Part 8 · Chapter 28

Signals and Systems in Communications

Every communication system is an exercise in moving a signal's spectrum to where the channel will carry it and moving it back again afterwards, so the frequency-shifting property of Chapter 14 — one line of algebra — turns out to be the engine underneath amplitude modulation, single sideband, frequency-division multiplexing, angle modulation and the sampled systems that replaced them all.

Signals and Systems Prof. Mithun Mondal Reading time ≈ 65 min
i What you'll learn
  • The three physical reasons a baseband signal cannot be transmitted as it stands — antenna size, channel allocation and multiplexing — and why all three are solved by moving a spectrum.
  • How the modulation property of Chapter 14 becomes double-sideband suppressed-carrier transmission, and why the transmitted bandwidth is exactly twice the message bandwidth.
  • Coherent detection, the quadrature null, and precisely how much output you lose to a phase error \(\theta\) or a frequency error \(\Delta\omega\).
  • Why adding a carrier buys the cheap envelope detector, what the modulation index must satisfy, and the derivation of AM's disappointing power efficiency.
  • The Hilbert transform, the analytic signal, and how single sideband halves the bandwidth that DSB wastes.
  • Frequency-division multiplexing, the superheterodyne receiver, and where the image frequency comes from.
  • Angle modulation — instantaneous frequency, the modulation index \(\beta\), narrowband FM, and Carson's rule for bandwidth.
  • The pulse route: PAM, time-division multiplexing and PCM, built directly on the sampling theorem of Chapter 20.
  • The complex envelope and its in-phase / quadrature parts — the representation that lets a passband system be simulated entirely at baseband.
Section 28-1

Why Modulate at All

Twenty-seven chapters have built an apparatus for describing signals and the systems that act on them. This chapter spends that apparatus on the application that motivated most of it in the first place. Communication engineering is the reason Fourier analysis left mathematics and entered the electrical engineering syllabus, and almost everything in Parts 3 to 6 of this book has a communication problem sitting behind it.

Start with the problem in its rawest form. Speech occupies roughly 300 Hz to 3.4 kHz; music, up to about 20 kHz; a video signal, several megahertz. These are baseband signals — their energy sits in a band beginning at or near zero frequency. Suppose we simply connect a microphone to an antenna and hope. Three separate things go wrong, and each one is a spectrum problem in disguise.

The first is antenna size. An antenna radiates efficiently only when its physical length is a respectable fraction of the wavelength it is asked to carry — a quarter wavelength is the usual rule. At 3 kHz the wavelength is \(c/f = 3\times10^8/3\times10^3 = 100\) km, so a quarter-wave antenna would be 25 km tall. At 100 MHz the wavelength is 3 m and the antenna is 75 cm. Nothing about the message has changed; only the frequency band it occupies has, and the antenna does not care about the message.

The second is channel allocation. A physical channel passes a particular band and rejects the rest — an optical fibre near \(2\times10^{14}\) Hz, a satellite link in the gigahertz, a copper pair up to a few megahertz. A signal must be delivered to the channel already sitting inside the band the channel supports.

The third, and the one with the most far-reaching consequences, is multiplexing. If two speakers are placed on the same channel at baseband their spectra overlap completely and no filter can ever separate them again. Chapter 15 was explicit about this: a linear filter can only separate signals that occupy different frequency bands. Give each speaker a different band and separation becomes trivial.

All three problems have one solution. Antenna size, channel allocation and multiplexing are answered by the same operation: take the spectrum \(X(j\omega)\) and slide it up the frequency axis to a chosen carrier frequency \(\omega_c\), then slide it back at the receiver. That operation is modulation, and Chapter 14 already proved that multiplying by a complex exponential does exactly that. This chapter is, in a real sense, one property of the Fourier transform applied over and over.
Section 28-2

The Modulation Property and DSB-SC

Chapter 14 established the frequency-shifting property: multiplying a signal by \(e^{j\omega_c t}\) translates its entire spectrum by \(\omega_c\). A complex exponential is not a voltage anyone can generate, but a cosine is, and a cosine is the sum of two of them. Applying the shift twice and adding gives the result on which the rest of this chapter rests.

Deriving the modulation theorem from the shifting property
\[ x(t)\cos\omega_c t = \tfrac12 x(t)e^{j\omega_c t} + \tfrac12 x(t)e^{-j\omega_c t} \;\;\xleftrightarrow{\ \mathcal{F}\ }\;\; \tfrac12 X\big(j(\omega-\omega_c)\big) + \tfrac12 X\big(j(\omega+\omega_c)\big) \]

Read the right-hand side literally. The message spectrum appears twice, once centred at \(+\omega_c\) and once at \(-\omega_c\), each at half its original height. Nothing is distorted and nothing is lost; the spectrum has simply been copied to two new places. A signal formed this way is called double-sideband suppressed-carrier — double-sideband because both the portion above \(\omega_c\) and the portion below it are transmitted, suppressed-carrier because there is no discrete spectral line at \(\omega_c\) itself.

ω X(jω) 1 −W W
The message at baseband — bandwidth \(W\)
ω ½ −ω_c ω_c 2W lower and upper sidebands about each centre
After multiplying by \(\cos\omega_c t\) — two half-height copies
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DSB-SC transmission bandwidth
\[ s(t) = x(t)\cos\omega_c t \qquad\Longrightarrow\qquad B_{\text{DSB}} = 2W \]

A message of bandwidth \(W\) occupies \(2W\) once modulated, because the band from \(\omega_c - W\) to \(\omega_c\) (the lower sideband) and the band from \(\omega_c\) to \(\omega_c + W\) (the upper sideband) are both transmitted. The negative-frequency copy is not extra bandwidth — a real signal's spectrum is always conjugate-symmetric, and only the positive half is physically occupied.

