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\documentclass[aspectratio=169]{beamer}
\def\stylepath{../styles}
\usepackage{\stylepath/com303}
\begin{document}
\begin{frame}
\frametitle{Discrete-Time Fourier Transform}
\[
X(e^{j\omega}) = \sum_{n=-\infty}^{\infty} x[n]\, e^{-j\omega n}
\]
\vspace{2em}
\[
x[n] = \frac{1}{2\pi} \int_{-\pi}^{\pi} X(e^{j\omega}) e^{j\omega n}d\omega, \qquad n \in \mathbb{Z}
\]
\end{frame}
\begin{frame}
\frametitle{Review: DFT}
\begin{center}
\begin{figure}
\psset{linewidth=1pt}
\begin{dspBlocks}{4}{0.2}
\circlenode{A}{~~$\mathbb{C}^{N}$~~} & & \circlenode{B}{~~$\mathbb{C}^{N}$~~}
\end{dspBlocks}
\end{figure}
\ncarc[nodesep=3pt,arcangle=45]{->}{A}{B}{\naput{$X[k] = \langle e^{j\frac{2\pi}{N}nk}, x[n] \rangle$}}
\ncarc[nodesep=3pt,arcangle=45]{->}{B}{A}{\naput{$x[n] = (1/N)\sum X[k]\,e^{j\frac{2\pi}{N}nk}$}}
\ncline[linestyle=none]{A}{B}{\ncput{basis: $\{e^{j\frac{2\pi}{N}nk}\}_k$}}
\end{center}
\end{frame}
\begin{frame}
\frametitle{Review: DFS}
\begin{center}
\begin{figure}
\psset{linewidth=1pt}
\begin{dspBlocks}{4}{0.2}
\circlenode{A}{~~$\tilde{\mathbb{C}}^{N}$~~} & & \circlenode{B}{~~$\mathbb{C}^{N}$~~}
\end{dspBlocks}
\end{figure}
\ncarc[nodesep=3pt,arcangle=45]{->}{A}{B}\naput{$\tilde{X}[k] = \langle e^{j\frac{2\pi}{N}nk}, \tilde{x}[n] \rangle$}
\ncarc[nodesep=3pt,arcangle=45]{->}{B}{A}\naput{$\tilde{x}[n] = (1/N)\sum \tilde{X}[k]\,e^{j\frac{2\pi}{N}nk}$}
\ncline[linestyle=none]{A}{B}\ncput{basis: $\{e^{j\frac{2\pi}{N}nk}\}_k$}
\end{center}
\end{frame}
\begin{frame}
\frametitle{DTFT}
\begin{center}
\begin{figure}
\psset{linewidth=1pt}
\begin{dspBlocks}{4}{0.2}
\circlenode{A}{~~$\ell_2(\mathbb{Z})$~~} & & \circlenode{B}{$L_2([-\pi,\pi])$}
\end{dspBlocks}
\end{figure}
\ncarc[nodesep=3pt,arcangle=45]{->}{A}{B}\naput{$X(e^{j\omega}) = \langle e^{j\omega n}, x[n] \rangle$}
\ncarc[nodesep=3pt,arcangle=45]{->}{B}{A}\naput{$x[n] = (1/2\pi)\int X(e^{j\omega})e^{j\omega n}d\omega$}
\ncline[linestyle=none]{A}{B}\ncput{``basis'': $\{e^{j\omega n}\}_\omega$}
\end{center}
\end{frame}
\begin{frame}
\frametitle{DTFT as basis expansion}
Some things are OK:
\begin{itemize}[<+->]
\item $\DFT{\delta[n]} = 1$
\item $\DTFT{\delta[n]} = \langle e^{j\omega n}, \delta[n] \rangle = 1$
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{DTFT as basis expansion}
Some things aren't:
\begin{itemize}
\item $\DFT{1} = \delta[n]$
\item $\DTFT{1} = \sum_{n=-\infty}^{\infty} e^{-j\omega n} = ?$
\vspace{2em}
\item problem: too many interesting sequences are {\em not} square summable!
\end{itemize}
\end{frame}
\intertitle{The Dirac delta}
\begin{frame}
\frametitle{The Dirac delta functional}
\centering
the functional $\delta(t)$ is defined by its ``sifting'' property:
\[
\int_{-\infty}^{\infty} \delta(t-s)f(t)dt = f(s)
\]
for all $f(t), \quad t \in \mathbb{R}$.
