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\documentclass[aspectratio=169]{beamer}
\def\stylepath{../styles}
\usepackage{\stylepath/com303}
\begin{document}
\intertitle{ideal filters and their approximation}
\begin{frame}
\frametitle{What is the best lowpass we can think of?}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=1,ylabel={$H(e^{j\omega})$}]{-1,1}{0,1.5}
\moocStyle
\dspFunc{x abs 0.3 ge {0} {1} ifelse}
\dspCustomTicks[axis=x]{0.3 $\omega_c$ -0.3 $-\omega_c$}
\psbrace[rot=-90,ref=tC,nodesepB=-7pt,linewidth=0.1pt](0.3,1.1)(-0.3,1.1){$\omega_b = 2\omega_c$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Ideal lowpass filter}
\[
H(e^{j\omega}) = \begin{cases}
1 & \mbox{for $|\omega| \le \omega_c$} \\
0 & \mbox{otherwise}
\end{cases} \qquad \mbox{($2\pi$-periodicity implicit)}
\]
\begin{itemize}
\item perfectly flat passband
\item infinite attenuation in stopband
\item zero-phase (no delay)
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Ideal lowpass filter: impulse response}
\begin{align*}
h[n] &= \IDTFT{H(e^{j\omega})} \\
&= \ldots \\
&= \frac{\sin\omega_c n}{\pi n}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Ideal lowpass filter: impulse response}
\begin{center}
\begin{figure}
\begin{dspPlot}[sidegap=1,xticks=10,yticks=custom,xout=true]{-24,24}{-0.3,1.2}
\moocStyle
{\dspSignal{x \dspSinc{0}{5}}}
{\dspFunc[xmin=-25,xmax=25,linecolor=green!80,linewidth=2pt]{x \dspSinc{0}{5}}}
\dspCustomTicks[axis=y]{0 0 1 $\omega_c/\pi$}
\end{dspPlot}
\end{figure}
\end{center}
\end{frame}
\begin{frame} \frametitle{The Sinc and the Rect!}
The sinc-rect pair:
\begin{eqnarray*}
\mbox{rect}(x) & = & \left\{
\begin{array}{ll}
1 & |x| \leq 1/2 \\
0 & |x| > 1/2
\end{array}
\right. \\ \\ \\
\mbox{sinc}(x) & = & \left\{
\begin{array}{ll}
\displaystyle \frac{\sin(\pi x)}{\pi x} & x \neq 0 \\
1 & x = 0
\end{array}
\right.
\end{eqnarray*}
\centering
(note that $\sinc(x) = 0$ when $x$ is a nonzero integer)
\end{frame}
\begin{frame}
\frametitle{The ideal lowpass in canonical form}
\Large
\[
\hlBox{a}{blue!30}{\displaystyle \rect\left(\frac{\omega}{2\omega_c}\right)} \,\, \xleftrightarrow{\mbox{\footnotesize ~DTFT~}} \,\,
\hlBox[\displaystyle \rect\left(\frac{\omega}{2\omega_c}\right)]{b}{red!30}{\displaystyle \frac{\omega_c}{\pi}\sinc\left(\frac{\omega_c}{\pi}n\right)}
\]
\end{frame}
\def\cof{0.3333 }
\begin{frame}
\frametitle{Example}
\center
$\omega_c = \pi/3$: $H(e^{j\omega}) = \mbox{rect}(3\omega/2\pi)$, $h[n] = (1/3)\mbox{sinc}(n/3)$
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=3,ylabel={$H(e^{j\omega})$},height=2cm]{-1,1}{0,1.1}
\moocStyle
\dspFunc{x abs \cof ge {0} {1} ifelse}
\end{dspPlot}