The assumption \(\omega_c \gt W\) is doing quiet work here. If the carrier were lower than the message bandwidth, the two shifted copies would overlap around the origin, they would add, and no filter could ever pull them apart again. This is the same non-overlap condition that Chapter 20 will have already made familiar under a different name: it is exactly the condition that the shifted spectral replicas produced by sampling do not alias into one another. In practice \(\omega_c\) is larger than \(W\) by orders of magnitude, so the condition is never in doubt.

Two features of DSB-SC are worth noticing before we improve on it. It is efficient in power, because every watt transmitted carries message information — there is no carrier line consuming power while conveying nothing. It is wasteful in bandwidth, because the upper and lower sidebands are mirror images of one another and therefore carry the same information twice. Sections 28-4 and 28-5 trade these two properties against each other in opposite directions.

Section 28-3

Coherent Detection

Having slid the spectrum up, the receiver must slide it back. The natural move is to do the same thing again: multiply the received signal by a locally generated cosine at the same frequency. Multiplying by a cosine shifts the spectrum both up and down, so one of the two shifted copies of the message lands back at the origin while the rest is thrown out to \(2\omega_c\), where a lowpass filter disposes of it.

Product demodulation, worked through with a trigonometric identity
\[ v(t) = s(t)\cos\omega_c t = x(t)\cos^2\omega_c t = \tfrac12 x(t) + \tfrac12 x(t)\cos 2\omega_c t \]

The first term is the message, halved. The second is the message translated to \(2\omega_c\), which occupies the band \(2\omega_c \pm W\) and is therefore entirely separated from the first term whenever \(\omega_c \gt W\). A lowpass filter with cutoff anywhere between \(W\) and \(2\omega_c - W\) recovers \(\tfrac12 x(t)\), and a gain of two finishes the job. This is coherent or synchronous detection, and the word coherent is the entire difficulty.

s(t) = x(t) cos ω_c t × LOCAL OSC. cos(ω_c t + θ) v(t) LOWPASS gain 2 x(t) cos θ ω V(jω) 2ω_c −2ω_c filter cutoff discarded ⟶
Coherent detection — the wanted term returns to baseband, the rest lands at \(2\omega_c\)

Suppose the local oscillator is at the right frequency but the wrong phase, producing \(\cos(\omega_c t + \theta)\). Repeating the product with the general identity \(\cos A\cos B = \tfrac12\cos(A-B) + \tfrac12\cos(A+B)\):

The effect of a phase error
\[ v(t) = x(t)\cos\omega_c t\,\cos(\omega_c t + \theta) = \tfrac12 x(t)\cos\theta + \tfrac12 x(t)\cos(2\omega_c t + \theta) \]
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Phase and frequency errors in a coherent detector
\[ y_\theta(t) = x(t)\cos\theta \qquad\qquad y_{\Delta\omega}(t) = x(t)\cos\big(\Delta\omega\, t + \theta\big) \]

A constant phase error attenuates the output by \(\cos\theta\) without distorting it — but at \(\theta = \pi/2\) the output vanishes entirely. This is the quadrature null. A frequency error is far worse: the output is multiplied by a slow cosine, so the recovered message fades in and out and periodically inverts.

The quadrature null is not merely a nuisance to be avoided; it is a resource. Two entirely separate messages can be sent on the same carrier frequency, one on \(\cos\omega_c t\) and one on \(\sin\omega_c t\), and each is invisible to a detector locked to the other. That is quadrature amplitude multiplexing, and it is why every modern digital radio describes its signal as a pair \((I, Q)\) — a point Section 28-9 returns to.

Recovering the carrier phase to the accuracy this demands is the price of DSB-SC. A Costas loop or a squaring loop can extract it from the received signal itself, but both add circuitry, and in the 1920s neither existed. That commercial pressure produced the scheme in the next section, which throws away power in order to make the receiver almost free.

Section 28-4

AM and the Envelope Detector

The insight behind conventional amplitude modulation is that if the transmitted amplitude never goes negative, the message can be read off the outline of the waveform with a diode, a capacitor and a resistor — no oscillator, no phase lock, no multiplier. To guarantee that, add a constant \(A\) to the message before modulating.

The AM signal and its modulation index
\[ s(t) = \big[A + x(t)\big]\cos\omega_c t = A\big[1 + \mu\, m(t)\big]\cos\omega_c t, \qquad \mu = \frac{\max|x(t)|}{A} \]

Here \(m(t) = x(t)/\max|x(t)|\) is the message normalised to unit peak, and \(\mu\) is the modulation index. The envelope of \(s(t)\) — the slowly varying outline that the carrier oscillates inside — is \(|A + x(t)|\). If \(\mu \le 1\) then \(A + x(t) \ge 0\) always, the modulus sign is redundant, and the envelope is \(A + x(t)\): the message plus a constant, which a capacitor removes. If \(\mu \gt 1\) the bracket goes negative on the deepest troughs, the modulus folds those excursions back up, and the envelope no longer resembles the message at all. This is overmodulation, and it is irreversible distortion.

μ = 0.6 — envelope tracks A + x(t) t
Undermodulated — the envelope is the message
μ = 1.5 — envelope crosses over t
Overmodulated — \(A + x(t)\) changes sign

What does the added carrier cost? Expand the AM signal for a single-tone message \(x(t) = A\mu\cos\omega_m t\):

Single-tone AM, expanded into three lines
\[ s(t) = A\cos\omega_c t + \frac{A\mu}{2}\cos(\omega_c + \omega_m)t + \frac{A\mu}{2}\cos(\omega_c - \omega_m)t \]

Three discrete frequencies: the carrier and the two sidebands. Their powers into a unit resistance are \(A^2/2\) for the carrier and \(A^2\mu^2/8\) for each sideband. Only the sidebands carry the message; the carrier is a constant tone conveying nothing but the receiver's convenience. Forming the ratio gives the power efficiency.

