The imported EPS files are
ActualEstimatedResponse.eps, BSAR2IIR2_Model.eps, EqualisedResponse.eps, Reverberation1.eps, Reverberation2.eps, SingleChannelDeconvolution.eps, UniversityLogo.eps, ar2blk1.eps, ar2blk7.eps, ar2blkF.eps,
\documentclass[pdf,slideColor,contemporain,colorBG,accumulate,total]{prosper}
\usepackage{amsmath, amsfonts, amsbsy, pstricks, pst-node, pst-text, pst-3d}
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\title{Blind Single Channel Deconvolution using Nonstationary Signal Processing}
\subtitle{Reverberation Cancellation in Acoustic Environments}
\author{James R. Hopgood}
\email{http://www-sigproc.eng.cam.ac.uk/\~{}jrh1008/\\[-2mm]
jrh1008@eng.cam.ac.uk}
\institution{Signal Processing Laboratory, Department of Engineering, University of Cambridge}
\slideCaption{Blind Deconvolution using Nonstationary Signal Processing. \today}
\Logo(8.3,7.78){\includegraphics[scale=0.58]{UniversityLogo}}
\DefaultTransition{Box}
% ==================================
% Start of Commands Used in Document
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\newcommand{\mc}[1]{{\ensuremath{\mathcal{#1}}}}
\newcommand{\mco}[1]{{\ensuremath{\mathcalold{#1}}}}
\newcommand{\mr}[1]{{\ensuremath{\mathrm{#1}}}}
\newcommand{\mb}[1]{{\ensuremath{\mathbf{#1}}}}
\newcommand{\bs}[1]{\ensuremath{\boldsymbol{#1}}}
\newcommand{\ie}{\emph{i.e.}}
\newcommand{\Bayes}{Bayes's}
\newcommand{\field}[1]{\mathbb{#1}}
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% ================================
% End of Commands Used in Document
% ================================
\begin{document}
\maketitle
\part{Go to full-screen mode now by hitting CTRL-L}
\begin{slide}{Introduction to Blind Deconvolution}
\slidetextsize
\begin{itemize}
\item Blind deconvolution fundamental in signal processing
\item Observation, $\mb x = \{x(t),\,t\in\mc T\}$, modelled as the convolution of unknown source, $\{s(t),\,t\in\mc T\}$, with unknown distortion operator, $\mc A$; \ie\ $x(t) \defas s(t) \conv\ \mc A$
\begin{figure}[t]
\begin{center}
\includegraphics[width=0.5\textwidth]{SingleChannelDeconvolution}
\end{center}
\end{figure}
\item Estimate $\mc A$, \emph{or} $\hat s(t) = a\,s(t - \tau)$, a scaled shifted version of $s(t)$, where $a,\,\tau\in\real{}$, given only the observations, $\mb x$
\end{itemize}
\end{slide}
\overlays{3}{
\begin{slide}[Box]{Acoustic Reverberation Cancellation}
\slidetextsize
\setlength{\MiniPageLeft}{0.4\textwidth}
\setlength{\MiniPageRight}{\textwidth}\addtolength{\MiniPageRight}{-\MiniPageLeft}
\begin{minipage}[t]{\MiniPageLeft}
\begin{figure}[t]
\begin{center}
\onlySlide*{1}{\includegraphics[width=\MiniPageLeft]{Reverberation1}}
\fromSlide{2}{\includegraphics[width=\MiniPageLeft]{Reverberation2}}
\end{center}
\end{figure}
\end{minipage}%
\begin{minipage}[t]{\MiniPageRight}
\begin{itemize}
\item Normal hearing: can concentrate on original sound despite:
\begin{itemize}
\item reverberation
\fromSlide{2}{\item environmental noise}
\end{itemize}
\fromSlide{3}{\item Hearing aid users unable to distinguish one voice from another
\item Sensori-neuro loss cannot be compensated for by simple
