EECS 216
SOLUTIONS TO PROBLEM SET #6
Winter 2008
1.
#4.5
Throughout
F{
x
(
t
)
}
=
X
(
ω
) =
rect
((
ω
−
1)
/
2) =
braceleftBig
1
for 0
≤
ω
≤
2
0
otherwise
.
Note:
X
(
ω
)
negationslash
=
X
(
−
ω
)
*
so
x
(
t
) is NOT real–DON’T use conjugate symmetry!
1a.
F{
x
(
−
t
)
}
=
X
(
−
ω
) =
rect
((
ω
+ 1)
/
2).
1b.
F{
tx
(
t
)
}
=
j
dX
dω
=
jδ
(
ω
)
−
jδ
(
ω
−
2).
1c.
F{
x
(
t
+ 1)
}
=
X
(
ω
)
e
jω
=
e
jω
rect
((
ω
−
1)
/
2).
1f.
F{
dx
dt
}
=
jωX
(
ω
) =
jω
for 0
≤
ω
≤
2 and 0 otherwise.
1g.
F{
t
dx
dt
}
=
j
d
dω
[
jωX
(
ω
)] = 2
δ
(
ω
−
2)
−
rect
((
ω
−
1)
/
2).
1i.
F{
x
(
t
)
e
-
j
2
t
}
=
X
(
ω
+ 2) =
rect
((
ω
+ 1)
/
2)=answer to (a).
2a.
#4.14
X
(
ω
) =
F{
e
-
at
u
(
t
)
}
=
1
jω
+
a
,
H
(
ω
) =
F{
sin
(2
t
)
πt
}
=
rect
(
ω/
4)=0 outside
|
ω
|
<
2.
Y
(
ω
) =
H
(
ω
)
X
(
ω
) =
rect
(
ω/
4)
jω
+
a
.
2b. Input energy=
integraltext
∞
-∞
|
x
(
t
)
|
2
dt
=
integraltext
∞
0
e
-
2
at
dt
=
1
2
a
.
Output energy=
1
2
π
integraltext
2
-
2
|
1
jω
+
a
|
2
dω
=
1
2
π
integraltext
2
-
2
1
ω
2
+
a
2
dω
=
1
aπ
tan
-
1 2
a
.
So:
1
aπ
tan
-
1 2
a
=
1
2
1
2
a
→
tan
-
1 2
a
=
aπ
4
a
=
π
4
→
a
= 2.
3.
#4.24
X
(
ω
) =
braceleftbigg
2
π
for
|
ω
|
< ω
1
π
for
ω
1
<
|
ω
|
< ω
2
and 0 otherwise.
3a.
Y
(
ω
) = 2
π
for
|
ω
|
< ω
f
→
y
(
t
) = 2
π
sin(
ω
f
t
)
πt
= 2
sin(
ω
f
t
)
t
.
3b.
Y
(
ω
) =
braceleftbigg
2
π
for
|
ω
|
< ω
1
π
for
ω
1
<
|
ω
|
< ω
f
→
y
(
t
) =
sin(
ω
1
t
)
t
+
sin(
ω
f
t
)
t
3c. Since
H
(
ω
) passes all of
x
(
t
),
y
(
t
) =
x
(
t
) =
sin(
ω
1
t
)
t
+
sin(
ω
2
t
)
t
4.
#4.19
Mult. by
p
(
t
) modulates
x
(
t
): shifts spectrum up and down by
ω
0
and halves.
Filtering by
H
1
(
ω
) doesn’t affect this if
ω
B
> ω
m
; clips this if
ω
B
< ω
m
.
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- Winter '08
- Yagle
- Gate, Sin, G protein coupled receptors, Amadeus IT Group
-
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