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AM wave power and bandwidth numerical

Short answer

Ac=40 V, μ=0.8, fm=1 kHz, R=30 Ω. Carrier power Pc = Ac²/(2R) = 26.67 W. Total sideband power Psb = (μ²/2)Pc = 8.53 W. Bandwidth = 2fm = 2 kHz.

The equation of an amplitude modulated wave is given by s(t) = 40[1+0.8cos(2π×10³t)]cos(4π×10⁵t). Find the carrier power, the total sideband power and bandwidth of the signal. The value of the resistor is given as 30 ohm.

s(t) = 40 [1 + 0.8 cos(2π×10³t)] cos(4π×10⁵t)

Compare with the standard form s(t) = Ac [1 + μ cos ωmt] cos ωct:

  • Ac = 40 V, μ = 0.8
  • ωm = 2π×10³, so fm = 1 kHz
  • ωc = 4π×10⁵ = 2π×(2×10⁵), so fc = 200 kHz
  • R = 30 Ω

Carrier power

Pc = Ac² / (2R) = 40² / (2 × 30) = 1600/60 = 26.67 W

Total sideband power

Psb = (μ²/2) Pc = (0.64/2) × 26.67 = 0.32 × 26.67 = 8.53 W (each sideband carries 4.27 W)

Total power Pt = Pc + Psb = 26.67 + 8.53 = 35.2 W.

Bandwidth

BW = 2fm = 2 × 1 kHz = 2 kHz

Answer: Pc = 26.67 W, sideband power = 8.53 W, bandwidth = 2 kHz.

✓ Answer checked by Claude Sonnet 5.5

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