Difference between revisions of "QE637 sol2013 Q2" - Rhea

Revision as of 12:55, 2 May 2017

Solution 1:

a) Since $ {{f}_{k}}(\lambda ),\ for\ k=0,\ 1,\ 2 $ are the spectral response functions for the three color outputs of a color camera, and the negative spectrum can’t be produced, they must be nonnegative.

b) Since $ {{r}_{0}}(\lambda ),\ {{g}_{0}}(\lambda ),\ and\ {{b}_{0}}(\lambda ) $ are the CIE color matching functions, they can be both positive and negative. The color matching function are given by

where $ {{r}^{+}},\ {{r}^{-}},\ {{g}^{+}},\ {{g}^{-}},\ {{b}^{+}},\ {{b}^{-}} $are the response to photons and must be positive, while the color matching function can be negative to produce a saturated color.

c)
$ \begin{align} & F=\left[ \begin{matrix} {{F}_{1}} \\ {{F}_{2}} \\ {{F}_{3}} \\ \end{matrix} \right]=\int\limits_{-\infty }^{\infty }{\left[ \begin{matrix} {{f}_{1}}(\lambda ) \\ {{f}_{2}}(\lambda ) \\ {{f}_{3}}(\lambda ) \\ \end{matrix} \right]}\ I(\lambda )\ d\lambda =\int\limits_{-\infty }^{\infty }{\left( M\left[ \begin{matrix} {{r}_{0}}(\lambda ) \\ {{g}_{0}}(\lambda ) \\ {{b}_{0}}(\lambda ) \\ \end{matrix} \right] \right)}\ I(\lambda )\ d\lambda=M\left( \int\limits_{-\infty }^{\infty }{\left[ \begin{matrix} {{r}_{0}}(\lambda ) \\ {{g}_{0}}(\lambda ) \\ {{b}_{0}}(\lambda ) \\ \end{matrix} \right]}\ I(\lambda )\ d\lambda \right)=M\left[ \begin{matrix} r \\ g \\ b \\ \end{matrix} \right]\ \\ & \Rightarrow\ \left[ \begin{matrix} r \\ g \\ b \\ \end{matrix} \right]={{M}^{-1}}\left[ \begin{matrix} {{F}_{1}} \\ {{F}_{2}} \\ {{F}_{3}} \\ \end{matrix} \right]={{M}^{-1}}_{{}}^{{}}{{F}^{t}} \\ \end{align} $

missed transpose sign on F. It should be [r,g,b]^t = M ^{− 1}[F₁,F₂,F₃]^t.

d) Yes, they do exist, like CIE XYZ. CIE XYZ is defined in terms of CIE RGB so that $ \left[ \begin{matrix} {{x}_{0}}(\lambda ) \\ {{y}_{0}}(\lambda ) \\ {{z}_{0}}(\lambda ) \\ \end{matrix} \right]=M\ \left[ \begin{matrix} {{r}_{0}}(\lambda ) \\ {{g}_{0}}(\lambda ) \\ {{b}_{0}}(\lambda ) \\ \end{matrix} \right],\ where\ M=\left[ \begin{matrix} 0.490 & 0.310 & 0.200 \\ 0.177 & 0.813 & 0.010 \\ 0.000 & 0.010 & 0.990 \\ \end{matrix} \right] $.

Solution 2:

a) Because for real pixels, measured energy from incident photons is always positive.

b) $ {{r}_{0}}(\lambda ),\ {{g}_{0}}(\lambda ),\ and\ {{b}_{0}}(\lambda ) $are the CIE color matching functions, and therefore can be negative. They go negative to match certain reference colors which are beyond the r, g, b primaries.

c)

