# Response of a linear array

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 8.6a

Under what conditions is the response of a linear array of ${\displaystyle n}$ evenly spaced geophones zero for a wave traveling horizontally (such as ground roll)?

### Background

In Figure 8.6a, a plane wave is approaching a linear array of ${\displaystyle n}$ identical geophones spaced at intervals of ${\displaystyle \Delta x}$. The wave arrives at the left-hand end of the array at time ${\displaystyle t}$ giving the output ${\displaystyle A\sin \omega t}$ (note that the plane of Figure 8.6a is not necessarily vertical).

Figure 8.6a.  Wavefront approaching array.

The path difference for adjacent geophones is ${\displaystyle \Delta x\sin \alpha }$, the time difference is ${\displaystyle \Delta t=(\Delta x\sin \alpha )/V}$ and the phase difference is ${\displaystyle \gamma =2\pi (\Delta t/T)=\omega \Delta t}$ The output of the ${\displaystyle r^{\rm {th}}}$ geophone is ${\displaystyle A\sin(\omega t-r\gamma )}$.

Summing the outputs of the ${\displaystyle n}$ geophones gives

 {\displaystyle {\begin{aligned}h(t)=\mathop {\sum } \limits _{r=0}^{n-1}A\sin(\omega t-r\gamma )\end{aligned}}} (8.6a)

 {\displaystyle {\begin{aligned}=A\left({\frac {\sin(n\gamma /2)}{\sin(\gamma /2)}}\right)\sin \left(\omega t-{\frac {1}{2}}(n-1)\gamma \right)\end{aligned}}} (8.6b)

(see Sheriff and Geldart, 1995, problem 15.12c). The amplitude is ${\displaystyle A}$ times the expression in the first bracket. If the ${\displaystyle n}$ geophones were located at one point, the amplitude would be ${\displaystyle nA}$. Dividing the amplitude of ${\displaystyle h(t)}$ by ${\displaystyle nA}$ gives the array response ${\displaystyle F}$:

 {\displaystyle {\begin{aligned}F={\frac {\sin(n\gamma /2)}{n\sin(\gamma /2)}}.\end{aligned}}} (8.6c)

But ${\displaystyle \gamma =\omega \Delta t=(\omega /V)\Delta x\sin \alpha =2\pi (\Delta x/\lambda )\sin \alpha }$, so equation (8.6c) can be written as

 {\displaystyle {\begin{aligned}F={\frac {\sin[n\pi (\Delta x/\lambda )\sin \alpha ]}{n\sin[\pi (\Delta x/\lambda )\sin \alpha ]}}\end{aligned}}} (8.6d)

The graph of ${\displaystyle |F|}$ is shown in Figure 8.6b(i) for five uniformly spaced geophones and in Figure 8.6b(ii) for nine geophones. Note that the abscissa can be expressed in a number of ways; apparent wavelength ${\displaystyle =\lambda _{a}=V_{a}/f=V/f\sin \alpha =\lambda /\sin \alpha }$, where ${\displaystyle \alpha }$ is the angle of approach in Figure 8.6a. Figure 8.6b(iii) shows the response for a tapered array.

Ground roll is a Rayleigh wave (see problem 2.14) and usually has low velocity and low frequency. It is nondispersive (see problem 8.11) if the medium is uniform, but it is dispersive on an inhomogeneous earth where velocity and other parameters change with depth.

### Solution

For a horizontally traveling wave, the plane of Figure 8.6a is horizontal and equation (8.6d) is unchanged. The only variable at our disposal in equation (8.6d) is ${\displaystyle \alpha }$ (the other quantities all being fixed by the nature of the array), and the obvious values to check are ${\displaystyle \alpha =0^{\circ }}$ and ${\displaystyle \alpha =90^{\circ }}$ For ${\displaystyle \alpha =0^{\circ }}$, the geophone outputs add up [see part (c)], so we try ${\displaystyle \alpha =90^{\circ }}$ Equation (8.6d) now reduces to

{\displaystyle {\begin{aligned}F={\frac {\sin[n\pi (\Delta x/\lambda )]}{n\sin(\pi \Delta x/\lambda )}}.\end{aligned}}}

For ${\displaystyle F}$ to be zero, the numerator must vanish and the denominator must be nonzero [see part (b) for the case where both numerator and denominator are zero]. Thus the arguments of the sines must equal ${\displaystyle m\pi }$ in the numerator and not equal ${\displaystyle n\pi }$ in the denominator, ${\displaystyle m}$ and ${\displaystyle n}$ being integers. Therefore ${\displaystyle (\Delta x/\lambda )=m/n}$ where ${\displaystyle m}$ is not a multiple of ${\displaystyle n}$, i.e., ${\displaystyle m=\ldots ,(n-1),(n+1)\ldots }$,

