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158 ADVANCED TOPICS Thus, for MAPSD, we can also form an LLR by taking the log of the ratio of the two symbol metrics. Now for an example. Referring to Table 7.1, the signed soft value for s% would be log(0.969975/1.4529) = -0.4043. Similarly, the signed soft value would be 0.5120 for «2· How do we use LLRs in decoding? Because the likelihood of the symbol being + 1 is in the numerator, we can think of the LLR as the log of the symbol metric for the symbol being +1. The symbol metric for the symbol being —1 is simply the negative of the LLR. As a result, we can perform decoding by correlating the soft values to the hypothetical message values. A correlation is obtained by forming products of corresponding values and summing. Thus, for the MAPSD case, the message metric for message 1 would be (+l)(-0.4043) + (-1)(0.5120) = -0.9163. 7.2.2.2 The (7,4) Hamming code There are different kinds of FEC codes. One kind is a block code, in which the message symbols are divided up into blocks, and each block is encoded separately. For example, consider a Hamming code in which 4 information bits (¿j, ¿2, ¿3 and ¿4) are encoded into 7 modem bits (a 7-bit codeword), sometimes called a (7,4) Hamming code. This is done by also transmitting 3 check bits (ci, C2 and C3), which are computed as C\ = ¿l¿2¿4 C 2 = íi¿3¿4 C.-j = ¿2*3*4- ( 7·10) If no errors occur during detection, then the detected values will also satisfy these equations. Otherwise, an error will have been detected. Note that there can be errors in the check bits as well as the information bits. If we are sure only one error has occurred, there is a way to detect it without soft information. This particular Hamming code has the property that it can correct single errors using only the hard decisions. It works like this. We form a syndrome by forming products Si = Cl¿l¿2¿4 Si = C2¿ii:s¿4 S-Λ = C;iÍ2Í\iÍA· (7.11) If there are no errors, then all three syndrome values should be +1. Why? Consider Si. We know that ¿i¿2¿4 = Ci, so that ci¿i¿2¿4 should equal c\ which is always +1. Similar arguments apply to the other syndromes. If one or more of the syndrome elements are —1, we know there has been an error. It turns out that the Hamming code is designed to tell us the location of the error, assuming only one error occurred. First, we need to think of the bits as having the positions shown in Table 7.5. Second, we need to map the +1 and —1 syndrome values to 0 and 1, i.e., Third, we can determine the position of the error using Bfc = (l-S fc )/2. (7.12) p = Bi + 2B 2 + 4B.S. (7.13)
MORE DETAILS 159 While it seems like magic, it's really clever engineering by Richard Hamming. Table 7.5 (7,4) Hamming code bit positions Position Ï Bit c7 2 C 2 3 ¿i 4 c 3 5 ¿2 6 ¿3 7 ¿4 Let's do an example. Suppose i\ = i 2 = +1 and ¿ 3 = ¿4 = —1. From (7.10), ci = (+1)(+1)(-1) = -1 C2 = (+1)(-1)(-1) = +1
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Channel Equalization for Wireless C
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To my colleagues at Ericsson
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CONTENTS List of Figures List of Ta
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CONTENTS XI 4.2 4.3 4.4 4.5 4.6 4.1
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CONTENTS XIII 8 Practical Considera
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xvi LIST OF FIGURES 2.2 Matched fil
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XviÜ LIST OF FIGURES 6.18 Scatter
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XX Prologue Alice was nervous. Woul
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XXÜ PREFACE is introduced as neede
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ACRONYMS AC A/D ADC AMLD AMPS ASK A
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ML MLD MLPD MLSD MLSE MRC MSE MSB O
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2 INTRODUCTION Table 1.1 Possible m
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4 INTRODUCTION s o S 1 8 2 S 3 S 0
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6 INTRODUCTION A block diagram of t
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8 INTRODUCTION Figure 1.7 QPSK. tra
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10 INTRODUCTION phases (phase modul
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12 INTRODUCTION rollo« 0.22 0.8 Ί
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14 INTRODUCTION is the "channel" re
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16 INTRODUCTION 1.4 MORE MATH In th
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18 INTRODUCTION For K = 1 (order 0)
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20 INTRODUCTION The K = 4 main bloc
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22 INTRODUCTION A sample of the noi
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24 INTRODUCTION The receiver may ha
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26 INTRODUCTION 1.5.1 Reference sys
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28 INTRODUCTION a) What most likely
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CHAPTER 2 MATCHED FILTERING In this
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MORE DETAILS 33 2.2 MORE DETAILS So
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THE MATH 35 Now compare the output
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THE MATH 37 The correlation in (2.2
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THE MATH 39 correlation function f(
