NEET UG · Physics

Communication systems

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NEET UG · Physics

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Read AM envelopes and sidebands, combine power gains/losses and derive geometric horizon reach. Communication systems: connect the signal, spectrum and link

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Communication systems: connect the signal, spectrum and link

Read AM envelopes and sidebands, combine power gains/losses and derive geometric horizon reach.

What you will understand

For NEET UG Physics students using this existing chapter catalogue: connect an AM envelope to its sidebands, combine power gains and losses correctly and derive the approximate two-antenna line-of-sight reach.

This guide teaches communication concepts; the archived source and chapter route do not establish current exam inclusion.

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Quick answer: identify what the signal, channel and receiver do

A transmitter prepares a message-bearing signal, a channel carries it with loss and possible noise, and a receiver recovers an estimate of the message. Conventional full-carrier amplitude modulation changes a carrier’s amplitude using the message. The carrier frequency, message frequency, occupied sideband span and signal power are different quantities. A geometric horizon is only one constraint on a link; it does not certify adequate power or signal quality at the receiver.

AM: envelope extremes and sideband amplitudes share one model

Original single-tone conventional full-carrier amplitude-modulation diagram. Supplied carrier peak Ac=5 V and modulation index m=0.60 give nonnegative envelope maximum 8 V and minimum 2 V. Their ratio formula is m=(Amax-Amin)/(Amax+Amin)=0.60. For carrier fc=1000 kHz and message fm=2 kHz, the positive-frequency components occur at 998,1000 and1002 kHz. With a peak-amplitude convention, the carrier has5V and each sideband mAc/2=1.5V. Bars share a zero amplitude baseline; the displayed frequency window is local around the carrier rather than starting at zero frequency. Ideal sideband span is4kHz. This is a supplied conventional AM model, not FM, suppressed-carrier AM or measured broadcast data. — For a single-tone conventional full-carrier AM signal, carrier peak 5 V and modulation index 0.60 give envelope maximum 8 V and minimum 2 V. With carrier 1000 kHz and message 2 kHz, components appear at 998, 1000 and 1002 kHz. Each sideband peak is 1.5 V under the stated amplitude convention. The local frequency axis is centred near the carrier, and the ideal sideband span is 4 kHz.

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Recover modulation index before calculating the AM spectrum

Use the supplied conventional AM model

s(t)=A_c[1+m\cos(2\pi f_mt)]\cos(2\pi f_ct)

with fc much larger than fm and 0≤m≤1. Measured envelope extremes are Amax=8.0 V and Amin=2.0 V. Then

m=(A_{\max}-A_{\min})/(A_{\max}+A_{\min})=0.60

and Ac=(8+2)/2=5.0 V. For fc=1.000 MHz and fm=2.00 kHz, sidebands are 998 kHz and 1002 kHz, giving ideal span 4.00 kHz. Expanding the product gives each sideband peak mAc/2=1.50 V. Amax/Amin=4 is not the modulation index. A message with a frequency band needs the corresponding highest-frequency sideband extent rather than a single-tone line count.

Overmodulation and bandwidth belong to the stated modulation scheme

In conventional AM, keeping the envelope coefficient nonnegative allows a simple envelope detector to follow the message. With m>1, the coefficient crosses zero; the absolute envelope no longer reproduces the intended signed modulation without distortion in that detector. The ideal algebraic three-component single-tone signal does not acquire extra frequencies merely from choosing a larger m; clipping or other nonlinear processing can change that. The familiar 2fm span applies to this double-sideband model, not automatically to FM, single-sideband or every digital system.

Line-of-sight reach: derive one tangent triangle before adding both ends

Original tangent-geometry schematic for one antenna above a smooth spherical Earth. O is the Earth centre, R the radius, h the antenna height, H the tangency horizon and d the straight antenna-to-horizon distance. Radius OH is perpendicular to tangent TH, so d squared=(R+h) squared-R squared=2Rh+h squared; for h much smaller than R, d is approximately sqrt(2Rh). The drawn height is deliberately greatly exaggerated and is not the numerical height scale. In a supplied ideal two-end comparison with R=6.40e6 m, transmitter height80m gives about32km and receiver height20m gives about16km, for approximately48km combined geometric reach. The diagram shows the one-end triangle; the two-end approximation adds horizon contributions rather than heights. Terrain, atmospheric refraction, diffraction, obstructions, antenna gain and link-budget losses are excluded, so this is not a coverage guarantee. — The radius to the horizon point is perpendicular to the antenna’s tangent line. Hence d²=(R+h)²−R², approximated by 2Rh when h is small relative to R. For two elevated endpoints, add their approximate horizon contributions. The drawn height is greatly exaggerated. Terrain, atmosphere, diffraction and signal-power constraints are omitted from this geometric model.

