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\section{Exercises}
\begin{exercise}{Zero-order hold}
Consider a discrete-time sequence $x[n]$ with DTFT $X(e^{j\omega})$. Next, consider the continuous-time interpolated signal
\[
x_0(t) = \sum_{n=-\infty}^{\infty}x[n] \,\mbox{rect}\,(t-n)
\]
i.e.\ the signal interpolated with a zero-order hold and $T_s = 1$~sec.
\begin{enumerate}
\item Express $X_0(f)$ (the spectrum of $x_0(t)$) in terms of $X(e^{j\omega})$.
\item Compare $X_0(f)$ to $X(f)$, the Fourier transform of the signal
\[
x(t) = \sum_{n\in\mathbb{Z}} x[n]\, \textrm{sinc} (t-n)
\]
Comment on the result: you should point out two major problems.
\end{enumerate}
As it appears, interpolating with a zero-order hold introduces in-band distortion in the region $|f| < 1/2$ and out-of-band spurious components at higher frequencies. Both problems could however be fixed by a well-designed continuous-time filter $G(f)$ applied to the ZOH interpolation.
\begin{enumerate}
\setcounter{enumi}{2}
\item Sketch the frequency response $G(f)$
\item Propose two solutions (one in the continuous-time omain, and another in the discrete-time domain) to eliminate or attenuate the in-band distortion due to the zero-order hold. Discuss the advantages and disadvantages of each.
\end{enumerate}
\end{exercise}
\begin{exercise}{A bizarre interpolator}
Consider the local interpolation scheme of the previous exercise but assume that the characteristic of the
interpolator is the following:
\[
I(t) = \begin{cases}
1 - 2|t| & \mbox{ for } |t| \leq 1/2 \\
0 & \mbox{ otherwise}
\end{cases}
\]
This is a triangular characteristic with the same support as the zero-order hold. If we pick an interpolation interval $T_s$ and interpolate a discrete-time signal $x[n]$ with $I(t)$, we obtain the continuous-time signal:
\[
x(t) = \sum_n x[n]\, I\! \left( \frac{t-nT_s}{T_s} \right)
\]
which looks like this:
\begin{center}
\begin{dspPlot}[height=\dspHeightCol,xout=true]{0,5}{-1.2,1.2}
\psset{linewidth=1pt,linecolor=gray}
\psline(-.5, 0)(0, 1)(.5, 0)
\psline(.5, 0)(1, 0.7)(1.5, 0)
\psline(1.5, 0)(2, -0.2)(2.5, 0)
\psline(2.5, 0)(3, -1)(3.5, 0)
\psline(3.5, 0)(4, 0.5)(4.5, 0)
\psline(4.5, 0)(5, -0.5)(5.5, 0)
\psset{linecolor=black}
\dspTaps[linecolor=dspDTColor]{0 1 1 0.7 2 -0.2 3 -1 4 0.5 5 -0.5}%
\end{dspPlot}
\end{center}
Assume that the spectrum of $x[n]$ between $-\pi$ and $\pi$ is
\[
X(e^{j\omega}) =
\begin{cases}
1 & \mbox{ for } |\omega| \leq 2\pi/3 \\
0 & \mbox{ otherwise}
\end{cases}
\]
(with the obvious $2\pi$-periodicity over the entire frequency axis).
\begin{enumerate}
\item Compute and sketch the Fourier transform $I(f)$ of the interpolating function $I(t)$. (Recall that the triangular function can be expressed as the convolution of $\mbox{rect}(t/2)$ with itself).
\item Sketch the Fourier transform $X(f)$ of the interpolated signal $x(t)$; in particular, clearly mark the Nyquist frequency $f_N = 1/(2T_s)$.
\item The use of $I(t)$ instead of a sinc interpolator introduces two types of errors: briefly describe them.
\item To eliminate the error \emph{in the baseband} $[-f_N, f_N]$ we can pre-filter the signal $x[n]$ with a filter $h[n]$ \emph{before} interpolating with
$I(t)$. Write the frequency response of the discrete-time filter $H(e^{j\omega})$.
