資源簡(jiǎn)介
離散小波變換與離散小波反變換 快速小波變換(Mallat小波分解算法):對(duì)一幅圖像做2級(jí)小波分解(離散小波變換)與合成(離散小波反變換)
代碼片段和文件信息
x=imread(‘lena.gif‘);???%?載入圖像lena.gif,需要更改路徑到指定的文件
figure(1);
imshow(x);
title(‘原始圖像‘);
x=double(x);
%圖像分解
lpd=[1/2?1/2];hpd=[-1/4?2/3];???????????%?默認(rèn)的低通、高通濾波器
[rowcol]=size(x);??????????????????????%?讀取輸入矩陣的大小
%一級(jí)分解
for?j=1:row?????????????????????????????%?首先對(duì)輸入矩陣的每一行序列進(jìn)行一維離散小波分解
????tmp1=x(j:);
????[ca1cd1]=mydwt(tmp1lpdhpd1);
????x(j:)=ca1;???????????????????%?將分解系數(shù)序列再存入矩陣x中,得到[L|H]
end
for?k=1:col?????????????????????????????%?再對(duì)輸入矩陣的每一列序列進(jìn)行一維離散小波分解
????tmp2=x(:k);
????[ca2cd2]=mydwt(tmp2lpdhpd1);
????x(:k)=ca2;???????????????????%?將分解所得系數(shù)存入矩陣x中,得到[LLHl;LHHH]
end
%二級(jí)分解
for?j=1:row/2?????????????????????????????%?首先對(duì)輸入矩陣左上角四分之一矩陣的每一行序列進(jìn)行一維離散小波分解
????tmp1=x(j1:col/2);
????[ca1cd1]=mydwt(tmp1lpdhpd1);
????x(j1:col/2)=ca1;???????????????
?屬性????????????大小?????日期????時(shí)間???名稱
-----------?---------??----------?-----??----
?????文件?????359424??2010-05-09?21:54??實(shí)驗(yàn)二?圖像變換.doc
?????文件????????308??2010-05-08?14:15??downspl.m
?????文件???????1204??2010-05-08?14:17??myidwt.m
?????文件????????431??2010-05-08?14:18??upspl.m
?????文件????????846??2010-05-09?14:54??mydwt.m
?????文件???????4033??2010-05-09?21:54??2.m
?????文件???????4804??2010-05-09?21:19??wave2gray.m
?????文件???????1089??2004-12-16?13:36??wavecut.m
?????文件???????3528??2004-12-16?13:36??wavework.m
?????文件????????940??2004-12-16?13:36??wavecopy.m
-----------?---------??----------?-----??----
???????????????376607????????????????????10
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