---
title: "AM wave power and bandwidth numerical"
subject: "Data Communication"
chapter: 5
key: "semester_5/data_communication:am-numeric-40-carrier-power"
source: https://www.bctengineer.com/q/semester_5/data_communication/am-numeric-40-carrier-power
site: BCT Engineering
---

# AM wave power and bandwidth numerical

Data Communication · Chapter 5: Data Encoding and Modulation

- Asked in: 2079 Baisakh, 4+4+2 marks
- Video: [Frequency Spectrum, Bandwidth and Power of AM Wave](https://www.youtube.com/watch?v=ibhBiC47yAQ) — Sugandh Gupta
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/am-numeric-40-carrier-power
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-5

## Question

The equation of an amplitude modulated wave is given by s(t) = 40[1+0.8cos(2π×10³t)]cos(4π×10⁵t). Find the carrier power, the total sideband power and bandwidth of the signal. The value of the resistor is given as 30 ohm.

## Short answer

Ac=40 V, μ=0.8, fm=1 kHz, R=30 Ω. Carrier power Pc = Ac²/(2R) = 26.67 W. Total sideband power Psb = (μ²/2)Pc = 8.53 W. Bandwidth = 2fm = 2 kHz.

## Answer

s(t) = 40 [1 + 0.8 cos(2π×10³t)] cos(4π×10⁵t)

Compare with the standard form s(t) = Ac [1 + μ cos ωmt] cos ωct:
- Ac = 40 V, μ = 0.8
- ωm = 2π×10³, so fm = 1 kHz
- ωc = 4π×10⁵ = 2π×(2×10⁵), so fc = 200 kHz
- R = 30 Ω

#### Carrier power
Pc = Ac² / (2R) = 40² / (2 × 30) = 1600/60 = **26.67 W**

#### Total sideband power
Psb = (μ²/2) Pc = (0.64/2) × 26.67 = 0.32 × 26.67 = **8.53 W**
(each sideband carries 4.27 W)

Total power Pt = Pc + Psb = 26.67 + 8.53 = 35.2 W.

#### Bandwidth
BW = 2fm = 2 × 1 kHz = **2 kHz**

**Answer: Pc = 26.67 W, sideband power = 8.53 W, bandwidth = 2 kHz.**

## Check yourself

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

1. From s(t)=40[1+0.8cos(2π×10³t)]cos(4π×10⁵t), fc is:
   A. 400 kHz
   B. 200 kHz
   C. 100 kHz
   D. 1 kHz
   Answer: B. 4π×10⁵ = 2π×2×10⁵. (Key point: Parameters)

2. Pc with R = 30 Ω:
   A. 53.3 W
   B. 13.3 W
   C. 26.67 W
   D. 35.2 W
   Answer: C. 1600/60. (Key point: Carrier power)

3. Total sideband power:
   A. 26.67 W
   B. 4.27 W
   C. 8.53 W
   D. 35.2 W
   Answer: C. 0.32 x 26.67. (Key point: Sideband power)

4. Bandwidth is:
   A. 2 kHz
   B. 1 kHz
   C. 200 kHz
   D. 4 kHz
   Answer: A. 2fm. (Key point: Bandwidth)

## Related in Data Communication

- [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 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: 10 sin1000πt on 50 sin 2π×10⁵t](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-10sin1000pi) — Data Communication, chapter 5
- [Energy or power of DT cos(3πk/16)](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Adt-signal-cos-energy-power) — Data Communication, chapter 2
- [FDM bandwidth for 10 voice channels](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afdm-10-voice-channels) — Data Communication, chapter 6

## Related in other subjects

- [AM numerical: S(t) = 10(1 + 0.8cos25132.74t)cos(9424777.96t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-10-0-8-25132-numerical) — Communication Systems, chapter 2
- [AM numerical: s(t) = 5[1 + 0.6cos(6280t)]cos(2π10⁴t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-5-0-6cos6280t-numerical) — Communication Systems, chapter 2
- [AM numerical: s(t) = 5[1 + 0.6cos(3140t)]cos(2π10³t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-5-0-6cos3140t-numerical) — Communication Systems, chapter 2
- [AM numerical: S(t) = 10(1 + 0.2cos2π10³t)cos2π10⁶t](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-numerical-10-0-2) — Communication Systems, chapter 2