Communication Systems: Modulation, Bandwidth & the Carrier

Every wireless link you use is modulation at work. Your phone modulates data onto microwave carriers near 700 MHz to a few GHz; Wi-Fi rides carriers at 2.4 and 5 GHz; GPS, Bluetooth, garage remotes, and radar all stamp information onto a carrier and read it back off. Modern systems mostly use digital modulation — schemes like QAM that pack bits by juggling amplitude and phase together — but the physics here is exactly the foundation: a carrier, a message, sidebands, and a bandwidth set by how fast the message changes. The relentless demand for bandwidth (streaming video needs far more than Marconi's dots) is why engineers keep climbing to higher carrier frequencies, where more spectrum is available — the same $\lambda = c/f$ logic that first drove us off the audio band. The next step is to look more closely at the electromagnetic waves all of this rides on — their generation, speed, and nature.

Picture three waves stacked one above another. At the bottom is the message (or modulating, or baseband) signal — the thing you actually want to send. For speech and music it is a jumble of frequencies from about 20 Hz up to 15 or 20 kHz; to keep the arithmetic clean we model it as a single pure tone $f_m$, say a 5 kHz whistle. In the middle is the carrier wave: a clean, high-frequency sinusoid, perhaps $f_c = 1\ \text{MHz}$ for a medium-wave station, whose only job is to be a fast, radiatable vehicle — on its own it carries no information, a featureless hum. At the top is the modulated wave, the carrier after the message has been stamped onto it, and it is this composite that leaves the antenna. A sinusoidal carrier is fully described by three quantities — amplitude, frequency, and phase — so you can encode the message by wobbling any one of them in step. Wobble the amplitude and you have amplitude modulation (AM): the carrier keeps its steady frequency but its height swells and shrinks, tracing the shape of the message as an envelope. Wobble the frequency and you have frequency modulation (FM): the carrier keeps a constant height but bunches up when the message is high and spreads out when it is low. (Wobbling the phase gives phase modulation, a cousin of FM used in digital systems, which we mention only in passing.)

Whichever route you take, one idea is central: the information no longer lives in a slow wave, it lives as a variation of a fast wave. That is what makes it radiatable, shareable, and — as Marconi found — able to cross an ocean.

Why can you not simply feed a microphone's output straight into an antenna? Three reasons, each decisive on its own.

Antenna length. An antenna radiates efficiently only when its length is comparable to the wavelength — a good rule of thumb is a quarter-wavelength. Wavelength and frequency are tied by the wave relation

where: $c = 3.0\times10^{8}\ \text{m/s}$ is the wave speed (light) and $f$ its frequency. A 15 kHz audio tone has $\lambda = (3.0\times10^{8})/(15\times10^{3}) = 20\,000\ \text{m} = 20\ \text{km}$, needing an aerial some $5\ \text{km}$ tall — absurd. Shift it onto a $1\ \text{MHz}$ carrier and $\lambda = (3.0\times10^{8})/(10^{6}) = 300\ \text{m}$, a quarter of which is a buildable $75\ \text{m}$ mast. Push to a $100\ \text{MHz}$ FM carrier and $\lambda = 3\ \text{m}$, so a $0.75\ \text{m}$ rod suffices — which is why your phone's aerial is a few centimetres, not a few kilometres.

Multiplexing. If every station broadcast its raw audio, they would all pile into the same 20 Hz–20 kHz band and drown one another. Give each station its own carrier — one at 990 kHz, the next at 1000 kHz — and their signals sit in separate slices of the spectrum, tuned in one at a time. This sharing of the spectrum by frequency is called frequency-division multiplexing, possible only because modulation lifts each message to a different carrier.

Range and power. Low-frequency waves are radiated feebly and quickly absorbed; a high-frequency carrier, radiated from an efficient antenna, carries far and (via the ionosphere or line-of-sight) reaches over the horizon. The carrier is the workhorse; the message is merely its passenger.

