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TelecomunicazioniDocente: Andrea Baiocchi

Dip. INFOCOM - Stanza 35, 1 piano palazzina P. Piga

Sede Facolt S. Pietro in Vincoli

E-mail: [email protected]

University of RomaLa Sapienza

Corso di Laurea in Ingegneria Gestionale

A.A. 2010/2011

Telecomunicazioni - a.a. 2010/2011 - Prof. Andrea Baiocchi

Programma

1. SERVIZI E RETI DI TELECOMUNICAZIONE

2. FONDAMENTI DI COMUNICAZIONI

3. ARCHITETTURE DI COMUNICAZIONE

4. MODI DI TRASFERIMENTO

5. LO STRATO DA ESTREMO A ESTREMO: UDPE TCP

6. LO STRATO DI RETE IN INTERNET

7. TECNOLOGIE DI STRATO DI

COLLEGAMENTO

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Fundamentals of communicationsA roadmap! Digital Representation of Information! Digital Representation of Analog Signals! Why Digital Communications?! Characterization of Communication Channels! Fundamental Limits in Digital Transmission! Line Coding! Modems and Digital Modulation! Properties of Media and Digital Transmission Systems

! Error Detection and Correction

Chapter 3


Digital Networks

! Digital transmission enables networks tosupport many services

E-mail

Telephone

TV

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Questions of Interest

! Can we reduce all information to sequences of bits?How?

! How many bits do we need to represent a message(text, speech, image)?

! How fast does the network/system transferinformation? Under which quality constraints?

! How can we deal with errors?! How are errors introduced?

! How are errors detected and corrected?

! What transmission speed and coding of data is possibleover radio, copper cables, fiber, infrared, ?

Fundamentals ofcommunications

Digital Representation ofInformation

Adapted from slides of the book:A. Leon Garcia, I. Widjaja, Communication

networks, McGraw Hill, 2004

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Bits, numbers, information

! Bit: BInary digiT

! Either symbol belonging to a set of two elements ornumber with value 0 or 1! nbits: digital representation for 0, 1, , 2n1! Byte or Octet, n= 8! Computer word, typically n= 32, or 64

! nbits allows enumeration of 2npossibilities! n-bit field in a header! n-bit representation of a voice sample! Message consisting of nbits

! The number of bits required to represent a messageis a measure of its information content! More bits -> More information


Block vs. Stream Information

Block

! Information that occurs in a single, delimiteddata unit(bit string)! Text message, Data file, JPEG image, MPEG file

! Size = bits / block

Stream

! Information that is produced and possibly conveyedover a communication system continuously! Real-time voice (e.g. telephony)

! Streaming video

! Bit rate = bits / second

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1-8 Mbytes

(5-30)

38.4

Mbytes

8x10 in2 photo

4002 pixels/in2

JPEGColor

Image

5-54 kbytes(5-50)

256kbytes

A4 page200x100pixels/in2

CCITT Group 3Fax

(2-6)kbytes-Mbytes

ASCIIZip, compressText

Compressed(Ratio)

OriginalFormatMethodType

Examples of Block Information


H

W

= + +H

W

H

W

H

W

Color

image

Red

component

image

Green

component

image

Blue

component

image

Total bits = 3 ! H ! W pixels ! B bits/pixel = 3HWB bits

Example: 8!10 inch picture at 400 ! 400 pixels per inch2

400 ! 400 ! 8 ! 10 = 12.8 million pixels

8 bits/pixel/color12.8 megapixels ! 3 bytes/pixel = 38.4 megabytes

Color Image

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Th e s p ee ch s i g n al l e v el v a r ie s w i th t i m(e)

Stream Information

! A real-time voice signal must be digitized and

transmitted or recorded as it is produced! Analog signal level varies continuously in time


Analog signal

! In communications engineering signal refersto e physically measurable entity that can beused to carry information! E.g. e.m. field, voltage and current in lumped

circuits, air pressure

! An analog signal is a function x(t) defined overthe realaxis and taking values in an interval ofthe realline

! Information is carried by the values of x(t) ateach time t

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Examples of analog signal sources

! Images are natively carried by analog signals, underthe form of e.m. field in the visible bandwidth, i.e. the

range of frequencies that produces a reaction inhuman sight sensors

! Sounds are natively carried by analog signals, i.e.variation of air pressure that produces a reaction inhuman hearing sensors

! Analog nature is due to the fact that for our purposesall these phenomena are well described by classicalphysics models (Newton mechanics, Maxwell e.m. field

theory) and classical physics rests on classical analysisto describe its models (variables taking values on acontinuum, e.g. real axis).


Digitization of Analog Signal

! Sample analog signal in time and amplitude

! Find closest approximation

"/2

3"/2

5"/2

7"/2

#"/2

#3"/2

#5"/2

#7"/2

Original signal

Sample value

Approximation

Rs= Bit rate = # bits/sample x # samples/second

3bits/sample

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"/2

3"/2

5"/2

7"/2

-"/2-3"/2

-5"/2

-7"/2

(a) Original

waveform and

the sample

values

"/2

3"/2

5"/2

7"/2

-"/2

-3"/2

-5"/2

-7"/2

(b) Original

waveform and

the quantized

values

Figure 3.2


Example: Voice & Audio

Telephone voice

! Ws= 4 kHz -> 8000samples/sec

! 8 bits/sample! Rs=8 x 8000 = 64 kbps

! Cellular phones use morepowerful compressionalgorithms: e.g. 6.5-13kbps for GSM

CD Audio! Ws= 22 kHertz -> 44000

samples/sec!

16 bits/sample! Rs=16 x 44000= 704 kbps

per audio channel! MP3 uses more powerful

compression algorithms:50 kbps per audio channel

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Video Signal

! Sequence of picture frames! Each picture digitized &

compressed! Frame repetition rate

! 10-30-60 frames/seconddepending on quality

! Frame resolution! Small frames for

videoconferencing

! Standard frames for

conventional broadcast TV! HDTV frames

30 fps

Rate = M bits/pixel x (WxH) pixels/frame x Fframes/second


Video Frames

Broadcast TV at 30 frames/sec =

10.4 x 106 pixels/sec

720

480

HDTV at 30 frames/sec =

67 x 106 pixels/sec1080

1920

QCIF videoconferencing at 30 frames/sec =

760,000 pixels/sec

144

176

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Digital Video Signals

19-38 Mbps1.6Gbps

1920x1080@30 fr/sec

MPEG2HDTV

2-6 Mbps249Mbps

720x480 pix@30 fr/sec

MPEG2Full Motion

64-1544kbps

2-36Mbps

176x144 or352x288 pix

@10-30fr/sec

H.261VideoConference

CompressedOriginalFormatMethodType

More recent standards: H.264, MPEG4


Transmission of Stream Information

! Constant bit-rate! Signals such as digitized telephone voice produce a

steady stream: e.g. 64 kbps! Network must support steady transfer of

information, e.g. 64 kbps circuit! Variable bit-rate! Signals such as digitized video produce a stream

that varies in bit rate, e.g. according to motion anddetail in a scene

! Network must support variable transfer rate ofinformation with possibly a guaranteed minimumrate, e.g. packet switching with traffic engineering

functions

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Stream Service Quality Issues

Network Transmission Impairments!

Delay: Is information delivered in timely fashion?! E.g. mean e2e delay! Jitter: Is information delivered in smooth fashion?

! E.g. delay standard deviation

! Loss: Is information delivered without loss? If lossoccurs, is delivered signal quality acceptable?! Errored data! Undelivered data! Mis-ordered data

! Applications & application layer protocols developedto deal with these impairments


Transmission Delay

Use data compression to reduceL

Use higher speed modem to increaseRPlace far end system closer to reduced

L number of bits in messageR bps speed of digital communication systemL/R time to transmit the messagetprop time for signal to propagate across medium

d distance in metersc speed of light (3x108 m/s in vacuum)

Delay = tprop + L/R = d/c + L/R seconds

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Compression

! Information usually not representedefficiently

! Data compression algorithms! Represent the information using fewer bits than

provided natively! Noiseless: original information recovered exactly

! E.g. zip, compress, GIF

! Noisy: recover information approximately.Tradeoff: # bits vs. quality! E.g. JPEG, MPEG

! Compression Ratio#bits (original file) / #bits (compressed file)


Lempel-Ziv (LZ77) algorithm

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Digital Representation of AnalogSignals

Fundamentals of

communications




Digitization of Analog Signals

" Sampling: obtain samples of x(t) at uniformlyspaced time intervals:

xk=x(tk), tk=t0+kT, kinteger

Tis the sampling time, F=1/Tis the samplingrate." Quantization: map each sample xk into an

approximation value yk=f(xk) of finiteprecision

" Compression: to lower bit rate further, applyadditional compression method

! Differential coding: cellular telephone speech! Subband coding: MP3 audio

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Sampling Rate and Bandwidth

! A signal that varies faster needs to besampled more frequently

! Bandwidthmeasures how fast a signal varies

! What is the bandwidth of a signal?

