Digital Communication Chap 3

8/4/2019 Digital Communication Chap 3

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Communication System II

Chapter 3

Samrat SubediKantipur Engineering College

[email protected]

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Digital Data Communication

System

Communication System are Designed toTransmit Information.

The Purpose of Communication System isto transmit the output of Source toDestination.

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Two Fundamental Questions Of

Communication Theory are:

What is the rate at which given

source is emitting information.Information Theory

What is the maximum rate of

information transmitted over a noisychannel

Channel Capacity theorem

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Information Theory

A message is a sequence of Symbol

intended to reduce Uncertainty ofReceiver.

If a sequence of symbol does not change

the Uncertainty Level of the receiver, themessage does not contain anyinformation.

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Consider the following Cases

1. Tomorrow the Sun rises in the East.

2. United States invades Cuba.

3. Cuba invades United States

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From the Viewpoint of common sense

The first headline hardly contain Any information

The Second contain large amount of information

The third Convey the largest

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Therefore

The Information content of any messagesignal is closely related to

The past knowledge of Occurrence of event

andLevel of uncertainty it contains with respect to

the recipients of the message

Thus in general

The amount of information received fromknowledge of Occurrence of an event is relatedto the probability or the likelihood ofOccurrence of the Event.

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Measurement

Let m1,m2,m3mq be out of a possible

message emitted by a source withprobabilities p1,p2,pq Such that

p1+p2++pq=1

If I(mk) is the information Content of Kth

message Then,1. I(mk) > I(mj) for pk


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To Satisfy above Condition We canRelate I(mk) and pk

I(mk) = log (1/pk)

Unit of I(mk) depends on baseassigned to log

Base e nat

Base 10 Hartley or decitBase 2 bit

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Entropy

The average information content of asequence of symbol

Consider a memory less source( i.e. probability ofOccurrence of symbol does not depend upon previous and future Occurrence of

Symbol) emitting m possible Symbolss1,s2,s3sm with probabilities p1,p2,pmrespectively.

For a long message Sequence ContainingN Symbols, rate of Occurrence of siSymbol is

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P(si) = piN

And the information Content

I(si) = piNlog2(1/pi)

Then

Itot

=

We know entropy(H) is average information

contenti.e. Iavg =

=

m

i 1 pi)piNlog2(1/

N

Itot

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H = bits/Symbol

If Symbol rate(message rate) is Rsym then

information rate isRinfo = H X R sym

Bits/sec bits/ symbol symbol/sec

=

m

i 1pi)piNlog2(1/

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Baseband Digital Communication System

It refers to a system in which tx and Rx ofdigital Signal over a band limited channelwithout carrier modulation.

Since baseband Signal have sizablepower at low frequency, they are suitablefor transmission over a pair of cable,

coaxial cable, optical fiber etc.

Before Transmission some transformationin data waveform is needed.

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Baseband data communication system usingPAM have following Blocks

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The output of pulse generator is

x(t) = bk Pg(t- KTb)

Where Pg(t) is a basic Pulse whoseamplitude bK depends upon the input dataSequence

bk = +b if Kth bit is 1

bk = -b if Kth bit is 0

= -k

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The signal from pulse generator is thenpassed through Tx filter, Rx filter, The

channel with noise added at channel.

So the output of Receiver filter is

x(t) = Kc bk Pg(t- KTb-d) no(t)

Where,Kc is cumulative response of Tx filter, Channel and

Rx. Filter such that

Bk = Kc bK ; Pr(t) = Kc Pg(t) ;

dis time delay

=K

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For Simplicity , Let us take

no(t) = 0 and d =0

Therefore

x(t) = Kc bk Pg(t- KTb)

y(t) is then passed through decisionmaking device.

If y(t) is above threshold output is 1 else 0.

