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\section{Odds and Ends}
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\begin{example}[AM radio.]\label{ex:lf:AMradio}
\index{AM radio}\index{galena}
The earliest method for radio transmission is based on amplitude modulation (AM); an analog AM transmitter uses a transducer such as a microphone to convert sound into a baseband electrical signal, and a high-frequency oscillator to generate a carrier that is modulated by the audio. The frequency of the carrier is chosen to match a band in the electromagnetic spectrum where propagation is efficient: commercial AM radio, still in use today, uses carrier frequencies in the 500kHz to 1500kHz range, but AM broadcasts at frequencies above 3MHz, called shortwave radio, can travel around the entire world thanks to the reflecting properties of the ionosphere.
We can analyze the basics of an AM transmitter in discrete time as well, neglecting just the digital-to-analog interface with the power amplifier and the antenna at the transmitting station. Assume $\mathbf{s}$ is a zero-mean, baseband signal with sample values in the $[-1, 1]$ range; the AM transmitted signal $\mathbf{x}$ is defined by
\[
x[n] = (1 + s[n])\cos(\omega_c n)
\]
where $\omega_c$ is the carrier frequency for a specific AM radio channel (that is, the frequency one would dial a radio receiver to). Because of the assumed range for $\mathbf{s}$, the carrier is always multiplied by a positive value; Figure~\ref{fig:lf:AMsignal} shows an example where the envelope $\mathbf{1 + s}$ is highlighted on top of the modulated carrier.
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\def\envelope{x 360 mul cos x 720 mul sin 2 add add 4 div 2 mul 1 sub }
\begin{figure}[b!]
\center\small
\begin{dspPlot}[height=0.5\dspHeight,xout=true,xticks=none,sidegap=0]{0,1}{-1.1,1.1}
\dspFunc[linecolor=ColorDT]{\envelope}
\end{dspPlot}
\begin{dspPlot}[height=0.5\dspHeight,xout=true,xticks=none,sidegap=0]{0,1}{-2.1,2.1}
\dspFunc[linecolor=ColorDT2]{\envelope 1 add x 10000 mul cos mul}
\dspFunc[linecolor=ColorDT!20,linewidth=2pt]{\envelope 1 add}
\end{dspPlot}
\caption{Baseband signal (top) and AM signal with highlighted envelope (bottom).}\label{fig:lf:AMsignal}
\end{figure}
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In the frequency domain, we have seen that amplitude modulation induces a shift of the baseband spectrum to the carrier frequency $\omega_c$; for AM radio, the spectrum $X(e^{j\omega})$ is shown in Figure~\ref{fig:lf:AMspectrum} and we can immediately notice two undesirable features. First, as we already remarked in Section~\ref{sec:lf:modulation}, the spectral occupancy is doubled with respect to the original baseband signal and this, ultimately, impacts the quality of AM broadcasts. To understand why this is so, remember that the electromagnetic spectrum is a precious resource, and therefore it is strictly regulated by law to allow concurrent and interference-free use by entities as diverse as the military, emergency services, amateur radio and commercial broadcasts. The band for commercial AM is about 1MHz-wide and, to be able to allocate a significant number of radio stations in that range, neighboring channels are only 10KHz apart. This means that the maximum frequency of the original baseband signal must be limited to about 5kHz, which is rather low and explains the poor audio quality of AM music broadcasts. If the modulation scheme did not imply a doubling of the spectral occupancy, the baseband audio signal could reach 10KHz and sound much better. Incidentally, this explains the higher quality of more modern transmission methods such as FM.
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\begin{figure}[tb]
\center\small
\begin{dspPlot}[height=0.5\dspHeight,xtype=freq,xticks=1,yticks=none,ylabel={$|S(e^{j\omega})|$}]{-1,1}{0,2}
\dspCustomTicks[axis=x]{0.2 $\omega_p$}
\dspFunc[linecolor=ColorDF]{x \dspAudio{0}{0.15}}
\end{dspPlot}
\begin{dspPlot}[height=0.5\dspHeight,xtype=freq,xticks=1,yticks=none,ylabel={$|X(e^{j\omega})|$}]{-1,1}{0,2}
\dspCustomTicks[axis=x]{0.7 $\omega_c$ -0.7 $\omega_c$}
\dspFunc[linecolor=ColorDF]{x \dspAudio{0.7}{0.15} x \dspAudio{-0.7}{0.15} add}
\dspDiracs[linecolor=ColorDF]{0.7 1.7 -0.7 1.7}
\end{dspPlot}
\caption{Baseband spectrum (top) and AM spectrum (bottom).}\label{fig:lf:AMspectrum}
\end{figure}
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The second problem stems from the fact that, by making the modulating signal positive (the envelope is $(\mathbf{1 + s})$ and not $\mathbf{s}$) we can observe an explicit sinusoidal component in the spectrum at $\omega_c$. This reduces the maximum achievable energy efficiency of AM to 33\% since one third of the transmitted energy is used by the carrier and another third by the unnecessary spectral duplication. AM transmission methods such as Single-Sideband modulation (SSB) exist and are in use to eliminate both these problems but the interesting question is why classic AM thrived in spite of these shortcomings; the answer lies in the ease with which it can be demodulated by an inexpensive receiver.
