Chapter 7

1. Chapter 7 Fundamentals of Digital Transmission

2. Baseband Transmission (Line codes) ON-OFF or Unipolar (NRZ) Non-Return-to-Zero Polar (NRZ)

3. 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.

4. 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)

5. More Line Codes On-Off RZ Better synch., at extra bandwidth Bi-Polar Better synch., at same bandwidth

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

7. Spectra of Some Line Codes

8. 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).

9. Nyquist-Criterion for Zero ISI • Use a pulse that has the following characteristics • One such pulse is the sinc function.

10. The Sinc Pulse 1 t -6Tb -5Tb -4Tb -3Tb -Tb -2Tb Tb 2Tb 3Tb 4Tb 5Tb 6Tb p(t) 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)

11. Zero ISI

12. 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.

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

14. Spectrum of Raised-Cosine Pulses

15. Extremes of Raised-Cosine Spectra

16. Raised-Cosine Pulses

17. Information 1 0 1 1 0 1 +1 Amplitude Shift Keying t 6T 6T 6T 2T 2T 2T 4T 4T 4T 5T 5T 5T 3T 3T 3T T T T 0 0 0 -1 +1 Frequency Shift Keying t -1 +1 Phase Shift Keying t -1 Carrier Modulation of Digital Signals

18. 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 RbHz.

19. Transmission rates of Typical Services • Speech • Audio • Fax • Coloured Image • Video

20. 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

21. Audio • B = 16-24 kHz • Rs = 44 000 samples/sec • Encoding = 16 bits/sample • Stereo type = 2 channels • Transmission rate = 1.4 Mbps

22. 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.

23. 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 !

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

25. Solutions • Compression • reduces data size • M-ary communication • Expands channel ability to carry information

26. 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).

27. 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.

28. Multiple Amplitudes (PAM) 000 100 011 111 110 010 101 001 1 00 10 0 11 01 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

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

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

31. 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

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

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

34. x Ak Yi(t) = Akcos(wc t) cos(wc t) + Y(t) x Bk Yq(t) = Bksin(wc t) sin(wc t) The Modulation Process of QAM Modulatecos(wct)and sin (wct)bymultiplying them by Akand Bk respectively:

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