Lecture 1: Tue Jan 8, Lecture introduction and motivation

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Lecture 1: Tue Jan 8, 2019 Lecture introduction and motivation 1

ECE 6602: Digital Communications GEORGIA INSTITUTE OF TECHNOLOGY, SPRING 2019 PREREQUISITE: ECE 6601. Strong background in probability is a must. COURSE OBJECTIVE: To study the design and implementation of digital communications systems. OFFICIAL TEXT: Digital Communications, Fifth Edition, Proakis and Salehi, McGraw-Hill, 2007, ISBN 0072957166. SUPPLEMENTAL READING: Digital Communication, Third Edition, J. R. Barry, E. A. Lee and D. G. Messerschmitt, Kluwer, 2004. 2

INSTRUCTOR: John R. Barry Centergy 5136 404-894-1705 E-Mail: barry@ece.gatech.edu Office Hours: after class, or by appointment. WEBSITE: http://barry.ece.gatech.edu/6602/ HONOR CODE: You are expected to uphold the honor code (www.honor.gatech.edu). Homework: Homework is assigned roughly once a week. May or may not be graded, depending on whether we are assigned a GTA. It is due at the beginning of class one week after it is assigned. Collaboration on homeworks is encouraged. Copying homework is an honor code violation. 3

GRADING: Homework 10% Quiz 1 (Tue Feb 19) 20% Quiz 2 (Thu Mar 14) 20% Quiz 3 (Tue Apr 16) 20% Final Exam (Thu May 2) 30% GRADING (NO TA): Quiz 1 (Tue Feb 19) 20% Quiz 2 (Thu Mar 15) 25% Quiz 3 (Tue Apr 16) 25% Final Exam (Thu May 2) 30% 4

A Pic from Wednesday 5

How? 6

Relay 7

Coding and Modulation to Earth MESSAGE BITS (2 Mb/s) REED-SOLOMON 233 ENCODER 255 CODED BITS (14 Mb/s) RATE 1/6 CONVOLUTIONAL ENCODER BPSK f 0 3 GHz 8

Meanwhile Last Monday, 4 billion miles away... NEW HORIZON ULTIMA THULE 9

Rate-1/6 Turbo Code MESSAGE CODED BITS BITS (1 kb/s) (6 kb/s) BPSK f 0 = 8.4 GHz INTERLEAVER RATE 1/6 TURBO ENCODER 10

Coding + Deep Space A Marriage Made in Heaven plenty of bandwidth weak signal (path loss, transmit constraints) power-limited linear AWGN channel model each db is incredibly valuable (range, launch costs, scientific) detector complexity nearly unlimited sophisticated algorithms OK 11

Trading Complexity for Performance 12 10 1958 Explorer (uncoded) GAP TO CAPACITY (db) 8 6 4 2 1969 Mariner (rate-6/32 Reed-Muller/biorthogonal) 1977 Voyager (rate-1/3, = 6, conv. code) 512 b/s 1968 Pioneer (rate-1/2, = 20, sequential) 1981 Voyager: e-rate-1/2, = 6, + RS 1989 Galileo rate-1/4, =14 + RS 1997 Cassini rate-1/6, =14 + RS 0 1995 Galileo-S rate-1/4, =13 + RS 4 0 10 20 30 40 50 NORMALIZED COMPLEXITY (db) [F. Pollara, Descanso Workshop, 1998.] 12

Trading Complexity for Performance 12 10 1958 Explorer (uncoded) GAP TO CAPACITY (db) 8 6 4 2 1969 Mariner (rate-6/32 Reed-Muller/biorthogonal) 1977 Voyager (rate-1/3, = 6, conv. code) 512 b/s 1968 Pioneer (rate-1/2, = 20, sequential) 1981 Voyager: e-rate-1/2, = 6, + RS 1989 Galileo rate-1/4, =14 + RS 1997 Cassini rate-1/6, =14 + RS 0 1995 Galileo-S 2004 (rate-1/4 turbo, 2 16-state cc) rate-1/4, =13 + RS 4 LDPC 0 10 20 30 40 50 NORMALIZED COMPLEXITY (db) [F. Pollara, Descanso Workshop, 1998.] 13