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\begin{figure}[tbh]
\centering
\includegraphics[height=60mm]{\localpath{figs/galena.ps}}
\caption{Schematics for a galena radio receiver {\em (from Gernsback's book ``Radio For All", 1922)}.} \label{fig:lf:galena}
\end{figure}
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Figure~\ref{fig:lf:galena} shows some vintage schematics for a \textit{crystal radio} receiver, a popular DIY device in the early part of the 20th century; the receiver, which is entirely passive and therefore both robust and inexpensive, was a driving force in the popularization of commercial AM radio before being supplanted by amplifying devices using vacuum tubes. It includes a piece of wire as the antenna, a coil as a simple tunable bandpass filter, a galena detector and a pair of headphones. The antenna and the tuning coil form a variable LC bandpass filter used to select the AM channel of interest; the detector is composed of a galena crystal, a natural mineral that, when touched by a thin metal wire called the ``cat's whisker'', acts as a rudimentary semiconductor, that is, a rectifying diode. The bandpass filter's output is a signal like the one in the bottom panel of Figure~\ref{fig:lf:AMsignal} and after the galena detector, only the positive part of the signal will be preserved, as in the top panel of Figure~\ref{fig:lf:galenaDetection}-(a). This signal still contains a high frequency component but the mechanical inertia of the loudspeaker in the headphones will act as a lowpass filter to reveal the original envelope as in the bottom panel of Figure~\ref{fig:lf:galenaDetection}. Extremely ingenious, simple and, above all, cheap -- which justifies the purported inefficiency of the trasmission scheme.
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\begin{figure}[tbh]
\center\small
\begin{dspPlot}[height=0.5\dspHeight,xout=true,xticks=none,sidegap=0]{0,1}{0,2.1}
\dspFunc[linecolor=ColorDT2]{/s \envelope 1 add x 10000 mul cos mul def s 0 ge {s} {0} ifelse}
\end{dspPlot}
\begin{dspPlot}[height=0.5\dspHeight,xout=true,xticks=none,sidegap=0]{0,1}{0,2.1}
\dspFunc[linecolor=ColorDT]{\envelope 1 add x 10000 mul cos 20 div add}
\end{dspPlot}
\caption{Galena demodulation: the signal after the detector (top) and lowpass-filtered by the headphones (bottom).}\label{fig:lf:galenaDetection}
\end{figure}
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\end{example}
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\begin{example}[Phase, linear phase, and minimum phase.]\label{ex:lf:phases}
\index{phase!linear}
\index{phase!minimum}
The magnitude response of a filter is of immediate intuitive significance: after all, at one point or another, we have all fiddled with the tone knobs on an audio amplifier to boost the bass or reduce the treble. The phase response, on the other hand, has a more elusive meaning and part of this problem comes from the fact that the human ear is fundamentally insensitive to the phase content of \textit{sustained} sounds. What this means is that if we take the steady-state part of the note generated by, say, an organ or a clarinet, take a DFT, scramble randomly the phase, and go back to the time domain with an inverse DFT, then the synthesized sound will be indistinguishable from the original. It is important to note, however, that the ear's insensitivity to phase only applies to sustained signals. The timbre of a musical instrument, on the other hand, is defined also by the initial portion of each note, called the \textit{attack}; the attack is determined by the mechanics of the note-generation process; that's why, for instance, in a violin the sound of a plucked string (the so-called \textit{pizzicato}) is different from the sound obtained by using the bow. The acoustic quality of the attack phase of a sound is determined by the shape of the underlying signal in the time domain; as a consequence, if we scramble the phase of an entire audio excerpt including the attack, the result will sound completely different from the original.
