Chapter 3 Digital Transmission Fundamentals

School of Info. Sci. & Eng.Shandong Univ.

Information comes in a big variety.

How does it all get encoded into signals for transmission over a physical layer channel?

Outlinel 3.1 Digital Representation of Infol 3.2 Why Digital? Why not Analog?l The basic sinewave and pulse train carrier signalsl 3.3 Characterization of Communication Channelsl 3.4 Fundamental Limits l 3.5 Line Codingl 3.6 Modemsl 3.7 Transmission Media

DomainNames nw.com

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1996

1996

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0120

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ions

(Log

Sca

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Growth of the Internet

01020304050607080

1996 1997 1998 1999 2000

Dom

ain

Nam

es (M

illio

ns)

Forecasting Internet growth from nw.com data

Linear Scale. Tough to extrapolate since it is an exponential.

Log Scale. Draw a line or use Excel LOGEST

Role of Exponential Growth in Making

l Moore’s Law models progress in the IC chip business. Since 1960 number of bits on a chip has doubled every 18 months or so. How long will it continue?

l “Andreessen’s Law” models growth in Internet domain names. Since 1995 number of names has doubled every 18 months or so. Is this a coincidence?

l Both processes were foundations for rapid growth in jobs, products and services, and Wall St. speculation

l Can you estimate the doubling interval for optical communications from Fig. 1.9. Note the change in 1995-2000 with the invention of DWDM. Can this continue? Will there be another investment frenzy? Will there be another bust?

Receiver

Communication channel

Transmitter

Figure 3.5

The basic communications model

Why Digital Transmission?l Information comes in many forms, but we need to

encode it into either an analog (continuously varying) or digital (two or M-level) signal for transmission

l Currently the digital option is preferred for new systems for several reasons. Computer processing is more flexible (software) vs. analog signal processing by electronic devices. We’ll see some more on the next few slides.

l Legacy systems like AM / FM radio, NSTC TV are very difficult to change to digital – Why?

(a) Analog transmission: all details must be reproduced accurately

Sent

Sent

Received

Received

• e.g digital telephone, CD Audio

(b) Digital transmission: only discrete levels need to be reproduced

• e.g. AM, FM, TV transmission

Figure 3.6

Analog vs. digital signal designs

Source Repeater DestinationRepeater

Transmission segment

Figure 3.7

Often repeaters are used to send a signal over a long distance.

Attenuated & distorted signal +

noise

EqualizerRecovered signal

+residual noise

Repeater

Amp.

Figure 3.8

At each analog repeater, we amplify the received signal and noise. We amplify both: A[as(t) + n(t)] Suppose Aa =1.

After n repeaters we have: s(t) + nAn(t);

not good

AmplifierEqualizer

TimingRecovery

Decision Circuit.& SignalRegenerator

Figure 3.9

A digital repeater can perfectly reconstruct the transmitted signal

Murphy’s brother’s-in-law Law: “Nothing is perfect.” What can go wrong here?

signal noise signal + noise

signal noise signal + noise

HighSNR

LowSNR

SNR = Average Signal Power

Average Noise Power

SNR (dB) = 10 log10 SNR

t t t

t t t

Figure 3.12

SNR

∆/23∆/25∆/27∆/2

−∆/2−3∆/2−5∆/2−7∆/2

∆/23∆/25∆/27∆/2

−∆/2−3∆/2−5∆/2−7∆/2

(a)

(b)

Figure 3.2

Original and samples

Original and quantized values

Conversion of analog voice to digital.

H

W

= + +H

W

H

W

H

W

Color

Image

Red Component Image

Green Component Image

Blue Component Image

Total bits before compression = 3xHxW pixels x B bits/pixel = 3HWB

Figure 3.1

Conversion of still pictures to digital

(a) QCIF Videoconferencing

(b) Broadcast TV

(c) HDTV

@ 30 frames/sec =

760,000 pixels/sec

@ 30 frames/sec =

10.4 x 106 pixels/sec

@ 30 frames/sec =

67 x 106 pixels/sec

720

480

1080

1920

144

176

Figure 3.3

Conversion of video for digital transmission

communication channeld meters

0110101... 0110101...

