Reflection and transmission coefficients

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Problem 3.7a

3.7a Calculate the reflection and transmission coefficients, and of equations (3.6a,b), for a sandstone/shale interface (for a wave incident from the sandstone) for the following:

  1. km/s, km/s, g/cm, and g/cm;
  2. km/s, km/s,
      g/cm, and  g/cm.
    
  3. What are the corresponding values in nepers and decibels?

Background

A wave has kinetic energy due to the velocity of the medium and potential energy due to the strains in the medium. For a harmonic wave , the particle velocity is and the kinetic energy/unit volume is , being the density. As the wave progresses, the energy changes back and forth between kinetic and potential. When the potential energy is zero, the kinetic energy is a maximum and therefore equals the total energy. The maximum value comes when , so the total energy/unit volume, where is the energy density (problem 2.3).

The coefficients and in equations (3.6a,b) give ratios of the relative amplitudes of the reflected and transmitted waves. We denote the fractions of the incident energy that are reflected and transmitted by and ; “energy” as used here denotes the amount of energy flowing through a unit area normal to the wave direction per unit time (the intensity). The energy density (i.e., energy/unit volume) is given by equation (3.3k). The energy flowing through a unit area normal to the wave direction per unit time is equal to the energy density times the velocity, that is, for a P-wave. Therefore


(3.7a)

The coefficients and are sometimes referred to as reflection and transmission energy coefficients to distinguish them from and . Note that both and are independent of the direction of travel through the interface.

Solution

i) Using equations (3.6a,b), we have

(where the units are g. km/cm3. s). Then,

(The minus sign denotes a phase reversal; see problem 3.6.) Because the incident wave is in the sandstone, we have

Note that the amplitude of the transmitted wave is larger than that of the incident wave when is negative (see problem 3.6c).


ii) .


Then,


iii) From problem 2.17, we have nepers ln(amplitude ratio) and 1 neper dB. Therefore,

for (i),

For (ii),

Negative values of nepers and merely mean that the values are less than unity.

Problem 3.7b

3.7b Calculate the energy coefficients and for cases (i) and (ii) in part (a).

Solution

We use equation (3.7a) to get,

for (i),


For (ii),

Note that (within the accuracy of the calculations).

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