=k

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At Decision Making Instant

y(t=mTb) = Bk Pr(mTb- KTb)

= Bm + Pr(mTb- KTb)

The first Term is mth decoded bit and the secondterm represents the residue while decoding mth

bit due to all other transmitted bit called ISI

=k

=mk

k

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Design Criteria

ISI arises due to dispersion of pulse shapeby the filter and channel.

The major task of System Designer is to

optimally design transmitting and receivingfilter and the shape of basic pulse tominimize ISI.

Parameter Known to DesignerInput Bit StreamChannel Characteristics

Statistical Characteristics of Noise

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Selection Of Optimum HT(f)

Hc(f) Pg(t)

Criteria

Maximize data rate Optimize Bandwidth

Minimum Error Rate

Minimum Transmission Power

Maximize SNR

Simple Circuit Design

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Inter Symbol Interference (ISI)

Spreading of pulse beyond its interval Tb will

cause it to interfere with neighboring Pulse. Thisis called ISI

We need to Transmit a pulse every Time at

interval of Tb.We are considering a time limited pulse and

such pulse are not band limited.

Parts of their spectra are suppressed by bandlimited channel.

This cause pulse distortion and ConsequentlyISI

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Band Limited pulse can not be time

limited.

Pulse amplitude can be detected correctly

despite pulse spreading if there is no ISI atthe decision making Instant

This can be accomplished by properlyshaped band limited Pulse

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ISI Removing Methods

1. Nyquist Method

i. Ideal Solution (Zero ISI)

ii. Raised Cosine Spectrum Method

1. Correlative Coding (Partial ResponseSignaling)

i. Duobinary Signalling

ii. Modified Duobinary Signalling

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Nyquist Pulse Shaping Cr i t er ia / Nyquist Condi t ion for Zero I SI

We have;

= Bm + Pr(mTb- KTb)

For Zero ISI,

Pr [(m- K)Tb] = 1 k = m= 0 k = m

=mk

k

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i.e.

Pr (nTb) = 1 n = 0= 0 n = 0

In General

P(t) = 1 t = 0

= 0 t = nTb

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A Pulse Sat is fy ing above Cr i t er ia c auses Zero ISI at

Signal ing Instant

Sampling Instants

ISI occurs but,

NO ISI is present atthe sampling instants

0 Tb 2Tb

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Transmission of Rb bits/Sec require aminimum bandwidth of Rb/2 Hz.

The Pulse satisfying Nyquist criteria andhas B/W of Rb/2 Hz is

p(t) = sinc (2B0t) =Where,

B0 = 1 = Rb is absolute minimum B/W required

2Tb 2 for zero ISI

)2()2(

Bot

BotSin

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Although Nyquist Pulse Shaping shapes for p(t)achieves economy in B/W, it suffers from two majorproblem.

To generate a sync function Signal must

be passed through a Filter having flat

response from -B0 to +B0 and zero

elsewhere. This response are physicallyUnrealizable.

p(t) decreases too slowly at the

Rate of 1/t. If the nominal data rate of Rb bits/secrequired for this scheme deviates a little, the pulseamplitude will not vanish at other pulse centre. TheCumulative interference at any pulse centre from all theremaining pulse will be very high.

0f

P(f)

1/2B0

B0-B0

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Raised Cosine Approximation

The overall Frequency response P(f) decreases

towards zero gradually rather then abrupt. In particular p(f) consist of flat portion and roll off

portion that has the form of raised cosine function

as1/2B0 for 0 f f1

P(f)= 1+ cos (f- f1) for f1 f 2B0-f12B0-2f1

0 for f> 2B0-f129@Samrat Subedi


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Here, f1 and B0 are related as

= 1 f1/B0 , is roll off factor

For = 0;f1 = B0 is absolute minimum bandwidth required

for zero ISI (Nyquist B/W)

For =1 and taking Inverse Fourier Transform

P(t) = Sinc( 4B0t)

1- 16B02t2

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.