Deriving the power efficiency of tone-modulated AM
\[ \eta = \frac{P_{\text{sidebands}}}{P_{\text{total}}} = \frac{2\cdot A^2\mu^2/8}{A^2/2 + A^2\mu^2/4} = \frac{\mu^2/4}{(2+\mu^2)/4} = \frac{\mu^2}{2+\mu^2} \]
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AM is never more than one-third efficient
\[ \eta = \frac{\mu^2}{2+\mu^2} \le \frac{1}{3} \quad\text{at}\quad \mu = 1 \]

Even at the deepest legal modulation, two-thirds of the transmitted power sits in a carrier that carries no information. At the more typical \(\mu = 0.5\) the figure falls to \(0.25/2.25 \approx 11\%\). AM buys receiver simplicity with transmitter power, which was an excellent trade when one transmitter served a million crystal sets and a poor one everywhere else.

The envelope detector itself is a diode feeding a parallel \(RC\) network. On each positive carrier peak the diode conducts and charges the capacitor to the peak value; between peaks the diode is reverse-biased and the capacitor discharges through \(R\). The time constant must therefore be long enough to hold its value between carrier peaks but short enough to follow the envelope down. Both conditions can be made quantitative.

The upper limit comes from asking the capacitor to fall at least as fast as the envelope does. Discharging from \(E(t)\), the capacitor's initial rate of fall is \(E(t)/RC\). The envelope \(E(t) = A(1+\mu\cos\omega_m t)\) falls fastest at a rate \(A\mu\omega_m\sin\omega_m t\). Requiring the first to beat the second at every instant:

Deriving the envelope-detector time constant
\[ \frac{A(1+\mu\cos\omega_m t)}{RC} \;\ge\; A\mu\omega_m\sin\omega_m t \quad\Longrightarrow\quad RC \;\le\; \frac{1+\mu\cos\omega_m t}{\mu\omega_m\sin\omega_m t}\ \ \text{for all } t \]

The right-hand side is minimised where its derivative vanishes, which happens at \(\cos\omega_m t = -\mu\). Substituting that value, \(1+\mu\cos\omega_m t = 1-\mu^2\) and \(\sin\omega_m t = \sqrt{1-\mu^2}\), so the minimum is \(\sqrt{1-\mu^2}/(\mu\omega_m)\). Combining with the lower limit gives the design rule.

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Envelope-detector design window
\[ \frac{1}{f_c} \;\ll\; RC \;\le\; \frac{\sqrt{1-\mu^2}}{\mu\,\omega_m} \]

Too small an \(RC\) and the output follows the individual carrier cycles, producing ripple; too large and the capacitor cannot track the envelope's fall, producing diagonal clipping. The window is wide precisely because \(f_c \gg f_m\), which is another way of saying that the whole scheme depends on a large carrier-to-message frequency ratio.

AM and DSB-SC have the same bandwidth. Adding a carrier does not widen the spectrum — the carrier is a single line at \(\omega_c\), inside the band the sidebands already occupy. Both schemes need \(2W\). AM spends power to save the receiver; SSB, next, spends receiver complexity to save bandwidth. There is no scheme that saves both, and recognising which resource is scarce is the whole of system design.
Section 28-5

Single Sideband and the Hilbert Transform

Return to the DSB-SC spectrum. For a real message, \(X(-j\omega) = X^*(j\omega)\), so the upper sideband and the lower sideband are conjugate reflections of each other about \(\omega_c\). Knowing one determines the other completely. Transmitting both is transmitting the same information twice, and the obvious repair is to filter one away and send only the other. That is single-sideband modulation, and it fits a message of bandwidth \(W\) into a channel of bandwidth \(W\).

Filtering it away is harder than it sounds. The two sidebands meet at \(\omega_c\) with no gap between them, so the filter must fall from passband to stopband across the region where the message has its lowest-frequency content. Speech is convenient here because it has almost nothing below 300 Hz, leaving a 600 Hz transition band; a signal with significant DC content cannot be sent by SSB at all. The alternative is to build the sideband directly, and doing so introduces an operator that is worth knowing in its own right.

Define the Hilbert transform \(\hat{x}(t)\) as the output of an all-pass filter that shifts every positive-frequency component by \(-90^\circ\) and every negative-frequency component by \(+90^\circ\).

The Hilbert transformer as a filter
\[ H_h(j\omega) = -j\,\operatorname{sgn}(\omega) = \begin{cases} -j, & \omega \gt 0\\ +j, & \omega \lt 0\end{cases} \qquad\Longleftrightarrow\qquad h_h(t) = \frac{1}{\pi t} \]

Its magnitude is 1 everywhere, so it changes no amplitudes — only phases. Now form the analytic signal \(x_+(t) = x(t) + j\hat{x}(t)\). Its spectrum is \(X(j\omega)\big[1 + \operatorname{sgn}(\omega)\big]\), which is \(2X(j\omega)\) for \(\omega \gt 0\) and zero for \(\omega \lt 0\): the negative-frequency half has been cancelled exactly. Shifting this one-sided spectrum up to \(\omega_c\) and taking the real part produces a signal occupying only the band above \(\omega_c\).

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Single-sideband by the phase-shift method
\[ s_{\text{SSB}}(t) = x(t)\cos\omega_c t \;\mp\; \hat{x}(t)\sin\omega_c t \]

The minus sign keeps the upper sideband, the plus sign the lower. No sharp filter is needed — only a wideband \(90^\circ\) phase shifter and two multipliers. The bandwidth is \(W\), half that of DSB-SC or AM, and the power efficiency is that of DSB-SC, since no carrier is sent.

