| Line 48: | Line 48: | ||
<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 | <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}) | + | = \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> |
| − | <br> b) Similar to question a), | + | Therefore, |
| + | |||
| + | <math>P_0(e^{j\mu}) = X(e^{j\mu},e^{j\nu})\vert_{\nu = 0}</math><br> | ||
| + | |||
| + | b) Similar to question a), | ||
<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 | <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}) = X(e^{j\mu},e^{j\nu})\vert_{\mu = 0}</math> | + | = \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> | ||
c) | c) | ||
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d)No. ''P''<sub>''0 ''</sub>''only ''represents the <span class="texhtml">μ</span> axis on <span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>. ''P<sub>1</sub>'' only represents the <span class="texhtml">ν</span> axis on <span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>. It is not enough to represent <span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>. | d)No. ''P''<sub>''0 ''</sub>''only ''represents the <span class="texhtml">μ</span> axis on <span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>. ''P<sub>1</sub>'' only represents the <span class="texhtml">ν</span> axis on <span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>. It is not enough to represent <span class="texhtml">''X''(''e''<sup>''j''μ</sup>,''e''<sup>''j''ν</sup>)</span>. | ||
| + | |||
| + | For example, | ||
| + | |||
| + | <math>A = \left [ | ||
| + | \begin{array}{cc} | ||
| + | 3 & 4 \\ | ||
| + | 5 & 6 | ||
| + | \end{array} | ||
| + | \right ]</math> and <math>A = \left [ | ||
| + | \begin{array}{cc} | ||
| + | 4 & 3 \\ | ||
| + | 4 & 7 | ||
| + | \end{array} | ||
| + | \right ]</math> have the same <math>p_0 and p_1</math> | ||
---- | ---- | ||
Revision as of 16:30, 12 November 2014
Contents
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:
p0(ej'w) = X(ejμ,ejw) | μ = 0
b) Similarly to a), we have:
p1(ej'w) = X(ejw,ejν) | ν = 0
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 point of the image.
d) No, it can't provide sufficient information. From the expression in a) and b), we see that p0(ej'w)and <span class="texhtml" />p1(ejw) are only slices of the DSFT. It lost the information when μ and ν are not zero. A simple example would be: 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} $
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).
For example,
$ A = \left [ \begin{array}{cc} 3 & 4 \\ 5 & 6 \end{array} \right ] $ and $ A = \left [ \begin{array}{cc} 4 & 3 \\ 4 & 7 \end{array} \right ] $ have the same $ p_0 and p_1 $
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).
