physical layer propagation

© 2009 Pearson Education, Inc. Publishing as Prentice Hall

Physical Layer Propagation

Chapter 3Updated January 2009

Raymond Panko’s Business Data Networks and Telecommunications, 7th edition

May only be used by adopters of the book

© 2009 Pearson Education, Inc. Publishing as Prentice Hall3-2

3-1: Signal and Propagation

A signal is a disturbance in the media that propagates (travels) down the transmission medium to the receiver

If propagation effects are too large, the receiver will not be able to read the received signal


BinaryData Representation


Binary-Encoded Data

• Computers store and process data in binary representations– Binary means “two”– There are only ones and zeros– Called bits

Non-Binary Data Must Be Encoded into Binary

1101010110001110101100111

Hello 11011001…

© 2009 Pearson Education, Inc. Publishing as Prentice Hall 3-5

Encoding AlternativeBits (N) Alternatives (2N)

1 22 43 84 165 326 647 1288 256

10 1,02416 65,536

An N-bit field can represent 2N alternatives

Each additional bit doubles the number of possibilities

Start with one you know and double or halve until you have what you need

E.g., if you know 8 is 256, 10 must be 4 times as large or 1,024


3-3: Binary Encoding for a Number of Alternatives

Number of Bits in Field

Number of Alternatives that Can Be Encoded1

Specific Bit Sequences

Example

1 21 = 2 0, 1 Yes or No, Male or Female, etc.

2 22 = 4 00, 01, 10, 11 North, South, East, West

4 24 = 16 0000, 0001, 0010, …

Top 10 security threats (6 values go unused)

8 28 = 256 00000000, 00000001, …

ASCII text representation (128 values go unused)

1There are 2N alternatives with N bits


3-3: Binary Encoding for a Number of Alternatives

• Examples:1. You have N bits. How many alternatives can you

represent?

2. You have 4 bits. How many alternatives can your represent?

3. You need to represent 8 things. How many bits must you use?

4. You need to represent 6 things. How many bits must you use?


3-4: ASCII

• Purpose– To represent text (A, a, 3, $, etc.) as binary data for

transmission

• ASCII– Traditional code to represent text data in binary– Seven bits per character– 27 (128) characters possible– Sufficient for all keyboard characters (including shifted

values)


3-4: ASCII

• Each ASCII Character is Sent in a Byte– 8th Bit in Data Bytes Normally Is Not Used

1 0 1 0 0 1 1 1

Data Byte

ASCII Codefor Character Unused.

Value doesnot matter


3-4: ASCII

• To send “Hello world!” (without the quotes), how many bytes will you have to transmit?


3-6: Data Encoding and Signals

We have just seen this

We will nowsee this

Before transmission, two things must happenFirst, data must be converted into a bit stream

We have already seen thisSecond, the 1s and 0s need to be converted into

signals—disturbances that travel down the medium


3-11: Multistate Digital Signaling

• Concepts

– Bit rate: Number of bits sent per second

– Baud rate: Number of clock cycles per second

• If 1,000 clock cycles per second, 1 kbaud

• If each clock cycle is 1/1,000 second = 1,000 clock cycles/second = 1 kbaud

Box



• Computing the Bit Rate

Bit rate = Baud rate X Bits sent per clock cycle

EX:– If baud rate is 10,000 baud

– If two bits per clock cycle

– Then bit rate is 2 x 10,000 or 20,000 bps = 20 kbps

Box



• Computing the Bit Rate– Know the baud rate and the number of states

– Compute the number of bits from the number of states

– States = 2Bits per clock cycle

Bit rate = Baud rate X Bits sent per clock cycle

EX– If baud rate is 10,000 baud (not bauds)

– If four states, can send 2 bits per clock cycle

– Then bit rate is 2 x 10,000 or 20,000 bps = 20 kbps

Box



• Computing the Required Number of States– Know the required bit rate and baud rate

– Bits sent per clock cycle =Bit rate / Baud rate – Compute the required number of states

• EX:– Required bit rate is 4 Mbps

– Baud rate is 1 Mbaud

– Bit rate / baud rate = 4 bits per clock cycle

– 4 bits per clock cycle are required

Box


Bit Rate versus Baud Rate

Number of Possible States

Bits per ClockCycle

2 (Binary)

4

8

16

1

2

3

4

If a Baud Rate is 1,200 Baud,Bit Rate is

1,200 bps

2,400 bps

3,600 bps

4,800 bps

Each Doubling of States Gives One More Bit per Clock Cycle


Quiz:

3-17

There are eight states. Each clock cycle is 1/8000 of a second. What is the baud rate? What is the bit rate?


