Fundamentals of Digital Transmission 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 Bi-Polar Better synch., at extra bandwidth at same bandwidth

More Line Codes Polar RZ Manchester (Bi-Phase) Perfect synch 3 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 t p(t) P(f) 1 Tb 2Tb t 3Tb 4Tb 5Tb 6Tb -6Tb -5Tb -4Tb -3Tb -2Tb -Tb P(f) Note that such pulse has a bandwidth of Rb/2 Hz. Therefore, the minimum channel bandwidth required for transmitting pulses at a rate of Rb pulses/sec is Rb/2 Hz f -1/(2Tb) 1/(2Tb)

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 Rb/2.

Raised-Cosine Pulses where b is 2Rb 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 Rb pulses/sec using carrier modulation is Rb Hz.

Transmission rates of Typical Services Speech Audio Fax Coloured Image Video

Speech (PCM) B = 3.4 kHz Rs = 8000 samples/sec Encoding = 8 bits/sample Transmission rate = 64 kbps Required bandwidth (passband) = 64 kHz One hour of speech = 64000x3600 = 230.4 Mb

Audio B = 16-24 kHz Rs = 44 000 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 307.2 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 M-ary communication reduces data size 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 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) 1 00 10 11 01 000 100 110 010 011 111 101 001 8 “levels” 3 bits / pulse 3×B bits per second 2 “levels” 1 bits / pulse B bits per second 4 “levels” 2 bits / pulse 2×B bits per second

Same-maximum-power Scenario 4 signal levels 8 signal levels typical noise

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

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 Signals Information 1 +1 -1 T 2T 3T 4T 5T 6T Amplitude Shift Keying Frequency Phase t

Spectrum

TDM for Digital

Digital Hierarchy

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

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

The Modulation Process of QAM Modulate cos(wct) and sin (wct) by multiplying them by Ak and Bk respectively: Ak x cos(wc t) Yi(t) = Ak cos(wc t) Bk sin(wc t) Yq(t) = Bk sin(wc t) + Y(t)

QAM Demodulation x x Y(t) LPF Ak 2cos(wc t) 2cos2(wct)+2Bk cos(wct)sin(wct) = Ak {1 + cos(2wct)}+Bk {0 + sin(2wct)} x LPF Bk 2sin(wc t) 2Bk sin2(wct)+2Ak cos(wct)sin(wct) = Bk {1 - cos(2wct)}+Ak {0 + sin(2wct)}