---
title: "AM transmitter 10 kW/11.8 kW: modulation index and power"
subject: "Communication Systems"
chapter: 2
key: "semester_6/communication_systems:am-10kw-11-8kw"
source: https://www.bctengineer.com/q/semester_6/communication_systems/am-10kw-11-8kw
site: BCT Engineering
---

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

Communication Systems · Chapter 2: Amplitude Modulation

- Asked in: 2072 Asoj (CS I), 3+4 marks
- Video: [Modulation Index in Amplitude Modulation - Modern Electronic Communications](https://www.youtube.com/watch?v=rzznMfag2Fk) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/am-10kw-11-8kw
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-2

## Question

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.

## 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.

## Answer

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

#### Modulation index

$$
P_t = P_c\left(1 + \frac{\mu^2}{2}\right)
$$

$$
\begin{aligned}
\frac{11.8}{10} &= 1 + \frac{\mu^2}{2} \\
\mu^2 &= 2(1.18 - 1) = 0.36 \\
\mu &= 0.6
\end{aligned}
$$

#### Total power with an additional 30% tone

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

$$
\mu_t^2 = \mu_1^2 + \mu_2^2 = 0.36 + 0.09 = 0.45 \quad (\mu_t = 0.671)
$$

$$
P_t = P_c\left(1 + \frac{\mu_t^2}{2}\right) = 10\left(1 + \frac{0.45}{2}\right) = 10 \times 1.225 = 12.25\ \text{kW}
$$

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

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

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which relation is used to find mu from Pc and Pt?
   A. Pt = Pc(1 + mu)
   B. Pt = Pc mu^2
   C. Pt = Pc(1 + mu^2/2)
   D. Pt = Pc/(1 + mu^2)
   Answer: C. The total power of AM is the carrier power plus the sideband power: Pc(1 + mu^2/2). (Key point: Power relation)

2. For Pc = 10 kW and Pt = 11.8 kW, mu is:
   A. 0.36
   B. 0.6
   C. 0.18
   D. 0.8
   Answer: B. mu^2 = 2(1.18 - 1) = 0.36, so mu = 0.6. (Key point: Modulation index)

3. When a second tone is added, the effective index squared is:
   A. The sum of the squares, 0.36 + 0.09 = 0.45
   B. (0.6 + 0.3)^2 = 0.81
   C. 0.6 x 0.3 = 0.18
   D. 0.3^2 = 0.09
   Answer: A. Sideband powers add, so mu_t^2 = mu1^2 + mu2^2. (Key point: Combining tones)

4. The total radiated power with both tones is:
   A. 11.8 kW
   B. 12.25 kW
   C. 10.45 kW
   D. 14.1 kW
   Answer: B. Pt = 10(1 + 0.45/2) = 12.25 kW, which is 11.8 kW plus 0.45 kW from the second tone. (Key point: Final power)

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