Department of Electrical & Electronic Engineering — Egerton University

Introduction to Amplitude Modulation (AM)

EEEN 462 — Analog Communications  |  4th Year Study Guide

Module 1  •  Amplitude Modulation Fundamentals

1. Learning Objectives

By the end of this study unit, the student should be able to:

  1. Define amplitude modulation and state its purpose in analog communication systems.
  2. Identify and sketch the modulating (message), carrier, and amplitude-modulated signals in the time domain.
  3. Derive the mathematical expression of a DSB-FC (full-carrier) AM wave from first principles.
  4. Calculate the modulation index (ma) from maximum and minimum envelope amplitudes and classify it as under-, 100%-, or over-modulated.
  5. Determine the bandwidth of an AM signal from the message bandwidth.
  6. Analyse the frequency spectrum of an AM wave in terms of carrier, upper sideband, and lower sideband components.
  7. Compute the total transmitted power and power efficiency of an AM signal.
  8. Describe envelope detection as the standard method of AM demodulation and explain the conditions for distortion-free recovery.
  9. Solve quantitative problems involving AM parameters, power distribution, and bandwidth.

Contents at a Glance

2. Introduction and Motivation

Modulation is the process of varying one or more properties of a high-frequency sinusoidal carrier signal in accordance with the instantaneous amplitude of a low-frequency modulating (message) signal such as voice, music, or sensor data. In amplitude modulation (AM), it is the amplitude of the carrier that is varied in proportion to the message; the carrier frequency and phase remain constant.

Baseband voice signals occupy roughly 20 Hz – 15 kHz and cannot be radiated efficiently from practical antennas (which would need kilometre-scale dimensions at audio frequencies). Modulating a radio-frequency carrier solves this and provides several benefits:

Key idea: AM is the simplest analog modulation scheme. Its great virtue is extremely cheap, simple receiver design (envelope detector); its great weakness is poor power efficiency because most of the transmitted power sits in the carrier, which carries no message information.

3. The Three Key Signals

3.1 Modulating (Message) Signal — m(t)

A single-tone message is modelled as m(t) = Am cos(2πfmt), where Am is the message amplitude and fm its frequency (≪ fc).

t m(t) +A_m −A_m T_m = 1/f_m Message (modulating) signal
Figure 1 — Modulating (message) signal m(t) = Amcos(2πfmt), carrying the information.

3.2 Carrier Signal — c(t)

The carrier is a high-frequency, constant-amplitude sinusoid c(t) = Ac cos(2πfct) generated by a stable oscillator. It contains no information; it is the "vehicle" that transports the message.

t c(t) +A_c −A_c T_c = 1/f_c Carrier signal constant amplitude, high frequency (f_c ≫ f_m)
Figure 2 — Carrier signal c(t) = Accos(2πfct), a sinusoid of constant amplitude and frequency. Its amplitude and frequency are constant and it carries no information.

3.3 Amplitude-Modulated Signal — s(t)

In AM, the carrier amplitude is made to vary in step with m(t). The result is a carrier whose envelope (the outline joining successive peaks) has the same shape as the message.

t s(t) A_c(1+m_a) — envelope max at message peak A_c(1−m_a) — envelope min at message trough A_c (m(t)=0) Upper envelope: A_c + m(t) Lower envelope: A_c − m(t) RF oscillations at f_c (peaks touch the envelope)
Figure 3 — AM wave s(t) = Ac[1 + macos(2πfmt)]cos(2πfct): the carrier amplitude at every instant equals the envelope height, so the RF peaks touch the envelope. The envelope is largest (Ac(1+ma)) where m(t) peaks and smallest (Ac(1−ma)) where m(t) troughs — matching the phase of Figure 1.
Reading the waveform: Where the message m(t) is most positive, the envelope peaks at Ac(1+ma); where m(t) is most negative, the envelope dips to Ac(1−ma). The envelope faithfully reproduces the message as long as ma ≤ 1.

4. Mathematical Model of AM

Let the message and carrier be:

Message (modulating) signalm(t) = Am cos(2π fm t)
Carrier signalc(t) = Ac cos(2π fc t),   fc ≫ fm

AM varies the carrier amplitude proportionally to the message:

Instantaneous amplitude of AM waveA(t) = Ac + m(t) = Ac[1 + ma cos(2π fm t)]

where the modulation index (or modulation depth) is

Modulation indexma = Am / Ac

Hence the complete AM wave (DSB-FC: Double SideBand — Full Carrier) is:

Time-domain AM signals(t) = Ac[1 + ma cos(2π fm t)] cos(2π fc t)

Expanding with the product-to-sum identity cos A cos B = ½[cos(A+B) + cos(A−B)]:

Expanded form — three frequency componentss(t) = Ac cos(2π fc t) + (maAc/2) cos[2π(fc+fm) t] + (maAc/2) cos[2π(fc−fm) t]
ComponentFrequencyAmplitudeInformation content
CarrierfcAcNone (constant)
Upper Sideband (USB)fc + fmmaAc/2Full message
Lower Sideband (LSB)fc − fmmaAc/2Full message (redundant)
Interpretation: The message appears in the sidebands, both of which contain the complete information. The carrier exists purely to enable simple envelope detection at the receiver. This is why AM wastes power — the carrier may consume most of the transmitted energy yet carries no information.