! How is bandwidth related to sampling rate?

1 0 1 0 1 0 1 0

. . . . . .

t

1 ms

1 1 1 1 0 0 0 0

. . . . . .

t

1 ms


Periodic Signals

! A periodic signal with period Tcan be represented assum of sinusoids using Fourier Series:

DC

long-term

averagefundamental

frequency f0=1/T

first harmonic

kth harmonic

x(t) = a0 + a1cos(2!f0t+ "1) + a2cos(2$2f0t+ "2) + + akcos(2!kf0t+ "k) +

|ak|2 determines amount of power in kth harmonic

Amplitude specturm |a0|, |a1|, |a2|,

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Example Fourier Series

T1 = 1 ms

x2(t)

T2 =0.25 ms

x1(t)

Only odd harmonics have power

x1(t) = 0 + cos(2!4000t)

cos(2!3(4000)t)

+ cos(2!5(4000)t) +

4!

45!

43!

x2(t) = 0 + cos(2!1000t)

cos(2!3(1000)t)

+ cos(2!5(1000)t) +

4!

45!

43!

1 0 1 0 1 0 1 0

. . . . . .

t

1 1 1 1 0 0 0 0

. . . . . .

t


Spectra & Bandwidth

! Spectrum of a signal:magnitude of amplitudes as afunction of frequency

! x1(t)varies faster in time &

has more high frequencycontent than x2(t)

! Bandwidth Wis defined asrange of frequencies where asignal has non-negligiblepower, e.g. range of bandthat contains 99% of totalsignal power

0

0.2

0.4

0.6

0.8

1

1.2

0 3 6 9 12 15 18 21 24 27 30 33 36 39 42

frequency (kHz)

0

0.2

0.4

0.6

0.8

1

1.2

0 3 6 9 12 15 18 21 24 27 30 33 36 39 42

frequency (kHz)

Spectrum ofx1(t)

Spectrum ofx2(t)

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Bandwidth of General Signals

! Not all signals are periodic! E.g. voice signals varies according

to sound! Vowels are periodic, s is

noiselike! Spectrum of long-term signal

! Averages over many sounds, manyspeakers! Involves Fourier transform

! Telephone speech: 4 kHz! CD Audio: 22 kHz

s (noisy ) | p (air stopped) | ee (periodic) | t (stopped) | sh (noisy)

speech

f

W

X(f)

0


Sampling theorem

Sampling theorem (Nyquist):

Perfect reconstruction if sampling rate 1/T! 2W

Interpolation

filter t

x(t)

t

x(nT)

(b)

Samplert

x(t)

t

x(nT)(a)

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Quantization error:

noise = y(nT) x(nT)

Quantizer maps input

into closest of 2

m

representation values

"/23"/2

5"/2

7"/2

-"/2

-3"/2

-5"/2

-7"/2

Original signal

Sample value

Approximation

3bits/sa

mple

Quantization of Analog Samples

inputx(nT)

output y(nT)

0.5"1.5"

2.5"

3.5"

-0.5"

-1.5"

-2.5"

-3.5"

" 2" 3" 4"

#"#2"#3"#4"

Uniform

quantizer


M= 2m quantization levels

Dynamic range (V, V)Quantization interval " = 2V/M(uniform quantization)

If the number of levels Mis large, the errore(x)=y(x)xisapproximately uniformly distributed between ("/2, "/2) in

each quantization interval

Quantizer error

!

2

...

error = y(nT)x(nT) = e(nT)

input...

!"

2

3"" "#2" 2"

x(nT) V-V

y(nT)

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Quantizer performance

! Power of quantization error signal = average of

squared error


Signal-to-Noise Ratio (SNR) =

Let %x2 be the signal power, then

The SNR is usually stated in decibels:

SNR dB = 10 log10(#x2/#e2) = 6m+ 10 log10(3#x2/V2)

Example: SNR dB = 6m 7.27 dB for V/#x= 4.

Signal-to-quantization noise ratio

Average signal power

Average noise power

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Digital Transmission of AnalogInformation

Interpolationfilter

Displayor

playout

2Wsamples / sec

2W m bits/secx(t)

Bandwidth W

Sampling(A/D)

QuantizationAnalogsource

2Wsamples / sec m bits / sample

Pulse

generator

y(t)

Original

Approximation

Transmission

or storage


W= 4 kHz, so Nyquist sampling theorem

& 2W = 8000 samples/second

Suppose error requirement = 1% error

SNR = 10 log10(1/.01)2 = 40 dBAssume V/#x= 4, then

40 dB = 6m 7.27 & m= 8 bits/sample

PCM (Pulse Code Modulation):

Bit rate= 8000 x 8 bits/sec= 64 kbps

Example: Telephone Speech

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Why Digital Communications?

Fundamentals of

communications




A Transmission System

Transmitter! Converts information into signalsuitable for transmission

! Signal = measurable physical quantity that can be modifiedaccording to the value of the data to be transmitted, conveyed overa transmissin medium and detected by a receiving device.

! Injects energy into communications medium or channel

Receiver! Receives energy from medium! Converts received signal into form suitable for delivery to user

Receiver

Communication channel

Transmitter

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Transmission Impairments

Communication Channel! Pair of copper wires

! Coaxial cable

! Optical fiber

!

Radio! Including infrared

Transmission Impairments! Attenuation

! Distortion

! Noise

!

Interference! Timing errors

Transmitted

SignalReceived

Signal Receiver


Transmitter


Analog vs. Digital Transmission

Analog transmission: all details must be reproduced accurately

Sent

Sent

Received

Received

Distortion

Attenuation

Digital transmission: only discrete levels need to be reproduced

Distortion

AttenuationSimple Receiver:

Was original

pulse positive ornegative?

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Analog Long-Distance Communications

! Each repeater attempts to restore signal to its original form! Attenuation is removed (amplifier)

! Distortion is not completely eliminated

! In-band noise & interference can be removed only in part (out of band)

! Signal quality decreases with # of repeaters

Source DestinationRepeater

Transmission segment

Repeater. . .

Attenuated and

distorted signal + noise

Equalizer

Recovered signal +

residual noiseRepeater

Amp


Digital Long-Distance Communications

! Regenerator recovers original data (bit) sequence fromdegraded signal and retransmits on next segment byusing a clean signal! But timing recovery is required!

! All impairments are condensed into bit errors

Source DestinationRegenerator

Transmission segment

Regenerator. . .

Amplifier

equalizer

Timing

recovery

Decision circuit

and signal

regenerator

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Digital Binary Signal

For a given communications medium:! How do we increase transmission speed?! How do we achieve reliable communications?! Are there limits to speed and reliability?

+A

-A

0 T 2T 3T 4T 5T 6T

1 1 1 10 0

Bit rate = 1 bit / T seconds

Signal is meaningless without associated clock


Clock signal

Message bits

Baseband signalwith NRZ coding

+d

-d

1

0

0 1 0 1 1 0 1 0 0 0 1 1 0

Example of clock

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Many wavelengths>1600 GbpsOptical fiber

1 wavelength2.5-10 GbpsOptical fiber5 km multipoint radio1.5-45 Mbps28 GHz radio

IEEE 802.11b/a/g wireless LANFrom 1 to 54 Mbps2.4 GHz radio

Coexists with analog telephonesignal

Up 8 Mbps down (ADSL)20 Mbps down (ADSL2+)50 Mbps down (VDSL)

ADSL twistedpair

Shared CATV return channel500 kbps-4 MbpsCable modem

From a few m up to a fewhundreds m of unshieldedtwisted copper wire pair

10 Mbps, 100 Mbps, 1Gbps, 10 Gbps

Ethernettwisted pair

4 kHz telephone channel33.6-56 kbpsTelephone

twisted pair

ObservationsBit RateSystem

Bit rates of digital transmission systems

Characterization ofCommunication Channels




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Communications Channels

! A physical mediumis an inherent part of a

communications system! Copper wires, radio medium, or optical fiber

! Communications system includes electronic oroptical devices that are part of the pathfollowed by a signal! Transmitter, equalizers, amplifiers, filters, couplers,

detector, clocks and carrier generators

! By communication channelwe refer to thecombined end-to-end physical medium andattached devices


Communication system block scheme

SourceSource

encoder

Channel

encoder

Line

encoder

Modulator &

tx front end

Information source

Channel

Line

decoder

A/D

converter

Rx

front end

Timing

recovery

Channel

decoder

Information

rendering

Final

user

Information destination

Demodulation,

equalization,

symbol decision

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How good is a channel?

! Performance: What is the maximum reliabletransmission speed?! Speed: Bit rate, Rbps! Reliability: Bit error rate, BER=10k

! Focus of this section

! Cost: What is the cost of alternatives at a given levelof performance?! Wired vs. wireless?

! Electronic vs. optical?

! Standard A vs. standard B?


