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AM numerical: 10 sin1000πt on 50 sin200000πt

Short answer

μ=Am/Ac=0.2. Bandwidth=2fm=1 kHz. Pc=Ac²/(2R)=50²/1600=1.56 W; total power Pt=Pc(1+μ²/2)≈1.59 W. Efficiency η=μ²/(2+μ²)×100≈1.96%.

An audio signal of 10sin1000πt is used for AM with a carrier of 50sin200000πt. Calculate: i) Modulation index ii) Required bandwidth iii) Total power using load resistance of 800 ohm iv) Efficiency of AM

Message: Am = 10 V, ωm = 1000π, so fm = 500 Hz. Carrier: Ac = 50 V, ωc = 200000π, so fc = 100 kHz. R = 800 Ω.

i) Modulation index

μ = Am/Ac = 10/50 = 0.2

ii) Bandwidth

BW = 2fm = 2 × 500 = 1 kHz

iii) Total power

Pc = Ac²/(2R) = 2500 / 1600 = 1.5625 W

Pt = Pc (1 + μ²/2) = 1.5625 × 1.02 = 1.594 W

iv) Efficiency

η = (sideband power / total power) × 100 = μ² / (2 + μ²) × 100 = 0.04/2.04 × 100 = 1.96 %

(Sideband power = 0.03125 W.)

✓ Answer checked by Claude Sonnet 5.5

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