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= [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] in Communication Networks Signal and Image processing (CS) =  
 
= [[ECE_PhD_Qualifying_Exams|ECE Ph.D. Qualifying Exam]] in Communication Networks Signal and Image processing (CS) =  
= [[QE637_T|Question 5, August 2012]], Part 1 =
+
= [[ECE-QE_CS5-2015|August 2015]], Part 1 =
  
:[[ CS5_2015_Aug_prob1_solution | Part 1 ]],[[ CS5_2015_Aug_prob2_solution | 2 ]]
+
:[[CS5_2015_Aug_prob1| Part 1 ]],[[CS5_2015_Aug_prob2_solution| 2 ]]
 
----
 
----
== Solution: ==
+
===Solution 1===
 
+
a) <math>\gamma=1</math>
a) <math>
+
\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}}
+
</math>
+
 
+
b)
+
  
 +
b) 
 
<math>
 
<math>
(x_r,y_r)=(\frac{a}{a+d+g},\frac{d}{a+d+g})
+
\left( \begin{array}{c}
 +
X_r\\
 +
Y_r \\
 +
Z_r \end{array} \right)=
 +
\left( \begin{array}{ccc}
 +
a & b & c \\
 +
d & e & f \\
 +
g & h & i \end{array} \right)
 +
\left( \begin{array}{c}
 +
1 \\
 +
0 \\
 +
0 \end{array} \right)=
 +
\left( \begin{array}{c}
 +
a \\
 +
d \\
 +
g \end{array} \right)</math><br \>
 +
So
 +
<math>
 +
(x_r,y_r)=(\frac{X_r}{X_r+Y_r+Z_r}, \frac{Y_r}{X_r+Y_r+Z_r})=(\frac{a}{a+d+g},\frac{d}{a+d+g})
 
</math> <br \>
 
</math> <br \>
 +
Similarly
 
<math>
 
<math>
 
(x_g,y_g)=(\frac{b}{b+e+h},\frac{e}{b+e+h})
 
(x_g,y_g)=(\frac{b}{b+e+h},\frac{e}{b+e+h})
</math><br \>
+
</math>,
 
<math>
 
<math>
(x_b,y_b)=(\frac{c}{c+f+i},\frac{f}{c+f+i})
+
(x_b,y_b)(\frac{c}{c+f+i},\frac{f}{c+f+i})
 
</math>
 
</math>
  
c)
+
c) The white point of the device is when the input <math>[R, G, B] = [1, 1, 1]</math>
  
 
<math>
 
<math>
Line 38: Line 50:
  
 
d)  
 
d)  
If <math> (X,Y,Z)=(0,1/2,1/2) </math>, then <math> (x,y)=(0,1/2) </math>.  [[ Image:Pro1_d.PNG ]]<br />
+
[[ Image:Pro1_solution1_2015_Aug.jpg]]<br />
In the chromaticity diagram, this point is outside the horse shoe shape, so its RGB values are not all larger than 0 (<math>R<0,G>0,B>0</math>).
+
  
 
e) We are likely to see quantization artifact in dark region.
 
e) We are likely to see quantization artifact in dark region.
 +
== Solution 2: ==
  
----
+
a) <math>
===Related Problem===
+
\frac{R}{255}^\alpha=r_{linear}\\
Consider a color imaging device that takes input values of <math> (r,g,b) </math> and produces ouput <math> (X,Y,Z)</math> values given by
+
 
 +
\Rightarrow
 +
\gamma=log_{\frac{R}{255}}{(R^{\alpha})}=\frac{ln{(R^{\alpha})}}{ln{\frac{R}{255}}}=\frac{\alpha{ln{R}}}{ln{R}-ln{255}}
 +
</math>
 +
 
 +
<span style="color:green"> <math>\gamma </math> should be 1. </span>
 +
 
 +
b)
  
