Specifications

University of Pretoria etd – Combrinck, M (2006)
For the case where
A is singular, not square or a non-linear operator, equation 4.12 must
be solved with a method such as generalised least squares inversion or other equivalent
techniques (Cooper, 2004). Disadvantages of solving systems of equations in this way are:
• need for a good initial model
• no guarantee for convergence
• no guarantee for uniqueness of the solution
• mathematical intensity
However, if
A can be formulated as a non-singular, square and linear matrix, equation 4.12
becomes a system of linear equations with the same number of unknowns as equations
(implying the existence of an inverse operator and a unique solution) which can be solved
by multiplying both sides with
A
-1
.
(4.13)....0),(
1
ni
i
xf)
i
(xf =⋅
−
=
′
A
In this case equation 4.12 can also be solved by applying Gaussian elimination with
backward substitution. Depending on whether
A and A
-1
will be calculated only once (for
a fixed number of equally spaced data points) or whether it will have to be recalculated (for
unequally spaced data points). For 20 data points it will require 3060
multiplications/divisions and 2850 additions/subtractions to solve equation 4.12 using
Gauss elimination with backward substitution. The equivalent numbers for calculating
A
-1
are 10 660 and 10 070. Thereafter 400 multiplications/divisions and 380 additions/
subtractions are required to solve for the
)(xf
i
′
.
Construction of the integration matrix
Integration and differentiation can be done very simply in the frequency domain by
respective division or multiplication with the factor (i
ω ) where ω is angular frequency and
1−=i . However, the process of Fourier transformation on discrete data points adds
noise to data and time (or spatial) domain calculations, although more complex and time-
consuming, are preferred in order to obtain cleaner data.
The first point that has to be addressed in any integration procedure is that of the
integration constant, formulated as
(4.14),)()( cxfdxxf +=
′
∫
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