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<H3><A NAME="SECTION002045200000000000000">
The Reconstruction</A>
</H3>
If the wavelet is the difference between two resolutions,
an evident reconstruction for a wavelet transform
<!-- MATH: ${\cal W} = \{w_1, w_2, \dots, w_{n_p}, c_{n_p}\}$ -->
<IMG
WIDTH="263" HEIGHT="44" ALIGN="MIDDLE" BORDER="0"
SRC="img724.gif"
ALT="${\cal W} = \{w_1, w_2, \dots, w_{n_p}, c_{n_p}\}$">
is:
<BR>
<DIV ALIGN="CENTER">
<!-- MATH: \begin{eqnarray}
\hat c_0(\nu) = \hat c_{n_p}(\nu) + \sum_j \hat w_j(\nu)
\end{eqnarray} -->
<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
WIDTH="254" HEIGHT="70" ALIGN="MIDDLE" BORDER="0"
SRC="img725.gif"
ALT="$\displaystyle \hat c_0(\nu) = \hat c_{n_p}(\nu) + \sum_j \hat w_j(\nu)$"></TD>
<TD> </TD>
<TD> </TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.60)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
But this is a particular case and other wavelet functions can be
chosen. The reconstruction can be done step by step, starting from
the lowest resolution. At each scale, we have the relations:
<BR>
<DIV ALIGN="CENTER">
<!-- MATH: \begin{eqnarray}
\hat c_{j+1} = \hat h(2^j \nu) \hat c_j(\nu) \\
\hat w_{j+1} = \hat g(2^j \nu) \hat c_j(\nu)
\end{eqnarray} -->
<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
WIDTH="184" HEIGHT="53" ALIGN="MIDDLE" BORDER="0"
SRC="img726.gif"
ALT="$\displaystyle \hat c_{j+1} = \hat h(2^j \nu) \hat c_j(\nu)$"></TD>
<TD> </TD>
<TD> </TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.61)</TD></TR>
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
WIDTH="187" HEIGHT="51" ALIGN="MIDDLE" BORDER="0"
SRC="img727.gif"
ALT="$\displaystyle \hat w_{j+1} = \hat g(2^j \nu) \hat c_j(\nu)$"></TD>
<TD> </TD>
<TD> </TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.62)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
we look for <I>c</I><SUB><I>j</I></SUB> knowing <I>c</I><SUB><I>j</I>+1</SUB>, <I>w</I><SUB><I>j</I>+1</SUB>, <I>h</I> and <I>g</I>.
We restore
<!-- MATH: $\hat c_j(\nu)$ -->
<IMG
WIDTH="54" HEIGHT="44" ALIGN="MIDDLE" BORDER="0"
SRC="img728.gif"
ALT="$\hat c_j(\nu)$">
with a least mean square estimator:
<BR>
<DIV ALIGN="CENTER">
<!-- MATH: \begin{eqnarray}
\hat p_h(2^j\nu)|\hat c_{j+1}(\nu)-\hat h(2^j\nu)\hat c_j(\nu)|^2 +
\hat p_g(2^j\nu)|\hat w_{j+1}(\nu)-\hat g(2^j\nu)\hat
c_j(\nu)|^2
\end{eqnarray} -->
<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
WIDTH="629" HEIGHT="53" ALIGN="MIDDLE" BORDER="0"
SRC="img729.gif"
ALT="$\displaystyle \hat p_h(2^j\nu)\vert\hat c_{j+1}(\nu)-\hat h(2^j\nu)\hat c_j(\nu...
...t^2 +
\hat p_g(2^j\nu)\vert\hat w_{j+1}(\nu)-\hat g(2^j\nu)\hat
c_j(\nu)\vert^2$"></TD>
<TD> </TD>
<TD> </TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.63)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
is minimum.
<!-- MATH: $\hat p_h(\nu)$ -->
<IMG
WIDTH="57" HEIGHT="44" ALIGN="MIDDLE" BORDER="0"
SRC="img730.gif"
ALT="$\hat p_h(\nu)$">
and
<!-- MATH: $\hat p_g(\nu)$ -->
<IMG
WIDTH="55" HEIGHT="44" ALIGN="MIDDLE" BORDER="0"
SRC="img731.gif"
ALT="$\hat p_g(\nu)$">
are weight
functions which permit a general solution to the
restoration of
<!-- MATH: $\hat c_j(\nu)$ -->
<IMG
WIDTH="54" HEIGHT="44" ALIGN="MIDDLE" BORDER="0"
SRC="img732.gif"
ALT="$\hat c_j(\nu)$">.
