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

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sriharsha bulusu 2015-2-16

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https://ww2.mathworks.cn/matlabcentral/answers/178664-in-dct-watermarking-i-got-the-answer-below-what-was-that-mean-pls-answer-me-sir

评论： sriharsha bulusu 2015-2-17

    [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]

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Geoff Hayes 2015-2-16

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 2015-2-17

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

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