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
An AM wave has a maximum envelope of 15 V and a minimum of 5 V. What is the modulation index?
Show answer and explanation
Answer: 0.5μ = (15 − 5)/(15 + 5) = 0.5.
An AM broadcast receiver with IF = 455 kHz is tuned to 1000 kHz. What is the image frequency in kHz?
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Answer: 1910f_image = f_s + 2f_IF = 1910 kHz.
What is the maximum power efficiency of conventional AM (single tone), in percent? (Give 2 decimal places.)
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Answer: 33.33η = μ²/(2 + μ²) = 1/3 at μ = 1.
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?
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Answer: 0.5The output scales by cos φ = cos 60° = 0.5.