\end{frame}
\begin{frame} \frametitle{The Dirac delta functional}
\def\tempFun{x 5 div 180 mul dup 1.1 mul cos 1 add 2 div exch 4 mul sin 0.1 mul add 3 mul x 5 div add}
\psset{xunit=1.8cm,yunit=1cm,linewidth=0.8pt}
\centering
\begin{pspicture}(-.5,-.5)(5.5,3.5)
\psline{->}(0,-.1)(0,3.4)
\psline{->}(-.1,0)(5.4,0)
\rput[t](2,-.1){$s$}
\psplot[plotstyle=curve,linewidth=1pt,linecolor=blue!60]{-.5}{5.5}{\tempFun}
\psdot[linecolor=blue!60,linewidth=2pt](2,1.9)
\psline[linecolor=darkred,linewidth=2pt]{->}(2,0)(2,1.9)
\end{pspicture}
\[
\int_{-\infty}^{\infty} \delta(t-s)f(t)dt = f(s)
\]
\end{frame}
\begin{frame} \frametitle{The Dirac delta functional in physics}
\psset{xunit=1.8cm,yunit=1cm,linewidth=0.8pt}
\centering
\begin{pspicture}(-.5,-.5)(5.5,3.5)
\psline{->}(0,-.1)(0,3.4)
\psline{->}(-.1,0)(5.4,0)
\rput[t](2,-.1){$t_0$}
\rput[t](5.5,0){$t$}
\rput[t](-.3,3.5){$v(t)$}
\psline[linewidth=2pt,linecolor=blue!60]{-}(-.4,0)(2,0)(2,2)(5,2)
\end{pspicture}
\[
\only<1>{F(t) = ma(t) = m \frac{\partial v(t)}{\partial t}}
\only<2>{a(t_0) = \infty ?}
\only<3>{v(t) = \int_{-\infty}^{t} a(\tau) d\tau = \int_{-\infty}^{t} \delta(\tau-t_0)d\tau}
\only<4>{a(t) = \delta(t-t_0)}
\only<5>{F(t) \propto \delta(t-t_0) \approx \begin{cases}
\infty & \mbox{for $t=t_0$} \\ 0 & \mbox{otherwise}
\end{cases}}
\]
\end{frame}
\begin{frame}
\frametitle{Intuition}
\centering
consider a family of {\em localizing} functions $r_k(t)$ with $k \in \mathbb{N}$ and $t \in \mathbb{R}$ where:
\begin{itemize}[<+->]
\item support inversely proportional to $k$
\item constant area
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Intuition}
\[
\rect(t) = \begin{cases} 1 & \mbox{for $|t| < 1/2$} \\ 0 & \mbox{otherwise} \end{cases}
\]
\vspace{3em}
we can build a localizing family as $r_k(t) = k\rect(kt)$:
\begin{itemize}
\item nonzero over $[-1/2k, 1/2k]$, i.e. support is $1/k$
\item area is $1$
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{The family $r_k(t) = k\rect(kt)$}
\center
\begin{figure}
\begin{dspPlot}[yticks=20,sidegap=0]{-1,1}{0,41}
\smallStems
\only<2->{\dspFunc[linecolor=gray]{x abs 0.5 gt {0} {1} ifelse} }
\only<3->{\dspFunc[linecolor=gray]{x abs 0.1 gt {0} {5} ifelse} }
\only<4->{\dspFunc[linecolor=gray]{x abs 0.0333 gt {0} {15} ifelse}}
\only<1>{\dspFunc[linecolor=red]{x abs 0.5 gt {0} {1} ifelse} \uput[45](0.1,2){$k=1$}}
\only<2>{\dspFunc[linecolor=red]{x abs 0.1 gt {0} {5} ifelse} \uput[45](0.1,5){$k=5$}}
\only<3>{\dspFunc[linecolor=red]{x abs 0.0333 gt {0} {15} ifelse} \uput[45](0.05,15){$k=15$}}
\only<4>{\dspFunc[linecolor=red]{x abs 0.0125 gt {0} {40} ifelse} \uput[45](0.01,36){$k=40$}}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Remember the Mean Value Theorem?}
for any continuous function over the interval $[a, b]$ there exists $\gamma\in [a, b]$ s.t.
\[
\int_{a}^{b} f(t)dt = (b-a)\,f(\gamma)
\]
\end{frame}
\begin{frame} \frametitle{The Mean Value Theorem}
\savedata{\smooth}[
1.0000 1.0000 1.1579 1.0011 1.3158 1.0130 1.4737 1.0484
1.6316 1.1201 1.7895 1.2408 1.9474 1.4232 2.1053 1.6777
2.2632 1.9852 2.4211 2.3076 2.5789 2.6068 2.7368 2.8445
2.8947 2.9824 3.0526 2.9828 3.2105 2.8418 3.3684 2.6079
3.5263 2.3338 3.6842 2.0725 3.8421 1.8769 4.0000 1.8000]
\psset{xunit=1.8cm,yunit=1cm,linewidth=0.8pt}
\centering
\begin{pspicture}(-.5,-.5)(5.5,3.5)
\pscustom{
\dataplot[plotstyle=curve]{\smooth}
\gsave
\psline(4,0) \psline(1,0)
\fill[fillstyle=solid,fillcolor=blue!20]
\grestore
}
\dataplot[plotstyle=curve,linewidth=2pt]{\smooth}
\psline{->}(0,-.1)(0,3.4)
\psline{->}(-.1,0)(5.4,0)
\rput[t](1,-.1){$a$}
\rput[t](4,-.1){$b$}
%\psline[linecolor=darkgray,linestyle=dashed]{-}(0.5,1.983)(4.5,1.983)
\only<2->{
\psline[linestyle=dashed,linewidth=0.5pt]{-}(0,2)(4,2)
\psline[linestyle=dashed,linewidth=0.5pt]{-}(2.3,0)(2.3,2)
\rput[r](0,2){$f(\gamma)$~~}\rput[t](2.3,-.1){$\gamma$}
\pspolygon[fillstyle=crosshatch,linecolor=red!30,hatchcolor=red!30](1,0)(4,0)(4,2)(1,2)
}
\end{pspicture}
\end{frame}
\begin{frame}
\frametitle{Extracting a point value}
for our family of localizing functions:
\begin{align*}
\int_{-\infty}^{\infty} r_k(t)f(t)dt &= k \int_{-1/2k}^{1/2k} f(t) dt \\
&= f(\gamma)|_{\gamma \in [-1/2k, 1/2k]}
\end{align*}
and so:
\[
\lim_{k\rightarrow\infty} \int_{-\infty}^{\infty} r_k(t)f(t)dt = f(0)
\]
\end{frame}
\begin{frame}
\frametitle{The Dirac delta functional}
\centering
The delta functional is a shorthand. Instead of writing
\[
\lim_{k\rightarrow\infty} \int_{-\infty}^{\infty} r_k(t-s)f(t)dt
\]
we write
\[
\int_{-\infty}^{\infty} \delta(t-s)f(t)dt.