Figure 3.10

A digital channel

How fast we can send digital information depends on:

•Energy per bit

•Distance

•Noise

•Bandwidth

f0 W

A(f)

(a) Lowpass and idealized lowpass channel

(b) Maximum pulse transmission rate is 2W pulses/second

0 Wf

A(f)1

Channel

tt

Figure 3.11

Effect of Bandwidth on DistortionIt is possible to send 2W pulses per second through an ILPF, using the sin x / x function. For only two possible signals, this is R =2W bps.

Channelt t

Aincos 2πft Aoutcos (2πft + ϕ(f))

AoutAin

A(f) =

Figure 3.13

When a sinusoid (perhaps a harmonic component) is passed through a linear channel, its amplitude and phase is changed by the filter function at that frequency

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5 1

(b) 2 Harmonics

(c) 4 Harmonics

(a) 1 Harmonic

Figure 3.16

The digital signal on the previous slide contains such harmonics. This slide shows the effect of passing that signal through filters that pass only 1, 2, and 4 of these terms

Channel

t0t

h(t)

td

Figure 3.17

f

1A(f) = 1

1+4π2f2

Figure 3.14 - Part 1

Effect of Bandwidth on Distortion. Typical amplitude response of channel or filter.

f

0

ϕ(f) = tan-1 2πf

-45o

-90o

1/ 2π

Figure 3.14 - Part 2

Effect of Bandwidth on Distortion. Effect of delay or phase shift on distortion

Playout delay

Jitter due to variable delay(b)

(c)

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6

Original sequence(a)

1 2 3 4 5 6 7 8 9

Figure 3.4

1 0 0 0 0 0 0 1

. . . . . .

t

1 ms

Figure 3.15

Fourier Theorem: Any signal can be written as a series of sinusoids. There is a constant component, a fundamental frequency component, and a series of multiples of that fundamental frequency. Show onblackboard.

-0.4

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1

1.2

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

s(t) = sin(2πWt)/ 2πWt

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

Figure 3.18

+A

-A0 T 2T 3T 4T 5T

1 1 1 10 0

Transmitter Filter

Comm. Channel

Receiver Filter Receiver

r(t)

Received signal

t

Figure 3.19

-2

-1

0

1

2

-2 -1 0 1 2 3 4

-1

0

1

-2 -1 0 1 2 3 4

(a) 3 separate pulses for sequence 110

(b) Combined signal for sequence 110

t

tT T T T TT

T T T T TT

Figure 3.20

W (1+α )W(1-α )W0 f

Figure 3.21

4 signal levels 8 signal levels

typical noise

Figure 3.22

Noise limitation on digital transmission. If we have more than 2 possible signals sent in one interval b, we can increase the number of bits / interval. We could send the whole dictionary in one pulse, but noise keeps us from distinguishing among so many levels

m =2 bits per pulse

m = 3 bits per pulse

Now data rate R = 2Wm bits/sec

Maxium Data Ratesl With 2 levels no noise R = 2W bpsl With m bits/pulse R= 2Wm bpsl With noise Shannon limit is C = W log2 (1 + SNR)If R<C, Shannon showed that we can get as low an

error rate as desired by more elaborate coding. Current 56Kbps modems are about equal to this limit for typical 40dB noise levels and 3.4kHz bandwidths. Therefore, don't expect more from a dial- modem.

x

222

21 σ

σπxe−

0

Figure 3.23

Noise often has a Gaussian or normal pdf

This is the famous "bell-shaped" curve. It often appears in applications because of the Central Limit Theorem. This one hasa mean of zero, so the only parameter is sigma, the variance.

1.00E-121.00E-111.00E-101.00E-091.00E-081.00E-071.00E-061.00E-051.00E-041.00E-031.00E-021.00E-011.00E+00

0 2 4 6 8 δ/2σ

Figure 3.24

Probability of a bit error if Gaussian noise is the limiting factor. Delta is the spacing between signal levels.