Now build an AM wave. Let the carrier be $c(t) = A_c\cos(2\pi f_c t)$ and the message a single tone $m(t) = A_m\cos(2\pi f_m t)$. Amplitude modulation lets the carrier's amplitude rise and fall by the message, so its instantaneous amplitude becomes $A_c + A_m\cos(2\pi f_m t)$. The transmitted wave is therefore

where: $A_c$ and $A_m$ are the carrier and message amplitudes, $f_c$ the carrier frequency and $f_m$ the message frequency. The physical reading: the bracket is the envelope, and it traces exactly the shape of the message — which is why a simple envelope detector (a diode and a capacitor) can pull the message back out at the receiver. Multiply the bracket through and something remarkable happens. The product $\cos(2\pi f_m t)\cos(2\pi f_c t)$ is not a single frequency; the product-to-sum identity turns it into two. The full expansion (done step by step in the derivation below) is

where: the three cosines sit at three distinct frequencies — the carrier $f_c$, an upper sideband at $f_c+f_m$, and a lower sideband at $f_c-f_m$. The message frequency has vanished as a standalone term; the information now lives entirely in the two sidebands straddling the carrier. This is the heart of communication theory in one line: modulation moves the message up to sidebands around the carrier. Real speech spanning up to $f_m$ smears each single sideband into a band of width $f_m$ on either side. The consequence is the most examined number in the topic. The signal spans from the lower edge $f_c-f_m$ to the upper edge $f_c+f_m$, so the bandwidth — the width of spectrum occupied — is

where: $f_m$ is the highest frequency in the message. The bandwidth does not depend on the carrier at all — only on how fast the message wiggles. A $1\ \text{MHz}$ carrier bearing a $5\ \text{kHz}$ tone spreads from $995\ \text{kHz}$ to $1005\ \text{kHz}$: a $10\ \text{kHz}$ slot. This is why medium-wave stations are spaced about $10\ \text{kHz}$ apart, and why AM audio is limited to about $5\ \text{kHz}$ — narrow bandwidth to fit more stations, at the cost of tinny, treble-starved sound.

Equation (2) hides one more crucial number. How strongly is the carrier modulated — does its amplitude swing gently or violently? That depth is the modulation index (or modulation factor)

where: $A_m$ is the message amplitude and $A_c$ the carrier amplitude. If $A_m = 6\ \text{V}$ rides on $A_c = 10\ \text{V}$, then $\mu = 0.6$, usually quoted as $60\%$ modulation. You can read $\mu$ straight off an oscilloscope trace of the AM wave, because the envelope swings between a maximum $A_{\max} = A_c + A_m$ and a minimum $A_{\min} = A_c - A_m$, which rearranges to a second, equivalent formula:

where: $A_{\max}$ and $A_{\min}$ are the peak and trough of the envelope. For our example $A_{\max} = 16\ \text{V}$, $A_{\min} = 4\ \text{V}$, and $(16-4)/(16+4) = 12/20 = 0.6$ — the same $\mu$, now measurable from a screen without knowing $A_c$ and $A_m$ separately. The rule that matters: keep $\mu\le 1$. At $\mu = 1$ (100% modulation) the envelope just kisses zero at its troughs — the message is impressed as deeply as possible while the envelope still traces its shape. Push to $\mu>1$ — over-modulation — and the demanded envelope would go negative, which a real amplitude cannot do: it clips and folds, the audio is grossly distorted, and the sharp corners splatter energy into neighbouring channels. Under-modulation is safe but wasteful — the sidebands, which carry all the information, are then feeble compared with the carrier, which carries none.