\begin{dspPlot}[xout=true,ylabel={$h[n]$},height=2cm]{-30,30}{-0.1,0.4}
\moocStyle
\dspSignal{x 0 eq {\cof } {x dup 180 mul 3 div sin exch 3.14 mul div} ifelse}
\dspCustomTicks[axis=y]{0.3333 $1/3$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{From the ideal lowpass...}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=1,ylabel={$H(e^{j\omega})$}]{-1,1}{0,1.5}
\moocStyle
\dspFunc{x abs 0.3 ge {0} {1} ifelse}
\dspCustomTicks[axis=x]{0.3 $\omega_c$ -0.3 $-\omega_c$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{... to the ideal highpass}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=1,ylabel={$H(e^{j\omega})$}]{-1,1}{0,1.5}
\moocStyle
\dspFunc{x abs 0.6 ge {1} {0} ifelse}
\dspCustomTicks[axis=x]{0.6 $\omega_c$ -0.6 $-\omega_c$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Ideal highpass filter}
\[
H_{hp}(e^{j\omega}) = \begin{cases}
1 & \mbox{for $\pi \ge |\omega| \ge \omega_c$} \\
0 & \mbox{otherwise}
\end{cases} \qquad \mbox{($2\pi$-periodicity implicit)}
\]
\vspace{2em}
\[
H_{hp}(e^{j\omega}) = 1 - H_{lp}(e^{j\omega})
\]
\[
h_{hp}[n] = \delta[n] - \frac{\omega_c}{\pi}\mbox{sinc}(\frac{\omega_c}{\pi}n)
\]
\end{frame}
\begin{frame}
\frametitle{Ideal bandpass filter}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=1,ylabel={$H(e^{j\omega})$}]{-1,1}{0,1.5}
\moocStyle
\dspFunc{x \dspRect{0.5}{0.6} x \dspRect{-0.5}{0.6} add}
\dspCustomTicks[axis=x]{0.2 $\omega_0-\omega_c$ 0.5 $\omega_0$ 0.8 $\omega_0+\omega_c$ -0.5 $-\omega_0$}
\end{dspPlot}
\end{figure}
\end{frame}
\def\wc{0.5 }
\def\sshape{\dspRect{0}{.6}}
\begin{frame}
\frametitle{Ideal bandpass filter}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=1]{-1,1}{0,1.5}
\moocStyle
\pnode(0,0){o} \pnode(\wc,0){a} \pnode(-\wc,0){b}
\only<1>{
\dspFunc{x \dspPeriodize \sshape}
\dspCustomTicks[axis=x]{0.3 $\omega_c$ -0.3 $-\omega_c$}}
\only<2-3>{
\dspFunc[linecolor=lightgray,linestyle=dashed,linewidth=0.4pt]{x \dspPeriodize \sshape }}
\only<2-3>{
\dspFunc[linecolor=green]{x \wc sub \dspPeriodize \sshape 2 div}
\nccurve[linewidth=0.4pt,linecolor=red,ncurv=1,angleB=-90,angleA=-90]{->}{o}{a}
\dspCustomTicks[axis=x]{{\wc} $\omega_0$}}
\only<3>{
\dspFunc[linecolor=blue]{x \wc add \dspPeriodize \sshape 2 div}
\nccurve[linewidth=0.4pt,linecolor=red,ncurv=1,angleB=-90,angleA=-90]{->}{o}{b}
\dspCustomTicks[axis=x]{{-\wc} $-\omega_0$}}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Ideal bandpass filter}
\[
H_{bp}(e^{j\omega}) = \begin{cases}
1 & \mbox{for $|\omega \pm \omega_0| \le \omega_c$} \\
0 & \mbox{otherwise}
\end{cases} \qquad \mbox{($2\pi$-periodicity implicit)}
\]
\vspace{2em}
\[
h_{bp}[n] = 2\cos(\omega_0 n) \frac{\omega_c}{\pi}\mbox{sinc}(\frac{\omega_c}{\pi}n)
