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\documentclass[a4paper,13pt]{article}
\usepackage{defsDSPcourse}
\usepackage{dspTricks}
\usepackage{dspFunctions}
\usepackage{dspBlocks}
\begin{document}
\begin{center}%
\sffamily
{\LARGE \bfseries COM-303 - Signal Processing for Communications \\ Final Exam} \\
\vspace{1em}
{\large Wednesday, July 3 2013, 08:15 to 11:15}
\vspace{1em}
\end{center}
\begin{exercise}{}
\begin{enumerate}
\item By simple inspection we can write:
\[
x_1[n] = 2x_2[n+3] - x_2[n+2] + x_2[n+1];
\]
We know that the system is time invariant, therefore $\mathcal{H}\{x_2[n+N]\} = y_2[n+N]$. If the system were linear we would therefore have:
\[
\mathcal{H}\{x_1[n]\} = \hat{y}[n] = 2y_2[n+3] - y_2[n+2] + y_2[n+1]
\]
However it is easy to see that $\hat{y}[n] \neq y_1[n]$ (in fact, $\hat{y}[n] = \ldots, 0, 0, \underline{4}, 0, 1, 1, 0, 0, \ldots$). As a consequence, $\mathcal{H}$ is not linear.
\item $\delta[n] = x_2[n+3]$ so that, because of time invariance, $\mathcal{H}\{\delta[n]\} = y_2[n+3] = \ldots, 0, 0, \underline{2}, 1, 0, 0, \ldots$
\end{enumerate}
\end{exercise}
\begin{exercise}{}
The system can be simplified as
\begin{center}
\begin{dspBlocks}{1.2}{0.4}
$x[n]$ & \BDadd & \BDfilter{$D(z)$} & & \BDsplit & $y[n]$ \\%
& & \BDdelay & & \\%
\psset{arrows=->,linewidth=1.5pt}
\ncline{1,1}{1,2} \ncline{-}{1,2}{1,3} \ncline{1,3}{1,6}
\ncline{-}{1,5}{2,5}\ncline{2,5}{2,3}\taput{$\alpha$}
\ncline{-}{2,3}{2,2}\ncline{2,2}{1,2}
\end{dspBlocks}
\end{center}
From this we can write the input/output relation in the $z$ domain
\[
Y(z) = D(z)[X(z) + \alpha z^{-1} Y(z)]
\]
from which we obtain the transfer function
\[
H(z) = \frac{D(z)}{1-\alpha z^{-1}D(z)}
\]
If we write $D(z)$ as a ratio of polynomials, i.e. $D(z) = B(z)/A(z)$, we finally obtain
\[
H(z) = \frac{B(z)}{B(z)-\alpha z^{-1}A(z)}
\]
From the figure, it is immediate to see that $D(z)$ is an (incomplete) second order section in direct form II, incomplete since it has a single zero. Its transfer function is therefore
\begin{align*}
D(z) &= \frac{1 - (5/4)z^{-1}}{1 -2z^{-1} + 2z^{-2}} \\
&= \frac{1 - (5/4)z^{-1}}{(1 -(1+j)z^{-1})(1 -(1-j)z^{-1})}
\end{align*}
\begin{enumerate}
\item By letting $B(z) = 1 - (5/4)z^{-1}$ and $A(z) = 1 -2z^{-1} + 2z^{-2}$, the transfer function with $\alpha=-2$ becomes
\[
H(z) = \frac{1 - (5/4)z^{-1}}{1 -2z^{-1} + 2z^{-2} +2z^{-1}(1 - (5/4)z^{-1})} = \frac{1 - (5/4)z^{-1}}{1 -(1/2)z^{-2}}
\]
\item There is a zero in $z = 5/4$ and two poles in $z = \pm \sqrt{1/2}$
\begin{center}
\begin{dspPZPlot}[width=5cm,clabel={$1$}]{1.8}
\dspPZ[type=zero,label=none]{1.25,0}
\dspPZ[label=none]{-.7,0}
\dspPZ[label=none]{0.7,0}
\end{dspPZPlot}
\end{center}
\item since pole and zero on the positive real axis are almost equidistant from one, their effects cancel each other out; the pole in $z = -\sqrt{1/2}$ brings the magnitude of the frequency response up to create a highpass characteristic:
\begin{center}
\begin{dspPlot}[xtype=freq,xticks=4,yticks=none]{-1,1}{0,25}
\dspFunc{x \dspTFM{1 -1.25}{1 0 -0.5}}
\end{dspPlot}
\end{center}
\item The inverse transfer function is not stable because the zero of $H(z)$ is outside the unit circle. By choosing
\[
G(z) = \frac{1 -(1/2)z^{-2}}{(5/4) - z^{-1}}
\]
the product G(z)H(z) is the allpass term $(1 - (5/4)z^{-1})/((5/4) - z^{-1})$ whose frequency response magnitude is one.
