<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 3.2 Final//EN"> <!--Converted with LaTeX2HTML 98.1p1 release (March 2nd, 1998) originally by Nikos Drakos (nikos@cbl.leeds.ac.uk), CBLU, University of Leeds * revised and updated by: Marcus Hennecke, Ross Moore, Herb Swan * with significant contributions from: Jens Lippmann, Marek Rouchal, Martin Wilck and others --> <HTML> <HEAD> <TITLE>The Wiener-like filtering in the wavelet space</TITLE> <META NAME="description" CONTENT="The Wiener-like filtering in the wavelet space"> <META NAME="keywords" CONTENT="vol2"> <META NAME="resource-type" CONTENT="document"> <META NAME="distribution" CONTENT="global"> <META HTTP-EQUIV="Content-Type" CONTENT="text/html; charset=iso-8859-1"> <LINK REL="STYLESHEET" HREF="vol2.css"> <LINK REL="next" HREF="node331.html"> <LINK REL="previous" HREF="node329.html"> <LINK REL="up" HREF="node328.html"> <LINK REL="next" HREF="node331.html"> </HEAD> <BODY > <!--Navigation Panel--> <A NAME="tex2html5567" HREF="node331.html"> <IMG WIDTH="37" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="next" SRC="icons.gif/next_motif.gif"></A> <A NAME="tex2html5564" HREF="node328.html"> <IMG WIDTH="26" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="up" SRC="icons.gif/up_motif.gif"></A> <A NAME="tex2html5558" HREF="node329.html"> <IMG WIDTH="63" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="previous" SRC="icons.gif/previous_motif.gif"></A> <A NAME="tex2html5566" HREF="node1.html"> <IMG WIDTH="65" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="contents" SRC="icons.gif/contents_motif.gif"></A> <BR> <B> Next:</B> <A NAME="tex2html5568" HREF="node331.html">Hierarchical Wiener filtering</A> <B> Up:</B> <A NAME="tex2html5565" HREF="node328.html">Noise reduction from the</A> <B> Previous:</B> <A NAME="tex2html5559" HREF="node329.html">The convolution from the</A> <BR> <BR> <!--End of Navigation Panel--> <H2><A NAME="SECTION002062000000000000000"> </A> <A NAME="sec_filt_4"> </A> <BR> The Wiener-like filtering in the wavelet space </H2> Let us consider a measured wavelet coefficient <I>w</I><SUB><I>i</I></SUB> at the scale <I>i</I>. We assume that its value, at a given scale and a given position, results from a noisy process, with a Gaussian distribution with a mathematical expectation <I>W</I><SUB><I>i</I></SUB>, and a standard deviation <I>B</I><SUB><I>i</I></SUB>: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} P(w_i/W_i) = \frac{1}{\sqrt{2\pi}B_i} e^{- \frac{(w_i-W_i)^2} {2B_i^2}} \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="298" HEIGHT="91" ALIGN="MIDDLE" BORDER="0" SRC="img786.gif" ALT="$\displaystyle P(w_i/W_i) = \frac{1}{\sqrt{2\pi}B_i} e^{- \frac{(w_i-W_i)^2} {2B_i^2}}$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.76)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> Now, we assume that the set of expected coefficients <I>W</I><SUB><I>i</I></SUB> for a given scale also follows a Gaussian distribution, with a null mean and a standard deviation <I>S</I><SUB><I>i</I></SUB>: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} P(W_i) = \frac{1}{\sqrt{2\pi}S_i}e^{-\frac{W_i^2}{2S_i^2}} \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="217" HEIGHT="96" ALIGN="MIDDLE" BORDER="0" SRC="img787.gif" ALT="$\displaystyle P(W_i) = \frac{1}{\sqrt{2\pi}S_i}e^{-\frac{W_i^2}{2S_i^2}}$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.77)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> The null mean value results from the wavelet property: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} \int_{-\infty}^{+\infty} \psi^*(x) dx = 0 \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="179" HEIGHT="73" ALIGN="MIDDLE" BORDER="0" SRC="img788.gif" ALT="$\displaystyle \int_{-\infty}^{+\infty} \psi^*(x) dx = 0$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.78)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> We want to get an estimate of <I>W</I><SUB><I>i</I></SUB> knowing <I>w</I><SUB><I>i</I></SUB>. Bayes' theorem gives: <BR> <DIV ALIGN="CENTER"><A NAME="equa_baye"> </A> <!