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THE MATH 41 Thus, in the complex pl
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THE MATH 43 to the medium response
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THE MATH 45 We can interpret f(t) a
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MORE MATH 47 10" 1 Odeg 90 deg X 18
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MORE MATH 49 With despread-level Ra
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MORE MATH 51 Figure 2.6 OFDM exampl
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MORE MATH 53 2.4.3 MF in colored no
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PROBLEMS 55 period) is examined in
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CHAPTER 3 ZERO-FORCING DECISION FEE
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MORE DETAILS 59 1/c r m —► H i
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MORE DETAILS 61 where «i = m - (d/
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MORE MATH 63 Because there is no IS
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MORE MATH 65 In the later chapter o
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PROBLEMS 67 b) What is the detected
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CHAPTER 4 LINEAR EQUALIZATION With
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THE IDEA 71 estimate of symbol s 2
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THE IDEA 73 Notice that £2 (MSE) d
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MORE DETAILS 75 The first term is t
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MORE DETAILS 77 To obtain the unity
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THE MATH 79 and SINR becomes w T h
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THE MATH 81 To obtain the MMSE solu
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THE MATH 83 where *m n = ν ^ ^ ^ '
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THE MATH 85 power is much larger th
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MORE MATH 87 IQ' 1 OC 111 ω 10 2 1
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MORE MATH 89 values of h k m (f) fr
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MORE MATH 91 where C(d(),d 0 ) . C(
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AN EXAMPLE 93 between z and s. This
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PROBLEMS 95 Another aspect of cocha
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PROBLEMS 97 The math 4.11 Consider
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100 MMSE AND ML DECISION FEEDBACK E
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Page 130 and 131: 108 MMSE AND ML DECISION FEEDBACK E
Page 137 and 138: CHAPTER 6 MAXIMUM LIKELIHOOD SEQUEN
Page 139 and 140: MORE DETAILS 117 r 2 Ht, r, fcl + f
Page 141 and 142: MORE DETAILS 119 Figure 6.3 MLSD tr
Page 143 and 144: THE MATH 121 legs of the journey is
Page 145 and 146: THE MATH 123 Figure 6.8 MLSD trelli
Page 147 and 148: THE MATH 125 We would then identify
Page 149 and 150: THE MATH 127 Expanding the square,
Page 151 and 152: THE MATH 129 Consider the matched f
Page 153 and 154: THE MATH 131 6.3.5.1 M-algorithm Wi
Page 155 and 156: THE MATH 133 results, DFE and MLSD
Page 157 and 158: THE MATH 135 IQ" 1 oc LU CD 10 2 10
Page 159 and 160: THE MATH 137 sometimes inaccurate w
Page 161 and 162: MORE MATH 139 POWER CONTROL effecti
Page 163 and 164: MORE MATH 141 15 POWER CONTROL 10
Page 165 and 166: MORE MATH 143 6.4.3 More approximat
Page 167 and 168: THE LITERATURE 145 EDGE is an evolu
Page 169 and 170: PROBLEMS 147 the channel to minimum
Page 171: PROBLEMS 149 c) Suppose we know «o
Page 174 and 175: 152 ADVANCED TOPICS detecting the w
Page 176 and 177: 154 ADVANCED TOPICS Table 7.2 Examp
Page 178 and 179: 156 ADVANCED TOPICS Let's look at t
Page 182 and 183: 160 ADVANCED TOPICS Table 7.6 Examp
Page 184 and 185: 162 ADVANCED TOPICS If we assume no
Page 186 and 187: 164 ADVANCED TOPICS Now we can focu
Page 188 and 189: 166 ADVANCED TOPICS Notice we used
Page 190 and 191: 168 ADVANCED TOPICS to achieve spat
Page 192 and 193: 170 ADVANCED TOPICS c) What are the
Page 195 and 196: CHAPTER 8 PRACTICAL CONSIDERATIONS
Page 197 and 198: MORE DETAILS 175 8.2 MORE DETAILS I
Page 199 and 200: MORE DETAILS 177 One extreme is to
Page 201 and 202: THE MATH 179 8.3.1 Time-invariant c
Page 203 and 204: THE MATH 181 With recursive channel
Page 205 and 206: MORE PRACTICAL ASPECTS 183 timing).
Page 207 and 208: THE LITERATURE 185 8.6 THE LITERATU
Page 209 and 210: PROBLEMS 187 a) Draw the new receiv
Page 211 and 212: APPENDIX A SIMULATION NOTES Perform
Page 213 and 214: FADING CHANNELS 191 where JA = A 2
Page 215 and 216: APPENDIX B NOTATION Channel Equaliz
Page 217: NOTATION 195 Variable Meaning p(t)
Page 220 and 221: 198 REFERENCES [Ari98] [Ari99] [Ari
Page 222 and 223: 200 REFERENCES [Bra91] W. R. Braun
Page 224 and 225: 202 REFERENCES [Div98] [Dou95] [Due
Page 226 and 227: 204 REFERENCES [Gon68] R. A. Gonsal
Page 228 and 229: 206 REFERENCES [Kaa94] [Kai60] V.-P
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208 REFERENCES [ManOl] [Mar73] [Mar
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210 REFERENCES [Pic95] [P0088] [Pri
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212 REFERENCES [Sto93] [Sto96] [Sui
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214 REFERENCES [Wan06a] T. Wang, J.
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