\]
\end{frame}
\begin{frame} \frametitle{The bad news}
\begin{itemize}
\item impulse response is infinite support, two-sided \\
$\Longrightarrow$ cannot compute the output in a finite amount of time \\
$\Longrightarrow$ that's why it's called ``ideal''
\item impulse response decays slowly in time \\ $\Longrightarrow$ we need a lot of samples for a good approximation
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{How can we approximate an ideal lowpass?}
Idea \#1:
\begin{itemize}
\item pick $\omega_c$
\item compute ideal impulse response $h[n]$
\item truncate $h[n]$ to a finite-support $\hat{h}[n]$
\item $\hat{h}[n]$ defines an FIR filter
\[
\hat{h}[n] = \begin{cases}
\displaystyle \frac{\omega_c}{\pi}\sinc\left(\frac{\omega_c}{\pi}n\right) & |n| \le N \\ \\
0 & \mbox{otherwise}
\end{cases}
\]
\item symmetric impulse truncation minimizes the MSE
\end{itemize}
\end{frame}
% DTFT of the truncated ideal lowpass
% h[n] = wc/pi sinc(wc/pi n) = (sin(wc n)) / n
% H(e^jw)=wc/pi + 2/pi \sum_1^N (sin(wc n)/n) cos(w n)
\def\wc{0.5 }
\def\gibbs#1{%
180 mul
0 % accumulator a
1 1 #1 % start, inc, end for loop
{dup dup % stack: x a n n n
\wc 180 mul mul sin % x a n n sin(wc n)
exch div exch % x a sin/n n
3 index % x a sin/n n x
mul cos mul
add}
for
exch pop
2 mul 3.14 div
\wc add}
\def\plotApp#1#2{%
\only<#1>{%
\dspFunc{x \gibbs{#2}}
\FPupn\M{#2 2 * 1 + 0 trunc}
\dspText(0.7,0.8){$M=\M$}}}
\begin{frame} \frametitle{Why it's not such a good idea}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=2,ylabel={$\hat{H}(e^{j\omega})$},xout=true]{0,1}{-0.3,1.2}
\moocStyle
\dspFunc[linecolor=green,linewidth=1pt]{x \dspRect{0}{1}}
\only<2->{
\psline[linewidth=0.5pt,linecolor=blue](0,1.1)(0.55, 1.1)
\psline[linewidth=0.5pt,linecolor=blue](0.45,-0.1)(1, -0.1)}
\plotApp{2}{4}
\plotApp{3}{10}
\plotApp{4}{50}
\plotApp{5}{150}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Frequency sampling}
Idea \#2:
\begin{itemize}
\item draw desired frequency response $H(e^{j\omega})$
\item take $M$ samples $S[k] = H(e^{j\omega_k})$, $\omega_k = (2\pi/M)k$
\item compute inverse DFT of the $S[k]$ values
\item use $s[n]$ to build impulse response $\hat{h}[n]$
\end{itemize}
\end{frame}
\def\interp#1{\dspSincS{#1 11 div 2 mul}{5} 11 div #1 11 div 2 mul \dspRect{0}{1} mul}
\def\tComp#1#2{\only<#1->{\dspFunc[linewidth=0.5pt,linecolor=blue!50]{x \interp{#2}}}}
\begin{frame}
\frametitle{Frequency sampling: frequency response}
\begin{center}
\only<5>{$\hat{H}(e^{j\omega})$}\vphantom{$\hat{H}(e^{j\omega})$}
\begin{figure}
\begin{dspPlot}[xtype=freq,yticks=none,xout=true]{-1,1}{-.5, 1.5}
\moocStyle
\dspFunc[linecolor=green,linewidth=0.1pt]{x \dspRect{0}{1}}
