# Difference between revisions of "Ricker wavelet relations"

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 6.21a

Verify that the Ricker wavelet in Figure 6.21a(i),

 {\begin{aligned}g(t)=(1-2\pi ^{2}f_{m}^{2}t^{2})e^{-(\pi f_{m}t)^{2}},\end{aligned}} (6.21a)

$f_{m}$ , being the peak frequency, has the Fourier transform [Figure 6.21a(ii)]

 {\begin{aligned}G(f)=(2/{\sqrt {\pi }})(f/f_{m})^{2}e^{-(f/f_{m})^{2}},\;\gamma (f)=0,\end{aligned}} (6.21b)

where $\gamma (f)$ is the phase.

### Background

Fourier transforms are discussed in problem 9.3 and theorems on Fourier transforms in Sheriff and Geldart, 1995, section 15.2.6.

The transform of $e^{-at^{2}}$ is

 {\begin{aligned}e^{-at^{2}}\leftrightarrow (\pi /a)^{1/2}e^{-\omega ^{2}/4a}\end{aligned}} (6.21c)

[Papoulis, 1962: p. 25, equation (2-68)]. Figure 6.21a.  Ricker wavelet (i) in time domain and (ii) in frequency domain.

### Solution

The time-domain expression for the Ricker wavelet can be written in the form

 {\begin{aligned}g(t)=(1-2at^{2})e^{-at^{2}}=e^{-at^{2}}-2at^{2}e^{-at^{2}},\end{aligned}} (6.21d)

where $a=(\pi f_{M})^{2}$ . The transform of the first term is $(\pi /a)^{1/2}e^{-\omega ^{2}}/4a$ . To get the transform of the second term, we use Sheriff and Geldart, 1995, equation (15.142) which states that when $g(t)\leftrightarrow G(\omega )$ , then,

{\begin{aligned}(-jt)^{n}g(t)\leftrightarrow {\frac {{\rm {d}}^{n}G(\omega )}{{\rm {d}}\omega ^{n}}},\end{aligned}} that is, for $n=2$ ,

{\begin{aligned}-t^{2}g(t)\leftrightarrow {\frac {{\rm {d}}^{2}G(\omega )}{{\rm {d}}\omega ^{2}}}.\end{aligned}} The transform of the second term now becomes

{\begin{aligned}2a{\frac {{\rm {d}}^{2}}{{\rm {d}}\omega ^{2}}}\left[\left({\frac {\pi }{a}}\right)^{1/2}e^{-(\omega ^{2}/4a)}\right]=2(\pi a)^{1/2}{\frac {\rm {d}}{{\rm {d}}\omega }}\left[-\left({\frac {\omega }{2a}}\right)e^{-(\omega ^{2}/4a)}\right]\\=-(\pi /a)^{1/2}{\frac {\rm {d}}{{\rm {d}}\omega }}\left[\omega e^{-(\omega ^{2}/4a)}\right]=-(\pi /a)^{1/2}\left[1-\left({\frac {\omega ^{2}}{2a}}\right)\right]e^{-(\omega ^{2}/4a)}.\end{aligned}} Adding the two transforms, we have

{\begin{aligned}&g(t)\leftrightarrow (2/{\sqrt {\pi }})(\omega /\omega _{M})^{2}e^{-(\omega /\omega _{M})^{2}}=(\pi /a^{3})^{1/2}(\omega ^{2}/2)e^{-(\omega /\omega _{M})^{2}}\\&\leftrightarrow (2/{\sqrt {\pi }})(f^{2}/f_{M}^{3})e^{-(f/f_{M})^{2}}.\end{aligned}} ## Problem 6.21b

Show that $f_{M}$ is the peak of the frequency spectrum.

### Solution

To find the peak frequency, we set the derivative ${\rm {d}}G(f)/{\rm {d}}f$ equal to zero. Thus, omitting the constant factor,

{\begin{aligned}\left({\frac {\rm {d}}{{\rm {d}}f}}\right)\left[f^{2}e^{-(f/f_{M})^{2}}\right]=e^{-(f/f_{M})^{2}}\left[2f-f^{2}(2f/f_{M}^{2})\right]=0,\end{aligned}} so $f=f_{M}$ for a maximum.

## Problem 6.21c

Show that $T_{D}/T_{R}={\sqrt {3}}$ (see Figure 6.21a) and that $T_{D}f_{M}={\sqrt {6}}/\pi .$ ### Solution

Since $g(t)=0$ for $t=+T_{R}/2$ , we have

 {\begin{aligned}[1-2(\pi f_{M}T_{R}/2)^{2}]e^{-(\pi f_{M}T_{R}/2)^{2}}=0,\end{aligned}} (6.21e)

hence $T_{R}=\pm {\sqrt {2}}/(\pi f_{M})$ .

Moreover, $g(t)$ is a minimum for $t=\pm T_{D}/2$ , so $T_{D}/2$ is a root of

{\begin{aligned}({\rm {d}}/{\rm {d}}t)[(1-2at^{2})e^{-at^{2}}]=0,\end{aligned}} that is, of the equation

 {\begin{aligned}e^{-at^{2}}[-4at+(1-2at^{2})(-2at)]=0,\\{\mbox{so}}\qquad \qquad \ t=T_{D}/2=[{\sqrt {(3/2)}}]/\pi f_{M}.\end{aligned}} (6.21f)

Hence, $T_{D}/T_{R}={\sqrt {3}}$ and $T_{D}f_{M}={\sqrt {6/\pi }}$ .