numerical methods - sRGB's linear segment to avoid infinite slope - why? -

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in article srgb (https://en.wikipedia.org/wiki/srgb) stated, gamma transformation has linear portion near zero, "avoid having infinite slope @ k = 0, can cause numerical problems". i'd know what's problem that.

there 2 answers, usual gamma. modern variant is:

the problem infinite slope need "infinite" resolution (many bits of storage) in order arrive @ linear representation invertible gamma-encoded without loss. in other words, allows small lookup table produce invertible linear encoding (8bit -> 10 bit -> 8 bit).

the numerical problem understood on first step (8 bit -> 10 bit). infinite slope near zero, need bigger encoding range stay faithful , reversible, i.e. you'd need more 16 bit (assuming integer coding, halfs not have problem).

the linear equivalent of #010101 or 1/255th square (gamma = 2.0) coding 1/(255*255)th. need 16 bits represent faithfully, , using 2.2 not 2.0 exponent make worse. these quite small numbers corollary of coding function, , in practice don't need resolution in lightness range is, roughly, black. linear segment helps coding not wasting resolution detail around black (or near zero).

the older answer, taken from

http://www.poynton.com/notes/colour_and_gamma/gammafaq.html#gamma_correction

is in equipment linear segment less sensitive noise. true of analog signal path.


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