Chapter 2/1 (Physical Layer)

Preview:

DESCRIPTION

Chapter 2/1 (Physical Layer). Theoretical Basis for Data Communication Guided Transmission Media Wireless Transmission Communication Satellites PSTN - Public Switched Telephone System. The Theoretical Basis for Data Communication. Fourier Analysis Bandwidth-Limited Signals - PowerPoint PPT Presentation

Citation preview

09/13/2007 EETS 7304 1

• Theoretical Basis for Data Communication

• Guided Transmission Media• Wireless Transmission• Communication Satellites • PSTN - Public Switched Telephone

System

Chapter 2/1 (Physical Layer)

09/13/2007 EETS 7304 2

The Theoretical Basis for Data Communication

• Fourier Analysis• Bandwidth-Limited Signals• Maximum Data Rate of a Channel:

Nyquist TheoremShannon Capacity Theorem

09/13/2007 EETS 7304 3

Fourier Transform: periodic signals g(t)

dtnfttgT

bT

n )2cos(2

0

dttgT

cT

0

2

)2cos()2sin(2

1

11

nftbnftactgn

nn

n

nn

where: dtnfttgT

aT

n )2sin(2

0

From: )2cos()2sin()2sin( nftbnftanftc nnnn

22nnn bac

n

nn a

barctgFollows:

)2sin(2

1

1n

n

nn nftcctg

09/13/2007 EETS 7304 4

Successive approximations to the original signal

A binary signal and its root-mean-square Fourier amplitudes.

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6wt

appr

oxim

atio

n

original

1

2

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6wt

appr

oxim

atio

n original

4

7

0

0.1

0.2

0.3

0.4

0.5

0.6

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

n

Cn

09/13/2007 EETS 7304 5

Bandwidth-Limited Signals

09/13/2007 EETS 7304 6

Successive approximations to the original signal

09/13/2007 EETS 7304 7

Relation between data rate and harmonics

If b is bit rate (bps) then bit duration is 1/b s Time to send 8 bits T = 8/b s.Therefore for b = 300 bps T = 8/300 = 0.02667 s = 26.67 ms. Since T is the period the first harmonic has frequency 1/T = 300/8 = 37.5 Hz. Within the bandwidth of 3000 Hz you can send 3000/37.5 = 80 harmonics.

09/13/2007 EETS 7304 8

Fourier Transform: periodic signals

tjnn

nn

tnfjn

nn eCeCtx 002 w

where dtetXT

C tjn

T

Tn

0

0

0

2

20

1 w

Example: Pulse train

t

A

0

0

0

2

20

sin10

0

0Tfn

Tnf

T

ATdteA

TC tjn

T

Tn

w

T0

T

=>>n

A/2

-5 -4 -3 -2 -1 0 1 2 3 4 5 6

;0;3

;0;;2 443322110

CC

ACCCC

ACC

AC

for T0/T=2 =>

T0/T = 2

09/13/2007 EETS 7304 9

non-periodic signals

dfefStS ftj

2

2

1

dtetSfS ftj 2

Example: unit impulse function (t)

f

f

fj

eedtet

fjfjftj

2

)2sin(

222

1 222

t

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-12.5 -10 -7.5 -5 -2.5 0 2.5 5 7.5 10 12.5

2f

f = 1/(2)

09/13/2007 EETS 7304 10

Delta impulse

12

dtett ftj or dtet ftj

2

2

1

=>> white spectrum.

t

f

1

white spectrum

0

09/13/2007 EETS 7304 11

Nyquist (1924) analog to digitalThe minimum sampling rate to recover any bandlimited signal to H Hz is 2H (samples per second). Example: assume 2400 bps 10 harmonics bandwidth 3000 Hz. T = 3.33 ms.

Minimum sampling rate is 6000 samples/sec, or every 1/6 = 0.167 ms or 20 samples per period.

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0.0

0

0.1

6

0.3

2

0.4

8

0.6

4

0.7

9

0.9

5

1.1

1

1.2

7

1.4

3

1.5

9

1.7

5

1.9

1

2.0

7

2.2

3

2.3

8

2.5

4

2.7

0

2.8

6

3.0

2

3.1

8

t (ms)

09/13/2007 EETS 7304 12

Telephone example• Subscriber loop has a bandwidth over 1 MHz. However,

switching office filters every signal to 4000 Hz. Therefore, after entering a switch the analog voice is limited to 4000 Hz. Minimum sampling rate is then 8000 samples (pulses) per sec.

• Pulse per sec = baud.• Since the value of every sample is represented by 8 bits a

digital voice bit rate is: 8000 bauds * 8 bits/baud = 64000 bps = 64 kbps.

• Inversely Nyquist theorem also says that maximum sample (pulse) rate going through the filter H that can be recovered is 2H.

• In addition if max amplitude pulse is divided into V levels then the bit rate carried with 2H baud is 2H*log2V.

• For example modem with 2400 bauds and V = 64 has bit rate 14,400 bps.

