# Dependence of resolvable limit on frequency

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 6.18a

A wavelet has a flat frequency spectrum from 0 to $f_{u}$ above which no frequencies are present. Show that the Rayleigh criterion gives a resolvable limit $t_{r}$ , where $t_{r}=0.715/f_{u}$ .

### Background

A reflecting layer is said to be resolvable if we can distinguish between the reflections from the top and bottom of the layer, usually on the basis of a phase break in the superimposed reflections (see Figure 6.18a). The Rayleigh criterion for vertical resolution states that at least a small depression must appear between successive events in order to recognize that more than one event is present. For this to occur the two reflections must be separated by at least a half-cycle; this corresponds to a minimum thickness of $\lambda /4$ , the two-way thickness then being $\lambda /2$ . This thickness of ${\frac {1}{4}}\lambda$ is called the tuning thickness or resolvable limit.

A boxcar (see Figure 6.18b) is a function whose value is unity within a certain range and zero outside this range (see problem 9.3).

### Solution

The frequency spectrum is a boxcar extending from $-f_{u}\ {\rm {to}}+f_{u}$ , shown in Figure 6.18b, which we write as ${\rm {box}}_{2f_{u}}(f)$ . The inverse transform of the spectrum is given by equation (9.3d), namely

 {\begin{aligned}g(t)=(1/2\pi )\mathop {\int } \nolimits _{-\infty }^{+\infty }\ {\rm {box}}_{2f_{u}}(f)e^{j\omega t}\ {\rm {d}}\omega =\mathop {\int } \nolimits _{-f_{u}}^{+f_{u}}e^{j2\pi ft}\ {\rm {d}}f\\=(1/2\pi jt)(e^{j2\pi f_{u}{f}}-e^{-j2\pi f_{u}{f}})=(1/\pi t){\rm {\;sin\;}}(2\pi f_{u}t)\\=2f_{u}\ {\rm {sinc}}\ (2\pi f_{u}t),\end{aligned}} (6.18a)

where sinc $x=({\rm {\;sin\;}}x)/x.$ The Rayleigh criterion gives a resolvable limit $t_{r}$ corresponding to the first trough (minimum) of the time-domain representation of the boxcar. Hence, we equate the derivative of the sinc function to zero, obtaining

{\begin{aligned}{\frac {\rm {d}}{{\rm {d}}t}}[{\rm {sinc}}(2\pi f_{u}t)]={\frac {\rm {d}}{{\rm {d}}t}}\left[{\frac {{\rm {\;sin\;}}(2\pi f_{u}t)}{2\pi f_{u}t}}\right]=0={\frac {\rm {d}}{{\rm {d}}t}}\left[{\frac {{\rm {\;sin\;}}(2\pi f_{u}t)}{t}}\right],\end{aligned}} {\begin{aligned}{\mbox{so}}\qquad \qquad \ (2\pi f_{u}/t){\rm {\;cos\;}}(2\pi f_{u}t)-(1/t^{2}){\rm {\;sin\;}}(2\pi f_{u}t)=0;\end{aligned}} This gives ${\rm {\;tan\;}}(2\pi f_{u}t)=2\pi f_{u}t$ , that is, we must solve the equation ${\rm {\;tan\;}}x=x$ where $x=2\pi f_{u}t$ . A graphical solution gives $x=4.49$ , hence $t_{r}=0.715/f_{u}$ .

## Problem 6.18b

Show that the value of $t_{r}$ for a wavelet with a flat spectrum extending from $f_{L}$ to $nf_{L}$ (that is, $m$ octaves where $n=2^{m}$ ) is given by the solution of the equation,

{\begin{aligned}nx\ {\rm {cos}}\ nx-{\rm {\;sin\;}}nx\\-x{\rm {\;cos\;}}x+{\rm {\;sin\;}}x=0,\end{aligned}} where $x=2\pi f_{L}t_{r}$ .

