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\documentclass[a4paper,fleqn,11pt]{article}
\usepackage[utf8]{inputenc}
\usepackage[parfill]{parskip}
\usepackage{cancel}
\usepackage{defsDSPcourse}
\usepackage{dspTricks}
\usepackage{dspFunctions}
\usepackage{dspBlocks}
\usepackage{comment}
\usepackage{pst-plot}
\usepackage{enumitem}
\usepackage{wrapfig}
\usepackage{mathtools}
\makeatletter
\def\dspLegendStyle(#1,#2)#3{%
\rput[lt](#1,#2){{%
\begin{tabular}{ll}
\@dsplegendStyle#3 \@empty
\end{tabular}}}}
\def\@dsplegendStyle#1 #2 #3{%
{\psline[#1](0,0.2)(2,0.2)\rule{2.5em}{0pt}} & #2\\ \space % fbox here to have a visual test
\ifx #3\@empty\else
\expandafter\@dsplegendStyle
\fi
#3}
\makeatother
\newif\ifanswers
\answerstrue
\begin{document}
\newcounter{tmpc}
\framebox[12cm][l]{\rule{0cm}{1cm}\LARGE }% STUDENT-NAME}
\hfill
\framebox[3cm][l]{\rule{0cm}{1cm}\LARGE } %SCIPER}
\vspace{5em}
\begin{center}%
\sffamily
{\LARGE \bfseries COM-303 - Signal Processing for Communications \\ Final Exam} \\
\vspace{1em}
{\large Tuesday 05.07.2022 from 09h15 to 12h15 (CE4, CE6)} \\
\end{center}
\centerline{\rule{\textwidth}{.5pt}}
\vspace{1em}
{
\LARGE \bfseries
\begin{center}
Verify that this exam has YOUR last name on top \\
\vspace{1em}
DO NOT OPEN THE EXAM UNTIL INSTRUCTED TO DO SO
\end{center}
}
\vspace{1em}
\centerline{\rule{\textwidth}{.5pt}}
\vspace{1em}
\begin{itemize}
\item {\bf Write your name} on the top left corner of {\bf ALL the sheets you turn in}.
\item There are FOUR problems for a total of 90 points; the number of points is indicated for each problem and, in the case of multipart exercises, also for each sub-question. Read \textbf{all} the problem statements very carefully before you begin solving the exam!
\item Please \textbf{write your derivations clearly!}
\item You can have two A4 sheets of \emph{handwritten} notes (front and back). Please \textbf{no photocopies, no books and no electronic devices}. Turn off your phone and store it in your bag.
\vspace{1em}
\item When you are done, \textbf{simply leave your solution on your desk \textit{with this page on top} and exit the classroom}. Do NOT bring the exam to the main desk.
\end{itemize}
\vspace{1em}
\centerline{\rule{\textwidth}{.5pt}}
\vspace{1em}
\newpage
%\onebyone
\begin{tabular}{c|c|c}
\hspace{-3.5em}
\begin{dspPlot}[height=2.5cm,width=5cm,xout=true,xticks=5]{-11,11}{-1.2,1.2}
\dspSignal{x 0 eq {1} {0} ifelse}
\end{dspPlot}
&
\hspace{-1.5em}
\begin{dspPZPlot}[width=3cm,clabel=~]{1.5}
\dspPZ[label=none]{0.8,0}
\dspPZ[label=none]{0.856, 0.2781}
\dspPZ[label=none]{0.856, -0.2781}
\dspPZ[type=zero,label={$3$}]{-1,0}
\end{dspPZPlot}
&
\begin{dspPlot}[height=2.5cm,width=5cm,xout=true,xticks=5]{-11,11}{-1.2,1.2}
\dspSignalOpt{\dspSetFilter{0.2 0.2 0.2 0.2 0.2}{}}{x 0 eq {1} {0} ifelse \dspFilter}
\end{dspPlot}
\\[1em]
(X1) & (F1) & (Y1)
\\[3em]
%
\hspace{-3.5em}
\begin{dspPlot}[height=2.5cm,width=5cm,xout=true,xticks=5]{-11,11}{-1.2, 1.2}
\dspSignal{x 0 ge {1} {0} ifelse}
\end{dspPlot}
&
\hspace{-1.5em}
\begin{dspPZPlot}[width=3cm,clabel=~]{1.5}
\dspPZ[label=none]{0.856, 0.2781}
\dspPZ[label=none]{0.856, -0.2781}
