%%%%		-*- Mode: LaTeX -*-
%%%% Author:    J. Robert Buchanan
%%%% Address:   Department of Mathematics
%%%%		Millersville University
%%%%            P.O. Box 1002
%%%%            Millersville, PA 17551-0302
%%%% Phone:     717-871-7305
%%%% FAX:	717-871-7948
%%%%
%%%% Email:     Robert.Buchanan@millersville.edu
%%%%
%%%% Date:      03/19/2025
%%%%
%%%% Purpose:   Fourier Transform and its Properties
%%%%

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\title{Fourier Transform and its Properties}
\subtitle{\textit{Partial Differential Equations}} % (optional)


\author % (optional, use only with lots of authors)
{J~Robert~Buchanan}
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%   affiliation.

\institute[Millersville University of Pennsylvania]
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{
  Department of Mathematics
}
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% - Keep it simple, no one is interested in your street address.

\date{ % (optional)
Fall
%Spring
%Summer
\the\year}

\subject{Talks}
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\begin{document}

\begin{frame}
  \titlepage
\end{frame}

%\begin{frame}
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%  \tableofcontents
  % You might wish to add the option [pausesections]
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% Since this a solution template for a generic talk, very little can
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% of between 15min and 45min and the theme suggest that you stick to
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% - Exactly two or three sections (other than the summary).
% - At *most* three subsections per section.
% - Talk about 30s to 2min per frame. So there should be between about
%   15 and 30 frames, all told.

%\section{Introduction}

\begin{frame}\frametitle{Objectives}

In this lesson we will:
\begin{itemize}
\item define the Fourier transform,
\item define the inverse Fourier transform,
\item calculate the Fourier transform and inverse Fourier transform of
  some common functions,
\item state some properties of the Fourier transform, and
\item define the Fourier convolution.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Definition of Fourier Transform}

\begin{definition}
Let $f(x)$ be defined for all real numbers.
The Fourier transform of $f$ is denoted by $\mathcal{F}[f](\omega)$ or
often as $\hat{f}(\omega)$.
The Fourier transform
of $f$ is
\[
%\label{eq:FT:defn}
\mathcal{F}[ f ](\omega)
	= \hat{f}(\omega)
	= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) e^{-i \omega x} \, dx	,
\]
where $i = \sqrt{-1}$.
The Fourier transform of $f$ is meaningful if and only if the improper
integral converges.
\end{definition}
\textbf{Remarks}:
\begin{itemize}
\item The Fourier transform maps the original function $f$
of variable $x$ to a new function $\hat{f}$ of variable $\omega$ which
is often thought of as a
frequency.
\item There are other definitions of the Fourier Transform that differ
  only in a multiplicative constant.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Existence of Fourier Transform}

Sufficient conditions for the existence of the Fourier transform are
the following,
\begin{itemize}
\item $\displaystyle \int_{-\infty}^{\infty} | f(x) | \, dx$ converges,
\item $f$ and $f'$ are piecewise continuous on every interval of the
  form $[-M, M]$ for arbitrary $M > 0$.
\end{itemize}
The first condition is known as \textbf{absolute
  integrability} on the real number line.

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the Fourier transform of the piecewise-defined function
\[
f(x)
	= \left\{ \begin{array}{rl}
1	& \mbox{if $|x| \le 1$}	\\
0	& \mbox{otherwise.}
\end{array} \right.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution}

\begin{align*}
\hat{f}(\omega)
	&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x) e^{-i \omega x} \, dx \\
	&= \frac{1}{\sqrt{2 \pi}} \int_{-1}^{1} e^{-i \omega x} \, dx	\\
	&= -\frac{1}{i \omega \sqrt{2 \pi}} \left( e^{-i \omega} - e^{i \omega} \right)	\\
	&= \frac{\sqrt{2}}{\omega \sqrt{\pi}} \sin \omega
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Definition of Inverse Fourier Transform}

\begin{definition}
The inverse Fourier transform of $\hat{f}(\omega)$ is given by
\[
%\label{eq:FT:fourier-inv}
\mathcal{F}^{-1}[ \hat{f} ](x)
	= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \hat{f}(\omega) e^{i \omega x}
\, d\omega	,
\]
provided the improper integral converges.
\end{definition}
\textbf{Remarks}:
\begin{itemize}
\item The original function can be recovered from its
Fourier transform.
\item If $f(x)$ is absolutely integrable and piecewise smooth, then
\[
\frac{1}{2} \left( f(x-) + f(x+) \right)
	= \mathcal{F}^{-1} [\hat{f}](x)	.
\]
Therefore, at all points $x$ where $f$ is  continuous, $f(x) =
\mathcal{F}^{-1}[\hat{f}](x)$.
\end{itemize}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Suppose $a$ is a positive constant, and find the inverse Fourier transform
of $e^{-a |w|}$.

