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%TCIDATA{Created=Sunday, November 21, 2004 09:23:51}
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\newtheorem{theorem}{Theorem}
\newtheorem{acknowledgement}[theorem]{Acknowledgement}
\newtheorem{algorithm}[theorem]{Algorithm}
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\newtheorem{case}[theorem]{Case}
\newtheorem{claim}[theorem]{Claim}
\newtheorem{conclusion}[theorem]{Conclusion}
\newtheorem{condition}[theorem]{Condition}
\newtheorem{conjecture}[theorem]{Conjecture}
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\newtheorem{criterion}[theorem]{Criterion}
\newtheorem{definition}[theorem]{Definition}
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\newtheorem{lemma}[theorem]{Lemma}
\newtheorem{notation}[theorem]{Notation}
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\begin{document}


\begin{center}
\bigskip {\Large Convolution}

\bigskip Continuous
\end{center}

\bigskip $f\ast g(x)=\int f(y)g(x-y)dy$

\begin{center}
Discrete
\end{center}

\QTP{Body Math}
$f\ast g(x)=\frac{1}{N}\dsum\limits_{n=0}^{N-1}f(n)g(x-n)$

\begin{center}
\bigskip Fourier
\end{center}

$f(x,y)\ast g(x,y)\leftrightarrow F(u,v)\cdot G(u,v)$

\bigskip 

\begin{center}
{\Large Correlation}

Continuous
\end{center}

\bigskip $f\circ g(x)=\int f^{\ast }(y)g(x+y)dy=f^{\ast }(y)\ast g(-y)$

\begin{center}
Discrete
\end{center}

\QTP{Body Math}
$f\circ g(x)=\frac{1}{N}\dsum\limits_{n=0}^{N-1}f^{\ast }(n)g(x+n)$

\begin{center}
Fourier
\end{center}

$f(x,y)\circ g(x,y)\leftrightarrow F^{\ast }(u,v)\cdot G(u,v)$

$f^{\ast }(x,y)g(x,y)\leftrightarrow F(u,v)\circ G(u,v)$

\bigskip 

\begin{center}
{\Large Point Spread Function (PSF)}
\end{center}

Response of a filter to a point image.

Let $P_{mn}=1$ if $m=n=0$ and $0$ otherwise. Then

\[
P_{mn}^{^{\prime }}=\sum_{i=-r}^{r}\sum_{j=-r}^{r}H_{ij}P_{m-i,n-j}=H_{mn}
\]

Since convolution is linear when we know the effect of a filter on a point
image we know it on any image.

Theorem: A linear shift-invariant operator must be a convolution operator in
space.

\end{document}
