資源簡介
任意比例放大 縮小 濾波 旋轉
均衡化處理
闕值分割
代碼片段和文件信息
%-----子函數adp_spa_denoise_guassian---------
%-------求出整個差值矩陣,再求這個矩陣得均值和方差相加后得到域值M,再來做噪聲判斷-------------
function?[xnoise]=adp_spa_denoise_guassian(tr)
[depwide]=size(t);
x=ones(size(t));
noise=zeros(size(t));
delta=ones(size(t));
for?i=3:dep-2
????for?j=3:wide-2
????????b=sort([abs(t(i-2j-2)-t(ij))?abs(t(i-2j-1)-t(ij))?abs(t(i-2j)-t(ij))?abs(t(i-2j+1)-t(ij))?abs(t(i-2j+2)-t(ij))??abs(t(i-1j-2)-t(ij))?abs(t(i-1j-1)-t(ij))?abs(t(i-1j)-t(ij))?abs(t(i-1j+1)-t(ij))?abs(t(i-1j+2)-t(ij))?abs(t(ij-2)-t(ij))?abs(t(ij-1)-t(ij))?abs(t(ij)-t(ij))?abs(t(ij+1)-t(ij))?abs(t(ij+2)-t(ij))?abs(t(i+1j-2)-t(ij))?abs(t(i+1j-1)-t(ij))?abs(t(i+1j)-t(ij))?abs(t(i+1j+1)-t(ij))?abs(t(i+1j+2)-t(ij))?abs(t(i+2j-2)-t(ij))?abs(t(i+2j-1)-t(ij))?abs(t(i+2j)-t(ij))
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件???????2522??2012-05-22?10:26??adp_spa_denoise_guassian.m
?????文件????????249??2012-05-22?09:51??filter.m
?????文件???????1384??2012-05-07?19:00??rot.m
?????文件???????1215??2012-05-07?20:09??rot2.m
?????文件???????2110??2012-05-07?21:18??rot3.m
?????文件???????1131??2012-05-22?09:26??rot4.m
?????文件???????1271??2012-05-15?10:01??rot5.m
?????文件???????2166??2012-05-20?21:11??rot6.m
?????文件????????784??2012-05-22?10:33??rot8.m
?????文件???????1031??2012-05-22?08:30??zhongzhi.m
-----------?---------??----------?-----??----
????????????????13863????????????????????10
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