Chapter 9: The Fourier Transform
- Page ID
- 208482
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Suppose that \(\mathbb{T} \) is a time scale with forward jump operator and delta differentiation operator \(\sigma \) and \(\Delta \),respectively. Let also, \(\inf \mathbb{T}=-\infty \),\(\sup\mathbb{T}=\infty \) and \(s\in \mathbb{T} \). Denote
\[\begin{aligned}
\mu_*(s) =& \inf_{t\in [s, \infty)}\mu(t),\quad \mu^*(s)= \sup_{t\in (-\infty, s]}\mu(t),\quad
\overline{\mu}(s)= \inf_{t\in (-\infty, s]}\mu(t), \\
\overline{\overline{\mu}}(s) =& \mu^*(s)\quad \text{if}\quad \text{Re}_{\overline{\mu}(s)}(x)\leq 0, \\
\overline{\overline{\mu}}(s) =& \overline{\mu}(s)\quad \text{if}\quad \text{Re}_{\overline{\mu}(s)}(x)\geq 0,
\end{aligned}\notag\]
where \(x\in \mathbb{R} \).
Suppose that \(f: \mathbb{T}\to \mathbb{R} \) is regulated. Then the Fourier transform of the function \(f \) is defined by
\[
\mathcal{F}(f)(x, s)= \int_{-\infty}^{\infty} f(t) e_{\ominus ix}^{\sigma}(t, s) \Delta t
\notag\]
for \(x\in \mathbb{R} \) for which \(1+i x \mu(t)\ne 0 \) for any \(t\in \mathbb{T}^{\kappa} \) and the improper integral exists.
Let \(\alpha\in \mathcal{R}^+([s, \infty)) \),\(\gamma\in \mathcal{R}^+((-\infty, s]) \). We say that \((s, \alpha, \gamma) \) is a real admissible triple if
\[\begin{aligned}
R_{s, \alpha, \gamma} =& \bigg\{ x\in \mathbb{R}: \text{Re}_{\mu^*(s)}(ix)<\gamma,\quad \text{Re}_{\mu_*(s)}(ix)>\alpha, \\
&1+\overline{\overline{\mu}}(s)\text{Re}_{\overline{\mu}(s)}(ix)\ne 0\bigg\}\ne \emptyset.
\end{aligned}\notag\]
If \(f\in \mathcal{C}_{rd}(\mathbb{T}) \),then the triple \((s, \alpha, \gamma) \) is a real admissible triple and \(f \) is of double exponential order \((\alpha, \gamma) \),then by the computations for the bilateral Laplace transform, it follows that \(\mathcal{F}(f)(\cdot, s) \) exists on \(R_{s, \alpha, \gamma} \) and converges absolutely and uniformly on \(R_{s, \alpha, \gamma} \). Below we will list some of the properties of the Fourier transform.
Let \(f, g: \mathbb{T}\to \mathbb{R} \),\(\alpha, \beta\in \mathbb{C} \). Then
\[
\mathcal{F}(\alpha f+\beta g)(x, s)= \alpha \mathcal{F}(f)(x, s)+\beta \mathcal{F}(g)(x, s)
\notag\]
for those \(x\in \mathbb{R} \) for which \(1+x\mu(t)\ne 0 \),\(t\in \mathbb{T}^{\kappa} \),and the respective integrals exist.
Proof
We have
\[\begin{aligned}
\mathcal{F}(\alpha f+\beta g)(x, s) =& \int_{-\infty}^{\infty} (\alpha f +\beta g)(t) e_{\ominus ix}^{\sigma}(t, s)\Delta t \\
=& \alpha \int_{-\infty}^{\infty} f(t) e_{\ominus ix}^{\sigma}(t, s) \Delta t+\beta \int_{-\infty}^{\infty} g(t) e_{\ominus ix}^{\sigma}(t, s)\Delta t \\
=& \alpha \mathcal{F}(f)(x, s)+\beta \mathcal{F}(g)(x, s).
\end{aligned}\notag\]
This completes the proof.
Let \(f: \mathbb{T}\to \mathbb{R} \). Then
\[
\mathcal{F}\left(e_y^{\sigma}(\cdot, s)f(\cdot)\right)(x, s)= \mathcal{F}(f)(z, s),
\notag\]
where \(z=\frac{x+iy}{1+\mu y} \),for those \(x,y\in \mathbb{R} \) for which \(1+\mu(t)y\ne 0 \) for any \(t\in \mathbb{T}^{\kappa} \) and the respective integrals exist.
