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ECE Ph.D. Qualifying Exam in "Communication, Networks, Signal, and Image Processing" (CS)

Question 5, August 2011, Part 1

Part 1,2]

 $ \color{blue}\text{Consider the following discrete space system with input } x(m,n) \text{ and output } y(m,n). $

                $ \color{blue} y(m,n) = \sum_{k=-\infty}^{\infty}{\sum_{l=-\infty}^{\infty}{x(m-k,n-l)h(k,l)}}. $

$ \color{blue} \text{For parts a) and b) let} $
                $ \color{blue} h(m,n)=sinc(mT,nT) $
$ \color{blue} \text{where } T\leq1. $

$ \color{blue} \text{For parts c), d), and e) let} $
                $ \color{blue} h(m,n)=sinc\left( \frac{(n+m)T}{\sqrt[]{2}},\frac{(n-m)T}{\sqrt[]{2}} \right) $
$ \color{blue} \text{where } T\leq1. $

$ \color{blue}\text{a) Calculate the frequency response, }H \left( e^{j\mu},e^{j\nu} \right). $

$ \color{blue}\text{Solution 1:} $

$ \color{green} \text{Recall:} $

$ \color{green} f(am,bn) \overset{DTFT}{\Leftrightarrow } \frac{1}{|a||b|}F(\frac{\mu}{|a|},\frac{\nu}{|b|}) $

$ \color{green} sinc(m,n) \overset{DTFT}{\Leftrightarrow } rect(\mu,\nu) $

$ H(e^{j\mu},e^{j\nu}) = \frac{1}{T^2} rect(\frac{\mu}{T},\frac{\nu}{T}) $

$ \color{blue}\text{Solution 2:} $

$ sinc(m,n) \rightarrow rect(\mu)rect(\nu) $

$ \Rightarrow sinc(mT,nT) \rightarrow \frac{1}{T^2}rect(\frac{\mu}{T})rect(\frac{\nu}{T}) $

$ \color{green} \text{Here, the student uses the Separability property of the sinc and rect functions.} $

$ \color{blue}\text{b) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |nu| < 2\pi \text{ when } T = \frac{1}{2} $

$ \color{green} \text{Recall:} $

$ \color{green} rect(t) = \left\{\begin{matrix} 1, for |t|\leq \frac{1}{2} \\ 0, otherwise \end{matrix}\right. $

$ \text{Using the separability property for rect function, for } T = \frac{1}{2} { we have:} $

$ H(e^{j\mu},e^{j\nu}) = \frac{1}{T^2} rect(\frac{\mu}{T},\frac{\nu}{T}) $

$ = 4 rect(2\mu)rect(2\nu) $

$ \text{Note that the gain in this sketch will be } 4. $

$ \color{blue}\text{Solution 2:} $

$ T = \frac{1}{2}, H(e^{j\mu},e^{j\nu}) = 4rect(2\mu)rect(2\nu) $

$ \color{blue}\text{c) Calculate the frequency response, }H \left( e^{j\mu},e^{j\nu} \right). $

$ \color{blue}\text{Solution 1:} $

$ \color{green} \text{Recall:} $

$ \color{green} f \left ( A \begin{bmatrix} m \\ n \end{bmatrix} \right) \overset{DTFT}{\Leftrightarrow } \frac{1}{|A|^{-1}}F([\mu, \nu] A^{-1}) $

$ \text{ In this case, A}= \begin{bmatrix} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \end{bmatrix} \text{, hence:} $

$ H(e^{j\mu},e^{j\nu}) = \frac{1}{T^2} rect \left ( \frac{(\mu + \nu)}{\sqrt{2}T},\frac{(\nu - \mu)}{\sqrt{2}T} \right ) $

$ \color{blue}\text{Solution 2:} $

$ \left ( \frac{(n + m)T}{\sqrt{2}},\frac{(n - m)T}{\sqrt{2}} \right) = \begin{bmatrix} \frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} &\frac{1}{\sqrt{2}} \end{bmatrix} \cdot \begin{pmatrix} mT\\ nT \end{pmatrix} = A \cdot \begin{pmatrix} mT\\ nT \end{pmatrix} $

$ \text{As } |A| = 1, A^{-1} = A^T, sinc \left( A \begin{pmatrix} mT\\ nT \end{pmatrix} \right) \overset{\mathcal{F}}{\rightarrow} F \left( A \begin{pmatrix} \mu\\ \nu \end{pmatrix} \right) $

$ = \frac{1}{T^2} rect \left ( \frac{(\mu + \nu)}{\sqrt{2}T},\frac{(\nu - \mu)}{\sqrt{2}T} \right ) $

$ \color{blue}\text{d) Sketch the frequency response for } |\mu| < 2\pi \text{ and } |nu| < 2\pi \text{ when } T = \frac{1}{2} $

$ \color{blue}\text{Solution 1:} $

$ \color{blue}\text{Solution 2:} $

$ T = \frac{1}{2}, H(e^{j\mu},e^{j\nu}) = 4rect(\sqrt{2}(\mu + \nu))rect(\sqrt{2}(\nu - \mu)) $

$ \color{blue}\text{e) Calculate } y(m,n) \text{ when } x(m,n)=1. $

$ \color{blue}\text{Solution 1:} $

$ Y(e^{j\mu},e^{j\nu}) = \delta(e^{j\mu},e^{j\nu}) \cdot H(e^{j\mu},e^{j\nu}) $

$ = \frac{1}{T^2} rect (0,0) = 4 $

$ \Rightarrow y(m,n) = 4\delta(m,n) $

$ \color{blue}\text{Solution 2:} $

$ y(m,n) = x(m,n) \cdot H(e^{j0},e^{j0}) = 4 $

"Communication, Networks, Signal, and Image Processing" (CS)- Question 5, August 2011

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• Part 1: solutions and discussions
• Part 2: solutions and discussions

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