-- MATH: \begin{eqnarray} P(W_i/w_i) = \frac{P(W_i)P(w_i/W_i)}{P(w_i)} \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="290" HEIGHT="74" ALIGN="MIDDLE" BORDER="0" SRC="img789.gif" ALT="$\displaystyle P(W_i/w_i) = \frac{P(W_i)P(w_i/W_i)}{P(w_i)}$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.79)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> We get: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} P(W_i/w_i) = \frac{1}{\sqrt{2\pi}\beta_i}e^{-\frac{(W_i-\alpha_i w_i)^2}{2\beta_i^2}} \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="310" HEIGHT="91" ALIGN="MIDDLE" BORDER="0" SRC="img790.gif" ALT="$\displaystyle P(W_i/w_i) = \frac{1}{\sqrt{2\pi}\beta_i}e^{-\frac{(W_i-\alpha_i w_i)^2}{2\beta_i^2}}$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.80)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> where: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} \alpha_i = \frac{S_i^2}{S_i^2+B_i^2} \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="137" HEIGHT="77" ALIGN="MIDDLE" BORDER="0" SRC="img791.gif" ALT="$\displaystyle \alpha_i = \frac{S_i^2}{S_i^2+B_i^2}$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.81)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> the probability <!-- MATH: $P(W_i/w_i)$ --> <I>P</I>(<I>W</I><SUB><I>i</I></SUB>/<I>w</I><SUB><I>i</I></SUB>) follows a Gaussian distribution with a mean: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} m = \alpha_i w_i \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="98" HEIGHT="39" ALIGN="MIDDLE" BORDER="0" SRC="img792.gif" ALT="$\displaystyle m = \alpha_i w_i$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.82)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> and a variance: <BR> <DIV ALIGN="CENTER"> <!-- MATH: \begin{eqnarray} \beta_i^2 = \frac{S_i^2B_i^2}{S_i^2 + B_i^2} \end{eqnarray} --> <TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%"> <TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG WIDTH="140" HEIGHT="77" ALIGN="MIDDLE" BORDER="0" SRC="img793.gif" ALT="$\displaystyle \beta_i^2 = \frac{S_i^2B_i^2}{S_i^2 + B_i^2}$"></TD> <TD> </TD> <TD> </TD> <TD WIDTH=10 ALIGN="RIGHT"> (14.83)</TD></TR> </TABLE></DIV> <BR CLEAR="ALL"><P></P> The mathematical expectation of <I>W</I><SUB><I>i</I></SUB> is <!-- MATH: $\alpha_i w_i$ --> <IMG WIDTH="49" HEIGHT="39" ALIGN="MIDDLE" BORDER="0" SRC="img794.gif" ALT="$\alpha_i w_i$">. <P> With a simple multiplication of the coefficients by the constant <IMG WIDTH="26" HEIGHT="39" ALIGN="MIDDLE" BORDER="0" SRC="img795.gif" ALT="$\alpha_i$">, we get a linear filter. The algorithm is: <DL COMPACT> <DT>1. <DD>Compute the wavelet transform of the data. We get <I>w</I><SUB><I>i</I></SUB>. <DT>2. <DD>Estimate the standard deviation of the noise <I>B</I><SUB>0</SUB> of the first plane from the histogram of <I>w</I><SUB>0</SUB>. As we process oversampled images, the values of the wavelet image corresponding to the first scale (<I>w</I><SUB>0</SUB>) are due mainly to the noise. The histogram shows a Gaussian peak around 0. We compute the standard deviation of this Gaussian function, with a <IMG WIDTH="30" HEIGHT="21" ALIGN="BOTTOM" BORDER="0" SRC="img796.gif" ALT="$3\sigma$"> clipping, rejecting pixels where the signal could be significant; <DT>3. <DD>Set i to 0. <DT>4. <DD>Estimate the standard deviation of the noise <I>B</I><SUB><I>i</I></SUB> from <I>B</I><SUB>0</SUB>. This is done from the study of the variation of the noise between two scales, with an hypothesis of a white gaussian noise; <DT>5. <DD> <!-- MATH: $S_i^2 = s_i^2 - B_i^2$ --> <I>S</I><SUB><I>i</I></SUB><SUP>2</SUP> = <I>s</I><SUB><I>i</I></SUB><SUP>2</SUP> - <I>B</I><SUB><I>i</I></SUB><SUP>2</SUP> where <I>s</I><SUB><I>i</I></SUB><SUP>2</SUP> is the variance of <I>w</I><SUB><I>i</I></SUB>. <DT>6. <DD> <!-- MATH: $\alpha_i = \frac{S_i^2}{S_i^2+B_i^2}$ --> <IMG WIDTH="114" HEIGHT="62" ALIGN="MIDDLE" BORDER="0" SRC="img797.gif" ALT="$\alpha_i = \frac{S_i^2}{S_i^2+B_i^2}$">. <DT>7. <DD> <!-- MATH: $W_i = \alpha_i w_i$ --> <IMG WIDTH="106" HEIGHT="41" ALIGN="MIDDLE" BORDER="0" SRC="img798.gif" ALT="$W_i = \alpha_i w_i$">. <DT>8. <DD><I>i</I> = <I>i</I> + 1 and go to 4. <DT>9. <DD>Reconstruct the picture from <I>W</I><SUB><I>i</I></SUB>. </DL> <P> <HR> <!--Navigation Panel--> <A NAME="tex2html5567" HREF="node331.html"> <IMG WIDTH="37" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="next" SRC="icons.gif/next_motif.gif"></A> <A NAME="tex2html5564" HREF="node328.html"> <IMG WIDTH="26" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="up" SRC="icons.gif/up_motif.gif"></A> <A NAME="tex2html5558" HREF="node329.html"> <IMG WIDTH="63" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="previous" SRC="icons.gif/previous_motif.gif"></A> <A NAME="tex2html5566" HREF="node1.html"> <IMG WIDTH="65" HEIGHT="24" ALIGN="BOTTOM" BORDER="0" ALT="contents" SRC="icons.gif/contents_motif.gif"></A> <BR> <B> Next:</B> <A NAME="tex2html5568" HREF="node331.html">Hierarchical Wiener filtering</A> <B> Up:</B> <A NAME="tex2html5565" HREF="node328.html">Noise reduction from the</A> <B> Previous:</B> <A NAME="tex2html5559" HREF="node329.html">The convolution from the</A> <!--End of Navigation Panel--> <ADDRESS> <I>Petra Nass</I> <BR><I>1999-06-15</I> </ADDRESS> </BODY> </HTML>