Homec4science
Diffusion epfl (master) Browse

R2653/teaching/COM303/lectures/old notation/19 - quantization oversampling/1_quantization.tex
bb0ecb168febmaster

Last Change
Raw File

1_quantization.tex
View Options

\documentclass[aspectratio=169]{beamer}
\def\stylepath{../styles}
\usepackage{\stylepath/com303}
\begin{document}
\begin{frame} \frametitle{Overview:}
\begin{itemize}[<+->]
\item Quantization
\item Uniform quantization and error analysis
\item Clipping, saturation, companding
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Quantization}
\begin{itemize}[<+->]
\item digital devices can only deal with integers ($b$ bits per sample)
\item we need to map the range of a signal onto a finite set of values
\item irreversible loss of information $\rightarrow$ quantization noise
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Quantization schemes}
\begin{figure}
\center
\begin{dspBlocks}{1.2}{.11}
$x[n]$~~ & {\BDfilter{$\mathcal{Q}\{\cdot\}$}} & ~~$\hat{x}[n]$ \\
& & \\
\ncline{->}{1,1}{1,2}\ncline{->}{1,2}{1,3}
\end{dspBlocks}
\end{figure}
\vspace{1em}
Several factors at play:
\begin{itemize}[<+->]
\item storage budget (bits per sample)
\item storage scheme (fixed point, floating point)
\item properties of the input
\begin{itemize}[<+->]
\item range
\item probability distribution
\end{itemize}
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Scalar quantization}
\begin{figure}
\center
\begin{dspBlocks}{1.2}{.1}
$x[n]$~~ & {\BDfilter{$\mathcal{Q}\{\cdot\}$}} & ~~$\hat{x}[n]$ \\
& & \only<3->{$R$ bps} \\
\ncline{->}{1,1}{1,2}\ncline{->}{1,2}{1,3}
\end{dspBlocks}
\end{figure}
\vspace{1em}
The simplest quantizer:
\begin{itemize}[<+->]
\item each sample is encoded individually (hence {\em scalar})
\item each sample is quantized independently (memoryless quantization)
\item each sample is encoded using $R$ bits
\end{itemize}
\end{frame}
\def\btick#1{\psline[linewidth=0.6pt](#1,-.1)(#1,.1)}
\def\qtick#1#2{\btick{#1}\uput{10pt}[-90](#1,0){$i_{#2}$}}
\def\qdot#1#2{\pscircle*(#1,0){2pt} \uput[90](#1,0){$\hat{x}_{#2}$}}
\def\qbrace#1#2#3#4{%
\psbrace[braceWidth=0.5pt,braceWidthOuter=3pt,rot=90,nodesepB=10pt]%
(! #1 0.1 add -0.3)(! #2 0.1 sub -0.3) {$I_{#3}$}%
\uput[-90](! #1 #2 add 0.5 mul 0.7){$k=#4$}}
\def\dbrace#1#2#3{%
\psbrace[braceWidth=0.5pt,braceWidthOuter=3pt,rot=90,nodesepB=10pt]%
(! #1 0.1 add -0.3)(! #2 0.1 sub -0.3){#3}}
\begin{frame} \frametitle{Scalar quantization}
Input signal: $A \le x[n] \le B$ ($A, B$ can be $\infty$)
\begin{itemize}
\item<2-> each sample quantized over $2^R$ possible values $\Rightarrow$ $2^R$ intervals.
