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== Solution 1: == | == Solution 1: == | ||
| − | a) | + | a) <math> |
| − | + | ||
| − | <math> | + | |
{{P}_{0}}({{e}^{j\omega }})=\sum\limits_{n=-\infty }^{\infty }{{{p}_{0}}(n){{e}^{-jn\omega }}}=\sum\limits_{n=-\infty }^{\infty }{\left( \sum\limits_{m=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jn\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m0+n\omega )}}=X({{e}^{j0}},{{e}^{j\omega }})} | {{P}_{0}}({{e}^{j\omega }})=\sum\limits_{n=-\infty }^{\infty }{{{p}_{0}}(n){{e}^{-jn\omega }}}=\sum\limits_{n=-\infty }^{\infty }{\left( \sum\limits_{m=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jn\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m0+n\omega )}}=X({{e}^{j0}},{{e}^{j\omega }})} | ||
</math><br> | </math><br> | ||
| − | b) | + | b) <math> |
| − | + | ||
| − | <math> | + | |
{{P}_{1}}({{e}^{j\omega }})=\sum\limits_{m=-\infty }^{\infty }{{{p}_{1}}(m){{e}^{-jm\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\left( \sum\limits_{n=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jm\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m\omega +n0)}}=X({{e}^{j\omega }},{{e}^{j0}})} | {{P}_{1}}({{e}^{j\omega }})=\sum\limits_{m=-\infty }^{\infty }{{{p}_{1}}(m){{e}^{-jm\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\left( \sum\limits_{n=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jm\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m\omega +n0)}}=X({{e}^{j\omega }},{{e}^{j0}})} | ||
</math> | </math> | ||
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c) <br> <math> | c) <br> <math> | ||
| − | |||
\sum\limits_{n=-\infty }^{\infty }{{{p}_{0}}(n)}==\sum\limits_{n=-\infty }^{\infty }{\left( \sum\limits_{m=-\infty }^{\infty }{x(m,n)} \right)}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m0+n0)}}=X({{e}^{j0}},{{e}^{j0}})} | \sum\limits_{n=-\infty }^{\infty }{{{p}_{0}}(n)}==\sum\limits_{n=-\infty }^{\infty }{\left( \sum\limits_{m=-\infty }^{\infty }{x(m,n)} \right)}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m0+n0)}}=X({{e}^{j0}},{{e}^{j0}})} | ||
</math> | </math> | ||
| − | d) | + | d) No, they don’t. From part (a) and (b), we know that <math>{{P}_{0}}({{e}^{jw}})</math> and <math>{{P}_{1}}({{e}^{jw}})</math> represent the horizontal and vertical axes of the 2D DSFT <math>X({{e}^{j\mu }},{{e}^{j\upsilon }})</math>, which is not enough for reconstruction of x(m, n). |
| − | No, they don’t. From part (a) and (b), we know that <math>{{P}_{0}}({{e}^{jw}})</math> and <math>{{P}_{1}}({{e}^{jw}})</math> represent the horizontal and vertical axes of the 2D DSFT <math>X({{e}^{j\mu }},{{e}^{j\upsilon }})</math>, which is not enough for reconstruction of x(m, n). For example, <math>{{x}_{1}}(m,n)=\left( \begin{matrix} | + | For example, <math>{{x}_{1}}(m,n)=\left( \begin{matrix} |
1 & 3 \\ | 1 & 3 \\ | ||
2 & 4 \\ | 2 & 4 \\ | ||
| Line 47: | Line 42: | ||
4 \\ | 4 \\ | ||
6 \\ | 6 \\ | ||
| − | \end{matrix} \right]</math> | + | \end{matrix} \right]</math>. |
| − | + | ||
== Solution 2: == | == Solution 2: == | ||
| − | a) | + | a) <math> |
