Image Processing Reference
In-Depth Information
constitutes an
N
-element vector
4
ψ
n
, i.e.,
exp(
it
n
ω
N−
1
))
T
ψ
n
=(1
,
exp(
it
n
ω
1
)
,
exp(
it
n
ω
2
)
···
(6.40)
or
ψ
n
(
m
)=exp(
it
n
ω
m
)
,
with
m
∈
0
···
N
−
1
.
(6.41)
There are
N
such vectors
ψ
n
are orthog-
onal to each other under the conventional vector scalar product of Eq. (3.27). That
is,
ψ
n
because
n
∈
0
···
N
−
1. Furthermore,
N−
1
l
N
)=
Nδ
(
m
ψ
n
,
ψ
m
=
exp(
i
(
m
−
n
)
−
n
)
(6.42)
l
=0
where
δ
is the Kronecker delta, holds.
Exercise 6.3.
Show that
ψ
m
are orthogonal.
HINT: Identify
ψ
m
,
ψ
n
as a geometric series.
To obtain the
k
th element of the vector
F
(
ω
k
), we compose the scalar product be-
tween the vector (
f
(
t
0
, ··· ,f
(
t
N−
1
))
T
ψ
k
, i.e., we multiply both sides of Eq.
(6.39) with exp(
−iω
k
t
n
) and sum over the index
n
:
and
N−
1
N−
1
N−
1
2
π
T
f
(
t
n
) exp(
−
iω
k
t
n
)=
exp(
−
iω
k
t
n
)
F
(
ω
m
) exp(
iω
m
t
n
)
n
=0
n
=0
m
=0
N−
1
N−
1
2
π
T
=
F
(
ω
m
) exp(
−
iω
k
t
n
) exp(
iω
m
t
n
)
n
=0
m
=0
N−
1
N−
1
2
π
T
=
F
(
ω
m
)
exp(
−
iω
k
t
n
) exp(
iω
m
t
n
)
m
=0
n
=0
exp
exp
inm
2
π
N
N−
1
N−
1
2
π
T
−ink
2
π
N
=
F
(
ω
m
)
m
=0
n
=0
N−
1
2
π
T
=
F
(
ω
m
)
Nδ
(
k
−
m
)
m
=0
=
N
2
π
T
F
(
ω
k
)
(6.43)
where we have used
t
n
ω
m
=
nm
2
N
and the orthogonality of
ψ
m
as given by Eq.
(6.42). Consequently, we obtain
N−
1
F
(
ω
m
)=(
N
2
π
T
)
−
1
f
(
t
n
) exp(
−
iω
m
t
n
)
(6.44)
n
=0
4
Note the typographic difference that now
ψ
n
is in boldface because it is an array or a
conventional vector, in contrast to
ψ
n
, which represents a function.
