where Y == [yl,y2,y3 ]T , X == [Xl, X 2 , X 3 ]T , Z ==[Zl, Z2, Z3]T are the vectors containing the received symbols,the transmitted symbols and the zero mean unit varianceadditive white Gaussian noise symbols for users indicated bythe subscripts. The transmit power constraint for each user isE[Xf] ::; P, k == 1,2,3. Consider two different values of thechannel matrix,
It is shown in [4] that taken individually either channelmatrix H; or Hj, by itself results in a sum capacity oflog(l + 3P), so that separate coding can at most achieve acapacity 21og(1 + 3P). However, taken together, the capacityof the parallel interference channel is 31og(1 + 2P) which isachieved only by joint coding across both channel matrices.The key is the complimentary nature of the two channelmatrices, i.e. ~ (H a + H b ) == I which allows the receivers tocancel interference by simply adding the outputs of the parallelchannels, provided the transmitters send the same symbol overboth channels.
In this paper, we take this idea further by recognizingthat in the ergodic setting, for a broad class of channeldistributions, the channel states can be partitioned into suchcomplimentary pairings over which interference can be alignedso that each user is able to achieve (slightly more than) half ofhis interference-free ergodic capacity at any SNR. Prior workin [3] has shown that for fading channels every user is able toachieve half the channel degrees of freedom. In other words,each user achieves (slightly less than) half of his interferencefree capacity asymptotically as SNR approaches infinity. Fairlysophisticated interference alignment schemes are constructedto establish this achievability. However, in this work we showthat for a broad class of fading distributions, including e.g.Rayleigh fading, alignment can be achieved quite simply andmore efficiently. Note, however, that the stronger result isobtained at the cost of some loss of generality due to theassumption of ergodic fading and certain restrictions on theclass of fading distributions, that are not needed in [3].
(2)
(1)
11
-1
HX+Zy
n. == [ ~-1
interference channel with the channel matrix:
B. Nazer and M. Gastpar were supported by NSF grants CCR-0347298,CNS-0627024, and CCF-0830428. S. A. Jafar was supported by NSF grantCCF-0830809 and by ONR YIP under grant N00014-08-1-0872. S. Vishwanath was supported by ARO YIP under grant 52491CL
I. INTRODUCTION
The interference channel is one of the fundamental buildingblocks of wireless networks. Following several recent advances, the capacity region of the classical two-user Gaussianinterference channel is known exactly for some interestingspecial cases (e.g. very weak or strong interference), andapproximately (within one bit) for all channel conditions [1].There is also increasing interest in generalizations of the twouser Gaussian interference channel model to more than 2 usersand fading channels. However these generalizations tum outto be far from trivial, as they bring in new fundamental issuesnot encountered in the classical setting. Extensions to morethan 2 users have to deal with the possibility of interferencealignment [2], [3] while extensions to fading channels arefaced with the inseparability of parallel interference channels[4], [5]. Interference alignment refers to the consolidation ofmultiple interferers into one effective entity which can beseparated from the desired signal in time, frequency, space,or signal level dimensions. The inseparability of interferencechannels refers to the necessity for joint coding across channel states. In other words, for parallel Gaussian interferencechannels, the capacity cannot be expressed in general as thesum of the capacity of the sub-channels.
The following example presented in [4] to establish theinseparability of parallel interference channels forms the relevant background for this work. Consider the 3-user Gaussian
Abstract-Consider a K -user interference channel with timevarying fading. At any particular time, each receiver will see asignal from most transmitters. The standard approach to sucha scenario results in each transmitter-receiver pair achieving arate proportional to -I< the single user rate. However, given twowell chosen time indices, the channel coefficients from interferingusers can be made to exactly cancel. By adding up these twosignals, the receiver can see an interference-free version of thedesired transmission. We show that this technique allows eachuser to achieve at least half its interference-free ergodic capacityat any SNR. Prior work was only able to show that half theinterference-free rate was achievable as the SNR tended toinfinity. We examine a finite field channel model and a Gaussianchannel model. In both cases, the achievable rate region has asimpie description and, in the finite field case, we prove it is theergodic capacity region.
The next section presents the main problem statement,where we formulate both a finite-field and a Gaussian interference network model. In Section III, we derive an achievablescheme for the finite field model in Section III and in SectionV we show this matches the upper bound exactly. In SectionIV, we give an achievable scheme for the Gaussian modelwhich we show is quite close to the outer bound for the equalSNR case for any number of users. We conclude the paper inSection VI.
