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R2653/teaching/COM303/lectures/22 - data transmission - alt notation/4_transmitter.tex
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\documentclass[aspectratio=169]{beamer}
\def\stylepath{../styles}
\usepackage{\stylepath/com303}
\usepackage{pst-3dplot}
%\def\sa#1#2{ #1 sub #2 div dup abs 1 gt {pop 0} {1 exch sub } ifelse }
%\def\rp#1{ #1 sub dup abs 1 gt {pop 0} {dup 0 gt {1 exch sub} {pop 1} ifelse} ifelse }
%\def\ip#1{ #1 sub dup abs 1 gt {pop 0} {1 add 45 mul sin} ifelse 3 mul}
\begin{document}
\begin{frame} \frametitle{QAM transmitter, final design}
\begin{figure}
\footnotesize
\begin{dspBlocks}{0.7}{0.2}
data~~ & \BDfilterMulti{Scrambler \&\\ Mapper} & \BDupsmp{K} & \BDlowpass[0.5em] & \BDmul & \BDfilter{Re} & ~~$s[n]$ \\
& & & & $e^{j\omega_c n}$
\ncline{->}{1,1}{1,2}\ncline{->}{1,2}{1,3}^{~~~~~~~~$a[n]$} \ncline{->}{1,3}{1,4}
\ncline{->}{1,4}{1,5}^{~~~$b[n]$} \ncline{->}{1,5}{1,6}^{\!\!$c[n]$} \ncline{->}{1,6}{1,7}
\ncline{->}{2,5}{1,5}
\end{dspBlocks}
\end{figure}
\end{frame}
\begin{frame} \frametitle{QAM receiver, idealized design}
\begin{figure}
\footnotesize
\begin{dspBlocks}{0.6}{0}
& & \BDfilter{$H(z)$} & & $e^{-j\omega_c n}$ \\
$\hat{s}[n]$~~ & \BDsplit & & \BDadd & \BDmul & \BDlowpass[0.5em] & \BDdwsmp{K} & \BDfilterMulti{Slicer \&\\ Descrambler} & ~~data
%
\psset{arrows=->}
\ncline{2,1}{2,4}
\ncline{2,4}{2,5}^{$\hat{c}[n]$}
\ncline{2,5}{2,6}^{\hspace{-1em}$\hat{b}[n]$}
\ncline{2,6}{2,7}\ncline{2,7}{2,8}^{\hspace{-2.5em}$\hat{a}[n]$}
\ncline{2,8}{2,9}
\ncline{-}{1,2}{2,2}\ncline{-}{1,3}{1,4}^{~~~~$j$}
\ncline{1,2}{1,3}
\ncline{1,4}{2,4}\ncline{1,5}{2,5}
\end{dspBlocks}
\end{figure}
\end{frame}
\end{document}
\def\rp#1{ \dspTri{#1}{1} 2 mul }
\def\ip#1{ #1 sub dup abs 1 gt {pop 0} {180 mul sin} ifelse 3 mul}
\newenvironment{threeDspec}[1]{%
\psset{unit=#1mm}
\begin{pspicture}(-6,-3)(6,3)
\psset{Alpha=170,Beta=10}
\psset{xPlotpoints=300}
\pstThreeDCoor[linecolor=black,%IIIDticks,IIIDlabels,%
xMin=-5,xMax=5,nameX={$\omega$},
yMin=-10,yMax=10,nameY={$\Imag$},
zMin=-3,zMax=3,nameZ={$\Real$}]}{%
\end{pspicture}}
\begin{frame}
\frametitle{QAM transmitter design}
\begin{figure}
\psset{linearc=0.2}
\begin{dspBlocks}{1.0}{0.8}
\parbox{4ex}{\small \tt \bf ..01100\\ 01010...}\hspace{5ex} &
\BDfilter{Scrambler} & \BDfilter{Mapper} & [name=A,mnode=circle] $K \uparrow$ \\
& &
\ncline{->}{1,1}{1,2} \ncline{->}{1,2}{1,3}
\ncline{->}{1,3}{1,4}^{~~~~$a[n]$}
\end{dspBlocks}
\begin{dspBlocks}{1.2}{0.8}
& [name=B] \BDlowpass[0.5em] & $\ldots$ & & \\
\ncline{->}{1,2}{1,3}^{$b[n]$}
\end{dspBlocks}
\ncbarr[angleA=0,arm=1cm,linewidth=1.2pt]{A}{B}
\end{figure}
\centering
$b[n] = b_r[n] + jb_i[n]$ is a complex-valued baseband signal...
