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final.tex

\documentclass[a4paper,11pt]{article}
\usepackage{defsDSPcourse}
\usepackage{dspTricks,dspBlocks,dspFunctions}
\begin{document}
\title{COM-303 - Signal Processing for Communications\\Final Exam}
\date{Tuesday, July 3 2012, 08:15 to 11:15}
\framebox[12cm][l]{Name:\rule{0cm}{1cm}}\hfill\framebox[3cm][l]{\#\rule{0cm}{1cm}}
\vspace{1em}
\begin{center}%
\sffamily
{\LARGE \bfseries COM-303 - Signal Processing for Communications \\ Final Exam} \\
\vspace{1em}
{\large Tuesday, July 3 2012, 08:15 to 11:15}
\vspace{1em}
\end{center}
\centerline{\rule{\textwidth}{.5pt}}
\begin{itemize}
\item {\bf Write your name} on the top left corner of {\bf ALL sheets you turn in}, including this one. When you are done, \textbf{staple} all your sheets together \textbf{with this sheet on top}!
\item You can have two A4 sheet of \emph{handwritten} notes (front and back). Please \textbf{no photocopies, no books and no electronic devices}. Turn off your phone if you have it with you.
\item There are 9 problems for a total of 100 points.
\item Please write your derivations clearly, as there is partial credit.
\end{itemize}
\centerline{\rule{\textwidth}{.5pt}}
\vspace{1em}
\begin{exercise}[Problem]{(5 points)}
Consider a signal $x[n]$ whose DTFT is sketched below. Sketch $X(-e^{j\omega})$.
\begin{center}
\begin{dspPlot}[xtype=freq,width=8cm,height=3cm,xticks=4]{-1,1}{0,1.1}
\dspFunc{x \dspQuad{0.25}{0.1} x \dspQuad{-0.25}{0.1} add }
\end{dspPlot}
\end{center}
\end{exercise}
\begin{exercise}[Problem]{(5 points)}
Consider an FIR filter with the frequency response plotted below (positive frequencies only):
\begin{center}
\begin{dspPlot}[xtype=freq,width=8cm,height=3cm]{0,1}{0,1.3}
\dspFunc{x \dspFIRI{0.3501 0.2823 0.1652 -0.0215 -0.0776 -0.0868 0.0374} abs }
\end{dspPlot}
\end{center}
Is the FIR optimal in the Parks-McClellan sense? If yes, then specify the length of the filter; if no, then explain why.
\end{exercise}
\begin{exercise}[Problem]{(10 points)}
Consider the following basis for $\mathbb{C}^4$:
\begin{align*}
\mathbf{x}_0 &= \begin{bmatrix} 1 & -1 & 0 & 0 \end{bmatrix}^T \\
\mathbf{x}_1 &= \begin{bmatrix} 0 & 1 & -1 & 0 \end{bmatrix}^T \\
\mathbf{x}_2 &= \begin{bmatrix} 0 & 0 & 1 & -1 \end{bmatrix}^T \\
\mathbf{x}_3 &= \begin{bmatrix} 1 & 1 & 1 & 1 \end{bmatrix}^T
\end{align*}
\begin{enumerate}
\item how would you quickly check in Matlab that the above four vectors do indeed form a basis?
\item is the basis orthogonal?
\item compute the inner products of the vector $\mathbf{y} = \begin{bmatrix} 1 & 1 & -1 & -1 \end{bmatrix}^T$ with each basis vector.
\end{enumerate}
\end{exercise}
\begin{exercise}[Problem]{(10 points)}
Consider the noncausal constant-coefficient difference equation
\[
y[n] = y[n+1] + by[n-1] + x[n]
\]
\begin{enumerate}
\item make it causal, i.e., write a CCDE in causal form that implements the same transfer function as described by the equation above
\item for which values of $b \in \mathbb{R}R$ is the causal system stable?
