MATLAB image processing -- image transformation correction second-order fitting

Problem description

Two adjacent images in a known sequence of images A(up.jpg)、B(down.jpg), The coordinates are （u,v） And （x,y）, Try to find the coefficients of the second-order fitting equation for image transformation correction K. It is required to solve by the least square method , The transformation model is ：

up.jpg：

down.jpg：

Methods described

The process

Firstly, coordinates of the points corresponding to distortion and reference image before and after transformation are obtained ：
Distorted image coordinates ：

• x=[300 386 293 498 369 384 52]
• y=[483 441 285 288 125 38 138]
  The corresponding coordinates of the reference image ：
• u=[99 69 254 121 326 381 510]
• v=[63 141 174 312 331 398 93]
  The coefficients of the corrected second-order fitting equation are solved by the least square method , Solve to get K：
  
  namely ：
  X= 619 – 0.597u – 0.532v + 0.0001u^2 - 0.0001v^2 – 0.0004uv
  Y= 133+0.756u – 0.572v + 0.0004u^2 + 0.0007v^2 – 0.0004uv

result

First, the transformed image size is obtained through coordinate transformation , And initialize a matrix , As a transformed matrix , The image pixel value is mapped by the obtained corrected second-order fitting equation , The final image is as follows ：

As can be seen from the above image , To be changed down Images . After transformation matrix mapping , The general direction of the image and the image to be mapped up Almost the same , But the size of the mapped image has also changed , The changed image cannot be completely filled , The straight lines in the original drawing may not be straight lines after transformation .

Code

clc
clear all

img_down = imread('down.jpg');
subplot(1,3,1)
imshow(img_down);
title("down");
img_up = imread('up.jpg');
subplot(1,3,2);
imshow(img_up);
title("up");

[h,w,c] = size(img_down);

% Distorted image coordinates 
x=[300 386 293 498 369 384 52]'; y=[483 441 285 288 125 38 138]';

% Reference image coordinates 
u=[99 69 254 121 326 381 510]'; v=[63 141 174 312 331 398 93]';

% The least square method 
P=[x,y];
a=[1 1 1 1 1 1 1]'; U=[a, u, v, u.^2, v.^2, (u.*v)]; K_xy=inv(U'*U)*U'*P P=[u,v]; U=[a,x,y,x.^2,y.^2,(x.*y)]; K_uv=inv(U'*U)*U'*P

ver_xy = [1,1; 1,w; h,1; h,w]; % down.jpg Four vertices of   
ver_data = [ones(4, 1), ver_xy, ver_xy(:,1).^2, ver_xy(:,2).^2, ver_xy(:,1).*ver_xy(:,2)];  
ver_t = ver_data*K_xy; %  Vertex coordinates after coordinate conversion   
ver_t_max = ceil(max(ver_t));  
ver_t_min = floor(min(ver_t));  
size_t = ver_t_max - ver_t_min + 1;  
down_t = zeros(size_t(1), size_t(2), 3); % 0 Initialize the converted image   
% The output image position of this part is offset , Need to reconsider the size of the picture 
for u=ver_t_min(1):ver_t_max(1)
    for v=ver_t_min(2):ver_t_max(2)
        d = [1, u, v, u^2, v^2, u*v];
        pos = round(d*K_uv); %  Find the corresponding by inverse transformation down.jpg Middle coordinates 
        if pos(1)>=1 && pos(2)>=1 && pos(1) <= h && pos(2) <= w
            down_t(u-ver_t_min(1)+1, v-ver_t_min(2)+1, :) = img_down(pos(1), pos(2), :);
        end
    end
end
down_t = uint8(down_t);  
subplot(1,3,3)
imshow(down_t);
title("down to up");