# Ghosting as a notch filter

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 9 295 - 366 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 9.20a

The ghost reflection from the sea surface acts as a notch filter for receivers planted on the sea floor. Plot the notch frequency versus water depth.

### Background

A notch filter discriminates against a very narrow band of frequencies.

### Solution

When a reflected wave is recorded by receivers on the sea floor, a ghost produced by reflection at the surface will be superimposed on the primary reflection ${\displaystyle G(z)}$. The reflection coefficient at the surface is –1, so, from problem 9.19, the ghost is ${\displaystyle G(z)(-z^{n})}$, where the two-way traveltime through the water layer is ${\displaystyle n\Delta }$. Therefore, the recorded signal is ${\displaystyle G(z)(1-z^{n})}$.

Figure 9.20a.  Ghosting versus source depth.

The recorded signal will be zero whenever ${\displaystyle z^{n}=+1}$. Using Euler’s formula (Sheriff and Geldart, 1995, problem 15.12a), we get

{\displaystyle {\begin{aligned}z^{n}=e^{-2\pi fn\Delta }\\=\cos(2\pi fn\Delta )\\-j{\sin }(2\pi fn\Delta ),\end{aligned}}}

so ${\displaystyle z^{n}=+1}$ when ${\displaystyle 2\pi fn\Delta =2\pi }$, that is, when ${\displaystyle f=1/n\Delta }$ The two-way travel traveltime for a source depth ${\displaystyle d}$ is ${\displaystyle n\Delta =2d/V_{W}}$. Taking the water velocity as ${\displaystyle V_{W}=1500}$ m/s, we arrive at the result:

 {\displaystyle {\begin{aligned}f=V_{W}/2d=750/d,\end{aligned}}} (9.20a)

where ${\displaystyle d}$ is in meters. The graph of ${\displaystyle f}$ versus ${\displaystyle d}$ is shown in Figure 9.20a.

## Problem 9.20b

If air-gun sources are fired at 10-m depth, how will this affect the spectrum?

### Solution

Energy leaving the source and reflected at the surface will produce a ghost delayed by ${\displaystyle \Delta t=2\times 10/1500\approx 13\ {\hbox{ms}}}$. The source ghost will have opposite polarity to the primary wave, with a delay of 13 ms corresponding to a frequency of 77 Hz; thus, it will interfere destructively with the frequency of 77 Hz in the original signal. As a result, frequencies in a narrow band centered on 77 Hz will be attenuated.

Additional ghosting will occur at receivers located below the surface due to reflection at the surface of the upcoming wavelet; this effect can be calculated in the same way as above.