---
title: "AM numerical: 20 sin2π500t on 60 sin2π10⁶t"
subject: "Data Communication"
chapter: 5
key: "semester_5/data_communication:am-numeric-20sin500"
source: https://www.bctengineer.com/q/semester_5/data_communication/am-numeric-20sin500
site: BCT Engineering
---

# AM numerical: 20 sin2π500t on 60 sin2π10⁶t

Data Communication · Chapter 5: Data Encoding and Modulation

- Asked in: 2070 Chaitra, 4 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/am-numeric-20sin500
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-5

## Question

An audio frequency signal 20sin2π·500t is used for AM with a carrier of 60sin2π·10⁶t. Calculate: (a) modulation index (b) Sideband frequencies, (c) Bandwidth requirements, (d) Total power delivered with a load of 75 Ohm.

## Short answer

μ = Am/Ac = 20/60 = 0.333. Sidebands: 1,000,500 Hz and 999,500 Hz. Bandwidth = 2×500 = 1000 Hz. Carrier power Pc = 60²/(2×75) = 24 W. Total power Pt = 24×(1+0.333²/2) ≈ 25.33 W.

## Answer

Message: Am = 20 V, fm = 500 Hz. Carrier: Ac = 60 V, fc = 10⁶ Hz = 1 MHz. R = 75 Ω.

#### a) Modulation index
μ = Am/Ac = 20/60 = **0.333** (33.3%)

#### b) Sideband frequencies
- Upper sideband: fc + fm = 1,000,000 + 500 = **1,000,500 Hz**
- Lower sideband: fc - fm = 1,000,000 - 500 = **999,500 Hz**

#### c) Bandwidth
BW = 2fm = 2 × 500 = **1000 Hz**

#### d) Total power
Pc = Ac²/(2R) = 3600/150 = 24 W

Pt = Pc (1 + μ²/2) = 24 × (1 + 0.1111/2) = 24 × 1.0556 = **25.33 W**

(Sideband power = 1.33 W.)

## Check yourself

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

1. For message amplitude 20 V and carrier amplitude 60 V, what is the modulation index?
   A. 3
   B. 0.333
   C. 0.5
   D. 0.667
   Answer: B. mu = Am/Ac = 20/60 = 0.333 (33.3%). (Key point: Modulation index)

2. With fc = 1 MHz and fm = 500 Hz, what are the sidebands?
   A. 1,000,500 Hz and 999,500 Hz
   B. 1,500,000 Hz and 500,000 Hz
   C. 1,000,250 Hz and 999,750 Hz
   D. 1,001,000 Hz and 999,000 Hz
   Answer: A. USB = fc + fm and LSB = fc - fm. (Key point: Sideband frequencies)

3. What is the AM bandwidth requirement?
   A. 500 Hz
   B. 1000 Hz
   C. 2 MHz
   D. 1 MHz
   Answer: B. BW = 2fm = 2 x 500 = 1000 Hz. (Key point: Bandwidth)

4. With Ac = 60 V and R = 75 ohm, Pc = 24 W. What is the total power Pt?
   A. 24 W
   B. 26.67 W
   C. 25.33 W
   D. 48 W
   Answer: C. Pt = Pc(1 + mu^2/2) = 24 x 1.0556 = 25.33 W. (Key point: Total power)

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