\end{enumerate}
\end{exercise}
\begin{exercise}[Problem]{(5 points)}
Assume $x[n] = \cos \omega_c n$ and consider the following system where $H(z)$ is the ideal Hilbert filter:
\begin{center}
\vspace{1em}
\begin{dspBlocks}{0.8}{0}
& & [name=A1] \BDfilter{$H(z)$} & \\
$x[n]$~~ & [name=A,mnode=dot,linewidth=2pt] & & [name=B,mnode=circle] + & $y[n]$ \\
& & & \\
\psset{arrows=->}
\ncangle[angleA=90,angleB=180,linewidth=\BDwidth]{A}{A1}
\ncangle[angleA=0,angleB=90,linewidth=\BDwidth]{A1}{B}^{~~~~~~~$j$}
\ncbar[angleA=-90,angleB=-90,linewidth=\BDwidth]{A}{B}
\ncline{2,1}{2,2}\ncline{2,4}{2,5}
\end{dspBlocks}
\end{center}
\begin{enumerate}
\item Write an expression for $y[n]$
\item Plot the values of $y[n]$ on the complex plane for $\omega_c = 2\pi/8$ and $n = 1, 2, \ldots, 10$
\end{enumerate}
\end{exercise}
\newpage
\begin{exercise}[Problem]{(15 points)}
Consider a i.i.d. discrete-time random signal whose samples are uniformly distributed over the $[-5, 5]$ interval. Consider a 2-bit quantizer with the following characteristic:
\[
\hat{x}[n] = \mathcal{Q}\{x[n]\} = \begin{cases}
+1 & \mbox{if} \quad 0 \leq x[n] \leq 2 \\
+3 & \mbox{if} \quad x[n] > 2\\
-1 & \mbox{if} \quad -2 \leq x[n] < 0 \\
-3 & \mbox{if} \quad x[n] < -2 \\
\end{cases}
\]
Compute the SNR at the output of the quantizer.
\end{exercise}
\begin{exercise}[Problem]{(20 points)}
You have been hired as the DJ in residence by the coolest club in town. For the upcoming Saturday night bash you're preparing a great mix of dance songs, all of which are at 128~BPM (beats per minute). There is one song that you absolutely want to include in the mix but unfortunately it is at 112~BPM. Since you're not just a DJ but also a cool digital signal processing expert, you're going to put your hard-earned DSP knowledge to use:
\begin{enumerate}
\item describe a full signal processing scheme that brings up the song to 128~BPM; you can assume the analog signals are bandlimited to 20KHz. Don't forget to exactly specify the parameters of the components (sampling rate, cutoff frequencies, etc.)
\item if the pitch (i.e. the fundamental frequency) of the song is 440Hz, what will be the pitch of the processed song?
\end{enumerate}
\end{exercise}
\begin{exercise}[Problem]{(20 points)}
For a causal system, stability requires that all the poles be inside the unit circle; stability does not impose any condition on a system's zeros, but the location of the zeros becomes important if we want to build the system's inverse. Consider the causal system described by:
\[
3y[n] = 3x[n] - 8x[n-1] + 4x[n-2] - y[n-2].
\]
\begin{enumerate}
\item show that the system is stable
\item verify that the inverse system $1/H(z)$ is \emph{not} stable.
