A 1D example

Now I consider the Z-transforms of the data, signal and noise PEFs for a 1D case. For the data PEF ${\bf D}$, I assume that the filter has the form

$$D(Z)=\alpha(Z-Z_1)(Z-Z_2)(Z-Z_3)(Z-Z_4),$$ (9)

with $\alpha=1/(Z_1Z_2Z_3Z_4)$. The Z_i correspond to the roots of the filter. In this example I consider that Z₁ and Z₂ are the roots for the noise and Z₃ and Z₄ the roots for the signal. Now I assume that we have for the noise PEF ${\bf N}$

$$N(Z)=\beta(Z-Z_1)(Z-Z_2),$$ (10)

with $\beta=1/(Z_1Z_2)$. The Spitz estimate yields for the signal PEF ${\bf S}$

$$S(Z)=\gamma(Z-Z_3)(Z-Z_4),$$ (11)

with $\beta=1/(Z_3Z_4)$. We see that by construction, the signal and noise PEF annihilate different parts of the data space and can't overlap.

Now, If we assume that the noise PEF is a ``bad'' estimate of the noise with one erroneous root, i.e,

$$N(Z)=\beta(Z-Z_1)(Z-Z_5),$$ (12)

with $\beta=1/(Z_1Z_5)$, we find for the signal PEF ${\bf S}$

$$S(Z)=\frac{Z_5}{Z_2Z_3Z_4}\frac{(Z-Z_2)(Z-Z_3)(Z-Z_4)}{Z-Z_5}.$$ (13)

Because the PEFs are minimum phase, we can write

$$\begin{eqnarray} \frac{1}{Z-Z_5}&=&\frac{-1}{Z_5}\frac{1}{1-Z/Z_5}, \nonumber \\... ...ac{Z}{Z_5}\right), \\ &\approx&\frac{-1}{Z_5^2}(Z+Z_5). \nonumber\end{eqnarray}$$ (14)

Then we obtain for the signal PEF

$$S(Z)\approx\frac{-1}{Z_2Z_3Z_4Z_5}(Z-Z_2)(Z-Z_3)(Z-Z_4)(Z+Z_5).$$ (15)

The wrong root in the noise PEF leaks in the signal PEF but with an opposite sign. Again, the Spitz estimate makes it impossible for the signal and noise operators to overlap in the data space. This simple example in 1D can be easily expendable in 2D via the helical coordinates Claerbout (1998).