By
<!-- MATH: $\hat c_j(\nu)$ -->
<IMG
WIDTH="53" HEIGHT="44" ALIGN="MIDDLE" BORDER="0"
SRC="img733.gif"
ALT="$\hat c_j(\nu)$">
derivation we get:
<BR>
<DIV ALIGN="CENTER"><A NAME="restauration"> </A>
<!-- MATH: \begin{eqnarray}
\hat{c}_{j}(\nu)=\hat{c}_{j+1}(\nu) \hat{\tilde h}(2^{j}\nu)
+\hat{w}_{j+1}(\nu) \hat{\tilde g}(2^{j}\nu)
\end{eqnarray} -->
<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
WIDTH="375" HEIGHT="63" ALIGN="MIDDLE" BORDER="0"
SRC="img734.gif"
ALT="$\displaystyle \hat{c}_{j}(\nu)=\hat{c}_{j+1}(\nu) \hat{\tilde h}(2^{j}\nu)
+\hat{w}_{j+1}(\nu) \hat{\tilde g}(2^{j}\nu)$"></TD>
<TD> </TD>
<TD> </TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.64)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
where the conjugate filters have the expression:
<BR>
<DIV ALIGN="CENTER"><A NAME="eqnht"> </A><A NAME="eqngt"> </A>
<!-- MATH: \begin{eqnarray}
\hat{\tilde h}(\nu) & = {\hat{p}_h(\nu) \hat{h}^*(\nu)\over \hat{p}_h(\nu)
\mid \hat{h}(\nu)\mid^2 + \hat{p}_g(\nu)\mid \hat{g}(\nu)\mid^2}\\
\hat{\tilde g}(\nu) & = {\hat{p}_g(\nu) \hat{g}^*(\nu)\over \hat p_h(\nu)
\mid \hat{h}(\nu)\mid^2 + \hat{p}_g(\nu)\mid \hat{g}(\nu)\mid^2}
\end{eqnarray} -->
<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
WIDTH="47" HEIGHT="63" ALIGN="MIDDLE" BORDER="0"
SRC="img735.gif"
ALT="$\displaystyle \hat{\tilde h}(\nu)$"></TD>
<TD ALIGN="CENTER" NOWRAP><IMG
WIDTH="226" HEIGHT="63" ALIGN="MIDDLE" BORDER="0"
SRC="img736.gif"
ALT="$\textstyle = {\hat{p}_h(\nu) \hat{h}^*(\nu)\over \hat{p}_h(\nu)
\mid \hat{h}(\nu)\mid^2 + \hat{p}_g(\nu)\mid \hat{g}(\nu)\mid^2}$"></TD>
<TD> </TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.65)</TD></TR>
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
WIDTH="45" HEIGHT="52" ALIGN="MIDDLE" BORDER="0"
SRC="img737.gif"
ALT="$\displaystyle \hat{\tilde g}(\nu)$"></TD>
<TD ALIGN="CENTER" NOWRAP><IMG
WIDTH="226" HEIGHT="57" ALIGN="MIDDLE" BORDER="0"
SRC="img738.gif"
ALT="$\textstyle = {\hat{p}_g(\nu) \hat{g}^*(\nu)\over \hat p_h(\nu)
\mid \hat{h}(\nu)\mid^2 + \hat{p}_g(\nu)\mid \hat{g}(\nu)\mid^2}$"></TD>
<TD> </TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.66)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
It is easy to see that these filters satisfy the exact reconstruction
equation <A HREF="node316.html#eqn_exact">14.19</A>. In fact, equations
<A HREF="node323.html#eqnht">14.65</A> and <A HREF="node323.html#eqngt">14.66</A> give the
general solution to this equation. In this analysis, the
Shannon sampling condition is always respected. No aliasing
exists, so that the dealiasing condition <A HREF="node316.html#eqn_desa">14.18</A> is not
necessary.
<P>
The denominator is reduced if we choose:
<BR><P></P>
<DIV ALIGN="CENTER">
<!-- MATH: \begin{displaymath}
\hat{g}(\nu) = \sqrt{1 - \mid\hat{h}(\nu)\mid^2}
\end{displaymath} -->
<IMG
WIDTH="200" HEIGHT="46"
SRC="img739.gif"
ALT="\begin{displaymath}\hat{g}(\nu) = \sqrt{1 - \mid\hat{h}(\nu)\mid^2}\end{displaymath}">
</DIV>
<BR CLEAR="ALL">
<P></P>
This corresponds to the case where the wavelet is the difference between
the square of two resolutions:
<BR>
<DIV ALIGN="CENTER">
<!-- MATH: \begin{eqnarray}
\mid \hat \psi(2\nu)\mid^2 = \mid \hat \phi(\nu)\mid^2 - \mid \hat
\phi(2\nu)\mid^2
\end{eqnarray} -->
<TABLE ALIGN="CENTER" CELLPADDING="0" WIDTH="100%">
<TR VALIGN="MIDDLE"><TD NOWRAP ALIGN="RIGHT"><IMG
WIDTH="304" HEIGHT="53" ALIGN="MIDDLE" BORDER="0"
SRC="img740.gif"
ALT="$\displaystyle \mid \hat \psi(2\nu)\mid^2 = \mid \hat \phi(\nu)\mid^2 - \mid \hat
\phi(2\nu)\mid^2$"></TD>
<TD> </TD>
<TD> </TD>
<TD WIDTH=10 ALIGN="RIGHT">
(14.67)</TD></TR>
</TABLE></DIV>
<BR CLEAR="ALL"><P></P>
<P>
<BR>
<DIV ALIGN="CENTER"><A NAME="fig_diff_uv_phi_psi"> </A><A NAME="15765"> </A>
<TABLE WIDTH="50%">
<CAPTION><STRONG>Figure 14.10:</STRONG>
Left, the interpolation function
<!-- MATH: $\hat{\phi}$ -->
<IMG
WIDTH="18" HEIGHT="53" ALIGN="MIDDLE" BORDER="0"
SRC="img741.gif"
ALT="$\hat{\phi}$">
and right, the wavelet
<!-- MATH: $\hat{\psi}$ -->
<IMG
WIDTH="21" HEIGHT="53" ALIGN="MIDDLE" BORDER="0"
SRC="img742.gif"
ALT="$\hat{\psi}$">.</CAPTION>
<TR><TD><IMG
WIDTH="940" HEIGHT="334"
SRC="img743.gif"
ALT="\begin{figure*}
\centerline{
\hbox{
\psfig{figure=fig_diff_uv_phi_psi.ps,bbllx=0...
