ECE 4710: Lecture #11 1 Binary vs. Multi-Level 1 0 0 1 0 0 1 1 8-Bit Message: 10010011 t 5 V T s = 1...

ECE 4710: Lecture #11 1

Binary vs. Multi-Level

1 0 0 1 0 0 1 1

8-Bit Message: 10010011

t

5 V

Ts = 1 msec T0 = 8 Ts = 8 msec

R = (8/8 ms) = 1 kbps FNBW = 1/Ts = 1 kHz

Ts

T0

Binary Waveform




t

5 V3 V1 V

Ts = 2 msec T0 = 4 Ts = 8 msec D = (4/8 ms) = 500 sps

R = (8/8 ms) = 1 kbps FNBW = 1/Ts = 500 Hz

T

s

T0

L=4 Multi-Level WaveformSymbol

Key

00 = 0 V

01 = 1 V

10 = 3 V

11 = 5 V

1 00 1

0 0 1 1

Same Data Rate, One-Half BW




t

5 V3 V1 V

Ts = 1 msec T0 = 4 T = 4 msec D = (4/4 ms) = 1 ksps

R = (8/4 ms) = 2 kbps FNBW = 1/Ts = 1 kHz

Ts

T0

L=4 Multi-Level WaveformSymbol

Key

00 = 0 V

01 = 1 V

10 = 3 V

11 = 5 V

1 0 0 1 0 0 1 1

Same BW, 2 Data Rate



Reduced BW OR increased data rate significant advantage for multi-level signal Why not do this 100% of the time?? Why not increase to L = 8 or L = 16 levels??

Primary disadvantage: for same S/N ratio a multi-level signal will have higher probability of bit errors compared to binary signal Reduced ability to accurately discriminate between

different signal levels



1 0 0 1 0 0 1 1 t

5 VT

VS.5 V3 V1 V

T1 0

0 10 0 1 1

t


Channel Capacity

Shannon’s capacity formula

Use multi-level signal to decrease BW required S/N increases to maintain same capacity for same BER

User error coding to lower S/N requirement for same BER required bandwidth increases to handle additional coding bits while maintaining same capacity (data rate)

BW for S/N tradeoff is ** fundamental ** for all communication systems

)bps(1log2

N

SBC


Binary Line Coding

Line Codes : Binary 1’s and 0’s represented by a variety of serial-bit signaling formats

Two Major Categories

Non Return-to-Zero (NRZ)

» Signal waveform amplitude stays at one constant value for full duration of bit period

Return-to-Zero (RZ)

» Signal waveform amplitude returns to zero volt level for a portion of the bit period zero level portion is usually 0.5 Tb for “1”

t

A

Tb

01 0 1

t

A

Tb

0

1 0 1


Binary Line Coding

Four major sub-classifications based on the rules used to assign voltage levels to binary data (1/0) Unipolar Signaling : positive signaling has “1” = +A volts and “0” = 0

volts» Also called “On/Off Keying”

Polar Signaling : “1”= + A volts and “0” = A volts Bipolar (AMI) Signaling : binary “1” represented by alternating positive

and negative values while binary “0” is represented by constant zero volt level» Also called “Alternate Mark Inversion” = AMI

Manchester Signaling : “1” represented by positive/negative cycle in one bit period while “0” represented by negative/positive cycle» Also called “Split Phase Coding”


Common Line Codes


Common Line Codes

1 1 0 1 0 0 1


Shorthand Names

Book adopts shorthand naming convention that is common in industry Unipolar NRZ Unipolar Polar NRZ Polar Bipolar RZ Bipolar

Bipolar vs. Polar Polar NRZ is sometimes called Bipolar NRZ (Bipolar)

» Common in satellite communications» Book does NOT use this convention

Book uses telephone industry convention» T1 Bipolar RZ = Bipolar


Properties

Various line codes have advantages and disadvantages Signaling line code selected based on properties and

intended application Important properties

Self-synchronization» Enough timing information built-into code so bit synchronizers in

Rx can be designed to extract timing/clock signal from the code itself Clock signal needed to control sampling trigger in receiver

» Long series of binary 1’s or 0’s should NOT cause problem in recovery of clock signal


Properties

Important properties (continued) Probability of Bit Error

» Rx designed so that BER is low when signal is corrupted by ISI or channel noise

Spectrum/Bandwidth» Signal BW should be small relative to channel BW so no ISI» Spectrum suitable for baseband channels with AC or DC coupling

AC coupled channels like phone lines require line code signal PSD to have little or no energy at DC (f =0) If PSD has significant energy at DC then AC channel will

significantly attenuate signal, distort signal, and cause large amount of ISI


Properties

Important properties (continued) Error Detection

» Some line codes can have simple error detection built in» Channel codecs should be easy to implement for chosen line code

Transparency» Data protocol and line code designed so that every possible data

sequence is faithfully and transparently received» Code is NOT transparent if certain data sequences are reserved

for control purposesRandom data might accidentally generate control sequence

» Code is NOT transparent if long string of 1’s or 0’s result in loss of synchronization signalBipolar format is not transparent since long string of 0’s will

cause loss of clocking signal


Spectrum

PSD for deterministic waveform given previously as

Stochastic approach finds PSD for line code with random data sequence more realistic

Digital signal represented by

)( )( e.g. )]([|)(|

lim)(2

fRRT

fWf www

T

Tw PP

nn

ssn

a

T

tf

nTtfats

data random ofset }{

period symbol

shape pulse)(

where)()(


Spectrum

For unipolar NRZ line code :

General expression for PSD of digital signal is

F(f ) is FT of f (t) and R(k) is autocorrelation of the binary data given by

"0"for V 0

"1"for V and )(

n

n

b a

Aa

Tt

tf

k

Tfkj

ss

sekRTfF

fP 22

)(|)(|

)(

product ofy probabilit

symbol )( of level voltage

symbol of level voltage

th

th

th

where)()(1

knanaiiP

knkna

nnaI

iiiknn PaakR


Polar NRZ line code Possible levels are +A and A If data are independent (uncorrelated from bit to bit)

FT of pulse shape is

Polar NRZ Spectrum

!!041

41

)(41

)(41

)()(

0

21

21

)()0(

0

24

1

2

222

1

2

AAAAAAPaakR

k

AAAPaaR

k

iiknn

iinn

b

bb

b Tf

TfTfF

Tt

tfsin

)()(


Polar NRZ Spectrum

Substituting into2

222 sin

)(|)(|

)(

b

bb

k

Tfkj

ss Tf

TfTAekR

TfF

fP s

Normalized A = 1

A2

ECE 4710: Lecture #11 1 Binary vs. Multi-Level 1 0 0 1 0 0 1 1 8-Bit Message: 10010011 t 5 V T s = 1...

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Transcript of ECE 4710: Lecture #11 1 Binary vs. Multi-Level 1 0 0 1 0 0 1 1 8-Bit Message: 10010011 t 5 V T s = 1...