For a message of bandwidth B, AM uses bandwidth 2B (the message generates both upper and lower sidebands around the carrier). For voice (3 kHz) the AM channel needs about 6 kHz; broadcast AM is typically allocated 10 kHz.

What is noise immunity?

Poor. Atmospheric and electrical noise add to amplitude and are not separated from the message at the receiver. AM broadcast at night famously suffers from interference.

What is power efficiency?

Poor. Most of the transmitted power is in the carrier itself rather than the sidebands (the sidebands carry the information). Variants like SSB (single sideband) suppress the carrier and one sideband to save power and bandwidth.

Low. AM receivers are simple (envelope detector); AM transmitters are simple.

Long-distance broadcast at HF/MF (because lower-frequency AM signals propagate via the ionosphere over long distances), aviation voice radio (typically AM at VHF for safety reasons).

Two amplitude levels (typically 0 and full) represent 0 and 1. Simple but vulnerable to amplitude noise. Used in low-cost remote controls and short-range RF.

Two frequencies represent 0 and 1. Better noise immunity than ASK. Used in early modems (Bell 103, V.21), some legacy paging systems.

Discrete phase states represent symbols. BPSK uses 2 phases (0 and 180 degrees); QPSK uses 4 phases (0, 90, 180, 270 degrees). Used in Wi-Fi, cellular, satellite communications.

Combines amplitude and phase. The constellation diagram has multiple symbol points (16-QAM uses 16 points, 64-QAM uses 64, 256-QAM uses 256). Each symbol carries _2(N) bits, so 16-QAM carries 4 bits per symbol.

§-Syllabus dot point

NSWEngineering StudiesSyllabus dot point

How do modulation techniques (AM, FM, PM, digital) shape what a signal can carry and how robustly?

Investigate modulation techniques including amplitude modulation, frequency modulation, phase modulation, and digital modulation (ASK, FSK, PSK, QAM), and the engineering trade-offs between bandwidth, complexity, power efficiency and noise immunity

A focused HSC Engineering Studies answer on modulation. Defines the carrier wave; explains AM, FM and PM; covers digital schemes (ASK, FSK, PSK, QAM); compares bandwidth, noise immunity, power efficiency and complexity; and gives the engineering selection criteria.

Generated by Claude Opus 4.810 min answerUpdated 2026-07-03

Reviewed by: AI editorial process; not yet individually human-reviewed

Have a quick question? Jump to the Q&A page

Jump to a section

What this dot point is asking
The answer
Examples in context
Try this

In plain English

Imagine you need to send a message across a noisy, crowded room. You could shout louder or softer to spell out a message (that is amplitude modulation, AM), but a noisy room drowns out volume changes easily. You could instead whistle at a slightly higher or lower pitch for each letter (that is frequency modulation, FM); pitch is much easier for a listener to pick out over background noise, even though it takes a bit more "room" (a wider range of pitches) to do it clearly. Or you could hold up a card with a dot in one of several positions on a grid, each position meaning a different chunk of letters at once (that is digital modulation like QAM): more positions let you send more information per card, but the positions get closer together and are easier to misread if the room is noisy. Every modulation technique is really just a different way of trading "how much room I need" against "how well the listener can hear me over the noise", and an engineer's job is to pick the trade-off that matches the actual room (channel) and message.

What this dot point is asking

Modulation is the process of impressing information from a message signal onto a carrier wave so it can be transmitted through a physical channel. The Telecommunications Engineering module expects you to define each main technique (AM, FM, PM, plus digital schemes), explain why each has its place, and use the engineering trade-offs (bandwidth, noise immunity, power efficiency, complexity) to justify a choice for a given application.

The answer

A typical communication channel is at a frequency very different from the information signal. A voice signal sits around 0 to 3.4 kHz; a broadcast radio channel sits at hundreds of kHz to hundreds of MHz. The carrier wave (sinusoidal at the channel's frequency) is modulated to carry the information.

The carrier wave

A sinusoidal carrier is described by:

v(t) = A \cos(2\pi f_c t + \phi)

The three parameters are amplitude $A$ , frequency $f_c$ , and phase $\phi$ . Modulating any one of them in time with the message signal produces a different modulation type.

Amplitude modulation (AM)

The message signal modulates the carrier amplitude:

v_{AM}(t) = [A + m(t)] \cos(2\pi f_c t)

The modulated signal carries the message in the variations of the envelope.