II. PROBLEM STATEMENT
We consider both a finite-field model and a Gaussian model.First, we will give definitions common to both models. Wewill use bold lowercase to denote column vectors and bolduppercase to denote matrices. There are K transmitter-receiverpairs (see Figure 1). Let n denote the number of channel uses.Let each message Wk be chosen independently and uniformlyfrom the set {I, 2, ... , 2n R k } for some Rk 2: O. Message Wk isonly available to transmitter k. Let X be the channel input andoutput alphabet. Each transmitter has an encoding function, £k,that maps the message into n channel uses:
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Definition 1: We say that an ergodic rate tuple(R 1 , R2 , . . . , RK ) is achievable if for all E > 0 and nlarge enough there exist channel encoding and decodingfunctions £1, ... , EK, D1, ... , DK such that:
n, > Rk - E, k == 1,2, ... , K, (6)
Pr({wl # WI} U ... U {WK # WK}) < E. (7)
Definition 2: The ergodic capacity region is the closure ofthe set of all achievable ergodic rate tuples.
A. Finite Field Model
The channel alphabet is a finite field of size q, X == IFq. Thechannel coefficients for block n, hkR, are drawn independentlyand uniformly from IFq \ {O}.
Remark 1: Our results can be extended to the case wherethe channel coefficients are sometimes zero through simplecounting arguments. However, this considerably complicatesthe description of the capacity region.
The additive noise terms Zk (t) are i.i.d. sequences drawnfrom a distribution that takes values on uniformly on{I, 2, ... , q - I} with probability p and is zero otherwise.We define the entropy of Zk(t) to be 0 ::; H(Z) ::; log2 q.
Fig. 1. K -user interference channel with fading.
III. FINITE FIELD ACHIEVABLE SCHEME
We now develop an achievable scheme for the finite fieldcase that can approach the symmetric ergodic capacity. First,we need some tools from the method of types [7]. Let Hdenote the alphabet of the channel matrix so that H(t) E H.Let N(HIHn) be the number of times the channel matrixH E H occurs in the sequence H",
B. Gaussian Model
The channel inputs and outputs are complex numbers, X ==ceo Each transmitter must satisfy an average power constraint:
where 5NRk 2: 0 is the signal-to-noise ratio. The channelcoefficients are drawn independently of each other and acrosstime. They can be drawn from any distribution that is symmetric about zero (with P(hkR) == P( -hkR)). This includes manypopular fading models such as Rayleigh fading and uniformphase fading. The noise terms are i.i.d. sequences drawn froma Ray leigh distribution, Z k ( t) rv eN(0, 1).
Remark 2: Our choice of power constraint eliminates theneed to search for the optimal power allocation policy. A nonequal power allocation over channel states could certainly beincluded as part ofour scheme but for the sake of simplicity weexplicitly disallow it. See [6] for a study of power allocationfor fast fading 2-user interference channels.
Remark 3: We could also allow for different interferenceto-noise ratios between each transmitter and receiver (usuallywritten as INRkR). However, the achievable rate derived inSection IV would still only depend on the 5NRk parameters.
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H(t)
Dk: X n ~ {1,2, ... ,2 n R k}
and produces an estimate Wk of its desired message Wk.
K
Yk(t) == L hkR(t)XR(t)+ Zk(t)R=1
where Zk(t) is additive noise. Note that addition and multiplication are carried out over a finite field or the complex field,depending on the channel model.
Each receiver is equipped with a decoding function:
We focus on the fast fading scenario where the channelmatrix changes at every time step. Let H(t) == {hkR(t) }kRdenote the channel matrix at time t and let H" denote theentire sequence of channel matrices. We assume that beforeeach time step t, all transmitters and receivers are given perfectknowledge of the channel matrix H(t).
At time t, the channel output seen by receiver k is givenby:
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Definition 3: A sequence of channel matrices, H", IS 8typical if:
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First, we will give an equivalent description of this rate regionand then show that any rate tuple can be achieved by timesharing the symmetric rate point from Theorem 1 and a singleuser transmission scheme.
Theorem 1: For the K -user finite field interference channel,the rate tuple (RsYM, R sYM, ••• ,RsYM) is achievable where:
RR + u; ::; log2 q - H(Z), Vk i- .e. (17)
Proof For any E > 0, let 8 be a small positive constantthat will be chosen later to satisfy our rate requirement. UsingLemma 1, choose n large enough so that P(A8') 2:: 1 - ~.
