Amplitude Modulation (AM)

Amplitude Modulation is the oldest and simplest form of modulation where the amplitude of a high-frequency carrier signal is varied in proportion to the instantaneous value of the message signal. Despite its inefficiency, AM remains widely used in broadcast radio, aviation communication, and educational applications due to its simplicity.

Mathematical Representation

Standard AM Signal

The time-domain expression for an AM signal is:

s(t) = Ac[1 + m·cos(2πfm·t)]·cos(2πfc·t)

Where:

• Ac = unmodulated carrier amplitude
• m = modulation index (0 ≤ m ≤ 1 for no distortion)
• fm = message signal frequency
• fc = carrier frequency (fc >> fm)

Modulation Index

The modulation index (also called depth of modulation) is defined as:

m = Am / Ac

Or in terms of the envelope:

m = (Amax - Amin) / (Amax + Amin)

Where Amax and Amin are the maximum and minimum amplitudes of the AM envelope.

Frequency Spectrum of AM

Expanding the AM signal mathematically:

s(t) = Ac·cos(2πfc·t) + (mAc/2)·cos[2π(fc+fm)·t] + (mAc/2)·cos[2π(fc-fm)·t]

Key observations:

• Total bandwidth: BT = 2fm
• Carrier component carries NO information
• Each sideband is a frequency-shifted replica of the message
• Upper Sideband (USB) = fc + fm
• Lower Sideband (LSB) = fc - fm

Power Analysis in AM

Total Power

PT = Pc(1 + m²/2)

Where Pc = Ac²/(2R) is the unmodulated carrier power.

Power Distribution

Transmission Efficiency

η = (Useful power)/(Total power) = m² / (2 + m²)

For m = 1: η = 1/3 = 33.33% (maximum efficiency for single-tone)

This means at best, only one-third of the transmitted power carries information. This is the major disadvantage of conventional AM.

Solved Example 1: Power Calculation

Problem: An AM transmitter has an unmodulated carrier power of 10 kW. Calculate the total power, sideband power, and efficiency when the modulation index is 0.8.

Solution:

Given: Pc = 10 kW, m = 0.8

Total power: PT = Pc(1 + m²/2) = 10(1 + 0.64/2) = 10(1.32) = 13.2 kW

Total sideband power: Psb = m²Pc/2 = (0.64)(10)/2 = 3.2 kW

Each sideband: PUSB = PLSB = m²Pc/4 = 0.64 × 10/4 = 1.6 kW

Efficiency: η = m²/(2 + m²) = 0.64/(2.64) = 24.24%

Solved Example 2: Modulation Index from Waveform

Problem: An AM wave observed on an oscilloscope shows Amax = 180 V and Amin = 60 V. Find the modulation index, carrier amplitude, and message amplitude.

Solution:

Modulation index: m = (Amax - Amin)/(Amax + Amin) = (180 - 60)/(180 + 60) = 120/240 = 0.5

Carrier amplitude: Ac = (Amax + Amin)/2 = (180 + 60)/2 = 120 V

Message amplitude: Am = (Amax - Amin)/2 = (180 - 60)/2 = 60 V

Verification: m = Am/Ac = 60/120 = 0.5 ✓

Solved Example 3: Multi-tone Modulation

Problem: A carrier is simultaneously modulated by three signals with individual modulation indices m1 = 0.3, m2 = 0.4, and m3 = 0.5. Find the effective modulation index and total power if Pc = 5 kW.

Solution:

Effective modulation index: mt = √(m1² + m2² + m3²) = √(0.09 + 0.16 + 0.25) = √0.5 = 0.707

Total power: PT = Pc(1 + mt²/2) = 5(1 + 0.5/2) = 5(1.25) = 6.25 kW

Note: mt must remain ≤ 1 to avoid overmodulation.

AM Generation Methods

Square Law Modulator

Switching Modulator

Uses a diode switching circuit driven by the carrier to multiply the message signal. The carrier acts as a switch, alternately connecting and disconnecting the message signal.

High-Level (Collector/Plate) Modulation

Used in high-power AM transmitters. The modulating signal is applied to the final RF power amplifier stage, varying its supply voltage.

AM Demodulation

Envelope Detector (Non-Coherent Detection)

The envelope detector works by:

1. Rectifying the AM signal (diode removes negative half)
2. Filtering the RF component (capacitor smooths the ripple)
3. Outputting the envelope which is the original message

Design constraint: The RC time constant must satisfy:

• RC >> 1/fc (to filter out carrier ripple)
• RC << 1/fm (to follow the message variations)

Diagonal Clipping

If RC is too large, the detector output cannot follow the envelope during rapid decreases, causing diagonal clipping distortion. To avoid this:

RC ≤ (1/ωm) × √(1-m²)/m

Overmodulation Effects

When m > 1 (overmodulation):

Comparison of AM Variants

Interview Questions

Q1: Why is the efficiency of conventional AM limited to 33.33%?

In conventional AM, the carrier component consumes 2/3 of the total transmitted power but carries no information. Only the sidebands contain the message, and together they account for at most m²/(2+m²) of total power. With maximum m=1, this gives 1/3 = 33.33%. The carrier is included solely to enable simple envelope detection at the receiver.

Q2: What happens during overmodulation (m > 1)?

When m exceeds 1, the envelope of the AM signal crosses zero and phase reversals occur. This causes: (1) severe distortion in the demodulated output, (2) spectral splatter that increases bandwidth and causes adjacent channel interference, (3) the envelope detector fails completely since it cannot track phase reversals. The signal effectively becomes a mix of AM and phase modulation.

Q3: How does an envelope detector work and what are its limitations?

An envelope detector consists of a diode followed by an RC low-pass filter. The diode rectifies the signal, and the RC filter extracts the slowly varying envelope. Limitations include: (1) it only works for m ≤ 1, (2) diagonal clipping occurs if RC is too large, (3) it cannot demodulate suppressed-carrier signals, and (4) it introduces a DC offset requiring a blocking capacitor.

Q4: Why is AM still used in broadcast radio despite its inefficiency?

AM persists because: (1) extremely simple receiver design enables low-cost mass production, (2) AM signals propagate well via skywave reflection for long-distance coverage, (3) the infrastructure is well-established globally, (4) compatibility with billions of existing receivers, and (5) for voice-only applications, the bandwidth inefficiency is acceptable given the wide channel allocations in MF/HF bands.