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<font size="4"> Communication Networks Signal and Image processing (CS) </font>
 
<font size="4"> Communication Networks Signal and Image processing (CS) </font>
  
−
<font size="4"> [[QE637_sol2013|Question 5, August 2013(Published on May 2017)]],</font>  
+
<font size="4"> [[QE637_sol2013|Question 5, August 2013(Published on May 2017)]]</font>
 +
</center>  
  
 
<font size="4">[[ QE637_sol2013_Q1 | Problem 1]],[[ QE637_sol2013_Q2 |2]]</font>  
 
<font size="4">[[ QE637_sol2013_Q1 | Problem 1]],[[ QE637_sol2013_Q2 |2]]</font>  
−
</center>
 
  
 
----
 
----
Line 15: Line 15:
  
 
a)  
 
a)  
−
Since <math>{{f}_{k}}(\lambda ),\ for\ k=0,\ 1,\ 2</math>$ 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) The CIE color matching functions are not always positive. This is the case because, to match some reference color that is too saturated, colors have to be subtracted from the <span class="texhtml">''R'',''G'',</span> and <span class="texhtml">''B''</span> 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.  
+
Since <math>{{f}_{k}}(\lambda ),\ for\ k=0,\ 1,\ 2</math> 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.
  
−
c) <br> <math>\left[ {\begin{array}{*{20}{c}}
+
b)  
−
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}}
+
Since <math>{{r}_{0}}(\lambda ),\ {{g}_{0}}(\lambda ),\ and\ {{b}_{0}}(\lambda )</math> are the CIE color matching functions, they can be both positive and negative. The color matching function are given by
−
\int_{-\infty}^{\infty}  
+
<center>
−
\end{array}}  
+
<math>\left\{ \begin{matrix}
−
M
+
  {{r}_{0}}(\lambda )={{r}^{+}}-{{r}^{-}}  \\
−
\left[ {\begin{array}{*{20}{c}}
+
  {{g}_{0}}(\lambda )={{g}^{+}}-{{g}^{-}}  \\
−
r_0(\lambda)\\
+
  {{b}_{0}}(\lambda )=={{b}^{+}}-{{b}^{-}}  \\
−
g_0(\lambda)\\
+
\end{matrix} \right.</math>
−
b_0(\lambda)
+
</center>
−
\end{array}} \right]
+
where <math>{{r}^{+}},\ {{r}^{-}},\ {{g}^{+}},\ {{g}^{-}},\ {{b}^{+}},\ {{b}^{-}}</math>are the response to photons and must be positive, while the color matching function can be negative to produce a saturated color.
−
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
+
c)
−
\left[ {\begin{array}{*{20}{c}}
+
−
r\\
+
−
g\\
+
−
b
+
−
\end{array}} \right]</math>
+
  
−
So that, <span class="texhtml">[''r'',''g'',''b'']<sup>''t''</sup> = ''M''<sup> − 1</sup>[''F''<sub>1</sub>,''F''<sub>2</sub>,''F''<sub>3</sub>]</span>. <br>
+
<math>\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}
 +
</math>  
  
−
<span style="color:green"> missed transpose sign on F. It should be <span class="texhtml">[''r'',''g'',''b'']<sup>''t''</sup> = ''M''<sup> − 1</sup>[''F''<sub>1</sub>,''F''<sub>2</sub>,''F''<sub>3</sub>]<sup>''t''</sup></span>.
+
d)
 +
 
 +
Yes, they do exist, like CIE XYZ. CIE XYZ is defined in terms of CIE RGB so that
 +
<math>\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]</math> and
 +
<math>\left\{ \begin{matrix}
 +
  {{x}_{0}}(\lambda )\ge 0  \\
 +
  {{y}_{0}}(\lambda )\ge 0  \\
 +
  {{z}_{0}}(\lambda )\ge 0  \\
 +
\end{matrix} \right.</math>
  
  
−
</span>
 
−
<br>
 
−
d) It exists. CIE XYZ is one example. However, XYZ has problems with its primaries, since, the primary colors are imaginary.
 
  
 
== 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)
  
−
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; '''
+
Because for real pixels, measured energy from incident photons is always positive.
  