\multido{\n=-5+1}{11}{
\FPupn\f{2 11 \n{} / mul 3 trunc}
\dspTaps[linecolor=gray]{{\f} {\f0 \dspRect{0}{1}}}}
\tComp{2}{0}
\tComp{3}{1}
\tComp{4}{2}
\only<5->{
\dspFunc[linewidth=0.5pt,linecolor=blue!50]{x \interp{-1}}
\dspFunc[linewidth=0.5pt,linecolor=blue!50]{x \interp{-2}}
\dspFunc{x \interp{-2} x \interp{-1} x \interp{0} x \interp{1} x \interp{2} add add add add}}
\end{dspPlot}
\end{figure}
\end{center}
\end{frame}
\intertitle{mathematical digression \#1: the Gibbs phenomenon}
\begin{frame} \frametitle{The Gibbs phenomenon}
\centering
The maximum error around the cutoff frequency\\ is around 9\% of the height of the jump\\ {\em regardless of $N$}
\end{frame}
\begin{frame} \frametitle{Understanding the Gibbs phenomenon}
\[
h[n] = \frac{\omega_c}{\pi}\sinc\left(\frac{\omega_c}{\pi}n\right)
\]
\[
\hat{h}[n] = h[n]\, w[n]
\]
\[
w[n] = \begin{cases}
1 & |n| \le N \\
0 & \mbox{otherwise}
\end{cases}
\]
\vspace{1em}
\[
\hat{H}(e^{j\omega}) = (X \ast W)(e^{j\omega})
\]
\end{frame}
\begin{frame} \frametitle{Convolution of DTFTs}
\centering
in $\ell_2(\mathbb{Z})$:
\begin{align*}
(x \ast y)[n] &= \langle x^*[k], y[n-k] \rangle \\
&= \sum_{k=-\infty}^{\infty}x[k]y[n-k]
\end{align*}
\vspace{1em}
in $L_2([-\pi,\pi])$:
\begin{align*}
(X \ast Y)(e^{j\omega}) &= \langle X^*(e^{j\sigma}), Y(e^{j(\omega-\sigma)}) \rangle \\
&= \frac{1}{2\pi}\int_{-\pi}^{\pi}X(e^{j\sigma}) Y(e^{j(\omega-\sigma)}) d\sigma
\end{align*}
\end{frame}
\def\N{9 }
\begin{frame} \frametitle{Understanding the Gibbs phenomenon}
\centering
$ \hat{h}[n] = h[n]\, w[n] $
\begin{figure}
\begin{dspPlot}[xticks=10,ylabel={$h[n]$},height=2cm]{-20,20}{-.7,1.2}
\moocStyle
\dspSignal[linecolor=green]{x \dspSinc{0}{2}}
\end{dspPlot}
\end{figure}
\begin{figure}
\begin{dspPlot}[xticks=10,ylabel={$w[n]$},height=2cm]{-20,20}{0,1.2}
\moocStyle
\dspSignal[linecolor=blue]{x \dspRect{0}{\N}}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Understanding the Gibbs phenomenon}
\centering
$ \hat{H}(e^{j\omega}) = (H \ast W)(e^{j\omega}), \qquad W(e^{j\omega}) = \sin(\omega M/2)/\sin(\omega/2)\quad$
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=none,ylabel={$H(e^{j\omega})$},height=2cm]{-1,1}{0,1.2}
\moocStyle
\dspFunc[linecolor=green]{x \dspRect{0}{1}}
\end{dspPlot}
\end{figure}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=2,xout=true,yticks=10,ylabel={$W(e^{j\omega})$},height=2cm]{-1,1}{-7,23}
\moocStyle
\dspFunc[linecolor=blue]{x \dspSincS{0}{\N}}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Mainlobe and sidelobes}
\centering
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=1,ylabel={$W(e^{j\omega})$}]{-1,1}{-7,23}
\moocStyle