1 0 1 0 1 1 0 01UnipolarNRZ

NRZ-Inverted(DifferentialEncoding)

BipolarEncoding

ManchesterEncoding

DifferentialManchesterEncoding

Polar NRZ

Figure 3.25

Line coding permits self-synchrononization. Common designs.

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fT

pow

er d

ensi

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NRZ

Bipolar

Manchester

Figure 3.26

Spectra of these line coding designs. This is the distribution of power vs. frequency in the signals.

ff2f1 fc0

Figure 3.27

The previous signal designs were for low pass channels. Now consider a "band pass" channel.

Information 1 1 1 10 0

+1

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

AmplitudeShift

Keying

+1

-1

FrequencyShift

Keying

+1

-1

PhaseShift

Keying

(a)

(b)

(c)

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

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

t

t

t

Figure 3.28

3 common types of digital modulation methods

1 1 1 10 0(a) Information

(d) 2Yi(t) cos(2πfct)

+2A

-2A

+A

-A

(c) ModulatedSignal Yi(t)

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

+A

-A

(b) BasebandSignal Xi(t)

0 2T 3T 6T

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

T 4T 5T

t

t

t

Figure 3.29

AM via multiplication by a sinusoid

(a) Modulate cos(2πfct) by multiplying it by Ak for (k-1)T < t <kT:

Ak x

cos(2πfct)

Yi(t) = Ak cos(2πfct)

(b) Demodulate (recover) Ak by multiplying by 2cos(2πfct) and lowpass filtering:

x

2cos(2πfct)2Ak cos2(2πfct) = Ak {1 + cos(2π2fct)}

LowpassFilter withcutoff W Hz

Xi(t)Yi(t) = Akcos(2πfct)

Figure 3.30

Akx

cos(2πfc t)

Yi(t) = Ak cos(2πfc t)

Bkx

sin(2πfc t)

Yq(t) = Bk sin(2πfc t)

+ Y(t)

Modulate cos(2πfct) and sin (2πfct) by multiplying them by Ak and Bk respectively for (k-1)T < t <kT:

Figure 3.31

Quadrature Amplitude Modulation QAM

A and B are odd and even symbols with half the data rate of input

Y(t) x

2cos(2πfc t)2cos2(2πfct)+2Bk cos(2πfct)sin(2πfct)

= Ak {1 + cos(4πfct)}+Bk {0 + sin(4πfct)}

LowpassFilter withcutoff W/2 Hz

Ak

x

2sin(2πfc t)2Bk sin2(2πfct)+2Ak cos(2πfct)sin(2πfct)

= Bk {1 - cos(4πfct)}+Ak {0 + sin(4πfct)}

LowpassFilter withcutoff W/2 Hz

Bk

Figure 3.32

QAM demodulator

Ak

Bk

16 “levels”/ pulse4 bits / pulse4W bits per second

Ak

Bk

4 “levels”/ pulse2 bits / pulse2W bits per second

2-D signal2-D signal

Figure 3.33

Signal constellations. Packing more than one bit into a single pulse to get higher data rates R.

Achieves R = 2W rate of best LPF

Ak

Bk

4 “levels”/ pulse2 bits / pulse2W bits per second

Ak

Bk

16 “levels”/ pulse4 bits / pulse4W bits per second

Figure 3.34

Constellations used in practice

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 &

tele

phon

e

broa

dcas

tra

dio

mic

row

ave

radi

o

infra

red

light

visi

ble

light

ultra

viol

et li

ght

x ra

ys

gam

ma

rays

Figure 3.35

Properties of transmission media

t = 0t = d/c

communication channel

d meters

Figure 3.36

Atte

nuat

ion

(dB

/mi)

f (kHz)

19 gauge

22 gauge

24 gauge

26 gauge

6

12

3

9

15

18

21

24

27

30

1 10 100 1000

Figure 3.37

l l l l l l

Figure 3.38

Centerconductor

Dielectricmaterial

Braidedouter

conductor

Outercover

Figure 3.39

Coaxial Cable

35

30

10

25

20

5

15Atte

nuat

ion

(dB

/km

)

0.01 0.1 1.0 10 100 f (MHz)

2.6/9.5 mm

1.2/4.4 mm

0.7/2.9 mm

Figure 3.40

Attenuation of Coaxial Cable. Note error on f axis in book.

