---
title: "FM numerical: deviation, index, 98% bandwidth, power"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:fm-numerical-10cos-80cos-400hz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/fm-numerical-10cos-80cos-400hz
site: BCT Engineering
---

# FM numerical: deviation, index, 98% bandwidth, power

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2079 Chaitra (CS I), 4×2.5 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/fm-numerical-10cos-80cos-400hz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

A message signal m(t) = 10 cos(2π 4000 t) Volt is used to frequency modulate the carrier signal c(t) = 80 cos(2π 10⁷ t) Volt. Assuming the frequency sensitivity of the frequency modulator to be 400 Hz/Volt, calculate:
a) Peak frequency deviation
b) Modulation index
c) Bandwidth of modulated signal for over 98% of FM power
d) Total modulated signal power dissipated in unit impedance

## Short answer

Peak deviation = 4 kHz; modulation index = 1; bandwidth (over 98% power) = 16 kHz; total power in 1 Ω = 3200 W.

## Answer

**Given:** $A_m = 10$ V, $f_m = 4000$ Hz, $A_c = 80$ V, $f_c = 10^7$ Hz = 10 MHz, $k_f = 400$ Hz/V.

#### a) Peak frequency deviation

$$
\Delta f = k_fA_m = 400 \times 10 = 4000\ \text{Hz} = 4\ \text{kHz}
$$

#### b) Modulation index

$$
\beta = \frac{\Delta f}{f_m} = \frac{4000}{4000} = 1
$$

#### c) Bandwidth containing over 98% of power

Carson's rule is based on keeping about 98% of the power:

$$
B_T = 2(\Delta f + f_m) = 2(4 + 4) = 16\ \text{kHz}
$$

Check with Bessel functions for $\beta = 1$: $J_0 = 0.7652$, $J_1 = 0.4401$, $J_2 = 0.1149$, $J_3 = 0.0196$.

| Components kept | Power fraction $J_0^2 + 2\sum J_n^2$ |
|---|---|
| $n = 0, \pm1$ | 0.9728 (97.3%) |
| $n = 0, \pm1, \pm2$ | 0.9992 (99.9%) |

So sidebands up to $n = 2$ are needed for more than 98% power:

$$
B = 2nf_m = 2 \times 2 \times 4 = 16\ \text{kHz}
$$

which agrees with Carson's rule.

#### d) Total power in unit impedance

FM has constant amplitude, so

$$
P = \frac{A_c^2}{2R} = \frac{80^2}{2 \times 1} = 3200\ \text{W}
$$

**Answer:** $\Delta f = 4$ kHz; $\beta = 1$; $B = 16$ kHz; $P = 3200$ W.

## Check yourself

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

1. How is the peak frequency deviation found?
   A. Delta f = kf x fm = 400 x 4000
   B. Delta f = kf x Am = 400 x 10 = 4 kHz
   C. Delta f = Am / kf = 0.025 Hz
   D. Delta f = kf x Ac = 400 x 80
   Answer: B. Deviation is the frequency sensitivity times the message peak amplitude. (Key point: Peak deviation)

2. What is beta here?
   A. beta = 4000 x 4000
   B. beta = 10 / 80 = 0.125
   C. beta = 4000 / 4000 = 1
   D. beta = 4
   Answer: C. beta = Delta f / fm. (Key point: Modulation index)

3. Which bandwidth contains more than 98% power for beta = 1?
   A. 8 kHz, from n = 0, +-1 which has 99.9%
   B. 32 kHz, from n = 4
   C. 4 kHz, equal to Delta f
   D. 16 kHz, since sidebands up to n = 2 are needed (also Carson's rule)
   Answer: D. n = 0, +-1 gives only 97.3%; adding n = 2 gives 99.9%, so B = 2 x 2 x 4 = 16 kHz. (Key point: 98% power bandwidth)

4. What is the power in a unit impedance?
   A. Ac^2/2 = 3200 W, independent of modulation
   B. Ac^2 = 6400 W
   C. Ac^2/2 x J0^2(1) = 1874 W
   D. Ac^2/(2 beta) = 1600 W
   Answer: A. FM has constant amplitude, so P = 80^2/2. (Key point: Total power)

## Close variants set in other exams

- [FM numerical: m(t) = 20cos(2π2000t), 200 Hz/V](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Afm-numerical-20cos-50cos-200hz) — Communication Systems, chapter 3, asked 2×

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