Lecture ¡4 Capacity ¡of ¡Wireless ¡Channels I-Hsiang Wang ihwang@ntu.edu.tw 3/20, 2014
What ¡we ¡have ¡learned • So far: looked at specific schemes and techniques • Lecture 2: point-to-point wireless channel - Diversity: combat fading by exploiting inherent diversity - Coding: combat noise, and further exploits degrees of freedom • Lecture 3: cellular system - Multiple access: TDMA, CDMA, OFDMA - Interference management: orthogonalization (partial frequency reuse), treat-interference-as-noise (interference averaging) 2
Information ¡Theory • Is there a framework to … - Compare all schemes and techniques fairly? - Assert what is the fundamental limit on how much rate can be reliably delivered over a wireless channel? • Information theory! - Provides a fundamental limit to (coded) performance - Identifies the impact of channel resources on performance - Suggest novel techniques to communicate over wireless channels • Information theory provides the basis for the modern development of wireless communication 3
Historical ¡Perspective G. Marconi C. Shannon 1901 1948 • • First radio built 100+ years ago Information theory: every channel has a capacity • Great stride in technology • Provides a systematic view of all • But design was somewhat ad-hoc communication problems Engineering ¡meets ¡science New ¡points ¡of ¡view ¡arise 4
Modern ¡View ¡on ¡Multipath ¡Fading Channel quality Time • Classical view: fading channels are unreliable - Diversity techniques: average out the variation • Modern view: exploit fading to gain spectral efficiency - Thanks to the study on fading channel through the lens of information theory! 5
Plot • Use a heuristic argument (geometric) to introduce the capacity of the AWGN channel • Discuss the two key resources in the AWGN channel: - Power - Bandwidth • The AWGN channel capacity serves as a building block towards fading channel capacity: - Slow fading channel: outage capacity - Fast fading channel: ergodic capacity 6
Outline • AWGN Channel Capacity • Resources of the AWGN Channel • Capacity of some LTI Gaussian Channels • Capacity of Fading Channels 7
AWGN ¡Channel ¡Capacity 8
Channel ¡Capacity • Capacity := the highest data rate can be delivered reliably over a channel - Reliably ≡ Vanishing error probability • Before Shannon, it was widely believed that: - to communicate with error probability → 0 - ⟹ data rate must also → 0 • Repetition coding (with M -level PAM) over N time slots on AWGN channel: r ! - 6 N Error probability ∼ 2 Q M 3 SNR - = log 2 M Data rate N - As long as M ≤ N ⅓ , the error probability → 0 as N → ∞ - = log 2 N still → 0 as N → ∞ But, the data rate � � � 3 N 9
Channel ¡Coding ¡Theorem • For every memoryless channel, there is a definite number C that is computable such that: - If the data rate R < C , then there exists a coding scheme that can deliver rate R data over the channel with error probability → 0 as the coding block length N → ∞ - Conversely, if the data rate R > C , then no matter what coding scheme is used, the error probability → 1 as N → ∞ • We shall focus on the additive white Gaussian noise (AWGN) channel - Give a heuristic argument to derive the AWGN channel capacity 10
AWGN ¡Channel z [ n ] ∼ N (0 , σ 2 ) Power constraint: N | x [ n ] | 2 ≤ NP X x [ n ] y [ n ] = x [ n ] + z [ n ] n =1 • We consider real-valued Gaussian channel • As mentioned earlier, repetition coding yield zero rate if the error probability is required to vanish as N → ∞ • Because all codewords are spread on a single dimension in an N -dimensional space • How to do better? 11
Sphere ¡Packing ¡Interpretation • By the law of large numbers, y = x + z R N as N → ∞ , most y will lie inside the N -dimensional p N ( P + σ 2 ) p sphere of radius N ( P + σ 2 ) Also by the LLN, as N → ∞ , • y will lie near the surface of √ the N -dimensional sphere N σ 2 centered at x with radius √ N σ 2 • Vanishing error probability ⟹ non-overlapping spheres How many non-overlapping spheres can be packed into the large sphere? 12
Why ¡Repetition ¡Coding ¡is ¡Bad y = x + z R N It only uses one dimension out of N ! p N ( P + σ 2 ) 13