Headend

Unidirectionalamplifier

Figure 3.41

Original Cable TV Network

Headend

Upstream fiber

Downstream fiber

Fibernode

Coaxialdistribution

plant

Fibernode

BidirectionalSplit-BandAmplifier

Fiber Fiber

Figure 3.42

Modification for Computer Network Service

Downstream

54 MH

z

500 MH

zUpstream

Downstream

5 MH

z

42 MH

z

54 MH

z

500 MH

z

550 MH

z

750 MH

z

(a)Currentallocation

(b) Proposedhybridfiber-coaxialallocation

Proposed downstream

Figure 3.43

Cable TV Spectrum Allocations

core

cladding jacketlight

θc

(a) Geometry of optical fiber

(b) Reflection in optical fiber

Figure 3.44

Optical Fiber Cable

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)

Loss

(dB/

km) Infrared absorption

Rayleigh scattering

Figure 3.45

Optical Fiber Attenuation

(a) Multimode fiber: multiple rays follow different paths

(b) Single mode: only direct path propagates in fiber

direct path

reflected path

Figure 3.46

Much lower attenuation possible

Optical fiber

Opticalsource

ModulatorElectricalsignal Receiver Electrical

signal

Figure 3.47

104 106 107 108 109 1010 1011 1012

Frequency (Hz)

Wavelength (meters)

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

105

satellite & terrestrialmicrowave

AM radio

FM radio & TV

LF MF HF VHF UHF SHF EHF104

Cellular& PCS

Wireless cable

Figure 3.48

ChannelEncoderUserinformation

PatternChecking

All inputs to channelsatisfy pattern/condition

Channeloutput Deliver user

informationor

set error alarm

Figure 3.49

Calculate check bits

Channel

Recalculate check bits

Compare

Information bits Received information bits

Checkbits

Information accepted if check bits match

Received check bits

Figure 350

x = codewords o = non-codewords

x

x x

x

x

x

x

o

oo

oo

oo

o

oo

o

oxx x

x

xx

x

o o o

oo

ooooo

o

o

A code with poor distance properties A code with good distance properties(a) (b)

Figure 3.51

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

Figure 3.52

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

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 1 0 1

1 0 0 1 0 0

1 0 0 0 1 0

1 0 0 1 1 1

Two errors

One error

Three errors

Four errors

Arrows indicate failed check bits

Figure 3.53

unsigned short cksum(unsigned short *addr, int count){

/*Compute Internet Checksum for “count” bytes * beginning at location “addr”.*/

register long sum = 0;while ( count > 1 ) {

/* This is the inner loop*/ sum += *addr++; count -=2;}

/* Add left-over byte, if any */if ( count > 0 )

sum += *addr;

/* Fold 32-bit sum to 16 bits */while (sum >>16)

sum = (sum & 0xffff) + (sum >> 16) ;

return ~sum;}

Figure 3.54

(x7 + x6 +1) + (x6 + x5 ) = x7 + (1 +1)x6 + x 5 +1

= x7 + x5 +1

(x +1)(x2 + x +1) = x3 + x 2 + x + x2 + x +1 = x3 +1

Addition:

Multiplication:

Division: x3 + x + 1 ) x6 + x5

x3 + x2 + x

x6 + x4 + x3

x5 + x4 + x3

x5 + x3 + x2

x4 + x2

x4 + x2 + xx

= q(x) quotient

= r(x) remainder

divisordividend

35 ) 1223

10517

Figure 3.55

Steps:1) Multiply i(x) by xn-k (puts zeros in (n-k) low order positions)

2) Divide xn-k i(x) by g(x)

3) Add remainder r(x) to xn-k i(x) (puts check bits in the n-k low order positions):

quotient remainder

transmitted codewordb(x) = xn-ki(x) + r(x)

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

Figure 3.56

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

Encoding: x3i(x) = x6 + x5

1011 ) 1100000

1110

1011

11101011

10101011

010

x3 + x + 1 ) x6 + x5

x3 + x2 + x

x6 + x4 + x3

x5 + x4 + x3

x5 + x3 + x2

x4 + x2

x4 + x2 + x

xTransmitted codeword:

b(x) = x6 + x5 + xb = (1,1,0,0,0,1,0)

