Directivity of a source plus its ghost

Problems in Exploration Seismology and their Solutions

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

Problem 6.8

An air gun is fired at a depth of 10 m. The waveform includes frequencies in the range $10-80\ {\rm {Hz}}$, 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 $0^{\circ }$, $30^{\circ }$, $60^{\circ }$, and $90^{\circ }$ 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 $2A{\rm {\;sin\;}}(2\pi c{\rm {\;cos\;}}\theta )$, where the depth of the source is $c\lambda$. For the 10-Hz component, $c=10/150=0.067$; for the 80 Hz component, $c=10/19=0.53$. The ratio of the amplitude of the 10-Hz component to that of the 80-Hz component is

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

Table 6.8a. Ratios of ghost amplitudes of 10- and 80- Hz components.

$\theta$	0.42 ${\rm {\;cos\;}}\theta$	3.3 ${\rm {\;cos\;}}\theta$	Amplitude ratio
$0^{\circ }$	0.42	3.3	–2.6
$30^{\circ }$	0.36	2.9	1.5
$32^{\circ }$	0.30	2.3	1.04
$60^{\circ }$	0.21	1.6	0.21
$90^{\circ }$	—	—	0.13$*$

* For $\theta =90^{\circ }$, ${\rm {\;cos\;}}\theta =0$, so the ratio is 0/0. However, when $\theta$ is slightly less than $90^{\circ }$, the arguments of the sines are small and we can replace the sines with the angles; the cos factors cancel and the ratio is $0.42/3.3=0.13.$

Table 6.8a shows the results for the given values of $\theta$. 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.

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Table of Contents (book)

Also in this chapter

• Characteristics of different types of events and noise
• Horizontal resolution
• Reflection and refraction laws and Fermat’s principle
• Effect of reflector curvature on a plane wave
• Diffraction traveltime curves
• Amplitude variation with offset for seafloor multiples
• Ghost amplitude and energy
• Directivity of a source plus its ghost
• Directivity of a harmonic source plus ghost
• Differential moveout between primary and multiple
• Suppressing multiples by NMO differences
• Distinguishing horizontal/vertical discontinuities
• Identification of events
• Traveltime curves for various events
• Reflections/diffractions from refractor terminations
• Refractions and refraction multiples
• Destructive and constructive interference for a wedge
• Dependence of resolvable limit on frequency
• Vertical resolution
• Causes of high-frequency losses
• Ricker wavelet relations
• Improvement of signal/noise ratio by stacking

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