(-TE/P(2));
J(:,2)=P(1)*TE.*exp(-TE/P(2))/(P(2)^2);
end
  1 个评论
Walter Roberson
Walter Roberson 2012-7-19
Are you encountering an error message? If so, what message and where?

请先登录，再进行评论。

请先登录，再回答此问题。

回答（5 个）

Walter Roberson
Walter Roberson 2012-7-19
Mus Bohr
Mus Bohr 2012-7-19
Thank you for your response.
My example contains only two variables (parameters). The result of the fitting is exact when I turn Jacobian 'off'. Otherwise, the result is completely erroneous.
Best,
MB
  1 个评论
Walter Roberson
Walter Roberson 2012-7-19
编辑：Walter Roberson 2012-7-19
P=lsqnonlin(@test,P0,[],[],options,S,TE)
  1. @test
  2. P0
  3. []
  4. []
  5. options
  6. S
  7. TE
That is 7 parameters. lsqnonlin() does not accept anything after "options".

请先登录，再进行评论。

Mus Bohr
Mus Bohr 2012-7-19
Thank you for this precision, but lsqnonlin function accept parameters after "options".
Indeed, I found the solution to my problem. Actually, my scripts is correct, but I should to swap Ft and S (in 'test' function). In other words, F=Ft-S instead of F=S-Ft.
Anyway, thank you so much for you time.
Warm regards,
MB.
  2 个评论
Walter Roberson
Walter Roberson 2012-7-19
parameters after "options" has no defined result, and so is subject to change at any time, without notice. We repeatedly get Questions here from people who have attempted to pass extra parameters in a similar manner only to have the function fail because of it. Is there a point in relying on accidental behavior when a simple and well-documented adjustment is available? http://www.mathworks.com/help/toolbox/optim/ug/brhkghv-7.html
Mus Bohr
Mus Bohr 2012-7-19
I take note. Again thank you.
Best,
MB

请先登录，再进行评论。

Mus Bohr
Mus Bohr 2012-7-19
or provide Jacobian with the negative sign.
Star Strider
Star Strider 2012-7-19
Swapping S and Ft so that F = Ft - S will likely solve your problem. In the objective function you gave it, the lsqnonlin function uses the Jacobian of F in its calculation, not the Jacobian of Ft, and while they may look the same, the derivatives of F = S - Ft will be the negative of the ones you posted, while the derivatives of F = Ft - S will have the same signs as those you posted.
This is likely the reason that with the ‘Jacobian’ option ‘off’, your function converged.

类别

在 Help Center 和 File Exchange 中查找有关 Signal Modeling 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