1 f/B 0

2B0P(f) =0

= 0.25

= 0.50= 0.75= 1.00

23/2

t

0

p(t)

-TbTb

1

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P(t)

1

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Characteristics of p(f)

For = 0, it is ideal case For = 0.5 & 1 p(f) cut off gradually and

hence it is physically implementable

Value of p(f) at f/B0 =1 will be half of itsmaximum value for any

P(f) is real i.e. non negative

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Characteristics of p(t)

At t= Tb/2, p(t) has half amplitude for =1 additional zero crossing at

+3Tb/2, +5Tb/2 in addition to Tb, 2Tb,.

These are used for generating timingsignal for Synchronization.

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Transmission Bandwidth Consideration

B/WPAM = B0 (B0= 1/2Tb)

Ideal Case For Raised Cosine

We have = 1 f1/B0

B= 2B0- f1= 2B0 B0(1- )

For =1

B/W = 2B0In general

B/WPAM =B0(1+ ) = Rb /2(1+ )35@Samrat Subedi


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Take an Example of T1 System

24 independent voice input based on 8 bit

PCM Word

Tb= 0.647 sec and Rb= 1.544 Mbps

Ideal solution B0= 1/2Tb = 772 KHz

Raised Cosine B= 1/Tb = 1.544 MHz

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Correlative-Level Coding

For Rb data Rate, the absolute min B/W as perNyquist Criteria is Rb/2 and with raised Cosine

minimum bandwidth is Rb.

By adding ISI to the transmitted signal in a

controlled manner, it is possible to achieve a signaling

rate equal to the Nyquist rate of 2Wsymbols/sec in a

channel of bandwidth WHz.

Correlative-level coding may be regarded as apractical method of achieving the theoretical

maximum signaling rate of 2 Wsymbols/sec in a

bandwidth ofWHz

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Data Rate Bandwidth ISI Condition

Rb B0(Rb/2)Zero Ideal

Rb 2B0(Rb) Zero Raised Cosine ( =1 )Rb B0(Rb/2)

Controlled ISI Correlative Coding

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DUOBINARY CODING


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DUOBINARY CODING

Duobinary signaling implies doubling the transmission

capacity in a straight binary system. This particular formof correlative-level coding is also called class I partialresponse.

Consider a binary input sequence {bk} applied to a pulse-

amplitude modulator to produce a two-level sequence{ak} :

+A if symbol bk is 1

ak = {-A if symbol bk is 0

When this sequence is applied to a duobinary encoder, it is

converted into a three-level output, namely, -2A, 0, and +2A. 39@Samrat Subedi


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We may express the duobinary coder output ckas the

sum of the present input pulse akand its previous value

ak-1, as shown by

ck = ak + ak-1

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+2A If ak & ak-1 are both 1

Ck = 0 If ak & ak-1 are different-2A If ak & ak-1 are both 0

And We have

HI(f) = HNyquist(f) . HDB(f)

For Tb seconds delay element having frequency response

exp(-j2fTb), the frequency response of the delay-line is

1 + exp(-j2fTb)

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Hence, the overall frequency response of this filterconnected in cascade with an ideal Nyquist channel is

HI(f) =HNyquist(f) [l + exp(-j2fTb)]

=HNyquist(f) [exp(jfTb) + exp(-jfTb)] exp(-jfTb)

= 2HNyquist(f) cos(fTb) exp(-jfTb)

For an ideal Nyquist channel of bandwidth W= 1/2Tb, we have

1 |f| < 1/2Tb (= B0)HNyquist(f) =

0 elsewhere

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The overall frequency response of the duobinarysignaling scheme has the form of a half-cycle cosinefunction, as

2cos(fTb) exp(-jfTb) |f| < 1/2Tb (=B0)

HI(f) =

0 otherwise

AND

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Frequency response of the duobinary conversionfilter. (a) Magnitude response. (b) Phase response.

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Ifckis received without error and if the previousestimate ak-1 corresponds to a correct decision, thenthe current estimate akwill be correct too.

