Radon transform matrix representation - matlab

Question

I am looking for a MATLAB solution to generate the matrix representation of a discrete Radon transform (DRT). That is, given a vectorized version of an MxN image, X, I'd like to generate the matrix R such that R*X(:) is a DRT of the image. In MATLAB, I am expecting it to look something like the following:
>> X = 2D_Image_Of_Size_MxN;
>> R = DRT_Matrix_Of_Size_LPxMN;
>> DRT = reshape( R * X(:), L, P );
I know there are several ways to define a DRT, so I'll just say that I am looking for a normal or standard or not-too-out-of-the-ordinary implmentation.

Answer 1

function [ R rho theta ] = radonmatrix( drho, dtheta, M, N )
% radonmatrix - Discrete Radon Trasnform matrix
%
% SYNOPSIS
% [ R rho theta ] = radonmatrix( drho, dtheta, M, N )
%
% DESCRIPTION
% Returns a matrix representation of a Discrete Radon
% Transform (DRT).
%
% INPUT
% drho Radial spacing the the DRT.
% dtheta Angular spacing of the DRT (rad).
% M Number of rows in the image.
% N Number of columns in the image.
%
% OUTPUT
% R LP x MN DRT matrix. The values of the L and
% P will depend on the radial and angular spacings.
% rho Vector of radial sample locations.
% theta Vector of angular sample locations (rad).
%
% For each angle, we define a set of rays parameterized
% by rho. We then find the pixels on the MxN grid that
% are closest to each line. The elements in R corresponding
% to those pixels are given the value of 1.
% The maximum extent of the region of support. It's for
% rho = 0 and theta = pi/4, the line that runs caddy-corner.
W = sqrt( M^2 + N^2 );
rho = -W/2 : drho : W/2;
theta = 0 : dtheta : 180 - dtheta;
L = length( rho );
P = length( theta );
R = false( L*P, M*N );
% Define a meshgrid w/ (0,0) in the middle that
% we can use a standard coordinate system.
[ mimg nimg ] = imggrid( 1, 1, [ M N ] );
% We loop over each angle and define all of the lines.
% We then just figure out which indices each line goes
% through and put a 1 there.
for ii = 1 : P
phi = theta(ii) * pi/180;
% The equaiton is rho = m * sin(phi) + n * cos(phi).
% We either define a vector for m and solve for n
% or vice versa. We chose which one based on angle
% so that we never g4et close to dividing by zero.
if( phi >= pi/4 && phi <= 3*pi/4 )
t = -W : min( 1/sqrt(2), 1/abs(cot(phi)) ) : +W;
T = length( t );
rhom = repmat( rho(:), 1, T );
tn = repmat( t(:)', L, 1 );
mline = ( rhom - tn * cos(phi) ) ./ sin(phi);
for jj = 1 : L
p = round( tn(jj,:) - min( nimg ) ) + 1;
q = round( mline(jj,:) - min( mimg ) ) + 1;
inds = p >= 1 & p <= N & q >= 1 & q <= M;
R( (ii-1)*L + jj, unique( sub2ind( [ M N ], q(inds), p(inds) ) ) ) = 1;
end
else
t = -W : min( 1/sqrt(2), 1/abs(tan(phi)) ) : +W;
T = length( t );
rhon = repmat( rho(:)', T, 1 );
tm = repmat( t(:), 1, L );
nline = ( rhon - tm * sin(phi) ) ./ cos(phi);
for jj = 1 : L
p = round( nline(:,jj) - min( nimg ) ) + 1;
q = round( tm(:,jj) - min( mimg ) ) + 1;
inds = p >= 1 & p <= N & q >= 1 & q <= M;
R( (ii-1)*L + jj, unique( sub2ind( [ M N ], q(inds), p(inds) ) ) ) = 1;
end
end
end
R = double( sparse( R ) );
return;
Here is the imggrid function used in the above.
function [ m n ] = imggrid( dm, dn, sz )
% imggrid -- Returns rectilinear coordinate vectors
%
% SYNOPSIS
% [ m n ] = imggrid( dm, dn, sz )
%
% DESCRIPTION
% Given the sample spacings and the image size, this
% function returns the row and column coordinate vectors
% for the image. Both vectors are centered about zero.
%
% INPUT
% dm Spacing between rows.
% dn Spacing between columns.
% sz 2x1 vector of the image size: [ Nrows Ncols ].
%
% OUTPUT
% m sz(1) x 1 row coordinate vector.
% n 1 x sz(2) column coordinate vector.
M = sz(1);
N = sz(2);
if( mod( M, 2 ) == 0 )
m = dm * ( ceil( -M/2 ) : floor( M/2 ) - 1 )';
else
m = dm * ( ceil( -M/2 ) : floor( M/2 ) )';
end
if( mod( N, 2 ) == 0 )
n = dn * ( ceil( -N/2 ) : floor( N/2 ) - 1 );
else
n = dn * ( ceil( -N/2 ) : floor( N/2 ) );
end

Answer 2

image = phantom();
projection = radon(image);
R = linsolve(reshape(image,1,[]), reshape(projection,1,[]));