booklet format - inaf iasf bologna

A.A. 2011/2012
Temporal Data Analysis
A k = 2 [∫ 0
(
1 + 2t ) ( ) ∫ 2πkt
+T /2
(
cos dt + 1 − 2t ) ( 2πkt
cos
T −T /2 T T
0 T T
= 2 ∫ 0
( ) 2πkt
cos dt + 2 ∫ +T /2
( ) 2πkt
cos dt +
T −T /2 T T 0 T
} {{ }
∫
4 0
T 2 t cos
−T /2
= − 8 ∫ +T /2
T 2 t cos
0
( 2πkt
T
( 2πkt
T
=0
)
dt − 4
)
dt
∫ +T /2
( 2πkt
T 2 t cos
T
The last integral can be solved by parts:
∫
x cos ax dx = x a sin ax + 1 cos ax
a2 Therefore we finally have
0
)
dt
) ]
dt
2(1 − cosπk)
A k =
π 2 k 2
Remember that B k = 0 because f (t) is an even function. We can rewrite the A k coefficients in
the form
In formula, we have
⎧
1
for k = 0
2
⎪⎨
A k = 4
for
π 2 k 2 k odd
⎪⎩
0 for k even, k ≠ 0
f (t) = 1 2 + 4 (
π 2 cosωt + 1 9 cos3ωt + 1 cos5ωt + ···)
25
This is the function plotted in Figure 2.2 at page 15.
M.Orlandini 113