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finalSol.tex
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\documentclass[a4paper,11pt]{article}
\usepackage{defsDSPcourse}
\usepackage{dspTricks,dspBlocks,dspFunctions}
\begin{document}
\begin{center}%
\sffamily
{\LARGE \bfseries COM-303 - Signal Processing for Communications \\ Final Exam - Solution} \\
\vspace{1em}
{\large Tuesday, July 3 2012, 08:15 to 11:15}
\vspace{1em}
\end{center}
\begin{exercise}[Problem]{(5 points)}
\[
X(-e^{j\omega}) = X(e^{-j\pi}e^{j\omega}) = X(e^{j(\omega - \pi)}):
\]
\begin{center}
\begin{dspPlot}[xtype=freq,width=8cm,height=3cm,xticks=4]{-1,1}{0,1.1}
\dspFunc{x \dspQuad{0.75}{0.1} x \dspQuad{-0.75}{0.1} add }
\end{dspPlot}
\end{center}
\end{exercise}
\begin{exercise}[Problem]{(5 points)}
The filter is clearly not equiripple, therefore it's not optimal in the Parks-McClellan sense.
\end{exercise}
\begin{exercise}[Problem]{(10 points)}
\begin{enumerate}
\item simply check that {\tt det([x0'; x1'; x2'; x3'])} is nonzero
\item no, $\langle \mathbf{x}_0, \mathbf{x}_1 \rangle = -1 \neq 0$
\item $\langle \mathbf{y}, \mathbf{x}_1 \rangle = 2$, the other products are zero
\end{enumerate}
\end{exercise}
\begin{exercise}[Problem]{(10 points)}
\begin{enumerate}
\item by delaying all terms:
\[
y[n-1] = y[n] + by[n-2] + x[n-1]
\]
from which
\[
H(z) = \frac{-z^{-1}}{1 - z^{-1} + bz^{-2}}
\]
\item the poles of the transfer function are
\[
p_{1,2} = \frac{1}{2} \pm \sqrt{\frac{1}{4} - b}
\]
and we want $|p_{1,2}| < 1$.
For $b < 1/4$ the square root is real and therefore:
\[
|p_1| < 1 \mbox{ for } \sqrt{1/4 - b} < 1/2 \Rightarrow 0 < b < 1/4\\
\]
\[
|p_2| < 1 \mbox{ for } \sqrt{1/4 - b} < -3/2 \Rightarrow -2 < b < 1/4
\]
For $b > 1/4$ the square root is imaginary and $|p_{1,2}|^2 = b$ so that:
\[
|p_{1,2}| < 1 \mbox{ for } 1/4 < b < 1\\
\]
so that in the end the system is stable for $0 < b < 1$
\end{enumerate}
\end{exercise}
\begin{exercise}[Problem]{(5 points)}
\begin{enumerate}
\item The Hilbert filter rotates a sinusoid by $\pi/2$ so that $\cos(\omega n) \ast h[n] = \sin \omega n$. Therefore
\[
y[n] = \cos \omega_c n + j \sin \omega_c n = e^{j\omega_c n}
\]
\item
\begin{center}
\begin{dspPZPlot}[width=6cm]{1.5}
\multido{\n=1+1}{10}{%
\FPupn\X{\n{} 0.7854 * cos}\FPupn\Y{\n{} 0.7854 * sin}%
\dspPZ[type=zero,label={$x[\n]$}]{\X,\Y}}
\end{dspPZPlot}
\end{center}
\end{enumerate}
\end{exercise}
\newpage
\begin{exercise}[Problem]{(15 points)}
The power of the signal is simply the variance of the input distribution:
\[
P_x = \frac{10^2}{12}
\]
The power of the quantization error is
\[
P_e = \mbox{E}[(x-\hat{x})^2] = \int_{-5}^{5}p(x)(x - \mathcal{Q}\{x\})^2 dx
\]
Exploiting symmetries
\[
P_e = \frac{1}{5}\int_{0}^{2}(x-1)^2 dx + \frac{1}{5}\int_{2}^{3}(x-3)^2 dx = \frac{11}{15}
\]
so that
\[
\mbox{SNR} = \frac{125}{11}
\]
\end{exercise}
\begin{exercise}[Problem]{(20 points)}
\begin{enumerate}
\item The song, converted to digital, must be ``accelerated'' by a factor of $8/7$, i.e., upsampled by 7 and downsampled by 8. The following scheme can be used:
\begin{dspBlocks}{1}{0.4}
$x(t)$ & \BDsampler & \BDupsmp{7} & \BDfilter{LP$\{\pi/8\}$} & \BDdwsmp{8} & \BDfilter{Int} & $y(t)$
\BDConnHNext{1}{1}\BDConnHNext{1}{2}\BDConnHNext{1}{3}\BDConnHNext{1}{4}\BDConnHNext{1}{5}\BDConnHNext{1}{6}
\end{dspBlocks}
where sampling and interpolator work at least at $F_s = 40$KHz
\item $440\cdot (8/7) \approx 503$Hz, almost a full tone up.
