---
title: "Angle modulated voltage 10 sin(10⁸t + 3 sin10⁴t)"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:angle-mod-10sin-1e8t"
source: https://www.bctengineer.com/q/semester_6/communication_systems/angle-mod-10sin-1e8t
site: BCT Engineering
---

# Angle modulated voltage 10 sin(10⁸t + 3 sin10⁴t)

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2071 Magh (CS I), 2+2+2+2 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-10sin-1e8t
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

The equation of an angle modulation voltage is E = 10 sin(10⁸t + 3 sin 10⁴t). Calculate the carrier and modulating frequency, modulating index and power dissipated in 100 Ω resistor.

## Short answer

Compare E = 10 sin(10^8 t + 3 sin 10^4 t) to standard form: carrier amplitude 10 V, ω_c=10^8 rad/s, ω_m=10^4 rad/s, β=3. Carrier frequency f_c = 10^8/(2π) ≈ 15.92 MHz, modulating frequency f_m = 10^4/(2π) ≈ 1.59 kHz. Power in 100 Ω resistor: P = (10^2)/(2×100) = 0.5 W.

## Answer

**Given:** $E = 10\sin(10^8 t + 3\sin 10^4 t)$ V, $R = 100\ \Omega$.

Comparing with the standard form $E = A_c \sin(\omega_c t + \beta \sin \omega_m t)$:

$$
A_c = 10\ \text{V}, \quad \omega_c = 10^8\ \text{rad/s}, \quad \omega_m = 10^4\ \text{rad/s}, \quad \beta = 3
$$

#### Carrier frequency

$$
f_c = \frac{\omega_c}{2\pi} = \frac{10^8}{2\pi} = 15.915 \times 10^6\ \text{Hz} \approx 15.92\ \text{MHz}
$$

#### Modulating frequency

$$
f_m = \frac{\omega_m}{2\pi} = \frac{10^4}{2\pi} = 1591.5\ \text{Hz} \approx 1.59\ \text{kHz}
$$

#### Modulation index

$$
\beta = 3 \quad (\text{peak deviation } \Delta f = \beta f_m = 3 \times 1591.5 = 4774.6\ \text{Hz})
$$

#### Power in 100 Ω

The amplitude of an angle-modulated wave is constant, so its power equals the unmodulated carrier power:

$$
P = \frac{V_{rms}^2}{R} = \frac{(A_c/\sqrt2)^2}{R} = \frac{10^2}{2 \times 100} = 0.5\ \text{W}
$$

**Answer:** $f_c = 15.92$ MHz, $f_m = 1.59$ kHz, $\beta = 3$, $P = 0.5$ W.

## Check yourself

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

1. Comparing E = 10 sin(10^8 t + 3 sin 10^4 t) with Ac sin(wc t + beta sin wm t), what is the modulation index?
   A. 10
   B. 10^4
   C. 3
   D. 10^8
   Answer: C. The coefficient of the sin(wm t) term inside the argument is the modulation index, here beta = 3. (Key point: Reading the standard form)

2. What is the carrier frequency fc?
   A. 10^8 Hz
   B. 10^8/(2π) = 15.92 MHz
   C. 10^4/(2π) = 1.59 kHz
   D. 10^8 x 2π Hz
   Answer: B. wc = 10^8 rad/s, so fc = wc/(2π) = 15.92 MHz; the given number is angular frequency. (Key point: Carrier frequency)

3. What is the modulating frequency fm?
   A. 10^4 Hz
   B. 3 kHz
   C. 15.92 MHz
   D. 10^4/(2π) = 1.59 kHz
   Answer: D. wm = 10^4 rad/s, so fm = 10^4/(2π) = 1591.5 Hz. (Key point: Modulating frequency)

4. Why is the power in the 100 Ω resistor 0.5 W, independent of beta?
   A. Angle modulation has constant amplitude, so power equals the carrier power Ac^2/(2R) = 100/200
   B. The sidebands add power proportional to beta^2
   C. Power is Ac^2/R = 1 W
   D. Power depends on the modulating frequency
   Answer: A. The envelope of an angle-modulated wave is constant, so P = (10)^2/(2 x 100) = 0.5 W. (Key point: Power in angle modulation)

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