\end{enumerate}
\end{exercise}
\begin{exercise}{Another view of sampling}
One of the standard ways of describing the sampling operation relies on the concept of ``modulation by a pulse train''. Choose a sampling interval $T_s$ and define a continuous-time pulse train $p(t)$ as
\[
p(t) = \sum_{k=-\infty}^{\infty} \delta (t-kT_s)
\]
The Fourier Transform of the pulse train is
\[
P(f) = F_s \sum_{k=-\infty}^{\infty} \delta \! \left( f- k F_s \right)
\]
(where, as per usual, $F_s = 1/T_s$). This result is tricky to show, so just take it as is. The ``sampled'' signal can now be expressed as the modulation of an arbitrary-continuous time signal $x(t)$ by the pulse train:
\[
x_s(t) = p(t)\,x(t)
\]
Note that, now, this sampled signal is still continuous time but, by the properties of the delta function, is non-zero only at multiples of $T_s$; in a sense, $x_s(t)$ is a discrete-time signal ``embedded'' in the continuous time world.
Derive the Fourier transform of $x_s(t)$ and show that if $x(t)$ is bandlimited to $1/(2T_s)$ then we can reconstruct $x(t)$ from $x_s(t)$.
\end{exercise}
\begin{exercise}{Aliasing can be good}
Consider a real, continuous-time signal $x_c(t)$ with the following spectrum $X_c(f)$:
\begin{center}
\begin{dspPlot}[height=\dspHeightCol,xtype=freq,xticks=custom]{-4,4}{0, 1.2}
\psline[linecolor=dspCTFColor](-4,0)(-2,0)(-2,1)(-1,0)(1,0)(2,1)(2,0)(4,0)
\dspCustomTicks{-2 $-2f_0$ -1 $-f_0$ 0 0 1 $f_0$ 2 $2f_0$}
\end{dspPlot}
\end{center}
\begin{enumerate}
\item What is the bandwidth of the signal? What is the minimum sampling period in order to satisfy the sampling theorem?
\item Take a sampling period $T_s = 1/(2f_0)$; clearly, with this sampling period, there will be aliasing. Plot the DTFT of the discrete-time signal $x_a[n] = x_c(nT_s)$.
\item Suggest a block diagram to reconstruct $x_c(t)$ from $x_a[n]$.
\item With such a scheme available, we can therefore exploit aliasing to reduce the sampling frequency necessary to sample a bandpass signal. In general, what is the minimum sampling frequency to be able to reconstruct, with the above strategy, a real-valued signal whose frequency support on the positive axis is $[f_0, f_1]$ (with the usual symmetry around zero, of course)?
\end{enumerate}
\end{exercise}
\begin{exercise}{Digital processing of continuous-time signals}
For your birthday, you receive an unexpected present: a $4$~MHz sampler, complete with anti-aliasing filter. This means you can safely sample signals up to a frequency of $2$ ~MHz; since this frequency is above the AM radio frequency band, you decide to hook up the sampler to your favorite signal processing system and build an entirely digital radio receiver. In this exercise we will explore how to do so.
To simplify things a bit, assume that the AM radio spectrum extends from $1$~Mhz to $1.2$~Mhz and that in this band you have ten channels side by side, each one of which occupies $20$~KHz.
\begin{enumerate}
\item Sketch the digital spectrum at the output of the A/D converter, and show the bands occupied by the channels, numbered from $1$ to $10$, with their beginning and end frequencies.
\end{enumerate}
The first thing that you need to do is to find a way to isolate
the channel you want to listen to and to eliminate the rest.