Communications Channel

Bandwidth

! In order to transfer datafaster, a signal has to varymore quickly.

! A channel or medium has aninherent limit on how fastthe signals it passes can vary

! Channel bandwidth limitshow tightly input pulses canbe packed

Impairments! Signal attenuation! Signal distortion! Spurious noise! Interference from other

signals! Channel impairments limit

accuracy of measurements onreceived signal

Transmitted

Signal

Received

Signal

ReceiverCommunication channelTransmitter

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Communication channel model

! We often assume two basic properties of

channels! Linearity

y1(t)=Ch[x1(t)], y2(t)=Ch[x2(t)] =>

y1(t)+y2(t)=Ch[x1(t)+x2(t)]! Counterexample: amplifiers distorsion

! Stationarity

y1(t)=Ch[x1(t)] => y2(t+d)=Ch[x2(t+d)]for any d>0.! Counterexample: radiomobile channels


Channel

tt

x(t) = Aincos(2$ft) y(t) = Aout(f)cos(2$ft+ '(f))

Aout

AinA(f) =

Frequency Domain ChannelCharacterization

! Assumption. Channel is linear and stationary! Linear: superposition of effects holds! Stationary: input-output relationship does not vary over time

! Apply sinusoidal input at frequency f! Output is sinusoid at same frequency, but attenuated & phase-

shifted! Sinusoids are autofunctions of LTI systems

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Channel

t t

x(t) = Aincos(2$ft) y(t) = Aout(f)cos(2$ft+ '(f))

Frequency Domain ChannelCharacterization

! Apply sinusoidal input at frequency f! Measure amplitude of output sinusoid (of same frequency f) and

calculate amplitude response A(f) = ratio of output amplitude toinput amplitude! If A(f) ! 1, then input signal passes readily! If A(f) ! 0, then input signal is blocked

! Bandwidth Wc is range of frequencies passed by channel

H(f) =A(f)ei'(f)Transfer function:


Ideal Low-Pass Filter

! Ideal filter: all sinusoids with frequency f

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Example: Low-Pass Filter

! Simplest non-ideal circuit that provides low-pass filtering

H(f)=1/(1+i2$bf)! Example: RC circuit, b=RC.

f

1 A(f)= (1+4$2b2f2)1/2

Amplitude Response

f0

$(f)= arctan(2$bf)

-45o

-90o

1/2$

Phase Response

Wc

1/" 2


Example: Bandpass Channel

! Some channels pass signals within a band that excludeslow frequencies! ADSL modems, radio systems,

! Channel bandwidthis the width of the frequency bandthat passes non-negligible signal power

!

Example. 3dB bandwidth: frequency interval whereoutput power density is no less than 1/2 than peak value

f

Amplitude Response

A(f)Wc=f2f1

f1 f2

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Channel Distortion

! Channel has two effects:! If amplitude response is not flat, then different frequency

components of x(t) will be transferred by different amounts! If phase response is not linear, then different frequency

components of x(t) will be delayed by different amounts

! In either case, the shape of x(t) is altered

! Let x(t) be a digital signal bearing data information

! How well does y(t) follow x(t)?

y(t) = )kA(fk) akcos(2$fkt+ *k+ '(fk))

Channely(t)x(t) = )kakcos(2$fkt+ *k)


Example: Amplitude Distortion

! Let x(t)input to ideal lowpass filter that has zero delayand Wc= 1.5 kHz, 2.5 kHz, or 4.5 kHz

1 0 0 0 0 0 0 1

. . . . . .

t1 ms

x(t)

! Wc= 1.5 kHz passes only the first two terms

! Wc= 2.5 kHz passes the first three terms! Wc= 4.5 kHz passes the first five terms

f0=1/T=1000 Hz

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-1.5

-1

-0.5

0

0.5

1

1.5

0

0.1

25

0.2

5

0.3

75

0.5

0.6

25

0.7

5

0.8

75 1

-1.5

-1

-0.5

0

0.5

1

1.5

0

0.1

25

0.2

5

0.3

75

0.5

0.6

25

0.7

5

0.8

75 1

-1.5

-1

-0.5

0

0.5

1

1.5

0

0.125

0.25

0.375

0.5

0.625

0.75

0.875 1

Amplitude Distortion

! As the channelbandwidth

increases, theoutput of thechannel resemblesthe input moreclosely


Time-domain Characterization

! Time-domain characterization of a channel requiresfinding the impulse response h(t)

! Apply a very narrow pulse of amplitude ato a channelat time ( and observe the channel output at time t

! The output in case of a linear, stationary, causalchannel is

y(t) = 0, t< ( y(t) = ah(t(), t >(

Channel

t0

t

h(t)

td

a

+(t)

0

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Impulse response

! h(t) is the impulse response of the channel

INPUT OUTPUT

By definition +(t) h(t)Causality a+(t) h(t)=0 for t

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InterSymbol Interference (ISI)

! By sampling channel output at time nT(perfectsync

) we getyn= y(nT) = $kxkhnk+ zn

with hnk= h(nTkT) and zn= z(nT).

yn= xnh0 + $k%nxkhnk+ zn

! ISI is the undesired interference coming from tailsof pulses other than the n-th one and giving non-nullcontributions to the n-th output sample

noise sampleISIuseful term

output

sample


Nyquist condition for null ISI

! Given a channel response H(f), we can addreception and possibly transmission filters, sothat the overall (filtered) channel response is

Hc(f) = HTX(f) H(f) HRX(f)! To get null ISI at sampling rate 1/Tit must be

$kHc(fk/T) = cost

! For a low-pass channel with Hc(f)=0 for |f|>Wcthis is not possible unless 1/T

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! For channel with ideal low-pass amplitude response ofbandwidth Wc, the impulse response is a Nyquist pulse

h(t)=s(t(), where T=1/(2Wc), and

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-7 -6 -5 -4 -3 -2 -1 0 1 2 3 4 5 6 7t

T T T T T T T T T T T T T T

! s(t) has zero crossings at t = kT, k= 1, 2,

! Pulses can be packed every Tseconds with zero Inter-Symbol Interference(ISI)

Nyquist Pulse with Zero ISI


-2

-1

0

1

2

-2 -1 0 1 2 3 4

tT T T T T T

-1

0

1

-2 -1 0 1 2 3 4

tT T T T T T

Example of composite waveformThree Nyquist pulsesshown separately

! + s(t)

! + s(t-T)

! - s(t-2T)

Composite waveform

r(t) = s(t)+s(t-T)-s(t-2T)

Samples at kT

r(0)=s(0)+s(-T)-s(-2T)=+1

r(T)=s(T)+s(0)-s(-T)=+1

r(2T)=s(2T)+s(T)-s(0)=-1

Zero ISI at samplingtimes kT

r(t)

+s(t) +s(t-T)

-s(t-2T)

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0f

A(f)

Nyquist pulse shapes! If channel is ideal low pass with Wc, then pulses maximum

rate pulses can be transmitted without ISI is2Wc pulse/s

! s(t) is one example of class of Nyquist pulses with zero ISI! Problem: sidelobes in s(t) decay as 1/twhich add up quickly when

there are slight errors in timing

! Raised cosine pulse below has zero ISI! Requires slightly more bandwidth than Wc! Sidelobes decay as 1/t3, so more robust to timing errors

1

sin(!t/T)!t/T cos(

!,t/T)1 (2,t/T)2

(1,)Wc Wc (1+,)Wc


! 10 Gbit/s signalwithout dispersion(negligible ISI)

! 10 Gbit/s signal aftertransmission througha dispersive channel(with non negligibleISI)

Eye diagram

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Fundamental Limits in DigitalTransmission

Fundamentals of

communications




Transmitter

Filter

Communication

Medium

Receiver

Filter Receiver

r(t)

Received signal

+A

-A0 T 2T 3T 4T 5T

1 1 1 10 0

t

Signaling with Nyquist Pulses

! p(t)pulse at receiver in response to a single input pulse (takes intoaccount pulse shape at input, transmitter & receiver filters, andcommunications medium)

! r(t)waveform that appears in response to sequence of pulses

! If s(t)is a Nyquist pulse, then r(t)has zero intersymbol

interference (ISI) when sampled at multiples ofT

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Multilevel Signaling! Nyquist pulses achieve the maximum signalling rate with

zero ISI! 2Wcpulses per second or 2Wcpulses / WcHz = 2 pulses / Hz

! With two signal levels, each pulse carries one bit of thesource bit stream! Bit rate = 2Wc(bit/s)

! With M = 2msignal levels, each pulse carries mbit! Bit rate = 2Wc(pulse/s) m(bit/pulse) = 2Wcm(bit/s)

In the absence of noise, the bit rate can be increasedwithout limit by increasingm

BUTAdditive noise limits # of levels that can be usedreliably.


