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<H2><A NAME="SECTION002063000000000000000">&#160;</A>
<A NAME="sec_filt_3">&#160;</A>
<BR>
Hierarchical Wiener filtering
</H2>
In the above process, we do not use the information between the wavelet
coefficients at different scales. We modify the previous
algorithm by introducing a prediction <I>w</I><SUB><I>h</I></SUB> of the wavelet coefficient from
the upper scale. This prediction could be determined from the regression
[<A
 HREF="node370.html#antonini">2</A>] between the two scales but better results are obtained 
when we only set <I>w</I><SUB><I>h</I></SUB> to <I>W</I><SUB><I>i</I>+1</SUB>. Between the expectation
coefficient <I>W</I><SUB><I>i</I></SUB> and the prediction, a dispersion exists where we
assume that it is a Gaussian distribution:
<BR>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{eqnarray}
P(W_i/w_h) = \frac{1}{\sqrt{2\pi}T_i}e^{-\frac{(W_i- w_h)^2}{2T_i^2}}
\end{eqnarray} -->

<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
 WIDTH="300" HEIGHT="92" ALIGN="MIDDLE" BORDER="0"
 SRC="img799.gif"
 ALT="$\displaystyle P(W_i/w_h) = \frac{1}{\sqrt{2\pi}T_i}e^{-\frac{(W_i- w_h)^2}{2T_i^2}}$"></TD>
<TD>&nbsp;</TD>
<TD>&nbsp;</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.84)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
<P>
The relation which gives the coefficient <I>W</I><SUB><I>i</I></SUB> knowing <I>w</I><SUB><I>i</I></SUB> and <I>w</I><SUB><I>h</I></SUB> is:
<BR>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{eqnarray}
P(W_i/w_i \mbox{ and } w_h) = \frac{1}{\sqrt{2\pi}\beta_i}
e^{-\frac{(W_i-\alpha_i w_i)^2}{2\beta_i^2}} \frac{1}{\sqrt{2\pi}T_i}
e^{-\frac{(W_i-w_h)^2}{2T_i^2}}
\end{eqnarray} -->

<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
 WIDTH="553" HEIGHT="92" ALIGN="MIDDLE" BORDER="0"
 SRC="img800.gif"
 ALT="$\displaystyle P(W_i/w_i \mbox{ and } w_h) = \frac{1}{\sqrt{2\pi}\beta_i}
e^{-\f...
...i w_i)^2}{2\beta_i^2}} \frac{1}{\sqrt{2\pi}T_i}
e^{-\frac{(W_i-w_h)^2}{2T_i^2}}$"></TD>
<TD>&nbsp;</TD>
<TD>&nbsp;</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.85)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
with:
<BR>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{eqnarray}
\beta_i^2 = \frac{S_i^2B_i^2}{S^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="img801.gif"
 ALT="$\displaystyle \beta_i^2 = \frac{S_i^2B_i^2}{S^2 + B_i^2}$"></TD>
<TD>&nbsp;</TD>
<TD>&nbsp;</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.86)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
and:
<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="138" HEIGHT="77" ALIGN="MIDDLE" BORDER="0"
 SRC="img802.gif"
 ALT="$\displaystyle \alpha_i = \frac{S_i^2}{S_i^2+B_i^2}$"></TD>
<TD>&nbsp;</TD>
<TD>&nbsp;</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.87)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
<P>
This follows a Gaussian distribution with a mathematical expectation:
<BR>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{eqnarray}
W_i = \frac{T_i^2}{B_i^2+T_i^2+Q_i^2} w_i +
\frac{B_i^2}{B_i^2+T_i^2+Q_i^2} w_h
\end{eqnarray} -->

<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
 WIDTH="412" HEIGHT="77" ALIGN="MIDDLE" BORDER="0"
 SRC="img803.gif"
 ALT="$\displaystyle W_i = \frac{T_i^2}{B_i^2+T_i^2+Q_i^2} w_i +
\frac{B_i^2}{B_i^2+T_i^2+Q_i^2} w_h$"></TD>
<TD>&nbsp;</TD>
<TD>&nbsp;</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.88)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
with:
<BR>
<DIV ALIGN="CENTER">