\end{frame}
\begin{frame} \frametitle{Fulfilling the bandwidth constraint}
\begin{figure}
\begin{dspPlot}[height=3cm,xtype=freq,xticks=1,yticks=1]{-1,1}{0,1.2}
\moocStyle
\psframe[fillstyle=vlines,linecolor=green!70,hatchcolor=green!70,hatchangle=20](.25,-.05)(.6,0.05)
\psframe[fillstyle=vlines,linecolor=green!70,hatchcolor=green!70,hatchangle=20](-.25,-.05)(-.6,0.05)
\dspFunc{x \dspRect{.425}{.35} x \dspRect{-.425}{.35} add 2 div}
\dspCustomTicks[axis=x]{.25 $\omega_{\min}$ .425 $\omega_c$ .6 $\omega_{\max}$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame} \frametitle{QAM transmitter, final design}
\begin{figure}
\psset{linearc=0.2}
\begin{dspBlocks}{1.0}{0.8}
\parbox{4ex}{\small \tt \bf ..01100\\ 01010...}\hspace{5ex} &
\BDfilter{Scrambler} & \BDfilter{Mapper} & [name=A,mnode=circle] $K \uparrow$ \\
& &
\ncline{->}{1,1}{1,2} \ncline{->}{1,2}{1,3}
\ncline{->}{1,3}{1,4}^{~~~~$a[n]$}
\end{dspBlocks}
\begin{dspBlocks}{1.2}{0.4}
& [name=B] \BDlowpass[0.5em] & \BDmul & \BDfilter{Re} & \BDfilter{D/A} & $s(t)$ \\
& & $e^{j\omega_c n}$
\ncline{->}{1,2}{1,3}^{$b[n]$} \ncline{->}{1,3}{1,4}^{$c[n]$}
\ncline{->}{1,4}{1,5}^{$s[n]$} \ncline{->}{1,5}{1,6} \ncline{->}{1,6}{1,7}
\ncline{->}{2,3}{1,3}
\end{dspBlocks}
\ncbarr[angleA=0,arm=1cm,linewidth=1.2pt]{A}{B}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{The transmitted passband signal}
\begin{align*}
c[n] &= b[n]\,e^{j\omega_c n} \\ \\
s[n] &= \Real\{c[n]\} \\ \pause
&=\Real\{( b_r[n] + jb_i[n] ) (\cos\omega_c n + j\sin\omega_c n)\} \\ \pause
&= b_r[n]\cos\omega_c n - b_i[n]\sin\omega_c n
\end{align*}
\end{frame}
\begin{frame}
\frametitle{From complex baseband to real passband signal}
\centering
\only<1>{$b[n]$}
\only<2>{$c[n] = b[n]\,e^{j\omega_c n}$}
\only<3>{$s[n] = \Real\{c[n]\}$}
\begin{figure}
\begin{dspPlot}[height=3cm,xtype=freq,xticks=1,yticks=1]{-1,1}{0,1.2}
\moocStyle
\psframe[fillstyle=vlines,linecolor=green!70,hatchcolor=green!70,hatchangle=20](.25,-.05)(.6,0.05)
\psframe[fillstyle=vlines,linecolor=green!70,hatchcolor=green!70,hatchangle=20](-.25,-.05)(-.6,0.05)
\only<1>{\dspFunc{x \dspRaisedCosine{0}{.175}{0.2}}}
\only<2>{\dspFunc{x \dspRaisedCosine{.425}{.175}{0.2}}}
\only<3>{\dspFunc{x \dspRaisedCosine{.425}{.175}{0.2} x \dspRaisedCosine{-.425}{.175}{0.2} add 0.7 mul}}
\dspCustomTicks[axis=x]{.25 $\omega_{\min}$ .6 $\omega_{\max}$}
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Can we go back?}
\begin{center}
receiver obtains $s[n]$
can we recover the complex baseband $b[n]$ from $s[n]$?
