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)f(t) = ⁴/_t²+4 A(t) = ⁴/_t²+4 = - ^2}{tu(t)- A(t) = - ⁴/_t²+4R(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 unoZt = Zlt (1_{t = 0}) = iπ...

2) Dato il parametro a > 0

dato il parametro a>0, si consideri la funzione f_a: ℝ→ℝ definita dat = 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

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 ]^t₀R[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 ₂z 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

Dato il parametro a > 0, si consideri la funzione fa : ℝ -> ℝ data dafa(t) = {a + a, t ∈(-∞,0)a - t, t ∈ [0,∞)0, t > a}Calcolate fΛ(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)w2Esercizio Uno & Due∞0 1-cos(at)t2 -1-cos(at)t2 dt =-∫ ∞2t2 ∞ -1t2 dt =∞ ∫ (taΛ) fα(w)f0(w) dw =0 = ∞∞ = 1fa(w)f0(w) dw [ =]-∞ 2n fa(t) Λ(π π) Λadt ()1/q = ∫rfa (λt) e-Λ(1-10(30-θ)) dt

3) Utilizzando opportunamente le masse

Utilizzando opportunamente le masse, calcolate da trasformando fi(t):f(t)=^{cosΛ(bt)-cos(ct)}_t²Riconoscendo una forma notaf(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)_t²Formula di cA(t)=2 ^sin(2t-1)_t²Δ ^sin(2t)₂1/∫¹/^∫n(t) / ∇₁(t) ^sin2(t) - 2^sin2(t)_t²CCalcolo 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_tRicordiamoci di nuovo,f̂_w(u) = π(2 - [^w/₂])f+π(2 - [^w/₂])f_t

4) Verifica della funzione in R->R

Dopo aver verificato che la funzione in R->R definita dau(t) =^{e-1t - e2tH}/_{e^-1t = t > 0}{ 1 , t { t = 0appartiene 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 ⇒ convergeDunque u(t) ∈ L¹(R) ∩ L²(R):=∫ ¹/_e^e1t dt = ∫ ¹/_2tH dt = ∫ ¹/_4t^eH dt⇒ convergeDunque u(t) ∈ L¹(R) ∩ L²(R):=∫ [Teorema RL:] a^w(u) ∈ C⁰(R) continua, limitata e nullatemperata per ω→0Calcolo esplicito v̂_w(u)p=1H-p=2ıHu(t) == (l/tH - e^-2ıHπ ^{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

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

Data la funzione f : R → R, definita daf(t) = {0 , t ∈ [0,1]1 , t ∈ [1,2)t , t ≥ 2a) Verificare che f ∈ F trasformabile e trovare la formula di F alla Hovab) Calcolare F(w), confrontando il risultato con quanto trovato al punto precedentef(t) = {0 , t ∈ [0,1) ∪ k = 1t², t t = 1 , t = 1, se t ≥ 0Jungo 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) atmanF(G) è imparb degre quoque ⇒ f(w) ≈¾ ThawnCalcolo Espluico di F trasformatef(t) = { t² χ_0,1 t² χ_2+0,3Fᵧ{ - ... ( ∂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⁄_dw²sin(^w⁄₂)_w- ^e-ω2⁄_2i|_sub>dw]|_ω⁄w[]|^d2⁄_dw²sinc(^w⁄₂)|sin(w²)1