09/13/2007 EETS 7304 13

Shannon (1948) theorems

• Bit rate in a noisy channel. Shannon Capacity Theorem:

maximum bit rate = H log2(1 + S/N), where

H - channel bandwidth,

S – signal power,

N – noise power.

• For instance: Voice-grade line channel H = 3000 Hz and

S/N = 30 dB -> S/N = 1000, therefore,

maximum bit rate = 3000 log2(1001) = 30000 bps.

• Definition of dB = 10 log10(S/N).

09/13/2007 EETS 7304 14

Guided Transmission Media

• Twisted Pair

• Coaxial Cable

• Fiber Optics

09/13/2007 EETS 7304 15

Twisted Pair

(a) Category 3 UTP (Unshielded Twisted Pair), 16 Mbps.(b) Category 5 UTP, 100 Mbps.Category 6 and 7, 250 – 600 Mbps.

09/13/2007 EETS 7304 16

Coaxial Cable

50 Ohm and 75 ohm 1 GHz

09/13/2007 EETS 7304 17

Fiber Optics

(a) Three examples of a light ray from inside a silica fiber

impinging on the air/silica boundary at different angles.

(b) Light trapped by total internal reflection.

Multimode uses total reflection.Single mode 50 Gbps over 100 km.

09/13/2007 EETS 7304 18

Transmission of Light through Fiber

Attenuation = 10*log10(transmitted power/received power)

per km of light through fiber in the infrared region.

f = c/ -> f = 300,000,000/1.3*10^-6 = 2.31*10^14 = 231 THz.

Bandwidth = 300,000,000*(1/1.22 – 1/1.37)*10^6 = 0.27*10^14 = 27000 GHz

09/13/2007 EETS 7304 19

Fiber Cables

(a) Side view of a single fiber.(b) End view of a sheath with three

fibers.

09/13/2007 EETS 7304 20

A comparison of semiconductor laser and LEDs as light sources.

09/13/2007 EETS 7304 21

A fiber optic ring with active repeaters

09/13/2007 EETS 7304 22

A passive star connection in a fiber optics network

09/13/2007 EETS 7304 23

Fiber vs. copper

• Repeaters: copper every 5 km vs. fiber 50 km.

• 1000 twisted pairs 1 km long weights 8 Tones vs. 2 fibers with more capacity weight 100 kg.

• Fiber doesn’t leak the light -> excellent security.

• Fiber is much lighter to hang on the poles.

• Fiber is dug about 1 m (3 ft) underground or replaces copper in ducts.

• Fiber is new technology therefore more expensive parts.

09/13/2007 EETS 7304 24

Wireless Transmission

• The Electromagnetic Spectrum

• Radio Transmission

• Microwave Transmission

Above 100 MHz for relaying 80 km.

09/13/2007 EETS 7304 25

The Electromagnetic Spectrum

09/13/2007 EETS 7304 26

Wireless bit rateWith current technology it is possible to code about 8 bps/Hz of bandwidth. Examples:• Coax 750 MHz bandwidth -> 6 Gbps.• Fiber:

f = c/ f = c

For = 1.3 band = 0.17 that gives the bandwidth f = 40 THz to get about 300 Tbps.

Two modes of transmission: • narrow band (f/f << 1) notably frequency hopping

spread spectrum (to avoid fading).• direct sequence spread spectrum.

09/13/2007 EETS 7304 27

Radio Transmission

(a) In the VLF, LF, and MF bands, radio waves follow the curvature of the earth.(b) In the HF band, they bounce off the ionosphere.

Microwave transmission uses above 100 MHz. MCI (Microwave Communications Inc.) was a long distance carrier before merged with WorldCom.

09/13/2007 EETS 7304 28

Radio spectrum is owned by governments or FCC in US. FCC allocates the spectrum while governments sell it to carriers on auctions. Some spectrum is left unsold for unlicensed use.

Bluetooth and WiFi WiFi

ISM (Industrial, Scientific, Medical) are not allocated but limited by distance like garage door openers, cordless phones, radio controlled toys etc. It is mandated by power < 1 W.

The ISM bands in the United States

09/13/2007 EETS 7304 29

Communication Satellites

• Geostationary Satellites 1962• Medium-Earth Orbit Satellites – GPS

24 GPS satellites orbiting at 18,000 km

every 6 hours.• Low-Earth Orbit Satellites (Communications)

09/13/2007 EETS 7304 30

Communication Satellites

Orbital period proportional to radius^(3/2). Latency = round-trip delay time

Number of satellites needed for global coverage.

24 h

6 h

1.5 h

P

09/13/2007 EETS 7304 31

ITU allocates the orbits as well as satellite bands

09/13/2007 EETS 7304 32

Low-Earth Orbit Satellites Motorola Iridium launched 1997

• The Iridium: six necklaces by 11 = 66 satellites (750 km).

• Each satellite covers 48 cells = 1628 moving cells.

