---
title: "FDM bandwidth for 10 voice channels"
subject: "Data Communication"
chapter: 6
key: "semester_5/data_communication:fdm-10-voice-channels"
source: https://www.bctengineer.com/q/semester_5/data_communication/fdm-10-voice-channels
site: BCT Engineering
---

# FDM bandwidth for 10 voice channels

Data Communication · Chapter 6: Multiplexing and Spreading

- Asked in: 2069 Chaitra, 2+3 marks
- Video: [17_Numerical on FDM](https://www.youtube.com/watch?v=Xe8Tfr2UkaM) — Disha A. Shukla
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/data_communication/fdm-10-voice-channels
- Whole chapter: https://www.bctengineer.com/md/semester_5/data_communication/chapter-6

## Question

What is FDM? Assume a voice channel occupies a bandwidth of 4 kHz. We need to multiplex 10 voice channels with guard bands of 500 Hz using FDM. Calculate the required bandwidth.

## Short answer

FDM is an analog technique where multiple signals modulate different carrier frequencies and share a link simultaneously in non-overlapping bands with guard bands. For 10 voice channels of 4 kHz and 500 Hz guard bands, required bandwidth = 10×4 kHz + 9×0.5 kHz = 44.5 kHz.

## Answer

#### FDM
**Frequency division multiplexing (FDM)** is an analog technique in which the bandwidth of a link is divided into several non-overlapping frequency bands. Each signal modulates a separate carrier frequency and occupies its own band, and all signals are transmitted simultaneously. Narrow guard bands separate adjacent channels to avoid interference. Band-pass filters at the receiver separate the channels.

#### Bandwidth calculation
Given: n = 10 channels, each 4 kHz; guard band = 500 Hz = 0.5 kHz. Guard bands are needed between adjacent channels, i.e. (n − 1) = 9 guard bands.

Total bandwidth = n × (channel BW) + (n − 1) × (guard BW)
= 10 × 4 kHz + 9 × 0.5 kHz
= 40 + 4.5
= 44.5 kHz

```text
|ch1|g|ch2|g| ... |g|ch10|     10 channels + 9 guards
```

**Answer: 44.5 kHz** (if a guard band is also placed after the last channel, 10 guards give 45 kHz).

## Check yourself

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

1. What is FDM?
   A. Dividing time into slots for signals
   B. Dividing link bandwidth into non-overlapping bands, each signal on its own carrier
   C. Spreading a signal with chips
   D. Combining wavelengths in fibre
   Answer: B. FDM gives each signal its own carrier and band, all transmitted simultaneously. (Key point: FDM definition)

2. For 10 channels, how many guard bands lie between adjacent channels?
   A. 9
   B. 10
   C. 11
   D. 8
   Answer: A. n - 1 = 9 guard bands separate 10 channels. (Key point: Number of guard bands)

3. What is the required bandwidth?
   A. 40 kHz
   B. 45 kHz
   C. 44.5 kHz
   D. 49 kHz
   Answer: C. 10 x 4 + 9 x 0.5 = 44.5 kHz. (Key point: Required bandwidth)

4. If a guard band is also placed after the last channel, the bandwidth becomes
   A. 44.5 kHz
   B. 45 kHz
   C. 49.5 kHz
   D. 50 kHz
   Answer: B. 10 x 4 + 10 x 0.5 = 45 kHz. (Key point: Variation)

## Related in Data Communication

- [Channel capacity for 6 kHz, 6 dB](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Acapacity-6khz-6db) — Data Communication, chapter 1
- [Need for channel coding](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Achannel-coding-need) — Data Communication, chapter 8
- [Required SNR for 40 Mbps over 6 MHz](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Acapacity-40mbps-6mhz) — Data Communication, chapter 1
- [Capacity for 3.1 kHz, SNR 0 dB](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Acapacity-3-1khz-0db) — Data Communication, chapter 1
- [AM wave power and bandwidth numerical](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Aam-numeric-40-carrier-power) — Data Communication, chapter 5
- [Channel performance measures](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Achannel-performance-measures) — Data Communication, chapter 1, asked 6×

## Related in other subjects

- [TDM of 24 voice signals: PAM bandwidth and PCM sampling rate](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Atdm-24-voice-pam-pcm) — Communication Systems, chapter 5
- [10-bit DAC triangular wave at 500 Hz](https://www.bctengineer.com/question.md?key=semester_5%2Finstrumentation_ii%3Adac10-triangular-500hz) — Instrumentation II, chapter 4
- [TDM vs FDM for voice; E1 and T1 hierarchies](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Atdm-fdm-voice-e1-t1) — Communication Systems, chapter 5
- [Numerical: cellular vs non-cellular voice channels](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-vs-noncellular-channels) — Wireless Communication, chapter 2