Source-array directivity & ghost notch

Part 1, Sources

Learning objectives

Write the far-field response of a towed source array as the product of an array factor (Section 0.5) and a ghost factor (Sections 1.2 and 1.4)
Explain why both factors change with frequency, because the wavelength $\lambda = V_{\mathrm{w}}/f$ changes while the geometry does not
Predict the ghost notches $f = nV_{\mathrm{w}}/(2h\cos\theta)$ and choose a tow depth $h$ for the band you need
Size the array so that the widest take-off angle you record loses no more than 3 dB at the highest frequency, $Nd < 0.44\,V_{\mathrm{w}}/(f_{\max}\sin\theta_{\max})$ , well inside the first null at $\sin\theta = V_{\mathrm{w}}/(Ndf)$

Section 0.5 gave the response of a line of elements. Sections 1.2 and 1.4 gave the ghost: the copy of the source that the surface above it reflects back, inverted. A towed air-gun array has both at once. In the far field, a ray that leaves at take-off angle $\\theta$ from the vertical, in the vertical plane of the array, carries their product:

|H(\\theta, f)| = \\underbrace{\\left|\\frac{\\sin(N\\pi d f\\sin\\theta/V\_{\\mathrm{w}})}{N\\sin(\\pi d f\\sin\\theta/V\_{\\mathrm{w}})}\\right|}\_{\\text{array factor}} \\;\\times\\; \\underbrace{2\\left|\\sin\\frac{2\\pi f h\\cos\\theta}{V\_{\\mathrm{w}}}\\right|}\_{\\text{ghost factor}}

Here $N$ guns sit $d$ apart at depth $h$ in water of velocity $V\_{\\mathrm{w}} \\approx 1500$ m/s, the sea surface reflects with a coefficient of $-1$ , and $|H|$ is measured against one gun with no sea surface. The spacing and the depth are fixed in metres, but the wavelength $\\lambda = V\_{\\mathrm{w}}/f$ is not, so both factors change with frequency. In the figure, choose a frequency and a take-off angle and read how much weaker or stronger the wave sent in that direction is than the one sent straight down.

Two factors that pull in different directions

The figure opens at 80 Hz and 40° with six guns 3 m apart at 7.5 m: the wave at 40° is 2.1 dB weaker than the one sent straight down, the net of 6.2 dB lost to the array and 4.1 dB given back by the ghost. Straight down, the array factor is 1 at every frequency, so the vertical spectrum is all ghost: zero at 0 Hz, a peak of +6 dB at $V\_{\\mathrm{w}}/(4h)$ , and notches at $f = nV\_{\\mathrm{w}}/(2h)$ , every 100 Hz for a 7.5 m tow. Tilt the ray and the extra path to the ghost shrinks from $2h$ to $2h\\cos\\theta$ , so every notch climbs to $nV\_{\\mathrm{w}}/(2h\\cos\\theta)$ . Just below the vertical notch this makes an oblique ray louder than the vertical one: at 95 Hz, the ray at 35° is 5.1 dB stronger than the ray straight down (exercise 2). Toward the horizontal the extra path shrinks, and below the ghost peak $V\_{\\mathrm{w}}/(4h)$ the ghost cancels more of the source the flatter the ray, down to a null along the surface itself: a source under a free surface radiates as a vertical dipole. How near the surface that weakening reaches depends on the tow, because it needs $2h\\cos\\theta$ to be a small fraction of a wavelength. A deep tow keeps strong oblique energy in the band: towed at 20 m, one gun at 40 Hz sends 8.4 dB more toward 80° than straight down, where 40 Hz sits beside the 37.5 Hz notch (exercise 5).

The array factor works the other way. At low frequency the array is short against a wavelength and radiates almost equally in every direction. As $f$ rises its main lobe narrows, and its first null, at $\\sin\\theta = V\_{\\mathrm{w}}/(Ndf)$ , swings in from the horizontal. Far offsets and shallow targets are reached by rays with large take-off angles, so they lose the high frequencies first, and a signature measured or modelled straight below the array describes them poorly. Deghosting and designature therefore have to be done angle by angle.

Design rules of thumb

Tow depth sets the ghost. The first notch straight down is $V\_{\\mathrm{w}}/(2h)$ : keep it above the highest frequency you need, 125 Hz for a 6 m tow. Towing deeper strengthens the low frequencies (at 10 Hz, from $-7.6$ dB at 5 m to $+1.4$ dB at 15 m) and pays by pulling the notch down, from 150 Hz to 50 Hz (exercise 3).
Array length sets the directivity. Keeping the first null outside the widest take-off angle $\\theta\_{\\max}$ is not enough: an angle just inside the null has already lost most of its energy. Six guns 3.5 m apart, $Nd$ = 21 m, keep the null at 100 Hz out to 46°, yet lose 18.9 dB at 40°. To lose no more than 3 dB to the array at $\\theta\_{\\max}$ and the highest frequency $f\_{\\max}$ , keep $Nd < 0.44\\,V\_{\\mathrm{w}}/(f\_{\\max}\\sin\\theta\_{\\max})$ , about 10 m for 100 Hz and 40° (exercise 4).
Grating lobes need $d \\ge \\lambda$ . A spacing below $\\lambda\_{\\min} = V\_{\\mathrm{w}}/f\_{\\max}$ , 15 m at 100 Hz, keeps them out. Air-gun spacings of a few metres are far inside this limit; the overall length is what matters.
Real arrays are not one line. They spread sub-arrays crossline and mix gun sizes, so their directivity differs inline and crossline and is computed from notional source signatures rather than from this formula. The two rules above still hold for each direction.

References

Dragoset, B. (1990). Air-gun array specs: a tutorial. Geophysics, 55(11), 1426-1440.
Ziolkowski, A. (1970). A method for calculating the output pressure waveform from an air gun. Geophysical Journal International, 21(2), 137-161.
Vermeer, G. J. O. (2002). 3-D Seismic Survey Design. SEG Geophysical References 12.
Pritchett, W. C. (1990). Acquiring Better Seismic Data. Chapman & Hall.