\end{enumerate}
Systems whose zeros are inside the unit circle are called {\em minimum phase} and they always admit a stable inverse. If a system is not minimum phase, a common approach is the following: for every zero $z=a$ outside the unit circle, we cascade a section of the form
\[
\frac{a-z^{-1}}{a-z^{-1}}
\]
to the transfer function. Although the factors cancel each other out, by collecting the terms appropriately we can instead refactor $H(z)$ as
\[
H(z) = H_m(z)H_a(z)
\]
where $H_m(z)$ is minimum phase and $H_a(z)$ is all-pass. For the $H(z)$ in the first part of the exercise:
\begin{enumerate}
\item factor $H(z)$ as a minimum phase term times an allpass term
\item although $H(z)$ does not have a stable inverse, determine a stable $G(z)$ so that
\[
|H(z)G(z)| = 1
\]
\end{enumerate}
\end{exercise}
\begin{exercise}[Problem]{(10 points)}
Consider a bandlimited continuous-time signal $x(t)$ with a spectrum $X(j\Omega)$ as sketched in the following figure:
\begin{center}
\begin{dspPlot}[xtype=freq,xticks=custom,yticks=none,width=8cm,height=3cm]{-1,1}{-.5,1.4}
\dspFunc{x \dspQuad{0}{0.2} }
\dspCustomTicks[axis=x]{0 0 0.2 2KHz}
\end{dspPlot}
\end{center}
Sketch the DTFT of the output signal for each of the following systems, where the frequency of the raw sampler is indicated on each line:
\vspace{1em}
\renewcommand{\arraystretch}{5}
\begin{tabular}{ccc}
a) & \vphantom{20em} \begin{dspBlocks}{1}{0.4}
$x(t)$~
&
\raisebox{-1.4em}{\psframebox[linewidth=1.5pt]{%
\psset{xunit=1em,yunit=1em,linewidth=1.8pt}%
\pspicture(-3,-1.8)(2,1.8)%
\psline(-2.8,0)(-1.6,0)(1.2,1.4)
\psline(1.1,0)(1.8,0)
\psarc[linewidth=1pt]{<-}(-1.6,0){2em}{-10}{55}
\endpspicture}}
&
$x[n]$
\psset{linewidth=1.5pt}
\ncline{->}{1,1}{1,2}
\ncline{->}{1,2}{1,3}
\end{dspBlocks} & $T_s = 4$KHz \\
b) & \begin{dspBlocks}{1}{0.4}
$x(t)$~
&
\raisebox{-1.4em}{\psframebox[linewidth=1.5pt]{%
\psset{xunit=1em,yunit=1em,linewidth=1.8pt}%
\pspicture(-3,-1.8)(2,1.8)%
\psline(-2.8,0)(-1.6,0)(1.2,1.4)
\psline(1.1,0)(1.8,0)
\psarc[linewidth=1pt]{<-}(-1.6,0){2em}{-10}{55}
\endpspicture}}
&
$x[n]$
\psset{linewidth=1.5pt}
\ncline{->}{1,1}{1,2}
\ncline{->}{1,2}{1,3}
\end{dspBlocks} & $T_s = 2$KHz \\
c) & \begin{dspBlocks}{1}{0.4}
$x(t)$~
&
\raisebox{-1.4em}{\psframebox[linewidth=1.5pt]{%
\psset{xunit=1em,yunit=1em,linewidth=1.8pt}%
\pspicture(-3,-1.8)(2,1.8)%
\psline(-2.8,0)(-1.6,0)(1.2,1.4)
\psline(1.1,0)(1.8,0)
\psarc[linewidth=1pt]{<-}(-1.6,0){2em}{-10}{55}
\endpspicture}}
&
\BDdwsmp{2} & $x[n]$
\psset{linewidth=1.5pt}
\ncline{->}{1,1}{1,2}
\ncline{->}{1,2}{1,3}\ncline{->}{1,3}{1,4}
\end{dspBlocks} & $T_s = 4$KHz \\
d) & \begin{dspBlocks}{1}{0.4}
$x(t)$~
&
\raisebox{-1.4em}{\psframebox[linewidth=1.5pt]{%
\psset{xunit=1em,yunit=1em,linewidth=1.8pt}%
\pspicture(-3,-1.8)(2,1.8)%
\psline(-2.8,0)(-1.6,0)(1.2,1.4)
\psline(1.1,0)(1.8,0)
\psarc[linewidth=1pt]{<-}(-1.6,0){2em}{-10}{55}
\endpspicture}}
&
\BDfilter{LP$\{\pi/2\}$} & \BDdwsmp{2} & $x[n]$
\psset{linewidth=1.5pt}
\ncline{->}{1,1}{1,2}
\ncline{->}{1,2}{1,3}\ncline{->}{1,3}{1,4} \ncline{->}{1,4}{1,5}
\end{dspBlocks} & $T_s = 4$KHz \\
\end{tabular}
\vspace{3em}
\end{exercise}
\end{document}

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