Frequency domain

In the frequency domain, the problems of bandwidth equalization and phase matching are orthogonal, since they are concerned with the amplitude and argument of a complex signal.

The steps I applied to equalize the bandwidth were: Fourier transform the training windows, S₁ and S₂, on the time-axis, and calculate their spectra, $\hat{S}_1$ and $\hat{S}_2$, where

$$\hat{S}_i(\omega) = \sum_{\bf x} \left\vert S_i(\omega,{\bf x})\right\vert^2$$ (1)

I then designed the desired spectrum of the output, $A(\omega)$, by simply taking the minimum of the two spectra at each frequency.

$$A(\omega) = min\left(\hat{S}_1(\omega),\hat{S}_2(\omega)\right)$$ (2)

I applied this filter to shape the spectra of the regions of interest

$$S_{i}(\omega,{\bf x}) = S_{i}(\omega,{\bf x})\frac{\sqrt{A(\omega)}} {\left\vert S_{i}(\omega) \right\vert + \epsilon}$$ (3)

The $\epsilon$ in the denominator is purely for the event $S_1(\omega)=S_2(\omega)=0$, since in this formalism $A(\omega) \leq S(\omega)$.

As well as calculating a spectral operator, $A(\omega)$, I calculated the average difference in phase between the training windows for each frequency, $\Theta(\omega)$.

$$\Theta(\omega) = arg\left\{ \sum_{\bf x} S_{1}(\omega,{\bf x... ...ght\} - arg\left\{ \sum_{\bf x} S_{2}(\omega,{\bf x}) \right\}$$ (4)

This was then applied to map S₂ onto S₁

$$S_{2}(\omega,{\bf x}) = S_{2}(\omega,{\bf x}) \; e^{- i \Theta(\omega)}$$ (5)

Although the phase and bandwidth are orthogonal, in this approach it does matter the order in which the corrections are applied. This is due to the fact I calculate the average phase as the phase of the average value, as opposed to the average of the phases. With my approach it is correct to apply the bandwidth correction before the phase correction, although this is unlikely to be very important.

The results of the frequency domain cross-equalization are displayed in Figure 3. The differences due to the fluid movements are clearly visible, and the signal-to-noise level is high.

 

freq
Figure 3 Results of frequency domain cross-equalization. Left is filtered base survey, center is filtered monitor survey and right is the difference. The synthetic fluid layer is clearly visible in the difference section. S/N = 9.14.