The PHY layer Interface to the physical world APP PHY tools: modulation, line coding, scrambling, precoding, error-control coding, equalization, synchronization, channel estimation, interference cancellation, space-time coding, multiuser detection, MIMO detection, turbo processing... Tangible outcomes modems, baseband processors read channels 100G transceivers NET LINK PHY The OSI 7-Layer Model...101011... Bits Volts, Photons... 14

Figure 1 SOURCE CODER ERROR- BITS CONTROL MOD CODER SYMBOLS WAVEFORM NOISY, DISPERSIVE CHANNEL 6605 6255 6606 6602 6603 SOURCE DECODER ERROR- CONTROL DECODER DEMOD 15

Source Coding = Data Compression 699,000 bits 8 bits/pixel 326 rows 268 columns original (JPEG) 55,600 bits H bits/pixel 0.64 bits/pixel compressed entropy Associated with each discrete source is an entropy H bits/symbol. The compression limit: a compressor to B bits/symbol with P(err) 0 as N B > H. 16

Source-Channel Coding Separation Theorem compress source independent of channel channel encoder independent of source 17

A Communications System source encoder bits MOD s( t ) speaker microphone r( t ) DEMOD bits source deoder s( t ) Equivalent model: H( f ) n( t ) r( t ) 0 A typical frequency response: H( f ) 2 (db) -10-20 -30 W = 2000 Hz -40 0 4000 Hz FREQUENCY 18

How to Communicate Across This Channel? s( t ) W W f n( t ) r( t ) 19

How to Communicate Across This Channel? DAC s( t ) W W f n( t ) r( t ) rate = 2W 20

How to Communicate Across This Channel? a k A DAC s( t ) W W f n( t ) r( t ) rate = 2W 21

Before Shannon Nyquist [1924]: The maximum symbol rate is 2W symbols/second. If the symbol alphabet is A, then each symbol conveys log 2 A bits. R b =2W log 2 A. Hartley [1928]: With amplitude constraint A, and noise margin, the maximum alphabet size is A = 1 + A/ : R b = 2W log 2 1 + A ---. Modulation was instantaneous. To fight random noise: Increase BW, increase power, live with inevitable errors. 22

How Big Can the Alphabet Be? 2 2 2A 2A + 2 It s a counting exercise, A = --------------------- 2A + 2 = 1 2 A + --- 23

Turning Point: Claude Shannon, 1948 The father of info theory & modern communication theory 1948 paper A (The) Mathematical Theory of Communication quantified information; separation theorem; noisy channel theorem The capacity of an AWGN channel with bandwidth W is: C = W log P 1 + ------------ 2 N 0 W bits/second. Capacity is a speed limit: a modem achieving bit rate R b with no errors R b < C. How to get close? 24

The Data Rate is Limited by 3 Key Parameters signal power P [W] one-sided noise density N 0 [W/Hz] s( t ) W W f n( t ) r( t ) The 3 combine to determine: P signal-to-noise ratio: SNR = ------------ N 0 W P capacity: C = W log 1 + ------------ 2 N 0 W bandwidth W [Hz] 25

Pop Quiz Q1: How much SNR does Shannon need to communicate at R b = 2 Gb/s across a 200-MHz channel? 26

Pop Quiz Q1: How much SNR does Shannon need to communicate at R b = 2 Gb/s across a 200-MHz channel? A1: Solve capacity equation SNR = 2 R b/w 1. = 2 10 1 = 1023 = 30.1 db. 27

Pop Quiz Q1: How much SNR does Shannon need to communicate at R b = 2 Gb/s across a 200-MHz channel? A1: Solve capacity equation SNR = 2 R b/w 1. = 2 10 1 = 1023 = 30.1 db. Q2: What happens to capacity if we remove bandwidth constraint? 28

Pop Quiz Q1: How much SNR does Shannon need to communicate at R b = 2 Gb/s across a 200-MHz channel? A1: Solve capacity equation SNR = 2 R b/w 1. = 2 10 1 = 1023 = 30.1 db. Q2: What happens to capacity if we remove bandwidth constraint? A2: As W, capacity saturates to ------------- P/N 0 bits/s. ln2 29