Bandwidth: For a message of bandwidth B, AM uses bandwidth 2B (the message generates both upper and lower sidebands around the carrier). For voice (3 kHz) the AM channel needs about 6 kHz; broadcast AM is typically allocated 10 kHz.
Noise immunity: Poor. Atmospheric and electrical noise add to amplitude and are not separated from the message at the receiver. AM broadcast at night famously suffers from interference.
Power efficiency: Poor. Most of the transmitted power is in the carrier itself rather than the sidebands (the sidebands carry the information). Variants like SSB (single sideband) suppress the carrier and one sideband to save power and bandwidth.
Complexity: Low. AM receivers are simple (envelope detector); AM transmitters are simple.
Used for: Long-distance broadcast at HF/MF (because lower-frequency AM signals propagate via the ionosphere over long distances), aviation voice radio (typically AM at VHF for safety reasons).

Frequency modulation (FM)

The message signal modulates the carrier frequency:

v_{FM}(t) = A \cos(2\pi f_c t + 2\pi K_f \int m(t)\, dt)

The modulated signal carries the message in the variations of the instantaneous frequency.

Bandwidth: Larger than AM. Wideband FM uses bandwidth approximately equal to $2(f_m + \Delta f)$ (Carson's rule), where $\Delta f$ is the peak frequency deviation. Broadcast FM stereo uses around 200 kHz per channel.
Noise immunity: Excellent. Atmospheric noise adds to amplitude rather than frequency, and FM receivers include a limiter that strips amplitude variations. FM is the standard for high-quality broadcast audio.
Power efficiency: Good. The carrier amplitude is constant; transmitters can use highly efficient class C amplifiers.
Complexity: Moderate. FM receivers need a discriminator; transmitters need precise frequency deviation control.
Used for: Broadcast FM radio, two-way radio communications (police, ambulance), the audio in television transmission.

The diagram below contrasts the two waveforms directly: AM keeps the wiggle rate (frequency) of the carrier constant and varies how tall the wiggles are (envelope); FM keeps the wiggle height constant and varies how fast the wiggles come (instantaneous frequency).

Phase modulation (PM)

The message signal modulates the carrier phase. Mathematically related to FM (PM and FM differ by an integration of the message). In analog form, PM is less common than FM standalone but is important as a building block for digital modulation schemes.

Digital modulation

Digital modulation uses discrete symbol states to carry binary data. The four main schemes:

ASK (Amplitude Shift Keying): Two amplitude levels (typically 0 and full) represent 0 and 1. Simple but vulnerable to amplitude noise. Used in low-cost remote controls and short-range RF.
FSK (Frequency Shift Keying): Two frequencies represent 0 and 1. Better noise immunity than ASK. Used in early modems (Bell 103, V.21), some legacy paging systems.
PSK (Phase Shift Keying): Discrete phase states represent symbols. BPSK uses 2 phases (0 and 180 degrees); QPSK uses 4 phases (0, 90, 180, 270 degrees). Used in Wi-Fi, cellular, satellite communications.
QAM (Quadrature Amplitude Modulation): Combines amplitude and phase. The constellation diagram has multiple symbol points (16-QAM uses 16 points, 64-QAM uses 64, 256-QAM uses 256). Each symbol carries $\log_2(N)$ bits, so 16-QAM carries 4 bits per symbol. Used in cable modems, DOCSIS, modern Wi-Fi, LTE downlink, cellular.

The more symbol points, the more bits per symbol, but the closer the points sit in the constellation, the more vulnerable the scheme is to noise. Adaptive modulation in modern cellular and Wi-Fi switches between schemes (QPSK in poor channel conditions, 256-QAM in excellent conditions) to optimise throughput.

Engineering selection trade-offs

A typical decision framework:

Bandwidth budget. How much spectrum is available? Lower-bandwidth schemes (AM, BPSK) for limited spectrum; higher-bandwidth schemes (wideband FM, high-order QAM) when bandwidth is abundant.
Noise environment. Noisy channel? Use FM, FSK, or robust digital schemes (BPSK, QPSK). Clean channel? Higher-order QAM extracts more throughput.
Power budget. Mobile / satellite / IoT devices need power-efficient schemes. Constant-envelope schemes (FM, FSK) allow more efficient transmitter design.
Receiver complexity. Cost-sensitive consumer products favour simpler schemes. Performance-critical applications can afford complex receivers.
Spectral efficiency. Bits per second per Hz. Higher for higher-order schemes (256-QAM around 8 bits/symbol).

