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MSc Data Communications

By R. A. Carrasco. Professor in Mobile Communications School of Electrical, Electronic and Computing Engineering University of Newcastle-upon-Tyne 2006. MSc Data Communications. Recommended Text Books. “Essentials of Error-Control Coding”, Jorge Costineira, Patrick Guy Farrell

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MSc Data Communications

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  1. By R. A. Carrasco Professor in Mobile Communications School of Electrical, Electronic and Computing Engineering University of Newcastle-upon-Tyne 2006 MSc Data Communications

  2. Recommended Text Books • “Essentials of Error-Control Coding”, Jorge Costineira, Patrick Guy Farrell • “Digital Communications”, John G. Proakis, Fourth Edition

  3. Goals Of A Digital Communication System • Deliver/Data from the source to the user in a: • FAST • INEXPENSIVE (Efficient) • RELIABLE WAY

  4. Digital Modulation Schemes • Task: to compare different modulation schemes with different values of M • Choice of modulation scheme involve trading off • Bandwidth • Power • Complexity • Define: • Bandwidth • Signal-to-noise ration • Error probability

  5. Source Symbols 0 0 1 1 0 1 1 0 0 1 Ts A a t -A M=2 T=Ts S1(t) = A, 0t<T S2(t) = -A T 10 01 01 01 01 b t Sinusoids with 4 different phases M=4 T=2Ts 010 011 c t M=8 T = 3Ts 8 different amplitude levels T 011001 010101 d T Examples: Memoryless Modulation (Waveforms are chosen independently – each waveform depends only on mi) 101

  6. A crucial question is raised • what is the difference? • If T is kept constant, the waveforms of scheme C requires less bandwidth than those of 2, because the pulse duration is longer • In the presence of noise, and if the same average signal power is used, it is more difficult to distinguish among the waveforms of c. AM/AM = Amplitude Modulation – Amplitude Modulation conversion AM/PM = Amplitude Modulation – Phase Modulation conversion

  7. Notice: • Waveforms b have constant envelopes. • This choice is good for nonlinear radio channels A Output Envelope B Input Envelope A: Output envelop (AM/AM conversion) B: Output phase shift (AM/PM conversion)

  8. TRADE-OFF BETWEENBANDWIDTH AND POWER • In a Power-Limited Environment, use low values of M • In a Band-Limited Environment, use high values of M What if both Bandwidth and Power are Limited ? • Expand Complexity • DEFINE: BANDWIDTH SIGNAL-TO-NOISE RATIO ERROR PROBABILITY

  9. Performance of Different Modulation Schemes

  10. DIGITAL MODULATION TRADE-OFFS SHANNON CAPACITY LIMIT FOR AWGN C = W LOG (1 + S/N) • S = Signal Power = e/T • N = Noise Power = 2NoW • W = Bandwidth

  11. dB S(f) Different bandwidth definitions of the power density spectrum of (5.2). B1 is the half-power bandwidth: B2 is the equivalent noise bandwidth: B3 is the null-to-null bandwidth: B4 is the fractional power containment bandwidth at an arbitrary level: B5 is the bounded power spectral density at a level of about 18dB. Notice that the depicted bandwidths are those around the frequency f0. 0 -10 -20 -3 -1 0 1 2 3 -2 fT B1 • Half – power • Equivalent – noise • Null – to – null • Fractional power containment • Bounded power spectral density B2 B3 B4 B5 In general, W = /T Hz,  depends on modulation scheme and on bandwidth definition Define Bandwidth W

  12. Rate at which the source outputs binary symbols Average signal energy • Average Signal Power b = average energy per bit • Signal-to-Noise Ratio Noise power spectral density. Bits/sec Hz (Bandwidth DEFINE SNR

  13. BPS/HZ Comparison

  14. Digital Modulation Trade – Offs(Comparison among different schemes) 16 SHANNON CAPACITY BOUND 32 16 16 8 8 4 x 8 16 x 16 4 8 4 8 x 2 BANDWIDTH LIMITED REGION 4 2 2 x 2 1 1 2 4 8 POWER LIMITED REGION 0.5 2 4 16 8 32 0.25 16 0.125 64 32 -2 0 2 4 6 8 10 12 14 16 18 20 (dB) PAM (SSB) (COHERENT) PSK AM-PM DCPSK COHERENT FSK INCOHERENT FSK x

