lecture 7 multiple access channelhomepage.ntu.edu.tw/.../slides/it_lecture_07_v3.pdf · lecture...

Basic Bounds and Gaussian MACGeneral Discrete Memoryless MAC

Summary

Lecture 7Multiple Access Channel

I-Hsiang Wang

Department of Electrical EngineeringNational Taiwan University

[email protected]

December 10, 2014

1 / 49 I-Hsiang Wang NIT Lecture 7

[email protected]


Summary

Noiseless Graphical Network → Noisy Multi-User Network

We have shown that Shannon’s paradigm can be readily extended tonoiseless graphical networks. Moreover, in particular for single multicast,

Zero-error decoding at finite blocklength is feasible.Low-complexity explicit construction of network codes is possible.

Caveat: There are couples of distinct features that a single noiselessgraphical multicast problem possesses such that its solution is so elegant:

Noiseless: links are modeled as noiseless finite capacitated edges.Orthogonal: links can carry independent information withoutinterfering with one another.

In other words, it well models the overlay network beyond PHY layer.Beyond wireline: in many scenarios, the communication medium isshared among multiple users, and hence the above two features may befar from valid. As a first step, we investigate several kinds of simplesingle-hop multi-user noisy channels.



Summary

ComparisonLecture 6 Lecture 7,8,9

Noiseless graphicalmulticast

Single-hop multi-usernoisy channel

Topology General Single-hop

Linkage Noiseless, Orthogonal General (Gaussian asMain Example)

TrafficSingle unicast/multicast;Special cases of multiple

unicast (MAC, BC)

Multiple access channel,Broadcast channel,Interference channel



Summary

Key Features Missing in Noiseless Graphical Multicast

1 Superposition of signals at receiving terminals. The simplestone-hop model is the multiple access channel. (Lecture 7)

Encoder 1

Multiple Access Channel DecoderEncoder

2

Encoder K

X1

X2

XK

Source 1

Source 2

Source Kp (y|x1, . . . , xK)

Y

Destination

2 Broadcast of signals from transmitting terminals. The simplestone-hop model is the broadcast channel. (Lecture 8)

Broadcast Channel

Decoder 2Encoder

Source Destination 2X

Y1 Decoder 1

Destination1

Decoder K

Destination K

Y2

YKp (y1, . . . , yK |x)



Summary

Key Features Missing in Noiseless Graphical Multicast

3 Interference among independent information flows. The simplestone-hop model is the interference channel. (Lecture 9)

Interference Channel

Decoder 2

Destination 2

Y1 Decoder 1

Destination1

Decoder K

Destination K

Y2

YK

Encoder 1

Encoder 2

Encoder K

Source 1

Source 2

Source Kp

�y[1:K]|x[1:K]

�

We shall start with the multiple access channel (MAC), which answerspartially the following two kinds of questions:

1 How do multiple transmitters trade-off their rates when accessing asingle receiver? (Traffic Pattern)

2 How does a single receiver decode multiple data streams when theyare superimposed together? (Superposition)



Summary

Multiple Access Channel: Problem Formulation

ENC 1

DECENC 2

ENC K

X1

X2

XK

Y

......

......

......

W1

W2

WK

cW1, . . . ,cWKpY |X1,...,XK

1 K independent messages {W1, . . . ,WK}, each of which is onlyaccessible by one encoder. Wk ∼ Unif

[1 : 2NRk

], ∀ k ∈ [1 : K].

2 Channel:(X1, . . . ,XK, pY|X1,...,XK , Y

).

3 Rate tuple: (R1, . . . ,RK).



Summary


ENC 1

DECENC 2

ENC K

X1

X2

XK

Y

......

......

......

W1

W2

WK

cW1, . . . ,cWKpY |X1,...,XK

4 A(2NR1 , 2NR2 , . . . , 2NRK ,N

)MAC channel code consists of

∀ k ∈ [1 : K], an encoding function enck,N :[1 : 2NRk

]→ XN

k thatmaps message wk to a length N codeword xN

k .a decoding function decN : YN →×K

k=1

[1 : 2NRk

]that maps a

channel output yN to a reconstructed message tuple (w1, . . . , wK).



