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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
A.A. 2011/2012 Temporal Data Analysis Figure 2.2: The triangular function and consecutive approximations by a Fourier series with more and more terms M.Orlandini 15
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Chapter5 Procedures for Analyzing R
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Chapter6 Temporal Analysis in X-ray
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AppendixA Examples Shown at the Bla
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