\item<3-> each interval associated to a quantization value
\end{itemize}
\setbeamercovered{invisible}
\begin{figure}
\center
\psset{xunit=2.4cm,yunit=2cm}
\begin{pspicture}(0.5,-1)(5.5,1)
\psline[linewidth=1pt,tickwidth=2pt,](1,0)(5,0)%
\btick{1} \uput{8pt}[90](1,0){$A$} \btick{5} \uput{8pt}[90](5,0){$B$}
\uncover<2->{\btick{1} \btick{2} \btick{5} \btick{2.5} \btick{4}}
\uncover<3->{\qdot{4.4}{3}\qdot{1.5}{0}\qdot{2.4}{1}\qdot{3}{2}}
\end{pspicture}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Scalar quantization}
Example for $R=2$:
\begin{figure}[t]
\center
\psset{xunit=2.4cm,yunit=2cm}
\begin{pspicture}(0.5,-1)(5.5,1)
\psline[linewidth=1pt,tickwidth=2pt,](1,0)(5,0)
\uput{8pt}[90](1,0){$A$} \uput{8pt}[90](5,0){$B$}
\qtick{1}{0} \qtick{2}{1} \qdot{1.5}{0} \qbrace{1}{2}{0}{00}
\qtick{2.5}{2} \qdot{2.4}{1} \qbrace{2}{2.5}{1}{01}
\qtick{4}{3} \qdot{3}{2} \qbrace{2.5}{4}{2}{10}
\qtick{5}{4} \qdot{4.4}{3} \qbrace{4}{5}{3}{11}
\end{pspicture}
\end{figure}
\pause
\begin{itemize}
\item what are the optimal interval boundaries $i_k$?
\item what are the optimal quantization values $\hat{x}_k$?
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Optimal Quantization}
\centering
The optimal quantizer minimizes the energy of the quantization error:
\[
e[n] = \mathcal{Q}\{ x[n] \} - x[n] = \hat{x}[n] - x[n]
\]
\vspace{1em}
\begin{itemize}[<+->]
\item model $x[n]$ as a stochastic process
\item find the optimal $i_k$ and $\hat{x}_k$ that minimize $\sigma_e^2 = \expt{e^2[n]}$
\item optimal quantizer will depend on the input's statistics
\end{itemize}
\end{frame}
\begin{frame} \frametitle{Quantization MSE}
\centering
\begin{align*}
\sigma_e^2 &= \int_{-\infty}^{\infty} (x - \mathcal{Q}\{x\})^2\,f_x(x)\,dx \\ \pause
&= \sum_{k=0}^{2^R - 1} \int_{i_k}^{i_{k+1}} (x - \hat{x}_k)^2\,f_x(x)\,dx
\end{align*}
\pause
\vspace{2em}
find global minimum wrt $i_k$, $\hat{x}_k$
\end{frame}
\begin{frame} \frametitle{Simple example: optimal one-bit quantizer}
\centering
\begin{figure}[t]
\psset{xunit=2.4cm,yunit=2cm}
\begin{pspicture}(0.5,-1)(5.5,1)
\psline[linewidth=1pt,tickwidth=2pt,](1,0)(5,0)
\uput{8pt}[90](1,0){$A$} \uput{8pt}[90](5,0){$B$}
\qtick{1}{0} \qtick{2.5}{1} \qtick{5}{2}
\qdot{2.2}{0} \qdot{4.2}{1}
\end{pspicture}
\end{figure}
3 free parameters: $i_1, \hat{x}_0, \hat{x}_1$
\end{frame}
\begin{frame} \frametitle{Simple example: optimal one-bit quantizer}
\centering
\[
\sigma_e^2 = \int_{A}^{i_1} (x - \hat{x}_0)^2\,f_x(x)\,dx + \int_{i_1}^{B} (x - \hat{x}_1)^2\,f_x(x)\,dx
\]
\pause
\vspace{1em}
find $i_1, \hat{x}_0, \hat{x}_1$ such that
\[
\frac{\partial \sigma_e^2}{\partial i_1} = \frac{\partial \sigma_e^2}{\hat{x}_0} = \frac{\partial \sigma_e^2}{\hat{x}_1} = 0