Worked example

Why FM was chosen for music broadcast and AM for talkback radio, working through the modulation trade-offs.

Step 1: Characterise the AM broadcast band

The AM band (around 540 to 1700 kHz) allocates 10 kHz per station. Voice (a 3 kHz message) fits comfortably, modulators and receivers are cheap, and lower frequencies propagate well at night via the ionosphere for long-distance coverage. Sound quality is limited by the narrow channel and AM's noise vulnerability.

Step 2: Characterise the FM broadcast band

The FM band (88 to 108 MHz) allocates 200 kHz per station. The message bandwidth (around 53 kHz for stereo) plus deviation needs the wide channel. Higher frequency means line-of-sight propagation, but FM's noise immunity suits music; the receiver is more complex.

Step 3: Compare and justify the choice

A given range of spectrum supports about 20 times more AM stations (10 kHz each) than FM stations (200 kHz each). So AM suits talkback voice, where bandwidth efficiency matters and audio quality is acceptable, while FM suits music, where audio quality is paramount, exactly the modulation trade-off at the design moment.

Marker's note: full-mark answers name the actual bandwidth figures (10 kHz vs 200 kHz) rather than saying "AM uses less bandwidth", and close with an explicit trade-off statement linking bandwidth to the specific criterion (station density versus audio fidelity) the application values most.

Common traps

Confusing modulation with coding: Modulation impresses bits on a carrier; coding (e.g. forward error correction) adds redundancy bits before modulation. Both can improve robustness; they are distinct operations.
Reporting AM as low quality without context: AM is bandwidth-efficient; its noise vulnerability is a separate issue. Strong responses name multiple criteria.
Treating bandwidth and bit rate as the same: Bandwidth is the spectrum the signal occupies (Hz). Bit rate is the information rate (bps). They are related by the modulation scheme.
Saying "digital is always better than analog": Digital schemes can be more bandwidth-efficient and noise-robust, but only when channel conditions and complexity budget allow. Aviation voice is still AM at VHF because the AM receiver's response to interference (audible drop-out) is safer than a digital scheme's all-or-nothing failure.
Forgetting why FM uses class C amplifiers: FM has constant envelope. AM does not. Class C amplifiers are highly efficient but distort amplitude, which makes them suitable for FM but unsuitable for AM. The constant-envelope property is the engineering reason FM can use efficient amplification.
Confusing the QAM order with the noise tolerance: Higher-order QAM (e.g. 256-QAM) packs more bits per symbol but is more vulnerable to noise. Lower-order schemes (QPSK) are more robust but slower.

Examples in context

Example 1. Wi-Fi modulation evolution. Original 802.11 (1997) used DSSS at low rates. 802.11a/g (1999/2003) added OFDM with BPSK / QPSK / 16-QAM / 64-QAM. 802.11ac (2013) added 256-QAM. 802.11ax / Wi-Fi 6 (2019) added 1024-QAM. Each generation adds higher-order modulation to extract more throughput when channel conditions allow, with backwards compatibility at lower orders for poor channel conditions.

Example 2. NBN fixed-line variants. Australian NBN deployment uses different modulation by access technology. HFC (hybrid fibre coaxial) uses 256-QAM downstream and lower-order modulation upstream. FTTN / FTTC use VDSL2 with adaptive modulation across multiple subcarriers (essentially OFDM at telecoms frequencies). FTTP uses optical modulation at gigabit rates. The modulation choices reflect the channel quality of each access technology.

Try this

Q1. Compare AM and FM on three engineering criteria. [4 marks]

Cue. Bandwidth: AM lower (around 10 kHz for broadcast voice/music) vs FM higher (around 200 kHz for stereo broadcast). Noise immunity: AM poor vs FM excellent (because limiter strips amplitude variations). Power efficiency: AM poor (most power in carrier) vs FM excellent (constant envelope allows class C amplifiers). Complexity: AM simpler receiver vs FM more complex (discriminator). Use cases differ accordingly.

Q2. Explain how a 16-QAM scheme carries 4 bits per symbol and why higher-order QAM is more vulnerable to channel noise. [4 marks]

Cue. 16-QAM has 16 constellation points; each point represents a unique 4-bit pattern (log2(16) = 4). Higher-order QAM packs more bits per symbol (256-QAM = 8 bits per symbol; 1024-QAM = 10 bits) but the constellation points sit closer together. Noise that displaces a received symbol by a small amount may cross to a neighbouring constellation point, producing a bit error. Lower-order schemes (QPSK, 4 points) tolerate more noise per symbol but carry fewer bits.

