EE 7615
Solutions Problem Set # 2
Sept.
2010
1. The following fact regarding Gaussian random variables is used in the
solution.
Fact:
If
X
1
,X
2
,X
3
and
X
4
are four Gaussian random variables, then
E
[
X
1
X
2
X
3
X
4
] =
E
[
X
1
X
2
]
E
[
X
3
X
4
]+
E
[
X
1
X
3
]
E
[
X
2
X
4
]+
E
[
X
1
X
4
]
E
[
X
2
X
3
]
.
(a)
E
[
Y
(
t
)] =
E
[
X
2
(
t
)] =
R
X
(0) =
Z
∞
∞
S
X
(
f
)
df
= 2
E
[
Z
(
t
)] =
m
Y
Z
∞
∞
h
(
τ
)
dτ
where
h
(
τ
) is the impulse response of the ﬁlter. From Fourier
transforms we know
Z
∞
∞
h
(
τ
)
dτ
=
H
(0) = 1
Thus
E
[
Z
(
t
)] = 2.
(b)
R
Y
(
t,s
) =
E
[
Y
(
t
)
Y
(
s
)] =
E
[
X
2
(
t
)
X
2
(
s
)]
Using the fact above we get
R
Y
(
t,s
) =
E
[
X
2
(
t
)]
E
[
X
2
(
s
)]+2
{
E
[
X
(
t
)
X
(
s
)
}
2
=
R
X
(0)
R
X
(0)+2
{
R
X
(
t

s
)
}
2
Thus
R
Y
(
τ
) = 2
R
2
X
(
τ
) +
R
2
X
(0)
We see that the process
{
Y
(
t
)
}
is a WSS process.
S
Y
(
f
) =
F {
R
Y
(
τ
)
}
= 2
S
X
(
f
)
*
S
X
(
f
) +
R
2
X
(0)
δ
(
f
)
where
*
is the convolution operation. Therefore
S
Y
(
f
) =
G
(
f
) +
R
2
X
(0)
δ
(
f
) where
G
(
f
) is given by
G
(
f
) =
(
2
 
f
 
f
 ≤
2
0
otherwise.
1
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View Full Document(c) Since
Z
(
t
) is the response of a linear timeinvariant system to
a WSS input, it is WSS. We know that
S
Z
(
f
) =
S
Y
(
f
)

H
(
f
)

2
.
Thus
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 Fall '11
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 Cos, Jaguar Racing, All wheel drive vehicles

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