Example of Multilevel Signaling

! Four levels {-1, -1/3, 1/3, +1} for {00,01,10,11}

! Waveform for 11,10,01 sends +1, +1/3, -1/3

! Zero ISI at sampling instants

-0.6

-0.4

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

-1 0 1 2 3

Composite waveform

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Noise & Reliable Communications

! All physical systems have noise

! Electrons always vibrate at non-zero temperature:motion of electrons induces noise

! Presence of noise limits accuracy ofmeasurement of received signal amplitude

! Errors occur if signal separation is comparableto noise level

! Noise places a limit on how many amplitude

levels can be used in pulse transmission


Four signal levels Eight signal levels

Noise Limits Accuracy! Receiver makes decision based on (sampled) received

signal level = source pulse level + noise! Error rate depends on relative value of noise amplitude and

spacing between signal levels

! Large (positive or negative) noise values can cause wrong decision

Typical noise

+A

+A/3

-A/3

-A

+A

+5A/7

+3A/7

+A/7

-A/7

-3A/7

-5A/7

-A

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Noise! Noise signal is usually a zero-mean process z(t)

characterized by

! probability distribution of amplitude samples, i.e.Pr(z(t) > u)

! Time auto-correlation, i.e. Rzz(t)=E[z(h)z(h+t)]

! Thermal electronic noise is inevitable (due tovibrations of electrons); thermal Noise can bemodeled as a white Gaussian process! Probability distribution is Gaussian zero mean

! Time auto-correlation is a Dirac pulse at 0! Often interference from a large number of

scattered and similar sources can be modeledas white Gaussian noise


22

2

2

1!

!"

x

e#

x0

Gaussian noise

t

x

Pr(X(t)>x0) = area under graph on the right ofx0

x0

x0

%2 = Avg Noise Power

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Probability of Error! Error occurs if noise value exceeds the

information signal magnitude over the decision

threshold! With two-level signalling, +A and A, probability

of error is Q(A/%)

1.00E-121.00E-11

1.00E-101.00E-091.00E-08

1.00E-071.00E-06

1.00E-05

1.00E-041.00E-031.00E-02

1.00E-011.00E+00

0 2 4 6 8

x

Q(x)


Role of SNR

! With M=2m levels per symbol, the tx symbolvalues are ak=A+(2k1)A/M, with k=1,,M.

! With equiprobable symbols:E[ak2]=(M21)A2/(3M2)=PTX

! Received sample is (dispersive channel, zeroISI): yn=h0xn+zn.

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signal noise signal + noise

signal noise signal + noise

HighSNR

Low

SNR

SNR =Average Signal Power

Average Noise Power

SNR (dB) = 10 log10 SNR

virtually error-free

error-prone

Channel Noise affects Reliability


! If transmitted power is limited, then as M increasesspacing between levels decreases

! Presence of noise at receiver causes more frequenterrors to occur as M is increased

Shannon Channel Capacity:! The maximum reliabletransmission rate over an AWGN

bandlimited channel with bandwidth WcHz is

Cb= Wc log2(1+SNR) bit/s

Shannon Channel Capacity

X

Input symbol

(Gaussian)

Y=X+Z

Output symbol

Z

White gaussian noise

2Wcsymbols/s

Bandlimited channel (Wc)

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Capacity of AWGN channel

! It can be shown that in case of real input

signal the optimal source is gaussian and theAWGN capacity isC= 0.5 log2(1+P/PN) [bit/symbol]

where PN is the additive noise power, Pis theuseful signal received power

! A dispersive, additive noise channel can bereduced to AWGN if zero ISI is provided; to

that end it must be symbol rate< 2Wc. ThenCb= Wclog2(1+(Eb/N0)Rb/Wc) [bit/s]


! Reliable communications is possible if the tx rate Rb Cb, then reliable communications is not possible.

Reliable means the bit error rate (BER) can be madearbitrarily small through sufficiently complex coding.

! Bandwidth Wc& SNRdetermine Cb! Cbcan be used as a measure of how close a system design is to

the best achievable performance.

! SNR=P/(N0Wc), with

! P= average power of input signal

! N0=noise power spectral density=k-F, k=1.381023 J/K, -=noisetemperature, typically 300 K, F=noise figure, typically 6 dB

Shannon Channel Capacity

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Example

! Find the Shannon channel capacity for a telephonechannel with Wc= 3400 Hz and SNR= 10000

C= 3400 log2 (1 + 10000)

= 3400 log10 (10001)/log102 = 45200 bps

Note that SNR= 10000 corresponds to

SNR(dB) = 10 log10(10000) = 40 dB

Line Coding




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What is Line Coding?

! Mapping of binary information sequence into thedigital signal that enters the channel! Ex. 1 maps to +A square pulse; 0 to A pulse

! Line code selected to meet system requirements:! Transmitted power: Power consumption = $

! Bittiming: Transitions in signal help timing recovery

! Bandwidthefficiency: Excessive transitions wastes bw

! Lowfrequencycontent: Some channels block low frequencies

! long periods of +A or of A causes signal to droop

! Waveform should not have low-frequency content

! Errordetection: Ability to detect errors helps

! Complexity/cost: Is code implementable in chip at high speed?


Line coding examples1 0 1 0 1 1 0 01

Unipolar

NRZ

NRZ-inverted

(differential

encoding)

Bipolar

encoding

Manchester

encoding

DifferentialManchester

encoding

Polar NRZ

Figure 3.35

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-0.2

0

0.2

0.4

0.6

0.8

1

1.2

00.2

0.4

0.6

0.8 1

1.2

1.4

1.6

1.8 2

fT

po

werdensity

NRZ

Bipolar

Manchester

Spectrum of Line codes

! Assume 1s & 0s independent & equiprobable

!

NRZ has high content atlow frequencies! Bipolar tightly packed

around T/2! Manchester wasteful of

bandwidth


Unipolar & PolarNon-Return-to-Zero (NRZ)

Unipolar NRZ! 1 maps to +A pulse! 0 maps to no pulse! High Average Power

0.5*A2 +0.5*02=A2/2! Long strings of A or 0

! Poor timing! Low-frequency content

! Simple

Polar NRZ! 1 maps to +A/2 pulse! 0 maps to A/2 pulse! Better Average Power

0.5*(A/2)2 +0.5*(-A/2)2=A2/4! Long strings of +A/2 or A/2

! Poor timing! Low-frequency content

! Simple

1 0 1 0 1 1 0 01

Unipolar NRZ

Polar NRZ

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Bipolar Code

! Three signal levels: {-A, 0, +A}

! 1 maps to +A or A in alternation

! 0 maps to no pulse! Every +pulse matched by pulse so little content at low frequencies

! String of 1s produces a square wave! Spectrum centered at T/2

! Long string of 0s causes receiver to lose synch

! Zero-substitution codes

1 0 1 0 1 1 0 01

BipolarEncoding


Manchester code & mBnB codes

!

1 maps into A/2 first T/2, -A/2 last T/2! 0 maps into -A/2 first T/2,

A/2 last T/2! Every interval has transition in

middle! Timing recovery easy! Uses double the minimum

bandwidth! Simple to implement! Used in 10-MbpsEthernet &

other LAN standards

!

mBnB line code! Maps block of mbits into n

bits! Manchester code is 1B2B

code! 4B5B code used in FDDI LAN! 8B10B code used in Gigabit

Ethernet! 64B66B code used in 10G

Ethernet

1 0 1 0 1 1 0 01

Manchester

Encoding

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Differential Coding

! Errors in some systems cause transposition in polarity, +A becomeA and vice versa! All subsequent bits in Polar NRZ coding would be in error

! Differential line coding provides robustness to this type of error! 1 mapped into transition in signal level! 0 mapped into no transition in signal level

! Same spectrum as NRZ! Errors occur in pairs! Also used with Manchester coding

NRZ-inverted

(differential

encoding)

1 0 1 0 1 1 0 01

Differential

Manchester

encoding

Modems and Digital Modulation




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Bandpass Channels

! Bandpass channels pass a range of frequencies aroundsome center frequency fc! Radio channels, telephone & DSL modems

! Digital modulators embed information into waveformwith frequencies passed by bandpass channel! A sinusoidal signal with frequency fccentered in middle of

bandpass channel is named carrier! Modulators embed information into the carrier

fc Wc/2 fc0 fc+ Wc/2


Information 1 1 1 10 0

+1

-10 T 2T 3T 4T 5T 6T

Amplitude

Shift

Keying

+1

-1

Frequency

Shift

Keying 0 T 2T 3T 4T 5T 6T

t

t

Amplitude Modulation andFrequency Modulation

Map bits into amplitude of sinusoid: 1 send sinusoid; 0 no sinusoid

Demodulator looks for signal vs. no signal

Map bits into frequency: 1 send frequency fc+ + ; 0 send frequency fc- +

Demodulator looks for power around fc + + orfc- +

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Phase Modulation

! Map bits into phase of sinusoid:! 1 send A cos(2$ft) , i.e. phase is 0

! 0 send A cos(2$ft+$) , i.e. phase is $

! Equivalent to multiplying cos(2$ft) by +A or -A! 1 send A cos(2$ft) , i.e. multiply by 1

