COLOR MANAGEMENT- P9

COLOR MANAGEMENT- P9: ICC White Papers are one of the formal deliverables of the International Color Consortium, theother being the ICC specification itself – ISO 15076: Image technology color management –Architecture, profile format, and data structure. The White Papers undergo an exhaustiveinternal development process, followed by a formal technical review by the membership and aballot for approval by the ICC Steering Committee.
Nội dung trích xuất từ tài liệu:
COLOR MANAGEMENT- P9224 Profile Construction and Evaluation Positive gamma 1 0.9 0.8 0.7 0.6 0.5 0.5 y 1.0 0.4 0.3 2.0 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 x Figure 26.3 Power law with exponent g > 0 It is not obvious how (or even whether) a CMM should process data when g is zero ornegative. The profile is almost certainly corrupt or in error in such a case. There is clearlyno point in defining a constant or decreasing function which will be clipped over the entiredomain. The CMM developer may choose, in such a case, to reject the profile or to replace Negative gamma 2 1.8 1.6 before clipping 1.4 1.2 1 y after clipping 0.8 0.6 0.4 0.2 0 0 0.5 1 1.5 2 x Figure 26.4 Power law with exponent g ¼ À1/2Use of the parametricCurveType 225the parametric curve with the identity function, y ¼ x (which amounts to setting g ¼ 1), orsome other default. Ideally profile creators should abide by the condition g > 0, and CMMdevelopers should treat any occurrences of negative and zero values as errors and takeappropriate action.26.3 The Power-Law ArgumentFunction type 0 is just the basic power law described above: y ¼ f0 ðxÞ ¼ xg x ¼ F0 ðyÞ ¼ y1=g : In the other types, the argument to the power law is not simply x, but a linear expression in x.In types 1, 2, 3, and 4, the argument, which we will call s, takes the form s ¼ ax þ b;where a and b are additional parameters. The power law is then sg . Since there is nopractical restriction on the parameter values, s can take on any value as x varies between 0and 1. If s is negative, sg can be imaginary or complex. (It will be real for integer g, but gcannot be restricted to integer values.) In such a situation, a CMM might choose to takethe real part of the expression. Alternatively, it could take the absolute magnitude. Itcould arbitrarily set the expression to zero or one. Another option is simply to require s to benon-negative. In types 1, 2, 3, and 4, the domain is divided into two segments, and the power law isemployed only in the higher segment. For instance, the definition of type 1 is y ¼ f1 ðxÞ ¼ 0; 0 x < Àb=a ¼ sg ; Àb=a x 1:In normal usage, a will be positive and b will be negative, so that the segment boundary, Àb/a,occurs at a positive value of x. The function is identically zero in the lower segment(Figure 26.5). The argument s is non-negative throughout the higher segment, where the power law is ineffect: ðÀb=a xÞ ) ð0 ax þ b ¼ sÞ:This conclusion is verified by multiplying both sides of the first inequality by a and thenadding b; it holds only if a > 0, however. Indeed, the inequality is reversed for negative a.(And if a ¼ 0, the segment boundary itself is indeterminate.) It seems reasonable to impose226 Profile Construction and Evaluation Type 1: a > 0, b < 0 1 0.9 0.8 0.7 0.6 0.5 y 0.4 0.3 0.2 0.1 0 0 0.2 0.4 0.6 0.8 1 x Figure 26.5 f1(x), with boundary at 0.2the condition a > 0 as a requirement, so negative and zero values can then be treated as errors,and the CMM can take appropriate action – for instance, by substituting a ¼ 1 or some otherdefault. But, as in the case of g, the CMM developer needs to be reassured that profile creatorsdo not have a legitimate use for negative or zero values. Another reason to require that a be positive is that it compels the power-law function to bemonotonically non-decreasing, which is the normal case. Similarly, one might consider imposing the condition b < 0. However, this is probably notnecessary. Positive values of b simply mean that the segment boundary will occur at negative x.The power law will be in effect over the entire domain, and there will be no lower segment withy ¼ 0. In such a case, s will always be positive, and there will be ...