Metodi matematici - Foglio 8 - Esercizi Trasformata di Fourier in L1(R) e L2(R)

FOGLIO 8

TRASFORMATA DI FOURIER IN L₁(R) E L₂(R). TEOREMA DI PALEY-WIENER, TEOREMA PLANCHEREL.

1) Calcolare la trasformata di Fourier di f(t)

f(t) = ⁴/_t²+4

A(t) = ⁴/_t²+4 = - ^2}{tu(t)

- A(t) = - ⁴/_t²+4

R(w) = L[f(t)] = ^{d}{d\omega}(\cdots)}

A(t) = ^4}{\pi} e^(...)

d/dw ( ^Zt}{2}\cdot = - i ( e^...\cdot)

(d/dw...) = 2Z[...](1_t?)

raccompando in uno

Zt = Zlt (1_{t = 0}) = iπ...

2) dato il parametro a>0, si consideri la funzione f_a: ℝ→ℝ definita da

• t = a, ..., t ∈(−a,0)
• f_a(t) = a-t, ..t ∈(0,a)
• ..., t > a

FOGLIO 8

TRASFORMATA DI FOURIER IN L¹(ℝ) E L²(ℝ), TEOREMA DI PALLEY-WIENER, TEOREMA PLANCHEREL

1. CALCOLARE LA TRASFORMATA DI FOURIER DI f(t)

f(t) = ¹⁄_t²+4 [δ - 1/2 sin(2t)]

A(t) = ¹⁄_t²+4 = -¹⁄_tsin(2t)

∫ A(dt) = A(e^-t²/2)² + [ t^-1e^-t ]^t₀

R[inve] = d ⁄ dw Z [¹⁄_w²+4 e^-2t/t² ]

      =          1/t2 e^-t Wsgn(w)

AM_<IW(*t)

g'(t) = δ(t) - c₀ cos(t) [[ ²x^t W_sgn(t) &sup0;w(e^-t² - e^-2t)&supF;f ]]

    =                        [∑^∞⁄_{w + A<N^×}]

            = -4[&sub2_;g(t)](^šr) d ⁄ dw ₂

z e^-t/2 - 2e^-t²s — e^-t²/2sign(w - z) e^-2A(w+z(w))

2^IQ_u ⁄ ρ(t) [[ e^x(y-z)] -²]

NOOOL...

Ricompattando il tutto

2) Dato il parametro a > 0, si consideri la funzione fa : ℝ -> ℝ data da

Calcolate f_Λ(w) e calcolate

^{π cos(at)}₀ e^-λt cos(wt) dt

= ^1-cos(at)_t² -^1-cos(at)_t² dt

f(t) = {t-a}² e^-at ; g(t)=t(a+b)(e^-at)

[trasformata (IV)]

f(w)= 2 * ^λcos(at)w²

Esercizio Uno & Due

^∞₀ ^1-cos(at)_t² -^1-cos(at)_t² dt =

-∫ ∞2t2 ∞ -1t2 dt =

∞ ∫ ^(ta_Λ) f_α(w)f₀(w) dw =

0 = _{∞^∞ = 1}f_a(w)f₀(w) dw [ =]

-∞ 2n ^f_a(t) Λ(^{π _π}) Λ_adt

1/q = ∫_rfa (λt) e^{-Λ(1-10(30-θ))} dt

(3) Utilizzando opportunamente le masse, calcolate da trasformando fi(t):

f(t)=^{cosΛ(bt)-cos(ct)}_t²

Riconoscendo una forma nota

f(t):= ^{cos(at)+cos(t)}_t² : ^t

= ^cos(at)+1_t² + ^1-cos(at)

[ϕ (t) * ^1-cos(at) ₂

= ^1-cos(at)_t² + ^sin2(t)_t²

Formula di c

A(t)=2 ^sin(2t-1)_t²Δ ^sin(2t)₂

1/∫¹/^∫n(t) / ∇₁(t) ^sin2(t) - 2^sin2(t)_t²

C

Calcolo esercizio trasformata

_[

_L^Λ(w)= = 2 ∫_α(w)

f̂_w(u) = π(1-(^w/₂))x_{[0,2]}(u) + π(1-^w/2)x{0,2}}

= π(1-(^w/₂))x_{[0,2]}(u) + π(1-^w/2)x{0,2}}

= π √(1-(^w/₂))