### Solution

The time-domain function corresponding to the spectrum in Figure 6.18c is

{\begin{aligned}g(t)=(1/2\pi )\left(\mathop {\int } \nolimits _{-nf_{L}}^{-f_{L}}e^{1^{\omega t}}\ {\rm {d}}\omega +\mathop {\int } \limits _{f^{L}}^{nf_{L}}e^{j\omega t}\ {\rm {d}}\omega \right)\\=(1/2\pi jt)[(e^{-j2\pi f_{L}{t}})-(e^{-j2\pi nf_{L}{t}})+(e^{j2\pi f_{L}{t}}-e^{j2\pi nf_{L}{f}})]\\=(1/\pi t)[{\rm {\;sin\;}}(2\pi nf_{L}t)-{\rm {\;sin\;}}(2\pi f_{L}t)].\end{aligned}} To get the first trough, we write $x=2\pi f_{L}t$ , then equate to zero the derivative of $g(x)$ with respect to $x$ . This gives

 {\begin{aligned}g(x)=(2f_{L}/x)({\rm {\;sin\;}}nx-{\rm {\;sin\;}}x),\\dg(x)/dx=0=2f_{L}[(n{\rm {\;cos\;}}nx-{\rm {\;cos\;}}x)/x-({\rm {\;sin\;}}nx-{\rm {\;sin\;}}x)/x^{2}],\\{\mbox{so}}\qquad \qquad \ nx{\rm {\;cos\;}}nx-{\rm {\;sin\;}}nx-x{\rm {\;cos\;}}x+{\rm {\;sin\;}}x=0.\end{aligned}} (6.18c)

## Problem 6.18c

Solve the equation in part (b) for $m=3,2,1.5,$ and 1, that is, for bandwidths of 3, 2, 1.5, and 1 octaves, and compare the relation between $t_{r}$ and $m$ .

### Solution

For $m=3$ , $n=2^{3}=8$ and we have from equation (6.18c)

{\begin{aligned}8x{\rm {\;cos\;}}8x-{\rm {\;sin\;}}8x-{\rm {\;cos\;}}x+{\rm {\;sin\;}}x=0.\end{aligned}} For $x=0$ , we get $0=0$ , but for $x$ slightly greater than zero, ${\rm {d}}g(x)/{\rm {d}}x$ is negative and continues to be negative until it changes sign for $x$ between 0.5 and 0.6; The corresponding root is $x=0.560$ , giving $t_{r}=0.089/f_{L}.$ .

When $m=2$ , $n=4$ , the root of equation (6.18c) is $x=1.10$ , giving $t_{r}=0.175/f_{L}$ . For $m=1.5$ , $n=2{\sqrt {2}}=2.83$ and the root is $x=1.51$ , so $t_{r}=0.240/f_{L}$ . For $m=1.0$ , $n=2$ and the root is $x=2.015$ , so $t_{r}=0.321/f_{L}$ .

## Problem 6.18d

Noting that part (a) involves an infinite number of octaves, what bandwidth is required to give nearly the same result?

### Solution

The resolution in part (a) is expressed in terms of $f_{u}$ while those in (c) are in terms of $f_{L}$ . To compare the results we equate $nf_{L}$ to $f_{u}$ , so that we have four frequency bands, each with top frequency $f_{u}$ and extending downward 1, 1.5, 2, 3, and an infinite number of octaves. This means that the values of $t_{r}$ in (c) must be adjusted to get $nf_{L}$ in the denominator, e.g.,for $n=4$ , $t_{r}=4\times 0.175/(4f_{L})$ . The results are shown in Table 6.18a.

Table 6.18a. Resolution versus number of octaves.
$m$ $n$ $t_{r}$ /($nf_{L}$ )
$\infty$ $\infty$ 0.715
3 8 0.712
2 4 0.700
1.5 2.83 0.679
1 2 0.321

Thus, three octaves bandwidth ($m=3$ ) gives almost as good resolution as an infinite number of octaves but the resolution deteriorates for band-widths $<1.5$ octaves.