Ricker wavelet relations

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Problem 6.21a

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


(6.21a)

, being the peak frequency, has the Fourier transform [Figure 6.21a(ii)]


(6.21b)

where 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 is


(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


(6.21d)

where . The transform of the first term is . To get the transform of the second term, we use Sheriff and Geldart, 1995, equation (15.142) which states that when , then,

that is, for ,

The transform of the second term now becomes

Adding the two transforms, we have

Problem 6.21b

Show that is the peak of the frequency spectrum.

Solution

To find the peak frequency, we set the derivative equal to zero. Thus, omitting the constant factor,

so for a maximum.

Problem 6.21c

Show that (see Figure 6.21a) and that

Solution

Since for , we have


(6.21e)

hence .

Moreover, is a minimum for , so is a root of

that is, of the equation


(6.21f)

Hence, and .

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