Amplitude modulation has a fatal weakness: noise is amplitude. Lightning, sparking motors, and every electrical crackle add to the height of the wave — and since AM reads the message off the height, that crackle comes straight through as static, the hiss and pop of AM radio in a thunderstorm. Frequency modulation sidesteps it. In FM the message is written into the carrier's frequency, and the receiver is built to ignore amplitude (it "limits" the signal to a constant height before decoding), so amplitude noise is simply thrown away and FM arrives clean — which is why music radio migrated to FM. The price is bandwidth. An FM signal produces a whole ladder of sidebands, not a tidy two, and its practical channel width is estimated by Carson's rule (quoted here — its derivation needs the frequency spectrum of an FM wave, beyond this article):

where: $\Delta f$ is the peak frequency deviation (how far the carrier swings from centre) and $f_m$ the top message frequency. For broadcast FM, $\Delta f = 75\ \text{kHz}$ and $f_m = 15\ \text{kHz}$ give $\text{BW}_{\text{FM}} \approx 2(75+15) = 180\ \text{kHz}$ — which is why FM stations get $200\ \text{kHz}$ channels, twenty times wider than an AM slot. FM buys fidelity and noise immunity with spectrum; AM is frugal but fragile. Engineers quote signal quality as the signal-to-noise ratio (SNR), the ratio of wanted signal power to noise power, almost always in decibels, $\text{SNR(dB)} = 10\log_{10}(P_{\text{signal}}/P_{\text{noise}})$. A signal a thousand times stronger than the noise is $10\log_{10}(1000) = 30\ \text{dB}$; a hundred times stronger is $20\ \text{dB}$. Higher SNR means a cleaner message, and FM's amplitude-blindness is exactly what gives it a better SNR than AM for the same power in a noisy channel. Finally, how the message gets home. At the receiver the wanted carrier is selected by tuning (an $LC$ circuit resonant at $f_c$), amplified, and then demodulated (or detected) — the reverse of modulation, stripping the carrier away to leave the message. For AM this is a diode-and-capacitor envelope detector that follows the envelope; for FM a discriminator converts frequency swings back into an amplitude. The carrier reaches the receiver by one of three propagation modes, chosen by frequency: ground waves hug the Earth's surface (below about 2 MHz), sky waves bounce off the ionosphere (short wave, roughly 3–30 MHz — the mode Marconi unknowingly exploited), and space waves travel line-of-sight to the horizon (VHF and above — FM, TV, mobile), needing tall antennas or repeaters.

Board level — sidebands, bandwidth and the modulation index (ICSE/ISC pattern)

A medium-wave transmitter uses a carrier of frequency $f_c = 1\ \text{MHz}$ and amplitude $A_c = 10\ \text{V}$. It is amplitude-modulated by a single audio tone of frequency $f_m = 5\ \text{kHz}$ and amplitude $A_m = 6\ \text{V}$. Find (a) the sideband frequencies, (b) the bandwidth, and (c) the modulation index.

Step 1. The sidebands sit at $f_c\pm f_m$: the lower at $1000 - 5 = 995\ \text{kHz}$ and the upper at $1000 + 5 = 1005\ \text{kHz}$. The carrier itself stays at $1000\ \text{kHz}$.

Step 2. The bandwidth is the full span occupied, $\text{BW} = 2f_m = 2\times 5 = 10\ \text{kHz}$ (equivalently $1005 - 995 = 10\ \text{kHz}$).

Step 3. The modulation index is $\mu = A_m/A_c = 6/10 = 0.6$, i.e. $60\%$ modulation — safely below the $\mu = 1$ ceiling. Answers: sidebands $995\ \text{kHz}$ and $1005\ \text{kHz}$; bandwidth $10\ \text{kHz}$; $\mu = 0.6$.

Board level — reading the modulation index off an oscilloscope (ICSE/A-Level pattern)

An AM waveform on an oscilloscope has an envelope that swings between a maximum height of $16\ \text{V}$ and a minimum of $4\ \text{V}$. Find the modulation index, and state the carrier and message amplitudes.

Step 1. Use the envelope form of the index: $\mu = (A_{\max}-A_{\min})/(A_{\max}+A_{\min}) = (16-4)/(16+4) = 12/20 = 0.6$.