## Problem 8.6b

If ${\displaystyle n}$ geophones are distributed uniformly over one wavelength, show that the response is ${\displaystyle F=1/n}$

### Solution

The effective array length is ${\displaystyle (n-1)\Delta x}$, If the wavelength is ${\displaystyle n\Delta x}$, samples in the peak and trough of the wave can be paired up yielding a response of zero. If the wavelength ${\displaystyle \lambda =(n-1)\Delta x}$, i.e., if the distance between the first and last geophones in the group is ${\displaystyle \lambda }$, then the outputs of ${\displaystyle n-1}$ geophones will cancel and the output will be that of the remaining geophone ${\displaystyle n}$.

{\displaystyle {\begin{aligned}F={\frac {\sin[n\pi /(n-1)]}{n\sin[\pi /(n-1)]}}={\frac {\sin[\pi +\pi (n-1)]}{n\sin[\pi /(n-1)]}}=-1/n.\end{aligned}}}

## Problem 8.6c

What is the response of the array in part (a) when the waves arrive perpendicular to the line of geophones?

Figure 8.6b.  Array response to 30-Hz signal. Alternative scales are shown in (i). (i) Five geophones spaced 10 m apart; (ii) nine phones spaced 5.5 m; (iii) five locations as (i) but weighted 1, 2, 3, 2, 1.

### Solution

Since the wave is a plane wave, it will arrive at all geophones at the same time, which will give the same result as if all ${\displaystyle n}$ geophones were together; therefore, ${\displaystyle F=1}$.

If we set ${\displaystyle \alpha =0^{\circ }}$ in equation (8.6d), we get ${\displaystyle F=0/0}$. However, using the simpler equation (8.6b) we have

{\displaystyle {\begin{aligned}F=\mathop {\lim } \limits _{\alpha \to 0}\left[{\frac {\sin \left({\frac {1}{2}}n\gamma \right)}{n\sin \left({\frac {1}{2}}\gamma \right)}}\right]=\mathop {\lim } \limits _{\alpha \to 0}\left({\frac {1}{2}}n\gamma /n\times {\frac {1}{2}}\gamma \right)=1,\end{aligned}}}

where we have replaced the sines by their arguments.

## Problem 8.6d

What is the response of the array in part (a) when the waves arrive at ${\displaystyle 45^{\circ }}$ to the line and ${\displaystyle n=8}$? Repeat for ${\displaystyle n=16}$.

### Solution

We have ${\displaystyle \alpha =45^{\circ }}$, ${\displaystyle n=8}$, so equation (8.6d) becomes

{\displaystyle {\begin{aligned}F={\frac {\sin \left[8\pi \left(\Delta x/\lambda \right)\sin 45^{\circ }\right]}{8\sin \left[\left(\pi \Delta x/\lambda \right)\sin 45^{\circ }\right]}}={\frac {\sin \left[\left(8\pi /{\sqrt {2}}\right)\left(\Delta x/\lambda \right)\right]}{8\sin \left[\left(\pi /{\sqrt {2}}\right)\left(\Delta x/\lambda \right)\right]}}.\end{aligned}}}

Replacing 8 with 16 in the above equation gives

{\displaystyle {\begin{aligned}F={\frac {\sin \left[\left(16\pi /{\sqrt {2}}\right)\left(\Delta x/\lambda \right)\right]}{16\sin \left[\left(\pi /{\sqrt {2}}\right)\left(\Delta x/\lambda \right)\right]}}.\end{aligned}}}

## Problem 8.6e

Repeat part (d) when ${\displaystyle \Delta x/\lambda =1/8}$. Compare the results when ${\displaystyle n=8,16,32}$.

### Solution

When ${\displaystyle \Delta x/\lambda =1/8}$ in part (d) and ${\displaystyle n=8}$, the result is

{\displaystyle {\begin{aligned}F=\sin(\pi /{\sqrt {2}})/8\sin(\pi /8{\sqrt {2}})=0.796/(8\times 0.274)=0.363.\end{aligned}}}

For ${\displaystyle n=16}$, we get

{\displaystyle {\begin{aligned}F=\sin(2\pi /{\sqrt {2}})/16\sin(\pi /8{\sqrt {2}})=-0.961/16\times 0.274=-0.220.\end{aligned}}}

When ${\displaystyle n=32}$, ${\displaystyle \Delta x/\lambda =1/8}$ the result is

{\displaystyle {\begin{aligned}F=\sin(4\pi /{\sqrt {2}})/32\sin(\pi /8{\sqrt {2}})=0.859/32\times 0.274=0.059.\end{aligned}}}

The ratios are 1 : 0.61 : 0.16.