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Homework 3 Solutions
Total: 55 points
8.19
(10 points)
S
[
k
] =
Nx
[(

k
)
N
] =
Nx
[0]
,
k
= 0
Nx
[
N

k
]
,
k
= 1
...N

1
To get this, start with the deﬁnition:
S
[
k
] =
N

1
X
n
=0
X
[
n
]
e

j
2
π
N
kn
Substitute the analysis expression for
X
[
n
]:
S
[
k
] =
N

1
X
n
=0
X
[
n
]
e

j
2
π
N
kn
=
N

1
X
n
=0
N

1
X
m
=0
x
[
m
]
e

j
2
π
N
mn
!
e

j
2
π
N
kn
=
N

1
X
m
=0
x
[
m
]
N

1
X
n
=0
e

j
2
π
N
n
(
m
+
k
)
Here, we have the sum over one period of a complex exponential, which can be written
as
Nδ
[(
m
+
k
)
N
]. This expression will be
N
when (
m
+
k
) mod
N
= 0 and 0 for all
other values of
m
. The value for
m
for which the expression in the sum is non0 is thus
m
= 0 when
k
= 0 and
m
=
N

k
when
k
= 1
...N

1. We then have
S
[
k
] =
N

1
X
m
=0
x
[
m
]
N

1
X
n
=0
e

j
2
π
N
n
(
m
+
k
)
=
N
N

1
X
m
=0
x
[
m
]
δ
[(
m
+
k
)
N
] =
Nx
[(

k
)
N
]
=
Nx
[0]
,
k
= 0
Nx
[
N

k
]
,
k
= 1
...N

1
8.20
(10 points)
Start with the deﬁnition of
Y
[
k
]:
Y
[
k
] =
2
N

1
X
n
=0
y
[
n
]
e

j
2
π
2
N
kn
Since only the even samples of
y
[
n
] have non0 values, we can substitute
m
=
n/
2
→
n
= 2
m
as the index in the sum:
Y
[
k
] =
N

1
X
m
=0
y
[2
m
]
e

j
π
N
k
(2
m
)
=
N

1
X
m
=0
x
[
m
]
e

j
2
π
N
km
=
X
[
k
]
,k
= 0
...N

1
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View Full Document This resulting expression is the DFT of
x
[
n
] =
X
[
k
], except that
X
[
k
] is only deﬁned
for
k
= 0
...N

1. However, if we input values of
k > N
into the sum, by periodicity:
Y
[
k
] =
N

1
X
m
=0
x
[
m
]
e

j
2
π
N
km
=
N

1
X
m
=0
x
[
m
]
e

j
2
π
N
(
k

N
)
m
=
X
[
k

N
]
,k
=
N ...
2
N

1
Thus, the ﬁnal expression for
Y
[
k
] is
Y
[
k
] =
X
[
k
]
,
k
= 0
...N
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This document was uploaded on 05/26/2010.
 Spring '09

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