For this, you need a bandpass filter centered on the band of
interest. Of course, this filter must be {\em tunable} in
the sense that you must be able to change its spectral
location when you want to change station. An easy way to
obtain a tunable bandpass filter is by modulating a lowpass
filter with a sinusoidal oscillator whose frequency is
controllable by the user:
\begin{enumerate}\setcounter{enumi}{1}
\item As an example of a tunable filter, assume $h[n]$ is an ideal lowpass filter with cutoff frequency $\pi/8$. Plot the magnitude response of the filter $h_m[n] = \cos(\omega_m n)h[n]$, where $\omega_m = \pi/2$; $\omega_m$ is called the {\em tuning frequency}.
\item Specify the cutoff frequency of a lowpass filter which can be used to select one of the AM channels above.
\item Specify the tuning frequencies for channel~$1$, $5$ and $10$.
\end{enumerate}
Now that you know how to select a channel, all that is left to do is to demodulate the signal and feed it to an interpolator and then to a loudspeaker.
\begin{enumerate}\setcounter{enumi}{4}
\item Sketch the complete block diagram of the radio receiver, from the antenna going into the sampler to the final loudspeaker. Use only one sinusoidal oscillator. Do not forget the filter before the interpolator (specify its bandwidth).
\end{enumerate}
The whole receiver now works at a rate of $4$ MHz; since it outputs audio signals, this is clearly a waste.
\begin{enumerate}\setcounter{enumi}{5}
\item Which is the minimum interpolation frequency you can use? Modify the receiver's block diagram with the necessary elements to use a lower frequency interpolator.
\end{enumerate}
\end{exercise}
\begin{exercise}{Acoustic aliasing}
Assume $x(t)$ is a continuous-time pure sinusoid at $10$ KHz. It is sampled with a sampler at $8$ KHz and then interpolated back to a continuous-time signal with an interpolator at $8$ KHz. What is the perceived frequency of the interpolated sinusoid?
\end{exercise}
\begin{exercise}{Interpolation subtleties}
We have seen that any discrete-time sequence can be sinc-interpolated into a continuous-time signal which is $F_s$-bandlimited; $F_s$ depends on the
interpolation interval $T_s$ via the relation $F_s = 1/T_s$.
Consider the continuous-time signal $x_c(t) = e^{-t/T_s}u(t)$ and the discrete-time sequence $x[n] = e^{-n}u[n]$. Clearly, $x_c(nT_s) = x[n]$; but, can we also say that $x_c(t)$ is the signal we obtain if we apply sinc interpolation to the sequence $x[n] = e^{-n}$ with interpolation interval $T_s$?
Explain in detail.
\end{exercise}
\begin{exercise}{Time and frequency}
Consider a real continuous-time signal $x(t)$. All you know about the signal is that $x(t) = 0$ for $|t| > t_0$. Can you determine a sampling frequency $F_s$ so that when you sample $x(t)$, there is no aliasing? Explain.
\end{exercise}
\begin{exercise}{Aliasing in time}\label{aliasTimeEx}
Consider an $N$-periodic discrete-time signal
$\tilde{x}[n]$, with $N$ an \emph{even} number, and let
$\tilde{X}[k]$ be its DFS:
\[
\tilde{X}[k] = \sum_{n=0}^{N-1}\tilde{x}[n]
\, e^{-j\frac{2\pi}{N}nk}\, ,\qquad \quad k\in \mathbb{Z}
\]
Let $\tilde{Y}[m] = \tilde{X}[2m]$, i.e.\ a ``subsampled''
version of the DFS coefficients; clearly this defines a
$(N/2)$-periodic sequence of DFS coefficients.
%Consider
Now consider
the $(N/2)$-point inverse DFS of $\tilde{Y}[m]$ and call
this $(N/2)$-periodic signal $\tilde{y}[n]$:
\[
\tilde{y}[n] = \frac{2}{N}\sum_{k=0}^{N/2-1}\tilde{Y}[k]
\, e^{j\frac{2\pi}{N/2}nk},\qquad \quad n \in \mathbb{Z}
\]
Express $\tilde{y}[n]$ in terms of $\tilde{x}[n]$ and
describe in a few words what has happened to $\tilde{x}[n]$
and why.
\end{exercise}
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