2
EE 562a
Homework Solutions 6
November 5, 2009
=
∞
X
t
=
∞
x
(
t
)
Z
1
/
2

1
/
2
e

i
2
πft
Y
*
(
f
)
df
!
=
∞
X
t
=
∞
x
(
t
)
Z
1
/
2

1
/
2
e
i
2
πft
Y
(
f
)
df
!
*
=
∞
X
t
=
∞
x
(
t
)
y
*
(
t
)
3. Solution:
(a)
z
(
u, t
) =
ax
(
u, t
) +
by
(
u, t
) =
⇒
m
z
(
t
) =
m
z
=
am
x
+
bm
y
,
and using centered notation,
K
z
(
t
+
τ, t
) =
K
z
(
m
)
=
E
{
z
o
(
u, t
+
τ
)
z
*
o
(
u, t
)
}
=
E
{
[
ax
o
(
u, t
+
τ
) +
by
o
(
u, t
+
τ
)][
a
*
(
x
*
o
(
u, t
) +
b
*
y
*
o
(
u, t
)]
}
=

a

2
K
x
(
τ
) +

b

2
K
y
(
τ
) +
a
*
b
E
{
x
*
o
(
u, t
)
y
o
(
u, t
+
τ
)
}
+
ab
*
E
{
x
o
(
u, t
+
τ
)
y
*
o
(
u, t
)
}
=

a

2
K
x
(
τ
) +

b

2
K
y
(
τ
)
e
K
z
(
t
+
τ, t
)
=
e
K
z
(
τ
) =
E
{
(
z
(
u, t
+
τ
)

m
z
)(
z
(
u, t
)

m
z
)
}
=
a
2
e
K
x
(
τ
) +
b
2
e
K
y
(
τ
)
.
where the fact that
x
(
u, t
) and
y
(
u, t
0
) are independent for all integers
t, t
0
, was used factor
the expectations in the crosscovariance and psuedocrosscovariance derivations.
The
PSD is found by taking the discretetime Fourier transform of the correlation function:
S
z
(
f
)
=
F
{
R
z
(
τ
)
}
=
F
{
K
z
(
τ
) +

m
z

2
}
=
F
{
a

2
K
x
(
τ
) +

b

2
K
y
(
τ
) +

am
x
+
bm
y

2
}
=
F
{
a

2
(
K
x
(
τ
) +

m
x

2
) +

b

2
(
K
y
(
τ
) +

m
y

2
) + 2
Re
ab
*
m
x
m
*
y
}
=

a

2
S
x
(
f
) +

b

2
S
y
(
f
) + 2
Re
ab
*
m
x
m
*
y
δ
D
(
f
)
,
where the above holds for
f
∈
[

1
/
2
,
1
/
2] (periodic with period 1).
(b) This is the product of two random sequences
m
z
(
t
)
=
m
z
=
E
{
x
(
u, t
)
y
(
u, t
)
}
=
E
{
x
(
u, t
)
}
E
{
y
(
u, t
)
}
=
m
x
m
y
R
z
(
t
+
τ, t
)
=
R
z
(
τ
) =
E
{
(
x
(
u, t
+
τ
)
y
(
u, t
+
τ
))(
x
(
u, t
)
y
(
u, t
))
*
}
=
E
{
x
(
u, t
+
τ
)
x
*
(
u, t
)
}
E
{
y
(
u, t
+
τ
)
y
*
(
u, t
)
}
=
R
x
(
τ
)
R
y
(
τ
)
K
z
(
τ
)
=
R
z
(
τ
)
 
m
z

2
=
R
x
(
τ
)
R
y
(
τ
)
 
m
x
m
y

2
=
(
K
x
(
τ
) +

m
x

2
)(
K
y
(
τ
) +

m
y

2
)
 
m
x

2

m
y

2
=

m
y

2
K
x
(
τ
) +

m
x

2
K
y
(
τ
) +
K
x
(
τ
)
K
y
(
τ
)
.
Similarly,
e
K
z
(
τ
) =
m
2
y
e
K
x
(
τ
) +
m
2
x
e
K
y
(
τ
) +
e
K
x
(
τ
)
e
K
y
(
τ
)
.