Changing Z channel into BEC Show that code {01, 10} can transform a Z channel into a BEC. What is a lower bound to capacity of the Z channel?
Typewriter type 2: ½ ½ ½ ½ Sum channel: 2 C = 2 C1 + 2 C2 where C 1 = C 2 = 0.322 C = 1,322 bits/channel use How many noise free symbols?
Example: Noisy typewriter ½ A A ½ B B C C X Y D D • E • • Z Z C = max (H(Y) – H(Y|X)) = 𝑚𝑝 2 26 − 𝑚𝑝 2 2 = 𝑚𝑝 2 13 bits/transmission Achieved with uniform distribution on the inputs.
Remark: For this example, we can also achieve 𝐷 = 𝑚𝑝 2 13 bits/transmission with P(error)=0 and n = 1 by transmitting alternating input symbols, i.e., X = {A C E … Z}.
Differential Entropy Let 𝑌 be a continuous random variable with density 𝑔 𝑦 and support 𝑇 . The differential entropy of 𝑌 is ℎ 𝑌 = − 𝑔 𝑦 log 𝑔 𝑦 𝑒𝑦 (if it exists). 𝑇 Note: Also written as ℎ 𝑔 .
Examples: Uniform distribution Let 𝑌 be uniform in the interval 0, 𝑏 . Then 1 𝑔 𝑦 = 𝑏 in the interval and 𝑔 𝑦 = 0 outside. 𝑏 1 1 ℎ 𝑌 = − 𝑏 𝑚𝑝 𝑏 𝑒𝑦 = log 𝑏 0 Note that ℎ 𝑌 can be negative for 𝑏 < 1. However, 2 ℎ(𝑔) = 2 log 𝑏 = 𝑏 is the size of the support set, which is non-negative.
Example: Gaussian distribution −𝑦 2 1 Let 𝑌 ~ 𝑦 = 2 2 𝑓𝑦𝑞 ( 2 2 ) 𝑦 2 2 2 − 𝑚𝑜 2 2 ] 𝑒𝑦 Then ℎ 𝑌 = ℎ = − 𝑦 [− 𝐹𝑌 2 1 2 𝑚𝑜 2 2 = 2 2 + 1 2 𝑚𝑜 ( 2 e 2 ) nats = 1 2 𝑚𝑝 ( 2 e 2 ) bits Changing the base we have ℎ 𝑌 =
Relation of Differential and Discrete Entropies Consider a quantization of X, denoted by X Let X = 𝑦 𝑗 i nside the 𝑗 th interval. Then 𝐼(𝑌 ) = - 𝑞 𝑗 𝑚𝑝 𝑞 𝑗 𝑗 = - 𝑔(𝑦 𝑗 ) 𝑚𝑝 𝑔(𝑦 𝑗 ) - 𝑚𝑝 𝑗 ℎ 𝑔 − log
Differential Entropy So the two entropies differ by the log of the quantization level . We can define joint differential entropy, conditional differential entropy, K-L divergence and mutual information with some care to avoid infinite differential entropies.
K-L divergence and Mutual Information 𝑔 𝐸(𝑔 g) = 𝑔 𝑚𝑝 𝑔(𝑦,𝑧) 𝐽 𝑌; 𝑍 = 𝑔 𝑦, 𝑧 𝑚𝑝 𝑔 𝑦 𝑔(𝑧) 𝑒𝑦 𝑒𝑧 Thus , I(X;Y) = h(X) + h(Y) – h(X,Y). Note: h(X) can be negative, but I(X;Y) is always 0.
Differential entropy of a Gaussian vector Theorem: Let 𝒀 be a Gaussian n -dimensional vector with mean and covariance matrix 𝐿. Then 2 log((2 𝑓) 𝑜 𝐿 ) 1 ℎ 𝒀 = where 𝐿 denotes the determinant of 𝐿. Proof: Algebraic manipulation.
