Telecomunicazioni About bit and numbers É - uniroma1.itnetlab.uniroma1.it/netgroup/sites/default/files/TLC_p03_FundComm... · FONDAMENTI DI COMUNICAZIONI 4. ... Telecomunicazioni

TelecomunicazioniDocente: Andrea Baiocchi

DIET - Stanza 107, 1° piano palazzina “P. Piga”

Via Eudossiana 18

E-mail: [email protected]

University of Roma“Sapienza”

Corso di Laurea in Ingegneria Gestionale

A.A. 2014/2015

Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea Baiocchi

About bit and numbers…

There are 10 types of people.Those who understand binary numbers and thosethat do not.

[anonymous joke]

But let your communication be Yea, yea; Nay, nay.For whatsoever is more than these, cometh of evil.

[Matthew, 5:37]


Programma

1. SERVIZI E RETI DI TELECOMUNICAZIONE

2. ARCHITETTURE DI COMUNICAZIONE EMODI DI TRASFERIMENTO

3. FONDAMENTI DI COMUNICAZIONI

4. ACCESSO MULTIPLO

5. LO STRATO DI COLLEGAMENTO

6. LO STRATO DI RETE IN INTERNET

7. LO STRATO DI TRASPORTO IN INTERNET


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

Fundamentals ofcommunications

Digital Representation ofInformation

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

networks”, McGraw Hill, 2004


Bits, numbers, information

! Bit: BInary digiT! Either symbol belonging to a set of two elements or

number with value 0 or 1! n bits: digital representation for 0, 1, … , 2n–1! Byte or Octet, n = 8! Computer word, typically n = 32, or 64

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

! 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, delimited data 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


Information and …

! Information is a logical concept! Made precise by Shannon theory: more later on

! Information can appear under two basic forms! A SEQUENCE of values taken from a discrete and

finite set (DIGITAL INFORMATION)

! A REAL-VALUED FUNCTION of one or more REAL-VALUED variables (ANALOG INFORMATION)

! Examples! A book, a file, a digitally encoded speech/photo/movie

! Voice in the air, images as captured by camera sensors


… signals

! 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 real axis and taking values in an interval ofthe real line

! Examples! Voice signal: pressure(t)! Static image signal: luminance(x,y)! Moving picture: luminance(x,y,t)


Examples of analog information sources

! Images and sounds come natively in the formon analog information carried by analog signals! Images: e.m. field in the visible bandwidth, i.e., the

range of frequencies that produces a reaction inhuman sight sensors.

! Sounds: variation of air pressure that produces areaction in human hearing sensors.

! Why analog?! for our purposes all these phenomena are well

described by classical physics models (Newtonmechanics, Maxwell e.m. field theory)

! classical physics rests on classical analysis todescribe its models (variables taking values on acontinuum, namely the real axis).


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)

Voice

! Voice is carried by a signal (air pressure orcurrent induced in a microphone by airpressure)

! Analog voice signal level varies continuously intime


H

W

= + +H

W

H

W

H

W

Color

image

Red

component

image

Green

component

image

Blue

component

image

Color image

! A color image can be described by the linearcombination of three basic e.m. signals withgiven frequencies in the visible range! Example: intensity of Red, Green, Blue

I(x,y) = r(x,y) RED+g(x,y) GREEN+b(x,y) BLUE(0!x!W; 0!y!H)


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/color

12.8 megapixels ! 3 bytes/pixel = 38.4 megabytes

Color image


Moving picture

! Sequence of picture frames! Each picture digitized &

compressed

! Frame repetition rate! 10-30-60 frames/second

depending on quality

! Frame resolution! Small frames for

videoconferencing

! Standard frames forconventional broadcast TV

! HDTV frames

30 fps

Rate = M bits/pixel x (WxH) pixels/frame x F frames/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


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


Transmission Delay on a link

Use data compression to reduce LUse higher speed modem to increase RPlace far end system closer to reduce d

L number of bits in messageR bps speed of digital communication systemL/R time to transmit the messagetprop time for signal to propagate across mediumd distance in metersc speed of light (3x108 m/s in vacuum)

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


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

Digital Representation of AnalogSignals





Digitization of Analog Signals

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

xk=x(tk), tk=t0+kT, k integerT is the sampling time, F=1/T is the samplingrate.