\]
\end{frame}
\begin{frame} \frametitle{little calculus reminder}
\[
\frac{\partial}{\partial t} \int_{\alpha}^{t} f(\tau)\,d\tau = \frac{\partial}{\partial t} \left[ F(t) - F(\alpha)\right] = f(t)
\]
\end{frame}
\begin{frame} \frametitle{Optimal one-bit quantizer: threshold}
\centering
\begin{align*}
\frac{\partial \sigma_e^2}{\partial i_1} &= \frac{\partial}{\partial i_1}\left[\int_{A}^{i_1} (x - \hat{x}_0)^2\,f_x(x)\,dx + \int_{i_1}^{B} (x - \hat{x}_1)^2\,f_x(x)\,dx \right] \\ \pause
&= (i_1 - \hat{x}_0)^2\,f_x(i_1) - (i_1 - \hat{x}_1)^2\,f_x(i_1) = 0 \\ \\ \pause
&\Rightarrow (i_1 - \hat{x}_0)^2 - (i_1 - \hat{x}_1)^2 = 0 \\ \\ \pause
&\Rightarrow i_1 = \frac{\hat{x}_0 + \hat{x}_1}{2}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Optimal one-bit quantizer: values}
\centering
\begin{align*}
\frac{\partial \sigma_e^2}{\partial \hat x_0} &= \frac{\partial}{\partial x_0} \int_{A}^{i_1} (x - \hat{x}_0)^2\,f_x(x)\,dx \\ \pause
&= \int_{A}^{i_1} 2(\hat{x}_0 - x)\,f_x(x)\,dx = 0 \\ \\ \pause
&\Rightarrow \hat x_0 = \frac{\int_{A}^{i_1} x\,f_x(x)\,dx}{\int_{A}^{i_1} f_x(x)\,dx} \qquad\mbox{\it (center of mass)} \\
&\Rightarrow \hat x_1 = \frac{\int_{i_1}^{B} x\,f_x(x)\,dx}{\int_{i_1}^{B} f_x(x)\,dx}
\end{align*}
\end{frame}
\begin{frame} \frametitle{For uniformly-distributed input}
\centering
\begin{align*}
f_x(x) &= \frac{1}{B-A} \\ \\ \pause
\hat x_0 &= \frac{\int_{A}^{i_1} x\,dx}{\int_{A}^{i_1} dx} = \frac{A + i_1}{2} \\
\hat x_1 &= \frac{\int_{i_1}^{B} x\,dx}{\int_{i_1}^{B} dx} = \frac{i_1 + B}{2} \\ \\ \pause
i_1 &= \frac{\hat{x}_0 + \hat{x}_1}{2} = \frac{A + B}{2}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Optimal one-bit quantizer}
\centering
\begin{figure}[t]
\psset{xunit=2.4cm,yunit=2cm}
\begin{pspicture}(0.5,-1)(5.5,1)
\psline[linewidth=1pt,tickwidth=2pt,](1,0)(5,0)
\uput{8pt}[90](1,0){$A$} \uput{8pt}[90](5,0){$B$}
\qtick{1}{0} \qtick{3}{1} \qtick{5}{2}
\qdot{2}{0} \qdot{4}{1}
\end{pspicture}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Uniform quantization of uniform input}
\begin{itemize}[<+->]
\item simple but very general case
\item optimal subdivision: $2^R$ \emph{equal} intervals of width $\Delta = (B-A)2^{-R}$
\item optimal quantization values: interval's midpoint
\end{itemize}
\setbeamercovered{invisible}
\uncover<3->{
\begin{figure}[t]
\center
\psset{xunit=2.4cm,yunit=2cm}
\begin{pspicture}(0.5,-1)(5.5,1)
\psline[linewidth=1pt,tickwidth=2pt,](1,0)(5,0)
\uput{8pt}[90](1,0){$A$} \uput{8pt}[90](5,0){$B$}
\btick{1}\btick{2}\btick{3}\btick{4}\btick{5}
\qdot{1.5}{0}\qdot{2.5}{1}\qdot{3.5}{2}\qdot{4.5}{3}
\dbrace{1}{2}{$\Delta$}
\end{pspicture}