| − | + | {{P}_{1}}({{e}^{j\omega }})=\sum\limits_{m=-\infty }^{\infty }{\left( \sum\limits_{n=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jm\omega }}}=X({{e}^{j\omega }},{{e}^{j0}})} | |
| − | <math> | + | </math><br> |
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| − | + | ||
| + | b) <math> | ||
| + | {{P}_{0}}({{e}^{j\omega }})=\sum\limits_{n=-\infty }^{\infty }{\left( \sum\limits_{m=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jn\omega }}}=X({{e}^{j0}},{{e}^{j\omega }})} | ||
| + | </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> | <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> | ||
Revision as of 21:01, 1 May 2017
Contents
ECE Ph.D. Qualifying Exam in Communication Networks Signal and Image processing (CS)
= Question 5, August 2013(Published on May 2017),
- Problem 1,2
Solution 1:
a) $ {{P}_{0}}({{e}^{j\omega }})=\sum\limits_{n=-\infty }^{\infty }{{{p}_{0}}(n){{e}^{-jn\omega }}}=\sum\limits_{n=-\infty }^{\infty }{\left( \sum\limits_{m=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jn\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m0+n\omega )}}=X({{e}^{j0}},{{e}^{j\omega }})} $
b) $ {{P}_{1}}({{e}^{j\omega }})=\sum\limits_{m=-\infty }^{\infty }{{{p}_{1}}(m){{e}^{-jm\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\left( \sum\limits_{n=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jm\omega }}}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m\omega +n0)}}=X({{e}^{j\omega }},{{e}^{j0}})} $
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\limits_{n=-\infty }^{\infty }{{{p}_{0}}(n)}==\sum\limits_{n=-\infty }^{\infty }{\left( \sum\limits_{m=-\infty }^{\infty }{x(m,n)} \right)}=\sum\limits_{m=-\infty }^{\infty }{\sum\limits_{n=-\infty }^{\infty }{x(m,n)}{{e}^{-j(m0+n0)}}=X({{e}^{j0}},{{e}^{j0}})} $
d) No, they don’t. From part (a) and (b), we know that $ {{P}_{0}}({{e}^{jw}}) $ and $ {{P}_{1}}({{e}^{jw}}) $ represent the horizontal and vertical axes of the 2D DSFT $ X({{e}^{j\mu }},{{e}^{j\upsilon }}) $, which is not enough for reconstruction of x(m, n). For example, $ {{x}_{1}}(m,n)=\left( \begin{matrix} 1 & 3 \\ 2 & 4 \\ \end{matrix} \right),_{{}}^{{}}and_{{}}^{{}}{{x}_{2}}(m,n)=\left( \begin{matrix} 0 & 4 \\ 3 & 3 \\ \end{matrix} \right) $ have the same $ {{p}_{0}}(n)=\left[ \begin{matrix} 3 & 7 \\ \end{matrix} \right]_{{}}^{{}}and_{{}}^{{}}{{p}_{1}}(m)=\left[ \begin{matrix} 4 \\ 6 \\ \end{matrix} \right] $. So, x(m,n) can’t be reconstructed from $ {{p}_{0}}(n)=\left[ \begin{matrix} 3 & 7 \\ \end{matrix} \right]_{{}}^{{}}and_{{}}^{{}}{{p}_{1}}(m)=\left[ \begin{matrix} 4 \\ 6 \\ \end{matrix} \right] $.
Solution 2:
a) $ {{P}_{1}}({{e}^{j\omega }})=\sum\limits_{m=-\infty }^{\infty }{\left( \sum\limits_{n=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jm\omega }}}=X({{e}^{j\omega }},{{e}^{j0}})} $
b) $ {{P}_{0}}({{e}^{j\omega }})=\sum\limits_{n=-\infty }^{\infty }{\left( \sum\limits_{m=-\infty }^{\infty }{x(m,n)} \right){{e}^{-jn\omega }}}=X({{e}^{j0}},{{e}^{j\omega }})} $
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 p0 and 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.)