Summary


ENC 1

DECENC 2

ENC K

X1

X2

XK

Y

......

......

......

W1

W2

WK

cW1, . . . ,cWKpY |X1,...,XK

5 Error probability P(N)e := Pr

{(W1 . . . ,WK) =

(W1, . . . , WK

)}.

6 A rate tuple R := (R1, . . . ,RK) is said to be achievable if thereexist a sequence of

(2NR1 , 2NR2 , . . . , 2NRK ,N

)MAC channel codes

such that P(N)e → 0 as N → ∞.

7 The capacity region C := cl{

R ∈ [0,∞)K: R is achievable

}.



Summary

Lecture Overview

We mainly focus on the two-user case (K = 2) in this lecture. For MAC,the results in the two-user case can be extended to the K-user case in astraightforward manner.

1 First we extend the achievability of point-to-point channels to thetwo-user MAC, and establish a capacity inner bound.

2 Second we characterize the capacity region of Gaussian MAC byproving that the inner bound above is tight.

3 We then use Gaussian MAC as an example to introduce variousschemes including successive interference cancellation (SIC),time-sharing, etc.

4 Finally we characterize the capacity region for general MAC, byproviding an enhanced achievability and a general converse proof.



Summary

1 Basic Bounds and Gaussian MAC

2 General Discrete Memoryless MAC

3 Summary



Summary

Capacity Inner Bound

Let us begin with extending the achievability of point-to-point channel.Now the decoder has to decode two independent messages, and the keylies in how to analyze the error event in the appropriate way.

Lemma 1 (Achievability)If (R1,R2) ≥ 0 satisfies the following for some (X1,X2) ∼ pX1 · pX2 ,then (R1,R2) is achievable.

R1 < I (X1 ;Y |X2) (1)R2 < I (X2 ;Y |X1) (2)

R1 + R2 < I (X1,X2 ;Y) (3)

Remark: Note that the input distribution is chosen such that X1 ⊥⊥ X2,which is reasonable since the two encoders are not cooperating.



Summary

Proof of Achievability

pf: As in the point-to-point case, we use random coding argument toprove the existence of sequence of

(2NR1 , 2NR2 ,N

)-codes such that

limN→∞ P(N)e = 0 as long as (1) – (3) hold.

Our proof is divided into three parts (similar to the point-to-point case):(1) random codebook generation, (2) encoding and decoding, and(3) error probability analysis.Random codebook generation: ∀ k = 1, 2, randomly and independentlygenerate 2NRk sequences xN

k (wk), wk ∈[1 : 2NRk

], each i.i.d. over time

according to pXk (that is, XNk ∼

∏Ni=1 pXk (xk[i])).

Encoding: ∀ k = 1, 2, to send message wk, Encoder k transmits xNk (wk).

Decoding: To facilitate error probability analysis, use typicality decoder:(w1, w2) = a unique (w1,w2) ∈

[1 : 2NR1

]×[1 : 2NR2

]such that(

xN1 (w1) , xN

2 (w2) , yN) ∈ T (N)ϵ (X1,X2,Y).



Summary

Error Probability Analysis: By the symmetry of codebook generation,we can assume WLOG the actual message tuple is (W1,W2) = (1, 1)and focus on analyzing the “averaged-over-codebook” error probabilitygiven (W1,W2) = (1, 1): (E denotes the error event

(W1, W2

)= (W1,W2))

P(1,1) {E} := Pr {E| (W1,W2) = (1, 1)} .

