Chapter 7 Fundamentals of Digital Transmission

Baseband Transmission (Line codes) ON-OFF or 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 On-Off RZ Better synch., at extra bandwidth Bi-Polar Better synch., at same bandwidth

More Line Codes Polar RZ Perfect synch 3 levels Manchester (Bi-Phase) Perfect Synch. 2 levels

Spectra of Some Line Codes

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.

The Sinc Pulse 1 TbTb 2T b t 3T b 4T b 5T b 6T b -6T b -5T b -4T b -3T b -2T b -T b f 1/(2T b ) -1/(2T b ) p(t) P(f) Note that such pulse has a bandwidth of R b /2 Hz. Therefore, the minimum channel bandwidth required for transmitting pulses at a rate of R b pulses/sec is R b /2 Hz

Zero ISI

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 R b /2.

Raised-Cosine Pulses where  b is 2  R b and  x is the excess bandwidth. It defines how much bandwidth required above the minimum bandwidth of a sinc pulse, where

Spectrum of Raised-Cosine Pulses

Extremes of Raised-Cosine Spectra

Raised-Cosine Pulses

Bandwidth Requirement of Passband Transmission  Passband transmission requires double the bandwidth of baseband transmission.  Therefore, the minimum bandwidth required to transmit R b pulses/sec using carrier modulation is R b Hz.

Transmission rates of Typical Services  Speech  Audio  Fax  Coloured Image  Video

Speech (PCM)  B = 3.4 kHz  R s = 8000 samples/sec  Encoding = 8 bits/sample  Transmission rate = 64 kbps  Required bandwidth (passband) = 64 kHz  One hour of speech = 64000x3600 = Mb

Audio  B = kHz  R s = samples/sec  Encoding = 16 bits/sample  Stereo type = 2 channels  Transmission rate = 1.4 Mbps

Fax  Resolution 200x100 pixels/square inch  1 bit/pixel (white or black)  A4 Paper size = 8x12 inch  Total size = 1.92 Mb = 240 KB  Over 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 square  8 bits/pixel  3 colours/photo  A 8x10 inch picture is represented by Mb = 38.4 MB !

Video (moving pictures)  Size of still pictures  15 frames/sec  307 Mb/frame x 15 frames/sec = 4605 Mbps =4.6 Gbps !!

Solutions  Compression reduces data size  M-ary communication Expands 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 rate  We need to generate 2 m 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 / pulse 2×B bits per second 8 “levels” 3 bits / pulse 3 × B bits per second 2 “levels” 1 bits / pulse B bits per second

4 signal levels8 signal levels typical noise Same-maximum-power Scenario

signal noise signal + noise signal noise signal + noise High SNR Low SNR SNR = Average Signal Power Average Noise Power t t t t t t

Same-BER Scenario  Average power for binary case: ½ A 2 + ½ A 2 = A 2  Average power for 4-ary case: ¼ (9 A 2 + A 2 + A A 2 ) = 5 A 2

Carrier Modulation of Digital Signals Information T 2T2T 3T3T 4T4T5T5T 6T6T Amplitude Shift Keying +1 Frequency Shift Keying +1 Phase Shift Keying 0 T 2T2T 3T3T 4T4T5T5T 6T6T 0 T 2T2T 3T3T 4T4T5T5T 6T6T t t t

Spectrum

TDM for Digital

Digital Hierarchy

Multiple Phases (MPSK) 4 “phase” 2 bits / pulse 2 × B bits per second 8 “phases” 3 bits / pulse 3 × B bits per second

Quadrature Amplitude Modulation (QAM) AkAk BkBk 16 “levels” or pulses 4 bits / pulse 4xB bits per second AkAk BkBk 4 “levels”or pulses 2 bits / pulse 2xB bits per second QAM 16 QAM

The Modulation Process of QAM AkAk x cos(  c t) Y i (t) = A k cos(  c t) BkBk x sin(  c t) Y q (t) = B k sin(  c t) +Y(t) Modulate cos(  c t) and sin (  c t) by multiplying them by A k and B k respectively:

QAM Demodulation Y(t) x 2cos(  c t) 2cos 2 (  c t)+2B k cos(  c t)sin(  c t) = A k {1 + cos(2  c t)}+B k {0 + sin(2  c t)} LPF AkAk x 2sin(  c t) 2B k sin 2 (  c t)+2A k cos(  c t)sin(  c t) = B k {1 - cos(2  c t)}+A k {0 + sin(2  c t)} LPF BkBk