Q3. Justify the choice of QPSK rather than 256-QAM for a satellite communication system in poor weather conditions. [4 marks]

Cue. Satellite links suffer from path loss, atmospheric absorption (rain fade) and noise. Poor signal-to-noise ratio makes 256-QAM unusable (received symbols would land in wrong cells). QPSK has only 4 constellation points and tolerates a much lower SNR. The choice trades throughput for reliability. Many modern satellite systems use adaptive coding and modulation (ACM) that drops to QPSK in adverse conditions and rises to 16-/64-QAM when conditions improve.

Exam-style practice questions

Practice questions written in the style of NESA exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

HSC 20223 marksCompare amplitude modulation and frequency modulation in terms of bandwidth and noise immunity, and identify a typical application of each.

Show worked answer →

AM varies the carrier amplitude, uses a narrow bandwidth (about twice the message bandwidth) and has poor noise immunity because amplitude noise adds directly to the signal; it is used for long-range broadcast and aviation voice radio. FM varies the carrier frequency, uses a much wider bandwidth (Carson's rule), and has strong noise immunity because receivers can limit amplitude and recover only frequency changes; it is used for high-fidelity broadcast and two-way radio. Markers reward the amplitude-versus-frequency mechanism, the bandwidth contrast (narrow versus wide), the noise-immunity contrast, and a valid application of each.

HSC 20244 marksAn FM broadcast signal has a peak frequency deviation of 75 kHz and the highest audio frequency is 15 kHz. Use Carson's rule to determine the transmission bandwidth, and explain why FM trades bandwidth for noise performance.

Show worked answer →

Carson's rule estimates the bandwidth of an FM signal as twice the sum of the peak deviation and the highest message frequency:

B = 2(\Delta f + f_m) = 2(75 + 15) = 2 \times 90 = 180 \text{ kHz}

So the FM channel needs about 180 kHz, far more than the roughly 30 kHz an AM channel would need for the same audio. The trade-off: by spreading the information over a wide bandwidth and encoding it as frequency rather than amplitude, the receiver can hard-limit the signal to strip amplitude noise and then recover the frequency changes, giving much better signal-to-noise ratio. FM therefore spends bandwidth (a scarce resource) to buy noise immunity and audio fidelity. Markers reward the correct use of Carson's rule with units, the comparison to AM bandwidth, and the bandwidth-for-noise-immunity reasoning.

Practice questions

Original practice questions graded from foundation to exam level, each with a full worked solution. Try them before revealing the solution.

foundation3 marksAn FM two-way radio system has a peak frequency deviation of 5 kHz and a maximum message frequency of 3 kHz. Use Carson's rule to calculate the transmission bandwidth.

Show worked solution →

B = 2(\Delta f + f_m) = 2(5 + 3) = 2 \times 8 = 16\ \text{kHz}

Marking criteria: 1 mark for correctly quoting Carson's rule, 1 mark for correct substitution with units, 1 mark for the correct final answer (16 kHz).

foundation4 marksA voice message has bandwidth 4 kHz. Calculate the AM transmission bandwidth required, and state whether this fits within a broadcast AM channel allocation of 9 kHz.

Show worked solution →

Step 1: apply the AM bandwidth rule.

B_{AM} = 2B = 2 \times 4 = 8\ \text{kHz}

Step 2: compare to the allocation. The required 8 kHz is less than the 9 kHz channel allocation, so the message fits with 1 kHz of margin.

Marking criteria: 1 mark for correctly stating the AM bandwidth rule (2B), 1 mark for correct substitution, 1 mark for the correct bandwidth (8 kHz), 1 mark for the correct comparison and conclusion (fits, with margin stated).

core5 marksA deep-space probe has a severely limited transmitter power budget and must communicate over a channel affected by significant background noise, but has abundant allocated spectrum. The table below compares four candidate schemes. | Scheme | Constant envelope | Noise immunity | Spectral efficiency (bits/Hz) | |---|---|---|---| | ASK | No | Poor | Low | | FSK | Yes | Good | Low | | 256-QAM | No | Poor | High | | BPSK | Yes | Good | Low | Using the table, select the most suitable scheme for the probe and justify your choice against the mission constraints.