5. Modulation Index, Envelope, and Over-modulation

From the maximum and minimum values of the envelope (see Figure 3):

Modulation index from the envelopema = (Amax − Amin) / (Amax + Amin) = Am / Ac
Envelope limitsAmax = Ac(1 + ma),    Amin = Ac(1 − ma)
m_a = 0.54: envelope between A_c(1+m_a) and A_c(1−m_a) > 0
(a) Under-modulation (ma < 1): envelope varies between Ac(1+ma) and Ac(1−ma) and never reaches zero; sinusoidal carrier follows the envelope.
m_a = 1: envelope just touches zero at message troughs
(b) 100% modulation (ma = 1): envelope just touches zero at message troughs — the limit of distortion-free operation.
envelope crosses zero → 180° phase reversals carrier flips where A_c + m(t) < 0 m_a = 1.5 > 1: envelope goes negative — an envelope detector recovers a distorted message
(c) Over-modulation (ma > 1): the carrier is multiplied by a negative factor, producing 180° phase flips. The envelope no longer equals m(t), so an envelope detector recovers a distorted message.
Rule for distortion-free AM: keep ma ≤ 1 (i.e., Am ≤ Ac). In practice, transmitters are operated with ma between about 0.8 and 0.95 to allow for message peaks. Over-modulation also splatters energy into adjacent channels, violating spectral regulations.

6. Frequency-Domain Analysis: Spectrum and Bandwidth

For a single-tone message, the AM wave contains exactly three spectral lines. For a general message band-limited to W Hz (e.g., audio 20 Hz – 15 kHz), the sidebands become continuous bands of width W on either side of the carrier.

f S(f) A_c f_c m_a A_c /2 f_c − f_m m_a A_c /2 f_c + f_m BW = 2 f_m Upper Sideband (USB) Lower Sideband (LSB)
Figure 4 — Single-tone AM spectrum: carrier at fc plus two sidebands at fc ± fm. Transmission bandwidth = 2fm (for message bandwidth W: BW = 2W).
Transmission bandwidthBW = 2 fm   (single tone)  or  BW = 2 W   (message band-limited to W Hz)
Worked insight: An AM broadcast station with a 15 kHz audio message needs 30 kHz of channel bandwidth. This is twice the message bandwidth — AM is spectrally inefficient compared with SSB (Single SideBand), which transmits only one sideband (BW = W) at the cost of more complex receivers.

7. Power Distribution in AM

Power is measured across a resistive load R (conventionally R = 1 Ω for normalized analysis). Each sinusoid of amplitude A contributes A²/(2R).

ComponentPower expressionWith R = 1 Ω
CarrierPc = Ac² / (2R)Pc = Ac² / 2
Each sidebandPSB = (maAc/2)² / (2R) = ma² Ac² / (8R)ma² Pc / 4
Total (both sidebands)PSB,total = ma² Ac² / (4R)ma² Pc / 2
Total transmitted powerPT = Pc + ma²Pc/2 = Pc(1 + ma²/2)
Modulation efficiency (useful power fraction)η = (sideband power) / (total power) = (ma²/2) / (1 + ma²/2) = ma² / (2 + ma²)
Maximum efficiency of full-carrier AM is only 33.3%, attained at ma = 1. For typical speech with ma ≈ 0.3, efficiency falls to about 4%. This is the fundamental price paid for the simplicity of envelope detection. (DSB-SC and SSB eliminate the wasteful carrier and/or one sideband.)

8. AM Demodulation — The Envelope Detector

Because the message is preserved in the envelope of s(t), a very simple circuit can recover it. The standard envelope detector consists of a diode followed by an RC low-pass filter:

s(t) in D R C m̂(t) out (recovered message) Time constant must satisfy: 1/f_c ≪ RC ≪ 1/W fast enough to follow the envelope, slow enough to reject the carrier ripple
Figure 5 — Diode envelope detector. The diode rectifies the AM wave; the RC pair smoothes the carrier ripple leaving the message envelope.

