資源簡介
本程序主要實現低分辨率圖像的超分辨率重建，有助于我們理解圖像重建的過程和原理
代碼片段和文件信息
%?位移、角度估計后利用pocs法求超精度圖像
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s{1}=double（imread（‘tu1.jpg‘））;
[rowcolum]=size（s{1}）;
r2=1:2*row;c2=1:2*colum;
r2=r2/2;c2=c2/2;c2=c2‘;
s{1}=interp2（s{1}r2c2‘cubic‘）
img1=uint8（s{1}）;
for?k=2:4
s{2}=double（imread（strcat（‘tu‘num2str（k）‘.jpg‘）））;
[rowcolum]=size（s{2}）;
r2=1:2*row;c2=1:2*colum;
r2=r2/2;c2=c2/2;c2=c2‘;
s{2}=interp2（s{2}r2c2‘cubic‘）;
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%估計角度
img2=uint8（s{2}）;
[im1?des1?loc1]?=?sift（img1）;?
[im2?des2?loc2]?=?sift（img2）;
distRatio?=?0.75;???
des2t?=?des2‘;?????????????????????????
for?i?=?1:size（des11）
???dotprods?=?des1（i:）?*?des2t;???????
???[valsindx]?=?sort（acos（dotprods））;???
???if?（vals（1）??屬性????????????大小?????日期????時間???名稱
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?????文件???????4427??2006-12-05?20:39??rotation_estimate.m
?????文件???????4647??2006-12-09?16:31??pocs2.m
?????文件???????2533??2006-12-07?17:07??pocs4.m
?????文件???????3879??2006-12-12?14:09??result.m
?????文件???????1730??2006-12-09?16:39??rotation.m
-----------?---------??----------?-----??----
????????????????17434????????????????????6
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