Step 2. Recover the amplitudes from $A_{\max} = A_c + A_m$ and $A_{\min} = A_c - A_m$: adding gives $2A_c = 20$, so $A_c = 10\ \text{V}$; subtracting gives $2A_m = 12$, so $A_m = 6\ \text{V}$. As a check, $A_m/A_c = 6/10 = 0.6$, matching. Answer: $\mu = 0.6$, with $A_c = 10\ \text{V}$ and $A_m = 6\ \text{V}$.

Competitive level — Carson's rule and why FM eats spectrum (A-Level pattern)

A broadcast FM station has a peak frequency deviation $\Delta f = 75\ \text{kHz}$ and carries audio up to $f_m = 15\ \text{kHz}$. (a) Estimate its bandwidth by Carson's rule. (b) Compare with the AM bandwidth for the same $15\ \text{kHz}$ audio, and comment on why FM stations are given $200\ \text{kHz}$ channels.

Step 1. Carson's rule, equation (7): $\text{BW}_{\text{FM}} \approx 2(\Delta f + f_m) = 2(75 + 15) = 2\times 90 = 180\ \text{kHz}$.

Step 2. For AM, the same $15\ \text{kHz}$ audio would need only $\text{BW}_{\text{AM}} = 2f_m = 2\times 15 = 30\ \text{kHz}$. FM is therefore $180/30 = 6$ times wider for this deviation.

Step 3. Comment: the $180\ \text{kHz}$ estimate fits comfortably inside the standard $200\ \text{kHz}$ FM channel, leaving guard bands to prevent adjacent stations bleeding into one another. FM spends this extra spectrum to win noise immunity — the amplitude-borne static of AM is discarded. Answers: $\text{BW}_{\text{FM}} \approx 180\ \text{kHz}$, versus $30\ \text{kHz}$ for AM; the extra width is the price of FM's freedom from static. Timing note: about a minute. Carson's rule is a plug-in — the only trap is remembering it is $2(\Delta f + f_m)$, not $2\Delta f$ or $2f_m$. Do the AM comparison as $2f_m$ in your head and the factor-of-six falls out instantly.

Competitive level — antenna length and line-of-sight range (A-Level pattern)

(a) Estimate the quarter-wave antenna length for a $100\ \text{MHz}$ FM carrier. (b) A VHF space-wave transmitter sits on a mast of height $h = 100\ \text{m}$. Taking the Earth's radius as $R = 6.4\times10^{6}\ \text{m}$, estimate its line-of-sight range from the quoted formula $d = \sqrt{2Rh}$.

Step 1. Wavelength of the carrier: $\lambda = c/f = (3.0\times10^{8})/(100\times10^{6}) = 3.0\ \text{m}$, so a quarter-wave antenna is $\lambda/4 = 0.75\ \text{m}$ — a rod under a metre, exactly why FM whip aerials are short.

Step 2. Line-of-sight range: $d = \sqrt{2Rh} = \sqrt{2\times(6.4\times10^{6})\times 100} = \sqrt{1.28\times10^{9}} = 3.58\times10^{4}\ \text{m} \approx 36\ \text{km}$.

Step 3. Interpret: a $100\ \text{m}$ VHF mast reaches only about $36\ \text{km}$ to the horizon — which is why FM and TV need many transmitters or tall towers, whereas Marconi's low-frequency sky-wave signal bounced off the ionosphere and skipped clean across the Atlantic. Answers: antenna $\approx 0.75\ \text{m}$; range $\approx 36\ \text{km}$. Timing note: seconds for part (a) once $\lambda = c/f$ is automatic. For part (b), $2Rh = 1.28\times10^{9}$ and its root is $3.6\times10^{4}$ — estimate $\sqrt{1.28}\approx 1.13$ and shift the exponent rather than reaching for a calculator.

Saying AM changes the carrier frequency. In amplitude modulation the frequency is fixed and only the amplitude varies. Frequency changes belong to FM. This wording error is the most common single-mark loss on the topic in ICSE/ISC and A-Level.