f̂₀^w(u) = π(2 - ^w/₂)x_[0,1]} + π(2 - ^w/₂)x_{0,1}

= π(2-(^w/₂))x_[0,1]} + π(2-^w/₂)x_{0,1}

= π(2 - [^w/₂])f_t

Ricordiamoci di nuovo,

f̂_w(u) = π(2 - [^w/₂])f+π(2 - [^w/₂])f_t

4). Dopo aver verificato che la funzione in R->R definita da

u(t) =^{e-1t - e2tH}/_{e^-1t = t > 0}

{ 1 , t < 0

{ t = 0

appartiene a L¹(R) ∩ L²(R). Calcolare la sua trasformata di Fourier

:=∫ ¹/(¹/_e^2tH) dt = ∫ ¹/₂ e^2H dt = ∫ ¹/₂ e^1t dt ⇒ converge

Dunque u(t) ∈ L¹(R) ∩ L²(R)

:=∫ ¹/_e^e1t dt = ∫ ¹/_2tH dt = ∫ ¹/_4t^eH dt

⇒ converge

Dunque u(t) ∈ L¹(R) ∩ L²(R)

:=∫ [Teorema RL:] a^w(u) ∈ C⁰(R) continua, limitata e nullatemperata per ω→0

Calcolo esplicito v̂_w(u)

p=1H-p=2ıH

u(t) =

= (l/tH - e^-2ıH

π ^{t u(t) = e~ Ht - e~ 2ıHt}

[Trasformata (ɛ)]

i ∫ dt/dw = ²/₄

[Trasformata (ɛ)][σ(x0)]

∫ d/dw ²/_{1+ѡ² ѡ²} → ²/_{1+ѡ² 4ѡ³}

1/2 σ (w/2)_(1/2)

} \frac{1}{2} \frac{2}{\pi} \int_{-\infty}^{+\infty} \frac{\omega}{\omega^2 + \frac{1}{4}} \sin \Big(\frac{5\pi}{2} \Big) \frac{1}{2} \frac{2}{\pi} \int_{-\infty} \limits^{+\infty} \frac{\omega}{\omega^2 + \frac{1}{4}} \sin \Big(\frac{5\pi}{2} \Big) \, \mathrm{d}\omega = \frac{1}{2} \frac{2}{\pi} \int_{-\infty}^{+\infty} \frac{\omega}{\omega^2 + \frac{1}{4}} \sin \Big(\frac{5\pi}{2} \Big) \, \mathrm{d}\omega =

\, \mathrm{d}s = \frac{1}{2} \frac{2}{\pi} \, \, \mathrm{d}s \, \Big[ \frac{\omega}{\omega^2+ \frac{1}{4}} \Big]_{-\omega}^{\omega} \, \Rightarrow \, \lim_{n \rightarrow \infty} \, -\frac{1}{2} \, [\frac{2}{\pi} \arctan(2\omega)\vert_{-\omega}^{\omega}] \, \Rightarrow \frac{1}{2} \frac{2}{\pi} \, \arctan \left(\frac{\omega}{\omega^2+ \frac{1}{4}} \right) \, -\omega^n \, , \, - \, \omega^n \, =

\Rightarrow \lim_{n \to \infty} \, \arctan \, \rightarrow \, [\arctan \omega]_{{-\frac{\omega^n}{\omega}}^{\omega} \, =[\arctan \left( \frac{\omega}{\omega^2+ \frac{1}{4}} \right) \, -\omega^n \, \, \frac{\omega}{\omega} \Rightarrow \frac{2}{\pi} \, \Rightarrow \, \frac{\omega^n}{\omega} - [\frac{\omega}{\omega^2} -}

5) Dati a, b > 0 , osserviamo come

\int_0^{+\infty} \sin(at)\sin(bt)\frac{1}{t^2}\;dt

• \scriptsize{\int_{\mathbb{R}} \begin{bmatrix} f(t) \, + C\Big(\frac{\bigg|{t}\bigg|}{\sqrt{2}} \cos^{\mathbf{T}}_{2\sin(\frac{1}{2}\Big(\frac{3}{4}\Big))^{-1}}} \}}_{\mathbb{R}} \mathrm{dt} = \frac{1}{\pi^4}
• C\bigg(-2t\bigg|_{-\infty}^{+\infty}\bigg) \mid \, \, \frac{1}{4}
• C\Big(\frac{\chi_{[a,b]}}{b} \bigg|_{b}^{\infty}\Big) \bigg|_{- \infty}^{\ldots} = \frac{2}{\pi} \lim_{n \to \infty} =