amplifying hearing aids or surgery}
\end{itemize}
\end{minipage}
\end{slide}}
\part[Split]{Bayesian Blind Deconvolution}
\overlays{4}{
\begin{slide}[Replace]{\Bayes\ Theorem}
\slidetextsize
\items{The \emph{posterior probability}, \cpdf{\bs{\theta}}{\mb x,\,\model}, of the system parameters, \bs{\theta}, given the state of the system,
\mb x, and an underlying model, \model, is given by Bayes' theorem:
\begin{equation*}
\cpdf{\bs{\theta}}{\mb x,\,\model} = \frac{%
\onlySlide*{1}{\rnode{NA}{\cpdf{\mb x}{\bs{\theta},\,\model}}}%
\onlySlide*{2}{\rnode{NA}{\cpdf{\mb x}{\bs{\theta},\,\model}}}%
\fromSlide*{3}{\rnode{NA}{\cpdf{\mb x}{\bs{\theta},\,\model}}}%
\,%
\untilSlide*{2}{\rnode{NB}{\cpdf{\bs{\theta}}{\model}}}%
\onlySlide*{3}{\rnode{NB}{\cpdf{\bs{\theta}}{\model}}}%
\fromSlide*{4}{\rnode{NB}{\cpdf{\bs{\theta}}{\model}}}%
}{%
\untilSlide*{3}{\rnode{NC}{\cpdf{\mb x}{\model}}}%
\onlySlide*{4}{\rnode{NC}{\cpdf{\mb x}{\model}}}%
}
\end{equation*}}
\begin{itemize}
\item \fromSlide{2}{\rnode{MA}{\cpdf{\mb x}{\bs{\theta},\,\model}} is the \emph{likelihood}}
\item \fromSlide{3}{\rnode{MB}{\cpdf{\bs{\theta}}{\model}} represents prior knowledge}
\item \fromSlide{4}{\rnode{MC}{\cpdf{\mb x}{\model}} is the \emph{evidence} and, although usually
regarded as a normalising constant, is of interest for model selection}
\end{itemize}
\onlySlide*{2}{\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{MA}{NA}}
\onlySlide*{3}{\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{MB}{NB}}
\onlySlide*{4}{\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{MC}{NC}}
\end{slide}}
\overlays{6}{
\begin{slide}[Replace]{Likelihood Function for System}
\slidetextsize
The likelihood function for the observed signal, $\mb x$, is:
\begin{equation*}
\cpdf{\mb x}{%
\untilSlide*{5}{\rnode{NE}{\bs{\theta}}}%
\onlySlide*{6}{\rnode{NE}{\pscirclebox[linecolor=red,linewidth=1.5pt,boxsep=false]{\bs{\theta}}}}%
,\,\bs{\phi},\,\model} = \prod_{i=1}^M \frac
1{(\sqrt{2\pi}\sigma_i)^{T_i}}\exp{\left\{-\frac {(%
\onlySlide*{1}{\rnode{NA}{\mb s_i}}%
\onlySlide*{2}{\rnode{NA}{\pscirclebox[linecolor=red,linewidth=1.5pt,boxsep=false]{\mb s_i}}}%
\fromSlide*{3}{\rnode{NA}{\mb s_i}}%
+ %
\untilSlide*{2}{\rnode{NB}{\mr S_i}}%
\onlySlide*{3}{\rnode{NB}{\pscirclebox[linecolor=red,linewidth=1.5pt,boxsep=false]{\mr S_i}}}%
\fromSlide*{4}{\rnode{NB}{\mr S_i}}%
\mb b_i)^T\,(\mb s_i + \mr S_i%
\untilSlide*{3}{\rnode{NC}{\mb b_i}}%
\onlySlide*{4}{\rnode{NC}{\pscirclebox[linecolor=red,linewidth=1.5pt,boxsep=false]{\mb b_i}}}%
\fromSlide*{5}{\rnode{NC}{\mb b_i}}%
)}{2%
\untilSlide*{4}{\rnode{ND}{\sigma_i}}%
\onlySlide*{5}{\rnode{ND}{\pscirclebox[linecolor=red,linewidth=1.5pt,boxsep=false]{\sigma_i}}}%
\fromSlide*{6}{\rnode{ND}{\sigma_i}}%
^2}\right\}}
\end{equation*}
\hspace*{3mm}
\begin{itemize}
\onlySlide*{1}{\item Where $s(t) \equiv s(t, \mb a,\,\mb x)$ given by:
\begin{equation*}
s(t) = x(t) + \sum\limits_{p\in\mc P} a(p)\,x(t - p)
\end{equation*}}
\fromSlide{2}{\item Data vector, \rnode{MA}{$\mb s_i$}: $[\mb s_i]_{t-t_i+1} = s(t),\,t\in\mc T_i,\,i\in\mc M$}%