\dspFunc[linecolor=gray]{x \dspSincS{0}{\N}}
\only<2->{
% first zero crossing is at 2/(2N+1)
\FPupn\fzc{2 \N * 1 + 2 / 4 trunc}
\pscustom[linestyle=none,fillstyle=solid,fillcolor=blue!30]{
\dspFunc[xmin=-\fzc,xmax=\fzc]{x \dspSincS{0}{\N}}}}
\only<3->{
\FPupn\szc{2 \N * 1 + 4 / 4 trunc}
\pscustom[linestyle=none,fillstyle=solid,fillcolor=red!50]{
\dspFunc[xmin=\fzc,xmax=\szc]{x \dspSincS{0}{\N}}}
\pscustom[linestyle=none,fillstyle=solid,fillcolor=red!50]{
\dspFunc[xmin=-\szc,xmax=-\fzc]{x \dspSincS{0}{\N}}}}
\end{dspPlot}
\end{figure}
\end{frame}
\def\wRes#1#2{%
\only<#1>{%
\dspFunc{x \dspSincS{#2}{\N}}
\psline[linewidth=1pt,linecolor=gray](#2,80)(#2,-70)}}
\def\tRes#1#2{%
\only<#1>{\dspFunc[xmax=#2]{x \gibbs{\N}}}}
\def\animate{%
\doSlide{1}{-.8} \doSlide{2}{-.7} \doSlide{3}{-.6} \doSlide{4}{-.5} \doSlide{5}{-.4}
\doSlide{6}{-.3} \doSlide{7}{-.2} \doSlide{8}{-.1} \doSlide{9}{0} \doSlide{10}{.1}
\doSlide{11}{.2} \doSlide{12}{.3} \doSlide{13}{.4} \doSlide{14}{.5} \doSlide{15}{.6}
\doSlide{16}{.7} \doSlide{17}{.8}}
\begin{frame} \frametitle{Implicit frequency-domain convolution}
\begin{center}
$N=\N, M=2N+1$
\begin{figure}
\begin{dspPlot}[height=1.4cm,xtype=freq,xticks=none,yticks=1,ylabel={$H(e^{j\omega})$}]{-1,1}{0,1.2}
\moocStyle
\psset{linecolor=green}
\dspFunc{x \dspRect{0}{1}}
\end{dspPlot}
\begin{dspPlot}[height=1.4cm,xtype=freq,xticks=none,xout=true,yticks=10,ylabel={$W(e^{j\omega})$}]{-1,1}{-6,23}
\moocStyle
\psset{linecolor=blue}
\def\doSlide{\wRes}\animate
\end{dspPlot}
\begin{dspPlot}[height=1.4cm,xtype=freq,xticks=2,xout=true,yticks=1,ylabel={$\hat{H}(e^{j\omega})$}]{-1,1}{-.2,1.2}
\moocStyle
\def\doSlide{\tRes}\animate
\end{dspPlot}
\end{figure}
\end{center}
\end{frame}
\def\N{9 }
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 1}
Observation 1: integral of window's transform is independent of $N$:
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=custom,yticks=custom]{-1,1}{-.3,1.2}
\moocStyle
\pscustom[linestyle=none,fillstyle=solid,fillcolor=blue!30]{
\dspFunc{x -1 eq {0} {x 1 eq {0} {x \dspSincS{0}{\N} \N 2 mul 1 add div} ifelse} ifelse}}
\dspFunc[linecolor=blue]{x \dspSincS{0}{\N} \N 2 mul 1 add div}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 1}
Observation 1: integral of window's transform is independent of $N$:
\begin{align*}
\int_{-\pi}^{\pi} W(e^{j\omega}) &= \int_{-\pi}^{\pi} \sum_{n=-N}^{N}e^{-j\omega n}d\omega \\
&= \sum_{n=-N}^{N} \int_{-\pi}^{\pi}e^{-j\omega n}d\omega \\
&= 2\pi
\end{align*}
\end{frame}
\def\doSincS#1#2{%
\only<#1>{\dspFunc[linecolor=blue]{x \dspSincS{0}{#2} #2 2 mul 1 add div}%
\dspText(0.7,1){$N=#2$}}}
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 2}