  15. u1 k=1 Encoder for the (3, 1) repetition code x1= u1, x2=u1, x3=u3 0 000 1 111 x2 x3 x1 n=3 k=2 The whole word is defined by x1=u1, x2=u2 x3=u1+u2 Encoder for the (3,2) parity-check code u2 u1 00 000 01 011 10 101 11 110 x3 x2 x1 n=3 [1], pages 157 - 179

  16. The codeword is defined by xi = ui, i = 1,2,3,4 x5 = u1 + u2 + u3 x6 = u2 + u3 + u4 x7 = u1 + u2 + u4 (7, 4) Hamming Code Hamming Code (7,4) Source symbols encoded symbols Block Encoders Notice that only 16 of 128 sequences of length 7 are used for transmission

  17. Convolutionally encoding the sequence 101000... Rate , K = 3 From B Sklar, Digital Communiations. Prentice-Hall, 1988

  18. Convolutionally encoding the sequence 101000... Output sequence: 11 10 00 10 11

  19. 1 (7, 5) Convolutional Encoder c1 d1 d3 d2 1 Input Data d3 d2 d1 Constraint length K = 3 Code Rate = 1/2 Output 2 d1 d3 c2

  20. Example message: Input 0 1 1 0 1 0 0 1 0 1 Output 00 11 01 01 00 10 11 11 10 00 The Finite State Machine a=00 b=01 c=10 d=11 1/10 d 11 0/01 1/00 1/01 c 01 b 10 0/10 0/11 1/11 output bit a 00 0/00 input bit The 0 or 1 The coder in state a= 00 A 1 appearing at the input produces 11 at the output, the system moves to state b = 01

  21. If in state b, a 1 at the input produces 01 as the output bits. The system then moves to state d (11). • If a 0 appears at the input while the system is in state b, the bit sequence 10 will appear at the output, and the system will move to state c (10).

  22. Tree Representation 00 a b c d 00 11 00 10 11 00 01 11 a b 10 00 11 01 Time c d 0 01 10 00 11 a b 1 11 K=3 Rate= 10 Input data bits 10 c d 00 01 11 11 a b 01 00 01 01 c d 10 10 1 2 3 4 0 bit 1 bit Upward transition Downward transition

  23. D2 Signal-flow Graph

  24. Transfer Function T(D) Therefore,

  25. Transfer Function T(D) Performing long division gives: This gives the number of paths in the state diagram with their corresponding distances. In this case, the minimum distance of the code is 5

  26. s1=(00) s2=(01) s3=(10) s4=(11) 1 Block Encoders by Professor R.A Carrasco STATE 000 U Ui Ui-1 Ui-2 S1 00 001 111 110 S3 10 S2 01 011 010 100 X S4 11 0 Source Symbol 101 “State Diagram” of Code u= (11011.....) corresponds to the paths s1 s3 s4 s2 s3 s4 through the state diagram and the output sequence is x=(111100010110100)

  27. Tree Diagram The path corresponding to the input sequence 11011 is shown as an example.

  28. Signal Flow Graph Xa = S1 Xb = S3 Xc = S4 Xd = S2

  29. Transfer Function T(D)

  30. Transfer Function of the State Diagram We have dfree = 6, for error events: S1 to S3 to S2 to S1 and S1 to S3 to S4 to S2 to S1

  31. Trellis Diagram for Code (Periodic from time 2 on) Legend Input 0 Input 1 The minimum distance of the convolutional code l = N dmin = dc (N), the column distance. The free distance dfree of the convolutional code

  32. Trellis Diagram for the computation of dfree Trellis labels are the Hamming distances of encoder outputs and the all-zero sequence.