! 0 send A cos(2$ft+$) = - A cos(2$ft) , i.e. multiply by -1

+1

-1

PhaseShift

Keying 0 T 2T 3T 4T 5T 6Tt

Information 1 1 1 10 0


Modulatecos(2$fct)by multiplying byAkforTseconds:

Ak x

cos(2$fct)

Y(t) =Ak cos(2$fct), kT

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Demodulator

Demodulate (recoverAk) by multiplying by 2cos(2$fct)

forTseconds and lowpass filtering (smoothing):

x

2cos(2$fct)2Akcos

2(2$fct) =Ak {1 + cos(2$2fct)}

Lowpass

Filter

(Smoother)

X(t)Y(t) =Akcos(2$fct)

Received signal

during kth interval

Multiplication for the local carrier and integration over a

symbol interval takes place for every symbol

Synchronous demodulation

(perfect local carrier)


1 1 1 10 0

+A

-A0 T 2T 3T 4T 5T 6T

Information

Baseband

Signal

Modulated

Signal

x(t)

+A

-A0 T 2T 3T 4T 5T 6T

Example of Modulation

A cos(2$ft) -A cos(2$ft)

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1 1 1 10 0Recovered

Information

Baseband

signal discernable

after smoothing

After multiplication

at receiverx(t) cos(2$fct)

+2A

2A

0 T 2T 3T 4T 5T 6T

+A

-A

0 T 2T 3T 4T 5T 6T

Example of DemodulationA {1 + cos(4$fct)} -A {1 + cos(4$fct)}


Signaling rate andTransmission Bandwidth! Fact from modulation theory:

Baseband signalx(t)

with bandwidth B Hz

If

then B

fc+B

f

f

fc-B fc

Modulated signalx(t)cos(2$fct) has

bandwidth 2B Hz

! If bandpass channel has bandwidth WcHz,! Then baseband channel has Wc/2 Hz available, so modulation

system supports Wc/2 x 2 = Wc pulses/second

! That is, Wc

pulse/s per Wc

Hz = 1 pulse/Hz

! Recall baseband transmission system supports 2 pulses/Hz

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Ak x

cos(2$fct)

Yi(t) =Ak cos(2$fct)

Bk x

sin(2$fct)

Yq(t) = Bk sin(2$fct)

+ Y(t)

! Yi(t) and Yq(t) both occupy the bandpass channel

! QAM sends 2 pulses/Hz

Quadrature Amplitude Modulation (QAM)

! QAM uses two-dimensional signaling! Akmodulates in-phase cos(2$fct)

! Bk modulates quadrature phase cos(2$fct+ $/2) = sin(2$fct)

Transmitted

Signal


QAM Demodulation

Y(t) x

2cos(2!fct)2Akcos

2(2!fct)+2Bkcos(2!fct)sin(2!fct)

=Ak [1 + cos(4!fct)]+Bksin(4!fct)=Ak+ [Akcos(4!fct)+Bksin(4!fct)]

Lowpass

filter

(smoother)

Ak

2Bksin2(2!fct)+2Akcos(2!fct)sin(2!fct)

= Bk[1 cos(4!fct)]+Ak sin(4!fct)

= Bk [Bk cos(4!fct) Ak sin(4!fct)]

x

2sin(2!fct)

Bk

Lowpass

filter

(smoother)

smoothed to zero

smoothed to zero

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Complex numbers

! Tx signal with in-phase and quad modulation is

x(t) = Akcos(2$fct)+Bksin(2$fct)! Let Ck=Ak+iBk; then

x(t) = Re[Ckexp(i2$fct)]

! We can now generalize to any alphabet ofcomplex symbols to code information bits.! E.g. in case of polar coordinates:

Re[Mkexp(i'k)exp(i2$fct)] = Mkcos(2$fct+'k)! This is phase modulation ifMk is a constant or mixed

amplitude and phase modulation in the general case(QAM)


Signal Constellations

! Each pair (Ak, Bk) defines a point in the plane

! Signal constellationset of signaling points

4 possible points perTs

2 bits / pulse

Ak

Bk


4 bits / pulse

Ak

Bk

(A, A)

(A,-A)(-A,-A)

(-A,A)

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Ak

Bk


Ak

Bk


Other Signal Constellations

! Point selected by amplitude & phase


Signal constellations and errors

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Telephone Modem Standards

Telephone Channel for modulation purposes has Wc=2400 Hz hence 2400 pulses per second

Modem Standard V.32bis

! Trellis modulation maps m bits into one of 2m+1constellation points

! 14,400 bps Trellis 128 2400x6

! 9600 bps Trellis 32 2400x4

! 4800 bps QAM 4 2400x2

Modem Standard V.34 adjusts pulse rate to channel! 2400-33600 bps Trellis 960 2400-3429 pulses/sec

Properties of Media and DigitalTransmission Systems




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Fundamental Issues in TransmissionMedia

! Information bearing capacity! Amplitude response & bandwidth

! Susceptibility to noise & interference

! Propagation speed of signal! c = 3 x 108 m/s in vacuum

! . = c/"/ speed of light in medium where />1 is the dielectricconstant of the medium

! . = 2.3 x 108 m/s in copper wire; . = 2.0 x 108 m/s in optical fiber

t= 0t = d/c


dmeters


Communications systems &Electromagnetic Spectrum

! Frequency of communications signals

Analog

telephoneDSL Cell

phone

WiFiOptical

fiber

102 104 106 108 1010 1012 1014 1016 1018 1020 1022 1024Frequency (Hz)

Wavelength (meters)

106 104 102 10 10-2 10-4 10-6 10-8 10-10 10-12 10-14

Powerand

telephone

Broadcast

radio

Microwave

radio

Infraredlight

Visiblelight

Ultravioletlight

X-rays

Gammarays

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Wireless & Wired Media

Wireless Media

! Signal energy propagatesin space, limiteddirectionality

! Interference possible,so spectrum regulated

! Limited bandwidth

! Simple infrastructure:antennas & transmitters

! User/terminal can move

Wired Media

! Signal energy contained& guided within medium

! Spectrum can be re-usedin separate media (wiresor cables), more scalable

! Extremely highbandwidth

! Complex infrastructure:ducts, conduits, poles,right-of-way


Attenuation

! Attenuation varies with media! Dependence on distance of central importance

! Wired media has exponential dependence! Received power at d meters proportional to 100d

! Attenuation in dB = 0d, where 0 is dB/meter

! Wireless media has logarithmic dependence! Received power at d meters proportional to d-n

! Attenuation in dB = n log d, where nis path loss exponent; n=2in free space

! Signal level maintained for much longer distances

! Space communications possible

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Twisted PairTwisted pair

! Two insulated copper wiresarranged in a regular spiralpattern to minimizeinterference

! Various thicknesses, e.g.0.016 inch (24 gauge)

! Low cost

! Telephone subscriber loopfrom customer to CO

! Intra-building telephonefrom wiring closet todesktop

! LAN cabling (Fast Ethernet,GbE)

Attenuation(dB/mi)

f (kHz)

19 gauge

22 gauge

24 gauge

26 gauge

6

12

18

24

30

110 100 1000

Lower

attenuation rate

analog telephone

Higher

attenuation rate

for DSL


Twisted Pair Bit Rates

! Twisted pairs can providehigh bit rates at shortdistances

! Asymmetric DigitalSubscriber Loop (ADSL)!

High-speed Internet Access! Above 28 kHz

! Much higher rates possible atshorter distances! Strategy for telephone

companies is to bring fiberclose to home & then twistedpair

! Higher-speed access + video

Data rates of 24-gauge twisted pair

1000 feet,

300 m

51.840

Mbps

STS-1

3000 feet,0.9 km

25.920Mbps

1/2 STS-1

4500 feet,1.4 km

12.960Mbps

1/4 STS-1

12,000 feet,3.7 km

6.312Mbps

DS2

18,000 feet,

5.5 km

1.544

Mbps

T-1

DistanceDataRateStandard

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Ethernet LANs! Category 3 unshielded twisted pair (UTP):

ordinary telephone wires! Category 5 UTP: tighter twisting to improve

signal quality

! Shielded twisted pair (STP): to minimizeinterference; costly! 10BASE-T Ethernet

! 10 Mbps, Baseband, Twisted pair! Two Cat3 pairs! Manchester coding, 100 meters

! 100BASE-T4 FastEthernet! 100 Mbps, Baseband, Twisted pair! Four Cat3 pairs! Three pairs for one direction at-a-time! 100/3 Mbps per pair;! 3B6T line code, 100 meters

! Cat5 & STP provide other options

! ! ! ! ! !