For large $N$, the area is concentrated around the midpoint:
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=custom,yticks=custom]{-1,1}{-.3,1.2}
\moocStyle
\doSincS{1}{9}
\doSincS{2}{51}
\doSincS{3}{101}
\end{dspPlot}
\end{figure}
\end{frame}
\def\N{9 }
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 3}
\centering
maximum value of the convolution integral: \only<2->{$A$} \only<3->{$+ B$}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=custom,yticks=custom]{-1,1}{-.3,1.2}
\moocStyle
\only<2->{
\pscustom[linestyle=none,fillstyle=solid,fillcolor=blue!30]{
\dspFunc[xmax=0]{x -1 eq {0} {x 0 eq {0} {x \dspSincS{0}{\N} \N 2 mul 1 add div} ifelse} ifelse}}}
\only<3->{
\FPupn\fzc{2 \N * 1 + 2 / 4 trunc }
\pscustom[linestyle=none,fillstyle=solid,fillcolor=yellow!30]{
\dspFunc[xmin=0,xmax=\fzc]{x 0 eq {0} {x \fzc\space ge {0} {x \dspSincS{0}{\N} \N 2 mul 1 add div} ifelse} ifelse}}}
\dspFunc[linecolor=blue]{x \dspSincS{0}{\N} \N 2 mul 1 add div}
\dspFunc[linecolor=green]{x 1 \N div ge {0} {1} ifelse}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 4}
\centering
For large $N$, A is basically half the total area:
\[
A \approx \frac{1}{2}\int_{-\pi}^{\pi} W(e^{j\omega}) = \pi
\]
\end{frame}
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 5}
\centering
$W(e^{j\omega}) = \sin(\omega M/2)/\sin(\omega/2), \qquad M=2N+1$
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=custom,yticks=custom]{0,0.5}{-.3,1.2}
\moocStyle
\FPupn\fzc{2 \N * 1 + 2 / 4 trunc}
\pscustom[linestyle=none,fillstyle=solid,fillcolor=yellow!30]{
\dspFunc[xmin=0,xmax=\fzc]{x 0 eq {0} {x \fzc\space ge {0} {x \dspSincS{0}{\N} \N 2 mul 1 add div} ifelse} ifelse}}
\dspFunc[linecolor=blue]{x \dspSincS{0}{\N} \N 2 mul 1 add div}
\dspPlotFrame
\dspCustomTicks{{\fzc} $2\pi/M$}
\dspCustomTicks[axis=y]{1 $M$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 5}
\begin{align*}
\int_{0}^{2\pi/M} W(e^{j\omega}) &= \int_{0}^{2\pi/M} \frac{\sin(\omega M/2)}{\sin(\omega/2)}d\omega \\
&= \frac{2}{M}\int_{0}^{\pi} \frac{\sin t}{\sin (t/M)}dt \qquad (t=M\omega/2)\\
&\approx \frac{2}{M}\int_{0}^{\pi}\frac{\sin t}{(t/M)}dt \qquad (\sin t/M \approx t/M \mbox{ for $M$ large})\\
&= 2 \int_{0}^{\pi}\frac{\sin t}{t}dt \\
&\approx 2\pi \times 0.589 \quad \mbox{independent of $M$!}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 6}
\centering
maximum value of the convolution: $(A+B)/2\pi$
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=custom,yticks=custom]{-1,1}{-.3,1.2}
\moocStyle
\pscustom[linestyle=none,fillstyle=solid,fillcolor=blue!30]{
\dspFunc[xmax=0]{x -1 eq {0} {x 0 eq {0} {x \dspSincS{0}{\N} \N 2 mul 1 add div} ifelse} ifelse}}
\FPupn\fzc{2 \N * 1 + 2 / 4 trunc }
