我尝试将查找正解的方法应用于未确定的线性方程组
A=[0 0.0992 0.315 0.619 1; 0 0.248 0.315 0.248 0]; b=[0.1266 0.4363].
假设这个问题存在积极和有限的解决方案。最糟糕的是我有一个答案代码，但我不能使它工作，因为我的Matlab版本不能识别匿名函数调用，而且一些指令对我来说是模糊的。 这是代码：
% example_pt_source_atmos_setup.m
% determine geometry
D2 = 0.5; % diameter of the observation aperture [m]
wvl = 1e-6; % optical wavelength [m]
k = 2*pi / wvl; % optical wavenumber [rad/m]
Dz = 50e3; % propagation distance [m]
% use sinc to model pt source
DROI = 4 * D2; % diam of obs-plane region of interest [m]
D1 = wvl*Dz / DROI; % width of central lobe [m]
R = Dz; % wavefront radius of curvature [m]
% atmospheric properties
Cn2 = 1e-16; % structure parameter [m^-2/3], constant
% SW and PW coherence diameters [m]
r0sw = (0.423 * k^2 * Cn2 * 3/8 * Dz)^(-3/5);
r0pw = (0.423 * k^2 * Cn2 * Dz)^(-3/5);
p = linspace(0, Dz, 1e3);
% log-amplitude variance
rytov = 0.563 * k^(7/6) * sum(Cn2 * (1-p/Dz).^(5/6) ...
.* p.^(5/6) * (p(2)-p(1)));
% screen properties
nscr = 11; % number of screens
A = zeros(2, nscr); % matrix
alpha = (0:nscr-1) / (nscr-1);
A(1,:) = alpha.^(5/3);
A(2,:) = (1 - alpha).^(5/6) .* alpha.^(5/6);
b = [r0sw.^(-5/3); rytov/1.33*(k/Dz)^(5/6)];
% initial guess
x0 = (nscr/3*r0sw * ones(nscr, 1)).^(-5/3);
% objective function
fun = @(X) sum((A*X(:) - b).^2);
% constraints
x1 = zeros(nscr, 1);
rmax = 0.1; % maximum Rytov number per partial prop
x2 = rmax/1.33*(k/Dz)^(5/6) ./ A(2,:);
x2(A(2,:)==0) = 50^(-5/3)
[X,fval,exitflag,output] ...
= fmincon(fun,x0,[],[],[],[],x1,x2)
% check screen r0s
r0scrn = X.^(-3/5)
r0scrn(isinf(r0scrn)) = 1e6;
% check resulting r0sw & rytov
bp = A*X(:); [bp(1)^(-3/5) bp(2)*1.33*(Dz/k)^(5/6)]
[r0sw rytov]
感谢您的关注。 Carolina Rickenstorff