ISIT 2003, Yokohama, Japan, June 29 – July 4, 2003
The On-Off Fading Channel
Etienne Perron1
Mohammad Rezaeian2 and Alex Grant2
Swiss Federal Institute of Technology
Lausanne, Switzerland
Institute for Telecommunications Research
University of South Australia
Australia
Abstract — We introduce a simple on-off fading discrete memoryless multiple-access channel in which the
multiplicative fading for each user is a Bernoulli random process. We find the capacity of this channel in
the case that the receiver does not know the values of
the channel gains.
I. Summary
The capacity for fading channels, where neither transmitter or
receiver knows the fading realization is largely an open problem. For single user channels, the capacity achieving distribution is discrete [1], and numerical methods must be used to
find capacity. Asymptotic results for MIMO Rayleigh channels have been found in [2] and partial results for the Rayleigh
multiple-access channel (MAC) were obtained in [3]. Rather
than attempting extensions of the Rayleigh channel, consider
the following simple generalization of the M -user binary adder
channel [4].
Definition 1 (On-Off MAC). At time k,
Puser i transmits
Xi [k] ∈ X ⊆ R and the output is Y [k] = M
i=1 hi [k]Xi [k] ∈
Y ⊆ R. The hi [k] ∈ {0, 1} are i.i.d. Bernoulli and Pr(hi =
0) = qi is the fading probability for user i.
Let Π be the set of product distributions on X =
(X1 , . . . , XM ). Let Ri be the rate of user i. We wish to find
the sum rate constraint C = maxP (X)∈Π I(X; Y ).
Suppose M = 2 and |X | = 2 and consider X = {a, b},
a < b. Hence Y = {0, a, b, a + b, 2a, 2b}. There are only four
distinct cases: (A) a = 0, (B) a = −b, (C) b = 2a and (D)
a 6∈ {0, −b, b/2}.
Let pi = Pr(Xi = a), for i = 1, 2.
Solution of
∂I(X; Y )/∂pi = 0, i = 1, 2 for Channel A appears intractable,
although an adaptation of the Blahut-Arimoto algorithm may
be used to numerically compute capacity. We can also compare to the single user channel in which the user with the
smallest qi is given exclusive use of the channel. This is a Zchannel with capacity everywhere smaller than the two-user
capacity. The difficulty in computation for Channel A is due
to the fact that transmitting 0 disturbs the symmetry of the
channel. This motivates the use of a different source alphabet.
For Channel B we can easily verify that p1 = p2 = 1/2
is optimal. Let q = min{q1 , q2 }. Then the corresponding
single-user channel is the binary erasure channel with erasure
probability q, and capacity 1 − q. There exist values of q1 , q2
for which 1 − q is larger than the two-user capacity, implying
the superiority of TDMA at those points. However for small
enough values of q1 , q2 , the multiple-access channel capacity is
greater. The optimality of TDMA for i.i.d. fading processes,
as described for the Rayleigh channel in [3] does not hold at
all for Channel A and only in certain cases of Channel B.
1 This work was performed while E. Perron was visiting the Institute for Telecommunications Research
2 Supported by ARC Grant DP0209658.
Why does p1 = p2 = 1/2 achieves capacity for Channel B but not for Channel A? It turns out to be because the transition probability table for Channel B is pointsymmetric to its center. Interchange of the values of pi and
1 − pi for both i = 1 and 2, results in P (Y )|p1 =p01 ,p2 =p02 =
P (−Y )|p1 =1−p01 ,p2 =1−p02 and therefore: H(Y )|p1 =p01 ,p2 =p02 =
Another easily verified fact is
H(Y )|p1 =1−p01 ,p2 =1−p02 .
H(Y |X)|p1 =p01 ,p2 =p02 = H(Y |X)|p1 =1−p01 ,p2 =1−p02 Thus, for all
q1 , q2 ,
I(X; Y )|p1 =p01 ,p2 =p02 = I(X; Y )|p1 =1−p01 ,p2 =1−p02
(1)
It can also be verified (with some difficulty) that I(X; Y ) is
concave on the closed convex set of vectors (p1 , p2 ), with 0 ≤
p1 , p2 , ≤ 1. This is in addition to the usual concavity on the
joint distribution space. Suppose we have found a = (p∗1 , p∗2 )
that maximizes the mutual information. By symmetry (1),
there must be a second maximum corresponding to a0 = (1 −
p∗1 , 1 − p∗2 ), having the same value. Concavity on this product
space implies that all convex combinations, in particular (a +
a0 )/2 result in mutual information that are no smaller which
contradicts our assumption about a being a maximum, unless
a = (1/2, 1/2). For Channel A however, all of the above is not
true, because P (Y |X) does not have the required symmetry
properties (similarly for Channel C).
It turns out that the type of symmetry that we have described here leads to the definition of an interesting class of
multiple-access channels for which the uniform distribution is
always optimal. This class is a type of generalization of the
single-user symmetric channels in [5]. We now consider the
M ≥ 2, |X | ≥ 2 extension of Channel D.
Definition 2 (Unique Outputs). The M -user on-off fading
channel with unique outputs has X = {u1 , u2 , . . . , uN } such
every sum of m ≤ M input symbols is non-zero and unique.
Theorem 1. For a M -user on-off fading channel with unique
outputs, the uniform distribution achieves capacity.
References
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