\pscustom[linestyle=none,fillstyle=solid,fillcolor=yellow!30]{
\dspFunc[xmin=0,xmax=\fzc]{x 0 eq {0} {x \fzc\space ge {0} {x \dspSincS{0}{\N} \N 2 mul 1 add div} ifelse} ifelse}}
\dspFunc[linecolor=blue]{x \dspSincS{0}{\N} \N 2 mul 1 add div}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Quantifying the Gibbs overshoot -- 6}
\begin{align*}
A &\approx \frac{1}{2}\int_{-\pi}^{\pi} W(e^{j\omega}) = \pi \\
B &\approx 2\pi \times 0.589
\end{align*}
\[
(A+B)/2\pi \approx 1.09
\]
\end{frame}
\begin{frame}
\frametitle{Mainlobe and sidelobes}
\centering
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=1,ylabel={$W(e^{j\omega})$}]{-1,1}{-7,23}
\moocStyle
\dspFunc[linecolor=gray]{x \dspSincS{0}{\N}}
% first zero crossing is at 2/(2N+1)
\FPupn\fzc{2 \N * 1 + 2 / 4 trunc}
\pscustom[linestyle=none,fillstyle=solid,fillcolor=blue!30]{
\dspFunc[xmin=-\fzc,xmax=\fzc]{x \dspSincS{0}{\N}}}
\FPupn\szc{2 \N * 1 + 4 / 4 trunc}
\pscustom[linestyle=none,fillstyle=solid,fillcolor=red!50]{
\dspFunc[xmin=\fzc,xmax=\szc]{x \dspSincS{0}{\N}}}
\pscustom[linestyle=none,fillstyle=solid,fillcolor=red!50]{
\dspFunc[xmin=-\szc,xmax=-\fzc]{x \dspSincS{0}{\N}}}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{What if we change the window?}
We want:
\begin{itemize}
\item narrow mainlobe so that transition is sharp
\item small sidelobe so Gibbs error is small
\item short window so FIR is efficient
\end{itemize}
\centering
\vspace{1em}
{very conflicting requirements!}
\end{frame}
\begin{frame}
\frametitle{Triangular window}
\centering
\begin{figure}
\begin{dspPlot}[xticks=10,ylabel={$w[n]$}]{-20,20}{0,1.2}
\moocStyle
\dspSignal{x \dspTri{0}{\N}}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Rectangular vs Triangular Window}
\begin{figure}
\begin{dspPlot}[xtype=freq,xticks=2,yticks=none]{0,1}{-0.3,1.2}
\moocStyle
\only<1->{\dspFunc[linecolor=gray]{x \dspSincS{0}{9} 19 div}}
\only<2->{%
\dspFunc{x \dspSincS{0}{5} 11 div dup mul}
\dspLegend(0.5,1){gray {19-tap rectangular} darkred {19-tap triangular}}}
\end{dspPlot}
\end{figure}
\end{frame}
\intertitle{mathematical digression \#2: stability of the sinc}
\begin{frame}
\frametitle{Little-known fact}
\note<1>{Divergence of sinc sum (integral criterion): \\
\[
\int_0^\infty \left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x=\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\left\vert\frac{\sin x}{x}\right\vert \,\mathrm{d}x>\sum_{k=0}^\infty\int_{k\pi}^{(k+1)\pi}\frac{\left\vert\sin x\right\vert}{(k+1/2)\pi} \,\mathrm{d}x=\frac{2}{\pi}\sum_{k=0}^\infty\frac{1}{k+1/2}
\]}
\begin{itemize}
\item the sinc is not absolutely summable
\item the ideal lowpass is not BIBO stable!
