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Problem Set 2 – Solution
3.2) (15 Points)
Since
1
)
(
=
θ
f
X
it must be that
]
[
]
[
n
n
x
δ
=
.
Hence sampling
x(t)
must produce
]
[
n
and since
0
2
02
.
0

t
e
for
all
t
the defining term is
sinc(t)
. For the above result to hold we must have at least
T=1
.
3.5) (15 Points)
For (3.65) please note that:
)
0
(
)
(
)
(
0
F
t
j
X
dt
e
t
x
dt
t
x
∫
∫
∞
∞

∞
∞

=
=
. Similarly via (3.14 and 3.10) we get
)
0
(
)
2
0
(
)
(
0
f
k
F
n
nT
j
TX
T
k
X
e
nT
x
T
=

=
∑
∑
∞
∞
=
∞
∞
=
π
which is equivalent using eq. 3.19.
For (3.66) we use the Parseval’s theorem:
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 Fall '06
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