---
title: "Numerical: FDMA channels (12.5 MHz)"
subject: "Wireless Communication"
chapter: 7
key: "semester_7/wireless_communication:fdma-channels-12-5mhz"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/fdma-channels-12-5mhz
site: BCT Engineering
---

# Numerical: FDMA channels (12.5 MHz)

Wireless Communication · Chapter 7: Multiple Access in Wireless Communication

- Asked in: 2078 Chaitra, 4 marks
- Video: [Numerical on FDMA and TDMA - Mobile Radio Propagation - Mobile Communication System](https://www.youtube.com/watch?v=YAxsx6bUYWM) — Ekeeda
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/fdma-channels-12-5mhz
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-7

## Question

If total bandwidth is 12.5 MHz, guard bandwidth is 10 KHz and channel bandwidth is 30 KHz, find the number of channels available in an FDMA system.

## Short answer

Using N = (B_t - 2B_guard)/B_c with 12.5 MHz total, 10 kHz guard at each edge, and 30 kHz channels gives N = (12,500 - 20)/30 = 416 channels.

## Answer

In FDMA the total band is split into equal channels, with a guard band at each edge of the allocated spectrum.

```text
N = (B_t − 2·B_guard) / B_c
  = (12.5 MHz − 2 × 10 kHz) / 30 kHz
  = (12 500 − 20) kHz / 30 kHz
  = 12 480 / 30
  = 416
```

where B_t = 12.5 MHz total bandwidth, B_guard = 10 kHz, B_c = 30 kHz.

This is the classic AMPS case: each operator in the US had 12.5 MHz in each direction, giving 416 simplex channels each way (416 duplex channel pairs).

**Answer: 416 channels**

## Check yourself

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

1. Which formula applies with guard band at each edge?
   A. N = (Bt - 2Bg)/Bc
   B. N = Bt/Bc
   C. N = (Bt - Bg)/Bc
   D. N = (Bt - 2Bc)/Bg
   Answer: A. Subtract guard bands at both edges, then divide by channel width. (Key point: Formula)

2. What is 12.5 MHz - 2 x 10 kHz?
   A. 12 480 kHz
   B. 12 490 kHz
   C. 12 500 kHz
   D. 12 520 kHz
   Answer: A. 12 500 - 20 = 12 480 kHz. (Key point: Numerator)

3. Number of channels for Bc = 30 kHz?
   A. 416
   B. 417
   C. 415
   D. 420
   Answer: A. 12 480 / 30 = 416. (Key point: Result)

## Related in Wireless Communication

- [Numerical: FDMA channels (12.8 MHz)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afdma-channels-12-8mhz) — Wireless Communication, chapter 7, asked 2×
- [Non-linear effect in FDMA; channels numerical](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afdma-nonlinear-channels) — Wireless Communication, chapter 7
- [Numerical: 12-cell system capacity at 2% GOS](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-capacity-12-cell-system) — Wireless Communication, chapter 2

## Related in other subjects

- [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
- [FDM bandwidth for 10 voice channels](https://www.bctengineer.com/question.md?key=semester_5%2Fdata_communication%3Afdm-10-voice-channels) — Data Communication, chapter 6
- [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×
- [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