Proof
Note that
\[\begin{aligned}
iz =& i\frac{x+iy}{1+\mu y} \\
=& -\frac{x+iy}{i(1+\mu y)}
\end{aligned}\notag\]
and
\[\begin{aligned}
\ominus iz =& \frac{-iz}{1+i \mu z} \\
=& \frac{\frac{x+iy}{i(1+\mu y)}}{1-\mu \frac{x+iy}{i(1+\mu y)}} \\
=& \frac{x+iy}{i+i\mu y -\mu x-i\mu y} \\
=& \frac{x+iy}{i-\mu x} \\
=& \frac{i(x+iy)}{-1-i\mu x} \\
=& \frac{y-ix}{1+i\mu x} \\
=& y\ominus ix.
\end{aligned}\notag\]
Hence,
\[\begin{aligned}
\mathcal{F}\left(e_y^{\sigma}(\cdot, s)f(\cdot)\right)(x, s) =& \int_{-\infty}^{\infty} e_{\ominus ix}^{\sigma}(t, s) e_y^{\sigma}(t, s) f(t)\Delta t \\
=& \int_{-\infty}^{\infty} e_{y\ominus ix}^{\sigma}(t, s) f(t) \Delta t \\
=& \mathcal{F}(f)(z, s).
\end{aligned}\notag\]
This completes the proof.
Let \(f: \mathbb{T}\to \mathbb{R} \). For any \(k\in \mathbb{N} \),we have
\[
\mathcal{F}\left(f^{\Delta^k}\right)(x, s)= (ix)^k \mathcal{F}(f)(x, s)
\notag\]
for those \(x\in \mathbb{R} \) for which \(1+x\mu(t)\ne 0 \),\(t\in \mathbb{T}^{\kappa} \),and the respective integrals exist and
\[
\lim_{t\to \pm \infty} f^{\Delta^l}(t) e_{\ominus ix}(t, s)=0,\quad l\in \{0, \ldots, k-1\}.
\notag\]
Proof
We will use the principle of mathematical induction.
1. For \(k=1 \),we have
\[\begin{aligned}
\mathcal{F}\left(f^{\Delta}\right)(x, s) =& \int_{-\infty}^{\infty} f^{\Delta}(t) e_{\ominus ix}^{\sigma}(t, s) \Delta t \\
=& \lim_{t\to\infty} f(t) e_{\ominus ix}(t, s)-\lim_{t\to -\infty}f(t) e_{\ominus ix}f(t) e_{\ominus ix}(t, s) \\
&-\int_{-\infty}^{\infty}(\ominus ix)f(t) e_{\ominus ix}(t, s)\Delta t \\
=& ix \int_{-\infty}^{\infty} f(t) e_{\ominus ix}^{\sigma}(t, s) \Delta t \\
=& ix \mathcal{F}(f)(x, s).
\end{aligned}\notag\]
2. Assume that
\[
\mathcal{F}\left(f^{\Delta^k}\right)(x, s)= (ix)^k \mathcal{F}(f)(x, s)
\notag\]
for some \(k\in \mathbb{N} \).
3.
We will prove that
\[
\mathcal{F}\left(f^{\Delta^{k+1}}\right)(x, s)= (ix)^{k+1} \mathcal{F}(f)(x, s).
\notag\]
Really, we have
\[\begin{aligned}
\mathcal{F}\left(f^{\Delta^{k+1}}\right)(x, s) =& ix \mathcal{F}\left(f^{\Delta^k}\right)(x, s) \\
=& (ix)^{k+1} \mathcal{F}(f)(x, s).
\end{aligned}\notag\]
This completes the proof.
Let \(f: \mathbb{T}\to \mathbb{R} \). Then
\[
\overline{\mathcal{F}(f)(x, s)}= \mathcal{F}(f)(-x, s)
\notag\]
for those \(x\in \mathbb{R} \) for which \(1\pm x \mu(t)\ne 0 \),\(t\in \mathbb{T}^{\kappa} \),and the respective integrals exist.
