Difference between revisions of "QE637 sol2012 Q1" - Rhea

Revision as of 10:27, 2 May 2017

Solution1:

a)

The gamma of the device is equal 1.

b)

$ \begin{align} & \left[ \begin{matrix} {{X}_{r}} \\ {{Y}_{r}} \\ {{Z}_{r}} \\ \end{matrix} \right]=\left[ \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} \right]\left[ \begin{matrix} 1 \\ 0 \\ 0 \\ \end{matrix} \right]\Rightarrow \left\{ \begin{matrix} {{X}_{r}}=a \\ {{Y}_{r}}=d \\ {{Z}_{r}}=g \\ \end{matrix} \right. \\ & {{x}_{r}}={{\frac{{{X}_{r}}}{{{X}_{r}}+{{Y}_{r}}+Z}}_{r}}=\frac{a}{a+d+g},_{{}}^{{}}{{y}_{r}}=\frac{{{Y}_{r}}}{{{X}_{r}}+{{Y}_{r}}+{{Z}_{r}}}=\frac{d}{a+d+g} \\ \end{align} $

$ \begin{align} & \left[ \begin{matrix} {{X}_{g}} \\ {{Y}_{g}} \\ {{Z}_{g}} \\ \end{matrix} \right]=\left[ \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 1 \\ 0 \\ \end{matrix} \right]\Rightarrow \left\{ \begin{matrix} {{X}_{g}}=b \\ {{Y}_{g}}=e \\ {{Z}_{g}}=h \\ \end{matrix} \right. \\ & {{x}_{g}}=\frac{{{X}_{g}}}{{{X}_{g}}+{{Y}_{g}}+{{Z}_{g}}}=\frac{b}{b+e+h},_{{}}^{{}}{{y}_{g}}=\frac{{{Y}_{g}}}{{{X}_{g}}+{{Y}_{g}}+{{Z}_{g}}}=\frac{e}{b+e+h} \\ \end{align} $

$ \begin{align} & \left[ \begin{matrix} {{X}_{b}} \\ {{Y}_{b}} \\ {{Z}_{b}} \\ \end{matrix} \right]=\left[ \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} \right]\left[ \begin{matrix} 0 \\ 0 \\ 1 \\ \end{matrix} \right]\Rightarrow \left\{ \begin{matrix} {{X}_{b}}=c \\ {{Y}_{b}}=f \\ {{Z}_{b}}=i \\ \end{matrix} \right. \\ & {{x}_{b}}=\frac{{{X}_{b}}}{{{X}_{b}}+{{Y}_{b}}+{{Z}_{b}}}=\frac{c}{c+f+i},_{{}}^{{}}{{y}_{b}}=\frac{{{Y}_{b}}}{{{X}_{b}}+{{Y}_{b}}+{{Z}_{b}}}=\frac{f}{c+f+i} \\ \end{align} $

c)

$ \begin{align} & \left[ \begin{matrix} {{X}_{w}} \\ {{Y}_{w}} \\ {{Z}_{w}} \\ \end{matrix} \right]=\left[ \begin{matrix} a & b & c \\ d & e & f \\ g & h & i \\ \end{matrix} \right]\left[ \begin{matrix} 1 \\ 1 \\ 1 \\ \end{matrix} \right]\Rightarrow \left\{ \begin{matrix} {{X}_{w}}=a+b+c \\ {{Y}_{w}}=d+e+f \\ {{Z}_{w}}=g+h+i \\ \end{matrix} \right. \\ & A={{X}_{w}}+{{Y}_{w}}+{{Z}_{w}}=a+b+c+d+e+f+g+h+i \\ & {{x}_{w}}=\frac{{{X}_{w}}}{{{X}_{w}}+{{Y}_{w}}+{{Z}_{w}}}=\frac{a+b+c}{A},_{{}}^{{}}{{y}_{w}}=\frac{{{Y}_{w}}}{{{X}_{w}}+{{Y}_{w}}+{{Z}_{w}}}=\frac{d+e+f}{A} \\ \end{align} $

d)

As can be seen, since the point lies outside the horseshoe shape diagram, it’s doesn’t exist (imaginary color) while mathematically we can talk about it and write down the equation for it. For this point, R<0, G>0, and B>0.

e)

We see false contours in dark region, because small changes in quantization level leads to large contrast changes that cause visible contours.

Solution 2:

a)

$ \left[ {\begin{array}{*{20}{c}} R\\ G\\ B \end{array}} \right] $ are linear with energy $ \Rightarrow $ $ \gamma=1 $.

b)

$ \left[ {\begin{array}{*{20}{c}} X\\ Y\\ Z \end{array}} \right]=M\left[ {\begin{array}{*{20}{c}} 1\\ 0\\ 0\\ \end{array}} \right] \Rightarrow (x_r,y_r)=(\frac{a}{a+d+g},\frac{d}{a+d+g}) $

$ \left[ {\begin{array}{*{20}{c}} X\\ Y\\ Z \end{array}} \right]=M\left[ {\begin{array}{*{20}{c}} 0\\ 1\\ 0\\ \end{array}} \right] \Rightarrow (x_g,y_g)=(\frac{b}{b+e+h},\frac{e}{b+e+h}) $

$ \left[ {\begin{array}{*{20}{c}} X\\ Y\\ Z \end{array}} \right]=M\left[ {\begin{array}{*{20}{c}} 0\\ 0\\ 1\\ \end{array}} \right] \Rightarrow (x_b,y_b)=(\frac{c}{c+f+i},\frac{f}{c+f+i}) $

c)

$ \left[ {\begin{array}{*{20}{c}} X\\ Y\\ Z \end{array}} \right]=M\left[ {\begin{array}{*{20}{c}} 1\\ 1\\ 1\\ \end{array}} \right] \Rightarrow {{x}_{w}}=\frac{a+b+c}{a+b+c+d+e+f+g+h+i}, {{Y}_{w}}=\frac{d+e+f}{a+b+c+d+e+f+g+h+i} , {{Z}_{w}}=1-{{x}_{w}}-{{x}_{w}}. $

d)

R is negative and B and G are positive.

The student can be more specific about the positive or negative of each R,G,B value of this color.

e)

Artifacts in dark regions of gradient where human contrast sensitivity is higher (more sensitive to quantization error).

Related Problem

Consider the color display that produces the following color (X,Y,Z) when given the input is (r, g, b). $ \left[ \begin{matrix} X \\ Y \\ Z \\ \end{matrix} \right]=\left[ \begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{matrix} \right]\left[ \begin{matrix} {{(r/255)}^{\alpha }} \\ {{(g/255)}^{\alpha }} \\ {{(b/255)}^{\alpha }} \\ \end{matrix} \right] $

a) What is the gamma of the display? b) What is the white point of the display? c) What are the color primaries of the display? d) Can such a display be physically built? Why or why not? e) You produce two images, one with a checker board pattern of values 0 and 255, and a second with a constant value of (r, g, b) = (a, a, a). The value of a is then adjusted so that the two images match when viewed from a distance. What is the value of gamma for the display in terms of the value a? f) Imagine that $ \alpha =1 $ and the values of (r, g, b) are quantized to 8-bits. Describe the defects you would expect to see in the displayed image.

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