\item If we remove the feedback branch, the transfer function becomes $H(z) = D(z)$. The poles of $D(z)$ are larger than one in magnitude ($|z_{1,2}| = |1 \pm j| = \sqrt{2}$) and so the system would not be stable.
\end{enumerate}
\end{exercise}
\begin{exercise}{}
\begin{enumerate}
\item The easiest way to prove the result is to invoke the convolution theorem. In the $z$-transform domain, the filter's output can be expressed as $Y(z) = H(z)X(z)$. If $x[n]$ is finite-support and $h[n]$ is FIR, then both $X(z)$ and $H(z)$ are finite-degree polynomials in $z$ and so their product is also finite-degree.
Alternatively, you can consider the convolution sum for a FIR with support over $[M_1,M_2]$:
\[
y[n] = \sum_{k = -\infty}^{\infty}h[k]x[n-k] = \sum_{k = M_1}^{M_2}h[k]x[n-k]
\]
If $x[n]$ is zero outside of $[N_1,N_2]$, then all the terms in the sum will be zero for $n> N_2+M_2$ and for $n<N_1+M_1$.
\item The simplest example is to have an input that exactly cancels the filter's poles. As an IIR filter, take a simple leaky integrator with transfer function
\[
H(z) = \frac{1}{1-\lambda z^{-1}}
\]
and consider the finite-support input $x[n] = \ldots, 0, 0, \underline{1}, -\lambda, 0, 0, \ldots$. We have that $Y(z) = H(z)X(z) = 1$ so that $y[n] = \delta[n]$, which is finite-support.
\end{enumerate}
\end{exercise}
\begin{exercise}{}
\begin{enumerate}
\item The power of the good signal is simply
\[
\int_{-\pi}^{\pi}P_x(e^{j\omega})d\omega = 2\pi\sigma^2_x
\]
The power of the noise is
\[
\int_{-\pi}^{\pi}P_\eta(e^{j\omega})d\omega = 2\pi\sigma^2_0
\]
so that the SNR is simply $\sigma^2_x/\sigma^2_0$
\item Just as in oversampling, the idea is to send the signal more slowly as to occupy less bandwidth. If the signal has a smaller bandwidth, we can increase its amplitude without exceeding the power constraint, which will allow us to have a better SNR over the band of interest.
Consider the following preprocessing chain:
\begin{center}
\begin{dspBlocks}{2}{0.5}
$x[n]~$ & \BDupsmp{3} & \BDlowpass[0.6em] & $~x'[n]$
\psset{arrows=->}
\ncline{1,1}{1,2}\ncline{1,2}{1,3}\taput{$\sqrt{3}$~~~~~}\ncline{1,3}{1,4}\ncline{1,4}{1,5}
\end{dspBlocks}
\end{center}
where the lowpass filter has a cutoff frequency $\pi/3$. The power of $x'[n]$ is
\[
\int_{-\pi}^{\pi}P_{x'}(e^{j\omega})d\omega =
\int_{-\pi/3}^{\pi/3}3\,\sigma^2_x = 2\pi\sigma^2_x
\]
so that the power constraint is fulfilled.