Assume that 8 and n are chosen such that n( I~I - 8) is aneven integer. Now condition on the event that the sequenceof channel matrices, H", is 8-typical. Since the channelcoefficients are i.i.d. and uniform, the probability of anychannel H E H is I~I' Since H" is 8-typical we have thatfor every H E H:
n C~I -Ii) < N(HIHn
) <::: n C~I -Ii) (16)
Throw out all but the first n( I~I - 8) indices for eachchannel realization. This results in losing at most a 8 fractionof the total rate. Group together all time indices that havechannel realization H and call this set of indices TH . We willencode for each TH separately. For each channel realizationH, transmitter .e generates a message W RH E IF~ wherem == ~(I~I - 8)(log2q)-1(RsYM - ~). Using a computationcode from Lemma 3, each transmitter .e sends its messageW RH during the first ~ ( I~ I - 8) time indices in TH . Receiver
k makes an estimate UkH of UkH == L~1 hkRWRH.For each channel realization H E H, pair up the first
~ (I~I - 8) blocks with H with the last ~ (I~I - 8) blockswith g(H) using g(.) from Lemma 2. Since 9 is one-to-one,this procedure pairs up all of the channel indices. During thelast ~ (I~I - 8) indices with channel g(H), the transmittersuse the message, WRH, and a computation code from Lemma3. The receivers make an estimate VkH of VkH == VkH ==f(hkk)WkH - LR#k hkRWRH where f(·) is the function suchthat f(h kR) + hkR == 1.
For n large enough, the total probability of error for allcomputation codes is upper bounded by ~. Receiver k makesan estimate of WkH by simply adding up the two equationsto get WkH == UkH + VkH. Note that the transmitters do notknow a priori which time indices will be successfully paired.To deal with this, the transmitters use an erasure code with rateat least (1 - 8)RsYM - ¥ with probability of error no greaterthan ~ over all transmissions. By choosing 8 small enough,we finally get that each receiver can recover its message ata rate greater than ~ (log2 q - H (Z)) - E with probability oferror less than E as desired. •
Theorem 2: For the K -user finite field interference channel,any rate tuple (R 1 , ... ,RK ) , satisfying the following inequalities is achievable:
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(12)
(13)R == log2 q - H(Z)
g(H) ==
where -hkRis the additive inverse of hkR. Clearly, g(H)+H ==I and 9 (.) is one-to-one. •
The basic idea underlying our scheme is to add together twowell-chosen channel outputs such that the interference exactlycancels out. However, for the finite field model, if we do thisin an uncoded fashion, we risk accumulating noise. Thus, wedenoise the desired linear functions using computation codesprior to combining them together [8].
Lemma 3: Consider a K -user finite field interference channel with fixed channel coefficients hkR E IFq \ {O}:
K
Yk(t) == L hkRXR(t) + Zk(t)R=1
I~N(HIHn) - p(H)1 <::: Ii \/H E 1{ (9)
where P(H) is the probability of channel H E H under thechannel model. Let A8' denote the set of all 8-typical channelmatrix sequences.
Lemma 1 (Csiszar-Korner 2.12): For any i.i.d. sequence ofchannel matrices, H", the probability of the set of all 8-typicalsequences, A8" is lower bounded by:
Proof Sketch: Let G E IF~ x m be a good linear code foradditive noise channel at rate R. Each encoder transmitsXR == GWR. Each receiver observes:
K
Yt; == L GhkRWR + Zk == GUk + Zk (14)R=1
from which it can recover Uk reliably. See Theorem 1 in [8]for a full proof and extensions.
We will now show that all users can achieve half the singleuser rate simultaneously.
where Zk (t) is i.i.d. additive noise with entropy H (Z). Eachtransmitter has a message Wk E IF~. The maximum rate, R ==~ log2 q, at which each receiver can reliably recover the linear
function Uk == L~1 hkRWR is given by:
P(An) > 1 - J2iL
8 - 4n8 2
For a proof, see [7].Lemma 2: There exists a one-to-one map, 9 IF: X K --*
IF: X K such that H + g(H) == I, VH where I is the identitymatrix.
