lec6_datatransmission

8/12/2019 Lec6_DataTransmission

1/34

BEIT, 5th Semester 1/20

Data Transmission andLine Coding

Ms. Tuhina SamantaLecturer

Dept of I.T.


2/34

2/32BEIT, 5th Semester

Digital Data, Digital Signal

Digital signal

Discrete, discontinuous voltage pulses

Each pulse is a signal elementBinary data encoded into signal elements


3/34

BEIT, 5th Semester 3/20

Line Coding


4/34


Line Coding

UnipolarAll signal elements have same sign

Polar

One logic state represented by positive voltagethe other by negative voltage

Data rate

Rate of data transmission in bits per second Duration or length of a bit

Time taken for transmitter to emit the bit


5/34


Terms

Modulation rate

Rate at which the signal level changes

Measured in baud = signal elements per second

Mark and Space

Binary 1 and Binary 0 respectively


6/34


Requirements of Encoding Schemes

Signal Spectrum

Lack of high frequencies reduces required bandwidth

Lack of dc component allows ac coupling viatransformer, providing isolation

Concentrate power in the middle of the bandwidth

Clocking Synchronizing transmitter and receiver

External clock

Sync mechanism based on signal


7/34


Requirements of Encoding Schemes

Error detectionCan be built in to signal encoding

Signal interference and noise immunity

Transparency to 1s and 0s.

Some codes are better than others

Cost and complexity

Higher signal rate (& thus data rate) lead tohigher costs


8/34


Encoding Schemes (Line Coding)

Nonreturn to Zero-Level (NRZ-L)

Nonreturn to Zero Inverted (NRZI)

Bipolar -AMI

Pseudoternary

Manchester

Differential Manchester B8ZS

HDB3


9/34


Nonreturn to Zero-Level (NRZ-L)

Two different voltages for 0 and 1 bits

Voltage constant during bit interval

no transition i.e. no return to zero voltage

e.g. Absence of voltage for zero, constant

positive voltage for one

More often, negative voltage for one value andpositive for the other

This is NRZ-L


10/34


NRZ


11/34


NRZ pros and cons

Pros

Easy to engineer

Make good use of bandwidth Cons

dc component

Lack of synchronization capability

Used for magnetic recording

Not often used for signal transmission


12/34


Some Line Codes

1 1 0 0 1 0

ON-OFF(RZ)

POLAR (RZ)

BIPOLAR (RZ)


13/34


Power Spectral Density (Polar-RZ)

Let R()time correlation function of pulse

train.

akbe pulse amplitude. R0time average of square of the pulse

amplitude, (correlation of akwith ak)

ltN(1/N)kak2 =ltN(N)/(N) = 1


14/34


15/34


PSD expression

PSD Sx() is Four ier transform of AutocorrelationRx()

assuming R() is even function of .

PSD of the resulting line code is,

))cos(2(1)(1

0

n

bn

b

x TnRR

TS

))cos(2(|)(|

)(|)(|)(10

22

n bn

b

xy TnRRT

PSPS

n

bn

b

x nTR

TR )(

1)(


16/34


PSD of Polar Signaling

For polar coding let p(t) be a rectangular pulse

of width Tb/2.

Then,

Therefore,

PSD of polar signaling is,

)2

()

2/

()(

bb T

trect

T

trecttp

)4

(sin2

)( bb Tc

TP

)4

(sin4

|)(||)(|)( 2

2

0

2bb

bb

y

Tc

T

T

PR

T

PS

Tb

Tb/2.p(t)


17/34


Properties of Polar (RZ) Coding If Tbis pulse duration with pulse

width Tb/2 then bandwidth is 2Rb. With pulse width Tbbandwidth is

Rb. Power Efficient.

Clock extraction is easier.

Essential bandwidth requirement2Rbnot bandwidth efficient

No error detection-correctioncapabilities

At = 0, PSD is present, i.e. nod.c. null

-4Rb -2Rb 0 2Rb 4Rb

PSD


18/34


BIPOLAR (RZ)

1 1 0 0 1 0


19/34


Bipolar Coding

R0= ltN(1/N) [((N/2)(1)2+ ((N/2)(0)) ]

=

R1= ltN(1/N) [((N/4) (-1)) + ((3N/4)(0))]= -1/4

R2 = ltN(1/N) [ (N/8)(1) +(N/8)(-1) + (3N/4)(0)]

= 0 Remaining terms are zero. Rn= 0 for n > 1


20/34


PSD Expression

For half width pulse, PSD of bipolar line coding is,

)2

(sin)4

(sin4

)2

(sin2

|)(|]cos1[

2

|)(|

...]0)cos(22

1[

|)(|

)]cos(2[|)(|

)(

22

222

1

2

1

0

2

bbb

b

bb

b

b

b

b

n

n

b

y

TTc

T

T

T

PT

T

P

TRT

P

TnRRT

PS


21/34


Properties of Bipolar (RZ) Coding

D.C. null exists

Lesser bandwidth

Single error

detection possible

Twice as much

power as polarsignal

Not transparent tolong 0s and 1s

0 Rb 2Rb

PSD


22/34


23/34


High Density Bipolar Coding

HDBN coding with N = 1,2,3...Popular choice of N is 3; 000V B00V;

B=1 confirms bipolar rule; V=1 violates bipolar rule


24/34


Multilevel Binary

Use more than two levels


25/34


Bipolar-AMI and Pseudoternary


26/34


27/34


Exercise

Draw Manchester and Differential Manchester

line coding

Calculate PSD for pseudoternary and on-offsignaling


28/34


29/34


Rb-Rb

P()

-3/Rb -2/Rb -1/Rb 0 1/Rb 2/Rb

1

p(t)

1 for t = 0

p(t) = sinc(Rbt) =0 for t = nTb

Pulse satisfying Nyquist Criterion

tf

)2

(1

)(bb R

rectR

P


30/34


Pulse satisfying Nyquist Criterion

Sincpulse decays too slowly (as 1/t)

Pulse satisfying Nyquist criteria should

decay faster than 1/t Such pulse requires bandwidth kRb/2, with

1k 2


31/34


Roll-off factor

The b.w. of the spectrumshown aside is b/2 + x

Roll-off factor (r) = excessbandwidth/theoretical

minimum bandwidth= 2x/b

0r1

Let theoretical minimumb.w. Rb/2 Hz

Then, BT= (Rb/2) + (rRb/2)

Tb

Tb /2

b/2x

P()

xb


32/34


Raised Cosine function:

The characteristic is also

called full-cosine-rolloffcharacteristic.

Decays as 1/t3

The time domain representation

is (taking inverse Fourier Transform),

)(sin41

)cos()(

22 tRc

tR

tRRtp b

b

b

b

0

1

|P()|

x= 0

x= b/2

t

Tb

2Tb-T

b

-

2Tb

p(t)

[Ref. B.P.Lathi, Ch. 7]

)4

()4

(cos

)4()2cos1(2

1)(

2

bb

bb

Rrect

R

RrectRP

x= b/4


33/34

BEIT, 5th Semester

Problems

1) A baseband signal of frequency 5kHz.is applied to aproduct modulator together with a carrier wave of1MHz. The modulator output is next applied to aresonant circuit tuned to 50Hz. Determine the

modulated output signal.

2) For tone modulation with amplitude modulationindex = 1 and 0.5 and baseband signal m(t) =

Bcos(mt) draw the waveforms of amplitudemodulated signals both in time and frequencydomain.


34/34

BEIT 5th Semester

lec6_datatransmission

Documents