---
title: "Numerical: N-TDMA simultaneous users"
subject: "Wireless Communication"
chapter: 7
key: "semester_7/wireless_communication:ntdma-simultaneous-users"
source: https://www.bctengineer.com/q/semester_7/wireless_communication/ntdma-simultaneous-users
site: BCT Engineering
---

# Numerical: N-TDMA simultaneous users

Wireless Communication · Chapter 7: Multiple Access in Wireless Communication

- Asked in: 2080 Chaitra, 4 marks
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_7/wireless_communication/ntdma-simultaneous-users
- Whole chapter: https://www.bctengineer.com/md/semester_7/wireless_communication/chapter-7

## Question

In the N-TDMA system, it uses one way bandwidth of 25 MHz for the forward (or reverse) channel. The system bandwidth is divided into radio channels of 30 kHz. Guard bands with Bg = 20 kHz are used. Calculate number of simultaneously transmitting users that can be accommodated in each cluster.

## Short answer

With 25 MHz one-way bandwidth, 20 kHz guard at each edge, and 30 kHz radio channels, channels = (25,000,000 - 40,000)/30,000 = 832. With 3 TDMA users per channel, simultaneous users per cluster = 3 × 832 = 2496.

## Answer

N-TDMA (narrowband TDMA, IS-54/IS-136 "US Digital Cellular") divides the band into 30 kHz radio channels, and each radio channel is time-shared by **3 full-rate users** (assumed, as in IS-54/136).

**Number of radio channels** (guard band at both edges of the allocation):

```text
N_ch = (B_t − 2·B_g) / B_c
     = (25 MHz − 2 × 20 kHz) / 30 kHz
     = (25 000 − 40) kHz / 30 kHz
     = 24 960 / 30
     = 832 radio channels
```

**Number of simultaneous users** (m = 3 TDMA slots per channel):

```text
N_users = m × N_ch = 3 × 832 = 2496
```

Here B_t = 25 MHz one-way (forward or reverse) bandwidth, B_g = 20 kHz guard band, B_c = 30 kHz channel bandwidth. Since all these channels can be used inside one cluster (each cell gets a share of them), this is the number of users the cluster can serve at the same time.

**Answer: 832 radio channels; with 3 users per channel, 2496 simultaneous users per cluster** (if each radio channel carried only one user, it would be 832).

## Check yourself

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

1. How many full-rate users share each 30 kHz channel in N-TDMA (IS-54/136)?
   A. 3
   B. 8
   C. 1
   D. 16
   Answer: A. Each radio channel is time-shared by 3 full-rate users. (Key point: Slots per channel)

2. For 25 MHz, Bg = 20 kHz and Bc = 30 kHz, number of radio channels?
   A. 832
   B. 833
   C. 834
   D. 816
   Answer: A. (25000 - 40)/30 = 832. (Key point: Radio channels)

3. What is the number of simultaneous users per cluster?
   A. 2496
   B. 832
   C. 1664
   D. 8320
   Answer: A. 3 x 832 = 2496. (Key point: Users)

## 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×
- [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
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- [Forward and reverse channel](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aforward-reverse-channel) — Wireless Communication, chapter 1
- [Forward/reverse channel and features of 5G](https://www.bctengineer.com/question.md?key=semester_7%2Fwireless_communication%3Aforward-reverse-5g-features) — Wireless Communication, chapter 1

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