Show worked solution →

Step 1: identify the binding constraints: The probe has a limited power budget (favours a constant-envelope scheme, allowing efficient class C-style amplification) and a noisy channel (favours good noise immunity). Spectral efficiency is not a priority because spectrum is abundant.
Step 2: eliminate unsuitable schemes: 256-QAM is eliminated: it is not constant-envelope (requiring a less power-efficient linear amplifier) and has poor noise immunity, both wrong for this mission. ASK is eliminated for the same noise-immunity reason, despite being simple.
Step 3: select and justify: FSK or BPSK are both constant-envelope with good noise immunity; either is a reasonable answer if justified. BPSK is a common real-world choice for deep-space links (e.g. used historically in NASA missions) because it offers good noise immunity at very low signal-to-noise ratio with reasonably simple coherent detection, matching the probe's power-limited, noise-limited, spectrum-abundant profile.

Marking criteria: 1 mark for correctly identifying the power and noise constraints as binding, 1 mark for eliminating 256-QAM with a valid reason, 1 mark for eliminating ASK with a valid reason, 1 mark for selecting a constant-envelope, good-noise-immunity scheme (FSK or BPSK), 1 mark for justification tied explicitly to the mission constraints (not spectral efficiency, which is not limiting here).

core5 marksA digital modulation scheme uses a constellation diagram with 64 equally spaced symbol points arranged in an 8 by 8 grid on the in-phase/quadrature plane. (a) Name this scheme and state how many bits it carries per symbol. (b) Explain why this scheme is more vulnerable to channel noise than QPSK, and state the trade-off gained in return.

Show worked solution →

(a) Identify the scheme and bits per symbol.

A 64-point constellation on the I/Q plane is 64-QAM. Bits per symbol:

\log_2(64) = 6\ \text{bits per symbol}

(b) Noise vulnerability versus QPSK. QPSK has only 4 widely spaced constellation points, so a large noise displacement is needed before a received symbol crosses into a neighbouring decision region and causes a bit error. 64-QAM packs 64 points into the same power budget, so adjacent points sit much closer together; a much smaller noise displacement is enough to push a received symbol across a decision boundary into the wrong region, causing more bit errors at the same noise level.

Trade-off gained. In return for this reduced noise tolerance, 64-QAM carries 6 bits per symbol compared with QPSK's 2 bits per symbol, tripling the data rate for the same symbol (baud) rate when the channel is clean enough to support it.

Marking criteria: 1 mark for correctly naming 64-QAM, 1 mark for correctly calculating 6 bits per symbol with the log2 relationship shown, 1 mark for explaining the closer point spacing as the cause of higher error vulnerability, 1 mark for correctly contrasting with QPSK's wider spacing, 1 mark for stating the throughput trade-off gained.

core6 marksA cable modem must sustain a downstream data rate of 42 Mbps using 64-QAM. Calculate the minimum symbol (baud) rate required, and explain what would happen to this required symbol rate if the system instead used 256-QAM for the same data rate.

Show worked solution →

Step 1: bits per symbol for 64-QAM.

\log_2(64) = 6\ \text{bits/symbol}

Step 2: required symbol rate.

\text{Symbol rate} = \frac{\text{data rate}}{\text{bits per symbol}} = \frac{42 \times 10^6}{6} = 7 \times 10^6\ \text{symbols/s} = 7\ \text{Mbaud}

Step 3: effect of switching to 256-QAM. 256-QAM carries $\log_2(256) = 8$ bits per symbol, so the same 42 Mbps could be sent at a lower symbol rate:

\text{Symbol rate} = \frac{42 \times 10^6}{8} = 5.25 \times 10^6\ \text{symbols/s} = 5.25\ \text{Mbaud}

Switching to 256-QAM reduces the required symbol rate (and hence occupied bandwidth) for the same data rate, but only remains usable if the channel's signal-to-noise ratio is high enough to reliably distinguish the more closely packed constellation points; otherwise the reduced noise margin causes an unacceptable bit error rate.

Marking criteria: 1 mark for correctly finding 6 bits/symbol, 1 mark for correct symbol rate calculation with units (7 Mbaud), 1 mark for correctly finding 8 bits/symbol for 256-QAM, 1 mark for correct recalculated symbol rate (5.25 Mbaud), 1 mark for correctly stating the symbol rate/bandwidth reduction, 1 mark for the caveat that this only works if channel SNR supports the tighter constellation.

exam7 marksJustify, with reference to bandwidth, noise immunity and power efficiency, why the Australian NBN uses different modulation approaches across its HFC, FTTN/FTTC and FTTP access technologies.