\onlySlide*{2}{\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{MA}{NA}}
\fromSlide{3}{\item Data matrix, \rnode{MB}{$\mr S_i$}: $[\mr S_i]_{t-t_i + 1, q} = \mb s(t - q),\,t\in\mc T_i,q\in\mc Q_i,\,i\in\mc M$}%
\onlySlide*{3}{\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{MB}{NB}}
\fromSlide{4}{\item Source parameters, \rnode{MC}{$\mb b = \{\mb b_i,\,i\in\mc M\}$}: $[\mb b_i]_q = b_i(q),\,q\in \mc Q_i$}%
\onlySlide*{4}{\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{MC}{NC}}
\fromSlide{5}{\item Excitation Variances, \rnode{MD}{$\bs{\sigma} = \{\sigma_i^2,\,i\in \mc M\}$}}
\onlySlide*{5}{\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{MD}{ND}}
\fromSlide{6}{\item All source and channel parameters, \rnode{ME}{$\bs{\theta} = \{\mb a,\,\bs{\sigma}, \mb b\}$}}
\onlySlide*{6}{\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{ME}{NE}}
\end{itemize}
\end{slide}}
\overlays{5}{
\begin{slide}[Box]{Likelihood Function}
\slidetextsize
The likelihood function for the observed signal, $\mb x$, is:
\begin{equation*}
\cpdf{\mb x}{\bs{\theta},\,%
\rnode{NF}{\pscirclebox[linecolor=red,linewidth=1.5pt,boxsep=false]{\bs{\phi}}}%
,\,\model} = \prod_{i=1}^M \frac
1{(\sqrt{2\pi}\sigma_i)^{T_i}}\exp{\left\{-\frac {(\mb s_i + \mr S_i \mb b_i)^T\,(\mb s_i + \mr S_i \mb b_i)}{2\sigma_i^2}\right\}}
\end{equation*}
\begin{itemize}
\item Other parameters: $\rnode{MF}{\bs{\phi}} = \rnode{NG}{\{\bs{\tau}},\,\rnode{NH}{\bs{\Xi}},\,\rnode{NI}{\bs{\delta}},\,\rnode{NJ}{\bs{\nu}},\,\rnode{NK}{\bs{\gamma}\}}$\nccurve[linecolor=red,linewidth=1.5pt,angleA=90,angleB=270]{->}{MF}{NF}
\fromSlide{2}{\items{vector of changepoints: $\bs{\tau} = \{t_i,\,i\in\mc M\}$}}
\fromSlide{3}{\items{vector of model orders: $\bs{\Xi} = \{Q_i,\,i\in\mc M\}$}}
\fromSlide{4}{\items{vectors of hyperparameters: $\bs{\delta} = \{\delta_i,\,i\in\mc M\}$}}
\fromSlide{5}{\items{vectors of hyper-hyperparameters: $\bs{\nu} = \{\nu_i,\,i\in\mc M\}$, and $\bs{\gamma} = \{\gamma_i,\,i\in\mc M\}$}}%}
\end{itemize}
\end{slide}}
\overlays{2}{
\begin{slide}[Wipe]{Posterior Distribution for System}
\slidetextsize
\begin{itemize}
\item Apply \Bayes\ rule to obtain the posterior pdf for the unknown parameters $\bs{\theta}$ (assuming $\bs{\phi}$ is known):
\begin{equation*}
\cpdf{\bs{\theta}}{\mb x,\,\bs{\phi},\,\model} \propto\ \cpdf{\mb
x}{\bs{\theta},\,\bs{\phi},\,\model}\,\cpdf{\bs{\theta}}{\bs{\phi},\,\model}
\end{equation*}
\fromSlide*{2}{\item Assuming $\{b_j,\,\sigma_j\}$ are independent between blocks, the
assigned priors are:
\begin{equation*}
\mb b_j\,|\,\sigma_j^2 \simnorm{\mb 0_{Q_j}}{\sigma_j^2\,\delta_j^2\,\eye{Q_j}},\,\delta_j\in\real{+}\,\,\text{and}\, \sigma_j^2 \simIG{\frac{\nu_j}2}{\frac{\gamma_j}2}
\end{equation*}
\item Hence:
\begin{equation*}
\cpdf{\bs{\theta}}{\bs{\phi},\,\model} = \cpdf{\mb a}{\bs{\phi},\,\model}\,\cpdf{\mb
b}{\bs{\sigma},\,\bs{\phi},\,\model}\,\cpdf{\bs{\sigma}}{\bs{\phi},\,\model}
\end{equation*}}
\end{itemize}
\end{slide}}
\begin{slide}[Dissolve]{Posterior Distribution for Channel}
\slidetextsize