 
<math>
 
<math>
\left[ {\begin{array}{*{20}{c}}
+
P_r=
X\\
+
\left( \begin{array}{ccc}
Y\\
+
a & b & c \\
Z
+
d & e & f \\
\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}
+
g & h & i \end{array} \right)
a&b&c\\
+
\left( \begin{array}{ccc}
d&e&f\\
+
1 \\
g&h&i
+
0 \\
\end{array}} \right]\left[ {\begin{array}{*{20}{c}}
+
0 \end{array} \right)
r^\alpha\\
+
=
g^\alpha\\
+
\left( \begin{array}{ccc}
b^\alpha
+
a \\
\end{array}} \right]
+
d \\
 +
g \end{array} \right) 
 +
\\
 +
\Rightarrow
 +
x_r=\frac{a}{a+d+g}
 +
,
 +
y_r=\frac{d}{a+d+g}
 +
\\
 +
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>
  
a) Calculate the white point of the device in chromaticity coordinates.
+
c)
  
b) What are the primaries associated with the r,g, and b components respectively?
+
<math>
 +
W=
 +
\left( \begin{array}{ccc}
 +
a & b & c \\
 +
d & e & f \\
 +
g & h & i \end{array} \right) 
 +
\left( \begin{array}{ccc}
 +
1 \\
 +
1 \\
 +
1\end{array} \right) 
 +
=
 +
\left( \begin{array}{ccc}
 +
a+b+c \\
 +
d+e+f \\
 +
g+h+i \end{array} \right) 
 +
\\
  
c) What is the gamma of the device?
+
\Rightarrow
 +
x_g=\frac{a+b+c}{a+b+c+d+e+f+g+h+i}
 +
,
 +
y_g=\frac{d+e+f}{a+b+c+d+e+f+g+h+i}
 +
</math>
 +
 
 +
d)  
 +
[[ Image:Pro1_2015_Aug.PNG ]]<br />
 +
 
 +
e) Gamma correction a quantization will create an effect of dynamic range compression for pixels with small values. This will create dark block of shadings in a gradient region instead of a smooth transition.
  
d) Draw the region on the chromaticity diagram corresponding to <math> r < 0, g > 0, b > 0</math>.
 
 
----
 
----
 +
[[ECE-QE_CS5-2015|Back to QE CS question 1, August 2013]]
 +
 
[[ECE_PhD_Qualifying_Exams|Back to ECE QE page]]:
 
[[ECE_PhD_Qualifying_Exams|Back to ECE QE page]]:

Latest revision as of 23:34, 3 December 2015


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

August 2015, Part 1

Part 1 , 2

Solution 1

a) $ \gamma=1 $

b) $ \left( \begin{array}{c} X_r\\ Y_r \\ Z_r \end{array} \right)= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right)= \left( \begin{array}{c} a \\ d \\ g \end{array} \right) $
So $ (x_r,y_r)=(\frac{X_r}{X_r+Y_r+Z_r}, \frac{Y_r}{X_r+Y_r+Z_r})=(\frac{a}{a+d+g},\frac{d}{a+d+g}) $
Similarly $ (x_g,y_g)=(\frac{b}{b+e+h},\frac{e}{b+e+h}) $, $ (x_b,y_b)=(\frac{c}{c+f+i},\frac{f}{c+f+i}) $

c) The white point of the device is when the input $ [R, G, B] = [1, 1, 1] $

$ (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) Pro1 solution1 2015 Aug.jpg

e) We are likely to see quantization artifact in dark region.

Solution 2:

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}} $

$ \gamma $ should be 1.

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} \\ 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)

$ W= \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ g & h & i \end{array} \right) \left( \begin{array}{ccc} 1 \\ 1 \\ 1\end{array} \right) = \left( \begin{array}{ccc} a+b+c \\ d+e+f \\ g+h+i \end{array} \right) \\ \Rightarrow x_g=\frac{a+b+c}{a+b+c+d+e+f+g+h+i} , y_g=\frac{d+e+f}{a+b+c+d+e+f+g+h+i} $

d) Pro1 2015 Aug.PNG

e) Gamma correction a quantization will create an effect of dynamic range compression for pixels with small values. This will create dark block of shadings in a gradient region instead of a smooth transition.


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