$ \begin{align} & \int\limits_{-\infty }^{\infty }{\left[ \begin{matrix} {{f}_{1}}(\lambda ) \\ {{f}_{2}}(\lambda ) \\ {{f}_{3}}(\lambda ) \\ \end{matrix} \right]}\left[ \begin{matrix} I(\lambda )d\lambda & I(\lambda )d\lambda & I(\lambda )d\lambda \\ \end{matrix} \right]=\int\limits_{-\infty }^{\infty }{M\left[ \begin{matrix} {{r}_{0}}(\lambda ) \\ {{g}_{0}}(\lambda ) \\ {{b}_{0}}(\lambda ) \\ \end{matrix} \right]}\left[ \begin{matrix} I(\lambda )d\lambda & I(\lambda )d\lambda & I(\lambda )d\lambda \\ \end{matrix} \right] \\ & \Rightarrow \left[ \begin{matrix} \int\limits_{-\infty }^{\infty }{{{f}_{1}}(\lambda )I(\lambda )d\lambda } \\ \int\limits_{-\infty }^{\infty }{{{f}_{2}}(\lambda )I(\lambda )d\lambda } \\ \int\limits_{-\infty }^{\infty }{{{f}_{3}}(\lambda )I(\lambda )d\lambda } \\ \end{matrix} \right]=M\left[ \begin{matrix} \int\limits_{-\infty }^{\infty }{{{r}_{0}}(\lambda )I(\lambda )d\lambda } \\ \int\limits_{-\infty }^{\infty }{{{g}_{0}}(\lambda )I(\lambda )d\lambda } \\ \int\limits_{-\infty }^{\infty }{{{b}_{0}}(\lambda )I(\lambda )d\lambda } \\ \end{matrix} \right]\Rightarrow \left[ \begin{matrix} {{F}_{1}} \\ {{F}_{2}} \\ {{F}_{3}} \\ \end{matrix} \right]=M\left[ \begin{matrix} r \\ g \\ b \\ \end{matrix} \right]\Rightarrow \left[ \begin{matrix} r \\ g \\ b \\ \end{matrix} \right]={{M}^{-1}}\left[ \begin{matrix} {{F}_{1}} \\ {{F}_{2}} \\ {{F}_{3}} \\ \end{matrix} \right] \\ \end{align} $

d)

$ \begin{align} \left[ \begin{matrix} r \\ g \\ b \\ \end{matrix} \right]={{M}^{-1}}\left[ \begin{matrix} {X} \\ {Y} \\ {Z} \\ \end{matrix} \right] \\ \end{align} $ where X, Y, Z are the xyzzy tristimulus values (always positive): $ X=\frac{x}{x+y+z},Y=\frac{y}{x+y+z},Z=\frac{z}{x+y+z} $

The student can be more specific on the example of such case. I am not sure what is a good example either. Will consult Professor to figure it out.

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 $. (Assume that the color matching allows for color to be subtracted from the reference in the standard manner described in class.) At each wavelength $ \lambda $, the matching color is given by

$ \left[ \begin{matrix} R, & G, & B \\ \end{matrix} \right]\left[ \begin{matrix} r(\lambda ) \\ g(\lambda ) \\ b(\lambda ) \\ \end{matrix} \right] $

where

$ \begin{align} & 1=\int\limits_{0}^{\infty }{r(\lambda )d\lambda } \\ & 1=\int\limits_{0}^{\infty }{g(\lambda )d\lambda } \\ & 1=\int\limits_{0}^{\infty }{b(\lambda )d\lambda } \\ \end{align}$ Further define the white point $W=\left[ \begin{matrix} R, & G, & B \\ \end{matrix} \right]\left[ \begin{matrix} {{r}_{w}} \\ {{g}_{w}} \\ {{b}_{w}} \\ \end{matrix} \right] $.

Let $ I(\lambda) $ be the light reflected from a surface.

a) Calculate $ ({{r}_{e}},\ {{g}_{e}},\ {{b}_{e}}) $ the tristimulus values for the spectral distribution $ I(\lambda) $ using primaries R, G, B and an equal energy white point.

b) Calculate $ ({{r}_{c}},\ {{g}_{c}},\ {{b}_{c}}) $ the tristimulus values for the spectral distribution $ I(\lambda) $ using primaries R, G, B and white point $({{r}_{w}},{{g}_{w}},{{b}_{w}})$. c) Calculate $ ({{r}_{\gamma }},\ {{g}_{\gamma }},\ {{b}_{\gamma }}) $ the gamma corrected tristimulus values for the spectral distribution $ I(\lambda) $ using primaries R, G, B and white point $ ({{r}_{w}},\ {{g}_{w}},\ {{b}_{w}}) $, and $ \gamma =2.2 $.

(Refer to ECE638 Lecture note 3: Trichromatic theory of color.)

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