\vspace{3em}
easiest way: Hilbert demodulation
\end{center}
\end{frame}
\begin{frame}
\frametitle{Hilbert demodulation}
\begin{center}
\begin{figure}
\begin{dspBlocks}{1.5}{0.5}
%
& & \BDfilter{$H(z)$} & & $e^{-j\omega_c n}$ & \\
$s[n]$~~ & \BDsplit & & \BDadd & \BDmul & ~~$b[n]$
%
\ncline{->}{2,1}{2,4} \ncline{->}{2,4}{2,5} \ncline{->}{2,5}{2,6}
\ncline{2,2}{1,2} \ncline{->}{1,2}{1,3}
\ncline{1,3}{1,4}^{~~$j$} \ncline{->}{1,4}{2,4}
\ncline{->}{1,5}{2,5}
\only<2->{
\psset{unit=1em}%\psgrid(-30,-5)(40,20)
\psframe[linecolor=darkred,framearc=.3,linestyle=dashed](-28,-2)(-11,7)
\dspText(-20,-4){\color{darkred}~$H_A(e^{j\omega}) = 1+jH(e^{j\omega})$}}
\end{dspBlocks}
\end{figure}
\end{center}
\end{frame}
\begin{frame} \frametitle{Hilbert demodulation}
\[
H(e^{j\omega}) = \begin{cases}
-j & \omega \geq 0 \\
j & \omega < 0
\end{cases}
\]
\vspace{2em}
\[
H_A(e^{j\omega}) = 1+jH(e^{j\omega}) = \begin{cases}
2 & \omega \geq 0 \\
0 & \omega < 0
\end{cases}
\]
\end{frame}
\begin{frame}
\frametitle{Hilbert demodulation (recap)}
\begin{center}
\begin{figure}
\begin{dspBlocks}{1.1}{0.5}
%
& & \BDfilter{$H(z)$} & & \\
$x[n]\cos(\omega_c n)$~~ & \BDsplit & & \BDadd & ~~~$x[n]e^{j\omega_c n}$
%
\ncline{->}{2,1}{2,4} \ncline{->}{2,4}{2,5}
\ncline{2,2}{1,2} \ncline{->}{1,2}{1,3}
\ncline{1,3}{1,4}^{~~$j$} \ncline{->}{1,4}{2,4}
\end{dspBlocks}
\end{figure}
\end{center}
\end{frame}
\begin{comment}
\begin{frame}
\frametitle{Hilbert demodulation}
\only<1>{$y[n] = x[n]\cos(\omega_c n)$}
\only<2-3>{$jy[n]$}
\only<4->{$jy[n]\ast h[n]$}
\begin{center}
\begin{threeDspec}{10}
\only<1-2>{
\parametricplotThreeD[linecolor=blue!50](-4,4){t t \ip{-2} t \ip{2} add 0.7 mul 0 }
\parametricplotThreeD[linecolor=red!50 ](-4,4){t 0 t \rp{-2} t \rp{2} add 0.5 mul}}
\only<3-4>{
\parametricplotThreeD[linecolor=red!50](-4,4){t 0 t \ip{-2} t \ip{2} add -0.25 mul }
\parametricplotThreeD[linecolor=blue!50 ](-4,4){t t \rp{-2} t \rp{2} add 2 mul 0}}
\only<5>{
\parametricplotThreeD[linecolor=blue!50](-4,4){t t \ip{-2} t \ip{2} sub -0.7 mul 0 }
\parametricplotThreeD[linecolor=red!50 ](-4,4){t 0 t \rp{-2} t \rp{2} sub -0.5 mul}}
\only<2>{
\psset{linecolor=green,arrows=->}
\pstThreeDCircle[beginAngle=-220,endAngle=80](4,0,0)(0,0,1)(0,1,0)
\pstThreeDCircle[beginAngle=-220,endAngle=80](-4,0,0)(0,0,1)(0,1,0)}
\only<4->{
\psset{linecolor=green,arrows=->}
\pstThreeDCircle[beginAngle=220,endAngle=-100](4,0,0)(0,0,1)(0,1,0)
\pstThreeDCircle[beginAngle=-220,endAngle=80](-4,0,0)(0,0,1)(0,1,0)}
\end{threeDspec}
\end{center}
\end{frame}
\begin{frame}
\frametitle{Hilbert demodulation}
\begin{tabular}{cc}
\begin{threeDspec}{6}
\parametricplotThreeD[linecolor=blue!50](-4,4){t t \ip{-2} t \ip{2} add 0.7 mul 0 }
\parametricplotThreeD[linecolor=red!50 ](-4,4){t 0 t \rp{-2} t \rp{2} add 0.5 mul}
\end{threeDspec}
&
\begin{threeDspec}{6}
