---
title: "FM numerical: 10 kHz baseband, 100 MHz carrier, 75 kHz deviation"
subject: "Communication Systems"
chapter: 3
key: "semester_6/communication_systems:fm-10khz-100mhz-75khz"
source: https://www.bctengineer.com/q/semester_6/communication_systems/fm-10khz-100mhz-75khz
site: BCT Engineering
---

# FM numerical: 10 kHz baseband, 100 MHz carrier, 75 kHz deviation

Communication Systems · Chapter 3: Angle Modulation

- Asked in: 2070 Magh (CS I), 4+4 marks
- Video: [5. Frequency Modulation Examples | GATE FM Examples in Analog Communication by Engineering Funda](https://www.youtube.com/watch?v=tBAxCMNyoBw) — Engineering Funda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_6/communication_systems/fm-10khz-100mhz-75khz
- Whole chapter: https://www.bctengineer.com/md/semester_6/communication_systems/chapter-3

## Question

In an FM system a baseband signal band limited to 10 kHz modulates 100 MHz carrier wave so that the frequency deviation is 75 kHz. Find:
i) Carrier frequency swing in the FM signal and modulation index
ii) The practical bandwidth of the FM signal

## Short answer

Carrier swing is 2*75 = 150 kHz; modulation index is 75/10 = 7.5; Carson bandwidth is 2(75+10) = 170 kHz (220 kHz by the 1% rule).

## Answer

**Given:** $f_m = 10$ kHz (highest baseband frequency), $f_c = 100$ MHz, $\Delta f = 75$ kHz.

#### i) Carrier frequency swing and modulation index

The instantaneous frequency varies between

$$
f_{max} = 100\ \text{MHz} + 75\ \text{kHz} = 100.075\ \text{MHz}, \qquad f_{min} = 99.925\ \text{MHz}
$$

$$
\text{Carrier swing} = 2\Delta f = 2 \times 75 = 150\ \text{kHz}
$$

Modulation index (deviation ratio, since $f_m$ is the highest message frequency):

$$
\beta = \frac{\Delta f}{f_m} = \frac{75}{10} = 7.5
$$

#### ii) Practical bandwidth

By **Carson's rule** (contains about 98% of the FM power):

$$
B_T = 2(\Delta f + f_m) = 2(75 + 10) = 170\ \text{kHz}
$$

Check with Bessel functions: with $\beta = 7.5$, Carson's rule keeps sidebands up to $n = \beta + 1 \approx 8$; the power in these is $J_0^2 + 2\sum_{n=1}^{8}J_n^2(7.5) = 0.981$, i.e. 98.1% of the total.

If the stricter **1% rule** is used (keep sidebands with $|J_n(\beta)| > 0.01$), $|J_{11}(7.5)| = 0.015$ and $|J_{12}(7.5)| = 0.005$, so $n_{max} = 11$ and

$$
B = 2 n_{max} f_m = 2 \times 11 \times 10 = 220\ \text{kHz}
$$

**Answer:** swing = 150 kHz, $\beta = 7.5$, practical (Carson) bandwidth = 170 kHz (220 kHz by the 1% rule).

## Check yourself

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

1. For a deviation of 75 kHz, what is the carrier frequency swing?
   A. 75 kHz
   B. 300 kHz
   C. 150 kHz
   D. 100.075 MHz
   Answer: C. The carrier moves from fc - 75 kHz to fc + 75 kHz, so swing = 2 delta f = 150 kHz. (Key point: Carrier swing)

2. What is the modulation index (deviation ratio) when fm = 10 kHz?
   A. 7.5
   B. 15
   C. 0.133
   D. 75
   Answer: A. beta = delta f / fm = 75/10 = 7.5. (Key point: Modulation index)

3. What is the practical bandwidth by Carson's rule?
   A. 150 kHz
   B. 85 kHz
   C. 220 kHz
   D. 170 kHz
   Answer: D. BT = 2(75 + 10) = 170 kHz. (Key point: Carson's rule result)

4. Why does the 1% rule give a larger bandwidth (220 kHz) than Carson's rule?
   A. It uses a different carrier
   B. It keeps all sidebands with |Jn| > 0.01 (n up to 11), while Carson's keeps about 98% of the power
   C. It doubles the deviation
   D. It includes the carrier twice
   Answer: B. The 1% rule keeps sidebands to n = 11 so B = 2 x 11 x 10 = 220 kHz, stricter than Carson's roughly 98% power criterion. (Key point: 1% rule)

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