in dct watermarking i got the answer below what was that mean pls answer me sir

1 vue (au cours des 30 derniers jours)
sriharsha bulusu
sriharsha bulusu le 16 Fév 2015
Commenté : sriharsha bulusu le 17 Fév 2015
[16x16 double] [16x16 double] [16x16 double] [16x16 double]
[16x16 double] [16x16 double] [16x16 double] [16x16 double]
[16x16 double] [16x16 double] [16x16 double] [16x16 double]
[16x16 double] [16x16 double] [16x16 double] [16x16 double]
  2 commentaires
Geoff Hayes
Geoff Hayes le 16 Fév 2015
Sriharsha - you need to provide some context surrounding how you obtained the above results. What is the code that you are using? Presumably you broke down an image into 16x16 blocks and did the dct on each block, and now need to restore the image to its original size. See cell2mat on how you can convert the above cell array back to a numeric matrix of dimension of 64x64.
sriharsha bulusu
sriharsha bulusu le 17 Fév 2015
imgWMrked=imread('cameraman.tif'); [LL11, HL11, LH11, HH11] = dwt2(imgWMrked, 'db1'); lvl1 = [LL11, HL11;LH11, HH11]; figure,imshow(uint8(lvl1)); % figure(2), imshow(uint8(WTr_Im_L1)); [LL22, HL22, LH22, HH22] = dwt2(HL11, 'db1'); lvl2 = [LL22, HL22;LH22, HH22]; lvl3 = [LL11, lvl2; LH11, HH11]; subplot(1,2,2); imshow(uint8(lvl3)); title('After DWT Image') [R, C] = size(HL22); startR = 1; endR = R/4; for m = 1:4 startC = 1; endC = C/4; for n = 1:4 parts(m,n) = {HL22(startR:endR, startC:endC)}; startC = startC + (C/4); endC = endC + (C/4); end startR = startR + (R/4); endR = endR + (R/4); end for m = 1:4 for n = 1:4 tempData = cell2mat(parts(m,n)); dctPartsWMarkd(m,n) = {dct2(tempData)}; end end at the end we got the above answer and the code is for digital image watermarking by combined dct and dwt pls give me the explanation clearly

Connectez-vous pour commenter.

Réponses (0)

Catégories

En savoir plus sur Discrete Multiresolution Analysis dans Help Center et File Exchange

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by