booklet format - inaf iasf bologna

A.A. 2011/2012
Temporal Data Analysis
S(t) = f (t) ⊗ g (t) (2.45)
where S(t) is the experimental, smeared signal. It’s obvious that the rise at t = 0 will not be as
steep, and the peak of the exponential function will get “ironed out”. We’ll have to take a closer
look:
S(t) = 1
σ 2π
= 1
∫ +∞
0
(
e −ξ/τ exp − 1 (t − ξ) 2 )
2 σ 2 dξ
t 2 )∫ +∞
[
σ 2 exp − ξ τ + tξ
(−
σ 2π exp 1 2 0
σ 2 − 1 ξ 2 ]
2 σ 2 dξ
= 1 (−
σ 2π exp 1 t 2 ) ( 1 t 2 ) (
2 σ 2 exp
2 σ 2 exp − t ) ( σ
2
)
exp
τ 2τ 2 ×
∫ {
+∞
exp − 1 [ )] 2
}
0
2σ 2 ξ −
(t − σ2
dξ
τ
= 1 (−
σ 2π exp t ) ( σ
2
)
exp
τ 2τ 2 ×
∫ ( )
+∞
exp − 1 ξ ′ 2
)
−(t−σ 2 /τ) 2 σ 2 dξ ′ with ξ ′ = ξ −
(t − σ2
τ
= 1 (
2 exp − t ) ( σ
2
) ( σ
exp erfc
τ 2τ 2 τ 2 − t )
σ 2
Here, erfc(x) = 1 −erf(x) is the complementary error function, where
(2.46)
erf(x) = 2 ∫ x
e −t 2 dt (2.47)
π
Figure 2.6 shows the result of the convolution of the exponential function with the Gaussian.
The following properties immediately stand out: (i) The finite time resolution ensures that there
is also a signal at negative times, whereas it was 0 before convolution. (ii) The maximum is not
at t = 0 any more. (iii) What can’t be seen straight away, yet is easy to grasp, is the following:
the center of gravity of the exponential function, which was at t = τ, doesn’t get shifted at all
upon convolution.
Now we prove the extremely important Convolution Theorem:
Theorem 2.1 (Convolution theorem). Let be
0
Then
f (t) ↔ F (ω)
g (t) ↔ G(ω)
h(t) = f (t) ⊗ g (t) ↔ H(ω) = F (ω) ×G(ω) (2.48)
M.Orlandini 27