The key is to distinguish error event E into the following four cases Ea,E(1)

t , E(2)t , and E(1,2)

t , such that E = Ea ∪ E(1)t ∪ E(2)

t ∪ E(1,2)t , where

Ea :={(

XN1 (1) ,XN

2 (1) ,YN) /∈ T (N)ϵ

}E(1)

t :={(

XN1 (w1) ,XN

2 (1) ,YN) ∈ T (N)ϵ for some w1 = 1

}E(2)

t :={(

XN1 (1) ,XN

2 (w2) ,YN) ∈ T (N)ϵ for some w2 = 1

}E(1,2)

t :={(

XN1 (w1) ,XN

2 (w2) ,YN) ∈ T (N)ϵ for some w1 = 1,w2 = 1

}



Summary

Next, we would like to find a set of sufficient conditions under which theabove error events have vanishing probability as N → ∞.Following the point-to-point proof, let us define event

A(w1,w2) :={(

XN1 (w1) ,XN

2 (w2) ,YN) ∈ T (N)ϵ

},

and rewrite Ea = Ac

(1,1)

E(1)t =

∪w1 =1 A(w1,1)

E(2)t =

∪w2 =1 A(1,w2)

E(1,2)t =

∪w1 =1,w2 =1 A(w1,w2)

.

Hence,

E = Ac(1,1) ∪

( ∪w1 =1

A(w1,1)

)∪

( ∪w2 =1

A(1,w2)

)∪

( ∪w1 =1,w2 =1

A(w1,w2)

).

Next, we present a key lemma bounding the probability of these events.



Summary

Lemma 2P(1,1)

{A(1,1)

}≥ 1− ϵ for N large enough, and

P(1,1)

{A(w1,1)

}≤ 2−N(I(X1;Y|X2)−δ1(ϵ)) for all w1 = 1

P(1,1)

{A(1,w2)

}≤ 2−N(I(X2;Y|X1)−δ2(ϵ)) for all w2 = 1

P(1,1)

{A(w1,w2)

}≤ 2−N(I(X1,X2;Y)−δ1,2(ϵ)) for all w1 = 1, w2 = 1.

where δ1 (ϵ) , δ2 (ϵ) , δ1,2 (ϵ) → 0 as ϵ → 0.

With the above lemma and the union of events bound, we see that for Nsufficiently large, P(1,1) {Ea} ≤ ϵ, and

P(1,1)

{E(1)

t

}≤ 2NR12−N(I(X1;Y|X2)−δ1(ϵ))

P(1,1)

{E(2)

t

}≤ 2NR22−N(I(X2;Y|X1)−δ2(ϵ))

P(1,1)

{E(1,2)

t

}≤ 2N(R1+R2)2−N(I(X1,X2;Y)−δ1,2(ϵ))

∴ As long as (R1,R2) satisfies (1) – (3), limN→∞ P(1,1) (E) = 0.



Summary

Proof of Lemma 2

Proof of P(1,1)

{A(1,1)

}≥ 1− ϵ for N large enough:

Given (W1,W2) = (1, 1),(XN

1 (1),XN2 (1),YN) are distributed i.i.d. over

time according to pX1,X2,Y = pX1 · pX2 · pY|X1,X2. Hence by LLN and the

fact that the typicality decoder is based on pX1,X2,Y, proof complete.

Proof of P(1,1)

{A(w1,1)

}≤ 2−N(I(X1;Y|X2)−δ1(ϵ)) for all w1 = 1:

Due to the memoryless channel assumption and the fact that codewords aregenerated i.i.d. at random, XN

1 (w1)⊥⊥(XN

2 (1) ,YN), and

P(1,1)

{A(w1,1)

}=

∑(xN

1 ,xN2 ,yN)∈T (N)

ϵ

p(

xN1

)· p

(xN2 , yN

)≤ 2N(1+ϵ)H(X1,X2,Y) · 2−N(1−ϵ)H(X1) · 2−N(1−ϵ)H(X2,Y)

= 2−N(I(X1;Y|X2)−δ1(ϵ)),

where δ1(ϵ) = ϵ (H (X1,X2,Y) + H (X1) + H (X2,Y)) → 0 as ϵ → 0.

Proof of the other two statements follows similarly.



Summary

Gaussian MAC: Model

ENC 1

ENC 2

X1

X2

W1

W2

DECY

Z

g2

g1

cW1,cW2

1 Channel law: Y = g1X1 + g2X2 + Z. Z ∼ N(0, σ2

)⊥⊥ (X1,X2).

2 White Gaussian: {Z [t]} is an i.i.d. (white) Gaussian random process3 Memoryless: Z[t]⊥⊥

(W1,W2,Xt−1

1 ,Xt−12 ,Zt−1

).

