lecture 2-3: multi-channel communication aliazam abbasfar

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Capacity and Gap analysis One dimension channel with AWGN Capacity : c = ½ log 2 (1 + SNR) SNR = E /  2 Random coding, P e = 0 General systems b < c b = ½ log 2 (1 + SNR/) : gap for a given b and P e Constant gap for PAM/QAM  = 8.8 dB for uncoded, P e = 1e-6 Coding reduces the gap Margin : excess SNR for a given b SNR allows b max (for a given ) The target is b not b max

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Lecture 2-3: Multi-channel Communication Aliazam Abbasfar Outline Multi-channel communications Capacity and Gap analysis One dimension channel with AWGN Capacity : c = log 2 (1 + SNR) SNR = E / 2 Random coding, P e = 0 General systems b < c b = log 2 (1 + SNR/) : gap for a given b and P e Constant gap for PAM/QAM = 8.8 dB for uncoded, P e = 1e-6 Coding reduces the gap Margin : excess SNR for a given b SNR allows b max (for a given ) The target is b not b max Parallel channels N dimensions (sub-channels) All sub-channels have the same P e Constant gap for all subchannels Average bits/dimension Large SNRs : Water filling Maximize date rate : b n Subject to fixed total energy : E = E n H n : n th channel gain Define g n = |H n | 2 /s n 2 (unit energy SNR) Energy allocation Convex problem Unique solution Water filling (2) Minimize Energy : E n Subject to fixed total energy : b = b n H n : n th channel gain Energy allocation : Margin : E / E n constant margin for all used subchannels Bit Loading Rate-Adaptive (RA) loading Maximize b; subject to a constant total energy All positive E n Sort channels wrt g n Energy allocation : Optimized b Bit Loading (2) Margin-Adaptive (RA) loading Minimize total energy; subject to a constant b All positive E n Sort channels wrt g n Energy allocation : Optimized E Discrete Loading In practice b n cannot assume any value b n = k ; k is integer Chows algorithm On-off constant energy allocation Quantize bit assignment Energy scaling Levin-Compello (LC) algorithm Optimum discrete loading Define E n (b n ), e n (b n )=E n (b n )-E n (b n - ) Efficiency of bit distribution max[e n (b n )] < min[e n (b n +)] E-tightness 0< E E n (b n ) < min [e n (b n + Reading Cioffi Ch. 4