EE 478
Handout #20
Multiple User Information Theory
November 6, 2008
Homework Set #5 Solutions
1.
Solution:
(a) From the lecture notes, the capacity region C
BC
(
g
1
, g
2
, P
) of the Gaussian broadcast
channel is given by the set of rate pairs (
R
1
, R
2
) such that
R
1
≤
C(
αS
1
)
,
R
2
≤
C
parenleftbigg
(1

α
)
S
2
αS
2
+ 1
parenrightbigg
,
for some 0
≤
α
≤
1, where
S
1
=
g
2
1
P
and
S
2
=
g
2
2
P
.
For the multipleaccess channel assume
P
1
=
βP
and
P
2
= (1

β
)
P
.
The capacity
region C
MAC
(
P
) is given by the set of rate pairs (
R
1
, R
2
) such that
R
1
≤
C(
βS
1
)
,
R
2
≤
C ((1

β
)
S
2
)
,
R
1
+
R
2
≤
C(
βS
1
+ (1

β
)
S
2
)
,
for some 0
≤
β
≤
1.
(b) We show that the corner points of each region is in the other region. Consider (
R
1
, R
2
) =
(C(
αS
1
)
,
C
parenleftBig
(1
−
α
)
S
2
αS
2
+1
parenrightBig
)
∈
C
BC
(
g
1
, g
2
, P
). For
β
=
α
(
S
2
+ 1)
αS
2
+ 1
,
we have
C
parenleftbigg
(1

α
)
S
2
αS
2
+ 1
parenrightbigg
= C ((1

β
)
S
2
)
.
Note that the defined
β
satisfies
β
≥
α
, therefore
R
1
= C(
βS
1
)
<
C(
αS
1
). Finally, it
suffices to show that
C(
αS
1
) + C
parenleftbigg
(1

α
)
S
2
αS
2
+ 1
parenrightbigg
≤
C(
βS
1
+ (1

β
)
S
2
)
.
Note that
C(
αS
1
) + C
parenleftbigg
(1

α
)
S
2
αS
2
+ 1
parenrightbigg
=
1
2
log
(1 +
S
2
)(
αS
1
+ 1)
αS
2
+ 1
,
and
C(
βS
1
+ (1

β
)
S
2
) =
1
2
log(
α
(
S
2
+ 1)
S
1
αS
2
+ 1
+
(1

α
)
S
2
αS
2
+ 1
+ 1)
= C(
αS
1
) + C
parenleftbigg
(1

α
)
S
2
αS
2
+ 1
parenrightbigg
.
1
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Therefore, C
BC
(
g
1
, g
2
, P
)
⊂
C
MAC
(
g
1
, g
2
, P
). To prove the other direction, let (
R
1
, R
2
) =
(C(
βS
1
)
,
C(
βS
1
+ (1

β
)
S
2
)

C(
βS
1
)) be a corner point of the MAC region. Note that
C(
βS
1
+ (1

β
)
S
2
)

C(
βS
1
) =
1
2
log
parenleftbigg
βS
1
+ (1

β
)
S
2
+ 1
βS
1
+ 1
parenrightbigg
=
1
2
log(1 +
(1

β
)
S
2
βS
1
+ 1
)
= C(
(1

β
)
S
2
βS
1
+ 1
)
.
Therefore, (
R
1
, R
2
)
∈
C
BC
. The other possibility for a corner point of the MAC region
is (
R
1
, R
2
) = (C(
βS
1
+ (1

β
)
S
2
)

C ((1

β
)
S
2
)
,
C ((1

β
)
S
2
)). Note that
C(
βS
1
+ (1

β
)
S
2
)

C ((1

β
)
S
2
) =
1
2
log
parenleftbigg
βS
1
+ (1

β
)
S
2
+ 1
(1

β
)
S
2
+ 1
parenrightbigg
= C(
βS
1
(1

β
)
S
2
+ 1
)
≤
C(
βS
1
)
.
Hence, C
MAC
(
g
1
, g
2
, P
)
⊂
C
BC
(
g
1
, g
2
, P
) as well.
(c) Consider the point
(
R
1
, R
2
) = (C(
βS
1
)
,
C(
βS
1
+ (1

β
)
S
2
)

C(
βS
1
))
=
parenleftbigg
C(
βS
1
)
,
C(
(1

β
)
S
2
βS
1
+ 1
)
parenrightbigg
,
on the boundary of the multipleaccess region. The rate pair (
R
1
, R
2
) can be achieved
using successive cancelation by first decoding the message of user 2 assuming the received
sequence of user 1 as noise and then decoding message of user 1 after subtracting off the
received sequence of user 2. However,
R
1
= C(
βS
1
)
,
R
2
= C
parenleftbigg
(1

β
)
S
2
βS
1
+ 1
parenrightbigg
,
is also on the boundary of the the broadcast region. Hence the same codebook (with
proper power scaling) can be used for the broadcast channel and achieves the same point
on the boundary of the region.
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 Spring '11
 Kelly
 SEPTA Regional Rail, M1 motorway, M0 motorway, CBC, Y1 U, βS1

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