The best starting point for inverse problems is the Kunetz equation (19).

(50) |

We need also the expression for the escaping wave (41)

(51) |

We also need to recall that . With this (50) becomes

(52) |

Multiplying through by *A*(*Z*) we get

(53) |

Since *A*(*Z*) is minimum-phase,
*A*(*Z*) may be written as 1/*B*(*Z*) or *A*(1/*Z*) = 1/*B*(1/*Z*).
Thus (53) becomes

(54) |

Identifying coefficients of zero
and positive powers of *Z* as simultaneous equations,
we get a set of equations which for a three-layer
model looks like (*r _{0}* = 1).

(55) |

In (55) we see our old friend the Toeplitz matrix.
It used to work for factoring spectra
and predicting time series.
Notice that -*c _{3}* has been inserted in
(55) as the highest coefficient of

The first four equations in (55)
would normally be thought of as follows:
Given the first three reflected pulses
*r _{1}*,

Now suppose we begin by observation of the escaping wave *E*(*Z*).
One way to determine the reflection coefficients would be to
form 1 + *R*(*Z*) + *R*(1/*Z*) by the autocorrelation of *E*(*Z*);
then, the Levinson recursion could be used to solve for
the reflection coefficients.
The only disadvantage of this method is that
*E*(*Z*) contains an infinite number of coefficients
so that in practice some truncation must be done.
The truncation is avoided by an alternative method.
Given *E*(*Z*) polynomial division will find *A*(*Z*).
The heart of the Levinson recursion is the building
up of *A*(*Z*) by *A*_{k}(*Z*) = *A*_{k-1}(*Z*) - *c*_{k} *Z*^{k}*A*_{k-1}(1/*Z*).
In particular, from () we have

(56) |

which shows how to get
*A _{3}*(

(57) |

Next multiply (56) by
1/(1 - *c _{3}*

(58) |

Equation (58) is the desired result
which shows how to reduce *A*_{k+1}(*Z*) to *A*_{k}(*Z*)
while learning *c*_{k+1}.
A program to continue this process is

COMPLEX A,C,AL,BE,TOP,CONJG C(1)=-1.; R(1)=1.; A(1)=1.; V(1)=1. 300 DO 310 I=1,N 310 C(I)=A(I) DO 330 K=1,N J=N-K+2 AL=1./(1.-C(J)*CONJG(C(J))) BE=C(J)*AL JH=(J+1)/2 DO 320 I=1,JH TOP=AL*C(I)-BE*CONJG(C(J-I+1)) C(J-I+1)=AL*C(J-I+1)-BE*CONJG(C(I)) 320 C(I)=TOP 330 C(J)=-BE/AL

A program to compute
reflection coefficients *c*_{k} from the prediction-error
filter *A*(*Z*).
The complex arithmetic is optional.

An inverse program to get *R* and *A* from *c* is

COMPLEX C,R,A,BOT,CONJG C(1)=-1.; R(1)=1.; A(1)=1.; V(1)=1. 100 DO 120 J=2,N A(J)=0. R(J)=C(J)*V(J-1) DO 110 I=2,J 110 R(J)=R(J)-A(I)*R(J-I+1) JH=(J+1)/2 DO 120 I=1,JH BOT=A(J-I+1)-C(J)*CONJG(A(I)) A(I)=A(I)-C(J)*CONJG(A(J-I+1)) 120 A(J-I+1)=BOT

This program computes both *R* and *A* from *c*.

8-12
Earthquake seismogram geometry with
white light incident from below.
In the top layer,
the sum of the waves vanishes representing zero pressure
at the free surface.
The difference of up- and down- is
the observed vertical component of velocity.
Figure 12 |

Finally, let us see how to do a problem
where there are random sources.
Figure 12 shows the ``earthquake geometry."
However,
in order to introduce a statistical element,
the pulse incident from below has been convolved
with a white light series *w*_{t} of random numbers.
Consequently,
all the waves internal to Figure 12 are
given by the convolution of *w*_{t} with the
corresponding wave in the impulse incident model.
Now suppose we are given the top-layer waves
*D* = 1 -*U* = *XW* and wish to consider
downward continuation.
We have the layer matrix

(59) |

(60) |

The Burg prediction-error scheme can be written in the form

(61) |

which makes it equivalent within a scale factor to downward continuing surface waveforms. The remaining question is whether Burg's estimate of the reflection coefficient, namely,

(62) |

turns out to estimate the reflection coefficient
*c*_{k} in the physical model.
To see how Burg's is related to the *c*_{k}
arising in the Levinson recursion,
we define and for *k*=2 as

(63) |

Next from the dot product

(64) |

(65) |

(66) |

(67) |

- An impulse and the first part of a reflection seismogram,
that is,
1 + 2
*R*(*Z*) is .What are the first three reflection coefficients? Assuming there are no more reflectors what is the next point in the reflection seismogram? - A seismogram
*X*(*Z*) = 1/(1 - .1*Z*+ .9*Z*) is observed at the surface of some layers over a halfspace. Sketch the time function and indicate its resonance frequency and decay time. Find the reflection coefficients if^{2}*X*(*Z*) is due to an impulsive source of unknown magnitude in the halfspace below the layers. - A source
*b*+_{0}*b*deep in the halfspace produces a seismogram .What are the layered structure and the source time function?_{1}Z

10/30/1997