\item example for $\omega_c = \pi/3$: $h[n] = (1/3)\sinc(n/3)$
\item take $x[n] = \mbox{sign}\{\sinc(-n/3)\}$ and
\[
y[0] = (x \ast h)[0] = \frac{1}{3}\sum_{k=-\infty}^{\infty}|\sinc(k/3)| = \infty
\]
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Divergence is however very slow...}
%% matlab code:
% N=10000;W=3;
% s=zeros(1,N);
% s(1) = 1;
% for n=2:N
% s(n) = s(n-1) + 2*abs(sin(pi*n/W)/(pi*n/W));
% end
% s = s/W;
\centering
$s(n) = \displaystyle (1/3)\sum_{k=-n}^{n}|\sinc(k/3)|$
\begin{figure}
\begin{dspPlot}[xticks=custom]{1,100}{0,4.1}
\moocStyle
\dspFuncData{1 0.3333 2 1.8960 3 2.1453 4 2.2943 5 2.4001 6 2.4810 7 2.5482 8 2.6050 9 2.6535 10 2.6969 11 2.7358 12 2.7704 13 2.8025 14 2.8320 15 2.8590 16 2.8844 17 2.9082 18 2.9303 19 2.9514 20 2.9713 21 2.9900 22 3.0080 23 3.0252 24 3.0413 25 3.0570 26 3.0721 27 3.0864 28 3.1003 29 3.1137 30 3.1265 31 3.1390 32 3.1511 33 3.1627 34 3.1740 35 3.1850 36 3.1956 37 3.2060 38 3.2161 39 3.2258 40 3.2354 41 3.2447 42 3.2537 43 3.2626 44 3.2713 45 3.2797 46 3.2880 47 3.2961 48 3.3039 49 3.3117 50 3.3193 51 3.3267 52 3.3340 53 3.3411 54 3.3481 55 3.3550 56 3.3617 57 3.3683 58 3.3748 59 3.3813 60 3.3875 61 3.3937 62 3.3998 63 3.4057 64 3.4116 65 3.4174 66 3.4231 67 3.4287 68 3.4343 69 3.4396 70 3.4450 71 3.4504 72 3.4555 73 3.4607 74 3.4658 75 3.4707 76 3.4757 77 3.4806 78 3.4853 79 3.4901 80 3.4948 81 3.4994 82 3.5040 83 3.5085 84 3.5129 85 3.5173 86 3.5217 87 3.5260 88 3.5302 89 3.5344 90 3.5386 91 3.5427 92 3.5468 93 3.5507 94 3.5547 95 3.5587 96 3.5625 97 3.5664 98 3.5702 99 3.5740 100 3.5777}
\dspCustomTicks[axis=x]{1 $1$ 50 $5000$ 100 $10000$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Proof:}
integral criterion for convergence:
\[
\sum_{n\in\mathbb{N}^+}f(n) < \infty \Leftrightarrow \int_{1}^{\infty}f(t)dt < \infty
\]
\vspace{2em}
so:
\[
\sum_{n\in\mathbb{N}^+}\left|\frac{\sin n}{n}\right| < \infty \Leftrightarrow \int_{1}^{\infty}\left|\frac{\sin t}{t}\right|dt < \infty
\]
\end{frame}
\begin{frame}
\frametitle{Proof:}
\begin{align*}
% advance integration limit
\int_1^\infty \left|\frac{\sin t}{t}\right|dt & \geq \int_\pi^\infty \left|\frac{\sin t}{t}\right|dt \\
% split the integral into contiguous intervals
&= \sum_{k=1}^\infty \int_{k\pi}^{(k+1)\pi} \left|\frac{\sin t}{t}\right|dt \\
% replace denominator by its maximum value
&\geq \sum_{k=1}^\infty\int_{k\pi}^{(k+1)\pi} \frac{|\sin t|}{(k+1)\pi}dt \\
% integrate sin
&= \frac{1}{\pi}\sum_{k=1}^\infty \frac{1}{k+1} \int_{k\pi}^{(k+1)\pi} |\sin t|dt \\
&= \frac{2}{\pi}\sum_{k=1}^\infty\frac{1}{k+1} = \infty
\end{align*}
\end{frame}
\end{document}
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