" Quantization: map each sample xk into anapproximation value yk=f(xk) of finiteprecision

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

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


Sampling Rate and Bandwidth

! A signal that varies faster needs to besampled more frequently

! Bandwidth measures 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 T can 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|, …


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 W is 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 6 912

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 6 912

15

18

21

24

27

30

33

36

39

42

frequency (kHz)

Spectrum of x1(t)

Spectrum of x2(t)


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

filtert

x(t)

t

x(nT)

(b)

Samplert

x(t)

t

x(nT)(a)


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

3 b

its /

sa

mp

le


Quantization error:

“noise” = y(nT) – x(nT)

Quantizer maps input

into closest of 2m

representation values

#/2

3#/2

5#/2

7#/2

-#/2

-3#/2

-5#/2

-7#/2

Original signal

Sample value

Approximation

3 b

its /

sa

mp

le

Quantization of Analog Samples

input x(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 M is large, the error e(x)=y(x)–x isapproximately 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)


Quantizer performance

! Power of quantization error signal = average ofsquared 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/#e

2) = 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


Digital Transmission of AnalogInformation

Interpolationfilter

Displayor

playout

2W samples / sec

2W m bits/secx(t)

Bandwidth W

Sampling(A/D)

QuantizationAnalogsource

2W samples / 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 dB

Assume 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

Why Digital Communications?





A Transmission System

Transmitter! Converts information into signal suitable 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


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 or

negative?


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


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


Many wavelengths>1600 GbpsOptical fiber

1 wavelength2.5-10 GbpsOptical fiber

5 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 kbpsTelephonetwisted pair

ObservationsBit RateSystem

Bit rates of digital transmission systems

Characterization ofCommunication Channels





Communications Channels

! A physical medium is an inherent part of acommunications 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 channel we 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


How good is a channel?

! Performance: What is the maximum reliabletransmission speed?! Speed: Bit rate, R bps

! Reliability: Bit error rate, BER=10–k

! 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

SignalReceived

Signal

ReceiverCommunication channelTransmitter


Communication channel model

! We often assume two basic properties ofchannels

! 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

t t

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

Aout

Ain

A(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


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<Wc arepassed without attenuation and delayed by ( seconds;sinusoids at other frequencies are blocked

Amplitude Response

f

1

f0

'(f) = –2"f(

1/ 2"

Phase Response

Wc

y(t) = Aincos(2"ft – 2"f() = Aincos(2"f(t – ( )) = x(t–()


Example: Low-Pass Filter

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

H(f)=1/(1+i2"bf)! Example: RC circuit, b=R·C.

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 bandwidth is 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 =f2–f1

f1 f2


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 frequencycomponents 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) = )k A(fk) ak cos(2"fkt + *k + '(fk ))

Channely(t)x(t) = )k akcos(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


-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

1.50

0.1

25

0.2

5

0.3

75

0.5

0.6

25

0.7

5

0.8

75 1

Amplitude Distortion

! As the channelbandwidthincreases, 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 a to 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) = a·h(t–(), t > (

Channel

t0t

h(t)

td

a

+(t)

0


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<0

Time invariance +(t–T) h(t–T)

Linearity a·+(t–Ta)+b·+(t–Tb) a·h(t–Ta)+b·h(t–Tb)

! If a signal x(t) is applied to LTI channel input, then

y(t) = # x(()h(t–()d(! x(t) can be thought of as the sum of pulses at times ( (–$< (<$)

with amplitude x((), so y(t) is the sum of responses x(()h(t–()

! It can be shown that H(f) is the Fourier transform ofh(t) and thatY(f) = H(f)X(f)


Linear, stationary channels andpulse transmission

! Time invariance

! Linearity

! In case of pulse transmission in a linear, timeinvariant, additive noise channel, we get anoutput signal y(t):

y(t) = !k xk h(t–kT) + z(t)

Input Output

xk·+(t–kT) xk·h(t–kT)

#k xk·+(t–kT) #k xk·h(t–kT)

noise


InterSymbol Interference (ISI)

! By sampling channel output at time nT (perfectsync) we get

yn = y(nT) = %k xk hn–k + zn

with hn–k = h(nT–kT) and zn = z(nT).

yn = xnh0 + %k& n xkhn–k + 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 HCH(f), we can addreception and possibly transmission filters, sothat the overall (filtered) channel response is

H(f) = HTX(f) HCH(f) HRX(f)

! To get null ISI at sampling rate 1/T it must be

%k H(f–k/T) = cost

! For a low-pass channel with H(f)=0 for |f|>Wcthis is not possible unless 1/T <= 2Wc.

