i am dealing with gradieent descent based minimization problem , my matrix is heavily ill conditioned , leading to false recovery in forward inverse problem

1 次查看（过去 30 天）
asim asrar
asim asrar 2022-2-6
回答： Matt J 2022-2-10
clc;
clear all;
close all;
%% greens function between source to target
soux=[3,3,4,4,5,5]; % source x coordinate
souy=[3,4,3,4,3,4]; % source y coordinate
w2=zeros(6,6);
for k=1:6
hg=1;
for i=3:5 %target x coordinate
for j=3:4 %target y coordinate
p=soux(k);
q=souy(k);
temp1=(i-p)*(i-p)+(j-q)*(j-q)+(0.2-0)^2 ;
dt1=sqrt(temp1);
%kg=9.9260
kg=0.3897;
grf2(hg,:)=(exp(-(kg*dt1)))/(2.0673*dt1);
hg=hg+1;
end
end
w2(:,k)=grf2;
end
G=w2
%% greens function between target to detector
a=3
b=4
c=5
dtx=[a,a,b,b,c,c];
dty=[a,b,a,b,a,b];
% dtx=[3,3,4,4,5,5]; % detector x coordinate
% dty=[3,4,3,4,3,4]; % detector y coordinate
w1=zeros(6,6);
% greens function between targets to detectors
t=1;
for i=3:5 %target x coordinate
for j=3:4 %target y coordinate
for k=1:6
g=dtx(k);
m=dty(k);
temp=(i-g)*(i-g)+(j-m)*(j-m)+(0.4-0.2)^2;% euclidean distance
dt=sqrt(temp);
kk=0.3897;% this kk is in centimeters not in millimeters and 20.422 is in centimeters
grf1(k,:)=(exp(-(kk*dt)))/(2.0673*dt);% greens function computation from target to detectors
end
w1(:,t)=grf1;% output of the sensing matrix
% w2=normc(w1);
t=t+1;
%
end
%
end
H=w1;
%%
I = eye(6,6);
f =[1 1 0 0 0 0;
1 1 0 0 0 0;
0 0 1 1 0 0;
0 0 1 1 0 0;
0 0 0 0 0 0;
0 0 0 0 0 0]
%%
A = I - G*diag(f);
% here we are trying to minimize u and tring to find values for u
% A = I - G*diag(f)
%u_in = ones(6,6);
u_ini = zeros(6,6)
u_in=ones(6,6)
% u_in=[1.995 0.997 0.955 0.996 0.988 0.992;
% 0.978 1.985 0.956 0.968 0.9823 0.934;
% 0.987 0.999 1.991 0.991 0.956 0.925;
% 0.965 0.989 0.963 1.989 0.993 0.973;
% 0.968 0.934 0.982 0.996 1.945 0.924;
% 0.988 0.989 0.954 0.9786 0.969 1.991]
% imagesc(u_in)
% colormap(gray)
%% gradient and objective function
funval =@(u) 0.5 * ((norm(A*u - u_in))^2); % computing value of objective function
gradu = @(u) [A'*(A*u - u_in)] % computing the gradient for the trikhonov case
%u_ini = ones(6,6);
u_size =size(u_ini);
% A_size=size(A);
iterations =200;
stepsize = 0.001;
u_next = u_ini;
u_rec = [u_ini] ; % to record the u iterative value
ff = zeros(iterations,1)
for i = 1 : iterations
u_next = u_next - stepsize * gradu(u_next);
recordedguess_u = [u_rec , u_next ];
ff(i,1) =funval(u_next);
ii(i,1)=i;
end
% rec = reshape(u_next, u_size);
u_tilda=u_next;
funval(u_tilda)
figure()
plot(ii,ff)
%display(u_tilda);
%C = u -u_next
%% calculating filed at detector
Y = (H*diag(u_tilda)).*f;
%Y = abs(Y)
figure()
imagesc(Y)
colormap(gray)
pp=H*diag(u_tilda)
pp_inv = pinv(pp)
fff = pp_inv.* Y
figure()
imagesc(f)
colormap(gray)
figure()
imagesc(fff)
colormap(gray)

请先登录，再回答此问题。

采纳的回答

Matt J
Matt J 2022-2-10
Try Newton's method instead of steepest descent. If you have the Optimization Toolbox, fminunc() basically has Newton-like method's already impleneted for you.

更多回答（0 个）

类别

在 Help Center 和 File Exchange 中查找有关 Matrix Computations 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