Chapter 7

Fundamentals of Digital Transmission

Baseband Transmission (Line codes)0 1 0 1 1 1 0 0 1Bit Value

5 V

0 V

0 1 0 1 1 1 0 0 1Bit Value

–5 V

0 V

5 V

ON-OFFor

Unipolar(NRZ)

Non-Return-to-Zero

Polar(NRZ)

Performance Criteria of Line Codes

Zero DC value Inherent Bit-Synchronization Rich in transitions

Average Transmitted Power For a given Bit Error Rate (BER)

Spectral Efficiency (Bandwidth) Inversely proportional to pulse width.

Comparison Between On-Off and Polar

Zero DC value: Polar is better.

Bandwidth: Comparable

Power: BER is proportional to the difference between the two levels For the same difference between the two levels, Polar

consumes half the power of on-off scheme.

Bit Synchronization: Both are poor (think of long sequence of same bit)

More Line Codes

0 1 0 1 1 1 0 0 1Bit Value

5 V

0 V

–5 V

0 1 0 1 1 1 0 0 1Bit Value

5 V

0 V

On-Off RZBetter synch., at extra bandwidth

Bi-PolarBetter synch., at same bandwidth

More Line Codes0 1 0 1 1 1 0 0 1Bit Value

5 V

0 V

–5 V

0 1 0 1 1 1 0 0 1Bit Value

5 V

0V or -5V

Polar RZPerfect synch3 levels

Manchester(Bi-Phase)Perfect Synch.2 levels

Spectra of Some Line Codes

-0.2

0

0.2

0.4

0.6

0.8

1

1.2

0

0.2

0.4

0.6

0.8 1

1.2

1.4

1.6

1.8 2

fT

pow

er d

ensi

ty

On-Of f (NRZ)

Bipolar (NRZ)

Manchester

Pulse Shaping

The line codes presented above have been demonstrated using (rectangular) pulses.

There are two problems in transmitting such pulses: They require infinite bandwidth. When transmitted over bandlimited channels

become time unlimited on the other side, and spread over adjacent symbols, resulting in Inter-Symbol-Interference (ISI).

Nyquist-Criterion for Zero ISI

Use a pulse that has the following characteristics

One such pulse is the sinc function.

1 0( )

0 , 2 , 3 ,b b b

tp t

t T T T

The Sinc Pulse

1

Tb 2Tb

t3Tb 4Tb 5Tb 6Tb

-6Tb -5Tb -4Tb -3Tb -2Tb -Tb

f1/(2Tb)-1/(2Tb)

p(t)

P(f)Note that such pulse has a bandwidth of Rb/2 Hz.Therefore, the minimum channel bandwidthrequired for transmitting pulses at a rate of Rb pulses/sec is Rb/2 Hz

Zero ISI

-1

0

1

-2 -1 0 1 2 3 4

More on Pulse Shaping

The sinc pulse has the minimum bandwidth among pulses satisfying Nyquist criterion.

However, the sinc pulse is not fast decaying; Misalignment in sampling results in significant ISI. Requires long delays for realization.

There is a set of pulses that satisfy the Nyquist criterion and decay at a faster rate. However, they require bandwidth more than Rb/2.

Raised-Cosine Pulses

where b is 2Rb and x is the excess bandwidth. It defines how much bandwidth required above the minimum bandwidth of a sinc pulse, where

/ 211 sin

2 2 2

( ) 02

12

b bx

x

bx

bx

P

02

bx

Spectrum of Raised-Cosine Pulses

b/2=/Tb

P()

xx

b/2 + xb/2 – x

Extremes of Raised-Cosine Spectra

b/2=/Tb

P()

x = 0“Sinc”

b=2/Tb

x = b/2

b/2b/2

2Excess Bandwidth

Minimum Bandwidth / 2x x

b b

r

Raised-Cosine Pulses

Bandwidth Requirement of Passband Transmission

Passband transmission requires double the bandwidth of baseband transmission.