\dspPZ[type=zero,label={$3$}]{-1,0}
\end{dspPZPlot}
&
\begin{dspPlot}[height=2.5cm,width=5cm,xout=true,xticks=5]{-11,11}{-1.2,1.2}
\dspSignalOpt{\dspSetFilter{0.2}{-0.8}}{x 0 ge {1} {0} ifelse \dspFilter}
\end{dspPlot}
\\[1em]
(X2) & (F2) & (Y2)
\\[3em]
%
\hspace{-3.5em}
\begin{dspPlot}[height=2.5cm,width=5cm,xout=true,xticks=5]{-11,11}{-1.2,1.2}
\dspSignal{x 180 mul cos 0.75 mul 0.25 add}
\end{dspPlot}
&
\hspace{-1.5em}
\begin{dspPZPlot}[width=3cm,clabel=~]{1.5}
\dspPZ[label=none]{0.8,0}
\dspPZ[label=none]{0.856, 0.2781}
\dspPZ[label=none]{0.856, -0.2781}
\dspPZ[type=zero,label={$3$}]{-1,0}
\end{dspPZPlot}
&
\begin{dspPlot}[height=2.5cm,width=5cm,xout=true,xticks=5]{-11,11}{-1.2,1.2}
%\dspSignalOpt{\dspSetFilter{1 1}{}}{x 180 mul cos 0.75 mul 0.25 add \dspFilter}
% avoid border effect
\dspSignal{0.5}
\end{dspPlot}
\\[1em]
(X3) & (F3) & (Y3)
\\[3em]
%
\hspace{-3.5em}
\begin{dspPlot}[height=2.5cm,width=5cm,xout=true,xticks=5]{-11,11}{-1.2,1.2}
\dspSignal{x 360 10 div mul cos}
\end{dspPlot}
&
\hspace{-1.5em}
\begin{dspPZPlot}[width=3cm,clabel=~]{1.5}
% b = [1 3 3 1], a = [1.0000 -2.5119 2.1795 -0.6480]
% poles at 0.8 and 0.9exp(\pm i \pi/10)
\dspPZ[label=none]{0.8,0}
\dspPZ[label=none]{0.856, 0.2781}
\dspPZ[label=none]{0.856, -0.2781}
\dspPZ[type=zero,label={$3$}]{-1,0}
\end{dspPZPlot}
&
%\hspace{-1em}
\begin{dspPlot}[height=2.5cm,width=5cm,xout=true,xticks=5]{-11,11}{-1.2,1.2}
\dspSignalOpt{\dspSetFilter{0.1}{0.9}}{x 360 10 div mul cos \dspFilter}
\end{dspPlot}
\\[1em]
(X4) & (F4) & (Y4)
\\[3em]
\end{tabular}
\end{document}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{exercise}{(10 points)}
Compute and sketch the DTFT of the following signal
\[
x[n] = \delta[n] + u[n+1] - u[n-2] - \sinc(n/3).
\]
\ifanswers{\em\vspace{3em}\par{\bfseries Solution: }
First of all, note that the signal is real and symmetric and therefore its DTFT will also be real and symmetric. Now call
\[
a[n] = \delta[n] + u[n+1] - u[n-2] = \begin{cases} 2 & n = 0 \\ 1 & |n| = 1 \\ 0 & \mbox{otherwise} \end{cases}
\]
and therefore
\[
A(e^{j\omega}) = \sum_{n=-1}^{1} a[n]e^{-j\omega n} = e^{j\omega} + 2 + e^{-j\omega} = 2 + 2\cos(\omega).
\]
The DTFT of $\sinc(n/3)$ is a rect of height three and support $[-\pi/3, \pi/3]$ and so
\[
X(e^{j\omega}) = 2 + 2\cos(\omega) - 3\rect(3\omega/2\pi).
\]
\begin{center}
\begin{dspPlot}[xtype=freq,xticks=3,yticks=1]{-1,1}{0, 3.5}
\dspFunc{x 180 mul cos 2 mul 2 add x \dspRect{0}{0.666666} 3 mul sub}
\end{dspPlot}
\end{center}
}\else{\vspace{1em}}\fi
\end{exercise}
\ifanswers{}\else{\vspace{10em}}\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{exercise}{(10 points)}
In digital communication systems it is often necessary to precisely establish a \textit{timing reference} between transmitter and receiver. To this end, the transmitter will send a signal that contains a detectable feature at time $t = t_0$; the receiver will receive the signal after a propagation delay $t_d$ and detect the feature at time $t_1 = t_0 + t_d$. From this point on, with the transmitter using $t_0$ as the reference and the receiver using $t_1$, transmitter and receiver will be synchronized.