\items{Only interested in estimating the channel, \mb a, so marginalise the \emph{nuisance} parameters $\mb b$ and $\bs{\sigma}$ by integrating over $\bs{\theta}_{-\mb a}$:}
\begin{equation*}
\begin{split}
\cpdf{\bs{\theta}_{-\mb b}}{\mb x} &=
\int\limits_{\real{Q_1}}\!\!\cdots\!\!\int\limits_{\real{Q_M}}
\cpdf{\bs{\theta}_{-\mb b},\,\mb b}{\mb
x}\,d\mb b_M\,\dots\,d\mb b_1\\
\cpdf{\mb a}{\mb x} &=
\int\limits_0^{\infty}\!\!\cdots\!\!\int\limits_0^{\infty}
\cpdf{\mb a,\,\bs{\sigma}}{\mb
x}\,d\sigma_M^2\dots\,d\sigma_1^2
\end{split}
\end{equation*}
\end{slide}
\begin{slide}[Glitter]{Posterior Distribution for Channel}
\slidetextsize
\items{Yields the posterior density for the channel parameters $\mb a$:
\begin{equation*}
\begin{split}
&\cpdf{\mb a}{\mb x,\,\bs{\phi},\,\model} \propto
\cpdf{\mb a}{\bs{\phi},\,\model}\\
& \times \prod_{i=1}^{M}
\frac {\left\{\gamma_i + \mb s_i^T \mb s_i
- \mb s_i^T \mr S_i\left(\mr S_i^T \mr S_i + \delta_i^{-2} \mr
I_{Q_i}\right)^{-1}\! \mr S_i^T \mb s_i\right\}^{-R_i}}{\left|\mr S_i^T\,\mr S_i + \delta_i^{-2} \mr
I_{Q_i}\right|^{\frac 12}}
\end{split}
\end{equation*}
where $R_i = \frac{T_i + \nu_i + 1}2,\,i\in\mc M$}
\items{Written in terms of $\mb s(t)$ to emphasise that it can be efficiently calculated by
`inverse filtering' the data, $\mb x(t)$}
\items{MMAP estimate used \ie\ $\arg\limits_{\mb a} \max \cpdf{\mb a}{\mb x,\,\bs{\phi},\,P,\,\model}$}
\end{slide}
\overlays{4}{
\begin{slide}[Replace]{Principle Revisited}
\slidetextsize
\setlength{\MiniPageLeft}{0.5\textwidth}
\setlength{\MiniPageRight}{\textwidth}\addtolength{\MiniPageRight}{-\MiniPageLeft}
\begin{minipage}[t]{\MiniPageLeft}
\begin{figure}[t]
\begin{center}
\includegraphics[width = \MiniPageLeft]{BSAR2IIR2_Model}\\
\textbf{$P = 2,\,Q_i = 2,\, N=10$ and $T_i = 1000,\,\forall i\in\mc M$}
\end{center}
\end{figure}
\items{Phase and magnitude of the pole
locations for this \BSAR{2} process change linearly with block number}
\end{minipage}%
\begin{minipage}[t]{\MiniPageRight}
\begin{figure}[t]
\begin{center}
\onlySlide*{2}{\includegraphics[width=\MiniPageRight]{ar2blk1}\\
\textbf{$\ln \cpdf[\hat p_1]{\mb r_{\mb a}}{\mb x_1,\,\mb x_0,\,\bs{\phi},\,\model}$}}%
\onlySlide*{3}{\includegraphics[width=\MiniPageRight]{ar2blk7}\\
\textbf{$\ln \cpdf[\hat p_7]{\mb r_{\mb a}}{\mb x_7,\,\mb x_6,\,\bs{\phi},\,\model}$}}%
\onlySlide*{4}{\includegraphics[width=\MiniPageRight]{ar2blkF}\\
\textbf{$\ln \cpdf[\hat p]{\mb r_{\mb a}}{\mb x,\,\bs{\phi},\,\model}$}}
\end{center}
\end{figure}
\end{minipage}
\end{slide}}
\begin{slide}[Blinds]{A simple acoustic environment}
\slidetextsize
\setlength{\MiniPageLeft}{0.5\textwidth}
\setlength{\MiniPageRight}{\textwidth}\addtolength{\MiniPageRight}{-\MiniPageLeft}
\begin{minipage}[t]{\MiniPageLeft}
\begin{figure}[t]
\begin{center}
\includegraphics[width=0.9\MiniPageLeft]{ActualEstimatedResponse}\\
\end{center}
\end{figure}
\end{minipage}%
\begin{minipage}[t]{\MiniPageRight}
\begin{figure}[t]
\begin{center}
\includegraphics[width=0.9\MiniPageLeft]{EqualisedResponse}\\
\end{center}
\end{figure}
\end{minipage}
\items{$\nu_i = \gamma_i = 0, \delta_i \sim 10^6$, $Q_i = 100$, $T_i = 1000$ and $M=100$}
\end{slide}
\end{document}