% vim: tw=50
% 16/11/2022 10AM

\subsection*{8.4a Fourier Transform of Generalised
Functions}

(See discussion in section 8.3 of R Jorsa notes
(or D Skinner))

\subsubsection*{Dirac delta function $\delta(x)$}

Consider the inversion then (8.3):
\begin{align*}
    f(x)
    &= \mathcal{F}^{-1}(\mathcal{F}(f))(x)
    &= \frac{1}{2\pi} \int_{-\infty}^\infty \left[
    \int_{-\infty}^\infty f(u) e^{-iku} \dd u
    \right] e^{ikx} \dd k \\
    &= \int_{-\infty}^\infty f(u) \left[
    \frac{1}{2\pi} \int_{-\infty}^\infty
    e^{ik(x-u)} \dd k \right] \dd u
\end{align*}
so identify
\[ \delta(x - u) = \frac{1}{2\pi}
\int_{-\infty}^\infty e^{ik}(x - u) \dd k =
\frac{1}{2\pi} \int_{-\infty}^\infty e^{-iku}
e^{ikx} \dd k \]
\begin{itemize}
    \item If $f(x) = \delta(x - a)$, then
        $\tilde{f}(k) = e^{-ika}$ (8.21).
        \begin{hiddenflashcard}[FT-of-delta-function]
        \prompt{What is the Fourier transform of
        $\delta(x - a)$?}
        \[ f(x) = \delta(x - a) \iff \tilde{f}(k)
        = \cloze{e^{-ika}} \]
        \end{hiddenflashcard}
    \item If $f(x) = \delta(x)$ then
        \[ \tilde{f}(k) = \int_{-\infty}^\infty
        \delta(x) e^{ikx} \dd x = 1 \tag{8.22} \]
    \item If $f(x) = 1$, then
        \[ \tilde{f}(k) = \int_{-\infty}^\infty
        e^{-ikx} \dd x = 2\pi \delta(k) \tag{8.23} \]
        by duality (8.14).
\end{itemize}

\subsubsection*{Trig Functions}

\begin{flashcard}[FT-of-cos]
\prompt{What is the Fourier transform of $\cos
\omega x$?}
\[ f(x) = \cos \omega x \iff \tilde{f}(k) =
\cloze{\pi(\delta(k + w) + \delta(k - w))} \]
\end{flashcard}
\begin{flashcard}[FT-of-sin]
\prompt{What is the Fourier transform of $\sin
\omega x$?}%
\[ f(x) = \sin \omega x \iff \tilde{f}(k) =
\cloze{i\pi(\delta(k + w) - \delta(k - w))}
\tag{8.24} \]
\end{flashcard}
Exercise: Find $\mathcal{F}^{-1}$ for $\sin\omega
k$, $\cos\omega k$ using (8.14).

\subsubsection*{Heaviside function}

Subtle derivation requiring central value $H(0) =
\half$; then $H(x) + H(-x) = 1$ for all $x$ and
continuous at $x = 0$. By (8.8) and (8.23)
\[ \tilde{H}(k) + \tilde{H}(-k) = 2\pi \delta(k)
\tag{$*$} \]
Recall (6.7) $H'(x) = \delta(x)$ which implies
\[ ik\tilde{H}(k) = 1 \tag{\dag} \]
by (8.13) and (8.22). But $k\delta(k) = 0$, so
($*$) and ($\dag$) are consistent if
\begin{flashcard}[FT-of-Heaviside]
\prompt{What is the Fourier transform of the
Heaviside function?}
\[ \tilde{H}(k) = \cloze{\pi\delta(k) +
\frac{1}{ik}}
\tag{8.25} \]
\end{flashcard}
Dirichlet discontinuous formula (8.16):
Rewrite as
\[ \half \sgn(x) = \frac{1}{2\pi}
\int_{-\infty}^\infty \frac{e^{ikx}}{ik} \dd x
\]
so
\begin{flashcard}[FT-of-sgn]
\prompt{What is the Fourier transform of the sign
function?}
\[ f(x) = \half \sgn(x) \iff \tilde{f}(k) =
\cloze{\frac{1}{ik}} \tag{8.26} \]
\end{flashcard}
\begin{hiddenflashcard}[FT-of-1]
\prompt{What is the Fourier transform of $1$?}
\[ f(x) = 1 \iff \tilde{f}(k) = \cloze{2\pi
\delta(k)} \]
\end{hiddenflashcard}

\subsection{Applications of Fourier Transforms}

\subsubsection*{Motivation I: ODE for BVP}

Consider $y'' - y = f(x)$ with homogeneous
boundary conditions $y \to 0$ as $x \to \pm
\infty$. Take the Fourier Transform:
\[ (ik)^2 \tilde{y} - \tilde{y} = (-k^2 - 1)
\tilde{y} = \tilde{f} \]
by (8.13). SO the solution is
\[ \tilde{y}(k) = -\frac{\tilde{f}(k)}{1 + k^2}
\equiv \tilde{f}(k) \tilde{g}(k) \]
where $\tilde{g}(k) = -\frac{1}{1 + k^2}$ but this
is the Fourier Transform of $g(x) = -\half
e^{-|x|}$ (see (8.6)). Thus convolution theorem
(8.17) implies
\begin{align*}
    y(x)
    &= \int_{-\infty}^\infty f(x)g(x - u) \dd u \\
    &= -\half \int_{-\infty}^\infty f(u) e^{-|x -
    u|} \dd u \\
    &= -\half \int_{-\infty}^x f(u) e^{u - x} \dd
    u - \half \int_x^\infty f(u) e^{x - u} \dd u
\end{align*}
which is in the form of a BVP Green's function. \\
Exercise: Verify by constructing Green's function.

