Esercizio trasformata

29/01/2012

2)

f(t) = ^cos(t)/_{t² + 1} per t ∈ ℝ

APPARTIENE A L2?

cos(t)/1 + t2 |/ 1 + t2

converge se:

∃ n, m ∈ ℤ\{0}, DetL (SI)

non ha poli su Re(s) de ^∞/_{t² - 1} ^∞/ _{t = +i}

APPARTIENE A L2?

|f(t)| = ^|cos(t)|/_{(1 + t²)²} < ¹/_{(1 + t²)²} converge se:

∀ n ∈ ℤ\{0}∉N (SI)

no poli su Re(s)

Calcolo la trasformata:

β(s) = ∫_−∞^+∞ f(t) e^−st dt = ∫_−∞^{+∞ cos(t)}/_{t² + 1} e^−st dt = ^{(cos(t)e−ist)}/_{t² + 1} e^−it dt = ¹/₂ ∫_−∞^{+∞ ei(1−s)t}/_{t² + 1} dt + +¹/₂ ∫_−∞^{+∞ e−i(1+s)t}/_{t² + 1} dt

Caso A

∫₀^+∞ e^i(1−s)t dt = (1−s)i ⟩0 → (1−s)⟩0 ⟹

∫₀^+∞ e²+1 dt = (1−s)i ⟨0 → (1−s)⟩0 ⟹ s ⟩1

RICORDA:

∮ f(Ē)^β(T) dz = 2πi Σ_Res(f,z)2k indice avvicamento

Res( f ,1 ) = ^ei(1−s)/_{2 ı t} |_tz= = e^{i(1−s) 1 + s}/_{2 ı} = ^{e−1 + s}/_{2 i} = ^{−1 + s}/_{2 i} (s⟩≤1)

Res ( f , 1− ) = ^ei(1−s)t/ _{0 2} |_tz= = ^e−i(1−s)/_{−2 i} = ^{e+1 (1−s)}/_{−2 i} = ^e/_{−2 i} = ^{−1 + s}/_{2 i} (5⟩⟨1)

29/01/2023

2) f(t) = ^cos(t)/_{t² + 1} per t ∈ ℝ

appartiene a L²? ^cos(t)/_{1 + t²} 1

Ricorda:

∮f(^eiz/_z)dz = 2πi∑Res(f,z∈Z^K), indice avvolgimento

Res(^f/_z-i) = ^ei(1-s)t/_2t |_tz =

^e(1-s)/_2i - ^1+s/₂ = ^s-1/_2i (s < 1)