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AM transmitter 10 kW/11.8 kW: modulation index and power

Short answer

Using Pt = Pc(1 + μ²/2) with Pc=10 kW, Pt=11.8 kW gives μ = 0.6. With an additional 30% tone, total modulation index squared is 0.45, so total radiated power = 10(1 + 0.45/2) = 12.25 kW.

A certain AM transmitter radiates 10 kW with the carrier unmodulated and 11.8 kW when the carrier is sinusoidally modulated. Calculate the modulation index. If another sine wave, corresponding to 30 percent modulation, is transmitted simultaneously, determine the total radiated power.

Given: Pc=10P_c = 10 kW (unmodulated), Pt=11.8P_t = 11.8 kW (modulated).

Modulation index

Pt=Pc(1+μ22)P_t = P_c\left(1 + \frac{\mu^2}{2}\right) 11.810=1+μ22μ2=2(1.18−1)=0.36μ=0.6\begin{aligned} \frac{11.8}{10} &= 1 + \frac{\mu^2}{2} \\ \mu^2 &= 2(1.18 - 1) = 0.36 \\ \mu &= 0.6 \end{aligned}

Total power with an additional 30% tone

Second tone μ2=0.3\mu_2 = 0.3, transmitted together with μ1=0.6\mu_1 = 0.6:

μt2=μ12+μ22=0.36+0.09=0.45(μt=0.671)\mu_t^2 = \mu_1^2 + \mu_2^2 = 0.36 + 0.09 = 0.45 \quad (\mu_t = 0.671) Pt=Pc(1+μt22)=10(1+0.452)=10×1.225=12.25 kWP_t = P_c\left(1 + \frac{\mu_t^2}{2}\right) = 10\left(1 + \frac{0.45}{2}\right) = 10 \times 1.225 = 12.25\ \text{kW}

Check: extra sideband power due to second tone =10×0.09/2=0.45= 10 \times 0.09/2 = 0.45 kW; 11.8+0.45=12.2511.8 + 0.45 = 12.25 kW.

Answer: μ=0.6\mu = 0.6 (60%); total radiated power =12.25= 12.25 kW.

✓ Answer checked by Claude Opus 5.5 (high effort)

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