The effects of phase distortion are particularly important in data communication systems, where the information is conveyed by level transitions in the transmitted signal; since most communication channels act as lowpass filter, it is instructive to see how different phase responses affect the shape of the transmission. Consider for instance the signal in Figure~\ref{fig:lf:pulse}-(a), representing a short pulse with a transition from high to low. In Figure~\ref{fig:lf:pulse}-(b) we can see the effects of filtering the pulse with a Moving Average of length 5; although the edges have been spread in time, the original shape is still recognizable and detecting the $\pm 1$ levels via thresholding poses no difficulty. Figure~\ref{fig:lf:pulse}-(c) shows the pulse filtered by a Leaky Integrator with $\lambda = 0.8$; the mildly nonlinear phase response does modify the rising and falling edges of the pulse but, once again, the overall shape is still detectable. On the other hand, Figure~\ref{fig:lf:pulse}-(d) shows the result of filtering the pulse via the CCDE
\[
y[n] = \alpha y[n-1] -\alpha x[n] + x[n-1],
\]
with $\alpha = 0.9$. The filter implemented by the equation is an allpass filter, that is, its magnitude response is equal to one (see Exercise~\ref{ex:lf:allpassPhase}). The phase response is however highly nonlinear and the net effect is a complete scrambling of the shape of the pulse. First, we can notice that there is an initial negative portion although we are expecting a rising edge first; furthermore, although it remains zero-average, the phase-distorted signal exhibits a much higher peak value, which could cause saturation and clipping in practical communication circuits.
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\begin{figure}[t]
\center\small
\def\signal{x 10 lt {0} {x 20 lt {1} {x 30 lt {-1} {0} ifelse} ifelse} ifelse }
\begin{tabular}{cc}
\hspace{-2em}
\begin{dspPlot}[width=\dspHalfWidth,height=\dspHalfHeight,xout=true,xticks=10]{0,42}{-2.2,2.2}
\dspSignalOpt[linecolor=ColorDT]{}{\signal}
\end{dspPlot}
&
\hspace{-2em}
\begin{dspPlot}[width=\dspHalfWidth,height=\dspHalfHeight,xout=true,xticks=10]{0,42}{-2.2,2.2}
\dspSignalOpt[linecolor=ColorDT]{\dspSetFilter{0.2 0.2 0.2 0.2 0.2}{}}{\signal \dspFilter}
\end{dspPlot}
\\
(a) & (b)
\\
\hspace{-2em}
\begin{dspPlot}[width=\dspHalfWidth,height=\dspHalfHeight,xout=true,xticks=10]{0,42}{-2.2,2.2}
\dspSignalOpt[linecolor=ColorDT]{\dspSetFilter{0.2}{-0.8}}{\signal \dspFilter}
\end{dspPlot}
&
\hspace{-2em}
\def\signal{x 10 lt {0} {x 20 lt {1} {x 30 lt {-1} {0} ifelse} ifelse} ifelse }
\begin{dspPlot}[width=\dspHalfWidth,height=\dspHalfHeight,xout=true,xticks=10]{0,42}{-2.2,2.2}
\dspSignalOpt[linecolor=ColorDT]{\dspSetFilter{-0.9 1}{-0.9}}{\signal \dspFilter}
\end{dspPlot}
\\
(a) & (b)
\end{tabular}
\caption{Original pulse (a); pulse filtered by a Moving Average (b); pulse filtered by a Leaky Integrator (c); pulse filtered by an allpass filter (d).}\label{fig:lf:pulse}
\end{figure}
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Finally, let's consider again the concept of the overall delay introduced by a causal filter. For the Moving Average, a linear-phase FIR, we have seen that the delay (at the net of the lowpass effect) is equal to half the length of the impulse response. For the Leaky Integrator we can approximate the delay in the passband as the group delay in $\omega = 0$ which, according to~(\ref{eq:lf:LIgrd}), is equal to $\lambda/(1-\lambda)$. The overall delay therefore increases as $\lambda$ gets closer to one and note that, at the same time, as $\lambda$ gets closer to one the impulse response takes a longer and longer time to decay. In a sense, we are in a similar situation as in the case of the Moving Average: impulse responses that are more distributed in time lead to longer delays. We may ask therefore, given a desired magnitude response, what is the causal filter that provides the smallest possible delay? The answer is called the \textit{minimum-phase} solution and it's given by the filter whose impulse response is most concentrated in time. In other words, given the set of all impulse responses $\{\mathbf{h}_i\}_{i=0, \ldots, K}$ implementing the same magnitude response, the minimum phase solution $\mathbf{h}'$ is the one for which
\[
\sum_{n=0}^{N} |h'[n]|^2 \ge \sum_{n=0}^{N} |h_i[n]|^2 \quad \forall N;
\]
this means that the minimum-phase filter has the fastest decay of all equivalent impulse responses.
