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Predictive deconvolution

The problem is now to estimate a predictive filter f with the input traces. As the input traces have 512 samples, I took the length of the filter equal to 50 samples.

The residuals of L2 and L1 deconvolution on the pure trace are similar (Figure [*]). As the input wavelet is minimum-phase, the L2 deconvolution is efficient. However, the L1 deconvolution has still improved the result. Supposing the a priori statistical distribution of the residuals is exponential (and not Gaussian as in L2 deconvolution), it has forced the smallest values of the L2 residuals to come closer to 0, and has increased the highest values.

In this case, the IRLS algorithm converged quickly, because the initial condition number was low ($\approx 1000$): with real data, we can expect this number to be larger than 104. I took the initial estimate of the filter equal to the L2 filter. In Figure [*], I plotted the matrices ATA, ATW1A, ATW30A; W1 and W30 were respectively computed with the L2 residuals (after the first iteration of the IRLS algorithm) and the L1 residuals (after the 30th iteration of the IRLS algorithm). The non-Toeplitz structure of ATW1A is evident. However, ATW30A is close to a Toeplitz structure, because W is closer to the Identity matrix (to the factor $\varepsilon$): as the algorithm proceeds, more and more residuals become smaller than $\varepsilon$, and their weights get all equal to 1/$\varepsilon$.

Finally, I applied L2 and L1 predictive deconvolution to the noisy trace. The residuals are represented in Figure [*]; I also plotted the difference between these residuals and the residuals of the pure trace. In both cases, the residuals still contain the original noise bursts. This is not surprising, because the predictive filters could not predict them. It is more interesting to study the influence of the noise bursts on the rest of the trace. The two plots of the perturbation of the residuals after introduction of the noise show that the L2 residuals are much more perturbed than the L1 residuals. This means that the L2 predictive filter has tried to remove these noise bursts, and its modification has of course transformed the rest of the trace. On the contrary, the L1 filter, being insensitive to the spikes, has hardly been changed: its residuals are not very different from the initial ones, except for the noise bursts. Thus, the L1 deconvolution is more reliable than the L2 deconvolution.


next up previous print clean
Next: Other families of noise Up: SYNTHETIC EXAMPLE Previous: Deconvolution with a known
Stanford Exploration Project
1/13/1998