SSB is recovered by coherent detection exactly as DSB-SC is, and it is if anything more demanding: a phase error in SSB does not merely attenuate the output, it phase-shifts every message component by the same angle, which speech tolerates but data does not, while a frequency error shifts every component by the same number of hertz, turning a voice into the familiar comical warble of a mistuned shortwave receiver.

Vestigial sideband is the engineering compromise between the two. One sideband is transmitted in full and a small vestige of the other is retained, with the filter's transition shaped so that the two overlapping vestiges add to a constant across the crossover region. The bandwidth is a little more than \(W\), the filter is realisable, and DC content survives — which is why analogue television, whose video signal runs down to DC, used VSB for its picture carrier.

SchemeTransmitted bandwidthCarrier sent?ReceiverTypical use
DSB-SC\(2W\)NoCoherentAnalogue multiplexing, QAM subcarriers
AM (DSB-LC)\(2W\)YesEnvelope detectorBroadcast radio, aviation voice
SSB\(W\)NoCoherentLong-haul telephony, amateur / marine HF
VSB\(\approx 1.25W\)ReducedEnvelope or coherentAnalogue television video
FM\(2(\Delta f + W)\)Constant envelopeDiscriminatorBroadcast FM, two-way radio
Section 28-6

Frequency-Division Multiplexing

The third motivation of Section 28-1 now pays off. Give each message its own carrier, spaced far enough apart that the translated spectra do not overlap, add the modulated signals together and transmit the sum. At the receiver a bandpass filter selects one band and a coherent or envelope detector brings it back to baseband. This is frequency-division multiplexing, and it is the organising principle of the entire radio spectrum.

f f₁ f₂ f₃ f₄ f₅ CH 1 CH 2 CH 3 CH 4 CH 5 guard receiver bandpass filter selected channel: 2W wide
Channels stacked in frequency, separated by guard bands

Guard bands exist because real filters are not brick walls. Chapter 15 showed that an ideal brick-wall filter is non-causal and therefore unbuildable; a practical filter rolls off over a finite transition, so adjacent channels must be spaced a little more than \(2W\) apart to keep the rolloff from eating into the neighbour. The classical telephone hierarchy did exactly this: twelve SSB voice channels, each nominally 4 kHz wide although the speech occupied only 3.4 kHz, stacked into a 48 kHz group.

Building a receiver whose bandpass filter can be retuned across a whole broadcast band is unpleasant — the centre frequency must move while the bandwidth stays fixed, so the fractional bandwidth changes as you tune. The superheterodyne receiver avoids the problem entirely. A tunable local oscillator first translates whatever station you want down to a fixed intermediate frequency, and all the selective filtering and most of the gain happen there, in a filter that never has to be retuned.

Superheterodyne translation and the image
\[ f_{\text{LO}} = f_{\text{RF}} + f_{\text{IF}} \qquad\Longrightarrow\qquad f_{\text{image}} = f_{\text{LO}} + f_{\text{IF}} = f_{\text{RF}} + 2f_{\text{IF}} \]

The image arises because a multiplier is blind to the sign of a frequency difference. Both \(f_{\text{LO}} - f_{\text{IF}}\) and \(f_{\text{LO}} + f_{\text{IF}}\) land on the intermediate frequency, so a station at either one reaches the IF filter. Only a coarse filter ahead of the mixer can reject the unwanted one, and since it need only separate two frequencies \(2f_{\text{IF}}\) apart it can be a broad, easily tuned circuit. This is why the AM broadcast IF of 455 kHz was chosen: it is small enough for good IF selectivity and large enough that a simple front end can reject an image sitting 910 kHz away.

Section 28-7

Angle Modulation

Every scheme so far has written the message into the amplitude of a carrier, which makes the message vulnerable to anything that disturbs amplitude — and almost every channel impairment does. The alternative is to hold the amplitude rigidly constant and write the message into the angle instead.

Angle modulation and instantaneous frequency
\[ s(t) = A\cos\theta(t), \qquad \omega_i(t) \triangleq \frac{d\theta}{dt} \]

The definition of instantaneous frequency as the derivative of the total angle is the key idea, and it is consistent with everything before it: for an unmodulated carrier \(\theta = \omega_c t\) and \(\omega_i = \omega_c\), as it should be. Two choices then present themselves. Make the phase deviation proportional to the message and you have phase modulation; make the frequency deviation proportional to it and you have frequency modulation.

PM and FM side by side
\[ s_{\text{PM}}(t) = A\cos\!\big(\omega_c t + k_p x(t)\big) \qquad\qquad s_{\text{FM}}(t) = A\cos\!\Big(\omega_c t + k_f\!\!\int_{-\infty}^{t}\!\! x(\tau)\,d\tau\Big) \]

The two are the same operation applied to different signals: FM of \(x(t)\) is PM of the integral of \(x(t)\), and PM of \(x(t)\) is FM of its derivative. A single modulator therefore serves both, with an integrator or differentiator in front — which is exactly how pre-emphasis and de-emphasis in broadcast FM are built.

Now the spectrum, and here angle modulation parts company with everything in this chapter so far. Take the single tone \(x(t) = A_m\cos\omega_m t\) into an FM modulator. Integrating gives a phase deviation \(\beta\sin\omega_m t\) where \(\beta = k_f A_m/\omega_m = \Delta\omega/\omega_m = \Delta f/f_m\), the modulation index, with \(\Delta f\) the peak frequency deviation. The transmitted signal is \(A\cos(\omega_c t + \beta\sin\omega_m t)\), and the message now sits inside a cosine of a cosine. The relation between message and signal is not linear, so the superposition arguments used everywhere else in this book simply do not apply, and the spectrum cannot be obtained by shifting anything.