Coaxial Cable

! Cylindrical braided outerconductor surroundsinsulated inner wireconductor

! High interference immunity

! Higher bandwidth thantwisted pair

! Hundreds of MHz

! Cable TV distribution

! Long distance telephonetransmission

! Original Ethernet LANmedium

35

30

10

25

20

5

15

Attenuation(dB/km)

0.1 1.0 10 100

f (MHz)

2.6/9.5 mm

1.2/4.4 mm

0.7/2.9 mm

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UpstreamDownstream

5MHz

42MHz

54MHz

500MHz

550MHz

750

MHz

Downstream

Cable Modem & TV Spectrum

! Cable TV network originally unidirectional

! Cable plant needs upgrade to bidirectional

! 1 analog TV channel is 6 MHz, can support very high data rates! Cable Modem: sharedupstream & downstream

! 5-42 MHz upstream into network; 2 MHz channels; 500 kbps to 4 Mbps

! >550 MHz downstream from network; 6 MHz channels; 36 Mbps


Cable Network Topology

Head

end

Upstream fiber

Downstream fiber

Fiber

node

Coaxial

distribution

plant

Fiber

node

= Bidirectional

split-band

amplifier

Fiber Fiber

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Optical Fiber

! Light sources (lasers, LEDs) generate pulses of light that aretransmitted on optical fiber! Very long distances (>1000 km)

! Very high speeds ( up ro 40 Gbps/wavelength)

! Nearly error-free (BER of 10-15)

! Profound influence on network architecture! Dominates long distance transmission

! Distance less of a cost factor in communications

! Plentiful bandwidth for new services

Optical fiber

Optical

source

ModulatorElectrical

signalReceiver

Electrical

signal


Core

Cladding JacketLight

1c

Geometry of optical fiber

Total Internal Reflection in optical fiber

Transmission in Optical Fiber

! Very fine glass cylindrical core surrounded by concentric layer ofglass (cladding)

! Core has higher index of refraction than cladding! Light rays incident at less than critical angle 1c is completely

reflected back into the core

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! Multimode: Thicker core, shorter reach! Rays on different modes interfere causing dispersion & limiting bit rate

! Single mode: Very thin core supports only one mode! More expensive lasers, but achieves very high speeds

Multimode fiber: multiple rays follow different paths

Single-mode fiber: only direct path propagates in fiber

Direct path

Reflected path

Multimode & Single-mode Fiber


Optical Fiber Properties

Advantages

! Very low attenuation

! Immunity to external e.minterference

! Extremely high bandwidth! Security: Very difficult to

tap without breaking

! No corrosion

! More compact & lighter thancopper wire

! Wideband optical amplifiersavailable (EDFA, SOA)

Disadvantages

! New types of optical signalimpairments & dispersion

! Shot noise

! Polarization dependence! Non linear effects

! Limited bend radius

! If physical arc of cable toohigh, light lost or wont reflect

! Will break

! Difficult to splice

! Mechanical vibration becomes

signal noise

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100

50

10

5

1

0.5

0.1

0.05

0.010.8 1.0 1.2 1.4 1.6 1.8

Wavelength (m)

Loss(dB/km)

Infrared absorption

Rayleigh scattering

Very Low Attenuation

850 nm

Low-cost

LEDs LANs

1300 nm

Metropolitan Area

Networks: Short Haul

1550 nm

Long Distance

Networks: Long Haul

Water Vapor Absorption

(removed in new fiber

designs)


100

50

10

5

1

0.5

0.1

0.8 1.0 1.2 1.4 1.6 1.8

Loss(dB/km)

Huge Available Bandwidth

! Optical range from 21to21+"2 contains bandwidth

! Example: 21= 1450 nm21+"2 =1650 nm:

B = !19 THz

B = f1 f2 = v

21+"2

v

21

v"221

2= !"2/21

1 + "2/21

v

21

2(108)m/s 200nm

(1450 nm)2

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Wavelength Division Multiplexing

! Different wavelengths carry separate signals

! Multiplex into shared optical fiber

! Each wavelength like a separate circuit

! A single fiber can carry 160 wavelengths, 10 Gbps perwavelength: 1.6 Tbps!

21

22

2moptical

mux

21

22

2moptical

demux

21 22.

2m

optical

fiber


Coarse & Dense WDM

Coarse WDM

! Few wavelengths 4-8with very wide spacing

! Low-cost, simpleDense WDM

! Many tightly-packedwavelengths

! ITU Grid: 0.8 nmseparation for 10Gbpssignals

!

0.4 nm for 2.5 Gbps

155

0

156

0

154

0

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193.1 THz

1552.52 nm

wavelength

196.1 THz

1528.77 nm

193.2 THz

1551.72 nm

193.0 THz

1553.33 nm

192.1 THz

1560.61 nm

100

GHz

100

GHz

198.51.4 THz

151010 nm

Supervisionchannel

ITU standard optical carrier grid


band Sband S band Cband C band Lband L

OAOA ErbiumErbium::SilicaSilica100100 GHzGHz

5050 GHzGHz

4040 40404040

80808080 8080

1440 162015801560154015201460 16001480

1440 162015801560154015201460 16001480

1440 162015801560154015201460 16001480

Carriers and optical bands

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! Attenuation

! Dispersion: chromatic (guide + material), polarization(PMD)

! Non linear effect: Brillouin, Raman, Kerr (Self-PhaseModulation (SPM), Cross-Phase Modulation (CPM),Four Wave Mixing (FWM))

! Launched power of 20&mW (13 dBm) in the cross section of amonomodal fiber (50 'm2) corresponds to 40 kW/cm2.

! Crosstalk in (D)WDM systems

! Noise of optical amplifiers (ASE)! Laser phase noise

Transmission impairments of o.f.


! Optical fiber response to an input pulse x(t) is the sumof two terms:! a linear dispersive term yL(t) = "x(()h(t()d(! an instantaneous cubic non linear term yNL(t) = Ax3(t)

! Let x(t) be the sum of three sinusoidal terms with

frequencies f1, f2, f3! Model for DWDM channels (extremely narrowband!!!)

! Besides linear terms, at the output we can findsinusoidal signals at frequencies f1f2f3! Part of these spurious harmonics falls within signal band

OAOA

Low dispersion fiberLow dispersion fiberinput output

Four Wave Mixing

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Power spectrum of an 8x10 Gbps DWDM signal with UnequalChannel Spacing after 50 km of G.653 fiber (2 mW/ch).

Effect of FWM


Standard fiber (G.652)

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Dispersion shifted fiber (G.653)


Non-zero dispersion fiber (G.655)

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Regenerators & Optical Amplifiers

! The maximum span of an optical signal is determined bythe available power & the attenuation:! If 30 dB attenuation are allowed, then at 1550 nm, optical

signal attenuates at 0.25 dB/km, so max span = 30 dB/0.25km/dB = 120 km

! Optical amplifiers increase optical signal power (noequalization, no regeneration): Pout=GPin+PASE.! Noise is added by each amplifier

! SNRout < SNRin, so there is a limit to the number of OAs in anoptical path

!

Optical signal must be regenerated when this limit isreached! Requires optical-to-electrical (O-to-E) signal conversion,

equalization, detection and retransmission (E-to-O)

! Expensive


Regenerator

R R R R R R R R

DWDM

multiplexer

R

R

R

R

R

R

R

R

R

R

R

R

R

R

R

R

DWDM & Regeneration

! Single signal per fiber requires 1 regenerator per span

! DWDM system carries many signals in one fiber

! At each span, a separate regenerator required per signal

! Very expensive

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R

R

R

R

Optical

amplifier

R

R

R

R

OA OA OA OA

Optical Amplifiers

! Optical amplifiers can amplify the composite DWDMsignal without demuxing or O-to-E conversion

! Erbium Doped Fiber Amplifiers (EDFAs) boost DWDMsignals within 1530 to 1620 range

! Spans between regeneration points >1000 km

! Number of regenerators can be reduced dramatically withdramatic reduction in cost of long-distance communications


Radio Transmission

! Radio signals: antenna transmits sinusoidal signal(carrier) that radiates in air/space

! Information embedded in carrier signal usingmodulation, e.g. QAM

! Communications without tethering! Cellular phones, satellite transmissions, Wireless LANs

! Multipath propagation causes fading; iintercation ofe.m. field with obstacles causes shadowing

! Interference from other users

! Spectrum regulated by national & internationalregulatory organizations

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104 106 107 108 109 1010 1011 1012

Frequency (Hz)

Wavelength (meters)

103 102 101 1 10-1 10-2 10-3

105

Satellite and terrestrial

microwave

AM radio

FM radio and TV

LF MF HF VHF UHF SHF EHF104

Cellular

and PCS

Wireless cable

Radio Spectrum

Omni-directional applications Point-to-Point applications


Source: Motorola

1 KHz 1 MHz 1 GHz

AudioFrequencies

Short-Wave Radio

AM Broad-cast

FM Broadcast

Television Telecommunications

Cellular Telephone,SMR, Packet Radio

PCS

ExtremelyLow

V

eryLow

Medium

Low

High

V

eryHigh

UltraHigh

Infrared

VisibleLight

U

ltraViolet

X-rays

Co

smicRays

GammaRays

Microwave

2.4 GHz ISM (Industrial Scientific, Medical)

Band

1 THz

Spectrum allocation

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Unlicensed Operation RF Bands! 902 MHz

