Line 22: Line 22:
  
 
<math>
 
<math>
(x_r,y_r)=(\frac{a}{a+d+g},\frac{d}{a+d+g})
+
P_r=
</math> <br \>
+
\left( \begin{array}{ccc}
<math>
+
a & b & c \\
(x_g,y_g)=(\frac{b}{b+e+h},\frac{e}{b+e+h})
+
d & e & f \\
</math><br \>
+
g & h & i \end{array} \right)  
<math>
+
\left( \begin{array}{ccc}
(x_b,y_b)=(\frac{c}{c+f+i},\frac{f}{c+f+i})
+
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 19:03, 2 December 2015


ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)

Question 5, August 2012, Part 1

Part 1 , 2

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) $. Pro1 d.PNG
In the chromaticity diagram, this point is outside the horse shoe shape, so its RGB values are not all larger than 0 ($ 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 $.


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Ryne Rayburn