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** Up:** SYNTHETIC EXAMPLE
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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 *L*^{2} and *L*^{1} deconvolution on the pure trace are similar
(Figure ). As the input wavelet is minimum-phase, the *L*^{2}
deconvolution is efficient. However, the *L*^{1} deconvolution has still improved
the result. Supposing the a priori statistical distribution of the residuals
is *exponential* (and not Gaussian as in *L*^{2} deconvolution), it
has forced the smallest values of the *L*^{2} 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 (): with real data, we can expect this
number to be larger than 10^{4}. I took the initial estimate of the filter
equal to the *L*^{2} filter. In Figure ,
I plotted the matrices *A*^{T}*A*, *A*^{T}*W*^{1}A, *A*^{T}*W*^{30}A; *W*^{1}
and *W*^{30} were respectively computed with the *L*^{2} residuals (after the
first iteration of the IRLS algorithm) and the *L*^{1} residuals (after the
30^{th} iteration of the IRLS algorithm). The non-Toeplitz structure of
*A*^{T}*W*^{1}A is evident. However, *A*^{T}*W*^{30}A is close to a Toeplitz structure,
because W is closer to the Identity matrix (to the factor ): as
the algorithm proceeds, more and more residuals become smaller than
, and their weights get all equal to 1/.

Finally, I applied *L*^{2} and *L*^{1} 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 *L*^{2} residuals are much more perturbed than the *L*^{1}
residuals. This means that the *L*^{2} 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 *L*^{1} 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 *L*^{1} deconvolution is
more reliable than the *L*^{2} deconvolution.

** Next:** Other families of noise
** Up:** SYNTHETIC EXAMPLE
** Previous:** Deconvolution with a known
Stanford Exploration Project

1/13/1998