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Double-tone AM

Short answer

Double-tone AM uses two modulating sinusoids. Total modulation index μt=√(μ₁²+μ₂²), peak condition μ₁+μ₂≤1. Sideband power is Pc μt²/2, bandwidth 2×max(f₁,f₂). For Ac=10V, μ₁=0.4, μ₂=0.3, fc=100kHz, f₁=1kHz, f₂=2kHz, R=50Ω: μt=0.5, Pc=1W, Pt=1.125W, BW=4kHz, η=11.1%.

Write a short note on double-tone AM.

In double-tone (two-tone) AM, the carrier is amplitude-modulated by a message that contains two sinusoids of different frequencies f₁ and f₂ at the same time (for example a two-tone test signal).

Expression

m(t) = Am₁ cos(2πf₁t) + Am₂ cos(2πf₂t)

s(t) = Ac [1 + μ₁ cos(2πf₁t) + μ₂ cos(2πf₂t)] cos(2πfct)

where μ₁ = Am₁/Ac and μ₂ = Am₂/Ac. Expanding:

s(t) = Ac cos ωct + (Acμ₁/2)[cos(ωc+ω₁)t + cos(ωc-ω₁)t] + (Acμ₂/2)[cos(ωc+ω₂)t + cos(ωc-ω₂)t]

Spectrum

 amplitude
       |
 Ac    |           |
       |           |
 Acu1/2|    |      |      |
 Acu2/2| |  |      |      |  |
       +-+--+------+------+--+----> f
       fc-f2 fc-f1 fc  fc+f1 fc+f2

The signal has the carrier and four side frequencies (two for each tone).

Total modulation index

μt = √(μ₁² + μ₂²). To avoid over-modulation and allow distortion-free envelope detection, the peak condition μ₁ + μ₂ ≤ 1 must hold. This is stricter than μt ≤ 1.

Power

Pc = Ac²/2R and Pt = Pc (1 + μ₁²/2 + μ₂²/2) = Pc (1 + μt²/2)

Sideband power Ps = Pc μt²/2 and efficiency η = μt² / (2 + μt²).

Bandwidth

BW = 2·f_max = 2·max(f₁, f₂).

Example

Ac = 10 V, μ₁ = 0.4, μ₂ = 0.3, fc = 100 kHz, f₁ = 1 kHz, f₂ = 2 kHz, R = 50 Ω.

  • μt = √(0.16 + 0.09) = 0.5
  • Pc = 100/(2×50) = 1 W
  • Pt = 1 (1 + 0.25/2) = 1.125 W
  • BW = 2 × 2 kHz = 4 kHz
  • η = 0.25/2.25 = 11.1%

✓ Answer checked by Claude Sonnet 5.5

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