Max symb rate for zero ISI = 2 x channel bandwidth


! For channel with ideal low-pass amplitude response ofbandwidth Wc, the impulse response is a Nyquist pulseh(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 T seconds 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 TT

-1

0

1

-2 -1 0 1 2 3 4

tT T T T TT

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)


0f

A(f)

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

rate pulses can be transmitted without ISI is 2Wc pulse/s

! s(t) is one example of class of Nyquist pulses with zero ISI! Problem: sidelobes in s(t) decay as 1/t which 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

Fundamental Limits in DigitalTransmission





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 intersymbolinterference (ISI) when sampled at multiples ofT


Multilevel Signaling! Nyquist pulses achieve the maximum signalling rate with

zero ISI! 2Wc pulses per second or 2Wc pulses / Wc Hz = 2 pulses / Hz

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

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

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

BUTAdditive noise limits # of levels that can be used reliably.


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


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 amplitudelevels 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

Decision thresholds


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 ofscattered and similar sources can be modeledas white Gaussian “noise”


22

2

2

1 !

!"

xe#

x0

Gaussian noise

t

x

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

x0

x0

%2 = Avg Noise Power


Probability of Error! Error occurs if noise value exceeds the

information signal magnitude over the decisionthreshold

! With two-level signalling, +A and –A, probabilityof error is Q(A/%)

1.00E-121.00E-11

1.00E-101.00E-091.00E-08

1.00E-071.00E-061.00E-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+(2k–1)A/M, with k=1,…,M.

! With equiprobable symbols:E[ak

2]=(M2–1)A2/(3M2)=PTX

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


signal noise signal + noise

signal noise signal + noise

High

SNR

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 reliable transmission rate over an AWGN

bandlimited channel with bandwidth Wc Hz is

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

Shannon Channel Capacity

X

Input symbol

(Gaussian)

Y=X+Z

Output symbol

Z

White gaussian noise

2Wc symbols/s

Bandlimited channel (Wc)


Capacity of AWGN channel

! It can be shown that in case of real inputsignal the optimal source is gaussian and theAWGN capacity is

C = 0.5 log2(1+PR/PN) [bit/symbol]

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

! A dispersive, additive noise channel can bereduced to AWGN if zero ISI is provided; tothat end it must be symbol rate < 2Wc. Then

Cb = Wclog2(1+(Eb/N0)Rb/Wc) [bit/s]


! Reliable communications is possible if the tx rate Rb<Cb.

! If 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 & SNR determine Cb

! Cb can be used as a measure of how close a system design is tothe best achievable performance.

! SNR=P/(N0Wc), with

! P= average power of input signal

! N0=noise power spectral density=k-F, k=1.38·10–23 J/K, -=noisetemperature, typically 300 K, F=noise figure, typically 6 dB

Shannon Channel Capacity


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

Modems and Digital Modulation




Telecomunicazioni - a.a. 2014/2015 - Prof. Andrea BaiocchiTanenbaum, Wetherall, Reti di calcolatori © Pearson 201283

Codifiche di linea: (a) Bit, (b) NRZ, (c)NRZI, (d) Manchester, (e) Bipolare o AMI.


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 fc centered in middle of

bandpass channel is named carrier

! Modulators embed information into the carrier

fc – Wc/2 fc0 fc + Wc/2


(a) Un segnale binario. (b) Modulazione diampiezza. (c) Modulazione di frequenza.(d) Modulazione di fase.


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 + + or fc - +


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

Phase

Shift

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

Information 1 1 1 10 0


Modulate cos(2"fct) by multiplying by Ak for T seconds:

Akx

cos(2"fct)

Y(t) = Ak cos(2"fct), kT<t<(k+1)T

Transmitted signal

during kth interval

Modulator


Demodulator

Demodulate (recover Ak) by multiplying by 2cos(2"fct)

for T seconds and lowpass filtering (smoothing):

x

2cos(2"fct)2Ak cos2(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)


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

-A0 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 signal x(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 Wc Hz,! 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


Akx

cos(2"fct)

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

Bkx

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! Ak modulates 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)2Ak cos2(2!fct)+2Bk cos(2!fct)sin(2!fct)

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

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

Lowpass

filter

(smoother)

