# Determining vibroseis parameters

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 8 253 - 294 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 8.14a

8.14a Signal and noise characteristics determined from a previous dynamite survey are shown in Figure 8.14a. The principal objective is at 3000 m with a stacking velocity of 3000 m/s. What frequencies should be covered by a linear sweep?

### Background

The vibroseis method employs a vibrator to impart to the ground a long train of harmonic signals of varying frequency. The vibrator consists of a piston pressing against a steel plate which is held against the ground by the weight of the vehicle. For the usual linear sweep, the vibrator, which is actuated hydraulically, exerts a pressure ${\mathcal {P}}(t)$ against the plate of the form

 {\begin{aligned}{\mathcal {P}}(t)=A(t)\sin\{2\pi t[f_{0}+({\rm {d}}f/{\rm {d}}t)t]\},\end{aligned}} (8.14a)

where $f_{0}$ is the starting frequency and $({\rm {d}}f/{\rm {d}}t)$ is either positive (for an upsweep) or negative (downsweep). The amplitude $A(t)$ is constant (except for about 0.2 s at the beginning and end of the sweep, when it increases from or decreases to zero. The frequency varies between about 12 and 60 Hz and the duration of the sweep is usually 7 to 35 s.

Each sweep generates a signal train which is reflected at each of the reflectors; since reflections are much more closely spaced than the length of the sweep, the recorded signal is a complex superposition of many reflected wave trains. To interpret a vibroseis record, the input sweep is recorded and crosscorrelated (see problem 9.8) with the record; this compresses the reflected wavetrains into short wavelets, thereby removing much of the overlap of the lengthy reflected wavetrains to produce a more-or-less normal seismic record.

The response of the ground is not an exact reproduction of the motion of the piston and distortion introduces harmonics into the ground, principally the second harmonic. The second harmonic produces reflected wavetrains like the primary reflected wavetrains, but these correlate with the sweep signal to indicate different arrival times. This effect, called correlation ghosts, adds a spurious set of reflection events to the record (Sheriff and Geldart, 1995, 208). The arrival time of this ghost $t_{g}$ is

 {\begin{aligned}t_{g}=f_{L}T/(f_{i}-f_{f}),\end{aligned}} (8.14a)

where $f_{L}$ is the lowest sweep frequency, $T$ the sweep time, $f_{i}$ the initial sweep frequency, and $f_{f}$ the final sweep frequency. For an upsweep, the ghosts arrive before $t=0$ , so the ghost is no problem; to avoid this problem with downsweeps, we can use long sweeps so that the ghost is delayed until after the zone of interest is recorded.

If we have $n$ records on which the signal shape is essentially constant and the noise is random, stacking builds up the signal strength whereas random noise tends to cancel and the signal-to-noise ratio (S/N) varies as $n^{1/2}$ (Sheriff and Geldart, 1995, 184).

When multiple source units are used, the vibrators are actuated synchronously at locations a few meters apart. Generally, several sweeps at closely spaced locations constitute a single vibrator point.

### Solution

The signal and noise spectra in Figure 8.14a show that the signal spectrum is strong between 15 to 55 Hz while the noise spectrum is confined largely to the narrow peak centered around 15 Hz. A passband of 20–60 Hz would include most of the signal and exclude much of the noise. This is a bandwidth of roughly 1.5 octaves.

## Problem 8.14b

If a downsweep of 8 s is used, at what time will the correlation ghost appear for a 9-s 60–15 Hz linear sweep? Will it interfere with the objective?

### Solution

Assume a downsweep that lasts 8 s and goes from 60 Hz to 15 Hz, that is, 5.6 Hz/s. Fundamental frequencies may generate second harmonic correlation ghosts which fall within the passband. The ghosts of 30–15 Hz may interfere with desired reflections arriving after

{\begin{aligned}t=15\times 8/\left(60-15\right)=2.667{\mbox{ s}},\end{aligned}} according to equation (8.14b).

## Problem 8.14c

If a single vibrator sweep of 8 s yields an S/N of 0.2 at the objective depth for a 15 to 60 Hz sweep, how many sweeps will have to be stacked to give ${\mbox{S/N}}=2.0$ ?

### Solution

Assuming the noise is random, the S/N varies as $n^{1/2}$ . If we need to increase S/N by a factor of 10, we require $10^{2}=100$ sweeps. Note that the improvement of S/N by the factor $n^{1/2}$ does not apply to coherent noise.

## Problem 8.14d

Assume that recording continues for an additional time of 6 s (listen time) beyond the sweep time and that it takes 10 s to move the vibrators between sweep points; how long will be required for four vibrators to record one vibrator point?

### Solution

The sweep time plus the listening time is 14 s; adding the time to move the vibrators, we get a total of 24 s per sweep. To obtain 100 sweeps as required by part (c), we must move the four vibrators 25 times, taking $25\times 24=600{\mbox{ s}}=10$ minutes. However, the vibrators probably build up the signal more than they build up the noise and much of the noise is probably not random, so the number of required sweeps should be cut at least in half.