The Gaussian Channel Z~N (0, N I) X 𝑋 W Y Channel Channel + Decoder Encoder Power Constraint: EX 2 ≤P
The Gaussian Channel Z~N (0, N I) X Y 𝑋 W Channel Channel + Decoder Encoder Power constraint: EX 2 ≤P W {1,2,…, 2 𝑜𝑆 } = message set of rate R X = (x 1 x 2 … x n ) = codeword input to channel Y = (y 1 y 2 … y n ) = codeword output from channel = decoded message P(error) = P{ W 𝑋} 𝑋
The Gaussian Channel Using the channel n times: X n • Y n • • • • • • • • •
The Gaussian Channel C𝑏𝑞𝑏𝑑𝑗𝑢𝑧 𝐷 = max 𝐽(𝑌; 𝑍) f(x): EX 2 ≤P 𝐽 𝑌; 𝑍 = ℎ 𝑍 − ℎ 𝑍 𝑌 = ℎ 𝑍 − ℎ 𝑌 + 𝑎|𝑌 1 1 = ℎ 𝑍 − ℎ 𝑎 ≤ 2 log 2 e 𝑄 + 𝑂 − 2 log 2 e 𝑂 1 𝑄 = 2 log 1 + 𝑂 bits/transmission
The Gaussian Channel The capacity of the discrete time additive Gaussian channel: 1 𝑄 𝐷 = 2 log 1 + 𝑂 bits/transmission achieved with X ~ N(0 , P).
Bandlimited Gaussian Channel Consider the channel with continuous waveform inputs x(t) 𝑈 1 𝑈 𝑦 2 𝑢 𝑒𝑢 ≤ 𝑄) and Bandwidth with power constraint ( 0 limited to W. The channel has white Gaussian noise with power spectral density N 0 /2 watt/Hz. In the interval (0,T) we can specify the code waveform by 2WT samples (Nyquist criterion). We can transmit these samples over discrete time Gaussian channels with noise variance N 0 /2. This gives 𝑄 𝐷 = 𝑋 log ( 1+ 𝑂 0 𝑋 ) bit/second
Bandlimited Gaussian Channel 𝑄 𝐷 = 𝑋 log ( 1+ 𝑂 0 𝑋 ) bit/second Note: If W 𝑄 𝑂 0 𝑚𝑝 2 𝑓 bits/second. we have C =
Bandlimited Gaussian Channel 𝑆 𝑋 be the spectral density in bits per second Let per Hertz. Also let 𝑄 = 𝐹 𝑐 𝑆 where 𝐹 𝑐 is the available energy per information bit. We get 𝑆 𝐷 𝐹 𝑐 𝑆 𝑋 ≤ 𝑋 = log ( 1+ 𝑂 0 𝑋 ) bit/second. Thus 2 −1 𝐹 𝑐 𝑂 0 ≥ This relation defines the so called Shannon Bound.
The Shannon Bound 2 −1 𝐹 𝑐 𝑂 0 ≥ 𝐹 𝑐 𝐹 𝑐 𝑂 0 (dB) – 𝑂 0 Shannon Bound 0 0.69 -1.59 – 5 0.1 0.718 -1.44 – • 4 0.25 0.757 -1.21 0.5 0.828 -0.82 – 3 1 1 0 – • 2 2 1.5 1.76 𝐹 𝑐 – • 𝑂 0 (dB) 1 4 3.75 5.74 8 31.87 15.03 • -1 0 1 2 3 4 5 6
Shannon’s Water Filling Solution
Parallel Gaussian Channels 3 2.5 2 1
Example of Water Filling Channels with noise levels 2, 1 and 3. Available power = 2 1 0.5 1 1.5 1 0 Capacity= 2 log (1+ 2 ) + 2 log (1+ 1 ) + 2 log (1+ 3 ) Level of noise + signal power = 2.5 No power allocated to the third channel.
Parallel Gaussian Channels 3 2.5 2 1
Differential capacity Discrete memoryless channel as a band limited channel
Multiplex strategies (TDMA, FDMA) P j j 𝑘 = 1 (1 + 𝑄 𝐷 2 log Aggregate capacity: : 𝑂 )
Multiplex strategies (non-orthonal CDMA) P j 1 𝑄 2 log (1 + 𝑂+ 𝑘−1 𝑄 ) j Discrete memoryless channel as a band limited channel M 𝑘 = 1 (1 + 𝑁𝑄 𝐷 2 log 𝑂 ) Aggregate capacity: : j=1
TDMA or FDMA versus CDMA Orthogonal schemes: Bandwidth limitation (2WT dimensions) Number of Users Non-orthogonal CDMA (log has no cap) Aggregate Power
Multiple User Information Theory Building Blocks: Multiple Access Channels (MACs) Broadcast Channels (BCs) Interference Channels (IFCs) Relay Channels (RCs) Note: These channels have their discrete memoryless and Gaussian versions. For simplicity we will look at the Gaussian models.