\underbrace{\int_{-\infty}^{+\infty} \sin(at) \sin(at) \frac{\mathrm{d}t}{t^2}}{ \rightarrow \int_{-\infty}^{+\infty} \sin(at) \sin(bt) \, \, \frac{1}{t^2} {\mathrm{d}t}

Teorema Plancherel\end{bmatrix}||p(0)} \Big \frac{a_{1}}{1 - b^\chi} \int_{[b,a]}^{}\sin(\Gamma(\pi t))\right|_{\infty}\Big(\sin\bigg(\frac{1}{2}\right)\right\}\right|_{{t}} \begin{bmatrix} = C\bigg(\lim_{t \to \infty}\int_{-\infty}^{+\infty}} \end{bmatrix} - \frac{\mu}{a}t \Rightarrow =

\begin{pmatrix} \frac{1}{2} \sin(\omega t) \sum_{n=n}^{\Delta_\omega}(\pi) \, \, \int_{\mathbb{R}} \frac{c}{(b)} \, \, \frac{ \ldots [ }_{\omega(t [a,\infty] \, \, \left| \bigg|^{b}_{R} \Rightarrow \, \frac{\left.b + C\right\}, \, \Big[\frac{1}{2} \right.g} \Big|{b\right)

\int_0^{+\infty} \Bigg(\frac{1}{2} \frac{\sin \left(\left(\Gamma\Big|\right)\Big(\frac{1}{\det(\Gamma)}\right) [\frac{\omega}{\omega^2+ \frac{1}{4}}]_{\lim_{n \to \infty}}}{1-br^{+}} \mathrm{d} \sigma(t) \right|_{0} \right|_{0} \mathrm{d}t\Bigg)= \frac{c}{\int_{[a,\infty)} (0,0,\frac{1}{r(c)}} - \]

=\frac{1}{2}b, \frac{1}{2}b = \begin{bmatrix} c(x)\end{bmatrix}_{[\omega]}^{}

6)

Data la funzione f : R → R, definita da

f(t) = {

• 0 , t ∈ [0,1]
• 1 , t ∈ [1,2)
• t , t ≥ 2

a) Verificare che f ∈ F trasformabile e trovare la formula di F alla Hova

b) Calcolare F(w), confrontando il risultato con quanto trovato al punto precedente

f(t) = {

• 0 , t ∈ [0,1) ∪ k = 1
• t², t < k, se t ≥ 0
• t = 1 , t = 1, se t ≥ 0

Jungo peti unique di f(t)

• ∫_R f(t) dt = ∫₀¹ ... dt = ∫₀¹ ... dt = ∫₁ ... dt = ∫₁ ... dt converge

⇒ f(t) ∈ L (a∈) o P

• ∫_R |f(t)|² dt = ∫₀¹ f(t) ² dt = ∫₀¹ f(t) dt = ∫₁ fy dt = ∫₀¹ t ² fy dt converge

⇒ f(t) ∈ L² (R) ∨ ^L2(R)

[Teorema HL] f(w) è continua e: C(∣OR) , unawtha e M film shad o per i(t) ∞

F(t) ∉ C(L) atman

F(G) è imparb degre quoque ⇒ f(w) ≈¾ Thawn

Calcolo Espluico di F trasformate

f(t) = { t² χ_0,1 t² χ_2+0,3

Fᵧ{ - ... ( ∂² - b = a,sι ) (t '+b _1⁄2 )}

• Fᵧ { - v - b^-a/2 - b = a₋9 , (t ^2+b-1/₂)
• [TransFormatio generic(s²)]
• [TransFormatio generic (s)]
• [TransFormatio generic(1.3)]

F| B_w (w) = (i) = ∂²/_dw (⌠e^-10/∞ sin ( w_w/_...)

1/B_θ (w) = (i-1) ⌠ ∂/_dw ( e ^#w/ ) sin ( w_w/_w )

[

_fA(w) + _fB(w) = 2

_d²⁄_dw²

sin(^w⁄₂)

_w

- ^e-ω2⁄_2i

|_sub>dw

]

|_ω⁄w

[

]

|

^d2⁄_dw²

sinc(^w⁄₂)

|

sin(w²)

1