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Temporal Data Analysis A.A. 2011/2012
Proof.
3.3.3 Autocorrelation
H j = 1 N
= 1 N
= 1 N
Here we have {g k } = {f k }, which leads to
N−1 ∑
k=0
N−1
∑
l=0
N−1
∑
l=0
1
N
N−1 ∑
l=0
1
N−1 ∑
f l
N
k=0
f l G ∗ j W −j l
N
f l g ∗ l+k W −k j
N
g ∗ l+k W −k j
N
= F j G ∗ j
and
h k ≡ (f ⋆ f ) k = 1 N
N−1 ∑
l=0
f l × f ∗
l+k
(3.24)
{f k } ↔ {F j }
{h k } = {(f ⋆ f ) k } ↔ {H j } = {F j × F ∗ j } = {|F j | 2 }
(3.25)
In other words: the Fourier transform of the autocorrelation of {f k } is the modulus squared of
the Fourier series {F j } or its power representation.
3.3.4 Parseval Theorem
We use (3.24) for k = 0, that is h 0 , and get on the one side:
h 0 = 1 N
N−1 ∑
l=0
|f l | 2 (3.26)
On the other hand, the inverse transformation of {H j }, especially for k = 0, results in (see (3.11b))
Put together, this gives us the discrete version of Parseval theorem:
1
N
N−1 ∑
h 0 = |F j | 2 (3.27)
N−1 ∑
l=0
j =0
|f l | 2 N−1 ∑
=
j =0
|F j | 2 (3.28)
52 M.Orlandini