Show worked solution →

This 7-mark JUSTIFY question rewards linking each access technology's physical channel characteristics to its modulation choice, not a description of each technology alone.

HFC (hybrid fibre coaxial): The coaxial cable segment supports a wide, relatively high-SNR bandwidth over short distances from the node, so high-order 256-QAM is used downstream to maximise spectral efficiency and throughput. Upstream uses lower-order modulation because the upstream path combines signals from many premises (ingress noise accumulates), reducing achievable SNR.
FTTN/FTTC (fibre to the node/curb): The remaining copper run varies hugely in length and quality between premises. VDSL2 uses OFDM-like multi-carrier transmission with adaptive modulation independently per subcarrier: short, clean copper runs support higher-order constellations on more subcarriers, while long or noisy runs automatically fall back to lower-order, more robust constellations on the affected subcarriers. This per-subcarrier adaptivity directly trades bandwidth efficiency for noise immunity where the copper's frequency response demands it.
FTTP (fibre to the premises): Optical fibre has enormous available bandwidth and extremely high SNR compared with copper or coax, so gigabit-rate optical modulation (on/off keying or more advanced schemes at very high symbol rates) is used without needing the noise-robustness trade-offs that copper and coax require.
Judgement: Each technology's modulation choice reflects an engineering match between the physical channel's noise and bandwidth characteristics and the modulation order used: highest order where the channel is cleanest and most consistent (FTTP, then HFC downstream), most conservative and adaptive where the channel is most variable and noise-prone (FTTN/FTTC copper, HFC upstream).

Marking criteria: 1 mark for correctly describing HFC's modulation approach, 1 mark for correctly describing FTTN/FTTC's adaptive per-subcarrier approach, 1 mark for correctly describing FTTP's approach, 1 mark for correctly linking each choice to channel bandwidth, 1 mark for correctly linking each choice to channel noise/SNR, 1 mark for a coherent overall judgement connecting channel quality to modulation order across all three, 1 mark for accurate technical terminology (OFDM/subcarrier, SNR, spectral efficiency) used correctly.

exam8 marksEvaluate the claim that 'digital modulation schemes are always superior to analog modulation schemes for telecommunications', using specific engineering criteria and named applications.

Show worked solution →

This 8-mark EVALUATE question rewards a supported judgement, not a list of digital advantages.

Plan.

Thesis: the claim is an overstatement; digital is often superior on spectral efficiency and error-correction, but analog AM remains better in specific safety-critical, cost-constrained or long-range broadcast contexts, so superiority is context-dependent.
For digital: QAM packs multiple bits per symbol (16-QAM carries 4, 256-QAM carries 8), digital signals can be regenerated bit-for-bit at repeaters instead of merely amplified with noise, and forward error correction can be added, none of which analog offers.
Against absolute superiority: aviation voice radio at VHF still uses AM because its failure mode (audible, gradual degradation) is safer for pilots than a digital scheme's silent, all-or-nothing dropout at the same SNR.
Further: adaptive digital schemes (QPSK, 16-QAM) still fall back to simpler, more robust forms when the channel degrades, showing digital is not immune to the bandwidth-for-noise-immunity trade-off, it just has more schemes to pick from.
Cost: very simple, low-cost links (some remote controls) still use ASK/OOK because a full digital chain would be disproportionately complex for the data rate needed.
Judgement: digital wins on spectral efficiency, error resilience and adaptability in most modern high-throughput uses, but "always superior" is false where safety, cost or failure-mode requirements favour AM.

Model paragraph (excerpt). QAM schemes undeniably out-perform AM on spectral efficiency, since 256-QAM carries eight bits per symbol, and digital signals can be regenerated bit-for-bit at repeaters rather than amplified along with accumulated noise. However, the blanket claim fails against aviation voice radio, which remains AM at VHF because its degradation under interference is audible and gradual, a safer failure mode for a pilot than a digital link's abrupt, silent dropout at the same SNR. "Superior" modulation is therefore a function of the specific requirement, bandwidth, noise, cost, and the consequence of failure, not an intrinsic property of digital modulation.

Marking criteria: top-band answers (1) state a judgement engaging with and partially rejecting the claim, (2) support both sides with accurate evidence (bits/symbol, regeneration, adaptive fallback, failure mode), (3) name a real application for each side, (4) conclude with a criteria-dependent judgement, not a flat yes/no.