# 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),

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

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

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

where $\displaystyle \gamma(f)$ 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 $\displaystyle e^{-at^{2}}$ is

 \displaystyle \begin{align} e^{-at^{2}} \leftrightarrow (\pi /a)^{1/2} e^{-\omega^{2} /4a} \end{align} (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

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

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

\displaystyle \begin{align} (-jt)^{n} g(t)\leftrightarrow \frac{{\rm d}^{n} G(\omega )}{{\rm d}\omega ^{n}}, \end{align}

that is, for $\displaystyle n=2$ ,

\displaystyle \begin{align} -t^{2} g(t)\leftrightarrow \frac{{\rm d}^{2} G(\omega )}{{\rm d}\omega ^{2}}. \end{align}

The transform of the second term now becomes

\displaystyle \begin{align} 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{align}

Adding the two transforms, we have

\displaystyle \begin{align} &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{align}

## Problem 6.21b

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

### Solution

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

\displaystyle \begin{align} \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{align}

so $\displaystyle f=f_{M}$ for a maximum.

## Problem 6.21c

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

### Solution

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

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

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

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

\displaystyle \begin{align} ({\rm d}/{\rm d}t)[(1-2at^{2} )e^{-at^{2} } ]=0, \end{align}

that is, of the equation

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

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