A common method to establish a timing reference is to send a sinusoidal carrier at frequency $f_c$ and to flip its sign at $t=t_0$, as illustrated in the following figure:
\begin{center}
\begin{dspPlot}[height=2cm,width=10cm,xout=true,xticks=custom,yticks=none,sidegap=0]{0,8}{-1.2, 1.2}
\dspFunc{x 360 mul x 2 le {0} {180} ifelse add cos}
\dspCustomTicks[axis=x]{2 $t_0$}
\dspText(-0.5, 0){TX}
\end{dspPlot}
\vspace{-1.2em}
\begin{dspPlot}[height=2cm,width=10cm,xout=true,xticks=custom,yticks=none,sidegap=0]{0.3,8.3}{-1.6, 1.6}
\dspFunc[linecolor=gray]{x 360 mul x 5 le {0} {180} ifelse add cos}
\dspCustomTicks[axis=x]{5 $t_1=t_0+t_d$}
\dspText(-0.2, 0){RX}
\end{dspPlot}
\end{center}
% \[
% s(t) = \begin{cases}
% \cos(2\pi f_c t) & t < t_0 \\
% \cos(2\pi f_c t + \pi) = -\cos(2\pi f_c t) & t \ge t_0;
% \end{cases}
% \]
by detecting the instant $t_1$ in which the phase jump occurs, the receiver achieves synchronization.
In the following exercise consider a simplified all-digital setup where the reference time at the transmitter is $n_0=0$; the (infinite-length) timing signal at frequency $\omega_c > 0$ can be expressed as
\[
s[n] = p[n]\cos(\omega_c n), \quad n \in \mathbb{Z}
\]
where
\[
p[n] = \begin{cases}
+1 & n < 0 \\
-1 & n \ge 0
\end{cases}
\]
Assume that the propagation delay is equal to an integer number of samples so that the received digital signal is simply
\[
x[n] = s[n - n_d], \qquad n_d \in \mathbb{N}^+.
\]
The signal $x[n]$ is processed at the receiver by the following system:
\vspace{1ex}
\begin{center}
\begin{dspBlocks}{1.3}{0.5}
& & \BDfilter{$H(z)$} & & & \\
$x[n]$~~ & \BDsplit & & \BDadd & \BDfilter{$D(z)$} & \BDfilter{$|\cdot|$} & ~~~$y[n]$ \\
\ncline{->}{2,1}{2,4} \ncline{->}{2,4}{2,5}^{$a[n]$~~~~} \ncline{->}{2,5}{2,6}^{$b[n]$} \ncline{->}{2,6}{2,7}
\ncline{2,2}{1,2} \ncline{->}{1,2}{1,3}
\ncline{1,3}{1,4}^{~~$j$} \ncline{->}{1,4}{2,4}
\end{dspBlocks}
\end{center}
where $H(z)$ is an ideal Hilbert filter, $D(z)$ is a length-$N$ FIR with complex-valued impulse response
\[
d[n] = \begin{cases}
e^{j\omega_c n} & 0 \le n < N \\
0 & \mbox{otherwise}
\end{cases}
\]
and the last block computes the magnitude of the filter's output, that is, $y[n] = |b[n]|$.
\begin{enumerate}
\item Assume $n_d = 20$; plot $y[n]$ over the interval $0 \le n \le 50$ for $N = 10$ and $N = 30$.
\item Describe a method to determine the value $n_d$ from $y[n]$.
\item What is the minimum value of $N$ that that you can use in theory?
\item Why would you use a larger value than the theoretical minumum for $N$?
\end{enumerate}
Consider now a situation where the channel introduces an unwanted DC offset and so the received signal is now
\[
x[n] = s[n - n_d] + \epsilon,
\]
where $\epsilon$ is a positive constant. Assume the frequency of the timing signal is $\omega_c = 2\pi(A/B)$, with $A, B$ positive integers and $A<B$.
\begin{enumerate}
\setcounter{enumi}{4}
\item Determine a value for $N$ that minimizes the effects of the DC offset on the detection process for $n_d$.
