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The Reconstruction

If the wavelet is the difference between two resolutions, an evident reconstruction for a wavelet transform ${\cal W} = \{w_1, w_2, \dots, w_{n_p}, c_{n_p}\}$ is:
$\displaystyle \hat c_0(\nu) = \hat c_{n_p}(\nu) + \sum_j \hat w_j(\nu)$     (14.60)

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:
$\displaystyle \hat c_{j+1} = \hat h(2^j \nu) \hat c_j(\nu)$     (14.61)
$\displaystyle \hat w_{j+1} = \hat g(2^j \nu) \hat c_j(\nu)$     (14.62)

we look for cj knowing cj+1, wj+1, h and g. We restore $\hat c_j(\nu)$ with a least mean square estimator:
$\displaystyle \hat p_h(2^j\nu)\vert\hat c_{j+1}(\nu)-\hat h(2^j\nu)\hat c_j(\nu...
...t^2 +
\hat p_g(2^j\nu)\vert\hat w_{j+1}(\nu)-\hat g(2^j\nu)\hat
c_j(\nu)\vert^2$     (14.63)

is minimum. $\hat p_h(\nu)$ and $\hat p_g(\nu)$ are weight functions which permit a general solution to the restoration of $\hat c_j(\nu)$. By $\hat c_j(\nu)$ derivation we get:
 
$\displaystyle \hat{c}_{j}(\nu)=\hat{c}_{j+1}(\nu) \hat{\tilde h}(2^{j}\nu)
+\hat{w}_{j+1}(\nu) \hat{\tilde g}(2^{j}\nu)$     (14.64)

where the conjugate filters have the expression:
  
$\displaystyle \hat{\tilde h}(\nu)$ $\textstyle = {\hat{p}_h(\nu) \hat{h}^*(\nu)\over \hat{p}_h(\nu)
\mid \hat{h}(\nu)\mid^2 + \hat{p}_g(\nu)\mid \hat{g}(\nu)\mid^2}$   (14.65)
$\displaystyle \hat{\tilde g}(\nu)$ $\textstyle = {\hat{p}_g(\nu) \hat{g}^*(\nu)\over \hat p_h(\nu)
\mid \hat{h}(\nu)\mid^2 + \hat{p}_g(\nu)\mid \hat{g}(\nu)\mid^2}$   (14.66)

It is easy to see that these filters satisfy the exact reconstruction equation 14.19. In fact, equations 14.65 and 14.66 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 14.18 is not necessary.

The denominator is reduced if we choose:

\begin{displaymath}\hat{g}(\nu) = \sqrt{1 - \mid\hat{h}(\nu)\mid^2}\end{displaymath}

This corresponds to the case where the wavelet is the difference between the square of two resolutions:
$\displaystyle \mid \hat \psi(2\nu)\mid^2 = \mid \hat \phi(\nu)\mid^2 - \mid \hat
\phi(2\nu)\mid^2$     (14.67)


  
Figure 14.10: Left, the interpolation function $\hat{\phi}$ and right, the wavelet $\hat{\psi}$.
\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*}


  
Figure: On left, the filter $\hat{\tilde{h}}$, and on right the filter $\hat{\tilde{g}}$.
\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*}

We plot in figure 14.10 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 14.11.

The reconstruction algorithm is:

1.
We compute the FFT of the image at the low resolution.
2.
We set j to np. We iterate:
3.
We compute the FFT of the wavelet coefficients at the scale j.
4.
We multiply the wavelet coefficients $\hat{w}_j$ by $\hat{\tilde{g}}$.
5.
We multiply the image at the lower resolution $\hat{c}_j$ by $\hat{\tilde{h}}$.
6.
The inverse Fourier Transform of the addition of $\hat{w}_j\hat{\tilde{g}}$ and $\hat{c}_i\hat{\tilde{h}}$ gives the image cj-1.
7.
j = j - 1 and we go back to 3.

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.


next up previous contents
Next: Visualization of the Wavelet Up: Multiresolution with scaling functions Previous: The Wavelet transform using
http://www.eso.org/midas/midas-support.html
1999-06-15