\end{figure}}
\end{frame}
\begin{frame}
\frametitle{Uniform 3-Bit quantization function}
\begin{figure}
\center
\psset{xunit=3cm,yunit=3cm}
\begin{pspicture}(-1.2,-1.2)(1.2,1.2)
\tiny
\psaxes[linewidth=1pt,tickwidth=0.5pt,%
Dx=.25,Dy=.25,%
showorigin=false](0,0)(-1.1,-1)(1.1,1)
\uput[90](0,1){$\hat{x}[n]$}
\uput[90](1.1,0){$x[n]$}
\multido{\n=0.875+-.250}{8}{%
\pscircle*[linecolor=orange](0,\n){2pt}
}
\psline[linecolor=blue,linestyle=dotted](-1,-1)(1,1)
\psplot[plotpoints=800,%
linewidth=1pt,linecolor=orange]{-.99}{.99}%
{x abs 0.25 div floor 0.25 mul 0.125 add x abs x div mul}
\uput[90](! -1 0.125 add 0.25 0 mul add dup){\gray 000}
\uput[90](! -1 0.125 add 0.25 1 mul add dup){\gray 001}
\uput[90](! -1 0.125 add 0.25 2 mul add dup){\gray 010}
%\uput[90](! -1 0.125 add 0.25 3 mul add dup){\gray 011}
\uput[90](! -1 0.125 add 0.25 4 mul add dup){\gray 100}
\uput[90](! -1 0.125 add 0.25 5 mul add dup){\gray 101}
\uput[90](! -1 0.125 add 0.25 6 mul add dup){\gray 110}
\uput[90](! -1 0.125 add 0.25 7 mul add dup){\gray 111}
\end{pspicture}
\end{figure}
\end{frame}
\begin{frame} \frametitle{Uniform quantization of uniform input: error analysis}
\begin{align*}
\sigma_e^2 &= \int_A^B f_x(x)(\mathcal{Q}\{ x \} - x)^2\,dx \\ \pause
&= \sum_{k=0}^{2^R-1} \int_{I_k} f_x(x)(\hat{x}_k - x)^2\,dx \\ \\ \pause
f_x(s) &= \frac{1}{B-A} \\
\Delta &= \frac{B-A}{2^R} \\
I_k &= [A + k\Delta, A+(k+1)\Delta] \\
\hat x_k &= A+k\Delta + \Delta/2
\end{align*}
\end{frame}
\begin{comment}
\begin{frame} \frametitle{Uniform quantization}
\begin{itemize}[<+->]
\item simple but very general case
\item range is split into $2^R$ \emph{equal} intervals of width $\Delta = (B-A)2^{-R}$
%\item quantization value is interval's midpoint
\end{itemize}
\setbeamercovered{invisible}
\uncover<3->{
\begin{figure}[t]
\center
\psset{xunit=2.4cm,yunit=2cm}
\begin{pspicture}(0.5,-1)(5.5,1)
\psline[linewidth=1pt,tickwidth=2pt,](1,0)(5,0)
\uput{8pt}[90](1,0){$A$} \uput{8pt}[90](5,0){$B$}
\btick{1}\btick{2}\btick{3}\btick{4}\btick{5}
\dbrace{1}{2}{$\Delta$}
\end{pspicture}
\end{figure}}
\end{frame}
\begin{frame}
\frametitle{Uniform quantization}
Mean Square Error is the second moment of the error signal:
\begin{align*}
\sigma_e^2 &= \expt{|\mathcal{Q}\{ x[n] \} - x[n]|^2} \\ \pause
&= \int_A^B f_x(\tau)(\mathcal{Q}\{ \tau \} - \tau)^2\,d\tau \\ \pause
&= \sum_{k=0}^{2^R-1} \int_{I_k} f_x(\tau)(\hat{x}_k - \tau)^2\,d\tau \\ \pause
\end{align*}
\centering
error depends on the probability distribution of the input
\end{frame}
\begin{frame}
\frametitle{Uniform quantization of uniform input}
Uniform-input hypothesis:
\[
f_x(\tau) = \frac{1}{B-A}
\]
\vspace{1em}
\[
