Line 22: | Line 22: | ||
<math> | <math> | ||
− | ( | + | P_r= |
− | + | \left( \begin{array}{ccc} | |
− | + | a & b & c \\ | |
− | ( | + | d & e & f \\ |
− | + | g & h & i \end{array} \right) | |
− | + | \left( \begin{array}{ccc} | |
− | ( | + | 1 \\ |
+ | 0 \\ | ||
+ | 0 \end{array} \right) | ||
+ | = | ||
+ | \left( \begin{array}{ccc} | ||
+ | a \\ | ||
+ | d \\ | ||
+ | g \end{array} \right) | ||
+ | |||
+ | \Rightarrow | ||
+ | x_r=\frac{a}{a+d+g} | ||
+ | , | ||
+ | y_r=\frac{d}{a+d+g} | ||
+ | \par | ||
+ | P_g= | ||
+ | \left( \begin{array}{ccc} | ||
+ | a & b & c \\ | ||
+ | d & e & f \\ | ||
+ | g & h & i \end{array} \right) | ||
+ | \left( \begin{array}{ccc} | ||
+ | 0 \\ | ||
+ | 1 \\ | ||
+ | 0 \end{array} \right) | ||
+ | = | ||
+ | \left( \begin{array}{ccc} | ||
+ | b \\ | ||
+ | e \\ | ||
+ | h \end{array} \right) | ||
+ | |||
+ | |||
+ | \Rightarrow | ||
+ | $x_g=\frac{b}{b+e+h} | ||
+ | |||
+ | , | ||
+ | y_g=\frac{e}{b+e+h} | ||
+ | \\ | ||
+ | |||
+ | P_b= | ||
+ | \left( \begin{array}{ccc} | ||
+ | a & b & c \\ | ||
+ | d & e & f \\ | ||
+ | g & h & i \end{array} \right) | ||
+ | \left( \begin{array}{ccc} | ||
+ | 0 \\ | ||
+ | 0 \\ | ||
+ | 1\end{array} \right) | ||
+ | = | ||
+ | \left( \begin{array}{ccc} | ||
+ | c \\ | ||
+ | f \\ | ||
+ | i \end{array} \right) | ||
+ | \Rightarrow | ||
+ | x_g=\frac{c}{c+f+i} | ||
+ | , | ||
+ | y_g=\frac{f}{c+f+i} | ||
</math> | </math> | ||
Revision as of 18:03, 2 December 2015
Contents
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)
Question 5, August 2012, Part 1
Solution:
a) $ \frac{R}{255}^\alpha=r_{linear}\\ \Rightarrow \gamma=log_{\frac{R}{255}}{(R^{\alpha})}=\frac{ln{(R^{\alpha})}}{ln{\frac{R}{255}}}=\frac{\alpha{ln{R}}}{ln{R}-ln{255}} $
b)
$ P_r= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{ccc} 1 \\ 0 \\ 0 \end{array} \right) = \left( \begin{array}{ccc} a \\ d \\ g \end{array} \right) \Rightarrow x_r=\frac{a}{a+d+g} , y_r=\frac{d}{a+d+g} \par P_g= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{ccc} 0 \\ 1 \\ 0 \end{array} \right) = \left( \begin{array}{ccc} b \\ e \\ h \end{array} \right) \Rightarrow $x_g=\frac{b}{b+e+h} , y_g=\frac{e}{b+e+h} \\ P_b= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{ccc} 0 \\ 0 \\ 1\end{array} \right) = \left( \begin{array}{ccc} c \\ f \\ i \end{array} \right) \Rightarrow x_g=\frac{c}{c+f+i} , y_g=\frac{f}{c+f+i} $
c)
$ (x_w,y_w)=(\frac{a+b+c}{a+b+c+d+e+f+g+h+i},\frac{d+e+f}{a+b+c+d+e+f+g+h+i}) $
d)
If $ (X,Y,Z)=(0,1/2,1/2) $, then $ (x,y)=(0,1/2) $.
In the chromaticity diagram, this point is outside the horse shoe shape, so its RGB values are not all larger than 0 ($ R<0,G>0,B>0 $).
e) We are likely to see quantization artifact in dark region.
Related Problem
Consider a color imaging device that takes input values of $ (r,g,b) $ and produces ouput $ (X,Y,Z) $ values given by
$ \left[ {\begin{array}{*{20}{c}} X\\ Y\\ Z \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} a&b&c\\ d&e&f\\ g&h&i \end{array}} \right]\left[ {\begin{array}{*{20}{c}} r^\alpha\\ g^\alpha\\ b^\alpha \end{array}} \right] $
a) Calculate the white point of the device in chromaticity coordinates.
b) What are the primaries associated with the r,g, and b components respectively?
c) What is the gamma of the device?
d) Draw the region on the chromaticity diagram corresponding to $ r < 0, g > 0, b > 0 $.