\]
as if $\lim_{k\rightarrow\infty}r_k(t) = \delta(t)$,
\end{frame}
\begin{frame}
\frametitle{The ``pulse train''}
\centering
\[
\tilde{\delta}(\omega) = 2\pi \sum_{k = -\infty}^{\infty} \delta(\omega - 2\pi k)
\]
just a technicality to use the Dirac delta in the space of $2\pi$-periodic functions
\end{frame}
\begin{frame}
\frametitle{Graphical representation}
\begin{center}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=1,yticks=none,width=12cm,height=2cm]{-4.5,4.5}{0, 1.5}
\moocStyle
\dspDiracs{-4 1 -2 1 0 1 2 1 4 1}
\end{dspPlot}
\end{figure}
\end{center}
\end{frame}
\begin{frame}
\frametitle{Now let the show begin!}
\begin{align*}
\IDTFT{\tilde{\delta}(\omega)} &= \frac{1}{2\pi} \int_{-\pi}^{\pi} \tilde{\delta}(\omega) e^{j\omega n}d\omega \\ \pause
&= \int_{-\pi}^{\pi} \delta(\omega) e^{j\omega n}d\omega \\ \pause
&= e^{j\omega n}|_{\omega = 0} \\ \pause
& = 1
\end{align*}
\end{frame}
\begin{frame}
\frametitle{In other words}
\[
\DTFT{1} = \tilde{\delta}(\omega)
\]
\end{frame}
\begin{frame}
\frametitle{Does it make sense?}
Partial DTFT sum:
\[
S_k(\omega) = \sum_{n=-k}^{k} e^{-j\omega n}
\]
\end{frame}
%% partial DTFT sum
\def\pds#1{
0 0 % re im
-#1.0 cvi 1 #1.0 cvi % start inc end
{x mul 180 mul % wn (degrees)
dup sin
exch cos % stack: re im sin cos
4 1 roll % cos re im sin
add % cos re im
3 1 roll % im cos re
add
exch
} for % re im
dup mul exch
dup mul add
sqrt}
\def\pdsSlide#1#2{%
\FPupn\nn{#1 1 + 0 trunc}
\only<#1>{\dspFunc{\pds{#2}}\dspText(0,88){$k=#2$}}
\only<\nn->{\dspFunc[linewidth=1pt,linecolor=gray]{\pds{#2}}}}
\begin{frame}
\frametitle{Plotting $|S_k(\omega)|$}
\begin{center}
\begin{figure}
\begin{dspPlot}[xtype=freq,yticks=20]{-1,1}{0, 83}
\moocStyle
\pdsSlide{1}{5}
\pdsSlide{2}{10}
\pdsSlide{3}{15}
\pdsSlide{4}{30}
\only<5>{\dspFunc{\pds{40}}\dspText(0,88){$k=40$}}
\end{dspPlot}
\end{figure}
\end{center}
\end{frame}
\begin{frame}
\frametitle{Does it make sense?}
\centering
Partial DTFT sums look like a family of localizing functions:
\[
S_k(\omega) \rightarrow \tilde{\delta}(\omega)
\]
\end{frame}
\begin{frame}
\frametitle{Using the same technique}
\[
\IDTFT{\tilde{\delta}(\omega - \omega_0)} = e^{j\omega_0 n}
\]
So:
\begin{itemize}[<+->]
\item $\DTFT{1} = \tilde{\delta}(\omega)$
\item $\DTFT{e^{j\omega_0 n}} = \tilde{\delta}(\omega - \omega_0)$
\item $\DTFT{\cos \omega_0 n} = [\tilde{\delta}(\omega-\omega_0) + \tilde{\delta}(\omega+\omega_0)]/2 $
\item $\DTFT{\sin \omega_0 n} = -j[\tilde{\delta}(\omega-\omega_0) - \tilde{\delta}(\omega+\omega_0)]/2 $
\end{itemize}
\end{frame}
\end{document}
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