Ak

2Bk sin2(2!fct)+2Ak cos(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


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 mixedamplitude and phase modulation in the general case(QAM)


Signal Constellations

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

! Signal constellation set of signaling points

4 possible points per T s

2 bits / pulse

Ak

Bk


4 bits / pulse

Ak

Bk

(A, A)

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

(-A,A)


Ak

Bk


Ak

Bk


Other Signal Constellations

! Point selected by amplitude & phase


(a) QPSK. (b) QAM-16. (c) QAM-64


Codifica di Gray per QAM-16


Signal constellations and errors

Properties of Media and DigitalTransmission Systems





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


d meters


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


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 1024

Frequency (Hz)

Wavelength (meters)

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

Pow

er

and

tele

phone

Bro

adcast

radio

Mic

row

ave

radio

Infr

are

d lig

ht

Vis

ible

lig

ht

Ultra

vio

let lig

ht

X-r

ays

Gam

ma r

ays


Attenuation

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

! Wired media has exponential dependence! Received power at d meters proportional to 10–0d

! 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 n is path lossexponent; n=2 in free space


(a) Nelle bande VLF, LF ed MF le onde radioseguono la curvatura del pianeta.(b) Nella banda HF rimbalzano contro laionosfera.


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

Extremely Low

Ve

ry L

ow

Me

diu

m

Lo

w

Hig

h

Ve

ry H

igh

Ultra

Hig

h

Infrared

Vis

ible

Lig

ht

Ultra

Vio

let

X-rays

Co

sm

ic R

ays

Ga

mm

a R

ays

Microwave

2.4 GHz ISM (Industrial Scientific, Medical)

Band

1 THz

Spectrum allocation


Bande ISM e U-NII utilizzate daidispositivi wireless negli Stati Uniti.


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)


Examples

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

! 800, 900 MHz! Initially analog voice

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

Wireless LAN! 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-Multipoint Systems! Directional antennas at

microwave frequencies! High-speed digital

communications between sites! High-speed Internet Access

Radio backbone links for ruralareas

Satellite Communications! Geostationary satellite @

36000 km above equator! Relays microwave signals from

uplink frequency to downlinkfrequency

! Long distance telephone! Satellite TV broadcast


Wireless Internet Service Provider

UMTS Operator

Node B

•• UTRAN UTRAN

RNC

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 Protocols

AA: Ad-Hoc Protocols

WLAN: (IETF)

MM: Mobile IP, Cellular IP, etc…

AA: User Name/Pwd

Current 3G scenario


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)

Att

en

ua

tio

n (

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

51.840Mbps

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

1.544Mbps

T-1

DistanceDataRateStandard


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 minimize

interference; costly! 10BASE-T Ethernet

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

! 100BASE-T4 Fast Ethernet! 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 outer conductorsurrounds insulated innerwire conductor

! 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

Att

en

ua

tio

n (

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


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


(a) Vista laterale di una singolafibra.(b) Vista frontale di una guaina contre fibre.


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


! 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 totap 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 won’t reflect

! Will break

! Difficult to splice

! Mechanical vibration becomessignal noise


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 (µm)

Lo

ss (

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

Lo

ss (

dB

/km

)

Huge Available Bandwidth

! Optical range from 21 to21+#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


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

2m

optical

mux

21

22

2m

optical

demux

21 22. 2m

optical

fiber


Coarse & Dense WDM

Coarse WDM

! Few wavelengths 4-8with very wide spacing

! Low-cost, simple

Dense WDM

! Many tightly-packedwavelengths

! ITU Grid: 0.8 nmseparation for 10Gbpssignals

! 0.4 nm for 2.5 Gbps

15

50

15

60

15

40


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… …

198.5±1.4 THz

1510±10 nm

Supervision

channel

ITU standard optical carrier grid


band Sband S band Cband C band Lband L

OA OA ErbiumErbium::SilicaSilica100 100 GHzGHz

50 50 GHzGHz

4040 40404040

80808080 8080

1440 162015801560154015201460 16001480

1440 162015801560154015201460 16001480

1440 162015801560154015201460 16001480

Carriers and optical bands


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

! Besides linear terms, at the output we can findsinusoidal signals at frequencies ±f1±f2±f3! Part of these spurious harmonics falls within signal band

OAOA

Low dispersion fiberLow dispersion fiberinput output

Four Wave Mixing


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)


Dispersion shifted fiber (G.653)


Non-zero dispersion fiber (G.655)


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=G•Pin+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


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

Access networks



Ampiezza di banda e distanza perUTP di categoria 3 usati da DSL.