! 26 MHz BW! Crowded

! Worldwide limited! 2.4 GHz

! 83.5 MHz BW! Available worldwide! IEEE802.11 WLANs

! 5.1 GHz! 300 MHz BW

discontinuous! Developing

902 928

North & South America902MHz

2400 2500 24802440

Americas, most of Europe

France

Spain

Japan

2.4GHz

*Frequency Allocations are pending

U-NII: Unlicensed National Information Infrastructure5100 5200 5300 5400 5500 5600 5700 5800 5900

U-NII

Europe

HiperLAN1

U-NII

Europe

HiperLAN2*

Japan*

5.1GHz


Examples

Cellular Phone! Allocated (licensed) spectrum! First generation:

! 800, 900 MHz! Initially analog voice

! Second generation:! 1800-1900 MHz! Digital voice, messaging

WirelessLAN! Unlicenced ISM spectrum

! Industrial, Scientific, Medical! 902-928 MHz, 2.400-2.4835

GHz, 5.725-5.850 GHz! IEEE 802.11 LAN standard

! 11-54 Mbps

Point-to-MultipointSystems! Directional antennas at

microwave frequencies! High-speed digital

communications between sites

! High-speed Internet AccessRadio backbone links for ruralareas

SatelliteCommunications! Geostationary satellite @

36000 km above equator! Relays microwave signals from

uplink frequency to downlinkfrequency

! Long distance telephone

! Satellite TV broadcast

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GSM: radio characteristics


Radio spectrum allocated to UMTS

! Frequency Division Duplex.

! 1920-1980 for Uplink.(12x5MHz)

! 2110-2170 for Downlink.(12x5Mhz)

! Time Division Duplex.

! 1900-1920 (4x5Mhz)

! 2010-2025 (3x5Mhz)

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Wireless Internet Service Provider

UMTS Operator

Node B

UTRANUTRANRNC

UMTS

IPv4 Core Network

GGSN

SGSN

Node B

IPv4 over WAN

Wireless LANs IEEE 802.11Wireless LANs IEEE 802.11

Server farm and IMS WWW

E-mail.,

Diameter

SIP proxy

Dynamic DNS

.

,

Signaling

Gateway

Profiles

DBs

IPv4 backbone

UMTS: (3GPP)

MM:Ad-Hoc ProtocolsAA:Ad-Hoc Protocols

WLAN: (IETF)

MM: Mobile IP, Cellular IP, etc

AA: User Name/Pwd

Current 3G scenario


! Applications! Local wireless networking (infrastructured, ad hoc)

! Public hot spots

! Home Networking

! Sensor networks

WLAN and WPAN standards

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Error Detection and Correction

Fundamentals of

communications




Error Control

! Digital transmission systems introduce errors

! Applications require certain reliability level

! Data applications require error-free transfer

! Voice & video applications tolerate some errors, the less themore source coding removes redundancy from original signal

! Error control used when transmission system does notmeet application requirement

! Two basic approaches:

Error detection& retransmission (ARQ)

Forward error correction(FEC)

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Key Idea

! All transmitted data blocks (codewords) satisfy apattern

! If received block doesnt satisfy pattern, it is in error! If it satisfies pattern, it is assumedto be correct

! Redundancy: Only a subset of all possible blocks can becodewords

! Example: spell checking by taking dictionary words with sameinitial as letter to be spelled out

ChannelEncoderUser

information

Pattern

checking

All inputs to channel

satisfy pattern or condition

Channel

output

Deliver user

information or

set error alarm


Single Parity Check! Given a data block of kbits, add one more bit so as to

make the number of 1s even! Same as making 0 the xor of the coded (k+1)-bit block

Info Bits: b1, b2, b3, , bk

Check Bit: bk+1 = b13b23b33 3bkCodeword: (b1, b2, b3, , bk,, bk+1)

! Receiver checks to see if # of 1s is even! All error patterns that change an odd # of bits are detectable;

all even-numbered patterns are undetectable

! Parity bit used in ASCII code

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Example of Single Parity Code

! Information (7 bits): (0, 1, 0, 1, 1, 0, 0)

! Parity Bit: b8

= 0 + 1 +0 + 1 +1 + 0 = 1

! Codeword (8 bits): (0, 1, 0, 1, 1, 0, 0, 1)

! If single error in bit 3 : (0, 1, 1, 1, 1, 0, 0, 1)! # of 1s =5, odd

! Error detected

! If errors in bits 3 and 5: (0, 1, 1, 1, 0, 0, 0, 1)! # of 1s =4, even! Error not detected


Checkbits & Error Detection

Information

accepted if

check bits match

Calculate

check bitsChannel

Recalculate

check bitsCompare

Information bits

kbitappend

nkbit

n bit

Sent codeword

n bit

Received codeword

Received

info bits

Received

check bits

Systematic code concept

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How good is the single parity checkcode?

! Redundancy: Single parity check code adds 1 redundant

bit per kinformation bits: overhead = 1/(k+1)! Coverage: all error patterns with odd # of errors can

be detected

! An error patten is a binary (k+ 1)-tuple with 1s where errorsoccur and 0s elsewhere

! Of 2k+1 binary (k+1)-tuples, 1/2 have odd weight, so 50% of errorpatterns can be detected

! Is it possible to detect more errors if we add more

check bits?! Yes, with the right codes


What if bit errors are random?

! Many transmission channels introduce bit errors atrandom, independently of each other, with probability p

! Some error patterns are more probable than others:

! In any worthwhile channel p

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Single parity check code withrandom bit errors

! Undetectable error pattern if even # of bit errors:

! Example: Evaluate above for n= 32, p= 103

! For this example, roughly 1 in 2000 error patterns isundetectable

P[error detection failure] = P[undetectable error pattern]= P[error patterns with even number of 1s]

= p2(1 p)n-2 + p4(1 p)n-4 + n

2

n

4

P[undetectable error] = (103)2 (1 103)30 + (103)4 (1 103)28

$ 496 (106) + 35960 (1012) $ 4.96 (104)

32

232

4


Two-Dimensional Parity Check

1 0 0 1 0 0

0 1 0 0 0 1

1 0 0 1 0 0

1 1 0 1 1 0

1 0 0 1 1 1

Bottom row consists of

check bit for each column

Last column consists

of check bits for each

row

! More parity bits toimprove coverage

! Arrange information ascolumns! Add single parity bit to

each column

! Add a parity column

! Used in early errorcontrol systems

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1 0 0 1 0 0

0 0 0 1 0 1

1 0 0 1 0 0

1 0 0 0 1 0

1 0 0 1 1 1

1 0 0 1 0 0

0 0 0 0 0 1

1 0 0 1 0 0

1 0 0 1 1 0

1 0 0 1 1 1

1 0 0 1 0 0

0 0 0 1 0 1

1 0 0 1 0 0

1 0 0 1 1 0

1 0 0 1 1 1

1 0 0 1 0 0

0 0 0 0 0 11 0 0 1 0 0

1 1 0 1 1 0

1 0 0 1 1 1

Arrows indicate failed check bits

Two

errorsOne error

Three

errors Four errors(undetectable)

Error-detecting capability

1, 2, or 3 errors

can always be

detected; Not all

patterns >4 errors

can be detected


x = codewords

o = noncodewords

x

x x

x

x

x

x

o

oo

oo

oo

ooo

o

o

o

x

x x

x

xx

x

oo

oo

oooo

o

o

oPoor

distance

properties

What is a good code?

! Many channels havepreference for errorpatterns that have fewer# of errors

! These error patterns maptransmitted codeword tonearby n-tuple

! If codewords close toeach other then detectionfailures will occur

! Good codes shouldmaximize separation

between codewords

Good

distance

properties

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Other Error Detection Codes

! Many applications require very low error rate!

