MATLAB: Reading 10bit raw image in Matlab

bitfread

I have a 10bit raw image (input.yuv) and would like to read it in Matlab. To do so, I am using the following commands:

fid = fopen('input.yuv','r');

a = fread(fid,'uint16');

fclose(fid);

When I get the max(a(:)), it is 730. However, I have white pixels in the image. My question is that:

To get the image data between 0 and 1, should I divid them by (2^16-1) or (2^10-1). The original content is 10bit.
Why I don't get (2^10-1 or 2^16-1) for the white pixels?

Best Answer

You have not decoded the yuv to its 4:4:2 components.

YUV with 8 bits per channel does not use the full range of values for any component. Also, white is not encoded as the largest possible value: white is U component and V component 0 (they are signed) which gets stored as YCrCb with Cr and Cb = 128 each https://en.m.wikipedia.org/wiki/YUV

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IDL's lun stands for "logical unit number", and it serves the same purpose as MATLAB's fileid.

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"endian" in files and in memory refers to whether a memory address is to refer to the "Most Significant Byte" or to the "Least Significant Byte" of the data. So if you have two consecutive bytes in memory, such as hex 01 and then hex 02 in that order, and if you are reading 16 bits ("short", int16), is the 01 the most significant byte of the short or is it the least significant byte of the short? To put it a different way, if the file has hex 01 then 02, is the numeric value to be 1 * 256 + 2, or is it to be 1 + 256 * 2 ? Just like writing coefficients of a polynomial, do you write the coefficient of x^0 then the coefficient of x^1 then the coefficient of x^2 and so on, or do you write the coefficient of x^2 followed by the coefficient of x^1 followed by the coefficient of x^0 ? The two orders are not compatible with each other, but are both internally inconsistent. "Little endian" architectures like all of the x86 and x64 architectures, would say that the order in memory should correspond to 1 + 256 * 2 = 513, whereas "Big endian" architectures would say the order in memory should correspond to 1 * 256 + 2 = 258 . The Internet standards say that when data is transmitted between systems where the byte order is not specifically indicated, that "big endian" order should be used, but that is not the same as what Intel processors use. So if you got your file from outside, such as from a research site, it might be in big-endian order.

Best Answer

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