---
title: "Numerical: FDMA channels (12.8 MHz)"
subject: "Wireless Communication"
chapter: 7
key: "semester_7/wireless_communication:fdma-channels-12-8mhz"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/fdma-channels-12-8mhz
site: BCT Engineering
---

# Numerical: FDMA channels (12.8 MHz)

Wireless Communication · Chapter 7: Multiple Access in Wireless Communication

- Asked in: 2079 Chaitra, 4 marks; 2077 Chaitra, 4 marks (asked 2 times)
- 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-12-8mhz
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-7

## Question

Calculate the number of radio channels available in the FDMA system with following data. A US analog mobile phone system is allocated 12.8 MHz for every simplex band. The total spectrum allocated is 12.8 MHz, the guard bandwidth is 10 KHz and the channel bandwidth is 30 KHz.

## Short answer

Number of channels N = (total bandwidth − 2 × guard bandwidth) / channel bandwidth = (12.8×10⁶ − 2×10×10³) / (30×10³) = 426. So 426 radio channels per simplex band.

## Answer

In FDMA, the total band is divided into channels of equal width, after leaving a guard band at **each edge** of the allocated spectrum. The number of channels is

N = (B_t − 2·B_guard) / B_c

where B_t = total spectrum, B_guard = guard band at each edge, B_c = channel bandwidth.

**Given:** B_t = 12.8 MHz (per simplex band), B_guard = 10 kHz, B_c = 30 kHz.

```text
 N = (12.8×10⁶ − 2 × 10×10³) / (30×10³)
   = (12 800 000 − 20 000) / 30 000
   = 12 780 000 / 30 000
   = 426
```

**Answer: N = 426 radio channels** (in each simplex band; these pair up to give 426 duplex channels in the forward and reverse bands).

This matches the AMPS (US analog) system, which uses 30 kHz channels with FDMA/FDD.

## Check yourself

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

1. Which formula gives the number of FDMA channels with a guard band at each edge?
   A. N = (Bt - 2Bg)/Bc
   B. N = (Bt - Bg)/Bc
   C. N = Bt/(Bc + Bg)
   D. N = (Bt + 2Bg)/Bc
   Answer: A. A guard band is lost at both edges, so 2Bg is subtracted before dividing by channel width. (Key point: Formula)

2. For Bt = 12.8 MHz and Bg = 10 kHz, what is Bt - 2Bg?
   A. 12 780 kHz
   B. 12 790 kHz
   C. 12 800 kHz
   D. 12 820 kHz
   Answer: A. 12 800 kHz - 20 kHz = 12 780 kHz. (Key point: Numerator)

3. How many channels result with Bc = 30 kHz?
   A. 426
   B. 427
   C. 416
   D. 433
   Answer: A. 12 780 / 30 = 426. (Key point: Result)

## Related in Wireless Communication

- [Numerical: FDMA channels (12.5 MHz)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Afdma-channels-12-5mhz) — Wireless Communication, chapter 7
- [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
- [CCI/ACI, reuse ratio proof and channels per cell](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acci-aci-reuse-duplex-channels) — Wireless Communication, chapter 2
- [Numerical: trunked channels and blocking (Erlang B)](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Atrunking-five-channel-blocking) — Wireless Communication, chapter 2
- [Numerical: 32-cell system area and channels](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Acellular-32-cells-capacity) — Wireless Communication, chapter 2

## Related in other subjects

- [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: 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
- [Maximum error-free data rate for 1 MHz, 10 dB SNR](https://www.bctengineer.com/question.md?key=semester_6%2Fcommunication_systems%3Amax-data-rate-1mhz-10db) — Communication Systems, chapter 6