Part V. AWGN Channel Capacity AWGN Capacity Formula; Sphere Packing; Resources in AWGN Channel 61
Channel coding theorem • For every memoryless channel, there is a definite number C (computable) such that: ‣ If the data rate R < C , then there exists a coding scheme that can deliver data at rate R over the channel with vanishing error probability as the block length n → ∞ ‣ Conversely, if the data rate R > C , then no matter what coding scheme is used, the error probability will converge to 1 as n → ∞ • C is called the capacity of the channel, and it has a computable formula for all kinds of channels (depending on the channel statistics) • We focus on the additive white Gaussian noise (AWGN) channel in this course, and give a heuristic argument to derive the AWGN channel capacity 62
������ AWGN channel model (discrete-time, real-valued) Z m ∼ N (0 , σ 2 ) V m = u m + Z m , m = 1 , . . . , n, Z m u m C comprises 2 nR length- n codewords u ∈ R n Codebook: n Power constraint: 1 | u m | 2 ⌘ 1 n k u k 2 P, 8 u 2 C X n unit: joule per channel use m =1 Code design problem is equivalent to placing 2 nR n -dimensional vectors within a sphere of radius , so that the error probability is minimized √ nP 63
Sphere packing interpretation • By LLN, as , the received V will lie V = u + Z n → ∞ at the surface of the n -dimensional sphere R n centered at u with radius with √ n σ 2 p probability 1 n ( P + σ 2 ) • Also by LLN, as , the received V n → ∞ will lie within the n -dimensional sphere √ with radius with probability 1 n σ 2 p n ( P + σ 2 ) • Asymptotically, vanishing error probability is equivalent to non-overlapping spheres • How many non-overlapping spheres can be packed into the large sphere? 64
Necessary condition: capacity upper bound • maximum # of non-overlapping spheres = V = u + Z maximum # of codewords that can be R n reliably delivered • A necessary condition is p n ( P + σ 2 ) √ n n ( P + σ 2 ) 2 nR ≤ √ n σ 2 n � √ � n n ( P + σ 2 ) ⇒ R ≤ 1 1 � 1 + P � n log = 2 log √ ⇐ n σ 2 σ 2 √ n σ 2 n • The channel capacity is hence upper bounded by 1 1 + P � � 2 log σ 2 • How to achieve it? 65
Achieving capacity (1) • Prove the existence of good codebook C u -sphere by random coding , as we did before for linear block codes: ‣ Randomly generate 2 nR length- n codewords uniformly inside the “ u -sphere” of radius √ √ nP nP ‣ Goal: ensure the average-over-random-code average probability of error vanishes as n → ∞ • u 1 Decoding: P + σ 2 : the MMSE coefficient α , P u 2 → MMSE − → Nearest Neighbor − V − → α V − u → ˆ 66
Achieving capacity (2) • Due to symmetry, we can assume WLOG u -sphere the true codeword sent by Tx is u 1 ‣ By LLN, the distance between and : α V u 1 ∥ α V − u 1 ∥ 2 = ∥ α Z + ( α − 1) u 1 ∥ 2 √ nP ≈ α 2 n σ 2 + ( α − 1) 2 nP = n P σ 2 P + σ 2 ‣ As long as lies inside the sphere centered � u 1 n P σ 2 � P + σ 2 at with radius , decoding will be n P σ 2 α V P + σ 2 correct w.h.p. u 1 • Pairwise probability of error α V P {E u 1 → u 2 } ‣ The probability that a random falls inside u 2 u 2 the sphere! ‣ Ratio of the volume of the two spheres. 67
Achieving capacity (3) • Pairwise probability of error P {E u 1 → u 2 } u -sphere ‣ Ratio of the volume of the two spheres: � n/ 2 n √ � nP σ 2 / ( P + σ 2 ) σ 2 P {E u 1 → u 2 } = √ = P + σ 2 n nP • √ Union bound: nP � n/ 2 ≤ (2 nR − 1) P {E u 1 → u 2 } ≤ 2 nR � P ( n ) σ 2 e P + σ 2 � n P σ 2 P + σ 2 = 2 n ( R − 1 2 log ( 1+ P σ 2 )) u 1 • Su ffi cient condition for vanishing : P ( n ) α V e u 2 ✓ ◆ R < 1 1 + P 2 log σ 2 68
Continuous-time AWGN channel capacity • For the continuous-time (waveform) channel model considered in Lecture 03: ‣ Power constraint P watts ‣ White Gaussian noise PSD N 0 /2 joules per second per hertz ‣ Total bandwidth W hertz (symbol duration T = 1/ W ) • Recall we can convert the waveform channel to an equivalent discrete-time complex-valued AWGN channel: ‣ Power constraint is PT = P / W joules per channel use ‣ Variance of the circular symmetric complex Gaussian noise is N 0 joules per channel use ‣ � � 1 + P/ 2 W For each real dimension, its capacity is bits per channel use 1 2 log N 0 / 2 ‣ � � � � 1 + P/ 2 W The channel capacity is bits per channel use P 2 × 1 2 log = log 1 + N 0 / 2 W N 0 ✓ ◆ P bits per second = W log 1 + WN 0 69
������ AWGN channel capacity ✓ ◆ P C AWGN ( P, W ) = W log 1 + N 0 W “spectral e ffi ciency” = log (1 + SNR ) ��������� SNR � P N 0 W • The capacity formula provides a high-level way of thinking about how the performance fundamentally depends on the basic resources in the channel • No need to go into details of specific coding and modulation schemes • Basic resources: power P and bandwidth W 70
Resources in AWGN channel Fix P ✓ ◆ P N 0 W log 2 e = P P C ( W ) = W log 1 + log 2 e ≈ W N 0 W N 0 1.6 P N 0 log 2 e 1.4 Power limited region 1.2 1 0.8 Capacity C ( W ) 0.6 (Mbps) Limit for W → ∞ 0.4 Bandwidth limited region 0.2 0 0 5 10 15 20 25 30 Bandwidth W (MHz) 71
������ Bandwidth-limited vs. power-limited ✓ ◆ P C AWGN ( P, W ) = W log 1 + N 0 W “spectral e ffi ciency” = log (1 + SNR ) ��������� SNR � P N 0 W • When : power-limited regime SNR ⌧ 1 � � P P C AWGN ( P, W ) ≈ W log 2 e = N 0 log 2 e N 0 W ‣ Linear in power; Insensitive to bandwidth • When : bandwidth-limited regime SNR � 1 � � P C AWGN ( P, W ) ≈ W log N 0 W ‣ Logarithmic in power; Approximately linear in bandwidth 72
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