booklet format - inaf iasf bologna

Temporal Data Analysis A.A. 2011/2012
Using these identities we can demonstrate that the system of basis functions consisting of
(sinω k t, cosω k t) with k = 0,1,2,... is an orthogonal system. This means that
∫ +T /2
−T /2
∫ +T /2
−T /2
∫ +T /2
−T /2
cosω n t cosω m t dt =
sinω n t sinω m t dt =
⎧
⎨
⎩
0 for n ≠ m
T /2 for n = m ≠ 0
T for n = m = 0
{ 0 for n ≠ m, n = 0, m = 0
T /2 for n = m ≠ 0
(2.5)
(2.6)
sinω n t cosω m t dt = 0 (2.7)
Please note that our basis system is not an orthonormal system, i.e. the integrals for n = m are
not normalized to 1. What’s even worse, the special case of n = m = 0 in (2.5) is a nuisance, and
will keep bugging us again and again.
Using the above orthogonality relations, we are able to calculate the Fourier coefficients straight
away. We need to multiply both sides of (2.1) by cosω k t and integrate from −T /2 to +T /2.
Due to the orthogonality, only terms with k = k ′ will remain; the second integral will always
disappear. This gives us:
A k = 2 T
A 0 = 1 T
∫ +T /2
−T /2
∫ +T /2
−T /2
f (t)cosω k t dt for k ≠ 0 (2.8)
f (t)dt (2.9)
Please note the prefactors 2/T or 1/T , respectively, in (2.8) and (2.9). Equation (2.9) simply is
the average of the function f (t). Now let’s multiply both sides of (2.1) by sinω k t and integrate
from −T /2 to +T /2. We now have:
B k = 2 T
∫ +T /2
−T /2
f (t)sinω k t dt for all k (2.10)
Equations (2.8) and (2.10) may also be interpreted like: by weighting the function f (t) with
cosω k t or sinω k t, respectively, we “pick” the spectral components from f (t), when integrating,
corresponding to the even or odd components, respectively, of the frequency ω k .
Ex. 2.1 Calculation of Fourier coefficients: Constant and triangular
functions
14 M.Orlandini