---
title: "FM numerical: 5cos(18849.55t), 9 kHz/V, 25 V, 88.7 MHz (five parts)"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:fm-5cos18849-five-parts"
source: https://www.bctengineer.com/q/semester_6/communication_systems/fm-5cos18849-five-parts
site: BCT Engineering
---

# FM numerical: 5cos(18849.55t), 9 kHz/V, 25 V, 88.7 MHz (five parts)

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2069 Bhadra (CS I), 2×5 marks
- 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-5cos18849-five-parts
- 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 a frequency sensitivity of 9 kHz/V. The amplitude of the carrier is 25 V and the frequency is 88.7 MHz. Compute:
i) Peak frequency deviation
ii) Modulation index
iii) Frequency swing
iv) Carson's bandwidth and
v) Total power delivered in 10 Ω resistor.

## Answer

**Given:** $m(t) = 5\cos(18849.55\,t)$, $k_f = 9$ kHz/V, $A_c = 25$ V, $f_c = 88.7$ MHz, $R = 10\ \Omega$.

$$
A_m = 5\ \text{V}, \qquad f_m = \frac{18849.55}{2\pi} = 3000\ \text{Hz} = 3\ \text{kHz}
$$

#### i) Peak frequency deviation

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

#### ii) Modulation index

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

#### iii) Frequency swing

$$
\text{Swing} = 2\Delta f = 90\ \text{kHz}
$$

The carrier moves between $88.7 - 0.045 = 88.655$ MHz and $88.745$ MHz.

#### iv) Carson's bandwidth

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

#### v) Power in 10 Ω

FM has constant amplitude, so total power equals the carrier power:

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

**Answer:** $\Delta f = 45$ kHz, $\beta = 15$, 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. With kf = 9 kHz/V, Am = 5 V and fm = 3 kHz, what are delta f and beta?
   A. 45 kHz and 15
   B. 9 kHz and 3
   C. 45 kHz and 5
   D. 15 kHz and 45
   Answer: A. delta f = kf Am = 45 kHz; beta = 45/3 = 15. (Key point: Deviation and index)

2. What are the frequency swing and Carson's bandwidth?
   A. 45 kHz and 96 kHz
   B. 90 kHz and 48 kHz
   C. 90 kHz and 90 kHz
   D. 90 kHz and 96 kHz
   Answer: D. Swing = 2 delta f = 90 kHz; BT = 2(45 + 3) = 96 kHz. (Key point: Swing and Carson)

3. Why does the power in the 10 Ω resistor not depend on the modulation index?
   A. Power depends only on fc
   B. The resistor limits it
   C. FM has constant amplitude, so total power equals the carrier power
   D. The sidebands cancel out
   Answer: C. Modulation only redistributes power among carrier and sidebands; total remains Ac^2/(2R). (Key point: Power in FM)

4. What is the power for Ac = 25 V and R = 10 Ω?
   A. 62.5 W
   B. 31.25 W
   C. 625 W
   D. 12.5 W
   Answer: B. P = Ac^2/(2R) = 625/20 = 31.25 W. (Key point: Power value)

## Close variants set in other exams

- [FM numerical: m(t) = 5cos(18849.55t), 9 kHz/V](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-numerical-5cos18849) — 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
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- [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×