\end{enumerate}
{\footnotesize Hints: let $\mathcal{H}\{\cdot\}$ denote the result of applying a Hilbert filter to a signal; then you can assume the following:
\begin{itemize}
\item for a cosine-modulated signal: $\mathcal{H}\{w[n]\cos(\omega_0 n)\} = w[n]\sin(\omega_0 n)$
\item for a constant signal: $\mathcal{H}\{c\} = 0$
\end{itemize}}
\ifanswers{\em\vspace{3em}\par{\bfseries Solution: }
With a propagation delay of zero the received signal would be $x[n] = s[n]$ and so
\begin{align*}
a[n] &= x[n] + j(x \ast h)[n] \\
&= p[n]\cos(\omega_c n) + j p[n]\sin(\omega_c n) \\
& = p[n]e^{j\omega_c n}.
\end{align*}
Because of time invariance, with a nonzero propagation delay $n_d$, it is
\[
a[n] = p[n-n_d] e^{j\omega_c (n-n_d)}.
\]
The output of the FIR filter is
\begin{align*}
b[n] &= (a \ast d)[n] = \sum_{k=0}^{N-1} d[k]a[n-k] \\
&= \sum_{k=0}^{N-1} e^{j\omega_c k} e^{j\omega_c (n - n_d - k)} p[n - n_d - k] \\
&= e^{j\omega_c (n - n_d)}\sum_{k=0}^{N-1} p[n - n_d - k]
\end{align*}
and so
\[
y[n] = |b[n]| = \left| \sum_{k=0}^{N-1} p[(n - n_d) - k]\right|.
\]
We can now distinguish three cases:
\begin{itemize}
\item when $n - n_d < 0$, the argument to $p[\cdot]$, i.e. $(n - n_d) - k$, will be negative for all values of $k$ in the sum and so
\[
y[n] = \left| \sum_{k=0}^{N-1} 1\right| = N \quad \mbox{for $n < n_d$}
\]
\item similarly, when $n - n_d \ge N - 1$, the argument to $p[\cdot]$ will be non-negative for all $k$ and so
\[
y[n] = \left| \sum_{k=0}^{N-1} -1\right| = N \quad \mbox{for $n \ge n_d + N - 1$}
\]
\item for the other cases, i.e. for $n_d \le n < n_d + N - 1$ it is
\[
y[n] = \left| \sum_{k=0}^{n - n_d} 1 + \sum_{m=n - n_d + 1}^{N - 1} -1 \right| = |N - 2(n - n_d + 1)|.
\]
\end{itemize}
\begin{enumerate}
\item For a given value of $n_d$, the output $y[n]$ will be equal to $N$ for $n < n_d$ and for $n > n_d + N - 1$ while, in between these values, it will decrease linearly to a minumum value of zero for $n = n_d + N/2 - 1$ and then climb up back to $N$:
\def\yfun#1#2{x #1 1 sub le {#2 } {x #1 #2 1 sub add ge {#2 } {#2 x #1 sub 1 add 2 mul sub abs} ifelse} ifelse}
\begin{center}
\begin{dspPlot}[height=3cm,width=10cm,xout=true,xticks=10]{0,60}{0,12}
\dspFunc{\yfun{20}{10}}
\dspTaps{19 10 20 8}
\dspCustomTicks[axis=x]{24 24}
\dspText(5, 6){$N=10$}
\end{dspPlot}
\begin{dspPlot}[height=3cm,width=10cm,xout=true,xticks=10]{0,60}{0,32}
\dspFunc{\yfun{20}{30}}
\dspText(5, 15){$N=30$}
\dspTaps{19 30 20 28}
\dspCustomTicks[axis=x]{34 34}
\end{dspPlot}
\end{center}
\item In the ideal case, $y[n]$ starts to drop from its initial value of $N$ exactly for $n = n_d$, that is, $y[n] = N$ for $n < n_d$ and $y[n_d] = N-2$. So a simple detection scheme would place a threshold on $y[n]$ and look for its first ``dip'' from $N$ to $N-2$.
\item In the ideal case, for the method just described, we need a minimum FIR length $N=2$.
\item In non-ideal cases, the presence of noise would make $y[n]$ noisy as well. By using a filter of length $N$ the difference between the maximum and minimum values of $y[n]$ is equal to $N$ and so, if we use a long filter, we increase the noise margin in our estimation process.