ADSL che sfrutta DMT


Una configurazione ADSL tipica


FttH con fibra ottica passiva


(a) Le frequenze non sono riutilizzate nelle celle adiacenti.(b) Per aumentare il numero di utenti si possono utilizzarecelle più piccole.


Architettura di rete GSM

Error Detection and Correction





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)


Key Idea

! All transmitted data blocks (“codewords”) satisfy apattern

! If received block doesn’t satisfy pattern, it is in error

! If it satisfies pattern, it is assumed to be correct

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

! Example: in the set of all possible strings made of alphabetcharacters of length less or equal to 20 chars (to make the setfinite), current English dictionary words are “codewords”

ChannelEncoderUser

information

Pattern

checking

All inputs to channel

satisfy pattern or condition

Channel

output

Deliver user

information or

set error alarm


Codewords communication

All n-bit strings

2n

2k

n-bit strings thatare codewords

All n-bit strings

2n

2k

TX RX

Transfer of a givencodeword, received

as a corruptedstring


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

make the number of 1’s 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 …3bk

Codeword: (b1, b2, b3, …, bk,, bk+1)

! Receiver checks to see if # of 1’s 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


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 1’s =5, odd

! Error detected

! If errors in bits 3 and 5: (0, 1, 1, 1, 0, 0, 0, 1)! # of 1’s =4, even

! Error not detected


How good is the single parity checkcode?

! Redundancy: Single parity check code adds 1 redundantbit per k information bits: overhead = 1/(k+1)

! Coverage: all error patterns with odd # of errors canbe detected

! An error patten is a binary (k + 1)-tuple with 1s where errorsoccur and 0’s 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 morecheck 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 < 0.5, and so p/(1–p) < 1! It follows that patterns with 1 error are more likely than

patterns with 2 errors and so forth

! What is the probability that an undetectable errorpattern occurs?

P[10000000] = p(1 – p)7 = (1 – p)8 and

P[11000000] = p2(1 – p)6 = (1 – p)8

p

1 – p p 2

1 – p


Single parity check code withrandom bit errors

! Undetectable error pattern if even # of bit errors:

! Example: Evaluate above for n = 32, p = 10–3

! 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] = (10–3)2 (1 – 10–3)30 + (10–3)4 (1 – 10–3)28

$ 496 (10–6) + 35960 (10–12) $ 4.96 (10–4)

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


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 1

1 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


Checkbits & Error Detection

Information

accepted if

check bits match

Calculate

check bitsChannel

Recalculate

check bitsCompare

Information bits

k bitappend

n–k bit

n bit

Sent codeword

n bit

Received codeword

Received

info bits

Received

check bits

Systematic code concept


x = codewords

o = noncodewords

x

x x

x

x

x

x

o

oo

oo

oo

o

oo

o

o

o

x

x x

x

xx

x

oo

oo

oooo

o

o

oPoor

distance

properties

What is a good code?

! Error patterns with fewererrors are most probableones in most channels

! These error patterns maptransmitted codeword tonearby n-tuple

! If codewords close toeach other then detectionfailures will occur

! Good codes shouldmaximize separationbetween codewords

Good

distance

properties


Hamming distance

! Let w(x) be the number of 1’s in binary vectorx (weight of x)

! We define dH(x,y) = w(x3y), where 3 is themod2 sum (ex-or)! This a distance indeed

! dH(x,x) = w(x3x) = w(0) = 0

! It is obviously symmetric

! dH(x,y)+dH(y,z) ) dH(x,z) is a consequence of the obviousinequality w(x3y) ! w(x)+w(y) and x3y3y3z=x3z.

! Let c be the transmitted codeword and r=c3ethe received binary vector. To detect errors itmust be dH(r,c’) > 0 for any codeword c’.


Other Error Detection Codes

! Many applications require very low error rate! Single parity check codes do not detect enough

errors

! 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 andincluded in a special field.