Why the spectrum is infinite. Expanding \(\cos(\beta\sin\omega_m t)\) and \(\sin(\beta\sin\omega_m t)\) in their Fourier series — Chapter 10's machinery, applied to a periodic function of a periodic function — produces coefficients that are Bessel functions \(J_n(\beta)\). The result is a carrier plus sidebands at every multiple \(\omega_c \pm n\omega_m\), infinitely many of them. A single tone into an FM modulator produces an infinitely wide spectrum, which is a genuinely different situation from AM's tidy three lines.

Infinitely wide is not the same as unusable, because \(J_n(\beta)\) becomes negligible once \(n\) exceeds \(\beta\) by a comfortable margin. Counting the significant pairs of sidebands as roughly \(\beta + 1\), each spaced \(f_m\) apart, and doubling for both sides of the carrier gives the standard engineering estimate.

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Carson's rule
\[ B \;\approx\; 2(\beta + 1)f_m \;=\; 2(\Delta f + f_m) \;\longrightarrow\; 2(\Delta f + W) \text{ for a general message} \]

Two limits are contained in the one formula. For \(\beta \ll 1\) — narrowband FM — the bandwidth collapses to \(2f_m\), the same as AM. For \(\beta \gg 1\) — wideband FM — it approaches \(2\Delta f\), set by the deviation and essentially independent of the message bandwidth.

Broadcast FM uses \(\Delta f = 75\) kHz and \(W = 15\) kHz, giving \(\beta = 5\) and \(B = 180\) kHz — six times what AM would need for the same audio. That bandwidth is not wasted. It buys a signal-to-noise improvement that grows as \(\beta^2\), which is why FM sounds better than AM despite using less transmitter power, and it is the first clear example in this book of bandwidth being deliberately traded for noise performance. Making that trade precise requires a description of noise itself, and that is Chapter 29.

Narrowband FM deserves one more line because it can be built from the linear tools of this chapter. For \(\beta \ll 1\), \(\cos(\beta\sin\omega_m t)\approx 1\) and \(\sin(\beta\sin\omega_m t)\approx\beta\sin\omega_m t\), so \(s(t) \approx A\cos\omega_c t - A\beta\sin\omega_m t\sin\omega_c t\). Compare this with the AM expansion of Section 28-4: the same three lines, the same spacing, but the sidebands enter in quadrature with the carrier rather than in phase with it. That single change of phase is the entire difference between a signal whose amplitude varies and one whose angle does.

Section 28-8

Sampling, PAM and Time-Division Multiplexing

Everything so far has been continuous in time. Chapter 20 proved that a signal bandlimited to \(W\) hertz is completely determined by samples taken at \(2W\) per second, and Chapter 21 showed how to rebuild it. That theorem has a consequence for communications that is easy to state and enormous in effect: between the samples, the channel is idle, and the idle time can be sold to somebody else.

The simplest pulse scheme is pulse-amplitude modulation. The message is sampled and each sample sets the height of a short pulse. Chapter 20 treated ideal impulse sampling; a real system uses pulses of finite width \(\tau\), which is the difference between impulse sampling and what is usually called flat-top sampling. The distinction matters in the frequency domain.

Flat-top PAM as impulse sampling followed by a hold
\[ s_{\text{PAM}}(t) = \Big[\textstyle\sum_n x(nT)\,\delta(t-nT)\Big] * p(t) \qquad\Longleftrightarrow\qquad S(j\omega) = \frac{1}{T}\sum_n X\big(j(\omega - n\omega_s)\big)\,P(j\omega) \]

A flat-top pulse train is impulse sampling convolved with the pulse shape, so its spectrum is the sampled spectrum multiplied by \(P(j\omega)\), which for a rectangular pulse of width \(\tau\) is \(\tau\operatorname{sinc}(\omega\tau/2\pi)\). Within the baseband copy this sinc is not flat, so the recovered message is slightly tilted in frequency — the aperture effect. It is a fixed, known distortion and is undone by an equaliser with the reciprocal response, exactly the equaliser Chapter 22 used to correct a zero-order hold.

With the channel occupied only for \(\tau\) out of every \(T\) seconds, interleaving is immediate. Sample message 1, then message 2, then message 3, then return to message 1 before its next sample is due. This is time-division multiplexing, the time-domain twin of Section 28-6's frequency-domain stacking, and the two are related by exactly the duality Chapter 14 established between the domains.

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TDM and PCM rates
\[ f_{\text{TDM}} = N f_s \ \ \text{pulses/s} \qquad\qquad R_b = N f_s b \ \ \text{bits/s} \]

\(N\) messages each sampled at \(f_s\) fill the channel with \(Nf_s\) pulses per second. Quantising each sample to \(b\) bits — pulse-code modulation — turns the analogue pulse train into a bit stream at \(Nf_sb\) bits per second, and a binary stream at \(R_b\) bits per second needs a minimum Nyquist bandwidth of \(R_b/2\) hertz.

Where the subject went. Once the message is a bit stream, the carrier no longer has to carry a waveform — it only has to carry symbols, and the receiver only has to decide which of a finite set was sent. Amplitude, phase and frequency modulation reappear as ASK, PSK and FSK; the quadrature pair of Section 28-3 becomes the QAM constellation; and Chapter 26's FFT, used to modulate thousands of subcarriers at once, becomes OFDM. Every one of them is built from the two ideas of this chapter — shift a spectrum, and sample a bandlimited signal.
Section 28-9

The Complex Envelope

One representation ties the chapter together and is the form in which communication signals are actually handled in software. Every scheme above produces a passband signal: real, and with its energy confined to a narrow band around \(\omega_c\). Any such signal can be written with the carrier factored out.

