---
title: "AM numerical: 10 sin1000πt on 50 sin 2π×10⁵t"
subject: "Data Communication"
chapter: 5
key: "semester_5/data_communication:am-numeric-10sin1000pi"
source: https://www.bctengineer.com/q/semester_5/data_communication/am-numeric-10sin1000pi
site: BCT Engineering
---

# AM numerical: 10 sin1000πt on 50 sin 2π×10⁵t

Data Communication · Chapter 5: Data Encoding and Modulation

- Asked in: 2075 Asoj, 2×3 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/am-numeric-10sin1000pi
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-5

## Question

An audio frequency signal 10 sin1000πt is used for single tone amplitude modulation with a carrier of 50 sin2π×10⁵t. Calculate: (i) Modulation index (ii) Bandwidth requirement (iii) Total power delivered if load = 60 Ω

## Short answer

From message 10sin1000πt and carrier 50sin2π×10⁵t, Am=10 V, Ac=50 V, fm=500 Hz. Modulation index μ=Am/Ac=0.2. Bandwidth=2fm=1 kHz. Carrier power Pc=Ac²/(2×60Ω)=20.83 W; total power Pt=Pc(1+μ²/2)=21.25 W.

## Answer

Compare with the standard forms: message m(t) = Am sin(ωmt), carrier c(t) = Ac sin(ωct).
- Am = 10 V, ωm = 1000π, so fm = 1000π/2π = 500 Hz
- Ac = 50 V, ωc = 2π×10⁵, so fc = 100 kHz
- R = 60 Ω

#### (i) Modulation index
μ = Am/Ac = 10/50 = **0.2** (20%)

#### (ii) Bandwidth
BW = 2fm = 2 × 500 = **1000 Hz = 1 kHz** (the sidebands are at 99.5 kHz and 100.5 kHz)

#### (iii) Total power
Carrier power Pc = Ac²/(2R) = 2500/120 = 20.83 W

Pt = Pc (1 + μ²/2) = 20.83 × (1 + 0.04/2) = 20.83 × 1.02 = **21.25 W**

(Sideband power = 0.417 W.)

## Check yourself

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

1. Index = Am/Ac:
   A. 0.2
   B. 0.5
   C. 5
   D. 0.02
   Answer: A. 10/50. (Key point: Modulation index)

2. fm from ωm = 1000π:
   A. 1000 Hz
   B. 500 Hz
   C. 1000π Hz
   D. 250 Hz
   Answer: B. ω/2π. (Key point: fm)

3. Pt with R = 60 Ω:
   A. 20.83 W
   B. 42.5 W
   C. 21.25 W
   D. 0.417 W
   Answer: C. 20.83 x 1.02. (Key point: Total power)

4. BW:
   A. 500 Hz
   B. 2 kHz
   C. 100 kHz
   D. 1 kHz
   Answer: D. 2fm. (Key point: Bandwidth)

## Related in Data Communication

- [AM numerical: 10 sin1000πt on 50 sin200000πt](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-10sin1000pi-800) — Data Communication, chapter 5
- [AM numerical: 20 sin2π500t on 60 sin2π10⁶t](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-20sin500) — Data Communication, chapter 5
- [AM wave power and bandwidth numerical](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-40-carrier-power) — Data Communication, chapter 5
- [Single tone FM parameters](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afm-single-tone-numeric) — Data Communication, chapter 5

## Related in other subjects

- [AM numerical: 15 sin2π(1500t) audio on 60 sin2π(100,000t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-audio-15sin-60sin) — Communication Systems, chapter 2, asked 2×
- [Two-tone AM: net modulation index, power, spectrum](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-50-20-12) — Communication Systems, chapter 2, asked 2×
- [Two-tone AM: spectrum, power, net index (50, 15, 20)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-two-tone-50-15-20) — Communication Systems, chapter 2, asked 2×
- [Double-tone AM expression, bandwidth and modulation index](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Adouble-tone-am-bw-index) — Communication Systems, chapter 2