A major drawback of this detection procedure is thatonce errors are made, they tend to propagate throughthe output because a decision on the current akdepends on the correctness of the decision made onthe previous a

k-1. 45@Samrat Subedi


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A means to avoid the error-propagation is to useprecoding before the duobinary coding

The precoding operation performed on the binary datasequence {bk} converts it into another binary sequence{dk} defined by

dk = bkdk-1 Ck = dk + dk-1

+A If it is 1dk = {

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dK

If kth input bit bk=0Then dk=dk-1

Soif b

k

=0 Ck

= +2A or -2Aif bk =1 bk is complement of bk-1 ck=0

0 if data symbol bkis 1ck =

+ 2A if data symbol bkis 047@Samrat Subedi


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Take an Example

Input bit Sequence{bk}

0 0 1 0 1 0

dk1

(let)1 1 0 0 1 1

Representation of dk +A +A +A -A -A +A +A

Output of DB

Encoder{ck=dk+dk-1}

+2A +2A 0 -2A 0 +2

ADecoded Bit 0 0 1 0 1 0

Put dk-1 =0 anddo again

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MODIFIED DUOBINARY SIGNALING


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MODIFIED DUOBINARY SIGNALING

If we observe the frequency response of duobinary

signalling, we see that it has non zero frequency responseat t=0

This is not Suitable for circuitary with no DC path. Hencewe use Modified DB.

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The overall frequency response of the delay-line


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The overall frequency response of the delay linefilter connected with an ideal Nyquist channel, as inFigure 4.16, is given by

HIV(f) =HNyquist(f) [l - exp(-j4fTb)]

= 2j HNyquist(f) sin(2fTb) exp(-j2fTb)

2jsin(2fTb) exp(-j2fTb), |f| < 1/2TbHIV(f) = {

0, elsewhere

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Baseband M-ary Signalling


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Baseband M-ary Signalling

In a basebandM-ary PAM system, the pulse-amplitude

modulator produces one ofMpossible amplitude levelswithM> 2.

A signal alphabet inM-ary PAM system containsMequally likely and statistically independent symbols, withsymbol duration T seconds.

The signaling rate 1/T is expressed in symbols persecond, or bauds.

This form of pulse modulation is illustrated in Figure4.20a for the case of quaternary (M= 4) system and thebinary data sequence 0010110111.

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Signaling Rate and B/W Requirement:


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Signaling Rate and B/W Requirement:-

Let Rs(Symbol per sec or Baud) be the Symbol rate

emitted by the source If M symbols are emitted that are equiprobable and

spastically independent

Entropy (H) = pilog

2(1/p

i)

Since M are equiprobable pi = 1/MTherefore

H = log2MAnd

Information Rate(Rinfo) =Rs x H Rslog2M bps

Absolute B/W required is Rs/2 --------

=

m

i 1

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For Binary

M=2

SO Rinfo = Rb = Rslog22= Rs bps

B/W = Rs/2 ---------

From Equation 1 & 2 we see

B/W requirement for Mary andbinary system is same

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Eye Diagram/ Eye Pattern


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Eye Diagram/ Eye Pattern

When the Sequence is transmitted over abaseband binary data transmissionsystem, the signal obtained at the outputYr(t) is a continuous time signal

1 1110 0

bT

bT2

bT3

bT40 bT5

ResultantChannel OutputWaveform

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If the received Signal is cut in each interval Tb and placeover one another, the diagram obtained is known as eyeDiagram as it looks like an eye.

Data 1

bT 0 bT0bT bT

Data 0

bT 0 bT0bT bT

Channel Input

Pulse width Tb

Channel Output

Pulse width Tb

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If the received Signal is cut in each interval Tb and placeover one another, the diagram obtained is known as eyeDiagram as it looks like an eye.

This pattern can also be obtained on CRO if Yr(t) is appliedto vertical input and a saw tooth signal with duration T=Tb

at horizontal input59@Samrat Subedi

Information Of Eye Diagram


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Information Of Eye Diagram

60@S S b di

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