SESSION 3.DIGITAL TRANSMISSION FUNDAMENTALS

Stephen Kimdskim@iupui.edu

Network – The Transmission Infrastructure What to be transmitted? (object) What to transmit through? (medium)

TRANSMISSION SYSTEMS

3-2

Object Medium

Electricity Network Electrical Energy Copper Wire

Transportation Network

Vehicle Road

Data Network Information Copper wire, fiber optic, radio, Infrared

Propagation of energy in the form of pulses. Analog Communication

Transmit a waveform – varying continuously on time Reproduce the waveform as exactly as possible at a remote site.

Digital Communication Transmit a symbol from a finite set Thru transmission medium or channel Binary – {1, 0}

1 – send a positive voltage for a certain period of time (t) 0 – send a negative voltage for t

Receiver determine the input symbol with high probability

DATA COMMUNICATION

3-3

transmitter ReceiverCommunication Channel

Can a digital signal go longer than an analog signal? Can we make a digital signal to go longer than an analog

signal? Distortion of electrical signal along the distance

Attenuation Noise

from other electrical systems from other electrical cables from nowhere (random noise)

Solutions make the medium insensitive to noise

shielded wire eliminate the noise source – how? repeater – regenerate the signal

ADVANTAGE OF DIGITAL COMM.

3-4

Periodically regenerate the signal Amplifier for attenuated signal Equalizer to eliminate the distortion Accumulation of noise

REPEATER

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S Repeater Repeater D

Amp EqualizerAttenuated anddistorted signal+noise

Recovered signal+residual noise

The symbol is recovered every time by the repeaters. No accumulation of noise Quality of signal is independent of distance

REPEATER IN DIGITAL TRANSMISSION

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S Repeater Repeater D

Attenuated anddistorted signal+noise

Recovered signal

ReceptorBit

GeneratorTransmitter

Range of frequencies passed by a channel Attenuation is a function of frequency as well as distance

Amplitude-response function A(f) = Amplitude of output tone / Amplitude of input tone

BANDWIDTH

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f

A(f)

0

1

If a signal run thru a low-pass filter of bandwidth W, the filtered signal can be reconstructed by making only 2W samples per second.

The fastest pulse rate = 2W pulse/second. Example

A noiseless 3-kHz channel can transmit binary signals at most at a rate of 6000 bit/sec.

Multilevel Transmission Each pulse represents M levels, then the pulse can encode log2M binary. A noiseless W-Hz channel can transmit at a bit rate of 2W·log2M.

NYQUIST’S THEOREM

3-8

23

10 0 0 0

23

2 21

0 0

Without noise, we can send information at any bit rate. Large M

SNR = Signal power / Noise power dB = 10 log10(S/N)

SIGNAL-TO-NOISE RATIO (SNR)

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signal noise signal + noise

HighSNR

t t t

LowSNR

t t t

Maximum bit rate, C=W·log2(1+S/N) Example

Bandwidth 3400Hz, SNR = 40dB C= 3400 log2(1+10000)=45Kbps

No matter how often sample are taken or how many levels are encoded.

SHANNON CHANNEL CAPACITY

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Attenuation – Aout/Ain

Phase shift - (f) Cascade of filters, or channels from various sources

FREQUENCY DOMAIN

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Channelt

Aincos 2ft

t

Aoutcos (2ft + (f))

Output Signal of the Channel y(t)

))(2cos()()( 000 kftkfkfAaty k

FOURIER ANALYSIS – 1

3-12

A periodic function g(t) is described by

dtT

nttg

Tb

dtT

nttg

Ta

dttgT

a

T

ntb

T

nta

atg

T

n

T

n

T

N

nnn

N

0

0

00

1

0

)2

cos()(2

)2

sin()(2

)(2

2cos

2sin

2)( lim

Consider any 8-bit word, say 01100010 T=8, f=1/8, g(1)=0, g(2)=1, g(3)=1, g(4)=0, g(5)=0, g(6)=0, g(7)=1, g(8)=0

FOURIER ANALYSIS – 2

3-13

dtT

nttg

Tb

dtT

nttg

Ta

dttgT

a

T

ntb

T

nta

atg

T

n

T

n

T

N

nnn

N

0

0

00

1

0

)2

cos()(2

)2

sin()(2

)(2

2cos

2sin

2)( lim

)2/3cos()4/7cos()4/sin()4/3sin(1

)4/7cos()2/3cos()4/3cos()4/cos(1

4/30

nnnnn

b

nnnnn

a

a

n

n

FOURIER ANALYSIS – 3

3-14

0 1 2 3 4 5 6 7 8-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 1 2 3 4 5 6 7 8-0.2

-0.1

0

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0 1 2 3 4 5 6 7 8-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0 1 2 3 4 5 6 7 8-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0000 0111

T

N=1

N=2

N=4

N=8

How to convert a binary sequence into a digital signal Cosinderation

Maximizing the bit rate Easy to implementation

Cost of network systems Embedding synchronization Built-in error detection Immunity to noise and interference

Unipolar NRZ encoding simple

A binary 1 for +A voltage A binary 0 for 0 voltage

Disadvantage – power for transmission

LINE CODING – 1

3-15

001110101

Polar NRZ Encoding A binary 1 for +A/2 voltage A binary 0 for –A/2 voltage Efficient transmission power

Problem in NRZ Encodings If there is a long sequence of 0’s, or a long sequence of 1’s

the encodings generate a low frequency Some network components consider a low frequency as a noise, so

eliminate it. Example, telephone system filters out frequencies lower than 200Hz.

