Synthetics

I made some synthetic marine data and added 5% noise. This, along with an attempted deconvolution, is shown in Figure 2.

 

syn+.2 Figure 2 Top is the synthetic data yt. Middle is the deconvolved data xt. Bottom is the noise nt. This choice of $\epsilon=.2$ gave my preferred result.

The plot in Figure 2 is for the value of $\epsilon$ that I subjectively regarded as best. The result is pleasing because the doublets tend to be converted to impulses. Unfortunately, the low-frequency noise in x_t is significantly stronger than that in y_t, as expected.

Taking $\epsilon$ larger will decrease ||X|| but increase the explicit noise ||Y-BX||. To decrease the explicit noise, I chose a tiny value of $\epsilon=.001$.Figure 3 shows the result. The explicit noise n_t appears to vanish, but the low-frequency noise implicit to the deconvolved output x_t has grown unacceptably.

 

syn+.001 Figure 3 As before, the signals from top to bottom are yt, xt, and nt. Choosing a small $\epsilon=.001$ forces the noise mostly into the output xt. Thus the noise is essentially implicit.

Finally, I chose a larger value of $\epsilon=.5$ to allow more noise in the explicit n_t, hoping to get a lower noise implicit to x_t. Figure 4 shows the result.

 

syn+.5 Figure 4 Noise essentially explicit. $\epsilon=.5$.

Besides the growth of the explicit noise (which is disturbingly correlated to the input), the deconvolved signal has the same unfortunate wavelet on it that we are trying to remove from the input.

Results of a field-data test shown in Figure 5 do not give reason for encouragement.

In conclusion, all data recording has an implicit filter, and this filter is arranged to make the data look good. Application of a theoretical filter, such as $\omega^{-2}$,may achieve some theoretical goals, but it does not easily achieve practical goals.

 

field+.3 Figure 5 Field-data test. As before, the signals from top to bottom are yt, xt, and nt.

EXERCISES:

1. The (1,-2,1) signature is an oversimplification. In routine operations the hydrophones are at a depth of 7-10 meters and the airgun is at a depth of 5-6 meters. Assuming a sampling rate of .004 s (4 milliseconds) and a water velocity of 1500 m/s, what should the wavelet be?
2. Rerun the figures with the revised wavelet of the previous exercise.