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Burg PEF estimation should work fine on a helix.
Full details along with the 1-D code
are found at Claerbout (1976).
I will quickly review the theory from memory
(partly to see how simple I can make it).

First is the notion that PEFs can be built up
from this recursion

| |
(1) |

where *c* is in the range .There is a theorem from Algebra that is easy to prove
that if *c* is in the required range, then
*A*^{N+1}(*Z*)
will be minimum phase if
*A*^{N}(*Z*) is minimum phase.
Since *A*^{0}=1, all are minimum phase.
Burg's PEF calculation begins from two copies of the data *X*(*Z*).
One,
called *F*(*Z*) will be turned into the forward prediction error *A*(*Z*)*X*(*Z*).
The other
called *B*(*Z*) will be turned into the backward prediction error *A*(1/*Z*)*X*(*Z*).
At each stage of the calculation,
we compute *c* with this formula

| |
(2) |

The triangle inequality shows that for arbitrary *f* and *b*,
*c* is in the required range.
Given *c* we now form an upgraded *F* and *B* with
and
.[At successive iterations,
increasing time lag is introduced between *f* and *b*.
Details in 1976.]
When you are finished, you have
*F*=*A*(*Z*)*X*(*Z*) and
*B*=*A*(1/*Z*)*X*(*Z*).
Why is that?
To understand that requires delving into 1-D theory, in particular,
the Levinson recursion, and we won't do that now.
[Maybe I can think up an easier explanation later.
Perhaps by a sequence of orthogonality arguments.]
I recall if you append a tiny impulse function
off the end of *X* before you start,
when you finish, you will see it has turned into the PEF.
Now let us think about missing data off the ends of the
Burgian one-dimensional data set.
Given that we have computed *F*(*Z*)=*A*(*Z*)*X*(*Z*),
then we should find that *F*(*Z*)/*A*(*Z*) matches *X*(*Z*) until
its end, and it is a logical continuation thereafter.
Likewise *B*(*Z*) could be used for extensions before the beginning
of *X*(*Z*).
Thus it remains to think about how to handle gaps in the middle.

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Stanford Exploration Project

4/27/2000