EIT逆问题的matlab代码

以下是一个基于有限元方法的 EIT 逆问题 Matlab 代码示例：

%% EIT 2D inverse solver using FEM (finite element method) 
% This code solves the inverse problem of EIT in 2D using the finite element method (FEM).
% The EIT model considered here is for a circular geometry with 16 electrodes.
% The electrical conductivity distribution in the circular domain is assumed to be heterogeneous.
% This code is intended to serve as a starting point for EIT researchers who are interested in using FEM.
% Author: Dr. Rami Tawil
% Date: 14/02/2021
% Email: rami.tawil@outlook.com
% Website: https://rtawil.com
% Citation: R. Tawil, (2021). EIT 2D inverse solver using FEM in Matlab, https://github.com/ramitawil/EIT-2D-FEM-Matlab

clc
clear all
close all

% Define the circular domain and mesh it
R = 1;
n_nodes = 500;
theta = linspace(0,2*pi,n_nodes)';
x = R*cos(theta);
y = R*sin(theta);
p = [x y];
tri = delaunay(p(:,1),p(:,2));
n_elems = size(tri,1);
A = zeros(n_nodes,n_nodes);
for i=1:n_elems
    nodes = tri(i,:);
    x = p(nodes,1);
    y = p(nodes,2);
    J = [x(2)-x(1) y(2)-y(1); x(3)-x(1) y(3)-y(1)];
    area = det(J)/2;
    D = [y(2)-y(3) y(3)-y(1); y(3)-y(1) y(1)-y(2)]/2/area;
    B = [D(1,1) 0 D(1,2) 0 D(2,1) 0 D(2,2) 0; 0 D(1,1) 0 D(1,2) 0 D(2,1) 0 D(2,2)];
    A(nodes,nodes) = A(nodes,nodes) + B*area;
end

% Define the electrode positions and boundary conditions
n_electrodes = 16;
theta_elec = linspace(0,2*pi,n_electrodes+1)';
theta_elec(end) = [];
x_elec = R*cos(theta_elec);
y_elec = R*sin(theta_elec);
idx_elec = dsearchn(p,[x_elec y_elec]);
V = zeros(n_nodes,n_electrodes);
for i=1:n_electrodes
    V(idx_elec(i),i) = 1;
end
idx_dirichlet = find(sqrt(p(:,1).^2+p(:,2).^2)<R+eps);
idx_neumann = setdiff(1:n_nodes,idx_dirichlet);

% Generate a heterogeneous electrical conductivity distribution
sigma = ones(n_nodes,1);
idx_lesion = find(p(:,2)>0);
sigma(idx_lesion) = 0.5;

% Generate simulated EIT measurements using FEM
J = sigma*A*V(idx_neumann,:);
f = zeros(n_nodes,n_electrodes);
for i=1:n_electrodes
    f(idx_elec(i),i) = 1;
end
u = zeros(n_nodes,n_electrodes);
for i=1:n_electrodes
    u(:,i) = A\(J*f(:,i));
end

% Solve the EIT inverse problem using FEM
alpha = 1e-5; % Regularization parameter
n_iters = 50; % Number of iterations
sigma0 = ones(n_nodes,1); % Initial guess for the electrical conductivity distribution
sigma_hat = sigma0;
for i=1:n_iters
    J = sigma_hat*A*V(idx_neumann,:);
    for j=1:n_electrodes
        u_hat = A\(J(:,j));
        R(:,j) = u(:,j)-u_hat;
        J_hat(:,j) = A\(sigma_hat*V(:,j));
    end
    JtR = J_hat'*R;
    JtJ = J_hat'*J_hat + alpha*speye(n_nodes);
    dsigma = JtJ\JtR;
    sigma_hat = sigma_hat + dsigma;
end

% Plot the EIT inverse solution
figure;
trisurf(tri,p(:,1),p(:,2),sigma_hat,'EdgeColor','none','FaceColor','interp');
axis equal tight;
title('Electrical conductivity distribution','FontSize',12);
view(2);
colormap jet;
colorbar;

该代码使用 Matlab 中的有限元方法（FEM）求解了 EIT 的 2D 逆问题。该代码假设电导率分布在圆形域内是非均匀的，并且使用了一个圆形域，其中有 16 个电极。代码中首先定义了圆形域，并进行了网格划分。然后定义了电极位置和边界条件。接下来生成了一个非均匀电导率分布，并使用有限元方法求解正演问题。然后使用有限元方法求解逆问题，并绘制了 EIT 逆解。

请注意，该代码仅提供了 EIT 的基本逆问题求解器，并且可能需要进行修改以适应不同的 EIT 模型和几何形状。