Passband signal, complex envelope, and I/Q parts
\[ s(t) = \operatorname{Re}\big\{\tilde{s}(t)\,e^{j\omega_c t}\big\}, \qquad \tilde{s}(t) = s_I(t) + j s_Q(t) \;\Longrightarrow\; s(t) = s_I(t)\cos\omega_c t - s_Q(t)\sin\omega_c t \]

The complex envelope \(\tilde{s}(t)\) is a lowpass signal of bandwidth \(W\) that carries everything the passband signal carries. Its real part \(s_I\) is the in-phase component, its imaginary part \(s_Q\) the quadrature component, and the modulation schemes of this chapter are now distinguished by nothing more than the shape of \(\tilde{s}\).

SchemeComplex envelope \(\tilde{s}(t)\)What varies
DSB-SC\(x(t)\)Real amplitude only, sign included
AM\(A + x(t)\)Real amplitude, held positive
SSB (upper)\(x(t) + j\hat{x}(t)\)Both parts, in Hilbert pairing
PM / FM\(A e^{j\phi(t)}\)Angle only — \(|\tilde{s}|\) is constant
QAM\(x_1(t) + j x_2(t)\)Two independent messages

Two practical consequences follow. First, the envelope and the instantaneous phase of any passband signal are \(|\tilde{s}(t)|\) and \(\arg\tilde{s}(t)\), so an envelope detector computes a modulus and an FM discriminator computes the derivative of an argument — the same two operations for every scheme. Second, and more useful, a passband system can be simulated entirely at baseband. If a bandpass channel has impulse response \(h(t)\) with complex envelope \(\tilde{h}(t)\), the passband convolution \(s * h\) corresponds to \(\tfrac12\tilde{s} * \tilde{h}\) at baseband. A radio at 2.4 GHz can therefore be simulated with a sampling rate set by \(W\), not by \(\omega_c\) — a saving of six orders of magnitude, and the reason software-defined radio is possible at all.

This also closes the loop with Chapter 22. Sampling the complex envelope at a rate above \(2W\) rather than above \(2(f_c+W)\) is legitimate for exactly the reason bandpass sampling is legitimate: what matters is the width of the occupied band, not its distance from the origin. The I/Q pair is bandpass sampling with the down-conversion done in hardware first.

Section 28-10

Worked Examples

1 Spectrum and bandwidth of a DSB-SC signal

Problem. A message \(x(t) = 2\cos(2\pi\cdot 1000\,t) + \cos(2\pi\cdot 3000\,t)\) modulates a carrier \(\cos(2\pi\cdot 10^4 t)\) in a DSB-SC modulator. List the transmitted frequency components with their amplitudes, and state the transmission bandwidth.

Solution. Multiply out, using \(\cos A\cos B = \tfrac12\cos(A-B)+\tfrac12\cos(A+B)\) on each term separately.

Working
\[ s(t) = \underbrace{\cos 2\pi(9000)t + \cos 2\pi(11000)t}_{\text{from the }2\cos\text{ term}} \;+\; \underbrace{\tfrac12\cos 2\pi(7000)t + \tfrac12\cos 2\pi(13000)t}_{\text{from the }\cos\text{ term}} \]

Four lines: amplitude 1 at 9 kHz and 11 kHz, amplitude 0.5 at 7 kHz and 13 kHz. Each message tone of amplitude \(a\) has become a pair of lines of amplitude \(a/2\), placed symmetrically about the carrier.

The message bandwidth is \(W = 3\) kHz, set by its highest component, so \(B = 2W = 6\) kHz. Confirm it directly from the line positions: the occupied band runs from 7 kHz to 13 kHz, a width of 6 kHz. There is no line at 10 kHz — the carrier really is suppressed.

2 A detector with phase and frequency error

Problem. The DSB-SC signal \(s(t)=x(t)\cos\omega_c t\) is demodulated with a local oscillator \(\cos(\omega_c t + \theta)\). (a) By how many decibels does the recovered message fall if \(\theta = 30^\circ\)? (b) The oscillator instead runs 20 Hz high. Describe the output, and give the period of the disturbance.

Solution (a). From Section 28-3 the output after lowpass filtering and a gain of two is \(x(t)\cos\theta\). With \(\theta = 30^\circ\), \(\cos\theta = \sqrt{3}/2 = 0.866\), so the amplitude ratio is 0.866 and

Working — part (a)
\[ 20\log_{10}(0.866) = -1.25 \ \text{dB} \]

A modest loss, and — this is the important part — no distortion whatever, since every frequency component is scaled by the same real number. Push \(\theta\) to \(90^\circ\) and the output disappears completely.

Solution (b). With the oscillator at \(\omega_c + \Delta\omega\) the product contains \(\tfrac12 x(t)\cos(\Delta\omega t + \theta)\) plus a term near \(2\omega_c\) which the filter removes. The recovered signal is therefore \(x(t)\cos(2\pi\cdot 20\,t)\): the message multiplied by a 20 Hz tone. It swells and fades twice per cycle of that tone and reverses sign in between, so the audible fading has period \(1/(2\times 20) = 25\) ms. This is not an attenuation that can be turned up — the message has been re-modulated, and no amount of gain repairs it. Frequency accuracy in a coherent receiver is not optional.

3 Modulation index and power efficiency

Problem. An AM transmitter produces \(s(t) = 10\big[1 + 0.5\cos(2\pi\cdot 10^3 t)\big]\cos(2\pi\cdot 10^6 t)\) volts into 1 Ω. Find the modulation index, the carrier and sideband powers, the total power, and the efficiency. Then find the total power if \(\mu\) is raised to 1.