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A power link budget uses 10 log, with loss as negative gain

A supplied chain starts with 3.00 mW, followed by power gain +10.0 dB, channel loss 16.0 dB and receiver power gain +20.0 dB. Net gain is +14.0 dB because the loss contributes −16.0 dB. Using

G_\mathrm{dB}=10\log_{10}(P_\mathrm{out}/P_\mathrm{in})

, output power is

3.00\times10^{14/10}\approx75.4\,\mathrm{mW}

. Check stages: 30.0 mW after the first gain, about 0.754 mW after the channel and 75.4 mW after the receiver gain. The 20log voltage-ratio form needs the appropriate equal-impedance condition; do not use it directly for this power ratio. Increased gain alone does not guarantee restoration of information lost through noise or distortion.

Propagation labels describe mechanisms and conditions

A line-of-sight space-wave model differs from ground-wave propagation along the surface and ionospheric sky-wave paths. Which mechanism is useful depends on frequency, environment, antennas and the link requirements. Increasing a tower height changes a geometric horizon but does not by itself specify every propagation mode or received signal quality. Similarly, a quarter-wavelength antenna estimate is an elementary design scale, not a complete rule for every practical antenna geometry.

Add two horizon distances, not the two heights inside one square root

For a smooth spherical Earth use supplied $R=6.40\times10^6\,\mathrm m$ , transmitter height ht=80.0 m and receiver height hr=20.0 m. With heights much smaller than R and no refraction, $d\approx\sqrt{2Rh_t}+\sqrt{2Rh_r}$ . The contributions are 32.0 km and 16.0 km, so the approximate geometric reach is 48.0 km. Using √(2R(ht+hr)) would give a different, incorrect two-end expression. The exact one-end straight tangent distance is √(2Rh+h²); the neglected h² correction is tiny for these supplied heights. This is geometric reach, not a guarantee of coverage, received power or an unobstructed real path.

Amplification and demodulation solve different problems

An amplifier raises a signal level using an energy supply; demodulation extracts the message from a modulated signal. A receiver may need both, plus filtering and other processing. A larger output voltage does not prove that the original message was recovered faithfully. Describe the stage function before naming it from a familiar block-diagram label.

Common mistakes: keep frequency span, power and geometry separate

Do not call the carrier frequency the AM bandwidth or divide envelope maximum by minimum to obtain modulation index. Use consistent Hz, kHz and MHz units. A specified positive loss magnitude contributes negative gain in a signed budget. Distinguish a power ratio from a voltage ratio. Add the two horizon distances in the stated small-height model, and keep its geometric reach separate from a real received-power prediction.

Try it: move the message tone, retain the carrier

A conventional single-tone AM carrier remains at 1.000 MHz while the message tone rises from 2.00 to 5.00 kHz. What are the new sideband frequencies and ideal occupied span? Does the carrier have to move to 1.005 MHz?

Reveal the answer

The sidebands move to 995 kHz and 1005 kHz, and their span is 10.0 kHz. The carrier remains at 1000 kHz. A sideband is a distinct frequency component; it does not replace the carrier in this full-carrier AM model.

Try it: quadruple only the transmitter height

In the supplied small-height line-of-sight model, raise the transmitter from 80 m to 320 m while the receiver remains at 20 m. What happens to the two contributions and their sum? Does the total reach double?

Reveal the answer

The transmitter contribution doubles from 32 km to 64 km because distance scales as √h. The receiver contribution stays 16 km. Their sum becomes 80 km, not twice the original 48 km. Only one endpoint changed. All geometric-model omissions still apply.

Come back tomorrow: reconstruct the envelope and the link budget

Derive the AM sidebands from the product rather than recalling a line count. Then write a gain/loss chain and check a two-end horizon calculation. Revisit NEET Electromagnetic Waves for wave quantities, or NEET Semiconductor Devices for nonlinear device and rectifier models.

Chapter practice opens the existing broad Communication Systems pool.

Sources and review

Technical review: Codex (automated). Primary textbook and university teaching sources checked on 23 October 2026. Diagrams, worked examples and questions are original and use explicit models. The archived 2018 Communication Systems exemplar supports concepts; this existing chapter catalogue does not establish current exam inclusion. Human subject review is not claimed.

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Sources and reading

NCERT Exemplar Problems: Communication Systems — Archived 21 April 2018, Chapter 15: modulation, sidebands, envelope index, gain/attenuation and line-of-sight models. Concept evidence, not current exam inclusion.
Carnegie Mellon University: Amplitude Modulation — Primary university teaching text, Section 6.3: carrier/message product and sideband frequencies/amplitudes