Proof: Let f : F q --* F q be the one-to-one map such thatf(a) + a == 1 for all a E IFq . Since IFq is a finite field, f(·)is guaranteed to exist. Then, define g(.) as follows:
f(h 11) -h12 -h1K
-h21 f(h 22) -h2K
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Lemma 4: Assume, without loss of generality, that theusers are labeled according to rate in descending order, sothat R I 2: R2 2: ... 2: RK. The achievable rate region fromTheorem 2 is equivalent to the following rate region:
R I ::; log2 q - H(Z) (18)
1u; < minIlog., q - H(Z) - R I, 2(log2 q - H(Z))}, k 2: 2
Proof: The key idea is that only one user can achieve arate higher than ~ (Iog., q - H(Z)). From (17), we must havethat R I + R2 ::; log2 q - H(Z) so if R I > ~(log2 q - H(Z))all other users must satisfy u, ::; log2 q-H(Z) -RI. If R I ::;~ (log2 q - H(Z)), then we have that Rk ::; ~ (log2 q - H(Z))for all other users since the rates are in descending order. •
Proof of Theorem 2: We show that the equivalent rateregion developed by Lemma 4 is achievable by time-sharing.First, we consider the case where R I > ~ (log2 q - H (Z) ).Let a == 2(1 - log2 q~lH(Z))' We allocate an channel uses tothe symmetric scheme from Theorem 1. For, the remaining(1 - a)n channel uses, users 2 through K are silent, and user1 employs a capacity-achieving point-to-point channel code.This results in user 1 achieving its target rate R I :
a(log2 q - H(Z)) + (1 - a) (log2 q - H(Z)) (19)2
== log2 q - H(Z) - R I - log2 q + H(Z) + 2R I == R I
and users 2 through K achieving Rk == log2 q - H(Z) - R I.If R I ::; ~(log2 q-H(Z)), we can achieve any rate point withthe use of the symmetric scheme from Theorem 1. •
IV. GAUSSIAN ACHIEVABLE SCHEME
The scheme for the Gaussian case is quite similar to ourfinite field scheme. The key difference is that we need toquantize the channel alphabet so that we can deal with a finiteset of possible matrices. By decreasing the quantization binsize, we can approach the desired rate in the limit. Also, hereit is beneficial to transmit combine the channel outputs priorto decoding to exploit a power gain.
Definition 4: For, > 0, let Q'Y(hkRJ represent the closestpoint in ,(Z + jZ) to hkR in Euclidean distance. The ,quantized version of a channel matrix H E C KxK is givenby H'Y == {Q'Y(hkR) }kR.
Theorem 3: For the K-user Gaussian interference channel,the rate tuple (R I, R2, ... ,RK) is achievable for:
u; = ~E [log (1 + 2lhkkl2SNRk)] . (20)
Proof: For any E > 0, choose 7 > ° such thatP (UkR{lhkRI > 7}) < ~. Let, and 8 be small positiveconstants that will be chosen later to satisfy our rate requirement. Also, using Lemma 1, choose n large enough so thatP( A8') 2: 1 - ~. We will throw out any time index with achannel coefficient with magnitude larger than 7. This ensuresthat the -y-quantized version of the channel is of finite size.Specifically, the size of the channel alphabet H, is given byIHI == (22:.)2K
2• We assume that 7",8 and n are chosen so
"1that all the appropriate ratios only result in integers.
ISIr 2009, Seoul, Korea, June 28 - July 3, 2009
We condition on the event that the sequence of ,-quantizedchannel matrices, H~, is 8-typical. Unlike the finite field case,the channel matrix distribution is not uniform. For all H'Y E Hwe have that:
Throw out all but the first n(P(H'Y) - 8) blocks of eachchannel realization. This causes a loss of at most a 8 fractionin rate. Let hkR denote the elements of H'Y' We define thefollowing one-to-one map 9 : H ~ H:
hII -hI2-hil hi2
-hkl -hk2 hkKNote that due to the symmetry of the channel distributionP(g(H'Y)) == P(H'Y)' Group together all time indices thathave channel realization H'Y and call this set of indicesTHI'. For each channel realization H E H, pair up thefirst ~ (P(H'Y) - 8) blocks with channel H'Y with the last~ (P(H'Y) - 8) blocks with channel g(H'Y)' We ensure thatwe use the same channel inputs during time index i from THI'
for i == 1,2, ... , ~(P(H'Y) - 8) as we do during time indexi + ~(P(H'Y) - 8) from ~(HI')' Let tl denote the first timeand t2 denote the second time. We have the following channeloutputs:
Since tl has quantized channel H'Y and t2 has quantizedchannel 9 (H'Y) we have that the channel from X k (tI) toYk(tl) + Yk(t2) has a signal-to-noise ratio of at least:
Thus, there exist , and 7 such that we achieve Rk >~E[log (1 + 21hkk12SNRk)J - E with probability 1 - Eo
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Similar outer bounds hold for all receiver pairs k and f.Comparing these to the achievable region in Theorem 2 yieldsthe capacity region. •
Using the results from [6], we have the following outerbound on the ergodic capacity region of the K -user Gaussianinterference channel.
