資源簡介
從源代碼的角度實現了小波的haar分解,有利于初學者對小波分解的理解,不單純的使用matlab wavelet工具箱

代碼片段和文件信息
%非正交的二次樣條二進小波
h?=?[0?0?0.125?0.375?0.375?0.125?0?0].*sqrt(2);
g?=?[0?0?0?-1?1?0?0?0].*sqrt(2);
g1?=?[0?0.015625?0.109375?0.34375?-0.34375?-0.109375?-0.015625?0].*sqrt(2);
l?=?[1?6?15?84?15?6?1?0]./64;
n?=?4;
img=imread(‘735586_1278167235pR5y.jpg‘);
[a?d1?d2]?=?swt1_decomp(imgnhg);
figure;
imshow(img)?title(‘原始矩陣‘);
row?=?ceil(sqrt(n));
col?=?ceil(n/row);
%?aa?=?a;
%?figure?title(‘逼近矩陣‘);
%?for?i=1:n
%?????aa(::i)?=?aa(::i)./max(max(aa(::i)));
%?????subplot(rowcoli);
%?????imshow(aa(::i));
%?end
%?
%?
%?dd1?=?d1;
%?figure?title(‘水平細節信息矩陣‘);
%?for?i=1:n
%?????dd1(::i)?=?dd1(::i)./max(max(dd1(::i)));
%?????subplot(rowcoli);
%?????imshow(dd1(::i));
%?end
%?
%?dd2?=?d2;
%?figure?title(‘豎直細節信息矩陣‘);
%?for?i=1:n
%?????dd2(::i)?=?dd2(::i)./max(max(dd2(::i)));
%?????subplot(rowcoli);
%?????imshow(dd2(::i));
%?end
aa?=?a;
for?i=1:n
????aa(::i)?=?aa(::i)./max(max(aa(::i)));
end
aa?=?reshape(aa?[size(a1)?size(a2)?1?size(a3)]);
figure?montage(aa)?title(‘逼近矩陣‘);
dd1?=?d1;
for?i=1:n
????dd1(::i)?=?dd1(::i)./max(max(dd1(::i)));
end
dd1?=?reshape(dd1?[size(a1)?size(a2)?1?size(a3)]);
figure?montage(dd1)?title(‘水平細節信息矩陣‘);
dd2?=?d2;
for?i=1:n
????dd2(::i)?=?dd2(::i)./max(max(dd2(::i)));
end
dd2?=?reshape(dd2?[size(a1)?size(a2)?1?size(a3)]);
figure?montage(dd2)?title(‘豎直細節信息矩陣‘);
x?=?swt1_recon(ad1d2nhg1l);
%?
xx?=?x(::[2:n+1]);
%?row?=?ceil(sqrt(n));
%?col?=?ceil(n/row);
%?figure?title(‘重構矩陣‘);
for?i=1:n
????xx(::i)?=?xx(::i)./max([max(max(xx(::i)))?256]);
%?????subplot(rowcoli);
%?????imshow(xx(::i));
end
xx?=?reshape(xx?[size(xx1)?size(xx2)?1?size(xx3)]);
figure?montage(xx)?title(‘重構矩陣‘);
?屬性????????????大小?????日期????時間???名稱
-----------?---------??----------?-----??----
?????文件????????1771??2012-04-28?21:20??bat.m
?????文件?????????284??2012-03-22?16:35??harrwave.m
?????文件????????2136??2005-05-27?16:11??swt1_decomp.m
?????文件????????1891??2005-05-27?16:09??swt1_recon.m
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