\item If $x[n] = s[n - n_d] + \epsilon$ then, by linearity,
\[
a[n] = p[n-n_d] e^{j\omega_c (n-n_d)} + \epsilon
\]
and
\[
b[n] = e^{j\omega_c (n - n_d)}\sum_{k=0}^{N-1} p[n - n_d - k] + \epsilon\sum_{k=0}^{N-1}e^{j\omega_c k}
\]
Since $\omega_c = 2\pi(A/B)$, the effect of the DC offset will be eliminated for any value of $N$ that is a multiple of $B$
\[
N = mB, \quad m\in \mathbb{N}^+
\]
since the orthogonality of the roots of unity implies that
\[
\sum_{k=0}^{mB-1}e^{j\frac{2\pi}{B}Ak} = \frac{1 - e^{j\frac{2\pi}{B}AmB}}{1 - e^{j\frac{2\pi}{B}A}} = \frac{1 - e^{j 2\pi Am}}{1 - e^{j\frac{2\pi}{B}A}} = 0.
\]
\end{enumerate}
\vspace{1em}
}\else{\vspace{1em}}\fi
\end{exercise}
\ifanswers{}\else{\vspace{10em}}\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{exercise}{(10 points)}
Consider a causal LTI system with transfer function
\[
H(z) = \frac{1}{1-e^{j\omega_0} z^{-1}} \qquad \omega_0 \in \mathbb{R}.
\]
Prove that the system is not BIBO stable by finding a causal bounded input signal that makes the output grow to infinity.
\ifanswers{\em\vspace{3em}\par{\bfseries Solution: }
From simple inspection, the CCDE describing the system is
\[
y[n] = x[n] + e^{j\omega_0}\, y[n]
\]
so that the impulse response is the sequence
\[
h[n] = e^{j\omega_0 n}u[n].
\]
Since $|h[n]| = 1$ for $n \ge 0$ the impulse response is not absolutely summable and therefore the system is not BIBO stable. For a generic causal input $x[n]$ the output is
\[
y[n] = (x\ast h)[n] = \sum_{k=0}^{n} x[k]e^{j\omega_0(n-k)} = e^{j\omega_0 n}\sum_{k=0}^{n} x[k]e^{-j\omega_0 k}.
\]
By choosing $x[n] = e^{j\omega_0 n}\, u[n]$, the above result becomes
\[
y[n] = e^{j\omega_0 n}\sum_{k=0}^{n} e^{j\omega_0 k}e^{-j\omega_0 k} = (n+1)e^{j\omega_0 n}
\]
ans since $|y[n]| = n+1$, the output grows unbounded.
}\else{\vspace{1em}}\fi
\end{exercise}
\ifanswers{}\else{\vspace{10em}}\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{exercise}{(10 points)}
Although the harmonic series is divergent, i.e., $\sum_{k=1}^{\infty} 1/k = \infty$, very similar derived series are surprisingly convergent. The so-called Leibniz formula, for instance, uses only the odd-indexed terms with alternating sign to yield
\[
\sum_{k=0}^{\infty} \frac{(-1)^k}{2k + 1} = 1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \ldots = \frac{\pi}{4}.
\]
Consider now a continuous-time square wave $x(t)$ with period $P$, as shown in the figure below. Knowing that the signal can be expressed as the Fourier series
\[
x(t) = \frac{4}{\pi}\sum_{k=1}^{\infty}\frac{1}{2k-1}\sin(2\pi(2k-1)f_0 t)
\]
prove Leibniz formula by
a surprising result is that the alternating
\ifanswers{\em\vspace{3em}\par{\bfseries Solution: }
From simple inspection, the CCDE describing the system is
\[
y[n] = x[n] + e^{j\omega_0}\, y[n]
\]
so that the impulse response is the sequence
\[
h[n] = e^{j\omega_0 n}u[n].
\]
Since $|h[n]| = 1$ for $n \ge 0$ the impulse response is not absolutely summable and therefore the system is not BIBO stable. For a generic causal input $x[n]$ the output is
\[
y[n] = (x\ast h)[n] = \sum_{k=0}^{n} x[k]e^{j\omega_0(n-k)} = e^{j\omega_0 n}\sum_{k=0}^{n} x[k]e^{-j\omega_0 k}.
\]
By choosing $x[n] = e^{j\omega_0 n}\, u[n]$, the above result becomes
\[
y[n] = e^{j\omega_0 n}\sum_{k=0}^{n} e^{j\omega_0 k}e^{-j\omega_0 k} = (n+1)e^{j\omega_0 n}
\]
ans since $|y[n]| = n+1$, the output grows unbounded.
}\else{\vspace{1em}}\fi
\end{exercise}
\ifanswers{}\else{\vspace{10em}}\fi
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{exercise}{(20 points)}
In 1734 Euler famously solved the so-called \textit{Basel problem}\/ by proving that
\[
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.