\parametricplotThreeD[linecolor=blue!50](-4,4){t t \ip{-2} t \ip{2} sub -0.7 mul 0 }
\parametricplotThreeD[linecolor=red!50 ](-4,4){t 0 t \rp{-2} t \rp{2} sub -0.5 mul}
\end{threeDspec} \\
$y[n]$ & $j y[n]\ast h[n]$
\end{tabular}
\end{frame}
\end{comment}
\begin{frame} \frametitle{Hilbert demodulation}
$y[n] = x[n]\cos(\omega_c n)$
\begin{center}
\begin{threeDspec}{10}
\parametricplotThreeD[linecolor=blue!50](-4,4){t t \ip{-2} t \ip{2} add 0.7 mul 0 }
\parametricplotThreeD[linecolor=red!50 ](-4,4){t 0 t \rp{-2} t \rp{2} add 0.5 mul}
\end{threeDspec}
\end{center}
\end{frame}
\begin{frame} \frametitle{Hilbert demodulation}
$j y[n]\ast h[n] + y[n]$
\begin{center}
\begin{threeDspec}{10}
\parametricplotThreeD[linecolor=blue!50](-4,4){t t \ip{2} 0}
\parametricplotThreeD[linecolor=red!50 ](-4,4){t 0 t \rp{2}}
\end{threeDspec}
\end{center}
\end{frame}
\begin{frame}
\frametitle{Hilbert demodulation}
\begin{center}
\begin{figure}
\begin{dspBlocks}{1.1}{0.5}
%
& & \BDfilter{$H(z)$} & & \\
\only<1>{$x[n]\cos\omega_c n~~$}%
\only<2>{$x[n]\sin\omega_c n~~$}%
& \BDsplit & & \BDadd &
~~~%
\only<1>{$x[n]e^{j\omega_c n}$}%
\only<2>{$-jx[n]e^{j\omega_c n}$}%
%
\ncline{->}{2,1}{2,4} \ncline{->}{2,4}{2,5}
\ncline{2,2}{1,2} \ncline{->}{1,2}{1,3}
\ncline{1,3}{1,4}^{~~$j$} \ncline{->}{1,4}{2,4}
\end{dspBlocks}
\end{figure}
\end{center}
\end{frame}
\begin{frame}
\frametitle{The transmitted passband signal}
\begin{align*}
c[n] &= b[n]\,e^{j\omega_c n} \\
&= ( b_r[n] + jb_i[n] )e^{j\omega_c n} \\ \\
s[n] &= \Real\{c[n]\} \\
&= b_r[n]\cos\omega_c n - b_i[n]\sin\omega_c n
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Hilbert demodulation}
\begin{center}
\begin{figure}
\begin{dspBlocks}{1.1}{0.5}
%
& & \BDfilter{$H(z)$} & & \\
\only<1>{$b_r[n]\cos\omega_c n - b_i[n]\sin\omega_c n$}%
\only<2>{$s[n]$}~~ &
\BDsplit & & \BDadd &
~~~%
\only<1>{$b_r[n]e^{j\omega_c n} + jb_i[n]e^{j\omega_c n}$}%
\only<2>{$c[n]$}%
%
\ncline{->}{2,1}{2,4} \ncline{->}{2,4}{2,5}
\ncline{2,2}{1,2} \ncline{->}{1,2}{1,3}
\ncline{1,3}{1,4}^{~~$j$} \ncline{->}{1,4}{2,4}
\end{dspBlocks}
\end{figure}
\end{center}
\end{frame}
\begin{frame}
\frametitle{Hilbert demodulation}
\begin{center}
\begin{figure}
\begin{dspBlocks}{1}{0.5}
%
& & \BDfilter{$H(z)$} & & $e^{-j\omega_c n}$ & \\
$s[n]$~~ & \BDsplit & & \BDadd & \BDmul & ~~$b[n]$
%
\ncline{->}{2,1}{2,4} \ncline{->}{2,4}{2,5} \ncline{->}{2,5}{2,6}
\ncline{2,2}{1,2} \ncline{->}{1,2}{1,3}
\ncline{1,3}{1,4}^{~~$j$} \ncline{->}{1,4}{2,4}
\ncline{->}{1,5}{2,5}
\end{dspBlocks}
\end{figure}
\end{center}
\end{frame}
\begin{frame}
\frametitle{QAM transmitter, final design}
\begin{figure}
\psset{linearc=0.2}
\begin{dspBlocks}{1.0}{0.8}
\parbox{4ex}{\small \tt \bf ..01100\\ 01010...}\hspace{5ex} &
\BDfilter{Scrambler} & \BDfilter{Mapper} & [name=A,mnode=circle] $K \uparrow$ \\
