---
title: "Short note: Ethernet"
subject: "Telecommunications"
chapter: 8
key: "semester_8/telecommunications:short-note-ethernet"
source: https://www.bctengineer.com/q/semester_8/telecommunications/short-note-ethernet
site: BCT Engineering
---

# Short note: Ethernet

Telecommunications · Chapter 8: Data Communication

- Asked in: 2064 Poush (old course), 10 marks
- Video: [What is Ethernet?](https://www.youtube.com/watch?v=HLziLmaYsO0) — RealPars
- Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_8/telecommunications/short-note-ethernet
- Whole chapter: https://www.bctengineer.com/md/semester_8/telecommunications/chapter-8

## Question

Write a short note on ETHERNET.

## Short answer

Ethernet (IEEE 802.3) is a LAN technology using CSMA/CD with Manchester coding at 10 Mbps. Frames are 64 to 1518 bytes; the 64-byte minimum ensures collision detection across the maximum network. Modern switched full-duplex Ethernet eliminates collisions and reaches speeds beyond 100 Gbps.

## Answer

**Ethernet** is the most widely used LAN technology, standardised as **IEEE 802.3**. It was developed at Xerox PARC (Robert Metcalfe, 1973) and later specified by DEC, Intel and Xerox (DIX, 10 Mbps). All stations share a common channel and access it using **CSMA/CD (Carrier Sense Multiple Access with Collision Detection)**. It uses **Manchester coding** at 10 Mbps.

#### Physical types

| Standard | Medium | Max segment | Topology |
|---|---|---|---|
| 10Base5 (thick) | Thick coax | 500 m | Bus |
| 10Base2 (thin) | Thin coax | 185 m | Bus |
| 10Base-T | UTP Cat 3/5 | 100 m | Star (hub) |
| 10Base-F | Optical fibre | 2000 m | Star |
| 100Base-TX | UTP Cat 5 | 100 m | Star (Fast Ethernet) |
| 1000Base-T / SX | UTP / fibre | 100 m / 550 m | Gigabit Ethernet |

In 10BaseX, "10" = 10 Mbps, "Base" = baseband, the last part = cable type or length (×100 m).

```text
Bus (10Base5/10Base2):
 T|==+=====+=====+=====+==|T   T = terminator
     H     H     H     H
Star (10Base-T):  H--[Hub/Switch]--H
                        |
                        H
```

#### Frame format
```text
+--------+---+----+----+-----+---------+-----+
|Preamble|SFD| DA | SA | Len | Data+Pad| FCS |
|   7    | 1 | 6  | 6  |  2  | 46-1500 |  4  |
+--------+---+----+----+-----+---------+-----+
```
- Preamble (10101010…) for clock sync; SFD (10101011) marks start.
- 48-bit MAC destination and source addresses.
- Length/type field; data 46–1500 bytes (padded to 46).
- FCS: 32-bit CRC. Frame size 64–1518 bytes.

#### CSMA/CD operation
1. A station senses the channel; if busy it waits until idle (1-persistent).
2. It transmits and listens at the same time.
3. If a collision is detected, it stops, sends a 32-bit **jam** signal, and waits a random **binary exponential backoff** time: after the $n$-th collision it picks $K$ from $0$ to $2^{\min(n,10)} - 1$ and waits $K \times 51.2\ \mu\text{s}$ (slot time = 512 bit times at 10 Mbps).
4. After 16 failed attempts the frame is dropped.

The **64-byte minimum frame** makes sure a sender is still sending when a collision from the far end (up to 2500 m with 4 repeaters) reaches it, i.e. frame time ≥ round-trip time.

#### Advantages and limitations
- Simple, cheap, easy to install and expand; no central control.
- Good performance at light to moderate load.
- At heavy load collisions waste capacity; access delay is not deterministic (unlike token ring); no priority.

Modern Ethernet uses switches and full-duplex links, so each link is collision-free and CSMA/CD is no longer needed; speeds have grown from 10 Mbps to 100 Gbps and beyond.

## Check yourself

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

1. What does a station do on detecting a collision in Ethernet?
   A. Keeps sending the full frame
   B. Waits for a token
   C. Stops, sends a jam signal, and waits a random backoff time
   D. Sends a priority request
   Answer: C. It stops transmitting, sends a 32-bit jam signal, then backs off randomly before retrying. (Key point: CSMA/CD operation)

2. After the 3rd collision, from which range does a station pick K (waiting K x 51.2 us)?
   A. 0 to 3
   B. 0 to 7
   C. 0 to 2
   D. 0 to 15
   Answer: B. K is chosen from 0 to 2^n - 1; for n = 3 this is 0 to 7. (Key point: Binary exponential backoff)

3. Why is a 64-byte minimum frame required?
   A. To fit the 32-bit CRC
   B. So the sender is still transmitting when a collision from the far end returns (frame time >= round trip time)
   C. To keep the preamble synchronised
   D. To match the 1500-byte maximum
   Answer: B. The sender must still be sending to detect a collision, so frame time must be at least the round-trip time. (Key point: Minimum frame size)

4. What does 10Base-T mean?
   A. 10 m, broadband, twisted pair
   B. 10 Mbps, baseband, twisted pair
   C. 10 Mbps, broadband, thick cable
   D. 10 stations, baseband, terminator
   Answer: B. 10 = 10 Mbps, Base = baseband, T = twisted pair cable. (Key point: 10Base-T naming)

5. Why is Ethernet access delay not deterministic?
   A. Because it uses tokens
   B. Because frames are fixed at 1518 bytes
   C. Because collisions and random backoff can repeat under heavy load
   D. Because it uses priority levels
   Answer: C. Collisions and random backoff make waiting time unpredictable, unlike token ring's bounded access. (Key point: Ethernet limitation)