\subsubsection*{Motivation II: Signal processing
(IVP)}

Suppose (given) input $\mathcal{J}(t)$ acting on by linear
operator $\mathcal{L}_{\text{in}}$ to yield output
$\theta(t)$.
\[ \theta(t) = \mathcal{L}_{\text{in}} \mathcal{J}(t) \]
The Fourier Transform $\tilde{\mathcal{J}}(\omega)$ is
denoted \emph{the resolution}
\[ \tilde{\mathcal{J}}(\omega) =
\int_{-\infty}^\infty \mathcal{J}(t)
e^{-i\omega t} \dd t \tag{8.27} \]
In frequency domain $\mathcal{L}_{\text{in}} \mathcal{J}(t)$
means $\tilde{\mathcal{J}}(\omega)$ is multiplied by a
\emph{transfer function} $\tilde{R}(\omega)$ to
yield output
\[ \theta(t) = \frac{1}{2\pi}
\int_{-\infty}^\infty \tilde{R}(\omega)
\tilde{\mathcal{J}}(\omega) e^{i\omega t} \dd \omega
\tag{8.28} \]
with \emph{response function} given by
\[ R(t) = \frac{1}{2\pi} \int_{-\infty}^\infty
\tilde{R}(\omega) e^{i\omega t} \dd \omega
\tag{8.29} \]
By the convolution theorem (8.17), output is
\[ \theta(t) = \int_{-\infty}^\infty \mathcal{J}(u) R(t - u)
\dd u \]
We assume no input $\mathcal{J}(t) = 0$ for $t < 0$ and by
\emph{causality}, zero output for $R(t) = 0$ for
$t < 0$ (i.e. $R(t - u)$ has source $\delta(t -
u)$), so we require $0 < u < t$:
\[ \theta(t) = \int_0^r \mathcal{J}(u) R(t - u) \dd
u\tag{8.30} \]
i.e. the same form as IVP Green's function.

\subsubsection*{General transfer functions for ODEs}

Suppose input / output relation given by linear
ODE ($n$-th order)
\[ \mathcal{L} \theta(t) \equiv \left( \sum_{i =
0}^n a_i \dfrac[i]{}{t} \right) \theta(t) = \mathcal{J}(t)
\tag{8.31} \]
where $a_i$ are constant and here set
$\mathcal{L}_{\text{in}} = 1$. Take the Fourier
Transform:
\[ (a_0 + a_1(i\omega) + a_2(i\omega)^2 + \cdots +
a_n(i\omega)^n) \theta(\omega) = \tilde{\mathcal{J}}
(\omega) \]
so the transfer function (8.28) is
\[ \tilde{R}(\omega) = \frac{1}{a_0 + a_1(i\omega)
+ \cdots + a_n (i\omega)^n} \tag{8.32} \]
Factorise $n$-th degree polynomial into product of
$n$ roots $(i\omega - c_j)^{k_j}$ with $j = 1, 2,
\dots, J$ (with repeated roots $k_j > 1$) i.e.
$\sum_{j = 1}^J k_j = n$. Then
\begin{align*}
    \tilde{R}(\omega)
    &= \frac{1}{(i\omega -
    c_1)^{k_1} + \cdots + (i\omega - c_J)^{k_J}}
    \\
    &= \sum_{j = 1}^J \sum_{m = 1}^{k_j}
    \frac{\Gamma_{jm}}{(i\omega - c_j)^m}
    \tag{8.33}
\end{align*}
since it can be expanded in partial fractions
(constant $\Gamma_{jm}$). For repeated roots ($1
\le m \le k_j$):
\[ \frac{1}{(i\omega - c_j)^{k_j}} \to
\frac{\Gamma_{j1}}{(i\omega - c_j)} +
\frac{\Gamma_{j2}}{(i\omega - c_j)^2} + \cdots +
\frac{\Gamma_{jk}}{(i\omega - c_j)^{k_j}} \]
To solve we must invert $\frac{1}{(i\omega -
a)^m}$, $m \ge 1$. We know (8.6a)
\[ \mathcal{F}^{-1} \left( \frac{1}{i\omega - a}
\right) = \begin{cases}
    e^{at} & t > 0 \\
    0 & t < 0
\end{cases} \]
for $\Re(a) < 0$, so we assume $\Re(c_j) < 0$,
$\forall j$ (eliminate exponential growing modes).