\end{example}
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\begin{example}[BIBO stability of the ideal lowpass.]\label{ex:lf:BIBOlowpass}
Here's a curious and little-known fact: the ideal lowpass is, in fact, not BIBO stable! Remember from Section~\ref{sec:lf:stability} that a necessary condition for stability is the absolute summability of the impulse response. To show that the sinc sequence is not absolutely summable, we can use the well-known integral test for the convergence of a series: if $f$ is a monotonically decreasing function defined on the positive real line, then
\[
\sum_{n=N}^{\infty}f(n) < \infty \Leftrightarrow \int_{N}^{\infty}f(t)dt < \infty.
\]
In our case we want to prove the divergence of
\begin{equation}\label{eq:lf:divergenceSinc}
\sum_{n\in\mathbb{N}^+}\left|\sinc\left(\frac{\omega_p}{\pi} n\right)\right| = \sum_{n\in\mathbb{N}^+}\left|\frac{\sin \omega_p n}{\omega_p n}\right|, \qquad 0 < |\omega_p| < \pi
\end{equation}
which we will prove by showing that
\[
\int_{1}^{\infty}\left|\frac{\sin \omega_p t}{\omega_p t}\right|dt = \infty.
\]
With a change of variable in the integral we have
\begin{align*}
% advance integration limit
\int_{1/\omega_p}^\infty \left|\frac{\sin t}{t}\right|dt & \geq \int_\pi^\infty \left|\frac{\sin t}{t}\right|dt \\
% split the integral into contiguous intervals
&= \sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \left|\frac{\sin t}{t}\right|dt \\
% replace denominator by its maximum value
&\geq \sum_{k=1}^\infty\int_{k\pi}^{(k+1)\pi} \frac{|\sin t|}{(k+1)\pi}dt \\
% integrate sin
&= \frac{1}{\pi}\sum_{k=1}^\infty \frac{1}{k+1} \int_{k\pi}^{(k+1)\pi} |\sin t|dt \\
&= \frac{2}{\pi}\sum_{k=1}^\infty\frac{1}{k+1} = \infty
\end{align*}
Since the last term is equivalent to a harmonic series, the divergence is however very slow; as an example, Figure~\ref{fig:lf:divergenceSinc} shows the value of the sum in~(\ref{eq:lf:divergenceSinc}) for $\omega_p = \pi/3$ as the number of the terms in the sum increases.
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\begin{figure}
\center\small
\begin{dspPlot}[xticks=custom]{1,100}{0,4.1}
\dspFuncData{1 0.3333 2 1.8960 3 2.1453 4 2.2943 5 2.4001 6 2.4810 7 2.5482 8 2.6050 9 2.6535 10 2.6969 11 2.7358 12 2.7704 13 2.8025 14 2.8320 15 2.8590 16 2.8844 17 2.9082 18 2.9303 19 2.9514 20 2.9713 21 2.9900 22 3.0080 23 3.0252 24 3.0413 25 3.0570 26 3.0721 27 3.0864 28 3.1003 29 3.1137 30 3.1265 31 3.1390 32 3.1511 33 3.1627 34 3.1740 35 3.1850 36 3.1956 37 3.2060 38 3.2161 39 3.2258 40 3.2354 41 3.2447 42 3.2537 43 3.2626 44 3.2713 45 3.2797 46 3.2880 47 3.2961 48 3.3039 49 3.3117 50 3.3193 51 3.3267 52 3.3340 53 3.3411 54 3.3481 55 3.3550 56 3.3617 57 3.3683 58 3.3748 59 3.3813 60 3.3875 61 3.3937 62 3.3998 63 3.4057 64 3.4116 65 3.4174 66 3.4231 67 3.4287 68 3.4343 69 3.4396 70 3.4450 71 3.4504 72 3.4555 73 3.4607 74 3.4658 75 3.4707 76 3.4757 77 3.4806 78 3.4853 79 3.4901 80 3.4948 81 3.4994 82 3.5040 83 3.5085 84 3.5129 85 3.5173 86 3.5217 87 3.5260 88 3.5302 89 3.5344 90 3.5386 91 3.5427 92 3.5468 93 3.5507 94 3.5547 95 3.5587 96 3.5625 97 3.5664 98 3.5702 99 3.5740 100 3.5777}
\dspCustomTicks[axis=x]{1 $1$ 50 $5000$ 100 $10000$}
\end{dspPlot}
\caption{Divergence of $\sum_{1}^{N} |\sinc(n/3)|$, as a function of $N$.}\label{fig:lf:divergenceSinc}
\end{figure}
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\end{example}
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