\sigma_e^2 =\,\sum_{k=0}^{2^R-1} \int_{I_k}\frac{(\hat{x}_k - \tau)^2}{B-A}\,d\tau
\]
\end{frame}
\begin{frame}
\frametitle{Uniform quantization of uniform input}
Let's find the optimal quantization point by minimizing the error
\begin{align*}
\frac{\partial\sigma_e^2}{\partial \hat{x}_m} &= \frac{\partial}{\partial \hat{x}_m}\,\sum_{k=0}^{2^R-1} \int_{I_k}\frac{(\hat{x}_k - \tau)^2}{B-A}\,d\tau \\ \pause
&= \int_{I_m} \frac{2(\hat{x}_m - \tau)}{B-A}\,d\tau \\ \pause
&= \left.\frac{(\hat{x}_m - \tau)^2}{B-A}\right|_{A + m\Delta}^{A + m\Delta + \Delta}
\end{align*}
\end{frame}
\begin{frame}
\frametitle{Uniform quantization of uniform input}
Minimizing the error:
\[
\frac{\partial\sigma_e^2}{\partial \hat{x}_m} = 0 \quad \mbox{for $\hat{x}_m = A + m\Delta + \frac{\Delta}{2}$}
\]
\vspace{1em}
\centering
optimal quantization point is the interval's midpoint, for all intervals
\end{frame}
\end{comment}
\begin{frame} \frametitle{Uniform quantization of uniform input: error analysis}
\begin{align*}
\sigma_e^2 &= \sum_{k=0}^{2^R-1} \int_{A+k\Delta}^{A+k\Delta+\Delta}\frac{(A + k\Delta + \Delta/2 - x)^2}{B-A}\,dx \\ \pause
&= 2^R \int_{0}^{\Delta}\frac{(\Delta/2 - x)^2}{B-A}\,dx \\ \pause
&= \frac{\Delta^2}{12}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Error analysis of the quantization error}
fundamental assumptions:
\begin{itemize}
\item signal and quantization error are uncorrelated (ok-ish)
\item quantization error process is white (stretch)
\end{itemize}
\vspace{1em}
\centering
quantization noise acts as additive white noise
\end{frame}
\begin{frame} \frametitle{Error analysis}
\begin{itemize}[<+->]
\item error energy
\[
\sigma^2_e = \Delta^2/12, \qquad \Delta = (B-A)/2^R
\]
\item signal energy
\[
\sigma_x^2 = (B-A)^2/12
\]
\item signal to noise ratio
\[
\mbox{SNR} = 2^{2R}
\]
\item in dB
\[
\mbox{SNR}_{\mbox{dB}} = 10\log_{10} 2^{2R} \approx 6R \,\,\mbox{dB}
\]
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{The ``6dB/bit'' rule of thumb}
\begin{itemize}[<+->]
\item a compact disk has 16 bits/sample:
\[
\mbox{max SNR} = 96\mbox{dB}
\]
\item a DVD has 24 bits/sample:
\[
\mbox{max SNR} = 144\mbox{dB}
\]
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Rate/Distortion Curve }
\begin{figure}
\begin{dspPlot}[width=5cm,height=5cm,sidegap=0,xticks=none,yticks=none,xlabel={rate ($R$)},ylabel={distortion ($\sigma_e^2$)}]{0,6}{0,1}
\moocStyle
\dspFunc{x -2 mul 2 exch exp }
\end{dspPlot}
\end{figure}
\end{frame}
\begin{frame}
\frametitle{Other quantization errors}
If input is not bounded to $[A,B]$:
\begin{itemize}[<+->]
\item clip samples to $[A,B]$: linear distortion (can be put to good use in guitar effects!)