4 Average power constraint: 1N∑N

t=1 |xk[t]|2 ≤ Pk, k = 1, 2.

5 Signal-to-noise ratio: SNRk := |gk|2Pkσ2 , k = 1, 2.



Summary

Characterization of Gaussian MAC Capacity

Theorem 1 (Capacity of Gaussian MAC)If (R1,R2) ≥ 0 satisfies the following, then (R1,R2) is achievable.

Rk <1

2log (1 + SNRk) , k = 1, 2 (4)

R1 + R2 <1

2log (1 + SNR1 + SNR2) (5)

Conversely, if (R1,R2) ≥ 0 is achievable, then it must satisfy (4) – (5)with “<” replaced by “≤”.

pf: Achievability is proved by extending the inner bound in Lemma 1 tothe continuous setting with input cost constraint (omitted), and pickXk ∼ N (0,Pk), k = 1, 2. (Evaluation of mutual information is left as exercise.)

To prove the converse part, we make use of Fano’s inequality and dataprocessing inequality to obtain for k = 1, 2, (ϵk,N → 0 as N → ∞ below)



Summary

NRk = H (Wk) = I(

Wk; Wk

)+ H

(Wk

∣∣∣Wk

)≤ I

(Wk;YN)+ Nϵk,N.

Bound (4) on Individual Rate: let k = 1 (k = 2 can be similarlyproved), and we continue with the above inequality:

N (R1 − ϵ1,N) ≤ I(W1;YN) (a)

≤ I(W1;YN,W2

) (b)= I

(W1;YN|W2

)=∑N

t=1 I(W1;Y[t]|Yt−1,W2

)≤∑N

t=1 I(W1,Yt−1;Y[t]|W2

).

(a) is due to I(W1;W2|YN) ≥ 0; (b) is due to I (W1;W2) = 0.

Next we upper bound I(W1,Yt−1;Y[t]|W2

)by I (X1[t];Y[t]|X2[t])

I(W1,Yt−1;Y[t]|W2

) (c)= I

(W1,Yt−1,X1[t];Y[t]|W2,X2[t]

)≤ I

(W1,W2,Yt−1,X1[t];Y[t]|X2[t]

) (d)= I (X1[t];Y[t]|X2[t]) .

(c) is due to the fact that Xk[t]f= Wk for k = 1, 2. (d) is due to the

memorylessness of the channel:(W1,W2,Yt−1

)− (X1[t],X2[t])− Y[t].



Summary

Bound (5) on Sum Rate: set ϵN := ϵ1,N + ϵ2,N. Then,

N (R1 + R2 − ϵN) ≤ I(W1;YN|W2

)+ I(W2;YN)

= I(W1,W2;YN) =∑N

t=1 I(W1,W2;Y[t]|Yt−1

)(a)≤∑N

t=1 I (X1[t],X2[t];Y[t])(b)≤ N · 1

2 log (1 + SNR1 + SNR2)

(a) is similar to the previous proof (exercise). (b) is due to the followingand Jensen’s inequality (exercise):

I (X1[t],X2[t];Y[t]) = h (Y[t])− h (Y[t]|X1[t],X2[t])= h (g1X1[t] + g2X2[t] + Z[t])− h (Z[t])(c)≤ 1

2 log(1 + Var[g1X1[t]+g2X2[t]]

σ2

)(d)= 1

2 log(1 +

|g1|2P1,t+|g2|2P2,tσ2

)(c) is due to “Gaussian maximizes differential entropy”. (d) is due to thefact that X1[t]⊥⊥ X2[t] since W1 ⊥⊥ W2.



Summary

Capacity Region of Gaussian MAC

R1

R2

12 log (1 + SNR1)

12 log (1 + SNR2)

12 log

⇣1 +

SNR21+SNR1

⌘

12 log

⇣1 +

SNR11+SNR2

⌘

CGMAC =

((R1, R2) � 0

��

(Rk 1

2 log (1 + SNRk) , k = 1, 2

R1 +R2 12 log (1 + SNR1 + SNR2)

)



Summary

Successive Interference Cancellation

A natural scheme for the receiver to resolve multiple data streams issuccessive interference cancellation:

1 First decode one user’s data, treating the other user’s signal as noise.2 Remove the decoded data, and then decode the next user’s data.