Proof
Let \(t\in \mathbb{T}^{\kappa} \) and \(x\in \mathbb{R} \) be such that \(1\pm x\mu(t)\ne 0 \). Then, we have
\[\begin{aligned}
\left(\ominus (ix)\right)(t) =& -\frac{ix}{1+i \mu(t) x} \\
=& -\frac{ix (1-i\mu(t)x)}{(1+i\mu(t)x)(1-i\mu(t) x)} \\
=& -\frac{ix+\mu(t) x^2}{1+(\mu(t))^2x^2}, \\
1+\mu(t)\left(\ominus (ix)\right)(t) =& 1-\mu(t)\frac{ix+\mu(t) x^2}{1+(\mu(t))^2 x^2} \\
=& \frac{1+(\mu(t))^2 x^2-i\mu(t) x-(\mu(t))^2 x^2}{1+(\mu(t))^2 x^2} \\
=& \frac{1-i\mu(t) x}{1+(\mu(t))^2 x^2} \\
=& \frac{1}{\sqrt{1+(\mu(t))^2 x^2}}\left(\frac{1}{\sqrt{1+(\mu(t))^2 x^2}}-i \frac{\mu(t) x}{\sqrt{1+(\mu(t))^2 x^2}}\right)
\end{aligned}\notag\]
and
\[
\left(\ominus (i(-x))\right)(t)=\frac{1}{\sqrt{1+(\mu(t))^2 x^2}}\left(\frac{1}{\sqrt{1+(\mu(t))^2 x^2}}+i \frac{\mu(t) x}{\sqrt{1+(\mu(t))^2 x^2}}\right).
\notag\]
Let
\[
r(t)= \frac{1}{\sqrt{1+(\mu(t))^2 x^2}},\quad \cos\theta(t)= \frac{1}{\sqrt{1+(\mu(t))^2 x^2}}.
\notag\]
Then
\[
1+\mu(t)\left(\ominus (ix)\right)(t)=r(t) e^{-itheta(t)}
\notag\]
and
\[
1+\mu(t)\left(\ominus (i(-x))\right)(t)= r(t) e^{i\theta(t)}.
\notag\]
Also,
\[\begin{aligned}
\log\left(1+\mu(t)\left(\ominus (ix)\right)(t)\right) =& \ln(r(t))-i \left(\theta+2k\pi\right), \\
\log\left(1+\mu(t)\left(\ominus (i(-x))\right)(t)\right) =& \ln(r(t))+i \left(\theta+2k\pi\right),\quad k\in \mathbb{Z}.
\end{aligned}\notag\]
From here and from the definition of the Fourier transform, we have
\[\begin{aligned}
\mathcal{F}(f)(x, s) =& \int_{-\infty}^{\infty}f(t) e_{\ominus (ix)}^{\sigma}(t, s)\Delta t \\
=& \int_{-\infty}^{\infty}e^{\int_s^{\sigma(t)}\frac{1}{\mu(\tau)}\log\left(1+\mu(\tau)\left(\ominus (ix)\right)(\tau)\right)\Delta\tau}f(\tau)\Delta \tau \\
=& \int_{-\infty}^{\infty} e^{\int_s^{\sigma(t)}\frac{1}{\mu(\tau)}\left(\ln r(\tau)-i(\theta(\tau)+2k\pi)\right)\Delta\tau}f(t)\Delta t
\end{aligned}\notag\]
and
\[\begin{aligned}
\overline{\mathcal{F}(f)(x, s)} =& \overline{\int_{-\infty}^{\infty} e^{\int_s^{\sigma(t)}\frac{1}{\mu(\tau)}\left(\ln r(\tau)-i(\theta(\tau)+2k\pi)\right)\Delta\tau}f(t)\Delta t} \\
=& \int_{-\infty}^{\infty} \overline{e^{\int_s^{\sigma(t)}\frac{1}{\mu(\tau)}\left(\ln r(\tau)-i(\theta(\tau)+2k\pi)\right)\Delta\tau}f(t)}\Delta t\\ \\
=& \int_{-\infty}^{\infty} \overline{e^{\int_s^{\sigma(t)}\frac{1}{\mu(\tau)}\left(\ln r(\tau)-i(\theta(\tau)+2k\pi)\right)\Delta\tau}}f(t)\Delta t \\
=& \int_{-\infty}^{\infty} e^{\overline{\int_s^{\sigma(t)}\frac{1}{\mu(\tau)}\left(\ln r(\tau)-i(\theta(\tau)+2k\pi)\right)\Delta\tau}}f(t)\Delta t \\
=& \int_{-\infty}^{\infty} e^{\int_s^{\sigma(t)}\overline{\frac{1}{\mu(\tau)}\left(\ln r(\tau)-i(\theta(\tau)+2k\pi)\right)}\Delta\tau} f(t)\Delta t \\
=& \int_{-\infty}^{\infty} e^{\int_s^{\sigma(t)}\frac{1}{\mu(\tau)}\left(\ln r(\tau)+i(\theta(\tau)+2k\pi)\right)\Delta\tau}f(t)\Delta t \\
=& \int_{-\infty}^{\infty}e^{\int_s^{\sigma(t)}\frac{1}{\mu(\tau)}\log\left(1+\mu(\tau)\left(\ominus (i(-x))\right)(\tau)\right)\Delta\tau}f(\tau)\Delta \tau \\
=& \int_{-\infty}^{\infty}f(t) e_{\ominus (i(-x))}^{\sigma}(t, s)\Delta t \\
=& \mathcal{F}(f)(-x, s).
