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[[Category:ECE]]
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[[Category:QE]]
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[[Category:CNSIP]]
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[[Category:problem solving]]
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[[Category:image processing]]
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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 2013]], Part 1 =
+
 
 +
= [[ECE-QE CS5-2013|Question 5, August 2013]], Problem 1 =
 +
 
 +
:[[QE637 2013 Pro1|Problem 1 ]],[[QE637 2013 Pro2|Problem 2 ]]
  
:[[ QE637_T_Pro1 | Part 1 ]],[[ QE637_T_Pro2 | 2 ]]
 
 
----
 
----
== Solution: ==
 
  
a) <math>\gamma=1</math>
+
== Solution 1:  ==
  
b)  
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a) Since
  
<math>
+
<math>X(e^{j\mu},e^{j\nu}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}  
(x_r,y_r)=(\frac{a}{a+d+g},\frac{d}{a+d+g})
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x(m,n)e^{-j(m\mu+n\nu)}</math><br>  
</math> <br \>
+
<math>
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(x_g,y_g)=(\frac{b}{b+e+h},\frac{e}{b+e+h})
+
</math><br \>
+
<math>
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(x_b,y_b)=(\frac{c}{c+f+i},\frac{f}{c+f+i})
+
</math>
+
  
c)
+
and
  
<math>
+
<span class="texhtml"><math>p_0(e^{jw}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty}  
(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})
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x(m,n)e^{-jnw}</math>,&nbsp;</span>  
</math>
+
  
d)
+
we have:  
If <math> (X,Y,Z)=(0,1/2,1/2) </math>, then <math> (x,y)=(0,1/2) </math>.  [[ Image:Pro1_d.PNG ]]<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.
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<math> p_0(e^{jw}) = X(e^{j\mu}, e^{jw})\vert_{\mu=0} </math>
  
== Solution 2: ==
+
b) Similarly to a), we have:  
  
a) The gamma is 1
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<math> p_1(e^{jw}) = X(e^{jw}, e^{j\nu})\vert_{\nu=0} </math>
  
b)  
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<span style="color:green"> The solution used <math> v </math> and <math>\mu</math> to represent frequency axis. It used <math> w</math> to subuslitude  both <math> v </math> and <math>\mu</math> which is confusing. The solution should stated let <math> w=v </math> and <math> w=\mu </math> at (a) and (b).  </span><br>
 +
c) <br> <math> \sum_{n=-\infty}^{\infty} p_0(n) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n) = X(e^{j\mu}, e^{j\nu}) |_{\mu=0, \nu=0} </math>
 +
which is the DC component of the image.
  
<math>
+
d) No, it can't provide sufficient information. From the expression in a) and b), we see that <math>p_0(e^{jw}) </math> and <math> p_1(e^{jw}) </math> are only slices of the DSFT. The information when <math>\mu</math> and <math>\nu</math> are not zeros is lost. <br>
(x_r,y_r)=(\frac{a}{a+d+g},\frac{d}{a+d+g})
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A simple example would be as follows.<br>
</math> <br \>
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Let <br> <math>
<math>
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x(m,n) =  
(x_g,y_g)=(\frac{b}{b+e+h},\frac{e}{b+e+h})
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\left[ {\begin{array}{*{20}{c}}
</math><br \>
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1 ~ 2 \\
<math>
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3 ~ 4\\
(x_b,y_b)=(\frac{c}{c+f+i},\frac{f}{c+f+i})
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\end{array}} \right] </math>, so<br> <math> p_0(n) =[4~6], p_1(m) =  [3 ~7]^T </math>.
</math>
+
<br>
 +
With the above the information of the projection, the original form of the 2D signal cannot be determined. For example, <math>
 +
x(m,n) =  
 +
\left[ {\begin{array}{*{20}{c}}
 +
2 ~ 1 \\
 +
2 ~ 5\\
 +
\end{array}} \right] </math> gives the same projection. '''
  
c)
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== Solution 2:  ==
  
<math>
+
a) From the question,&nbsp;
(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})
+
</math>
+
  
d) This color is imaginary. At least one of R,G,B values must be negative. Cannot be produced by this device. [[ Image:Pro1_d2.PNG ]]<br />
+
<math>P_0(e^{j\mu}) = \sum_{n=-\infty}^{\infty}p_0(n)e^{-jn\mu} = \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty}x(m,n) e^{-jn\mu}\cdot1
 +
= \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty}x(m,n) e^{-jn\mu}e^{-jm\cdot0} = X(e^{j\mu},e^{j\cdot0})</math>  
  
<span style="color:green"> The student can be more specific about the positive or negative of each R,G,B value of this color. </span>
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Therefore,&nbsp;
  
e) Quantization artifacts in the dark area.
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<math>P_0(e^{j\mu}) = X(e^{j\mu},e^{j\nu})\vert_{\nu = 0}</math><br>
 +
 