Proposition 1 (Successive Decoding Achievability)SIC with decoding order W1 → W2 achieves (R1,R2) ≥ 0 satisfying (forsome (X1,X2) ∼ pX1 · pX2 in the first equalities below){

R1 < I (X1;Y) = 12 log

(1 + SNR1

1+SNR2

)R2 < I (X2;Y|X1) =

12 log (1 + SNR2)

The proof is quite straightforward. Decoding W1 first will be successfulas long as R1 < I (X1;Y). Decoding W2 with XN

1 (W1) known will besuccessful as long as R2 < I (X2;Y|X1).



Summary

Time Sharing: Capacity Region is Convex

Proposition 2 (Time Sharing Achievability)If rate tuples R(1) ∈ [0,∞)

K and R(2) ∈ [0,∞)K are both achievable,

then R(λ) := λR(1) + λR(2) is also achievable ∀λ ∈ [0, 1], λ := 1− λ.

pf: Since R(1) and R(2) are both achievable, there exist a sequence of(2NR(1)

,N)

-codes and a sequence of(2NR(2)

,N)

-codes, both of whichhave vanishing error probability. The main idea of achieving R(λ) is tosplit the blocklength N into two parts λN and λN, and split the data Winto two parts W(1) ∈

[1 : 2NλR(1)

]and W(2) ∈

[1 : 2NλR(2)

].

The proof is complete by using a(2λNR(1)

, λN)

-code to send W(1) in

the first part (of length λN) and using a(2λNR(2)

, λN)

-code to sendW(2) in the second part (of length λN).



Summary

R1

R2

I (X1;Y |X2)

I (X2;Y |X1)

I (X2;Y )

I (X1;Y )

B

A

SIC with decoding order W1 → W2 achieves A (the green region)SIC with decoding order W2 → W1 achieves B (the blue region)



Summary

R1

R2

I (X1;Y |X2)

I (X2;Y |X1)

I (X2;Y )

I (X1;Y )

B

A

With time sharing, all other rate pairs inside the inner bound region canbe achieved.



Summary



3 Summary



Summary

Enlarged Achievable Region by Time Sharing

By Lemma 1, for (X1,X2) ∼ pX1 · pX2 , an achievable rate region (innerbound of capacity region) Rinner (X1,X2) is defined by

Rinner (X1,X2) :=

(R1,R2) ≥ 0

∣∣∣∣∣ R1 ≤ I (X1;Y|X2)R2 ≤ I (X2;Y|X1)R1 + R2 ≤ I (X1,X2;Y)

By time sharing we obtain an enlarged region (conv: convex hull operation)

Rinner := conv

∪(X1,X2)∼pX1

·pX2

Rinner (X1,X2)

. (6)

Remark: For (scalar) Gaussian MAC, since a single distribution(Gaussian) maximizes three boundaries simultaneously, time sharing willnot enlarge the inner bound, and hence capacity region is characterizedwithout using time sharing.



Summary

R1

R2

R1



Summary

R1

R2

R1

R2



Summary

R1

R2

R1

R2

R3



Summary

R1

R2

R1

R2

R3

conv (R1 [ R2 [ R3)



Summary

It turns out that Rinner in (6) is equal to C .To prove this result, we first develop an outer bound region Router, andthen show that outer and inner bounds match, Router = Rinner = C .



Summary

Outer Bound Region of General MAC

Lemma 3 (Outer Bound)If (R1,R2) ≥ 0 is achievable, then it must satisfy the following rateconstraints for some (Q,X1,X2) ∼ pQ · pX1|Q · pX2|Q:

R1 ≤ I (X1;Y|X2,Q) (7)R2 ≤ I (X2;Y|X1,Q) (8)

R1 + R2 ≤ I (X1,X2;Y|Q) (9)

Router (Q,X1,X2) := {(R1,R2) ≥ 0 satisfying (7) – (9)}.