\end{enumerate}
\end{exercise}
\begin{exercise}[Problem]{(20 points)}
\begin{enumerate}
\item
\[
H(z) = \frac{3 - 8z^{-1} + 4z^{-2}}{3 + z^{-2}} = \frac{(1 - (2/3)z^{-1})(1 - 2z^{-1})}{(1-(j/\sqrt{3})z^{-1})(1+(j/\sqrt{3})z^{-1})}
\]
poles are inside the unit circle but one zero is not.
\item Since in the inverse systems zeros become poles, the pole in $z = 2$ makes the system unstable.
\item
\[
H(z) = \frac{(1 - (2/3)z^{-1})(1 - 2z^{-1})}{3 + z^{-2}}\frac{2-z^{-1}}{2-z^{-1}} =
\frac{(1 - (2/3)z^{-1})(2-z^{-1})}{3 + z^{-2}}\,\frac{1 - 2z^{-1}}{2-z^{-1}}
\]
where $(1 - 2z^{-1})/(2-z^{-1})$ is a classic allpass term.
\item
\[
G(z) = 1/H_m(z) = \frac{3 + z^{-2}}{(1 - (2/3)z^{-1})(2-z^{-1})}
\]
\end{enumerate}
\end{exercise}
\begin{exercise}[Problem]{(10 points)}
\def\oneShotSpec{\dspQuad{0}{\specHW}}
\def\aliasedSpec{%
x -6 sub \oneShotSpec %
x -4 sub \oneShotSpec %
x -2 sub \oneShotSpec %
x \oneShotSpec %
x 2 sub \oneShotSpec %
x 4 sub \oneShotSpec %
x 6 sub \oneShotSpec %
add add add add add add}
\begin{enumerate}
\item
\begin{center}
\begin{dspPlot}[xtype=freq,yticks=none,width=8cm,height=3cm]{-1,1}{0,2}
\def\specHW{1 }
\dspFunc{x \oneShotSpec}
\end{dspPlot}
\end{center}
\item
\begin{center}
\begin{dspPlot}[xtype=freq,yticks=none,width=8cm,height=3cm]{-1,1}{0,2}
\def\specHW{2 }
\dspFunc{\aliasedSpec}
\end{dspPlot}
\end{center}
\item
\begin{center}
\begin{dspPlot}[xtype=freq,yticks=none,width=8cm,height=3cm]{-1,1}{0,2}
\def\specHW{2 }
\dspFunc{\aliasedSpec}
\end{dspPlot}
\end{center}
\item
\begin{center}
\begin{dspPlot}[xtype=freq,yticks=none,width=8cm,height=3cm]{-1,1}{0,2}
\def\specHW{2 }
\dspFunc{x \oneShotSpec}
\end{dspPlot}
\end{center}
\end{enumerate}
\end{exercise}
\end{document}

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