The signal and the noise at the receiver after the sampler have the following psd's:
\begin{center}
\begin{dspPlot}[xtype=freq,xticks=3,yticks=custom]{0,1}{0,10}
\dspFunc{x \dspRect{0}{0.66666} 9 mul}
\dspFunc[linecolor=gray]{1}
\dspCustomTicks[axis=y]{1 $\sigma_0^2$ 3 $\sigma_x^2$ 6 $2\sigma_x^2$ 9 $3\sigma_x^2$}
\end{dspPlot}
\end{center}
At the receiver we can filter out the out-of-band noise with the following scheme:
\begin{center}
\begin{dspBlocks}{2}{0.5}
$\hat{x'}[n]~$ & \BDlowpass[0.6em] & \BDdwsmp{3} & $~\hat{x}[n]$
\psset{arrows=->}
\ncline{1,1}{1,2}\ncline{1,2}{1,3}\ncline{1,3}{1,4}\ncline{1,4}{1,5}
\end{dspBlocks}
\end{center}
where, once again, the lowpass has cutoff frequency $\pi/3$. After the filter the psd is
\begin{center}
\begin{dspPlot}[xtype=freq,xticks=3,yticks=custom]{0,1}{0,10}
\dspFunc{x \dspRect{0}{0.66666} 9 mul}
\dspFunc[linecolor=gray]{x \dspRect{0}{0.66666}}
\dspCustomTicks[axis=y]{1 $\sigma_0^2$ 3 $\sigma_x^2$ 6 $2\sigma_x^2$ 9 $3\sigma_x^2$}
\end{dspPlot}
\end{center}
the downsampler stretches the signal to full band again but reduces the spectral ``amplitude'' by a factor of 3, which means the psd gets divided by 9:
\begin{center}
\begin{dspPlot}[xtype=freq,xticks=3,yticks=custom]{0,1}{0,10}
\dspFunc{9}
\dspFunc[linecolor=gray]{1}
\dspCustomTicks[axis=y]{1 $\sigma_0^2/9$ 9 $\sigma_x^2/3$}
\end{dspPlot}
\end{center}
and the signal-to-noise ratio becomes
\[
\mbox{SNR}_{2} = 2\pi(\sigma_x^2/3)/2\pi(\sigma_0^2/9) = 3\mbox{SNR}_{1}
\]
\item 9 minutes.
\end{enumerate}
\end{exercise}
\begin{exercise}{}
Whenever we have a sum of the form $\sum_{n=-\infty}^{\infty} x[n]$ it is always useful to see if we can compute the DTFT of $x[n]$ since the sum is equal to the value of the DTFT in $\omega = 0$
\[
\sum_{n=-\infty}^{\infty} x[n] = X(e^{j\omega})|_{\omega=0}.
\]
In this case we have
\begin{align*}
x[n] &= \frac{\sin(2\pi n/3)}{4\pi n}\cos(\pi n/3) \\
&= \frac{1}{4}\left[\frac{2}{3}\sinc\left(\frac{2}{3}n\right)\right]\cos\left(\frac{\pi}{3}n\right)
\end{align*}
By using the modulation theorem, the DTFT of $x[n]$ is
\[
X(e^{j\omega}) = \frac{1}{2}\left[\frac{1}{4}\rect\left(\frac{\omega - (\pi/3)}{2\pi/3}\right) + \frac{1}{4}\rect\left(\frac{\omega +(\pi/3)}{2\pi/3}\right)\right]
\]
Graphically:
\begin{center}
\begin{dspPlot}[xtype=freq,xticks=3,yticks=0.25]{-1,1}{0,0.5}
\dspFunc{x \dspRect{0.333333}{1.333333} 0.25 mul 0.5 mul
x \dspRect{-0.333333}{1.333333} 0.25 mul 0.5 mul
add}
\end{dspPlot}
\end{center}
so that $X(e^{j\omega})|_{\omega=0} = 1/4$
\end{exercise}
\begin{exercise}{}
\begin{enumerate}
\item NO. The zeros are not in complex-conjugate pairs, so the filter's impulse response cannot be real-valued.
\item YES. The filter has three zeros therefore its impulse response has length 4.
\item NO. The zero $z_0$ on the positive axis does not have a reciprocal zero in $1/z_0$ so the filter cannot be linear-phase.
\item NO. The filter has a pole, so it cannot be FIR linear phase.
\end{enumerate}
\end{exercise}
\end{document}
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