  33. We want to compute Functions whose arguments l can take on a finite number of values Simplest situation arises when T2, T1…….are “independent” (The value taken on by each one of them does not influence the other variables) 1 Then D C 0 B A Viterbi Algorithm [1], pages 181 – 185

  34. 2. We have, for a binary symmetric channel: 1-P 0 0 P Rx Tx Irrelevant multiplicative constant Irrelevant additive constant P 1 1 1-P Hamming distance Between xl and yl Viterbi Decoding of Convolutional Codes is a maximum received sequence transmitted symbols Observe (memoryless channel) received no-tuple n0-tuple of coded digits

  35. Brute force approach: • Compute all the values of the function and choose the smallest one. • We want a sequential algorithm

  36. What if 0, 1, … are not independent? 0 = A 1 = C or 1 = D 1 D C 0 0 = B 1 = D B A What is the shortest route from Los Angeles to Denver? Viterbi Algorithm

  37. l=5 l=0 l=4 l=1 l=2 2 1 3 1 0 3 2 2 2 1 1 1 0 3 4 1 2 4 1 1 2 2 l=1 l=2 2 l=3 2 4 2 1 3 3 1 1 2 2 3 1 1 0 4 2 1 6 4 (a) 1 2 2 (c) (b) 4 2 l=5 l=4 4 0 4 1 4 5 3 1 6 5 (d) (e) 4 Viterbi Algorithm l=3

  38. Conclusion • We maximise P(y|x) by minimising , the Hamming distance between the received sequence and coded sequence. • Brute-Force Decoding • Compute all the distances between y and all the possible x’s. Choose x that gives the minimum distance. • Problems with Brute-Force Decoding • - Complexity • - Delay • Viterbi algorithm solves the complexity problem (complexity increases only linearly with sequence length) • Truncated Viterbi algorithm also solves delay problem

  39. Band-limited environment (Terrestrial radio communications) • Power-limited environment (Satellite radio communications) • Use higher-order modulation schemes. (8 PSK instead of 4 PSK) Same BW, More bit/s per hz, more power • Use coding: Less power, less bit/s per hz, BW expanded Trellis-Coded Modulation A.K.A - Ungerboeck Codes - Amplitude-Redundant Codes - Modulation How to increase transmission efficiency? Reliability or Speed [2], pages 522-532 G. Ungerboeck, "Channel coding with multilevel/phase signals," IEEE Trans. Inform. Theory, vol. IT-28, pp. 55-67, 1982.

  40. Construction of TCM • Constellation is divided into smaller constellations with larger euclidean distances between constellation points. • Construction of Trellis (Ungerboeck’s rules) • Parallel transitions are assigned members of the same partition • Adjacent transitions are assigned members of the next larger transitions • Signals are used equally often

  41. Model for TCM Memory Part n Select Constellation an Select Signal From Constellation

  42. 2 3 1 4 0 d’ 5 7 6 Some examples of TCM schemes Consider transmission of 2 bits/signal We examine TCM schemes using 8-PSK With uncoded 4-PSK we have We use the octogonary constellation

  43. The Free Distance of a convolutional code is the Hamming distance between two encoded signals • dfree is a measure of the separation among encoded sequences: the larger dfree, the better the code (at least for large enough SNR). • Fact: To compute dfree for a linear convolutional code we may consider the distances with respect to the all-zero sequence.

  44. Remerge Split (00) (00) An “error event” • A simple algorithm to compute dfree is • Compute dc(l) for l = 1,2,… • If the sequence giving dc(l) merges into the all-zero sequence, store its weight as dfree

  45. First Key Point Constellation size must be increased to get the same rate of information Minimum distance between sequences • We gain Minimum distance for uncoded transmission Energy with coding • We lose Energy without coding Gain is

  46. Second Key Point How to introduce the dependence among signals Transmitted symbol at time nT Source symbol at time nT Previous source Symbols = “state” n (Describes output as a function of input symbol + encoder state) We write (Describes transitions between states)

  47. TCM Example 1 Consider the 8-PSK TCM scheme, which involves the transmission of 2 bits/symbol using an uncoded 4-PSK constellation and the coded 8-PSK constellation for the TCM scheme as shown below Show that and

  48. TCM Example 1: Solution We have from the uncoded 4-PSK constellation We have from the coded 8-PSK constellation Using

  49. 0 0 0 4 5 6 1 2 3 1 7 Hence, we get the coding gain a0 a1 I 4PSK 8PSK Can this performance gain Trellis coded QPSK be improved? The answer is yes by going to more trellis states. TCM Scheme Based on 2-State Trellis

  50. TCM Example 2 • Draw the set partition of 8-PSK with maximum Euclidean distance between two points. • By how much is the distance between adjacent signal points increased as a result of partitioning?

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