• Satellite phones didn’t have much success in competition with terrestrial

and 5 b$ was sold for 25 m$ resumed service in 2001.

09/13/2007 EETS 7304 33

Globalstar

(a) Relaying in space (Iridium).(b) Relaying on the ground (Globalstar 48 LEOs).

09/13/2007 EETS 7304 34

Public Switched Telephone System

• Structure of the Telephone System

• The Politics of Telephones

• The Local Loop: Modems, ADSL and Wireless

09/13/2007 EETS 7304 35

Structure of the Telephone System

(a) Fully-interconnected network.

(b) Centralized switch.

(c) Two-level hierarchy.

09/13/2007 EETS 7304 36

Structure of the Telephone System (2)

A typical circuit route for a medium-distance call.

09/13/2007 EETS 7304 37

Major Components of the Telephone System• Local loops: analog twisted pairs going to houses and

businesses: the weakest link.

• Trunks: digital fiber optics connecting the switching offices.

Three different way of multiplexing: frequency,

time, and wavelength.

• Switching offices: where calls are moved from one trunk to another.

Types of switching: circuit switching vs. packet

switching.

09/13/2007 EETS 7304 38

The Politics of Telephones

1984 US was divided into 164 LATAs (Local Access and Transport Area equivalent to an area code). BOC has monopoly by its ILECs (Incumbent Local Exchange Carrier) within LATA. AT&T (IXC#1) and competitors (IXC#2) use IXCs (IntereXchange Carrier) for traffic between LATAs. Long Distance carrier also must build IXC POP switches at each LATA to connect it to IXC. BOCs are required to connect to each IXC POP. 1996 interference was allowed.

09/13/2007 EETS 7304 39

Present day Internet

POP – Point Of Presence are ISP (e.g. AOL) modems connected to Regional ISP network.Regional ISP network is connected to the backbone. Backbones are connected by NAP (Network Access Point) or by their own routers. Finally Server Farm (multiplicity of identical servers) are connected to the router.

09/13/2007 EETS 7304 40

The Local Loop: Modems, ADSL, and Wireless

ISP1 handles Internet call from the computer. Digital > modem>analog subscriber line>codec>digital trunk>codec>analog subscriber line>modem (bank) >ISP1 computer (POP). ISP2 handles it faster.

09/13/2007 EETS 7304 41

Handshaking between two modems in RS-232CDSR

DTR

RI

RTS

CTS

CD

TD

RD

DSR

DTR

RI

RTS

CTS

CD

TD

RD

DTE DCE

TD

RTS

DTR

RD

CTS

DSR

RI

CD

GRND

DTEDCE

TD

RTS

DTR

RD

CTS

DSR

RI

CD

GRND

Data Set Ready

Data Terminal Ready

Ring Indicator

Request To Send

Clear To Send

Transmit Data

Receive Data

Modem PC

Carrier Detect

DB9 bit connector

ModemPC

RI RTS RTS

09/13/2007 EETS 7304 42

RS232 electrical signals

Speed pulse/sec = baud1200/2400/4800/9600/19200 baud

Control charactersRTS – 0011110RI - 0000111

Start( “0”) + 7 data + parity + Stop (at least 1.5 “1”)

ASCII Data (binary) 0 0 1 1 1 1 0

Start “0”

Stop “1”

0 0 1 1 1 1 0

+15 V

-15 V

parityStop “1”

tline signals

Amplitude“0” +5/+15 V “1” -5/-15 V

09/13/2007 EETS 7304 43

Modems use carriers between 1 and 2 kHz

(a) A binary signal

(b) Amplitude modulation(c) Frequency modulation

(d) Phase modulation

09/13/2007 EETS 7304 44

Modems (2)

(a) QPSK.

(b) QAM-16: V.32 for 9600/2400 = 4.

(c) QAM-64: V32 bis for 14,400/2400 = 6.

09/13/2007 EETS 7304 45

Modems (3)

(a) 2400 *5 = 12000 bps (b) 2400 * 7 = 16800 bps

09/13/2007 EETS 7304 46

Digital Subscriber Lines

Maximum bit rate versus distance over category 3 UTP for DSL.

09/13/2007 EETS 7304 47

Digital Subscriber Lines

Ch – 0 voice; Ch 1-5 not used

32 Chs * 32 kbps = 1 Mbps upstream data + 1 ch for control

216 Chs * 32 kbps = 7 Mbps downstream data + 1 for control

Operation of ADSL using discrete multitone modulation.256 channels * 4312.5 Hz = 1.1040 MHz.

It works as 250 different frequency modems.

09/13/2007 EETS 7304 48

Digital Subscriber Lines

NID – Network Interface Device.DSLAM – Digital Subscriber Line Access Multiplexer.

09/13/2007 EETS 7304 49

Wireless Local Loops

Connection to the CLEC (Competitive Local Exchange Carrier): LMDS (Local Multipoint Distribution System).FCC allocated bandwidth 198 MHz at 2.5 GHz taken from Instructional TV.

Recommended