TRANSMISSION

Bits per second (bps) aka bit rate

A measure of transmission speed. The number of binary digits (0 or 1) which can be transmitted per second

Not to be mistaken for Bytes per second (Bps)

When converting Bps to bps, remember that there are 8 bits to a Byte

Therefore 10MB = 10 x 8 Mb = 80Mb

Baud Rate aka symbol rate

A measure of how fast a change of state occurs (i.e. a change from 0 to 1)

The number of signal events occurring each second along a communication channel

Because a signal can contain more than one bit of data, the baud rate and the bit rate may be different. E.g. 1200 baud might transmit at 4800 bps.

Baud Rate aka symbol rate

The number of different symbols required depends on the number of bits used in a single signal event

E.g. if there are 4 bits represented in each baud, there will need to be 16 different symbols.

Baud Rate aka symbol rate

E.g. 24 = 16

0000 0001 0010 00110100 0101 0110 01111000 1001 1010 10111100 1101 1110 1111

E.g. 22 = 4

00 01 10 11

How baud rate relates to bps

Example

An analogue signal carries 6 bits in each signal event. If 1000 signal events are sent per second, what is the baud rate and bit rate?

Solution: Baud rate = 1000 bauds per second Bit rate = 1000 x 6 = 6000bps

Modulation

The process of conveying a message signal, for example a digital bit stream or an analogue audio signal, inside another signal that can be physically transmitted

QAM

Most current connections to the Internet use Quadrature Amplitude Modulation (QAM).

This system represents different bit patterns by altering the amplitude and phase of the wave.

16QAM uses 16 different symbols to represent 4 bits/symbol, 32QAM uses 32 different symbols to represent 5 bits/symbol, etc.

Group Task Activity

Calculate the minimum transmission time required to transfer a 1kB packet at 10Mbps to a satellite located 40 000km above Earth.

3 x 108 m/s = 300 000 000 m/s Time for the first part of the signal to reach

the satellite will be

40 000 000 ÷ 300 000 000 = 0.1333 seconds

Group Task Activity Now we need to calculate the time taken

from when the first part of the signal is placed on the medium until the last signal is placed on the medium.

10Mb (megabits) of data are placed on the medium in 1 second. We are placing a 1kB (kilobyte) packet on the medium.

10Mb = ~ 10 000kb 1 kB = 8kb

∴ 8kb will take 8 ÷ 10 000 = 0.0008 seconds for the sender to place on the medium.

Group Task Activity

So the first part of the signal will arrive at the satellite in 0.1333 seconds; the remainder of the signal follows during the next 0.0008 seconds.

Therefore, the minimum time taken until the entire message reaches the satellite is

0.1333 + 0.0008 = 0.1341 seconds

Group Task Discussion

Why is the speed of wave propagation particularly significant over longer distances when each data packet must be acknowledged before the next one can be sent? Discuss

Group Task Activity

The time taken for the signal to physically propagate through the medium (signal latency) adds to the time between a packet being received by the destination and acknowledgements being received by the sender. Furthermore the propagation time must be added to the time taken for every packet to reach it’s destination.

Group Task Activity

As CPU speeds increase and motherboards transfer data faster, will the speed of wave propagation within and between motherboard components become significant? Discuss

Distances between motherboard components are miniscule compared to distances on a global scale.

Group Task Activity

Consider a distance of 300mm between motherboard components.

At 2 x 108 m/s the propagation time is just

0.3 ÷ 200 000 000 = 0.0000000015 seconds.

∴ CPU speeds would need to increase enormously before propagation speeds will become a significant motherboard issue.

Bandwidth Not to be confused with speed or bps The range of frequencies used by a

transmission channel (the difference between the highest and lowest)

Expressed in hertz (Hz) E.g. Standard telephone equipment used

for voice operates within a frequency range from about 200Hz to 3400Hz, so the available bandwidth is 3200Hz (3.2KHz).

Example

A fibre-optic cable has a high bandwidth. When cable television is transmitted through

fibre-optic cable, many different channels can be transmitted at the same time.

Group Task ActivityCalculate the time taken to transfer a 2MB file over each of the below communication methods. First we need to convert all numbers to bits 2MB = 2 000 000 Bytes 2 000 000 Bytes = 16 000 000 bits

1. 56kbps = 56 000bps. Time = 16 000 000 ÷ 56 000

= 286 seconds

Group Task Activity

2. 10 Mbps

16 000 000 ÷ 10 000 000 = 1.6 seconds

3. 100Mbps

16 000 000 ÷ 100 000 000 = 0.16 seconds

4. 1000Mbps

16 000 000 ÷ 1 000 000 000 = 0.016 seconds

Group Task Activity

5. 1.5Mbps

16 000 000 ÷ 1 500 000 = 10.7 seconds

6. 64QAM uses 64 symbols to represent 6 bits/symbol. Therefore 4Msym/s is equivalent to

4 x 6 = 24Mbps = 24 000 000bps

16 000 000 ÷ 24 000 000 = 0.67 seconds

Group Task Activity

Identify and discuss reasons why it is unlikely that the minimum times calculated above would be realised in reality.

Errors including the time taken to resend faulty packets.

Other simultaneous transmissions occurring. Protocol headers/trailers and time waiting for

acknowledgements Time to pass through other components such as

NICs, routers, etc.

Error Checking Methods

When an error is detected the receiver can respond in various ways depending on the rules of the particular protocol.

Parity bit check Check sum Cycle redundancy check (CRC)

Parity bit check The simplest form of error detection involves

adding a parity bit to the data bits The sender computes an additional bit based

on the given data bits The receiver performs the same computation

and verifies that the parity bits agree The computation is chosen so that a one-bit

alteration can be detected Parity can be even or odd, this is established

during handshaking E.g. For even parity, the sender sets the parity

bit so that the total number of 1 bits is even

Parity bit check

Checksums All the 1s and 0s in a block of data are

summed to make a total and the receiver calculates a checksum value. If the sent value does not agree, then an error is detected. If the count matches, it is assumed that a complete transmission was received.

The position of the error is unknown

Group Task Discussion Activity

IP datagrams include a 16-bit checksum calculated using just their header whilst TCP segments include a 16-bit checksum calculated using the entire message. Discuss possible reasons for this difference and describe likely differences between the accuracy of IP and TCP checksums.

Group Task Discussion Activity

IP aims to route messages and isn’t concerned with guaranteeing delivery. The header contains all the data needed for routing

TCP acknowledges each segment, therefore it checks the accuracy of the entire segment

IP checksums are calculated over a smaller number of bytes compare to TCP checksums, this tend to be more accurate.

Cyclic Redundancy Check (CRC)

The most accurate error check Calculated using division rather than addition The data is divided into predetermined lengths and

divided by a fixed number. The remainder of the calculation is attached and sent with the data.

When the data is received, the remainder is recalculated. If the remainders do not match, an error in transmission has occurred.

There are a number of different standards for CRC. A 32-bit CRC achieves a 99.99% detection of all possible errors.