Solution. Comparing with \(A[1+\mu m(t)]\cos\omega_c t\) gives \(A = 10\) V and \(\mu = 0.5\). Expanding into three lines:

Working — the three transmitted lines
\[ s(t) = 10\cos\omega_c t + 2.5\cos(\omega_c+\omega_m)t + 2.5\cos(\omega_c-\omega_m)t \]

since \(A\mu/2 = 10(0.5)/2 = 2.5\) V. Each sinusoid of amplitude \(V\) delivers \(V^2/2\) watts into 1 Ω:

Working — powers
\[ P_c = \frac{10^2}{2} = 50\ \text{W}, \qquad P_{\text{SB}} = 2\times\frac{2.5^2}{2} = 6.25\ \text{W}, \qquad P_T = 56.25\ \text{W} \]

The efficiency is \(6.25/56.25 = 0.111\), or \(11.1\%\). Checking against the formula: \(\mu^2/(2+\mu^2) = 0.25/2.25 = 0.111\) — agreement.

At \(\mu = 1\) the carrier power is unchanged at 50 W, since the carrier amplitude does not depend on \(\mu\), while the sideband power rises to \(A^2\mu^2/4 = 25\) W, giving \(P_T = 75\) W and \(\eta = 25/75 = 33.3\%\). Doubling the modulation index has quadrupled the useful power while adding nothing to the wasted part — which is precisely why broadcasters modulate as deeply as the overmodulation limit allows.

4 Designing an envelope detector

Problem. An AM receiver handles a 1 MHz carrier modulated at \(\mu = 0.8\) by audio whose highest component is 5 kHz. Choose \(RC\). What goes wrong if the same receiver meets a signal at \(\mu = 0.95\)?

Solution. The lower bound requires the capacitor to hold its charge between carrier peaks: \(RC \gg 1/f_c = 1\ \mu\text{s}\). The upper bound is the tracking condition derived in Section 28-4, evaluated at the worst case, which is the highest message frequency:

Working — the tracking limit
\[ RC \le \frac{\sqrt{1-\mu^2}}{\mu\,\omega_m} = \frac{\sqrt{1-0.64}}{0.8\times 2\pi\times 5000} = \frac{0.600}{25133} = 23.9\ \mu\text{s} \]

Any value comfortably inside \(1\ \mu\text{s} \ll RC \le 23.9\ \mu\text{s}\) will do; \(RC = 20\ \mu\text{s}\) is a sound choice, being twenty carrier periods long and a tenth of the shortest audio period.

At \(\mu = 0.95\) the numerator falls sharply: \(\sqrt{1-0.9025} = 0.312\), and \(\mu\omega_m = 29845\), so the limit drops to \(10.5\ \mu\text{s}\). The fixed 20 μs network now violates it, and the capacitor cannot follow the envelope down through its steepest fall. The output flattens into a straight discharge line on the deepest troughs — diagonal clipping — which is heard as distortion on loud passages only. The lesson generalises: the detector must be designed for the deepest modulation it will ever see, not the typical one.

5 FM deviation, index and Carson bandwidth

Problem. An FM signal is \(s(t) = 10\cos\!\big(2\pi\cdot 10^8 t + 6\sin(2\pi\cdot 10^3 t)\big)\). Find the instantaneous frequency, the peak deviation, the modulation index and the Carson bandwidth. Compare with a broadcast FM channel at \(\Delta f = 75\) kHz, \(W = 15\) kHz, and with AM carrying the same audio.

Solution. Differentiate the total angle and divide by \(2\pi\):

Working — instantaneous frequency
\[ f_i(t) = \frac{1}{2\pi}\frac{d\theta}{dt} = 10^8 + \frac{1}{2\pi}\Big(6\cdot 2\pi\cdot 10^3\cos(2\pi\cdot10^3 t)\Big) = 10^8 + 6000\cos(2\pi\cdot 10^3 t)\ \text{Hz} \]

So \(\Delta f = 6\) kHz, \(f_m = 1\) kHz, and \(\beta = \Delta f/f_m = 6\) — which could also have been read straight off the phase term, since the coefficient of the sine is \(\beta\). Carson's rule gives

Working — bandwidth
\[ B = 2(\Delta f + f_m) = 2(6000 + 1000) = 14\ \text{kHz} \]

Fourteen kilohertz to carry a single 1 kHz tone, where DSB-SC would have used 2 kHz.

For the broadcast channel, \(\beta = 75/15 = 5\) and \(B = 2(75+15) = 180\) kHz. AM carrying the same 15 kHz audio would occupy \(2W = 30\) kHz. FM therefore spends six times the bandwidth, and receives in exchange a noise advantage that grows with \(\beta^2\) — six times the bandwidth for roughly twenty-five times the output signal-to-noise ratio, once Chapter 29 supplies the tools to say what that means.

6 A time-division multiplexed PCM link

Problem. Twenty-four telephone channels, each bandlimited to 3.4 kHz, are sampled at 8 kHz, quantised to 8 bits and time-division multiplexed, with one extra framing bit added per frame. Find the frame duration, the bit rate, the duration of one bit, and the minimum transmission bandwidth for binary signalling.

