Amplitude Modulation: AM, DSB-SC and SSB

Conventional AM, modulation index, power and efficiency, envelope detection, DSB-SC and coherent detection, SSB and VSB, superheterodyne receivers and image frequency.

Amplitude modulation varies the amplitude of a high-frequency carrier with the message. It is simple, and appears in GATE as power, bandwidth and receiver questions.

Conventional AM (DSB-FC)

s(t) = A_c[1 + μ cos ω_m t] cos ω_c t

  • Modulation index μ = A_m/A_c (must be ≤ 1 to avoid over-modulation). From the envelope: μ = (A_max − A_min)/(A_max + A_min).
  • Spectrum: carrier at f_c and sidebands at f_c ± f_m. Bandwidth = 2f_m (2W for a message of bandwidth W).
  • Total power P_t = P_c(1 + μ²/2); sideband power = P_c μ²/2.
  • Power efficiency η = μ²/(2 + μ²): at most 33.3% (μ = 1).
  • Multiple tones: μ_t = √(μ₁² + μ₂² + …).
  • Envelope detector: simple; needs μ ≤ 1 and an RC time constant with 1/f_c ≪ RC ≪ 1/W.

DSB-SC

s(t) = m(t) cos ω_c t: no carrier power wasted, bandwidth 2W. Needs coherent (synchronous) detection: multiply by a local carrier and low-pass filter. A phase error φ scales the output by cos φ (zero at 90°). Carrier recovery by squaring loop or Costas loop.

SSB and VSB

  • SSB: only one sideband, bandwidth W: saves bandwidth and power. Generated by filtering or the phase-shift (Hilbert transform) method.
  • VSB: one sideband plus a vestige of the other; used for TV picture signals where the message has low-frequency content.

Superheterodyne receiver

All stations are converted to a fixed intermediate frequency (IF) (455 kHz for AM broadcast).

  • Local oscillator f_LO = f_s + f_IF (usually above the signal).
  • Image frequency f_image = f_s + 2f_IF; it must be rejected by the RF stage.
  • Image rejection ratio α = √(1 + Q²ρ²), with ρ = f_image/f_s − f_s/f_image.

Worked example

A 100 W carrier is amplitude modulated with μ = 0.5.

  • P_t = 100(1 + 0.125) = 112.5 W; sideband power 12.5 W; efficiency = 12.5/112.5 = 11.1%.

Practice

GATE-styleNAT · numerical answer

An AM wave has a maximum envelope of 15 V and a minimum of 5 V. What is the modulation index?

Show answer and explanationAnswer: 0.5

μ = (15 − 5)/(15 + 5) = 0.5.

GATE-styleNAT · numerical answer

An AM broadcast receiver with IF = 455 kHz is tuned to 1000 kHz. What is the image frequency in kHz?

Show answer and explanationAnswer: 1910

f_image = f_s + 2f_IF = 1910 kHz.

GATE-styleNAT · numerical answer

What is the maximum power efficiency of conventional AM (single tone), in percent? (Give 2 decimal places.)

Show answer and explanationAnswer: 33.33

η = μ²/(2 + μ²) = 1/3 at μ = 1.

GATE-styleNAT · numerical answer

A DSB-SC signal is detected with a local carrier that has a 60° phase error. The output amplitude becomes what fraction of the correct value?

Show answer and explanationAnswer: 0.5

The output scales by cos φ = cos 60° = 0.5.