Watermarking in the Frequency Domain

I am new in Matlab and I have an assignment that is asking for watermarking an image using DCT transform:

• Read Lin.jpg color image and apply DCT.
• Threshold the logo.jpg (watermark) into binary and ten times of its strength, and then add it to the coefficient of transformed Lin image.

Here are the two images:

I have three questions:

1. Am I supposed to divide Lin.jpg into 8x8 blocks and logo.jpg into 2x2 blocks or that is not necessary?
2. what does it mean by : "ten times of its strength"? Is that just multiplying by 10?
3. How can I get the coefficient of transformed Lin.jpg image?

Here is what I tried:

img = imread('Lin.jpg');
wImg = imread('njit_logo.jpg');

wImgBinary = imbinarize(wImg) * 10;
[rows, cols] = size(img(:,:,1));
[Wrows, Wcols] = size(wImgBinary);

% make the watermark image as large as the original
watermark = zeros(size(img), 'uint8');
for column = 1:cols
    for row = 1:rows
        watermark(row, column) = wImgBinary(mod(row,Wrows)+1, mod(column,Wcols)+1);
    end
end
watermark = watermark(1:rows, 1:cols);

% apply dct and add with watermark at each channel
for i = 1:3
    imgDct = dct2(img(:,:,i));
    C = imgDct + double(watermark);
    Iw(:,:,i) = round(real(idct2(C)));
end

IIw = uint8(Iw);
figure, imshow(IIw), title('watermarked image');

[Walter Roberson]

I would tend to think the ten times would be multiplying the binary (0, 1) by 10, getting either 0 or 10.

wImgDct = dct2(wImgBinary);

That is not asked for. Just add the wImgBinary to imgDct .

[Hadi Ghahremannezhad]

ok I fixed that part. but still not sure how to do this.

The resulting watermarked image doesn't have any difference with the original.

[Walter Roberson]

The watermarked image does differ from the original. Do

imshowpair(IIw, img)

and

imshowpair(IIw, img, 'diff')

[Walter Roberson]

https://inis.iaea.org/search/search.aspx?orig_q=RN:45019480

https://www.mathworks.com/help/images/image-quality-metrics.html

[Hadi Ghahremannezhad]

Yes, I asked the same question here and I think the problem was I should have applied DCT to the logo as well.

Our professor said no, but without applying DCT to both the image and watermark before addition, I got strange results.

[Jevan Watson]

Hadi if you look when you do the side by side it kind of looks like his note in HW4R in the top left, so I think its strange to us but correct. Have you figured out the inverse or part B?

[Hadi Ghahremannezhad]

Yes I think I had it wrong. This is what Professor said:

"You take the Lin image by DCT. Then the NJIT watermark is thresholded into a binary image.

Then add the binary watermark to the DCT of Lin point-by-point starting at the upper-left corner of the image."

For part B, I did this:

• noisy = added noist to the watermarked image from part A
• noisyDCT = dct2(noisy)
• sub = noisyDCT - 10 * dct2(binaryLogo)
• de-watermarked = uint8(real(idct2(sub)))

but I am sure this one is wrong as well, especially the dct2(binaryLogo) part.

[Jevan Watson]

So wait because he confuses me too... isn't idct supposed to be one of the five images in part A too? or does part B provide the last 2 images he asked for in A. Because in the above I didn't notice where you did the inverse dct on Lin

[Hadi Ghahremannezhad]

part B it says first DCT the noisy image. Then subtract the (logo*10) and at the end IDCT the result. For me, it is in the last line above (uint8(real(idct2(sub)))). But I got black images from this so I applied DCT on the (logo) as well.