Synthetic example

Figure 1 shows inline sections through two synthetic surveys, which were created to test 4D seismic processing algorithms. Although they were modeled post-stack, they contain many features derived from actual 4D field datasets, including different wavelets, random noise, spatially varying static and phase shifts, different time-varying gain functions, and different mute and dead-trace zones. The two surveys have different bandwidths and signal-to-noise levels to simulate multi-vintage 3D seismic.

 

raw
Figure 1 Inline sections through two synthetic datasets. d₁ is on the left and d₂ is on the right.

 

perfect
Figure 2 Best achievable cross-equalization difference result

To simulate the effect of production changes, we created base and monitor reflectivity functions by introducing changes along a `reservoir' interval. The best achievable result is shown in Figure 2. This contains only the coherent fluid changes and residual random noise.

A simple cross-equalization flow consists of matched-filtering and amplitude balancing. For this synthetic example, we designed the matched-filters on a trace-to-trace basis, in order to best estimate the spatially-varying static and phase shifts.

Figure 3 shows the filtered survey, ${\bf A d_1}$ at the same scale as ${\bf d_2}$. The low amplitude of the figure on the left illustrates the characteristic low value of a due to the noise in ${\bf d_1}$. The result of the low value of a is coherent geological signal leaking through into the difference plot, which is shown in Figure 4. The amplitude of ${\bf A d_1}$are equivalent to those obtained by renormalizing the data and scaling by $\nu = 0.61$.

 

matchamp
Figure 3 ${\bf A d_1}$ (left) and ${\bf d_2}$ (right) plotted at same scale

 

matchdiff
Figure 4 Difference using matched-filter amplitudes, $\nu \approx 0.61$.

Equalizing the energy in this case (using $\nu=1$), produces the poor result shown in Figure 5. The brightest event in this image is no longer the reservoir interval, but rather the event below it. This would cause problems for interpretation.

For this synthetic, we can calculate a value for $\nu=0.808$ based on the known signal-to-noise levels in the two surveys. This gave us the results shown in Figure 6. Considerably less coherent energy is present in this image than in Figures 4 and 5.

 

eneqdiff
Figure 5 Difference after equalizing the energy between surveys, $\nu=1$.

 

bestdiff
Figure 6 Difference using $\nu=0.808$.