UTP PropagationUnshielded Twisted Pair wiring


• Two main categories:– wires, cables– wireless transmission, e.g. radio, microwave, infrared,

…

• Wired– Twisted-Pair cables: – Coaxial cables– Fiber-optic cables

Transmission Media


3-12: Unshielded Twisted Pair (UTP) Wiring

• UTP Characteristics– Inexpensive and to purchase and install– Dominates media for access links between computers

and the nearest switch



• Cord Organization– A length of UTP wiring is a cord– Each cord has eight copper wires– The wires are organized as four pairs

• Each pair’s two wires are twisted around each other several times per inch

– There is an outer plastic jacket that encloses the four pairs



• Connector– RJ-45 connector is the standard

connector– Plugs into an RJ-45 jack in a NIC, switch, or wall jack

RJ-45Jack

RJ-45Jack

8-pin RJ-45 connectors


3-14: Attenuation and Noise

Power

Distance

3.Noise Floor

(Average Noise level)

2.Noise

4.Noise Spike

1.Signal

6.Signal-

to-NoiseRatio (SNR)

5.Error

1. The signal attenuates (falls in power) as it propagates2. There is noise (random energy) in the wire that adds to the signal3. The average noise level is called the noise floor4. Noise is random. Occasionally, there will be large noise spikes5. Noise spikes as large as the signal cause errors6. You want to keep the signal-to-noise ratio high


Limiting UTP Cord Length

• Limit UTP cord length to 100 meters

– This keeps the signal-to-noise ration (SNR) high

– This makes attenuation and noise problems negligible

– Note that limiting cord lengths limits BOTH noise and attenuation problems

100 Meters MaximumCord Length


UTP Wiring

• Electromagnetic Interference (EMI)– Electromagnetic interference is electromagnetic

energy from outside sources that adds to the signal• From fluorescent lights, electrical motors,

microwave ovens, etc.


3-16: Electromagnetic Interference (EMI) and Twisting

Interference on the Two Halves of a Twist Cancels Out

TwistedWire

ElectromagneticInterference (EMI)

UTP is twistedto reduce EMI


3-16: Crosstalk Interference and Terminal Crosstalk Interference

Untwistedat Ends Signal

Terminal CrosstalkInterference

Crosstalk Interference

Terminal crosstalk interferencenormally is the biggest EMI problem for UTP


UTP Limitations

• Limit cords to 100 meters– Limits BOTH noise AND attenuation problems to an

acceptable level

• Do not untwist wires more than 1.25 cm (a half inch) when placing them in RJ-45 connectors– Limits terminal crosstalk interference to an acceptable

level

• Neither completely eliminates the problems but they usually reduce the problems to negligible levels

2


Optical Fiber Transmission

Light through Glass

Spans Longer Distances than UTP


3-20: Optical Fiber Transceiver and Strand

An optical fiber strand has a thin glass coreThis core is 8.3, 50, or 62.5 microns in diameter

This glass core is surrounded by a tubular glass claddingThe outer diameter of the cladding is 125 microns,

regardless of the core’s diameterThe transceiver injects laser light into the core


3-20: Optical Fiber Transceiver and Strand

When a light wave ray hits the core/cladding boundary,there is perfect internal reflection. There is no signal loss


3-21: Roles of UTP and Optical Fiber in LANs


Two-Strand Full-Duplex Optical Fiber Cord with SC and ST Connectors

A fiber cord has two-fiber strands

for full-duplex (two-way) transmission

SC Connectors

ST Connectors

TwoStrands

Cord


Radio Propagation


Radio Propagation

Radio signals also propagate as waves.Radio waves are measured in hertz (Hz),

which is a measure of frequency.Radio usually operates in the MHz and GHz range.

Hertz (Hz) is the term for cycles per second


3-27: Omnidirectional and Dish Antennas


3-28: Wireless Propagation Problems

UTP and optical fiber propagation are fairly predictable.However, radio suffers from many propagation effects.

This makes radio transmission difficult to manage.We will look at these problems one at a time.



The first propagation problem is electromagneticinterference (EMI) from nearby radio sources

This includes other wireless devicesIt can include microwave ovens an other devices



Another problem is inverse square law attenuation.As a signal propagates, its energy spreads out over the

Surface of an ever-expanding sphere.


LaptopComm. Tower

ShadowZone


NoSignal


MultipathInterference

LaptopComm. Tower


Signals Arriving by Different PathsMay Cancel Out


TopologyNetwork topology is the physical

arrangement of a network’s computers,

switches, routers, and transmission lines

It is a physical layer concept


3-29: Major Topologies

The simplest topology is the point-to-point topology



Ethernet uses a star topologyNote that the switch does not have to be in the middle of the star



Mesh (Routers, Frame Relay, ATM)

AB

CD

PathABD

PathACD

In a mesh topology, there are many connectionsbetween switches or routers

Consequently, there are many alternative routes between hosts



In the ring topology, messages travel around a loop



The bus topology uses broadcasting.The message receives each host at almost the same time.

All wireless transmission uses a bus topology.

physical layer propagation

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