! Checksum recalculated at every router, so algorithm selectedfor ease of implementation in software

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

b0, b1, b2, ..., bL–1

! The algorithm appends a 16-bit checksum bL


The checksum bL is calculated as follows:

! Treating each 16-bit word as an integer, find

x = b0 + b1 + b2 + ... + bL–1 modulo 216–1

! The checksum is then given by:

bL = –x modulo 216–1

Thus, the headers must satisfy the following pattern:

0 = b0 + b1 + b2 + ... + bL–1 + bL modulo 216–1

! 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 24-1 arithmetic

! 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


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 polynomialcodes for error detection

! Polynomial codes also basis for powerful error-correctionmethods


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


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 + x5

x6 + 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 binary generating polynomial of degree n–k

! k information bits define polynomial of degree !k–1

! Find remainder polynomial of at most degree n–k–1

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

q(x)

r(x)

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

! Define the codeword polynomial of degree !n–1

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

n bits k bits n-k bits

g(x) = xn-k + gn-k-1xn-k-1 + … + g2x

2 + g1x + 1

i(x) = ik-1xk-1 + ik-2x

k-2 + … + i2x2 + i1x + i0


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, n–k=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 g

1=1

!

M(x) = x3+ x

Encoder for

Reg 1 Reg 2


The Pattern in Polynomial Coding

! All codewords satisfy the following pattern:

! All codewords are a multiple of g(x)!

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

! If remainder is nonzero, then received n-tupleis 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) by g(x)! Blindspot: If e(x) is a multiple of g(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 most

common error patterns can be detected.

b(x)

e(x)

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

(Receiver)(Transmitter)

Error polynomial(Channel)


Designing good polynomial codes

! Select generator polynomial so that likely errorpatterns are not multiples of g(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

! If g(x) is a primitive polynomial, it cannot divide xm+1 for allm<2n-k-1 (Need to keep codeword length less than 2n-k-1)

! Primitive polynomials can be found by consulting coding theory books


Designing good polynomial codes

! Detecting Odd Numbers of Errors

! Define g(x) to be a multiple of x+1

! This implies x+1 must be a factor of all codewords b(x)

! For e(x) to be an undetectable error pattern, itmust be e(x) = q’(x)g(x) for some q’(x)

! Evaluate this identity at x=1 and get e(1) = 0 (itmust be g(1)=0; why?); then e(x) cannot have ad oddnumber of 1’s (why?)

! All odd numbers of errors are detectable!

! Pick g(x)=(x+1)p(x), where p(x) is primitive


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-correcting codes

! Capable of correcting all single-error patterns

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

5726114

k = n–m

6/636365/313154/151543/773m/nn = 2m–1m

Redundancy


m = 3 Hamming Code

! Information bits are b1, b2, b3, b4

! Equations for parity checks b5, b6, b7

! 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 0

41 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


Parity Check Equations

! Rearrange parity check equations:

! All codewords mustsatisfy these equations

! Note: each nonzero 3-tuple appears once as acolumn in check matrix H

! In matrix form:

0 = b5 + b5 = b1 + b3 + b4 + b5

0 = 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 = H bt = 0

0 = 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 0

1 1 0 1 0 1 0

0 1 1 1 0 0 1

1

1

1

0

0

0

0

s = H e = = + + = 0 1

1

0

Triple error not

detected

0

1

1

1

0

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


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 bechanged to convert one codeword into another codeword

b1 b2o o

o

o

oo

o

o

Set of n-tuples

within distance

1 of b1

Set of n-tuples

within distance

1 of b2

! Spheres of distance 1 around each codeword do notoverlap

! If a single error occurs, the resulting n-tuple will be in aunique sphere around the original codeword

Distance 3


General Hamming Codes

! For m > 2, the Hamming code is obtainedthrough the check matrix H:! Each nonzero m-tuple appears once as a column of H

! The resulting code corrects all single errors

! For each value of m, there is a polynomial codewith g(x) of degree m that is equivalent to aHamming code and corrects all single errors! For m = 3, g(x) = x3+x+1


Error-correction using Hamming Codes

! The receiver first calculates the syndrome:s = HR = H (b + e) = Hb + He = He

! If s = 0, then the receiver accepts R as the transmittedcodeword

! If s is nonzero, then an error is detected! Hamming decoder assumes a single error has occurred! Each single-bit error pattern has a unique syndrome! The receiver matches the syndrome to a single-bit error pattern

and corrects the appropriate bit

b

e

R+ (Receiver)(Transmitter)

Error pattern


Error correcting codes

! Very powerful correcting codes have beendevised over time! Used especially for wireless communications,

satellite, deep space communications

! Two common code types are

! Convolutional codes

! Turbo-codes

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