RGB2YCBCR image convertion

i have this code i need to center Y,CB,CR around the center please show me what can i do.is truo?

I = imread('lena.tiff');
J2 = rgb2ycbcr((I));;
J2 =double(J2);
Y =((J2(:,:,1)));
cb=((J2(:,:,2)));
cr=((J2(:,:,3)));
Y=Y-128;
cb=cb-128;
cr=cr-128
T=dctmatx(8);

[DGM]

What do you mean "center" it? Cb and Cr should be centered about 128, since they're uint8. Also since they're uint8, subtracting 128 will truncate half the data.

[eng m]

mean around to zero ,thanks

Answer 1

[Walter Roberson]

Y=Y-128;
cb=cb-128;
cr=cr-128

If you read the documentation for rgb2ycbcr

Converted YCbCr color values, returned as a numeric array of the same size as the input.

• If the input is double or single, then Y is in the range [16/255, 235/255] and Cb and Cr are in the range [16/255, 240/255].
• If the input is uint8, then Y is in the range [16, 235] and Cb and Cr are in the range [16, 240].
• If the input is uint16, then Y is in the range [4112, 60395] and Cb and Cr are in the range [4112, 61680].

So if you wanted to center Y around 0 you would subtract (235-16)/2 + 16 = 125.5 and for Cb and Cr you would subtract (240-16)/2 + 16 = 128

[Walter Roberson]

I = randi([0 255], 640, 480, 3, 'uint8');
J2 = rgb2ycbcr((I));;
J2 = double(J2);
Y = ((J2(:,:,1)));
cb = ((J2(:,:,2)));
cr = ((J2(:,:,3)));
Y = Y-128;
cb = cb-128;
cr = cr-128;

[min(cr(:)), max(cr(:))]

ans = 1×2

-111 111

That is evenly distributed around 0

[min(cb(:)), max(cb(:))]

ans = 1×2

-111 111

That is evenly distributed around 0

[min(Y(:)), max(Y(:))]

ans = 1×2

-111 106

That is not evenly distributed around 0: it is about 5 more negative than goes positive. To even that out you would need to increase about about (5/2) = 2.5 . That would give you a base of 128-2.5 = 125.5 . Which just happens to be the exact 125.5 that I indicated in my Answer. So my Answer is experimentally justified.

[eng m]

thanks mr walter ,

and then can add again in the last code to repeat it ?

like this

Y = Y+125.5;

cb = cb+128;

cr = cr+128;

[Walter Roberson]

yes

Answer 2

[Image Analyst]

Do you mean like

cb = cb - mean(cb(:));
cr = cr - mean(cr(:));

or possibly rescale so the max is at 127 and the min is at -127?

cb = rescale(cb, -127, 127);
cr = rescale(cr, -127, 127);

Those are two different things of course. They shift and scale the gamut by different amounts. I don't really know what your intent is, or why it's even necessary.

[eng m]

i need it to do compression for jpeg color image by zigzag and huffman later .

Y how can i do the mean ,

[Image Analyst]

I dont' know what "Y how can i do the mean" means. You do (get) the mean by using the mean() function of course.

[eng m]

please @Image Analyst i need to center y cb, cr around to zero

to do best compession and reduse storge .to do jpeg color image

[Image Analyst]

@eng m, as Walter discussed "centered" is not a clear cut definition. You need to describe what centered means. We can think of at least 3 definitions based on range, mean, or median. But it looks like you are all set now (whatever the definition you used was) because you've accepted an answer.