Single parity check codes do not detect enougherrors

! Two-dimensional codes require too many check bits

! The following error detecting codes are usedin practice:

! Internet Check Sums

! CRC Polynomial Codes


Internet Checksum

! Several Internet protocols (e.g. IP, TCP, UDP) usecheck bits to detect errors in the IP header(or in theheader and data for TCP/UDP)

! A checksum is calculated for header/segment contents and

included in a special field.! Checksum recalculated at every router, so algorithm selected

for ease of implementation in software

! Let header consist of L, 16-bit words,

b0, b1, b2, ..., bL1

! The algorithm appends a 16-bit checksum bL

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The checksum bL is calculated as follows:

! Treating each 16-bit word as an integer, findx = b0 + b1 + b2 + ... + bL1 modulo 2161

! The checksum is then given by:

bL= x modulo 2161

Thus, the headers must satisfy the followingpattern:

0 = b0 + b1 + b2+ ...+ bL1 + bL modulo 2161

! The checksum calculation can be carried out insoftware at speed compatible with router operations

Checksum Calculation


Internet Checksum Example

Use Modulo Arithmetic

! Assume 4-bit words

! Use mod 24-1arithmetic

! b0=1100 = 12! b1=1010 = 10

! b0+b1=12+10=7 mod15

! b2 = -7 = 8 mod15

! Therefore

! b2=1000

Use Binary Arithmetic

! Note 16=1 mod15

! So: 10000 = 0001 mod15

! leading bit wraps around

b0+b1 =1100+1010

=10110

=10000+0110

=0001+0110

=0111 (=7)

Take 1-complement

b2 = 0111 =1000

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Polynomial Codes

! Binary codewords can be associated to polynomials

! Coefficients of polynominal are orderly equal to bits of thecodeword, e.g. MSB being the coefficient of the leading powerof the polynomial

! Polynomial arithmetic instead of check sums

! Implemented using shift-register circuits

! Also called cyclic redundancy check (CRC)codes

! Most data communications standards use polynomial

codes for error detection! Polynomial codes also basis for powerful error-correction

methods


Addition:

Multiplication:

Binary Polynomial Arithmetic

! Binary vectors map to polynomials

(bk-1 ,bk-2 ,, b2 , b1 , b0) #

bk-1xk-1 + bk-2x

k-2 + + b2x2 + b1x+ b0

(x7 +x6 + 1) + (x6 + x5) =x7 +x6 + x6 +x5 + 1

=x7 +(1+1)x6+x5 + 1

=x7 +x5 + 1 since 1+1=0 mod2

(x+ 1) (x2 + x + 1) =x(x2 +x + 1) + 1(x2 +x+ 1)

=x3

+x2

+ x) + (x2

+x

+ 1)=x3 + 1

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Binary Polynomial Division

! Division with Decimal Numbers

32

35 | 1222

3

105

17 2

4

140divisor

quotient

remainder

dividend1222 = 34 x 35 + 32

dividend = quotient x divisor + remainder

! Polynomial Division x3 +x+ 1 |x6 +x5x6 + x4 +x3

x5 +x4 +x3

x5 + x3 +x2

x4 + x2

x4 + x2 +x

x

= q(x) quotient

= r(x) remainder

divisordividend

+x+x2x3

Note: Degree of r(x) is less

than degree of divisor


Polynomial Coding

! Code has binarygenerating polynomialof degree nk

! k information bitsdefine polynomial of degree (k1

! Find remainder polynomialof at most degree nk1

g(x) |xn-k i(x)

q(x)

r(x)

xn-ki(x) = q(x)g(x) + r(x)

! Define the codeword polynomialof degree (n1

b(x) =xn-k

i(x) + r(x)n bits kbits n-kbits

g(x) =xn-k+ gn-k-1xn-k-1 + + g2x

2 + g1x+ 1

i(x) = ik-1xk-1

+ ik-2xk-2

+ + i2x2

+ i1x+ i0

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Transmitted codeword: b(x) = x6+ x5+ x b = (1,1,0,0,0,1,0)

1011 | 1100000

1110

1011

1110

1011

1010

1011

010

x3 + x+ 1 | x6+ x5

x3 + x2+ x

x6+ x4 + x3

x5+ x4 + x3

x5+ x3 + x2

x4 + x2

x4 + x2+ x

x

Polynomial example: k=4, nk=3

Generator polynomial: g(x) = x3 + x+ 1

Information: (1,1,0,0): i(x) = x3 + x2

Encoding: x3i(x) = x6 + x5


Calcolo con shift register

G(x) = x3+ x+1

Reg 0 ++

g3= 1

M(x)

g0= 1 g1 =1

M(x)= x3+ x

Encoder for

Reg 1 Reg 2

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The Patternin Polynomial Coding

! All codewords satisfy the following pattern:

! All codewords are a multiple ofg(x)!

! Receiver should divide received n-tuple byg(x)and check if remainder is zero

! If remainder is nonzero, then received n-tuple

is not a codeword

b(x) =xn-ki(x) + r(x) = q(x)g(x) + r(x) + r(x) = q(x)g(x)


Undetectable error patterns

! e(x)has 1s in error locations & 0s elsewhere! Receiver divides the received polynomial R(x)byg(x)! Blindspot: If e(x) is a multiple ofg(x), that is, e(x) is

a nonzero codeword, then R(x) = b(x) + e(x) = q(x)g(x) + q(x)g(x)! The set of undetectable error polynomials is the set

of nonzero code polynomials!

Choose the generator polynomial so that mostcommon error patterns can be detected.

b(x)

e(x)

R(x)=b(x)+e(x)+

(Receiver)(Transmitter)

Error polynomial(Channel)

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Designing good polynomial codes

! Select generator polynomial so that likely errorpatterns are not multiples ofg(x)

! Detecting Single Errors

! e(x) = xi for error in location i+1

! If g(x)has more than 1 term, it cannot divide xi

! Detecting Double Errors

! e(x) = xi + xj= xi(xj-i+1) where j> i

! If g(x)has more than 1 term, it cannot divide xi

! Ifg(x) is a primitivepolynomial, it cannot divide xm+1 for allm

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Standard Generator Polynomials

! CRC-8:

! CRC-16:

! CCITT-16:

! CCITT-32:

CRC = cyclic redundancy check

HDLC, XMODEM, V.41

IEEE 802, DoD, V.42

Bisync

ATM= x8+ x2+ x + 1

= x16+ x15+ x2+ 1

= (x+ 1)(x15+ x + 1)

= x16+ x12+ x5+ 1

= x32+ x26+ x23 +x22+ x16+ x12+ x11

+ x10+ x8+x7+ x5+ x4 + x2+ x+ 1


Hamming Codes

! Class of error-correctingcodes

! Capable of correcting all single-errorpatterns

! For each m> 2, there is a Hamming code of lengthn= 2m 1 with n k= mparity check bits

57

26

11

4

k= nm

6/63636

5/31315

4/15154

3/773

m/nn= 2m1m

Redundancy

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m= 3 Hamming Code

! Information bits are b1, b2, b3, b4! Equations for parity checks b

5, b

6, b

7

! There are 24 = 16 codewords

! (0,0,0,0,0,0,0) is a codeword

b5 = b1 + b3 + b4

b6 = b1 + b2 + b4

b7 = + b2 + b3 + b4


Hamming (7,4) code

71 1 1 1 1 1 11 1 1 1

31 1 1 0 0 0 01 1 1 041 1 0 1 0 1 01 1 0 1

41 1 0 0 1 0 11 1 0 0

41 0 1 1 1 0 01 0 1 1

41 0 1 0 0 1 11 0 1 0

31 0 0 1 0 0 11 0 0 1

31 0 0 0 1 1 01 0 0 0

40 1 1 1 0 0 10 1 1 1

40 1 1 0 1 1 00 1 1 0

30 1 0 1 1 0 00 1 0 1

30 1 0 0 0 1 10 1 0 0

30 0 1 1 0 1 00 0 1 1

30 0 1 0 1 0 10 0 1 0

40 0 0 1 1 1 10 0 0 1

00 0 0 0 0 0 00 0 0 0

w(b)b1 b2 b3 b4 b5 b6 b7b1 b2 b3 b4

WeightCodewordInformation

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Parity Check Equations

! Rearrange parity check equations:

! All codewords must

satisfy these equations! Note: each nonzero 3-

tuple appears once as acolumn in check matrix H

! In matrix form:

0 = b5 + b5 = b1 + b3 + b4 + b50 = b6 + b6 = b1 + b2 + b4 + b6

0 = b7 + b7 = + b2 + b3 + b4 + b7

b1

b2

0 = 1 0 1 1 1 0 0 b3

0 = 1 1 0 1 0 1 0 b4 = Hbt

= 00 = 0 1 1 1 0 0 1 b5

b6

b7


0

0

1

0

0

0

0

s = H e= =1

0

1

Single error detected

0

1

0

0

1

0

0

s = H e= = + =0

1

1

Double error detected1

0

0

1 0 1 1 1 0 01 1 0 1 0 1 0

0 1 1 1 0 0 1

1

1

1

0

0

0

0

s = H e= = + + = 0110

Triple error notdetected

01

1

10

1

1 0 1 1 1 0 0

1 1 0 1 0 1 0

0 1 1 1 0 0 1

1 0 1 1 1 0 0

1 1 0 1 0 1 0

0 1 1 1 0 0 1

1

1

1

Error Detection with Hamming Code

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Minimum distance of Hamming Code

! With Hamming (7,4) code undetectable error patternmust have 3 or more bits, i.e. at least 3 bits must be

changed to convert one codeword into another codeword

b1 b2o o

o

o

oo

o

o

Set ofn-tuples

within distance

1 of b1

Set ofn-tuples

within distance

1 of b2

! Spheres of distan

TLC p02 FundComm 2011

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