LINE CODING – 2

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001110101

Bipolar Encoding A binary 0 is mapped to 0 voltage. A consecutive 1s are alternatively mapped to +A/2 and –A/2. No low frequency signal

Synchronization Determine the boundary between bits. Using NRZ encodings or Bipolar encoding, need of separate synchronization

signal

LINE ENCODING – 3

3-17

001110101

Manchester Encoding A binary 1 is denoted by a transition from +A/2 to –A/2 in the middle of the bit time

interval. A binary 0 is denoted by a transition from –A/2 to +A/2. Every bit contributes a transition that is used for synchronization. Doubled pulse rate Applied to Ethernet LAN

Differential Manchester Encoding A transition in the middle of every bit time Transition in the beginning of the bit interval represents a binary 0. No transition in the beginning of the bit interval represents a binary 1. Applied to Token-ring

LINE ENCODING – 4

3-18

001110101Manchester

encoding

Differential Manchester

encoding

The Manchester code is an 1B2B code scheme. 1 bit information is encoded 2-bit signal. 0 is mapped to 01, 1 is mapped to 10.

FDDI (Fiber Distributed Data Interface) use 4B5B 4 bit information is encoded 5-bit signal

LINE ENCODING – 5

3-19

Propagation speed c/f c – speed of light (3x108m/sec), f – frequency

Guided Media – Wired media Point-to-point, so discrete network topologies Unlimited resource

simply install more wires, expensive though. Attenuation 10^<distance>, exponential Twisted pair, Coaxial cable, Fiber optic

Unguided Media – wireless media Broadcast with limited directionality Broad and continuous network topologies Limited resource, so regulated by some organizations Inexpensive to be installed Attenuation <distance>^2, quadratical Radio, Infrared light

PROPERTY OF COMMUNICATION MEDIATHE BIG PICTURE

3-20

Twisted Pair Two parallel insulated wires

one for signal, another for reference (ground) Crosstalk

Many cables are usually bundled A cable talks to other adjacent cable.

Twist the pair of wires Let them (signal&reference) catch same noise

Coaxial cable Better immunity to noise and interference

Optical fiber On-Off pulse on the guided medium No interference with electromagnetic Multimode mode and Single mode

WIRED MEDIA

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Direct path

Reflected path

Radio Frequency

Low frequency – can pass obstacles, but quick power dissipation High frequency – can travel in straight; bounce off obstacles; absorbed

by rain; Multipath fading Multipath fading – bounced signal are cancelled direct signal. Regulated by government agency

Infrared Light Cannot penetrate walls Directional No regulation

WIRELESS MEDIA

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SUMMARY OF MEDIA

3-23

Twisted Pair

CoaxialCable

Optical Fiber

Radio Infrared

Telephone UTP3

Ethernet UTP5 Thin, Thick

CATV, Cable modem

75-ohm

Trunk,Backbone,LAN

X

ISDN, ADSL UTP3

Cellular x

Cordless Phone X

Wireless LAN-802.11

Unlicensed

Bandwidth

Entertainment TV Satellite

remote controller X

IrDA X

Highly reliable modern communication Non-zero error probability Error-sensitive digital communication

Compressed information, no redundancy, distinct roles of data

How to handle errors ARQ

Automatic Retransmission reQuest detect and send a request for retransmission bandwidth redundancy

FEC Forward Error Correction detect and correct embedded redundancy

OVERVIEW OF ERROR DETECTION AND CORRECTION

3-24

Codeword = information bits & check bit Two Choices

Even parity - # of 1s in the codeword is 2n. Odd parity - # of 1s in the codeword is 2n+1. mostly use the even parity method.

Modulo k arithmetic - concern with remainder alone. k 2k 3k …, k+1 2k+1 3k+1 … ex) (modulo 3) 3693n, 47103n+1, 58113n+2

The parity is (modulo 2) operation for individual bits in the codeword. ex) 1001010 (1+0+0+1+0+1+0) modulo 2 = 1

A sender attaches the result as an additional bit. A recipient computes the even # of 1s in every codeword. Ability of the single parity check

ERROR DETECTION – SINGLE PARITY CHECK

3-25

A Checksum is calculated for the header The header is modified by an Internet

router, sothe checksum must be re-calculated at every router

Seek for an efficient software Assume a header contain L words(b0 to

bL-1), and then add a word (checksum word, bL). x = b0 + b1 + … + bL-1 modulo 2L – 1 bL = -x

INTERNET CHECKSUM

Hamming Distance Hypercube Valid codeword or Invalid codeword

PRINCIPLE OF ERROR DETECTION

3-27

S + R

error

data

data+redn (data . redn)error

data/alarm

A HYPERCUBE

3-28

d+c V/I

0000 V

0001 I

0010 I

0011 V

0100 I

0101 V

0110 V

0111 I

1000 I

1001 V

1010 V

1011 I

1100 V

1101 I

1110 I

1111 V0000

0010

0100

1000

0001

0110

1010

1110

1100

0101

0111 1111

0011 1011

1001

1101

The minimum distance is d , so as to detect d-1 errors

Can we make codeword to detect 3 bit errors? 00000011, 00001100, 00110000, 11000000

Eg) 8-bit word => 256 possible codes Select codewords as many as possible so that the

minimum distance is 4.

