Communication Networks Signal and Image processing (CS)
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
$ \left\{ \begin{matrix} {{r}_{0}}(\lambda )={{r}^{+}}-{{r}^{-}} \\ {{g}_{0}}(\lambda )={{g}^{+}}-{{g}^{-}} \\ {{b}_{0}}(\lambda )=={{b}^{+}}-{{b}^{-}} \\ \end{matrix} \right. $
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
where
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.)