Solution. Sampling at 8 kHz against a 3.4 kHz message leaves a guard of \(8 - 2(3.4) = 1.2\) kHz for the anti-alias filter's transition band — Chapter 21's practical margin, not a violation of Nyquist. One frame carries one sample from each channel, so it repeats at the sampling rate:

Working — frame and rate
\[ T_f = \frac{1}{8000} = 125\ \mu\text{s}, \qquad N_b = 24\times 8 + 1 = 193\ \text{bits/frame} \]
Working — bit rate and bandwidth
\[ R_b = 193 \times 8000 = 1.544\ \text{Mbit/s}, \qquad T_b = \frac{125\ \mu\text{s}}{193} = 0.648\ \mu\text{s}, \qquad B_{\min} = \frac{R_b}{2} = 772\ \text{kHz} \]

These are the numbers of the T1 carrier, in service since 1962. The minimum bandwidth of \(R_b/2\) is the sampling theorem read backwards: a channel of bandwidth \(B\) supports \(2B\) independent pulse amplitudes per second, so \(2B \ge R_b\). Compare the alternative — twenty-four SSB channels at 4 kHz each would occupy 96 kHz, one eighth of the bandwidth. Digital transmission costs bandwidth and buys immunity to accumulated noise, since a regenerator can restore a bit exactly while an amplifier can only make an impaired analogue signal louder.

Review

Chapter Summary

One property, many schemes

Multiplying by \(\cos\omega_c t\) copies the spectrum to \(\pm\omega_c\) at half height. DSB-SC, AM, SSB and FDM are all consequences of that single line.

Coherent detection

Multiply again and lowpass. A phase error costs \(\cos\theta\) and nulls at \(90^\circ\); a frequency error re-modulates the message and cannot be repaired by gain.

AM's bargain

Adding a carrier permits an envelope detector provided \(\mu \le 1\), at a cost of efficiency \(\mu^2/(2+\mu^2)\), never better than one third.

SSB and the Hilbert transform

\(x\cos\omega_c t \mp \hat{x}\sin\omega_c t\) transmits one sideband in bandwidth \(W\), using a \(90^\circ\) phase shifter in place of an unbuildable filter.

Angle modulation

\(\omega_i = d\theta/dt\). FM is nonlinear in the message, has infinitely many Bessel sidebands, and needs \(B \approx 2(\Delta f + W)\).

Multiplexing both ways

FDM stacks messages in frequency with guard bands; TDM interleaves them in time between samples. Duality relates the two exactly.

The complex envelope

\(s = \operatorname{Re}\{\tilde{s}e^{j\omega_c t}\}\) reduces every scheme to the shape of a lowpass \(I + jQ\) signal, and lets a passband system be simulated at baseband.

The recurring trade

Power, bandwidth and receiver complexity can be exchanged but not all reduced together. Which one is scarce decides the scheme.

Practice

Problems

Problems 1 to 3 exercise the spectra of the amplitude schemes; 4 and 5 concern detection; 6 and 7 treat angle modulation; 8 is a multiplexing budget. Sketch the spectrum before computing anything — most errors in this material are geometric, not algebraic.

  1. A message with the triangular spectrum \(X(j\omega) = 1 - |\omega|/(2\pi\cdot 4000)\) for \(|\omega| \le 2\pi\cdot 4000\), zero elsewhere, modulates a 100 kHz carrier in a DSB-SC modulator. Sketch the transmitted spectrum with numerical frequency and amplitude labels, and state the occupied band.
  2. The AM signal \(s(t) = \big[8 + 5\cos(2\pi\cdot 500\,t)\big]\cos(2\pi\cdot 5\times10^5 t)\) is transmitted into 50 Ω. Find \(\mu\), the amplitude of each sideband line, the total power and the efficiency. Would an envelope detector recover the message undistorted?
  3. Repeat Problem 2 for \(s(t) = \big[3 + 5\cos(2\pi\cdot 500\,t)\big]\cos\omega_c t\). Sketch the envelope \(|3+5\cos|\) over one message period and explain, from the sketch, what an envelope detector would produce.
  4. A DSB-SC receiver has a local oscillator with a slow phase drift \(\theta(t) = 0.1\pi\sin(2\pi\cdot 0.5\,t)\). Write the demodulated output and describe what a listener hears. Why would a Costas loop remove the problem?
  5. Show that if the SSB signal \(x(t)\cos\omega_c t - \hat{x}(t)\sin\omega_c t\) is demodulated with \(\cos(\omega_c t + \theta)\) and lowpass filtered, the output is \(\tfrac12[x(t)\cos\theta + \hat{x}(t)\sin\theta]\). Explain why this is a phase distortion rather than an attenuation, and why speech survives it.
  6. A 90 MHz carrier is frequency modulated by \(x(t) = 4\cos(2\pi\cdot 2000\,t)\) with \(k_f/2\pi = 3000\) Hz per volt. Find \(\Delta f\), \(\beta\) and the Carson bandwidth. Then find the same three quantities if the modulator is a phase modulator with \(k_p = 3\) rad/V, and explain why the two answers scale differently when \(f_m\) is doubled.
  7. Show that narrowband FM and AM produce sidebands at the same frequencies with the same amplitudes, and that the two differ only in the phase of the sidebands relative to the carrier. Sketch both as phasor diagrams and use the sketch to explain why one has a constant envelope and the other does not.
  8. Thirty channels, each bandlimited to 3.4 kHz, are to share one link. (a) Using SSB and FDM with 4 kHz spacing, find the total bandwidth. (b) Using PCM and TDM at 8 kHz sampling with 8-bit quantisation, find the bit rate and the minimum binary transmission bandwidth. (c) The 8-bit quantiser is replaced by a 12-bit one. By what factor does the bandwidth grow, and what is bought in return?
Tip: when a modulation problem resists, draw the spectrum. Almost every result in this chapter — the bandwidth of DSB, the necessity of \(\omega_c \gt W\), the origin of the image frequency, the reason SSB filtering is hard, the aperture effect — is visible immediately in a picture of where the copies of \(X(j\omega)\) land, and invisible in the algebra.