 +
b) Similar to question a),&nbsp;
 +
 
 +
<math>P_1(e^{j\nu}) = \sum_{m=-\infty}^{\infty}p_1(m)e^{-jm\mu} = \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty}x(m,n) e^{-jm\nu}\cdot1
 +
= \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty}x(m,n) e^{-jn\cdot0}e^{-jm\nu} = X(e^{j\cdot0},e^{j\nu})</math>
 +
 
 +
Therefore,
 +
 
 +
<math>P_0(e^{j\mu}) = X(e^{j\mu},e^{j\nu})\vert_{\mu = 0}</math><br>
 +
 
 +
<span style="color:green"> To be consistent with the problem statement, frequency notation<math>\mu</math> corresponds to the spatial notation <math>m </math> and is the first parameter. As a result, the solution of the (a) and (b) can be switched. </span>
 +
 
 +
 
 +
c)
 +
 
 +
<math>\sum_{n = -\infty}^{\infty}p_0(n) = \sum_{n = -\infty}^{\infty} \sum_{m = -\infty}^{\infty} x(m,n)
 +
=\sum_{n = -\infty}^{\infty} \sum_{m = -\infty}^{\infty} x(m,n) e^{-jn\cdot0}e^{-jm\cdot0} = X(e^{-jn\cdot0},e^{-jm\cdot0}) = X(e^{j\mu},e^{j\nu})\vert_{\mu = 0, \nu = 0}</math>
 +
 
 +
d)No. ''P''<sub>''0 &nbsp;''</sub>''only&nbsp;''represents the&nbsp;<span class="texhtml">μ</span>&nbsp;axis on <span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>. ''P<sub>1</sub>''&nbsp;only represents the <span class="texhtml">ν</span>&nbsp;axis on <span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>. It is not enough to represent&nbsp;<span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>.
 +
 
 +
<span style="color:green"> Similarly to the above comment, <math>P_0(e^{jw}) </math>is the slice along <math>\nu </math> axis. </span>
 +
 
 +
For example, assume two different array x''<sub>1</sub>'' and x''<sub>2</sub>''.
 +
 
 +
<math>x_1 = \left [
 +
    \begin{array}{cc}
 +
    3 & 4 \\
 +
    5 & 6
 +
    \end{array}
 +
    \right ]</math>&nbsp;and&nbsp;<math>x_2 = \left [
 +
    \begin{array}{cc}
 +
    4 & 3 \\
 +
    4 & 7
 +
    \end{array}
 +
    \right ]</math>&nbsp;have the same&nbsp;<span class="texhtml">''p''<sub>0&nbsp;</sub>and''&nbsp;p''<sub>1</sub>'''<sub>.&nbsp;</sub>'''</span>
 +
 
 +
<span class="texhtml">'''<sub></sub>'''</span>Therefore, ''P''<sub>''0''</sub> and ''P''<sub>''1''</sub> will be the same for ''X<sub>0</sub>'' and ''X<sub>1</sub>''. We will not be able to recover x<sub><span style="font-size: 11px;">''0''</span></sub>&nbsp;and x<sub><span style="font-size: 11px;">''1''</span></sub>&nbsp;based on ''P''<sub>''0''</sub> and ''P''<sub>''1''</sub>.&nbsp;
 +
 
 +
<br>
 +
 
 +
<span class="texhtml"><sub></sub></span>
  
 
----
 
----
===Related Problem===
 
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
 
  
<math>
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=== Related Problem  ===
\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]
+
</math>
+
  
a) Calculate the white point of the device in chromaticity coordinates.
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1.Let <span class="texhtml">''g''(''x'',''y'') = ''s''''i''''n''''c'''''<b>(</b>'''''x'' / 2,''y'' / 2)'''</span>''''', and let &lt;span class="texhtml" /&gt;''s''(''m'',''n'') = ''g''(''''''<i>T</i>,''n''''T'''''<b>) where T = 1.<br> </b>
  
b) What are the primaries associated with the r,g, and b components respectively?
+
a) Calculate <span class="texhtml">''G''(μ,ν)</span> the CSFT of <span class="texhtml">''g''(''x'',''y'')</span>. <br> b) Calculate <span class="texhtml">''S''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span> the DSFT of <span class="texhtml">''s''(''m'',''n'')</span>. <br>
  
c) What is the gamma of the device?
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2. Assume that we know (or can measure) the function
 +
 
 +
<math>p(x) = \int_{-\infty}^{\infty}f(x,y)dy</math>
 +
 
 +
Using the definitions of the Fourier transform, derive an expressoin for&nbsp;<span class="texhtml">''F''(''u'',0)</span>&nbsp;in terms of the function&nbsp;<span class="texhtml">''p''(''x'')</span>.
 +
 
 +
(Refer to ECE637 2008 Exam1 Problem2.)
  
d) Draw the region on the chromaticity diagram corresponding to <math> r < 0, g > 0, b > 0</math>.
 