Figure 3.57

g (x ) = x 3 + x +1

reg 0 reg 1 reg 2++

g3 = 1

i (x )

g0 = 1 g1 =1 i (x ) = x 3 + x 2

Encoder for

clock input reg 0 reg 1 reg 20 - 0 0 01 1=i3 1 0 02 1=i2 1 1 03 0=i1 0 1 14 0=i0 1 1 15 0 1 0 16 0 1 0 07 0 0 1 0

check bits: r0 = 0 r1 = 1 r2 = 0r(x) = x

Figure 3.58

b(x)

e(x)

R(x)+ (Receiver)(Transmitter)

Error pattern

Figure 3.59

1. Single errors: e(x) = xi 0 ≤ i ≤ n-1If g(x) has more than one term, it cannot divide e(x)

2. Double errors: e(x) = xi + xj 0 ≤ i < j ≤ n-1= xi (1 + xj-i )

If g(x) is primitive, it will not divide (1 + xj-i ) for j-i ≤ 2n-k−1

3. Odd number of errors: e(1) =1 If number of errors is odd.If g(x) has (x+1) as a factor, then g(1) = 0 and all

codewords have an even number of 1s.

Figure 3.60

4. Error bursts of length b: 0000110 • • • • 0001101100 • • • 0

e(x) = xi d(x) where deg(d(x)) = L-1g(x) has degree n-k; g(x) cannot divide d(x) if deg(g(x))> deg(d(x))

l L = (n-k) or less: all will be detectedl L = (n-k+1): deg(d(x)) = deg(g(x))

i.e. d(x) = g(x) is the only undetectable error pattern,fraction of bursts which are undetectable = 1/2L-2

l L > (n-k+1): fraction of bursts which are undetectable = 1/2n-k

Lith

position

error pattern d(x)

Figure 3.61

b

e

r+ (receiver)(transmitter)

error pattern

b

e

r+ (receiver)(transmitter)

error pattern

(a) Single bit input

(b) Vector input

Figure 3.62

0010000

s = H e = =101

single error detected

0100100

s = H e = = + =011

double error detected100

1 0 1 1 1 0 01 1 0 1 0 1 00 1 1 1 0 0 1

1110000

s = H e = = + + = 0110

triple error notdetected

011

101

1 0 1 1 1 0 01 1 0 1 0 1 00 1 1 1 0 0 1

1 0 1 1 1 0 01 1 0 1 0 1 00 1 1 1 0 0 1

111

Figure 3.63

s = H r = He

s = 0 s = 0

no errors intransmission

undetectableerrors

correctableerrors

uncorrectableerrors

(1-p)7 7p3

1-3p 3p

7p

7p(1-3p) 21p2

Figure 3.64

If dmin= 2t+1, non-overlapping spheres of radius tcan be drawn around each codeword; t=2 in the figure

b1 b2o o o oset of all n-tupleswithin distance t

set of all n-tupleswithin distance t

Figure 3.66

b1 b2 b3 b4 bL-3 bL-2 bL-1 bL. . .

L codewordswritten verticallyin array; thentransmitted rowby row

b1 b2 b3 b4 bL-3 bL-2 bL-1 bL

. . .

A long error burst produceserrors in two adjacent rows

Figure 3.66

DTE DCE

Protective Ground (PGND)Transmit Data (TXD)

Receive Data (RXD)

Request to Send (RTS)

Clear to Send (CTS)

Data Set Ready (DSR)

Ground (G)

Carrier Detect (CD)

Data Terminal Ready (DTR)

Ring Indicator (RI)

12

34

56

78

2022

12

34

56

78

2022

(b)

(a)• • • • • • • • • • • • •

• • • • • • • • • • • •

1 13

2514

Figure 3.67

Startbit

Stopbit1 2 3 4 5 6 7 8

Data bits

Lineidle

3T/2 T T T T T T T

Receiver samples the bits

Figure 3.68