Hence, an outer bound region of C can be defined as

Router :=∪

(Q,X1,X2)∼pQ·pX1|Q·pX2|Q

Router (Q,X1,X2) . (10)



Summary

Single Letterization

pf: Recall that in the converse proof of Gaussian MAC, without makinguse of the assumption of Gaussian, we arrive at:If (R1,R2) ≥ 0 is achievable (i.e., ∃ a sequence of

(2NR1 , 2NR2 ,N

)-codes with

limN→∞ P(N)e = 0), then (R1,R2) must satisfy the following 3 inequalities

for some(XN

1 ,XN2

),XN

1 ⊥⊥ XN2 :

R1 ≤ 1N∑N

t=1 I (X1[t];Y[t]|X2[t]) + ϵ1,N (11)R2 ≤ 1

N∑N

t=1 I (X2[t];Y[t]|X1[t]) + ϵ2,N (12)R1 + R2 ≤ 1

N∑N

t=1 I (X1[t],X2[t];Y[t]) + ϵN (13)

Observe that the right-hand-side of (11) – (13) are the average of mutualinformation terms over time. In the point-to-point case, since there’sonly one rate constraint, just find a maximizing distribution to upperbound the mutual information terms at all time slots.



Summary

Auxiliary Random Variable Q

R1 ≤ 1N∑N

t=1 I (X1[t];Y[t]|X2[t]) + ϵ1,N (11)R2 ≤ 1

N∑N

t=1 I (X2[t];Y[t]|X1[t]) + ϵ2,N (12)R1 + R2 ≤ 1

N∑N

t=1 I (X1[t],X2[t];Y[t]) + ϵN (13)

However, here we cannot do this, simply because such simultaneouslymaximizing distribution may not exist for all rate constraints (except forsome special cases such as Gaussian MAC).Instead, we introduce an auxiliary random variable Q ∼ Unif [1 : N] and(X1,X2,Y) such that (X1,X2,Y) |{Q = t} d

= (X1[t],X2[t],Y[t]).

=⇒ I (X1;Y|X2,Q) =∑N

t=1 Pr {Q = t} I (X1;Y|X2,Q = t)= 1

N∑N

t=1 I (X1[t];Y[t]|X2[t])



Summary

Since XN1 ⊥⊥ XN

2 , the newly introduced (Q,X1,X2) ∼ pQ · pX1|Q · pX2|Q.Finally, let us rewrite the righ-thand-side of inequalities (11) – (13) intoI (X1;Y|X2,Q) + ϵ1,N, I (X2;Y|X1,Q) + ϵ2,N, and I (X1,X2;Y|Q) + ϵNrespectively, and obtain

R1 ≤ I (X1;Y|X2,Q) + ϵ1,N

R2 ≤ I (X2;Y|X1,Q) + ϵ2,N

R1 + R2 ≤ I (X1,X2;Y|Q) + ϵN

Hence, if (R1,R2) ≥ 0 is achievable, then (7) – (9) must hold for some(Q,X1,X2) ∼ pQ · pX1|Q · pX2|Q.

Q here is usually called the “time-sharing” random variable, for reasonsthat will become clear later.Remark: In multi-user information theory, in order to obtain single-lettercapacity bounds, introduction of auxiliary random variable is ofteninevitable, except for some special cases such as Gaussian networks.



Summary

So far, by Lemma 1, time sharing, and Lemma 3, we have shown

Rinner ⊆ C ⊆ Router.

Next, we shall prove Router ⊆ Rinner to complete the proof for

Router = Rinner = C .



Summary

Capacity Region for General MAC

Theorem 2 (Capacity of General MAC)For a general multiple access channel pY|X1,X2

, its capacity region C canbe characterized in the following two equivalent forms:

C = Rinner as in (6) = Router as in (10).

pf: Note that for any (Q,X1,X2) ∼ pQ · pX1|Q · pX2|Q, bounds (7) – (9)defines a “pentagon” with a 45° side, with two corner points

A:{

R1 = I (X1;Y|Q)

R2 = I (X2;Y|X1,Q), B:

{R1 = I (X1;Y|X2Q)

R2 = I (X2;Y|Q).