\end{aligned}\notag\]
This completes the proof.
Let \(a\in \mathbb{T} \). Then
\[
\mathcal{F}_a(\delta_a^H)(x, s)= e_{\ominus ix}^{\sigma}(a, s)
\notag\]
for those \(x\in \mathbb{R} \) for which \(1+x\mu(t)\ne 0 \),\(t\in \mathbb{T}^{\kappa} \).
Proof
We have
\[\begin{aligned}
\mathcal{F}(\delta_a^H)(x, s) =& \int_{-\infty}^{\infty} \delta_a(t-a) e_{\ominus ix}^{\sigma}(t, s) \Delta t \\
=& e_{\ominus ix}^{\sigma}(a, s)
\end{aligned}\notag\]
for those \(x\in \mathbb{R} \) such that \(1+x\mu(t)\ne 0 \),\(t\in \mathbb{T}^{\kappa} \).
1. If \(\mathbb{T}=\mathbb{R} \),then we take \(s=a \) and we get
\[
\mathcal{F}_a(\delta_a^H)(x, s)=1.
\notag\]
2. If \(a\in \mathbb{T} \) is right-scattered and \(s=a \),then
\[\begin{aligned}
e_{\ominus ix}^{\sigma}(a, a) =& e^{\log \frac{1}{1+i\mu(a) x}} \\
=& \frac{1}{1+i\mu(a) x}
\end{aligned}\notag\]
for those \(x\in \mathbb{R} \) for which \(1+i\mu(a) x\ne 0 \).
Let \(f: \mathbb{T}\to \mathbb{R} \) be regulated and
\[
F(t)= \int_a^t f(\tau)\Delta \tau,\quad t\in \mathbb{T},
\notag\]
for some fixed \(a\in \mathbb{T} \). Then
\[
\mathcal{F}(F)(x, s)=-\frac{i}{x}\mathcal{F}(f)(x, s)
\notag\]
for those \(x\in \mathbb{R} \),\(x\ne 0 \), for which
\[
\lim_{t\to \pm \infty} F(t) e_{\ominus ix}(t, s)=0.
\notag\]
Proof
We have
\[\begin{aligned}
\mathcal{F}(F)(x, s) =& \int_{-\infty}^{\infty} F(t) e_{\ominus ix}^{\sigma}(t, s)\Delta t \\
=& \int_{-\infty}^{\infty}F(t)\left(1+\mu(t) (\ominus (ix))(t)\right)e_{\ominus ix}(t, s)\Delta t \\
=& \int_{-\infty}^{\infty}F(t) \frac{1}{1+i\mu(t) x}e_{\ominus ix}(t, s)\Delta t \\
=& -\frac{1}{ix}\int_{-\infty}^{\infty}F(t)\frac{-ix}{1+i\mu(t) x}e_{\ominus ix}(t, s)\Delta t \\
=& \frac{i}{x}\int_{-\infty}^{\infty}F(t) (\ominus ix)(t) e_{\ominus ix}(t, s)\Delta t \\
=& \frac{i}{x}\int_{-\infty}^{\infty} F(t) e_{\ominus ix}^{\Delta}(t, s)\Delta t \\
=& \frac{i}{x}\left(\lim_{t\to \infty} F(t) e_{\ominus ix}(t, s)-\lim_{t\to -\infty} F(t) e_{\ominus ix}(t, s)\right) \\
&-\frac{i}{x}\int_{-\infty}^{\infty}f(t) e_{\ominus ix}^{\sigma}(t, s)\Delta t \\
=& -\frac{i}{x} \mathcal{F}(f)(x, s)
\end{aligned}\notag\]
for those \(x\in \mathbb{R} \),\(x\ne 0 \), for which
\[
\lim_{t\to \pm \infty} F(t) e_{\ominus ix}(t, s)=0.
\notag\]
This completes the proof.