<!-- MATH: \begin{eqnarray}
Q_i^2 = \frac{T_i^2B_i^2}{S_i^2}
\end{eqnarray} -->

<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
 WIDTH="119" HEIGHT="77" ALIGN="MIDDLE" BORDER="0"
 SRC="img804.gif"
 ALT="$\displaystyle Q_i^2 = \frac{T_i^2B_i^2}{S_i^2}$"></TD>
<TD>&nbsp;</TD>
<TD>&nbsp;</TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.89)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P><I>W</I><SUB><I>i</I></SUB> is the barycentre of the three values <I>w</I><SUB><I>i</I></SUB>, <I>w</I><SUB><I>h</I></SUB>, 0 with the
weights <I>T</I><SUB><I>i</I></SUB><SUP>2</SUP>, <I>B</I><SUB><I>i</I></SUB><SUP>2</SUP>, <I>Q</I><SUB><I>i</I></SUB><SUP>2</SUP>.  The particular cases are:
<UL>
<LI>If the noise is large (
<!-- MATH: $S_i \ll B_i$ -->
<IMG
 WIDTH="83" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
 SRC="img805.gif"
 ALT="$S_i \ll B_i$">)
and even if the correlation 
 between the two scales is good (<I>T</I><SUB><I>i</I></SUB> is low), we get 
<!-- MATH: $W_i \rightarrow 0$ -->
<IMG
 WIDTH="78" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
 SRC="img806.gif"
 ALT="$W_i \rightarrow 0$">.
<LI>if 
<!-- MATH: $B_i \ll S_i \ll T$ -->
<IMG
 WIDTH="132" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
 SRC="img807.gif"
 ALT="$B_i \ll S_i \ll T$">
then 
<!-- MATH: $W_i \rightarrow w_i$ -->
<IMG
 WIDTH="90" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
 SRC="img808.gif"
 ALT="$W_i \rightarrow w_i$">.
<LI>if 
<!-- MATH: $B_i \ll T_i \ll S$ -->
<IMG
 WIDTH="131" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
 SRC="img809.gif"
 ALT="$B_i \ll T_i \ll S$">
then 
<!-- MATH: $W_i \rightarrow w_i$ -->
<IMG
 WIDTH="90" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
 SRC="img810.gif"
 ALT="$W_i \rightarrow w_i$">.
<LI>if 
<!-- MATH: $T_i \ll B_i \ll S$ -->
<IMG
 WIDTH="131" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
 SRC="img811.gif"
 ALT="$T_i \ll B_i \ll S$">
then 
<!-- MATH: $W_i \rightarrow w_h$ -->
<IMG
 WIDTH="94" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
 SRC="img812.gif"
 ALT="$W_i \rightarrow w_h$">.
</UL>
<P>
At each scale, by changing all the wavelet coefficients <I>w</I><SUB><I>i</I></SUB> of the
plane by the estimate value <I>W</I><SUB><I>i</I></SUB>, we get a Hierarchical Wiener
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>.
<DT>3.
<DD>Set i to the index associated with the last plane: <I>i</I> = <I>n</I><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>.
<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>Set <I>w</I><SUB><I>h</I></SUB> to <I>W</I><SUB><I>i</I>+1</SUB> and compute the standard deviation <I>T</I><SUB><I>i</I></SUB>of <I>w</I><SUB><I>i</I></SUB> - <I>w</I><SUB><I>h</I></SUB>.
<DT>7.
<DD>
<!-- MATH: $W_i = \frac{T_i^2}{B_i^2+T_i^2+Q_i^2} w_i + \frac{B_i^2}{B_i^2+T_i^2+Q_i^2} w_h$ -->
<IMG
 WIDTH="329" HEIGHT="62" ALIGN="MIDDLE" BORDER="0"
 SRC="img814.gif"
 ALT="$W_i = \frac{T_i^2}{B_i^2+T_i^2+Q_i^2} w_i + \frac{B_i^2}{B_i^2+T_i^2+Q_i^2} w_h$"><DT>8.
<DD><I>i</I> = <I>i</I> - 1. If <I>i</I> &gt; 0 go to 4
<DT>9.
<DD>Reconstruct the picture 
</DL>
<P>
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<ADDRESS>
<I>Petra Nass</I>
<BR><I>1999-06-15</I>
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