Line 87: Line 87:
 
== Solution 2:  ==
 
== Solution 2:  ==
  
a)&nbsp;<span class="texhtml">''f''<sub>1</sub>(λ)</span>,&nbsp;<span class="texhtml">''f''<sub>2</sub>(λ)</span>&nbsp;and&nbsp;<span class="texhtml">''f''<sub>3</sub>(λ)</span>&nbsp;are the spectral functions for the three color outputs of color camera. It must be positive because we cannot produce negative spectrum.&nbsp;
+
a)
 +
Because for real pixels, measured energy from incident photons is always positive.
  
 
b) No.&nbsp;<span class="texhtml">''r''<sub>''o''</sub>(λ),''g''<sub>''o''</sub>(λ)''a''''n''''d''''b'''''<b><sub>''o''</sub>(λ)</b></span>'''&nbsp;are CIE color matching. It takes negative value in order to substract some color to be saturated.&nbsp; '''  
 
b) No.&nbsp;<span class="texhtml">''r''<sub>''o''</sub>(λ),''g''<sub>''o''</sub>(λ)''a''''n''''d''''b'''''<b><sub>''o''</sub>(λ)</b></span>'''&nbsp;are CIE color matching. It takes negative value in order to substract some color to be saturated.&nbsp; '''  

Revision as of 10:16, 2 May 2017


ECE Ph.D. Qualifying Exam

Communication Networks Signal and Image processing (CS)

Question 5, August 2013(Published on May 2017),

Problem 1,2


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) No. ro(λ),go(λ)a'n'd'bo(λ) are CIE color matching. It takes negative value in order to substract some color to be saturated. 

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_2(\lambda)\\ f_3(\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] $

d)

Yes. They exist. If there is a matrix M that exist to satisfy this equation  $ \left[ {\begin{array}{*{20}{c}} f_1(\lambda)\\ f_2(\lambda)\\ f_3(\lambda) \end{array}} \right] = M \left[ {\begin{array}{*{20}{c}} r_0(\lambda)\\ g_0(\lambda)\\ b_0(\lambda) \end{array}} \right] $

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

1. In a color matching experiment, the three primaries R, G, B are used to match the color of a pure spectral component at wavelength λ. Here the color matching allows for color to be subtracted from the reference color. At each wavelength λ, 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] $

where r(λ), g(λ), and b(λ) are normalized to 1.

Further define the white point

$ W = \left[ {\begin{array}{*{20}{c}} R, G, B \end{array}} \right] \left[ {\begin{array}{*{20}{c}} r_w\\ g_w\\ b_w \end{array}} \right] $

Let I(λ) be the light reflected from a surface.

a) Calculate (re,ge,be) the tristimulus values for the spectral distribution I(λ) using primaries R,G,B and an equal energy white point.

b) Calculate (rc,gc,bc) the tristimulus values for the spectral distribution I(λ) using primaries R,G,B and white point (rw,gw,bw).

(Refer to ECE637 2004 Final Problem 4.)

2. Consider the two channel sensors with response function QS(λ) anQL(λ). Suppose that we have two primaries PL(λ) = σ(λ − 0.6) and PS(λ) = σ(λ − 0.5).

QE637 2013 P2 F1.PNG

Find the color matching function $ \bar{l}(\lambda) $ and $ \bar{s}(\lambda) $ for these two primaries.

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


Back to ECE QE page:

Alumni Liaison

Basic linear algebra uncovers and clarifies very important geometry and algebra.

Dr. Paul Garrett