−
c)
+
<span style="color:green"> The student should mention the non-negativity inherence of the spectrum.</span>
  
−
&nbsp;<math>\left[ {\begin{array}{*{20}{c}}
+
b) <math>{{r}_{0}}(\lambda ),\ {{g}_{0}}(\lambda ),\ and\ {{b}_{0}}(\lambda )</math>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.
−
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}}
+
<span style="color:green"> The student should mention the saturated colors, which need negative color matching function .</span>
−
\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
+
c)  
−
{\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
+
<math>\begin{align}
−
\left[ {\begin{array}{*{20}{c}}
+
  & \int\limits_{-\infty }^{\infty }{\left[ \begin{matrix}
−
r\\
+
  {{f}_{1}}(\lambda ) \\
−
g\\
+
  {{f}_{2}}(\lambda )  \\
−
b
+
  {{f}_{3}}(\lambda )  \\
−
\end{array}} \right]</math>  
+
\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}
 +
</math>  
  
 
d)  
 
d)  
  
−
Yes. They exist. If there is a matrix M that exist to satisfy this equation &nbsp;<math>\left[ {\begin{array}{*{20}{c}}
+
<math>
−
f_1(\lambda)\\
+
\begin{align}
−
f_2(\lambda)\\
+
\left[ \begin{matrix}
−
f_3(\lambda)
+
  r  \\
−
\end{array}} \right]
+
  g  \\
−
= M \left[ {\begin{array}{*{20}{c}}
+
  b  \\
−
r_0(\lambda)\\
+
\end{matrix} \right]={{M}^{-1}}\left[ \begin{matrix}
−
g_0(\lambda)\\
+
  {X} \\
−
b_0(\lambda)
+
  {Y} \\
−
\end{array}} \right]</math>.&nbsp;
+
  {Z}  \\
 +
\end{matrix} \right] \\  
 +
\end{align}
 +
</math>
 +
where X, Y, Z are the xyz tristimulus values (always positive):
 +
<math> X=\frac{x}{x+y+z},Y=\frac{y}{x+y+z},Z=\frac{z}{x+y+z}</math>  
  
−
<span style="color:green"> 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.</span>
+
<span style="color:green"> The three written formulas for tristimulus values are not correct, actually chromaticity ccordinates can be written as a function of tristimulus values (X, Y, Z) as follows: <math> x=\frac{X}{X+Y+Z},y=\frac{Y}{X+Y+Z},z=\frac{Z}{X+Y+Z}</math>. </span>
 +
 
 +
<span style="color:green">Also, the student should describe the CIE XYZ system. </span>
  
 
----
 
----
Line 145: Line 173:
 
=== Related Problem  ===
 
=== 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 <span class="texhtml">λ</span>. Here the color matching allows for color to be subtracted from the reference color. At each wavelength <span class="texhtml">λ</span>, the matching color is given by  
+
In a color matching experiment, the three primaries R, G, B are used to match the color of a pure spectral component at wavelength <math>\lambda</math>. (Assume that the color matching allows for color to be subtracted from the reference in the standard manner described in class.)
−
 
+
At each wavelength <math>\lambda </math>, the matching color is given by  
−
<math>
+
−
\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>
+
−
 
+
−
where <span class="texhtml">''r''<sub>(</sub>λ)</span>, <span class="texhtml">''g''<sub>(</sub>λ)</span>, and <span class="texhtml">''b''<sub>(</sub>λ)</span> are normalized to 1.
+
−
 
+
−
Further define the white point
+
−
 
+
−
<math> 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]
+
−
</math>
+
  
−
Let <span class="texhtml">''I''(λ)</span> be the light reflected from a surface.
+
<center><math>\left[ \begin{matrix}
 +
  R, & G, & B  \\
 +
\end{matrix} \right]\left[ \begin{matrix}
 +
  r(\lambda ) \\
 +
  g(\lambda )  \\
 +
  b(\lambda )  \\
 +
\end{matrix} \right]</math></center>
 +
where
 +
<center><math>\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}</math>
 +
</center>
  
−
a) Calculate <span class="texhtml">(''r''<sub>''e''</sub>,''g''<sub>''e''</sub>,''b''<sub>''e''</sub>)</span> the tristimulus values for the spectral distribution <span class="texhtml">''I''(λ)</span> using primaries <span class="texhtml">''R'',''G'',''B''</span> and an equal energy white point.
+
Further define the white point
  