As long as point A and B ∈ Rinner, we prove that Router ⊆ Rinner.This is simple to prove since point A is a convex combination of points(I (X1;Y) , I (X2;Y|X2)) achieved in Lemma 1. Similarly, B ∈ Rinner.



Summary

Coded Time Sharing

We establish the capacity region of general MAC in a pretty convolutedway. The main reason is that, the inner bound by Lemma 1 and (6), hasa different form from the outer bound by Lemma 3 and (10).Question: Can we directly prove an inner bound that has the same formas the outer bound by Lemma 3 and (10)?The answer is YES, by a new coding technique called coded time sharing.

Lemma 4 (Inner Bound by Coded Time Sharing)If (R1,R2) ≥ 0 satisfies the following rate constraints for some(Q,X1,X2) ∼ pQ · pX1|Q · pX2|Q, then it is achievable:

R1 ≤ I (X1;Y|X2,Q)

R2 ≤ I (X2;Y|X1,Q)

R1 + R2 ≤ I (X1,X2;Y|Q)



Summary

Proof Sketch of Coded Time Sharing Inner Bound

The key idea is to generate a time-sharing sequence qN to control the“configuration” of coding schemes at the two distributed encoders.Hence, qN should be thought of as part of the codebook, and it isrevealed to ALL terminals (all encoders and the decoder).

ENC 1

DEC

ENC 2

yN

w1

w2

bw1, bw2pY |X1,X2

qN

QNt=1 pQ (q[t])

QNt=1 pX1|Q (x1[t]|q[t])

QNt=1 pX2|Q (x2[t]|q[t])

x

N2 (w2)

x

N1 (w1)

�q

N, x

N1 ( bw1) , x

N2 ( bw2) , y

N�

2 T (N)✏ (Q,X1, X2, Y )



Summary

R1

R2

Q = {1, 2} , |Q| = 2

2Y

k=1

pXk|Q (·|Q = 1)

2Y

k=1

pXk|Q (·|Q = 2)



Summary

R1

R2

Q = {1, 2} , |Q| = 2

2Y

k=1

pXk|Q (·|Q = 1)

2Y

k=1

pXk|Q (·|Q = 2)

pQ (1) = 34



Summary

R1

R2

Q = {1, 2} , |Q| = 2

2Y

k=1

pXk|Q (·|Q = 1)

2Y

k=1

pXk|Q (·|Q = 2)

pQ (1) = 12

pQ (1) = 34



Summary

R1

R2

Q = {1, 2} , |Q| = 2

2Y

k=1

pXk|Q (·|Q = 1)

2Y

k=1

pXk|Q (·|Q = 2)

pQ (1) = 14

pQ (1) = 12

pQ (1) = 34



Summary

R1

R2

Q = {1, 2} , |Q| = 2

2Y

k=1

pXk|Q (·|Q = 1)

2Y

k=1

pXk|Q (·|Q = 2)

No need to take the convex hull!



Summary

Remarks

1 For the capacity region in (10) to be computable, the time-sharingrandom variable Q has to take value in a finite set Q.For a K-user MAC, it suffices to evaluate (10) for |Q| ≤ K. This iscalled the cardinality bound on the auxiliary random variable.

2 The capacity region in (10), without any convex hull operation, isalready convex and closed.

3 For MAC, it turns out that coded time sharing does not improve theinner bound. This is in general not true for other multi-userchannels, such as the interference channel.

4 The single-letterization technique presented in the converse proofwill be useful in other multi-user channel coding problems.



Summary



3 Summary



Summary

Comparison between point-to-point and multi-user channels:1 Capacity vs. Capacity Region; Supremum vs. Closure

2 Single-letterization in converse proofs:single maximizing distribution vs. auxiliary random variables

3 Achievability: both use random coding arguments; extendsstraightforwardly

Time sharing

Successive decoding (successive interference cancellation SIC)

Coded time sharing


lecture 7 multiple access channelhomepage.ntu.edu.tw/.../slides/it_lecture_07_v3.pdf · lecture...

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