\item smoothly saturate input: this simulates the saturation curves of analog electronics
\end{itemize}
\end{frame}
\begin{frame}
\frametitle{Clipping vs saturation}
\begin{columns}
\begin{column}{0.45\paperwidth}
\begin{figure}
\begin{dspPlot}[width=5cm,height=5cm,sidegap=0]{-2.5,2.5}{-1.5,1.5}
\moocStyle
\dspFunc{x abs 1 gt {x abs x div} {x} ifelse}
\end{dspPlot}
\end{figure}
\end{column}
\begin{column}{0.45\paperwidth}
\begin{figure}
\begin{dspPlot}[width=5cm,height=5cm,sidegap=0]{-2.5,2.5}{-1.5,1.5}
\moocStyle
\dspFunc{2.7 2 x mul exp dup 1 sub exch 1 add div}
\end{dspPlot}
\end{figure}
\end{column}
\end{columns}
\end{frame}
\begin{frame}
\frametitle{Other quantization errors}
If input is not uniform:
\begin{itemize}[<+->]
\item use uniform quantizer and accept increased error. \\
For instance, if input is Gaussian:
\[
\sigma^2_e = \frac{\sqrt{3} \pi}{2} \, \sigma^2 \, \Delta^2
\]
\item use ``companders''
\item design optimal quantizer for input distribution, if known (Lloyd-Max algorithm)
\end{itemize}
\end{frame}
\def\muval{255 }
\def\mulaw{ dup dup abs div exch abs \muval mul 1 add ln 1 \muval add ln div mul }
\def\imulaw{ abs 1 \muval add exch exp 1 sub \muval div }
\begin{frame}
\frametitle{$\mu$-law compander}
\begin{columns}
\begin{column}{0.3\paperwidth}
\[
\mathcal{C}\{x[n]\} = \mbox{sgn}(x[n]) \frac{\ln(1+ \mu |x[n]|)}{\ln(1+\mu)}
\]
\end{column}
\begin{column}{0.5\paperwidth}
\begin{figure}
\center
\psset{xunit=3cm,yunit=3cm}
\begin{pspicture}(-1.2,-1.2)(1.2,1.2)
\psgrid[gridlabels=0,subgriddiv=10,subgriddots=5](0,0)(-1,-1)(1,1)
\psaxes[linewidth=1pt,labels=none,showorigin=false](0,0)(-1,-1)(1,1)
\psaxes[linewidth=1pt,axesstyle=frame,yAxis=false,ticks=none,labels=none](0,0)(-1,-1)(1,1)
\uput[90](0,1){$\mathcal{C}\{x\}$}
\uput[0](1,0){$x$}
\psplot[plotpoints=800,linewidth=2pt,linecolor=darkred]{-.99}{.99}{x \mulaw }
\end{pspicture}
\end{figure}
\end{column}
\end{columns}
\end{frame}
\begin{frame} \frametitle{Lloyd-Max Quantizer design}
\begin{align*}
\sigma_e^2 &= \sum_{k=0}^{2^R - 1} \int_{i_k}^{i_{k+1}} (x - \hat{x}_k)^2\,f_x(x)\,dx \\ \\
\mbox{A)}\quad \frac{\partial \sigma_e^2}{\partial \hat x_k} = 0 &\Rightarrow \hat x_k = \frac{\displaystyle\int_{i_{k-1}}^{i_k} x\,f_x(x)\,dx}{\displaystyle\int_{i_{k-1}}^{i_k} f_x(x)\,dx} \\
\mbox{B)}\quad\frac{\partial \sigma_e^2}{\partial i_k} = 0 &\Rightarrow i_k = \frac{\hat{x}_{k-1} + \hat{x}_k}{2}
\end{align*}
\end{frame}
\begin{frame} \frametitle{Lloyd-Max Quantizer design}
\begin{itemize}
\item start with a guess for the $i_k$
\item solve A and B iteratively until convergence
\end{itemize}
\end{frame}
\end{document}
c4science · Help