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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 $.
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(e^jμ,e^jν). P₁ only represents the ν axis on X(e^jμ,e^jν). It is not enough to represent X(e^jμ,e^jν).

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 p_1. 

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(e^jμ,e^jν) 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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