資源簡介
SSR算法,自己已經(jīng)測試過了,且我自己增加了直方圖顯示
代碼片段和文件信息
clear;
close?all;
I?=?imread(‘E:\project\color\msr.bmp‘);
subplot(231);
imshow(I);
f=I(::1);
ff=I(::2);
fff=I(::3);
%%%%%%%%%%%%%構(gòu)造高斯核%%%%%%%%%%%%%%%%%%%%%%%
k1=8;
k2=8;
r=161;
alf=1600;
nn=floor((r+1)/2);??%?nn=81為尺度C,取80最合適
for?i=1:r???%從1循環(huán)到81
????for?j=1:r
????????b(ij)?=exp(-((i-nn)^2+(j-nn)^2)/(k1*alf))/(k2*pi*alf*10000);????????%高斯函數(shù)
???end
end
%%%%%%%%%%%對(duì)R分量的處理%%%%%%%%%%%%%
Img?=?double(f);
K=imfilter(Imgb);??%?中值濾波后的圖像
G=log(Img+1)-log(K+1);??%在對(duì)數(shù)域中用原始圖像減高斯平滑后的圖像?得到反射圖像
mi=min(min(G));
ma=max(max(G));
???????L=(G-mi)*255/(ma-mi);?%?對(duì)比度拉伸處理
%%%%%%%%%%%%%%對(duì)G分量的處理%%%%%%%%%%%%%
Img?=?double(ff);
K=imfilter(Imgb);??
G=log(Img+1)-log(K+1);
???
mi=min(min(G));
ma=max(max(G));
???????LL=(G-mi)*255/(ma-mi);
%%%%%%%%%%%%%隨B分量的處理%%%%%%%%%%%
?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????文件????????2053??2012-06-11?17:49??SSR.m
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