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ECE 562
Fall 2011
HW 3 SOLUTIONS
1. For a complex random vector
Y
, show the following:
Σ = (Σ
I
+ Σ
Q
) +
j
(Σ
QI

Σ
IQ
)
and
ˇ
Σ = (Σ
I

Σ
Q
) +
j
(Σ
QI
+ Σ
)
Ans:
Σ =
E
h
(
Y

m
Y
) (
Y

m
Y
)
†
i
=
E
±
²
Y
I
+
j
Y
Q

(
m
Y
I
+
jm
Y
Q
)
³²
Y
I
+
j
Y
Q

(
m
Y
I
+
Y
Q
)
³
†
´
=
E
±
²
Y
I

m
Y
I
+
j
²
Y
Q

m
Y
Q
³³²
Y
I

m
Y
I
+
j
²
Y
Q

m
Y
Q
³³
†
´
=
E
±
²
Y
I

m
Y
I
+
j
²
Y
Q

m
Y
Q
³³
µ
(
Y
I

m
Y
I
)
>

j
²
Y
Q

m
Y
Q
³
>
¶´
=
E
±
(
Y
I

m
Y
I
)(
Y
I

m
Y
I
)
>
+
²
Y
Q

m
Y
Q
Y
Q

m
Y
Q
³
>
´
+
j
E
±µ
²
Y
Q

m
Y
Q
³
(
Y
I

m
Y
I
)
>

(
Y
I

m
Y
I
)
²
Y
Q

m
Y
Q
³
>
¶´
= (Σ
I
+ Σ
Q
) +
j
(Σ
QI

Σ
)
˘
Σ =
E
h
(
Y

m
Y
) (
Y

m
Y
)
>
i
=
E
±
²
Y
I
+
j
Y
Q

(
m
Y
I
+
Y
Q
)
Y
I
+
j
Y
Q

(
m
Y
I
+
Y
Q
)
³
>
´
=
E
±
²
Y
I

m
Y
I
+
j
²
Y
Q

m
Y
Q
Y
I

m
Y
I
+
j
²
Y
Q

m
Y
Q
³³
>
´
=
E
±
²
Y
I

m
Y
I
+
j
²
Y
Q

m
Y
Q
³³
µ
(
Y
I

m
Y
I
)
>
+
j
²
Y
Q

m
Y
Q
³
>
¶´
=
E
±
(
Y
I

m
Y
I
Y
I

m
Y
I
)
>

²
Y
Q

m
Y
Q
Y
Q

m
Y
Q
³
>
´
+
j
E
±µ
²
Y
Q

m
Y
Q
³
(
Y
I

m
Y
I
)
>
+
(
Y
I

m
Y
I
)
²
Y
Q

m
Y
Q
³
>
¶´
= (Σ
I

Σ
Q
) +
j
(Σ
QI

Σ
)
c
±
V.V. Veeravalli, 2011
1
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View Full Document 2. Consider the signal
s
(
t
) = [sin(
πt
) +
j
cos(
πt
)]11
0
≤
t
≤
1
. Suppose this signal is corrupted
by complex WGN
w
(
t
) with PSD
N
0
= 2 to form the received signal
r
(
t
) =
s
(
t
) +
w
(
t
)
.
Further suppose we form the random variable
Z
as:
Z
=
Z
1
0
r
(
t
) sin(
πt
)
dt
(a) Find
P
{
Z
I
>
1
}
.
(b) Find
P
{
Z
I
+
Z
Q
>
1
}
.
(c) Find
P
(
{
Z
I
>
1
} ∩ {
Z
Q
>
1
}
)
Ans:
For a complex WGN process
w
(
t
) and a function
ψ
(
t
) by deﬁnition
h
w,ψ
i ∼ CN
(
0
,N
o
k
ψ
k
2
)
Then in the case of
Z
Z
=
Z
1
0
r
(
t
) sin
πtdt
=
Z
1
0
(
s
(
t
) +
w
(
t
)) sin
=
Z
1
0
(
sin
2
πt
+
j
sin
πt
cos
πt
)
dt
+
Z
1
0
w
(
t
) sin
=
1
2
+
h
i
where
ψ
(
t
) = sin
πt
. Since
k
ψ
k
2
=
1
2
,
h
i ∼ CN
(0
,
N
o
2
). So
Z
∼ CN
(
1
2
,
N
o
2
),
Z
I
∼ N
(
1
2
,
N
o
4
), and
Z
Q
∼ N
(0
,
N
o
4
) with
Z
I
and
Z
Q
independent.
(a) We have
P
{
Z
I
>
1
}
=
Q
±
1

1
2
p
N
o
/
4
!
=
Q
²
1
√
2
³
(b) Since
Z
I
and
Z
Q
are independent,
Z
I
+
Z
Q
∼ N
(
1
2
,
N
o
2
). Then
P
{
Z
I
+
Z
Q
>
1
}
=
Q
±
1

1
2
p
N
o
/
2
!
=
Q
²
1
2
³
(c) Again since
Z
I
and
Z
Q
are independent
P
{
Z
I
>
1
,Z
Q
>
1
}
=
P
{
Z
I
>
1
}
P
{
Z
Q
>
1
}
=
Q
±
1

1
2
p
N
o
/
4
!
Q
±
1

0
p
N
o
/
4
!
=
Q
²
1
√
2
³
Q
´
√
2
µ
c
±
V.V. Veeravalli, 2011
2
3.
Unequal Priors.
Describe how the optimal decision regions for BPSK signaling get
modiﬁed when the priors on the messages are not the same, say
π
1
= 2
/
3 and
π
2
= 1
/
3.
Ans:
For BPSK we have points in the constellation at
±
√
E
. The conditional pdfs of
r
are given by
p
1
(
r
) =
1
πN
o
exp
{
r

√
E
2
/N
o
}
p
2
(
r
) =
1
πN
o
exp
{
r
+
√
2
o
}
Then we decide on 1 if
π
1
p
1
(
r
)
> π
2
p
2
(
r
)
or equivalently
π
1
p
1
(
r
)
π
2
p
2
(
r
)
>
1
Subsitituting the deﬁnitions gives
π
1
π
2
exp
±

r
+
√
2
 
r

√
2
N
o
²
>
1
⇒
π
1
π
2
exp
±
4
√
E
r
I
N
o
²
>
1
Taking the log of both sides and solving for
r
I
gives
r
I
>

N
o
4
√
E
ln
³
π
1
π
2
´
=

N
o
4
√
E
ln2
4.
P
e
for PAM.
Compute (the exact) P
e
as a function of
E
s
for 4ary PAM. Note that
P
e
,m
is not the same for all
m
in this case.
Figure 1: 4ary PAM Constellation
Ans:
The average signal energy is
E
s
=
³
2
d
2
4
+ 2
9
d
2
4
´
1
4
=
5
d
2
4
c
²
V.V. Veeravalli, 2011
3
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View Full Document so
d
=
q
4
E
s
5
. We need to consider the case of exterior and interior points separately.
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This document was uploaded on 02/08/2012.
 Fall '09

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