imaginary value from solution nonlinier least square
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im trying to fitting data from reflectance measurement with lavenberg marquardt algorithm , ive got this code LMFnlsq in file exchange but i dont understand how to use it in my case
here is the equation
R = (0.6/(a + b))*(sqrt(3*a*(a + b))+(1/c))*(exp(-sqrt(3*a*(a + b ))*c))/(c^2)
a and b is parameter that im trying to solve
R is known vector from measurement reflectance
c is known vector from distance in measurement
im using the code like this
r = [0.8 0.9 1.0 1.1 1.2 1.3 1.4 1.6]';
data = [0.787357544 0.63199586 0.393109753 0.271890049 0.179074593 0.133498607 0.112568249 0.05274273]';
res = @(x) ((0.6./(x(1)+x(2))).*(sqrt(3.*x(1).*(x(1)+x(2)))+(1./r)).* (exp(-sqrt(3*x(1).*(x(1)+x(2)))*r))./(r.^2)) - data;
x0 = [0.1,10];
[x,ssq,cnt] = LMFnlsq(res,x0)
hold on
plot(r,data)
plot(r,res(x)+ data,'y'), grid
the problem is it sure did the iteration but the final value is in imaginary solution like this
x =
0.0528 - 0.2730i
1.1469 + 0.6248i
ssq =
0.0091
cnt =
29
anything wrong with the code?
采纳的回答
Miroslav Balda
2013-3-12
Let us reformulate the problem:
data = (q + 1./r).*exp(-r*q).*(0.6/(p(1)*r.^2)),
where
q = sqrt(p(2));
The problem may be solved without any troubles by introducing a penalty function
(q<=0)*w
into vector of residuals. Thus, the main script could be
% Surpan.m Application of LMFnlsq
%%%%%%%%%%% 12. 3. 2013
% Answers: Imaginary value from solution nonlinear least square
clc
clear all
close all
%
global data r p0 w
%
data = [0.779543143 0.624771688 0.387776261 0.266986051 0.174732672 ...
0.130313689 0.110540421 0.051412959]'; % column vector
r = [0.8 0.9 1 1.1 1.2 1.3 1.4 1.6]'; % column vector
w = 10;
%
while 1
% function inp from www.mathworks.com/matlabcentral/fileexchange/9033
w = inp('weight', w); % weight of penalty function
if w<=0, break, end
p0 = rand(2,1);
[p,ssq,cnt] = LMFnlsq(@resid,ones(2,1),'Display',-100,'MaxIter',1e3);
p = p.*p0;
a = p(2)/(3*p(1))
b = p(1)-a
res = resid(p./p0);
plot(r,data,'o-', r,res(1:end-1)+data,'-or')
end
The function for evaluating residuals has the form:
function res = resid(p) % Residuals of Surpans' equations
%%%%%%%%%%%%%%%%%%%%%%%%%
global data r w p0
%
p = p.*p0; % p(1) = (a+b)
q = sqrt(p(2)); % q = sqrt(3*a*(a+b))
%
res = [(q + 1./r).*exp(-r*q).*(0.6./(p(1)*r.^2)) - data
(q>=0)*w % penalty function
];
The nonlinear fit is made by the function LMFnlsq, which may be found on
www.mathworks.com/matlabcentral/fileexchange/17534
更多回答（1 个）
Matt J
2013-2-28
0 个投票
You have square roots in your function, which will produce complex values when their argument goes negative. You must use another algorithm, one that lets you constrain 3*a*(a + b))+(1/c) to a strictly positive lower bound.
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