---
title: "FM numerical: m(t) = 5cos(18849.55t), 9 kHz/V"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:fm-numerical-5cos18849"
source: https://www.bctengineer.com/q/semester_6/communication_systems/fm-numerical-5cos18849
site: BCT Engineering
---

# FM numerical: m(t) = 5cos(18849.55t), 9 kHz/V

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2076 Baisakh (CS I), 8 marks; 2076 Bhadra (CS I), 2+2+2+2 marks (asked 2 times)
- Video: [Carson's Rule in Frequency Modulation - Calculating Bandwidth of FM - Solved Problems on Carson Rule](https://www.youtube.com/watch?v=dj5emtMrMY0) — 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/fm-numerical-5cos18849
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

A sinusoidal modulating signal m(t) = 5cos(18849.55t) is applied to an FM modulator that has frequency sensitivity of 9 kHz/V. The amplitude of carrier is 25 V and the carrier frequency is 88.7 MHz. Compute a) modulation index b) Carrier frequency swing c) Bandwidth using Carson rule d) total power delivered to 10 Ω resistor.

## Short answer

f_m=18849.55/(2π)=3 kHz. Δf=k_f A_m=9×5=45 kHz. Modulation index β=Δf/f_m=15. Carrier swing=2Δf=90 kHz. Carson bandwidth=2(Δf+f_m)=96 kHz. Total power in 10 Ω is A_c²/(2R)=25²/20=31.25 W.

## Answer

**Given:** $m(t) = 5\cos(18849.55t)$, so $A_m = 5$ V and

$$
f_m = \frac{18849.55}{2\pi} = 3000\ \text{Hz} = 3\ \text{kHz}
$$

$k_f = 9$ kHz/V, $A_c = 25$ V, $f_c = 88.7$ MHz, $R = 10\ \Omega$.

Peak frequency deviation:

$$
\Delta f = k_fA_m = 9 \times 5 = 45\ \text{kHz}
$$

#### a) Modulation index

$$
\beta = \frac{\Delta f}{f_m} = \frac{45}{3} = 15
$$

(wide-band FM, since $\beta \gg 1$)

#### b) Carrier frequency swing

The carrier swings from $f_c - \Delta f$ to $f_c + \Delta f$, i.e. from 88.655 MHz to 88.745 MHz:

$$
\text{Carrier swing} = 2\Delta f = 2 \times 45 = 90\ \text{kHz}
$$

#### c) Bandwidth by Carson's rule

$$
B_T = 2(\Delta f + f_m) = 2(45 + 3) = 96\ \text{kHz}
$$

#### d) Total power in 10 Ω

The FM wave has constant amplitude, so its power does not depend on modulation:

$$
P = \frac{A_c^2}{2R} = \frac{25^2}{2 \times 10} = \frac{625}{20} = 31.25\ \text{W}
$$

**Answer:** $\beta = 15$; carrier swing = 90 kHz; $B_T = 96$ kHz; $P = 31.25$ W.

## Check yourself

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

1. For m(t) = 5cos(18849.55t) and kf = 9 kHz/V, what are fm and Δf?
   A. fm = 18.85 kHz, Δf = 45 kHz
   B. fm = 3 kHz, Δf = 9 kHz
   C. fm = 3 kHz, Δf = 45 kHz
   D. fm = 6 kHz, Δf = 90 kHz
   Answer: C. fm = 18849.55/2π = 3 kHz and Δf = kf·Am = 9 × 5 = 45 kHz. (Key point: Message frequency and deviation)

2. What is β = Δf/fm?
   A. 15
   B. 5
   C. 135
   D. 0.067
   Answer: A. 45 kHz / 3 kHz = 15, wide-band FM. (Key point: Modulation index)

3. What is the Carson bandwidth?
   A. 90 kHz
   B. 6 kHz
   C. 96 kHz
   D. 48 kHz
   Answer: C. BT = 2(Δf + fm) = 2(45 + 3) = 96 kHz. Note the carrier swing 2Δf = 90 kHz is a different quantity. (Key point: Carson bandwidth)

4. What is the power delivered to 10 Ω, and why does it not depend on modulation?
   A. 62.5 W; the amplitude doubles
   B. 31.25 W; FM has constant amplitude so power is Ac^2/2R
   C. 625 W; power equals Ac^2
   D. 3.125 W; power depends on β
   Answer: B. P = 25^2/(2×10) = 31.25 W, independent of modulation. (Key point: FM power)

## Close variants set in other exams

- [FM numerical: 5cos(18849.55t), 9 kHz/V, 25 V, 88.7 MHz (five parts)](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-5cos18849-five-parts) — Communication Systems, chapter 3, asked 3×

## Related in Communication Systems

- [FM numerical: 5cos(18849.55t), 9 kHz/V — deviation, swing, Carson](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-5cos18849-peak-carson) — Communication Systems, chapter 3
- [FM numerical: 102.4 MHz carrier, 75 kHz deviation](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-102-4mhz-75khz) — 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