Theorem 5: For the K-user Gaussian interference channelwith i.i.d. Rayleigh fading, the following constraints are anouter bound to the ergodic capacity region:
V. UPPER BOUNDS
We now briefly describe upper bounds for both the finitefield case and the Gaussian case. The finite field upper boundmatches the achievable performance thus yielding the ergodiccapacity region. For the Gaussian case, we demonstrate thatour achievable performance is very close to the upper boundwhen the transmitters have equal power constraints.
Theorem 4: For the K -user finite field interference channel,the ergodic capacity region is:
[1] R. H. Etkin, D. N. C. Tse, and H. Wang, "Gaussian interference channelcapacity to within one bit," IEEE Transactions on Information Theory,vol. 54, pp. 5534-5562, December 2008.
[2] M. A. Maddah-Ali, A. S. Motahari, and A. K. Khandani, "Communicationover mimo x channels: Interference alignment, decomposition, and performance analysis," IEEE Transactions on Information Theory, vol. 54,pp. 3457-3470, August 2008.
[3] V. R. Cadambe and S. A. Jafar, "Interference alignment and the degreesof freedom for the K user interference channel," IEEE Transactions onInformation Theory, vol. 54, pp. 3425-3441, August 2008.
[4] V. Cadambe and S. A. Jafar, "Multiple access outerbounds and theinseparability of parallel interference channels," Dec. 2008.
[5] L. Sankar, X. Shang, E. Erkip, and H. V. Poor, "Ergodic two-userinterference channels: Is separability optimal?," 46th Annual AllertonConf. on Comm., Control, and Computing, 2008.
[6] D. Tuninetti, "Gaussian fading interference channels: Power control," inProceedings of the 42nd Asilomar Conference on Signals, Systems andComputers, (Monterey, CA), October 2008.
[7] I. Csiszar and 1. Komer, Information Theory: Coding Theorems forDiscrete Memoryless Systems. New York: Academic Press, 1982.
[8] B. Nazer and M. Gastpar, "Computation over multiple-access channels,"IEEE Transactions on Information Theory, vol. 53, pp. 3498-3516,October 2007.
[9] S. A. Jafar, "The ergodic capacity of interference networks," tech. rep.,October 2009. See http://arxiv.org/abs/0902.0838.
REFERENCES
case, uniform phase fading channels with a large number ofusers [9].
Fig. 2. Ergodic symmetric rate per user and upper bound for the K -userGaussian interference channel with i.i.d. Rayleigh fading and equal transmitpowers: SNRI = SNR2 = ... = SNRK .
VI. CONCLUSIONS
We developed a new communication strategy, ergodic interference alignment, that codes efficiently across parallelinterference channels. With this strategy, every user in thechannel can attain at least half the rate available to them inthe single-user setting. Moreover, we showed that for a finitefield channel model this achievable scheme matches the outerbound exactly, thus yielding the ergodic capacity region. Thekey to the achievable strategy was perfect channel knowledgeat the transmitters. An interesting direction for future workis developing an ergodic alignment scheme for the case oflimited channel state information.
(27)Vk -I f.
»:+ RR < E [log (1 + IhkRI2SNRR+ 1~1~~:~2~~kRk) ]
+ E [log (1 + IhRkl2SNRk + 1~I~~R~~S~RJ ] Vk # £
In Figure 2, we plot the performance of our scheme versusthe upper bound from Theorem 5 for the equal SNR, equalrate per user case. The plot is for i.i.d. Rayleigh fadingand is valid for any number of users K. This shows thatergodic interference alignment can provide close-to-optimalperformance for any number of users so long as they have thesame SNR constraint. In very recent work, Jafar has shown thatergodic interference alignment achieves capacity whenever anetwork is in a "bottleneck state." This includes, as a special
Proof The required upper bound follows from stepssimilar to those in Appendix II of [3]. Without loss ofgenerality, we upper bound the rates of users 1 and 2. Notethat the capacity of the interference channel only dependson the noise marginals. Thus, we can assume that ZI(t) ==hI2(t)(h22(t))-1 Z2(t). Let Y2(t) == hI2(t)(h22(t))-1Y2(t).
We give the receivers full access to the messages from users3 through K as this can only increase the outerbound. FromFano's inequality, we have that n(R I +R2 - En ) where E;; ~ 0as n ~ 00 is upper bounded as follows:
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