# Directivity of a source plus its ghost

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 6 181 - 220 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 6.8

An air gun is fired at a depth of 10 m. The waveform includes frequencies in the range ${\displaystyle 10-80\ {\rm {Hz}}}$, the amplitudes of the 10- and 80-Hz components being the same near the source. Compare their amplitudes for the wave plus ghost at considerable distance from the source in the directions ${\displaystyle 0^{\circ }}$, ${\displaystyle 30^{\circ }}$, ${\displaystyle 60^{\circ }}$, and ${\displaystyle 90^{\circ }}$ to the vertical.

### Background

Air guns are described in problem 7.7.

### Solution

We take the velocity in water as 1.5 km/s so that the wavelengths are 150 m and 19 m for the 10-Hz and 80-Hz components. From equation (6.7a) the amplitude of the ghost is ${\displaystyle 2A{\rm {\;sin\;}}(2\pi c{\rm {\;cos\;}}\theta )}$, where the depth of the source is ${\displaystyle c\lambda }$. For the 10-Hz component, ${\displaystyle c=10/150=0.067}$; for the 80 Hz component, ${\displaystyle c=10/19=0.53}$. The ratio of the amplitude of the 10-Hz component to that of the 80-Hz component is

{\displaystyle {\begin{aligned}{\frac {{\rm {\;sin\;}}(2\pi \times 0.067{\rm {\;cos\;}}\theta )}{{\rm {\;sin\;}}(2\pi \times 0.53{\rm {\;cos\;}}\theta )}}={\frac {{\rm {\;sin\;}}(0.42{\rm {\;cos\;}}\theta )}{{\rm {\;sin\;}}(3.3{\rm {\;cos\;}}\theta )}}.\end{aligned}}}

Table 6.8a. Ratios of ghost amplitudes of 10- and 80- Hz components.
${\displaystyle \theta }$ 0.42 ${\displaystyle {\rm {\;cos\;}}\theta }$ 3.3 ${\displaystyle {\rm {\;cos\;}}\theta }$ Amplitude ratio
${\displaystyle 0^{\circ }}$ 0.42 3.3 –2.6
${\displaystyle 30^{\circ }}$ 0.36 2.9 1.5
${\displaystyle 32^{\circ }}$ 0.30 2.3 1.04
${\displaystyle 60^{\circ }}$ 0.21 1.6 0.21
${\displaystyle 90^{\circ }}$ 0.13${\displaystyle *}$
 * For ${\displaystyle \theta =90^{\circ }}$, ${\displaystyle {\rm {\;cos\;}}\theta =0}$, so the ratio is 0/0. However, when ${\displaystyle \theta }$ is slightly less than ${\displaystyle 90^{\circ }}$, the arguments of the sines are small and we can replace the sines with the angles; the cos factors cancel and the ratio is ${\displaystyle 0.42/3.3=0.13.}$

Table 6.8a shows the results for the given values of ${\displaystyle \theta }$. The values in columns 2 and 3 are in radians and the column headed “Amplitude ratio” is the sine of the values in column two divided by the sine of the values in column three.

Thus, the 10-Hz component is stronger than the 80-Hz component as the direction approaches the vertical. The minus sign in the first ratio is due to a phase reversal of the 80-Hz component.