DISTANCE BETWEEN 2 VALID CODEWORD

3-29

d

If recipient receives an invalid code, make it go to the nearest valid code.

To correct t errors, the distance is at least 2t+1. Correction of single bit errors

m message bits, r check bits Each valid codeword is surrounded by m+r invalid codeword. So, there are at least (m+r+1)2m codeword In addition, m+r bits can make 2 (m+r) combinations (m+r+1) 2 r

m=7, then 8+r 2 r , the smallest r is 4.

PRINCIPLE OF ERROR CORRECTION

3-30

To measure the effectiveness of error detection/correction,need to know the behavioral property of the errors.

Error is considered as a binary vector of length n (size of codeword)

Model Random error vector model

All 2m possible error vectors are equally likely to occur. P(0000)=P(0001)=P(1111)

Random bit error model The bit errors occur independently of each other. p = a single bit error, P(j errors) = nCj pj (1-p)n-j

P(no error) = (1-p)n, P(any error) = 1–(1-p)n np Burst error model

low error period + high error period random bit error model & random error vector model Close to most communication channel

ERROR MODELS

3-31

In Random error vector model

In Random bit error model

In Burst error model Hard to say

ERROR DETECTION FAILURE FOR SINGLE PARITY

3-32

...)1(4

)1(2

)1s ofnumber even containsor error vect(

4422

nn pp

npp

n

P

2

1

2

42

)1s ofnumber even containsor error vect(

n

n

nnn

P

S and R agree upon a generator function g(x) of degree n.

Use binary and modulo-2 arithmetic no carry for addition, no borrow for subtraction addition = subtraction = exclusive OR. n is the degree of g(x).

CYCLIC REDUNDANCY CHECK (CRC)

3-33

Sg(x)

Rg(x)

+

f(x)

xnf(x)-r(x)

e(x)

xnf(x)-r(x)+e(x)

xnf(x)=g(x)*s(x)+r(x)

Is xnf(x)-r(x)+e(x) divisible by g(x) ?

g(x)*s(x) g(x)*s(x)+e(x)

g(x) = x3+x+1 = 1011 f(x) = x3+x2 = 1100 xnf(x) = x6+x5 = 1100000

EXAMPLE OF CRC

3-34

1011 1100000

111

10111110001011

101001011

10

xnf(x)+r(x)=x6+x5+x=1100010

Transmit

No error1100010 is divisible by 1011

Detect all single errors if g(x) cannot divide xi. Detect all double errors if g(x) cannot divide 1+xm , where m is

the bit distance b/w two error bits. For a primitive polynomial of degree N, the smallest m is 2N-1, for

which 1+xm is divisible by the primitive. ex) g(x) has a factor of a primitive polynomial of degree 15, it will

not divide 1+xm for any m below 32768. x15+x+1 is such a polynomial.

Detect all triple error if g(x) contains x+1 as a factor. Detect all burst error of length n, where n is the degree of the

generator function. e(x) is not divisible by g(x) if deg[e(x)] < deg[g(x)]

All generator function is (x+1)p(x). (CRC-8) x8+x2+x+1 (CRC-16) x16+x15+x2+1

CAPABILITY OF CRC

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The error detection (or correction) problem can be mapped to a graph coloring problem.

General Graph Coloring For an undirected graph G=(V,E) Assign a color to each vertex such that the color is different from

its neighbors with the minimum number of color used. The color index will be the redundancy.

It is known to unsolvable. For a planar graph, the number of colors seems to be 4.

Nobody proved it yet, many believed so. For a hypercube, the number of colors is 2 (0 or 1).

n-D hypercube (a binary number of length n) a=[an an-1…a1] The color is a*[1111…1]T

ADVANCED TOPIC: GRAPH COLORING

3-36

Xk = the minimum number of colors in a vertex-coloring such that any two vertices with distance exactly k.

Yk = the minimum number of colors in a vertex-coloring such that any two vertices within distance k.

For a binary a of length n, consider nlog(n) m

EXPANDED GRAPH COLORING

3-37

111..

010..

001..

000..

][ 11

aaama nn

If the distance of a and b is 2, then am bmProof? (do it yourself if you are interesting at)

The matrix T will assign different colors within distance 3. So, we solved Xk and Yk for k3. Why?

How about Xk and Yk for k>3 ? There is a solution, but it is complicated.

EXPANDED GRAPH COLORING

3-38

1111..

0101..

0011..

0001..

]1...11[||

TmT