---
title: "FM numerical: u(t) = 10cos(100,000t + 10sin5000t)"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:fm-numerical-2cos5000t-10sin"
source: https://www.bctengineer.com/q/semester_6/communication_systems/fm-numerical-2cos5000t-10sin
site: BCT Engineering
---

# FM numerical: u(t) = 10cos(100,000t + 10sin5000t)

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2064 Shrawan (CS I), 4×2 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/fm-numerical-2cos5000t-10sin
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

A low frequency signal m(t) = 2cos(5000t) modulates the carrier signal c(t) = 10cos(100,000t) to produce the modulated signal u(t) = 10cos(100,000t + 10sin5000t). Calculate: the total modulated signal power; modulation index; peak frequency deviation; the bandwidth of modulated signal.

## Short answer

Modulation index from signal is β = 10. Peak frequency deviation Δf = β f_m = 10×(5000/(2π)) ≈ 7.96 kHz. Bandwidth B_T = 2(β+1)f_m ≈ 17.51 kHz. Total power P = A_c²/2 =50 W.

## Answer

**Given:** $m(t) = 2\cos(5000t)$, $c(t) = 10\cos(100000t)$, $u(t) = 10\cos(100000t + 10\sin 5000t)$.

So $A_m = 2$ V, $\omega_m = 5000$ rad/s, $A_c = 10$ V, $\omega_c = 10^5$ rad/s, $\beta = 10$.

$$
f_m = \frac{5000}{2\pi} = 795.77\ \text{Hz}, \qquad f_c = \frac{10^5}{2\pi} = 15.915\ \text{kHz}
$$

#### Total modulated signal power

FM has constant amplitude; normalised to 1 Ω:

$$
P = \frac{A_c^2}{2} = \frac{10^2}{2} = 50\ \text{W}
$$

#### Modulation index

From the expression: $\beta = 10$.

#### Peak frequency deviation

$$
\Delta f = \beta f_m = 10 \times 795.77 = 7957.7\ \text{Hz} \approx 7.96\ \text{kHz}
$$

($\Delta\omega = 10 \times 5000 = 50000$ rad/s; $k_f = \Delta\omega/A_m = 25000$ rad/s/V, i.e. 3978.9 Hz/V.)

#### Bandwidth (Carson's rule)

$$
B_T = 2(\Delta f + f_m) = 2(\beta + 1)f_m = 2 \times 11 \times 795.77 = 17507\ \text{Hz} \approx 17.51\ \text{kHz}
$$

(In rad/s: $2 \times 11 \times 5000 = 110000$ rad/s.)

**Answer:** $P = 50$ W, $\beta = 10$, $\Delta f = 7.96$ kHz, $B_T \approx 17.51$ kHz.

## Check yourself

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

1. For u(t) = 10 cos(100000 t + 10 sin 5000 t), what are beta and fm?
   A. beta = 10, fm = 795.8 Hz
   B. beta = 5000, fm = 10 Hz
   C. beta = 10, fm = 5000 Hz
   D. beta = 2, fm = 795.8 Hz
   Answer: A. beta = 10 from the expression and fm = 5000/(2π) = 795.77 Hz. (Key point: Reading beta and fm)

2. What is the peak frequency deviation?
   A. 50 kHz
   B. 10 Hz
   C. 7.96 kHz
   D. 795.8 Hz
   Answer: C. delta f = beta fm = 10 x 795.77 = 7957.7 Hz (50000 rad/s). (Key point: Peak deviation)

3. What is the total power (1 Ω)?
   A. 100 W
   B. 25 W
   C. 10 W
   D. 50 W
   Answer: D. P = Ac^2/2 = 100/2 = 50 W, independent of beta. (Key point: Power)

4. What is the Carson bandwidth?
   A. 15.9 kHz
   B. 17.51 kHz
   C. 8.75 kHz
   D. 7.96 kHz
   Answer: B. BT = 2(beta + 1) fm = 2 x 11 x 795.77 = 17507 Hz. (Key point: Bandwidth)

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