Design condition: RC must be large compared with the carrier period 1/fc (to filter ripple) yet small compared with the smallest message period 1/W (to track the fastest envelope variations). If RC is too large, the detector fails to follow rapid envelope falls — a defect called diagonal clipping.

Why AM survived: an entire AM broadcast receiver can be built with a handful of passive components (tuned antenna coil, diode, resistor, capacitor, earpiece). This is why AM dominated early radio and remains in medium-wave broadcasting, two-way aviation radios, and low-cost remote-control toys.

9. Applications, Advantages, and Limitations

Advantages

  • Simplest analog modulation scheme to generate (a nonlinear device or multiplier plus carrier source).
  • Demodulation requires no carrier recovery circuit — envelope detector is trivially cheap.
  • Constant RF bandwidth of 2W regardless of modulation depth.
  • Robust reception with simple superheterodyne receivers.

Limitations

  • Power efficiency ≤ 33% — most power wasted in the information-free carrier.
  • Bandwidth = 2W is double that of SSB — spectrally inefficient.
  • Susceptible to amplitude noise (impulse noise, fading) since the message lives in amplitude.
  • Over-modulation causes distortion and adjacent-channel interference.

Typical Applications

VariantCarrier transmitted?Sidebands transmittedBandwidthEfficiency (max)
DSB-FC (standard AM)YesBoth2W33%
DSB-SCSuppressedBoth2W100%
SSBSuppressedOneW100%
Vestigial SB (TV video)Partial carrierOne full + vestige of otherW + vestigeHigh

10. Worked Examples

Example 10.1 — Modulation index from the envelope

An AM wave has a maximum envelope amplitude of 15 V and a minimum of 5 V. Find the modulation index and carrier amplitude.

ma = (Amax − Amin)/(Amax + Amin) = (15 − 5)/(15 + 5) = 10/20 = 0.5 (50%)
Ac = (Amax + Amin)/2 = (15 + 5)/2 = 10 V;   Am = maAc = 5 V

Example 10.2 — Bandwidth and spectrum

A 1 MHz carrier is modulated by a 5 kHz tone with ma = 0.8. Sketch the spectrum and find the bandwidth.

Example 10.3 — Power and efficiency

An AM transmitter radiates 10 kW of carrier power with ma = 0.6. Find the total power and the power in each sideband.

PT = Pc(1 + ma²/2) = 10 000 × (1 + 0.36/2) = 10 000 × 1.18 = 11 800 W
Each sideband: PSB = ma²Pc/4 = 0.36 × 10 000/4 = 900 W (both sidebands: 1800 W)
η = ma²/(2 + ma²) = 0.36/2.36 ≈ 15.3%

11. Summary of Key Results

QuantityFormula
AM wave (time domain)s(t) = Ac[1 + macos(2πfmt)]cos(2πfct)
Modulation indexma = Am/Ac = (Amax − Amin)/(Amax + Amin)
Envelope limitsAmax = Ac(1+ma),   Amin = Ac(1−ma)
Distortion-free conditionma ≤ 1
Component frequenciesfc,   fc+fm (USB),   fc−fm (LSB)
BandwidthBW = 2fm (single tone) or 2W (message band W)
Total powerPT = Pc(1 + ma²/2),   Pc = Ac²/2R
Sideband power (each)PSB = ma²Pc/4
Efficiencyη = ma²/(2 + ma²) ≤ 33.3%
DemodulationEnvelope detector: 1/fc ≪ RC ≪ 1/W

12. Review Questions

  1. Define amplitude modulation and explain why baseband audio signals cannot be radiated directly.
  2. A carrier of 100 V peak is amplitude-modulated by a tone such that the envelope varies between 160 V and 40 V. Determine ma, Am, and state whether the modulation is acceptable.
  3. Derive the three-component expansion of the AM wave using a product-to-sum identity, and sketch its spectrum labelling all amplitudes.
  4. Explain, with a waveform sketch, what happens when ma > 1 and why an envelope detector then produces distortion.
  5. An FM-broadcast-quality audio signal (W = 15 kHz) amplitude-modulates a 1 MHz carrier. Find the required channel bandwidth.
  6. A transmitter delivers 5 kW total power at ma = 1. Find Pc, the power in each sideband, and the efficiency.
  7. Why is standard AM said to be power-inefficient yet receiver-simple? Compare with DSB-SC and SSB on bandwidth and efficiency.
  8. State the RC design condition for an envelope detector and explain the consequences of choosing RC too large (diagonal clipping) and too small (excessive ripple).
Selected answers: Q2 — ma = 0.6, Am = 60 V (acceptable). Q5 — 30 kHz. Q6 — Pc = 5 kW/1.5 = 3.33 kW; each sideband = 833 W; η = 33.3%.

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