−
b) Calculate <span class="texhtml">(''r''<sub>''c''</sub>,''g''<sub>''c''</sub>,''b''<sub>''c''</sub>)</span> the tristimulus values for the spectral distribution <span class="texhtml">''I''(λ)</span> using primaries <span class="texhtml">''R'',''G'',''B''</span> and white point <span class="texhtml">(''r''<sub>''w''</sub>,''g''<sub>''w''</sub>,''b''<sub>''w''</sub>)</span>.  
+
<center>
 +
<math>W=\left[ \begin{matrix}
 +
  R, & G, & B \\
 +
\end{matrix} \right]\left[ \begin{matrix}
 +
  {{r}_{w}}  \\
 +
  {{g}_{w}}  \\
 +
  {{b}_{w}}  \\
 +
\end{matrix} \right]</math>.
 +
</center>
 +
Let <math>I(\lambda)</math> be the light reflected from a surface.
  
−
(Refer to ECE637 2004 Final Problem 4.)  
+
a) Calculate <math>({{r}_{e}}, {{g}_{e}}, {{b}_{e}})</math> the tristimulus values for the spectral distribution <math>I(\lambda)</math> using primaries R, G, B and an equal energy white point.
  
−
2. Consider the two channel sensors with response function&nbsp;<span class="texhtml">''Q''<sub>''S''</sub>(λ)&nbsp;''a'''n'''d&nbsp;'''Q'''''<b><sub>''L''</sub>(λ). Suppose that we have two primaries&nbsp;<span class="texhtml">''P''<sub>''L''</sub>(λ) = σ(λ − 0.6)</span>&nbsp;and&nbsp;<span class="texhtml">''P''<sub>''S''</sub>(λ) = σ(λ − 0.5)</span>.</b></span>
+
b) Calculate <math>({{r}_{c}}, {{g}_{c}}, {{b}_{c}})</math> the tristimulus values for the spectral distribution <math>I(\lambda)</math> using primaries R, G, B and white point <math>({{r}_{w}}, {{g}_{w}}, {{b}_{w}})</math>.
  
−
<span class="texhtml">'''[[Image:QE637 2013 P2 F1.PNG]]'''</span>  
+
c) Calculate <math>({{r}_{\gamma }}, {{g}_{\gamma }}, {{b}_{\gamma }})</math> the gamma corrected tristimulus values for the spectral distribution <math>I(\lambda)</math> using primaries R, G, B and white point <math>({{r}_{w}}, {{g}_{w}}, {{b}_{w}})</math>, and <math>\gamma =2.2</math>.
  
−
Find the color matching function&nbsp;<math>\bar{l}(\lambda)</math>&nbsp;and&nbsp;<math>\bar{s}(\lambda)</math>&nbsp;for these two primaries.
 
  
−
(Refer to ECE638 <u>[https://engineering.purdue.edu/~ece638/lectures/03.%20Trichromatic%20theory%20-%202011.pdf Lecture note 3: Trichromatic theory of color].</u>)  
+
(Refer to <u>[https://engineering.purdue.edu/~bouman/ece637/previous/ece637S2004/exams/final/exam.pdf ECE 637 Spring 2004 Final Exam Problem 4].</u>)  
  
 
----
 
----

Latest revision as of 21:06, 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} $

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] $ and $ \left\{ \begin{matrix} {{x}_{0}}(\lambda )\ge 0 \\ {{y}_{0}}(\lambda )\ge 0 \\ {{z}_{0}}(\lambda )\ge 0 \\ \end{matrix} \right. $


Solution 2:

a)

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

The student should mention the non-negativity inherence of the spectrum.

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.

The student should mention the saturated colors, which need negative color matching function .

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 xyz tristimulus values (always positive): $ X=\frac{x}{x+y+z},Y=\frac{y}{x+y+z},Z=\frac{z}{x+y+z} $

The three written formulas for tristimulus values are not correct, actually chromaticity ccordinates can be written as a function of tristimulus values (X, Y, Z) as follows: $ x=\frac{X}{X+Y+Z},y=\frac{Y}{X+Y+Z},z=\frac{Z}{X+Y+Z} $.

Also, the student should describe the CIE XYZ system.

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 ECE 637 Spring 2004 Final Exam Problem 4.)

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