Therefore, the minimum bandwidth required to transmit Rb pulses/sec using carrier modulation is Rb Hz.

Transmission rates of Typical Services

SpeechAudioFaxColoured ImageVideo

Speech (PCM)B = 3.4 kHzRs = 8000 samples/secEncoding = 8 bits/sampleTransmission rate = 64 kbpsRequired bandwidth (passband) = 64 kHzOne hour of speech = 64000x3600 = 230.4 Mb

Audio

B = 16-24 kHzRs = 44 000 samples/sec

Encoding = 16 bits/sampleStereo type = 2 channelsTransmission rate = 1.4 Mbps

Fax

Resolution 200x100 pixels/square inch1 bit/pixel (white or black)A4 Paper size = 8x12 inchTotal size = 1.92 Mb = 240 KBOver a basic telephone channel (3.4 kHz, baseband) it

takes around 4.7 minutes to send one page.

Colour Image (still pictures)

Resolution 400x400 pixels/inch square8 bits/pixel3 colours/photoA 8x10 inch picture is represented by

307.2 Mb = 38.4 MB !

Video (moving pictures)

Size of still pictures15 frames/sec307 Mb/frame x 15 frames/sec = 4605 Mbps =4.6 Gbps !!

Solutions

Compression reduces data size

M-ary communicationExpands channel ability to carry information

M-ary Transmission

In the binary case one pulse carries one bit.Let each pulse carry (represent) m bits.Bit rate becomes m multiples of pulse rateWe need to generate 2m different pulses.They can be generated based on:

Multiple Amplitudes (baseband and passband)Multiple Phases (passband)Multiple frequencies (passband) Some combination (Amplitude and Phase).

Signal Constellation

Signal constellation is a convenient way of representing transmitted pulses.

Each pulse is represented by a point in a 2-dimensional space.

The square of the distance to the origin represents the pulse energy.

The received signals form clouds around the transmitted pulses.

A received points is decoded to the closest pulse point.

Multiple Amplitudes (PAM)

4 “levels”2 bits / pulse2×B bits per second

8 “levels”3 bits / pulse3×B bits per second

2 “levels”1 bits / pulseB bits per second

0 1 00 10 11 01 000100110 010 011111101 001

4 signal levels 8 signal levels

typical noise

Same-maximum-power Scenario

signal noise signal + noise

signal noise signal + noise

HighSNR

LowSNR

SNR = Average Signal Power

Average Noise Power

t t t

t t t

Same-BER Scenario

Average power for binary case:½ A2 + ½ A2 = A2

Average power for 4-ary case:¼ (9 A2 + A2 + A2 + 9 A2 ) = 5 A2

Carrier Modulation of Digital SignalsInformation 1 1 1 10 0

+1

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

T

AmplitudeShift

Keying

+1

-1

FrequencyShift

Keying

+1

-1

PhaseShift

Keying

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

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

t

t

t

Spectrum

TDM for Digital

Digital Hierarchy

Multiple Phases (MPSK)

4 “phase”2 bits / pulse2×B bits per second

8 “phases”3 bits / pulse3×B bits per second

Quadrature Amplitude Modulation (QAM)

Ak

Bk

16 “levels” or pulses4 bits / pulse4xB bits per second

Ak

Bk

4 “levels”or pulses2 bits / pulse2xB bits per second

QAM 16 QAM

The Modulation Process of QAM

Akx

cos(c t)

Yi(t) = Ak cos(c t)

Bkx

sin(c t)

Yq(t) = Bk sin(c t)

+ Y(t)

Modulate cos(ct) and sin (ct) by multiplying them by Ak and Bk respectively:

QAM Demodulation

Y(t) x

2cos(c t)2cos2(ct)+2Bk cos(ct)sin(ct) = Ak {1 + cos(2ct)}+Bk {0 + sin(2ct)}

LPF Ak

x

2sin(c t)2Bk sin2(ct)+2Ak cos(ct)sin(ct) = Bk {1 - cos(2ct)}+Ak {0 + sin(2ct)}

LPF Bk