 
----
 
----
[[ECE_PhD_Qualifying_Exams|Back to ECE QE page]]:
+
 
 +
[[ECE PhD Qualifying Exams|Back to ECE QE page]]:  
 +
 
 +
[[Category:ECE]] [[Category:QE]] [[Category:CNSIP]] [[Category:Problem_solving]] [[Category:Image_processing]]

Latest revision as of 21:10, 2 December 2015


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

Question 5, August 2013, Problem 1

Problem 1 ,Problem 2

Solution 1:

a) Since

$ X(e^{j\mu},e^{j\nu}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n)e^{-j(m\mu+n\nu)} $

and

$ p_0(e^{jw}) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n)e^{-jnw} $, 

we have:

$ p_0(e^{jw}) = X(e^{j\mu}, e^{jw})\vert_{\mu=0} $

b) Similarly to a), we have:

$ p_1(e^{jw}) = X(e^{jw}, e^{j\nu})\vert_{\nu=0} $

The solution used $ v $ and $ \mu $ to represent frequency axis. It used $ w $ to subuslitude both $ v $ and $ \mu $ which is confusing. The solution should stated let $ w=v $ and $ w=\mu $ at (a) and (b).
c)
$ \sum_{n=-\infty}^{\infty} p_0(n) = \sum_{m=-\infty}^{\infty} \sum_{n=-\infty}^{\infty} x(m,n) = X(e^{j\mu}, e^{j\nu}) |_{\mu=0, \nu=0} $ which is the DC component of the image.

d) No, it can't provide sufficient information. From the expression in a) and b), we see that $ p_0(e^{jw}) $ and $ p_1(e^{jw}) $ are only slices of the DSFT. The information when $ \mu $ and $ \nu $ are not zeros is lost.
A simple example would be as follows.
Let
$ x(m,n) = \left[ {\begin{array}{*{20}{c}} 1 ~ 2 \\ 3 ~ 4\\ \end{array}} \right] $, so
$ p_0(n) =[4~6], p_1(m) = [3 ~7]^T $.
With the above the information of the projection, the original form of the 2D signal cannot be determined. For example, $ x(m,n) = \left[ {\begin{array}{*{20}{c}} 2 ~ 1 \\ 2 ~ 5\\ \end{array}} \right] $ gives the same projection.

Solution 2:

a) From the question, 

$ P_0(e^{j\mu}) = \sum_{n=-\infty}^{\infty}p_0(n)e^{-jn\mu} = \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty}x(m,n) e^{-jn\mu}\cdot1 = \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty}x(m,n) e^{-jn\mu}e^{-jm\cdot0} = X(e^{j\mu},e^{j\cdot0}) $

Therefore, 

$ P_0(e^{j\mu}) = X(e^{j\mu},e^{j\nu})\vert_{\nu = 0} $

b) Similar to question a), 

$ P_1(e^{j\nu}) = \sum_{m=-\infty}^{\infty}p_1(m)e^{-jm\mu} = \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty}x(m,n) e^{-jm\nu}\cdot1 = \sum_{n=-\infty}^{\infty} \sum_{m=-\infty}^{\infty}x(m,n) e^{-jn\cdot0}e^{-jm\nu} = X(e^{j\cdot0},e^{j\nu}) $

Therefore,

$ P_0(e^{j\mu}) = X(e^{j\mu},e^{j\nu})\vert_{\mu = 0} $

To be consistent with the problem statement, frequency notation$ \mu $ corresponds to the spatial notation $ m $ and is the first parameter. As a result, the solution of the (a) and (b) can be switched.


c)

$ \sum_{n = -\infty}^{\infty}p_0(n) = \sum_{n = -\infty}^{\infty} \sum_{m = -\infty}^{\infty} x(m,n) =\sum_{n = -\infty}^{\infty} \sum_{m = -\infty}^{\infty} x(m,n) e^{-jn\cdot0}e^{-jm\cdot0} = X(e^{-jn\cdot0},e^{-jm\cdot0}) = X(e^{j\mu},e^{j\nu})\vert_{\mu = 0, \nu = 0} $

d)No. P0  only represents the μ axis on X(ejμ,ejν). P1 only represents the ν axis on X(ejμ,ejν). It is not enough to represent X(ejμ,ejν).

Similarly to the above comment, $ P_0(e^{jw}) $is the slice along $ \nu $ axis.

For example, assume two different array x1 and x2.

$ x_1 = \left [ \begin{array}{cc} 3 & 4 \\ 5 & 6 \end{array} \right ] $ and $ x_2 = \left [ \begin{array}{cc} 4 & 3 \\ 4 & 7 \end{array} \right ] $ have the same pand p1

Therefore, P0 and P1 will be the same for X0 and X1. We will not be able to recover x0 and x1 based on P0 and P1



Related Problem

1.Let g(x,y) = s'i'n'c(x / 2,y / 2), and let <span class="texhtml" />s(m,n) = g('T,n'T) where T = 1.

a) Calculate G(μ,ν) the CSFT of g(x,y).
b) Calculate S(ejμ,ejν) the DSFT of s(m,n).

2. Assume that we know (or can measure) the function

$ p(x) = \int_{-\infty}^{\infty}f(x,y)dy $

Using the definitions of the Fourier transform, derive an expressoin for F(u,0) in terms of the function p(x).

(Refer to ECE637 2008 Exam1 Problem2.)


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