Getting the bandwidth wrong. For AM it is $2f_m$ — twice the highest message frequency, never anything to do with the carrier. Students who write $2f_c$ or $f_c\pm f_m$ as the bandwidth lose the mark. For FM use Carson's rule $2(\Delta f + f_m)$, not $2\Delta f$.

Forgetting the $\mu\le 1$ ceiling. A modulation index above 1 means over-modulation, envelope clipping and distortion — a frequent short-answer question. Quoting $\mu>1$ as if it were fine, or confusing the two formulae for $\mu$, are classic errors.

Believing the message travels as its own wave. After modulation the baseband frequency $f_m$ is not radiated; the information lives in the sidebands at $f_c\pm f_m$. The message reappears only after demodulation at the receiver.

Board scope — read this carefully. Communication systems were removed from the CBSE Class 12 syllabus in 2020–21 (they were the old Chapter 15), so a CBSE board candidate will not be examined on this at all. The topic is, however, live on ICSE/ISC and on several A-Level specifications. The quantitative sideband, bandwidth and modulation-index work here is squarely ICSE/ISC and A-Level; treat the whole chapter as enrichment if you sit CBSE, JEE or NEET, where it no longer appears. Do not cite CBSE, JEE, NEET or AP Physics for this topic.

Texts. ISC Physics (Class 12), the Communication / Electronics chapter · S. L. Arora, Physics for Class XII, "Communication Systems" (still a clear treatment even though CBSE dropped it) · A-Level: Cambridge International AS & A Level Physics · B. P. Lathi, Modern Digital and Analog Communication Systems, for the proper treatment of AM/FM spectra and Carson's rule.

Provenance. Anchor values from LANE_SPEC: AM bandwidth $= 2f_m$; carrier $1\ \text{MHz}$ with $5\ \text{kHz}$ tone gives sidebands $0.995$ and $1.005\ \text{MHz}$, bandwidth $10\ \text{kHz}$; modulation index $\mu = A_m/A_c = 6/10 = 0.6$; $\lambda = c/f$ gives $300\ \text{m}$ at $1\ \text{MHz}$ and $3\ \text{m}$ at $100\ \text{MHz}$. Recompute list: $\lambda(15\ \text{kHz}) = 20\ \text{km}$ and quarter-wave $5\ \text{km}$; sidebands $f_c\pm f_m = 995,\,1005\ \text{kHz}$; $\text{BW}_{\text{AM}} = 2f_m = 10\ \text{kHz}$; $\mu = 6/10 = 0.6 = (16-4)/(16+4)$; $A_{\max}=16$, $A_{\min}=4$, $A_c=10$, $A_m=6$; Carson $\text{BW}_{\text{FM}} = 2(75+15) = 180\ \text{kHz}$ versus $2\times 15 = 30\ \text{kHz}$ AM (ratio 6); quarter-wave at $100\ \text{MHz}$ $= 0.75\ \text{m}$; line-of-sight $d = \sqrt{2Rh} = \sqrt{1.28\times10^{9}} = 35.8\ \text{km}$; $\text{SNR} = 10\log_{10}(1000) = 30\ \text{dB}$ and $10\log_{10}(100) = 20\ \text{dB}$; medium-wave stations across $530$–$1700\ \text{kHz}$ at $10\ \text{kHz}$ spacing $\approx 117$ channels.

Prerequisites. The Electromagnetic Spectrum — where radio, microwave and the relation $\lambda = c/f$ across the spectrum are established; this article puts a message onto those radio waves. Leads to. The Nature of Electromagnetic Waves, which returns to the carrier itself — how an accelerating charge radiates an electromagnetic wave, its speed, and its transverse structure. For the full spectral analysis of modulated signals — the frequency content of an FM wave and the Bessel-function sideband ladder behind Carson's rule — the university communication-theory seam is where to go deeper (enrichment only; never needed for the boards above, and requiring the Fourier analysis this article deliberately avoids).