Boston University
Department of Electrical and Computer Engineering
ENG EC515 Digital Communication (
Fall 2009
)
Solution to Problem set 5
Posted:
Mon 26 Oct
Suggested reading:
Classnotes, handouts, and Ch. 4 textbook
Problem 5.1
(Irrelevant data, dimensionality reduction)
One of three (
K
=
3) messages is transmitted over a vector (real, zeromean) AWGN channel with
noise variance
σ
2
. The messages need not be equally likely. Speciﬁcally,
M
∈ M
=
{
1
,
2
,
3
}
and
M
∼
p
M
where
p
M
need not be uniform. The channel codewords (or symbols) are given by
C
3
=
{
x
1
,
x
2
,
x
3
}
where
x
1
=
(0
.
5
,
1
,
10
,
2)
T
,
x
2
=
(0
.
5
,
2
,
10
,
3)
T
, and
x
3
=
(0
.
5
,
3
,
10
,
4)
T
. Notice
that all three codewords agree in their ﬁrst and third components. Let
X
=
x
M
be the transmitted
vector,
Y
=
X
+
Z
be the received vector, and
Z
be real, zeromean, AWGN with variance
σ
2
.
(a) Show that the optimum receiver can ignore
Y
(1) and
Y
(3) which are the ﬁrst and third com
ponents of
Y
.
(b) Now suppose that the noise is
not
white. Speciﬁcally, let all 4 components of the noise be
equal to
Z
where
Z
∼ N
(0
, σ
2
)(
z
) is independent of the message
M
. Thus,
Z
(1)
=
Z
(2)
=
Z
(3)
=
Z
(4)
=
Z
. (i) [1pt] Can the optimum receiver ignore
Y
(1) and
Y
(3)? (ii) [1pt] Why?
(iii) Design an optimum receiver for this situation; (iv) Compute the probability of error of
the optimum receiver.
Solution:
(a)
[2pts]
b
m
MAP
(
y
)
=
arg max
m
∈M
±
p
Y

X
(
y

x
m
)
p
M
(
m
)
²
=
arg max
m
∈M
{
p
Z
(
y

x
m
)
p
M
(
m
)
}
=
arg min
m
∈M
{
ln(
p
Z
(
y

x
m
))

ln(
p
M
(
m
))
}
=
arg min
m
∈M
K
X
i
=
1
1
2
σ
2

y
(
i
)

x
m
(
i
)

2

ln(
p
M
(
m
))
=
arg min
m
∈M
X
i
=
2
,
4
1
2
σ
2

y
(
i
)

x
m
(
i
)

2

ln(
p
M
(
m
))
+
X
i
=
1
,
3
1
2
σ
2

y
(
i
)

x
m
(
i
)

2
=
arg min