---
title: "Angle modulated signal 12 sin(10⁶t + 5 sin10⁴t)"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:angle-mod-12sin-numerical"
source: https://www.bctengineer.com/q/semester_6/communication_systems/angle-mod-12sin-numerical
site: BCT Engineering
---

# Angle modulated signal 12 sin(10⁶t + 5 sin10⁴t)

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2080 Chaitra (CS I), 4×1.5 marks
- Video: [Numerical related to Angle modulation| An angle modulated signal with carrier frequency](https://www.youtube.com/watch?v=2SAyNo_m9Wo) — Technically Explained
- 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-12sin-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

The equation of an angle modulated signal is v(t) = 12 sin(10⁶t + 5 sin 10⁴t). Determine the following:
a) Carrier frequency
b) Modulating frequency
c) Modulation index
d) Power dissipated in 100 Ω resistor

## Short answer

Identify A_c=12, ω_c=10^6 rad/s, ω_m=10^4 rad/s, β=5. Carrier frequency f_c = ω_c/(2π) = 159.15 kHz, modulating frequency f_m = ω_m/(2π) ≈ 1.59 kHz, modulation index β = 5. Power dissipated in 100 Ω resistor P = A_c²/(2R) = 12²/(2×100) = 0.72 W.

## Answer

Compare with the standard single-tone angle-modulated wave $v(t) = A_c\sin(\omega_ct + \beta\sin\omega_mt)$:

$A_c = 12$ V, $\omega_c = 10^6$ rad/s, $\omega_m = 10^4$ rad/s, $\beta = 5$.

#### a) Carrier frequency

$$
f_c = \frac{\omega_c}{2\pi} = \frac{10^6}{2\pi} = 159.15\ \text{kHz}
$$

#### b) Modulating frequency

$$
f_m = \frac{\omega_m}{2\pi} = \frac{10^4}{2\pi} = 1591.55\ \text{Hz} \approx 1.59\ \text{kHz}
$$

#### c) Modulation index

The peak phase deviation is the coefficient of $\sin 10^4t$:

$$
\beta = 5\ \text{rad}
$$

(Treated as FM, the peak deviation is $\Delta f = \beta f_m = 5 \times 1591.55 = 7957.7$ Hz.)

#### d) Power in 100 Ω

The amplitude of an angle-modulated wave is constant, so

$$
P = \frac{A_c^2}{2R} = \frac{12^2}{2 \times 100} = \frac{144}{200} = 0.72\ \text{W}
$$

**Answer:** $f_c = 159.15$ kHz, $f_m = 1.59$ kHz, $\beta = 5$, $P = 0.72$ W.

## Check yourself

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

1. How are the parameters read from v(t) = 12 sin(10^6 t + 5 sin 10^4 t)?
   A. Ac = 12 V, fc = 10^6 Hz, fm = 10^4 Hz, beta = 5
   B. Ac = 12 V, wc = 10^6 rad/s, wm = 10^4 rad/s, beta = 5
   C. Ac = 6 V, wc = 10^6 rad/s, wm = 10^4 rad/s, beta = 5
   D. Ac = 12 V, wc = 10^4 rad/s, wm = 10^6 rad/s, beta = 5
   Answer: B. Matching with Ac sin(wc t + beta sin wm t) gives the values directly; the frequencies are in rad/s. (Key point: Compare with standard form)

2. What is the carrier frequency in Hz?
   A. 10^6 Hz = 1 MHz
   B. 10^6 x 2 pi = 6.28 MHz
   C. 10^6 / (2 pi) = 159.15 kHz
   D. 10^6 / pi = 318.3 kHz
   Answer: C. The coefficient of t is angular, so f_c = w_c/(2 pi). (Key point: Carrier frequency)

3. What is the modulation index?
   A. 5 x 159.15 kHz
   B. 10^4 / 10^6 = 0.01
   C. 12 / 5 = 2.4
   D. 5 rad, the coefficient of sin 10^4 t
   Answer: D. The peak phase deviation in the argument is beta. (Key point: Modulation index)

4. Why does the power in 100 ohm not depend on the modulation?
   A. Angle modulation has constant amplitude, so P = Ac^2/(2R) = 0.72 W
   B. Power rises with beta, so P = 0.72 W is only the minimum
   C. Power equals Ac^2/R = 1.44 W
   D. Power is shared among sidebands so P falls below the carrier power
   Answer: A. The envelope is constant at 12 V, so P = 144/200 = 0.72 W. (Key point: Power in resistor)

## Related in Communication Systems

- [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
- [Angle modulated signal 20cos(6×10⁸t + 7sin1250t)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Aangle-mod-20cos-6e8t) — Communication Systems, chapter 3
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- [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
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