...blly=13.5cm,bburx=20.5cm,bbury=27cm,height=5cm,width=17cm,clip=}}}
\end{figure*}"></TD></TR>
</TABLE>
</DIV>
<BR>
<BR>
<DIV ALIGN="CENTER"><A NAME="fig_diff_uv_ht_gt"> </A><A NAME="15838"> </A>
<TABLE WIDTH="50%">
<CAPTION><STRONG>Figure:</STRONG>
On left, the filter
<!-- MATH: $\hat{\tilde{h}}$ -->
<IMG
WIDTH="19" HEIGHT="33" ALIGN="BOTTOM" BORDER="0"
SRC="img746.gif"
ALT="$\hat{\tilde{h}}$">,
and on right the filter
<!-- MATH: $\hat{\tilde{g}}$ -->
<IMG
WIDTH="17" HEIGHT="52" ALIGN="MIDDLE" BORDER="0"
SRC="img747.gif"
ALT="$\hat{\tilde{g}}$">.</CAPTION>
<TR><TD><IMG
WIDTH="940" HEIGHT="338"
SRC="img748.gif"
ALT="\begin{figure*}
\centerline{
\hbox{
\psfig{figure=fig_diff_uv_ht_gt.ps,bbllx=0.5...
...blly=13.5cm,bburx=20.5cm,bbury=27cm,height=5cm,width=17cm,clip=}}}
\end{figure*}"></TD></TR>
</TABLE>
</DIV>
<BR>
<P>
We plot in figure <A HREF="node323.html#fig_diff_uv_phi_psi">14.10</A> the chosen scaling function
derived from a B-spline of degree
3 in the frequency space and
its resulting wavelet function. Their
conjugate functions are plotted in figure <A HREF="node323.html#fig_diff_uv_ht_gt">14.11</A>.
<P>
The reconstruction algorithm is:
<DL COMPACT>
<DT>1.
<DD>We compute the FFT of the image at the low resolution.
<DT>2.
<DD>We set <I>j</I> to <I>n</I><SUB><I>p</I></SUB>. We iterate:
<DT>3.
<DD>We compute the FFT of the wavelet coefficients at the scale j.
<DT>4.
<DD>We multiply the wavelet coefficients <IMG
WIDTH="29" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img749.gif"
ALT="$\hat{w}_j$">
by
<!-- MATH: $\hat{\tilde{g}}$ -->
<IMG
WIDTH="18" HEIGHT="52" ALIGN="MIDDLE" BORDER="0"
SRC="img750.gif"
ALT="$\hat{\tilde{g}}$">.
<DT>5.
<DD>We multiply the image at the lower resolution <IMG
WIDTH="23" HEIGHT="41" ALIGN="MIDDLE" BORDER="0"
SRC="img751.gif"
ALT="$\hat{c}_j$">
by
<!-- MATH: $\hat{\tilde{h}}$ -->
<IMG
WIDTH="19" HEIGHT="33" ALIGN="BOTTOM" BORDER="0"
SRC="img752.gif"
ALT="$\hat{\tilde{h}}$">.
<DT>6.
<DD>The inverse Fourier Transform of the addition of
<!-- MATH: $\hat{w}_j\hat{\tilde{g}}$ -->
<IMG
WIDTH="41" HEIGHT="52" ALIGN="MIDDLE" BORDER="0"
SRC="img753.gif"
ALT="$\hat{w}_j\hat{\tilde{g}}$">
and
<!-- MATH: $\hat{c}_i\hat{\tilde{h}}$ -->
<IMG
WIDTH="35" HEIGHT="63" ALIGN="MIDDLE" BORDER="0"
SRC="img754.gif"
ALT="$\hat{c}_i\hat{\tilde{h}}$">
gives the
image <I>c</I><SUB><I>j</I>-1</SUB>.
<DT>7.
<DD><I>j</I> = <I>j</I> - 1 and we go back to 3.
</DL>
<P>
The use of a scaling function with a cut-off frequency
allows a reduction of sampling at each scale, and limits the
computing time and the memory size.
<P>
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<ADDRESS>
<I>Petra Nass</I>
<BR><I>1999-06-15</I>
</ADDRESS>
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