\end{frame}
%%

\begin{frame}\frametitle{Solution}

\begin{align*}
\mathcal{F}^{-1}[ e^{-a |w|} ](x)
	&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{-a |w|} e^{i \omega x} \, d\omega
	\\
	&= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{0} e^{(a + i x) \omega} \, d\omega
	+ \frac{1}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{(-a + i x) \omega} \, d\omega
	\\
	&= \frac{1}{(a + i x) \sqrt{2 \pi}} + \frac{1}{(a - i x) \sqrt{2 \pi}}	\\
	&= \frac{a \sqrt{2}}{(a^{2} + x^{2}) \sqrt{\pi}}	.
\end{align*}

\end{frame}
%%

\begin{frame}\frametitle{Linear Transformation}

\begin{theorem}
%\label{thm:FT:linearity}
Let $f(x)$ and $g(x)$ be functions having Fourier transforms
$\hat{f}(\omega)$ and $\hat{g}(\omega)$ and let $a$ and $b$ be any
complex numbers, then $a f(x) + b g(x)$ has a Fourier transform and
\[
%\label{eq:FT:linearity}
\mathcal{F}[ a f + b g ](\omega)
	= a \hat{f}(\omega) + b \hat{g}(\omega)	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Fourier Transform of a Derivative}

\begin{theorem}
%\label{thm:FT:deriv}
If $f(x)$, $f'(x)$, \ldots, $f^{(n-1)}(x)$ are all Fourier
transformable, $\lim_{x \to \pm\infty} f^{(k)}(x) = 0$ for $k = 0, 1,
\ldots, n-1$, and if $f^{(n)}(x)$ exists then
$\mathcal{F}[ f^{(n)} ](\omega) = (i \omega)^{n} \hat{f}(\omega)$.
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Derivative of a Fourier Transform}

\begin{theorem}
%\label{thm:FT:xn}
If function $f(x)$ is Fourier transformable and such that the
operations of integration and differentiation can be interchanged, then
\[
%\label{eq:FT:xn}
\mathcal{F}[ x^{n} f(x) ](\omega)
	= i^{n} \frac{d^{n}}{d\omega^{n}} \left[ \hat{f}(\omega) \right]	.
\]
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the Fourier transform of the ``sign''
function,
\[
\sgn(x)
	= \left\{ \begin{array}{rl}
-1	& \mbox{if $x < 0$,}	\\
1	& \mbox{if $x > 0$.}
\end{array} \right.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution}

The sign function is not absolutely integrable, but the Dirac delta
function and the previous theorem can overcome this drawback.
Making use of the integral,
\[
-1 + 2 \int_{-\infty}^{x} \delta(\tau) \, d\tau
	= \sgn(x)	.
\]
Thus $\sgn'(x) = 2 \delta(x)$.
Therefore,
\[
i \omega \mathcal{F}[\sgn](\omega)
	= \mathcal{F}[\sgn'](\omega)
	= \mathcal{F}[2 \delta](\omega)
	= \frac{2}{\sqrt{2 \pi}}
\implies
\mathcal{F}[\sgn](\omega)
	= \frac{2}{i \omega \sqrt{2 \pi}}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Example}

Find the Fourier transform of the Heaviside
function,\index{Heaviside function}
\[
H(x)
	= \left\{ \begin{array}{rl}
0	& \mbox{if $x < 0$,}	\\
1	& \mbox{if $x > 0$.}
\end{array} \right.
\]

\end{frame}
%%

\begin{frame}\frametitle{Solution}

The Heaviside function can be built from the sign function of the
previous example.
\[
H(x)
	= \frac{1}{2} (1 + \sgn(x))
\]
This implies,
\[
\mathcal{F}[H](\omega)
	= \frac{1}{2} \mathcal{F}[1](\omega) + \frac{1}{2} \mathcal{F}[\sgn](\omega)
	= \frac{1}{2} \mathcal{F}[1](\omega) + \frac{1}{i \omega
          \sqrt{2 \pi}}	.
\]
The constant function $1$ is not absolutely integrable, so instead
find a Fourier transform $\hat{f}(\omega)$ whose inverse Fourier
transform is $1$.
\[
1
	= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty}
        \hat{f}(\omega) e^{i \omega x} \, d\omega
\implies
\hat{f}(\omega)
	= \sqrt{2 \pi} \delta(\omega)
\]
Together these yield,
\[
\mathcal{F}[H](\omega)
	= \sqrt{\frac{\pi}{2}} \delta(\omega) - \frac{i}{\omega
          \sqrt{2 \pi}}	.
\]

\end{frame}
%%

\begin{frame}\frametitle{Shifting Property}

\begin{theorem}
%\label{thm:FT:shifting}
Suppose function $f$ has Fourier transform $\hat{f}$ and that $a$ is a
constant, then $f(x-a)$ and $f(x) e^{-i a x}$ have Fourier transforms and
\begin{align*}
%\label{eq:FT:shiftf}
\mathcal{F}[ f(x-a) ](\omega)
	&= e^{-i \omega a} \hat{f}(\omega)	\\
%\label{eq:FT:shiftfhat}
\mathcal{F}[ f(x) e^{-i a x} ](\omega)
	&= \hat{f}(\omega+a)	.
\end{align*}
\end{theorem}

\end{frame}
%%

\begin{frame}\frametitle{Fourier Convolution}

\begin{definition}
The \textbf{Fourier convolution}
of two functions $f$ and $g$ is
\[
%\label{eq:FT:conv}
(f \ast g)(x) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} f(x - z) g(z) \, dz	,
\]
provided the improper integral converges.
\end{definition}

\begin{theorem}[Fourier Convolution Theorem]
%\label{thm:FT:conv}
Suppose functions $f$ and $g$ have Fourier transforms $\hat{f}$ and
$\hat{g}$ respectively, then $(f \ast g)(x)$ has a Fourier transform and
\[
%\label{eq:FT:transconv}
\mathcal{F}[ f \ast g ](\omega)
	= \hat{f}(\omega) \hat{g}(\omega)	.
\]
\end{theorem}
\textbf{Remark}: the Fourier transform of the Fourier
convolution of $f$ and $g$ is the product of the Fourier transforms of
$f$ and $g$.

\end{frame}
%%

%\begin{frame}\frametitle{Homework}

%\begin{itemize}
%\item Read Section $x.y$
%\item Exercises: 
%\end{itemize}

%\end{frame}
%%

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