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Cross-equalization via an intermediate model

It is possible to construct many other operators which try to maximize the frequency content of the difference section.

For example another approach may be to find filters, A1 and A2, that when applied to unchanged areas of the two datacubes, s1 and s2, give the same result, m. In addition we wish the survey with the higher bandwidth to remain as unchanged as possible. These constraints are contained in the regression equations:
{\bf A_1 s_1} - {\bf m} \approx {\bf 0} \\  
{\bf A_2 s_2} - {\bf m} \approx {\bf 0} \\ {\bf m} - {\bf s_2} \approx {\bf 0} \end{eqnarray} (8)
The unknowns, ${\bf m}$, ${\bf a_1}$ and ${\bf a_2}$ can be packed into a single vector, allowing the above equations to be rewritten in matrix form:
\left( \begin{array}
-{\bf I} & {\bf S_1} & {\bf 0} \\... 0} \\  \lambda {\bf s_2}\end{array} \right) \approx {\bf 0} \end{displaymath} (11)
where $\lambda$ controls the relative weights of the regression equations, and has to be chosen by the user.

Since we are estimating an intermediate model simultaneously with the pair of filters, a good initial model speeds convergence. The initial models chosen were ${\bf m_0} = {\bf s_2}$ and ${\bf A_1} = {\bf A_2}
= {\bf I}$.

With two sets of regression, a good choice of $\lambda$ is important. A choice of $\lambda$ that is too large results in ${\bf m} = {\bf s_2}$ and does not do a good job of match-filtering. Alternatively, a choice of $\lambda$ that is too small does a good job of match-filtering, but a lot of the resolution is lost and the result approaches that of the simple inverse filtering above. Unfortunately an intermediate choice of $\lambda$ is the worst of both worlds, as it does a poor job of match-filtering and the bandwidth of the signals are reduced. The results of the inversion with $\lambda=0.1$ are shown in Figures 8. By playing around with different values of $\lambda$ and different filter lengths, I was not able to get a successful result.

Figure 8
Unsatisfactory attempt at `optimal' cross-equalization via an intermediate model - result with $\lambda=0.1$.

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Although this approach looks like it may be successful on paper, I believe its failure stems from the size of the model space. Although time domain match-filtering uses a similar least squares formulation, it has only about a dozen elements in the model space, as opposed to a model the size of the data space which has to be estimated in this intermediate model approach.

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