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%TCIDATA{BibliographyScheme=Manual}
%TCIDATA{Created=Thursday, December 22, 2005 10:09:22}
%TCIDATA{LastRevised=Thursday, December 22, 2005 14:30:05}
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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
\newtheorem{axiom}[theorem]{Axiom}
\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
\newtheorem{corollary}[theorem]{Corollary}
\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
\newtheorem{example}[theorem]{Example}
\newtheorem{exercise}[theorem]{Exercise}
\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
\newtheorem{problem}[theorem]{Problem}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{remark}[theorem]{Remark}
\newtheorem{solution}[theorem]{Solution}
\newtheorem{summary}[theorem]{Summary}
\newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}}
\input{tcilatex}

\begin{document}

\title{The Title of a Standard LaTeX Article}
\author{A. U. Thor \\
%EndAName
The University of Stewart Island}
\maketitle

\begin{abstract}
We study the effects of warm water on the local penguin population. The
major finding is that it is extremely difficult to induce penguins to drink
warm water. The success factor is approximately $-e^{-i\pi }-1$.
\end{abstract}

\begin{center}
Approximation Theory
\end{center}

\bigskip 

Given a Hilbert space H with an inner product ( , ) we define%
\begin{equation*}
d(\varphi _{1},\varphi _{2})=||\varphi _{1}-\varphi _{2}||=\sqrt{(\varphi
_{1}-\varphi _{2},\varphi _{1}-\varphi _{2})}
\end{equation*}

\bigskip 

Given a subspace $S\subset H$ with a basis $(\psi _{1},\psi _{2},,...,\psi
_{n})$

Any $\widehat{\varphi }\in S$ can be written as $\widehat{\varphi }%
=\tsum\limits_{i=1}^{n}\alpha _{i}\psi _{i}$.

\bigskip 

Problem: Given $\varphi \in H$ find $\widehat{\varphi }\in S$ so that $%
||\varphi -\widehat{\varphi }||$ is a minimum

\bigskip 

Solution: Choose $\varphi $ so that%
\begin{equation*}
(\varphi -\widehat{\varphi },v)=0\text{ \ \ for every }v\in S
\end{equation*}%
\newpage 

\begin{center}
Wiener Filter
\end{center}

\bigskip 

Notation:

$f(x,y)$ \ = "original" picture

$g(x,y)$= "actual" picture

$\widehat{f}(x,y)$ = estimate to $f(x,y)$

\bigskip 

Assumption: the noise is additive. So

\begin{equation*}
\text{(*) \ \ \ }g(x,y)=\tiint h(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)f(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
+n(x,y)
\end{equation*}

We \textbf{assume }$h(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)$ is known (?)

Assumption:

\begin{enumerate}
\item $E\left[ n(x,y)\right] =0$

\item $\left. 
\begin{array}{c}
E\left[ f(x,y)f(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)\right] =R_{ff}(x,y,x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
) \\ 
E\left[ n(x,y)n(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)\right] =R_{nn}(x,y,x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)%
\end{array}%
\right\} $ known
\end{enumerate}

\bigskip 

Problem:

Given $g(x,y)$ find $\widehat{f}(x,y)$ so that%
\begin{equation*}
\varepsilon ^{2}=E\left[ \left( f(x,y)-\widehat{f}(x,y)\right) ^{2}\right] 
\text{ is a minimum}
\end{equation*}

\bigskip 

Solution:

\bigskip 

To get a meaningful solution we need to assume that $\widehat{f}(x,y)$ is a
linear function of $g(x,y)$.

\bigskip 
\begin{equation*}
\widehat{f}(x,y)=\tiint m(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)g(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
\end{equation*}

\textbf{We need to calculate }$m(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)$\textbf{\ as a function of }$g$\textbf{\ and }$h$

We wish that%
\begin{equation*}
E\left[ \left( \widehat{f}(x,y)-\tiint m(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)g(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
\right) ^{2}\right] \text{ be a minimum}
\end{equation*}

Using the orthogonality theorem we have%
\begin{equation*}
E\left[ \left( \widehat{f}(x,y)-\tiint m(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)g(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,g(\widetilde{x},\widetilde{y})\right) \right] =0\text{ \ \ \ all }%
\widetilde{x},\widetilde{y}
\end{equation*}

Interchanging the order of the integrals%
\begin{equation*}
E\left[ \widehat{f}(x,y)g(\widetilde{x},\widetilde{y})\right] =\tiint E\left[
m(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)g(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)g(\widetilde{x},\widetilde{y})dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
\right] \text{ }
\end{equation*}

or using the cross-correlation%
\begin{eqnarray*}
R_{fg}(x,y,\widetilde{x},\widetilde{y}) &=&E\left[ \widehat{f}(x,y)g(%
\widetilde{x},\widetilde{y})\right]  \\
R_{gg}(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,\widetilde{x},\widetilde{y}) &=&E\left[ g(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)g(\widetilde{x},\widetilde{y})\right] 
\end{eqnarray*}%
\begin{equation*}
R_{fg}(x,y,\widetilde{x},\widetilde{y})=\tiint_{\text{image}}R_{gg}(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,\widetilde{x},\widetilde{y})m(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
\end{equation*}

This is still very difficult so we assume that all the statistics are
homogeneous and invariant. So%
\begin{eqnarray*}
R_{fg}(x,y,\widetilde{x},\widetilde{y}) &=&R_{fg}(x-\widetilde{x},y-%
\widetilde{y}) \\
R_{gg}(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,\widetilde{x},\widetilde{y}) &=&R_{gg}(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
-\widetilde{x},y-\widetilde{y}) \\
R_{nn}(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,\widetilde{x},\widetilde{y}) &=&R_{nn}(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
-\widetilde{x},y-\widetilde{y}) \\
m(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
) &=&m(x-x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
}y-y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)
\end{eqnarray*}

Then%
\begin{eqnarray*}
R_{fg}(x-\widetilde{x},y-\widetilde{y}) &=&\tiint_{\text{image}}R_{gg}(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
-\widetilde{x},y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
-\widetilde{y})m(x-x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
},y-y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4} }%
%BeginExpansion
{\acute{}}
%EndExpansion
\\
&=&R_{gg}(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
-\widetilde{x},y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
-\widetilde{y})\ast m(x,y)
\end{eqnarray*}

Fourier transform yields%
\begin{equation*}
S_{fg}(u,v)=S_{gg}(u,v)H(u,v)
\end{equation*}

but from (*) we have

\begin{equation*}
g(x,y)=\tiint h(x,x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
};y,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)f(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
+n(x,y)=\tiint h(x-x^{%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
},y-y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)f(x%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
,y%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
)dx%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
dy%
%TCIMACRO{\U{b4}}%
%BeginExpansion
{\acute{}}%
%EndExpansion
+n(x,y)
\end{equation*}

and in Fourier space%
\begin{equation*}
G=HF+N
\end{equation*}

\end{document}