\]
As a signal processing expert, re-derive Euler's result by using the fact that the Hilbert filter, whose frequency response is allpass, has impulse response
\[
h[n] = \begin{cases}
0 & \mbox{$n$ even} \\
\frac{2}{\pi n} & \mbox{$n$ odd}.
\end{cases}
\]
\vspace{1em}
{\footnotesize \it Hint: at some point in the proof you may find it useful to split a series into even and odd terms: $\sum_{n=1}^{\infty} a_n = \sum_{n=1}^{\infty} a_{2n} + \sum_{n=1}^{\infty} a_{2n - 1}$.}
\ifanswers{\em\vspace{3em}\par{\bfseries Solution: }
Since the Hilbert filter is allpass, its $L_2$ norm is one:
\[
\|H(e^{j\omega})\|^2 = \frac{1}{2\pi}\int_{-\pi}^{\pi} |H(e^{j\omega})|^2 d\omega = \frac{1}{2\pi}\int_{-\pi}^{\pi} d\omega = 1.
\]
By Parseval's theorem we therefore have that
\[
\|h[n]\|^2 = \|H(e^{j\omega})\|^2 = 1
\]
and so
\[
\sum_{n=-\infty}^{\infty} |h[n]|^2 = 2\sum_{n=1}^{\infty}\left(\frac{2}{\pi (2n-1)}\right)^2 = \frac{8}{\pi^2}\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = 1
\]
from which
\[
\sum_{n=1}^{\infty}\frac{1}{(2n-1)^2} = \frac{\pi^2}{8}.
\]
Now we can write
\[
\sum_{n=1}^{\infty} \frac{1}{n^2} = \sum_{n=1}^{\infty} \frac{1}{(2n)^2} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{n^2} + \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2}
\]
so that
\[
\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{4}{3} \sum_{n=1}^{\infty} \frac{1}{(2n-1)^2} = \frac{\pi^2}{6}.
\]
}\else{\vspace{1em}}\fi
\end{exercise}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\begin{exercise}{(20 points)}
Determine the range of the real-valued feedback parameter $r$ for which the causal system implemented by the following block diagram is stable:
\vspace{2em}
\begin{center}
\begin{dspBlocks}{1.2}{0.4}
& & & & \\
$x[n]$~~~~ & \BDadd & \BDsplit & \BDsplit & ~~~~~$y[n]$ \\%
& & \BDdelay & & \\%
& \BDadd & \BDsplit & & \\%
& & \BDdelay & & \\%
& & & &
\psset{arrows=->,linewidth=1.5pt}
\ncline{2,1}{2,2} \ncline{2,2}{2,5}
\ncline{-}{4,3}{4,2} \taput{$1/2$}
\ncline{-}{6,3}{6,2}\taput{$-1$}
\ncline{2,3}{3,3} \ncline{3,3}{5,3} \ncline{-}{5,3}{6,3}
\ncline{6,2}{4,2}
\ncline{4,2}{2,2}
\ncline{1,2}{2,2}\ncline{-}{2,4}{1,4}\ncline{-}{1,4}{1,2}\taput{$r$}
\end{dspBlocks}
\end{center}
\ifanswers{\em\vspace{3em}\par{\bfseries Solution: }
The block diagram can be redrawn as
\vspace{2em}
\begin{center}
\begin{dspBlocks}{1.2}{0.4}
& & & & \\
$x[n]$~~~~ & \BDadd & \BDfilter{$G(z)$} & \BDsplit & ~~~~~$y[n]$ \\%
\psset{arrows=->,linewidth=1.5pt}
\ncline{2,1}{2,2} \ncline{2,2}{2,3}\ncline{2,3}{2,5}
\ncline{1,2}{2,2}\ncline{-}{2,4}{1,4}\ncline{-}{1,4}{1,2}\taput{$r$}
\end{dspBlocks}
\end{center}
where $G(z)$ is a pure-feedback second order system with transfer function
\[
G(z) = \frac{1}{1 + z^{-1} + z^{-2}} = \frac{1}{A(z)}.
\]
The overall input-output relation is thus $Y(z) = G(z)[X(z) + rY(z)]$ and so the system's transfer function is
\[
H(z) = \frac{G(z)}{1 - rG(z)} = \frac{1}{A(z) - r}
\]
}\else{\vspace{1em}}\fi
\end{exercise}
\end{document}
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