---
title: "Angle modulated signal 20cos(6×10⁸t + 7sin1250t)"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:angle-mod-20cos-6e8t"
source: https://www.bctengineer.com/q/semester_6/communication_systems/angle-mod-20cos-6e8t
site: BCT Engineering
---

# Angle modulated signal 20cos(6×10⁸t + 7sin1250t)

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2073 Magh (CS I), 2.5×4 marks
- Video: [Frequency Modulation (FM Modulation)- FM Wave Equation- Modulation Index of FM - Frequency Deviation](https://www.youtube.com/watch?v=32DPN8Bn62I) — Engineering Made Easy
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/angle-mod-20cos-6e8t
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

The angle modulated signal is given by S(t) = 20 cos(6×10⁸t + 7 sin 1250t). Determine:
i) The carrier and modulating frequency
ii) The modulation index
iii) Maximum frequency deviation
iv) Power dissipated in 10 Ω resistor

## Short answer

Carrier f_c = 6×10⁸/(2π) ≈ 95.49 MHz, modulating f_m = 1250/(2π) ≈ 198.94 Hz. Modulation index β = 7, maximum deviation Δf = β f_m ≈ 1.39 kHz. Power in 10 Ω = 20²/(2×10) = 20 W.

## Answer

Compare with $s(t) = A_c\cos(\omega_ct + \beta\sin\omega_mt)$:

$A_c = 20$ V, $\omega_c = 6 \times 10^8$ rad/s, $\omega_m = 1250$ rad/s, $\beta = 7$.

#### i) Carrier and modulating frequency

$$
\begin{aligned}
f_c &= \frac{6 \times 10^8}{2\pi} = 95.49 \times 10^6\ \text{Hz} = 95.49\ \text{MHz} \\
f_m &= \frac{1250}{2\pi} = 198.94\ \text{Hz}
\end{aligned}
$$

#### ii) Modulation index

$$
\beta = 7
$$

#### iii) Maximum frequency deviation

$$
\Delta f = \beta f_m = 7 \times 198.94 = 1392.6\ \text{Hz} \approx 1.39\ \text{kHz}
$$

#### iv) Power in 10 Ω

$$
P = \frac{A_c^2}{2R} = \frac{20^2}{2 \times 10} = 20\ \text{W}
$$

**Answer:** $f_c = 95.49$ MHz, $f_m = 198.94$ Hz; $\beta = 7$; $\Delta f = 1392.6$ Hz; $P = 20$ W.

## Check yourself

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

1. What is fc for 20 cos(6x10^8 t + 7 sin 1250 t)?
   A. 6x10^8/(2 pi) = 95.49 MHz
   B. 600 MHz
   C. 6x10^8 x 2 pi
   D. 95.49 kHz
   Answer: A. Angular frequency divided by 2 pi. (Key point: Carrier frequency)

2. How is the maximum deviation found?
   A. Delta f = beta / fm
   B. Delta f = beta fm = 7 x 198.94 = 1392.6 Hz
   C. Delta f = 1250 Hz
   D. Delta f = 7 x 1250 = 8750 Hz
   Answer: B. Delta f = beta fm with fm = 1250/2pi. (Key point: Deviation)

3. What power in 10 ohm?
   A. 20^2/10 = 40 W
   B. 20/10 = 2 W
   C. 20^2/(2 x 10) = 20 W
   D. 20^2 x 10/2 = 2000 W
   Answer: C. Constant amplitude 20 V. (Key point: Power)

## Related in Communication Systems

- [Angle modulated signal 12 sin(10⁶t + 5 sin10⁴t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-12sin-numerical) — Communication Systems, chapter 3
- [Angle modulated voltage 10 sin(10⁸t + 3 sin10⁴t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-10sin-1e8t) — Communication Systems, chapter 3
- [FM numerical: deviation, index, 98% bandwidth, power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-numerical-10cos-80cos-400hz) — Communication Systems, chapter 3, asked 2×
- [Modulated signal Z(t) = 50cos{98.6×10⁶×2πt + 15sin(2×10³×2πt)}](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-50cos-98-6mhz) — Communication Systems, chapter 3
- [FM numerical: 10 kHz baseband, 100 MHz carrier, 75 kHz deviation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-10khz-100mhz-75khz) — Communication Systems, chapter 3
- [AM transmitter 10 kW/11.8 kW: modulation index and power](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aam-10kw-11-8kw) — Communication Systems, chapter 2

## Related in other subjects

- [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/power signals; periodicity of cos + sin](https://www.bctengineer.com/question.md?key=semester_7%2Fdigital_signal_analysis_and_processing%3Aenergy-power-periodic-2) — Digital Signal Analysis and Processing, chapter 1