Revision as of 20:10, 10 November 2014

ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)

Question 5, August 2013, Part 2

part1, part 2

Solution 1:

a) If the color matching functions $ f_k(\lambda) $ has negative values, it will result in negative values in $ F_k $. In this case, the color can not be reproduced by this device.

b) The CIE color matching functions are not always positive. $ r_0(\lambda) $ takes negative values. This is the case because, to match some reference color that is too saturated, colors have to be subtracted from the $ R, G, $ and $ B $ primaries. This results in negative values in tristimulus values r, g, and b. So the color matching functions at the corresponding wavelength have negative values.

c)
$ \left[ {\begin{array}{*{20}{c}} F_1\\ F_2\\ F_3 \end{array}} \right] = {\begin{array}{*{20}{c}} \int_{-\infty}^{\infty} \end{array}} \left[ {\begin{array}{*{20}{c}} f_1(\lambda)\\ f_1(\lambda)\\ f_1(\lambda) \end{array}} \right] I(\lambda)d\lambda = {\begin{array}{*{20}{c}} \int_{-\infty}^{\infty} \end{array}} M \left[ {\begin{array}{*{20}{c}} r_0(\lambda)\\ g_0(\lambda)\\ b_0(\lambda) \end{array}} \right] I(\lambda)d\lambda = M {\begin{array}{*{20}{c}} \int_{-\infty}^{\infty} \end{array}} \left[ {\begin{array}{*{20}{c}} r_0(\lambda)\\ g_0(\lambda)\\ b_0(\lambda) \end{array}} \right] I(\lambda)d\lambda = M \left[ {\begin{array}{*{20}{c}} r\\ g\\ b \end{array}} \right] $

So that, $ [r, g, b]^t = M^{-1} [F_1, F_2, F_3] $.

d) It exists. CIE XYZ is one example. However, XYZ has problems with its primaries, since, the primary colors are imaginary.

Related problem

In a color matching experiment, the three primaries R, G, B are used to match the color of a pure spectral component at wavelength $ \lambda $. Here the color matching allows for color to be subtracted from the reference color. At each wavelength $ \lambda $, the matching color is given by $ \left[ {\begin{array}{*{20}{c}} R G B \end{array}} \right] \left[ {\begin{array}{*{20}{c}} r(\lambda)\\ g(\lambda)\\ b(\lambda) \end{array}} \right] $ </math>