& &
\ncline{->}{1,1}{1,2} \ncline{->}{1,2}{1,3}
\ncline{->}{1,3}{1,4}^{~~~~$a[n]$}
\end{dspBlocks}
\begin{dspBlocks}{1.2}{0.4}
& [name=B] \BDlowpass[0.5em] & \BDmul & \BDfilter{Re} & \BDfilter{D/A} & $s(t)$ \\
& & $e^{j\omega_c n}$
\ncline{->}{1,2}{1,3}^{$b[n]$} \ncline{->}{1,3}{1,4}^{$c[n]$}
\ncline{->}{1,4}{1,5}^{$s[n]$} \ncline{->}{1,5}{1,6} \ncline{->}{1,6}{1,7}
\ncline{->}{2,3}{1,3}
\end{dspBlocks}
\ncbarr[angleA=0,arm=1cm,linewidth=1.2pt]{A}{B}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{QAM receiver, idealized design}
\begin{figure}[tp]
\begin{dspBlocks}{0.8}{0}
& & & [name=A1] \BDfilter{$H(z)$} & & $e^{-j\omega_c n}$ \\
$\hat{s}(t)$~~ & \BDsampler
& [name=A,mnode=dot,linewidth=2pt] &
& [name=B,mnode=circle] + & \BDmul & [name=C] \\ \\ \\ \\ \\
%
\psset{arrows=->}
\ncangle[angleA=90,angleB=180,linewidth=\BDwidth]{A}{A1}
\ncangle[angleA=0,angleB=90,linewidth=\BDwidth]{A1}{B}^{~~~~$j$}
\ncbar[angleA=-90,angleB=-90,linewidth=\BDwidth]{A}{B}
%
\ncline{-}{2,1}{2,2}
\ncline{-}{2,2}{2,3}^{~~~~~~$\hat{s}[n]$}
\ncline{2,5}{2,6}^{$\hat{c}[n]$}
\ncline{-}{2,6}{2,7}^{~~~$\hat{b}[n]$}
\ncline{1,6}{2,6}
\end{dspBlocks}
\vspace{7ex}
\begin{dspBlocks}{1.2}{0}
[name=D,mnode=circle] $K \downarrow$ & \BDfilter{Slicer} &
\BDfilter{Descrambler} &
\hspace{1ex}\parbox{4ex}{\small \tt \bf ..01100\\ 01010...}
\psset{arrows=->}
\ncline{1,1}{1,2}^{$\hat{a}[n]$}
\ncline{1,2}{1,3}
\ncline{1,3}{1,4}
\end{dspBlocks}
\ncbarr[angleA=0,linewidth=1.2pt,linearc=0.2]{C}{D}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Example: the V.32 voiceband modem}
\begin{itemize}[<+->]
\item analog telephone channel: $F_{\min}=450$Hz, $F_{\max} = 2850$Hz
\item usable bandwidth: $W=2400$Hz, center frequency $F_c=1650$Hz
\item {\color{blue!70} pick $F_s = 3\cdot 2400 = 7200$Hz, so that $K=3$}
\item {\color{blue!70} $\omega_c = 0.458\pi$}
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Example: the V.32 voiceband modem}
\begin{itemize}[<+->]
\item maximum SNR: 22dB
\item {\color{blue!70} pick $P_{\mbox{err}} = 10^{-6}$}
\item {\color{blue!70} using QAM, we find
\[
M = \log_2\left(1 -\frac{3}{2} \frac{10^{22/10}}{\ln(10^{-6})} \right) \approx 4.1865
\]
so we pick $M=4$ and use a 16-point constellation}
\item {\color{green!70} final data rate is $WM = 9600$ bits per second}
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Theoretical channel capacity}
\begin{itemize}[<+->]
\item we used very specific design choices to derive the throughput
\item what is the best one can do?
\item Shannon's capacity formula is the upper bound
\[
C= W\,\log_2\left(1 + SNR \right)
\]
\item for instance, for the previous example $C\approx 17500$ bps
\item the gap can be narrowed by more advanced coding techniques
\end{itemize}
\end{frame}
\end{document}
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