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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 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.)