Multiple Access Channel A well understood model. Models the uplink channel in wireless comm. X 1 W 1 Encoder ^ ^ W 1 ,W 2 Y P(y|x 1 ,x 2 ) Decoder X 2 Encoder W 2 Capacity region obtained by Ahlswede (1971) and Liao (1972)
Capacity region - MAC C = closure of convex hull of { (R1,R2) s.t. R 1 ≤ I(X 1 ; Y | X 2 ), R 2 ≤ I(X 2 ; Y | X 1 ), R 1 + R 2 ≤ I(X 1 , X 2 ; Y ) for all p 1 (x 1 ) . p 2 (x 2 ) }
Multiple Access Channel (MAC)
Broadcast Channel (Cover, 1972) Still open in discrete memoryless case. Models the downlink channel in wireless comm. Y 1 ^ W 1 Decoder X W 1 ,W 2 Encoder P(y 1 ,y 2 |x) Y 2 ^ W 2 Decoder Cover introduced superposition coding.
Superposition coding Message Clouds Y 1 can see the message, Y 2 can see the cloud center
Gaussian Broadcast Channel
Superposition coding (1- )P N 2 P 1
Superposition coding (1- )P N 2 P 1
Interference Channel Gaussian Interference Channel - standard form Brief history Z-Interference channel Symmetric Interference channel
Standard Gaussian Interference Channel Power P1 ^ W 1 W 1 a b ^ W 2 W 2 Power P2
Symmetric Gaussian Interference Channel Power P Power P
Z-Gaussian Interference Channel
Interference Channel: Strategies Things that we can do with interference: Ignore (take interference as noise (IAN) 1. Avoid (divide the signal space (TDM/FDM)) 2. Partially decode both interfering signals 3. Partially decode one, fully decode the other 4. Fully decode both (only good for strong inter- 5. ference , a≥1)
Interference Channel: Brief history Carleial (1975): Very strong interference does not reduce capacity (a 2 ≥ 1+P) Sato (1981), Han and kobayashi (1981): Strong interference (a 2 ≥ 1) : IFC behaves like 2 MACs Motahari, Khandani (2007), Shang, Kramer and Chen (2007), Annapureddy, Veeravalli (2007): Very weak interference (2a(1+a 2 P) ≤ 1 ) : Treat interference as noise – (IAN)
Interference Ch.: History (continued) Sason (2004): Symmetrical superposition to beat TDM Etkin, Tse, Wang (2008): capacity to within 1 bit Polyanskiy and Wu, 2016: Corner points established.
Summary: Z interference Channels Z-Gaussian Interference Channel as a degraded interference channel Discrete Memoryless Channel as a band limited channel Multiplex Region: growing Noisebergs Overflow Region: back to superposition
Z-Gaussian Interference Channel
Degraded Gaussian Interference Channel
Differential capacity Discrete memoryless channel as a band limited channel
Interference x Broadcast Channels
Superposition coding (1- )P N 2 P 1
Superposition coding (1- )P N 2 P 1
Degraded Interference Channel - One Extreme Point
Degraded Interference Channel - Another Extreme Point
Intermediary Points (Multiplex Region)
Admissible region for ( , h)
Intermediary Point (Overflow Region)
Admissible region h Q 1 =1 Q 2 = 1 a = 0.5 N 2 = 3
The Z-Gaussian Interference Channel Rate Region R 2 Q 1 =1 Q 2 = 1 a = 0.5 N 2 = 3 R 1
Admissible region h Q 1 =1 Q 2 = 1 a = 0.99 N 2 = 0.02
The Z-Gaussian Interference Channel Rate Region R 2 Q 1 =1 Q 2 = 1 a = 0.99 N 2 = 0.02 R 1
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