Complex coefficient of reflection

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Problem 3.6a

3.6a Using the expression to represent a plane wave incident on a plane interface, show that a complex coefficient of reflection,

R [defined by equation (3.6a) below] corresponds to a reduction in amplitude by the factor and an advance in phase by .

Background

When a plane P-wave is incident perpendicularly on a plane interface, the tangential displacements and stresses vanish, so equations (3.2f,i) are not constraining and we are left with equations (3.2e,h). Moreover, , so the equations reduce to

, , being impedances (see problem 3.2). The solution of these equations is


(3.6a)


(3.6b)

and being the coefficient of reflection and coefficient of transmission, respectively. Although equations (3.6a,b) hold only for normal incidence, the definitions and are valid for all angles of incidence. A negative value of means that in equation (3.6a) is opposite in sign to . Since , the minus sign is equivalent to adding to the phase of the waveform in part (a), that is, reversing the phase. Note that (except for phase reversal) is independent of the direction of incidence on the interface; however, the magnitude of depends upon this direction, and when necessary, we shall write and to distinguish between the two values. Note the following relations:


(3.6c)

where is the fraction of energy transmitted as defined in equation (3.7a).

Euler’s formulas (see Sheriff and Geldart, 1995, p. 564) express and as


(3.6d)

The hyperbolic sine and cosine are defined by the relations


(3.6e)

Solution

Writing , we have

But where (see Sheriff and Geldart, 1995, section 15.1.5), so

Since , the amplitude is reduced by this factor and the phase is advanced by

Problem 3.6b

3.6b Show that an imaginary angle of refraction (see Figure 3.1b) in equations (3.2e,f,h,i) leads to a complex value of and hence to phase shifts.

Solution

Let , where is real. Then, using equations (3.6d,e), we have

hence some of the coefficients in equations (3.2e,f,h,i) are imaginary and and will in general be complex, so phase shifts will occur.

Problem 3.6c

3.6c Show that when is negative, .

Solution

If is negative, in equation (3.6a), so . Therefore, from equation (3.6b) we see that , and since , .

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