Capacity ¡Upper ¡Bound Maximum # of non-overlapping y = x + z R N spheres = Maximum # of codewords that can be reliably delivered p N ( P + σ 2 ) N p N ( P + σ 2 ) 2 NR ≤ √ N σ 2 N √ N p ! N σ 2 N ( P + σ 2 ) ⇒ R ≤ 1 = N log √ N σ 2 N = 1 ✓ ◆ 1 + P 2 log σ 2 This is hence an upper bound of the capacity C . How to achieve it? 14
Achieving ¡Capacity ¡(1/3) • (random) Encoding: randomly generate 2 NR codewords { x 1 , x 2 , ...} lying inside the “ x -sphere” of radius √ NP • Decoding: P α := P + σ 2 → b → MMSE − → Nearest Neighbor − y − → α y − x • Performance analysis: WLOG let x 1 is sent - By the LLN, P σ 2 || α y − x 1 || 2 = || α w + ( α − 1) x 1 || 2 ≈ α 2 N σ 2 + ( α − 1) 2 NP = N P + σ 2 - As long as x 1 lies inside the uncertainty sphere centered at α y r P σ 2 with radius � � � , decoding will be correct N P + σ 2 ◆ N/ 2 - σ 2 ✓ Pairwise error probability (see next slide) = P + σ 2 15
Achieving ¡Capacity ¡(2/3) When does an error occur? x -sphere Ans : when another codeword x 2 falls inside the uncertainty sphere of α y √ NP What is that probability (pairwise error probability)? r P σ 2 Ans : the ratio of the volume of N P + σ 2 the two spheres N x 1 p NP σ 2 / ( P + σ 2 ) Pr { x 1 → x 2 } = √ N α y NP ◆ N/ 2 σ 2 ✓ x 2 = P + σ 2 Union bound: ◆ N/ 2 σ 2 ✓ Total error probability ≤ 2 NR P + σ 2 16
Achieving ¡Capacity ¡(3/3) • Total error probability (by union bound) !! ◆ N/ 2 R + 1 1 σ 2 N ✓ 2 log 1+ P Pr {E} ≤ 2 NR = 2 σ 2 P + σ 2 • As long as the following holds, Pr {E} → 0 as N → ∞ ✓ ◆ R < 1 1 + P 2 log σ 2 • Hence, indeed the capacity is ✓ ◆ C = 1 1 + P 2 log bits per symbol time σ 2 17
Resources ¡of ¡ AWGN ¡Channel 18
Continuous-‑Time ¡AWGN ¡Channel • System parameters: - Power constraint: P watts; Bandwidth: W Hz - Spectral density of the white Gaussian noise: N 0 /2 • Equivalent discrete-time baseband channel (complex) z [ n ] ∼ CN (0 , N 0 W ) Power constraint: N | x [ n ] | 2 ≤ NP X x [ n ] y [ n ] = x [ n ] + z [ n ] n =1 - 1 complex symbol = 2 real symbols • Capacity: C AWGN ( P, W ) = 2 × 1 ✓ P/ 2 ◆ 2 log 1 + bits per symbol time N 0 W/ 2 SNR := P / N 0 W SNR per complex symbol ✓ ◆ P bits/s = log (1 + SNR ) bits/s/Hz = W log 1 + N 0 W 19
Complex ¡AWGN ¡Channel ¡Capacity ¡ ✓ ◆ P C AWGN ( P, W ) = W log 1 + bits/s N 0 W = log (1 + SNR ) bits/s/Hz Spectral Efficiency • The capacity formula provides a high-level way of thinking about how the performance fundamentally depends on the basic resources available in the channel • No need to go into details of specific coding and modulation schemes • Basic resources: power P and bandwidth W 20
Power 7 P SNR = 6 N 0 W 5 4 log (1 + SNR ) 3 2 1 0 Fix W : 0 20 40 60 80 100 SNR • High SNR: C = log(1 + SNR ) ≈ log SNR - Logarithmic growth with power • Low SNR: C = log(1 + SNR ) ≈ SNR log 2 e - Linear growth with power 21
Bandwidth ✓ ◆ P N 0 W log 2 e = P P Fix 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) 22
Bandwidth-‑limited ¡vs. ¡Power-‑limited ✓ ◆ P C AWGN ( P, W ) = W log 1 + bits/s N 0 W P SNR = N 0 W • When SNR ≪ 1: (Power-limited regime) ✓ ◆ P log 2 e = P C AWGN ( P, W ) ≈ W log 2 e N 0 W N 0 - Linear in power; Insensitive to bandwidth • When SNR ≫ 1: (Bandwidth-limited regime) ✓ ◆ P C AWGN ( P, W ) ≈ W log N 0 W - Logarithmic in power; Approximately linear in bandwidth 23
Capacity ¡of ¡Some LTI ¡Gaussian ¡Channels 24
SIMO ¡Channel y = h x + w ∈ C L Power constraint: P ↓ � 0 , σ 2 I L � w ∼ CN h y x h ∗ MRC , || h || ↓ � 0 , σ 2 � y = || h || x + e w ∼ CN e e w, • MRC is a lossless operation: we can generate y from : e y y = e y ( h / || h || ) • Hence the SIMO channel capacity is equal to the capacity of the equivalent AWGN channel, which is 1 + || h || 2 P ✓ ◆ C SIMO = log Power gain due to σ 2 Rx beamforming 25
MISO ¡Channel x , h ∈ C L y = h ∗ x + w ∈ C , ↓ Power constraint: Tx Beamforming h y x N || x || 2 ≤ NP X x = x h / || h || ↓ n =1 ⇥ h 1 ⇤ h ∗ = h 2 y = x || h || + w • Goal: maximize the received power || h * x || 2 - The answer is || h || 2 P ! (check. Hint: SVD) • Achieved by Tx beamforming - Send a scalar symbol x on the direction of h - Power constraint on x : still P • Capacity: 1 + || h || 2 P ✓ ◆ C MISO = log σ 2 26
Recommend
More recommend
