# Reflection and transmission coefficients

Series Geophysical References Series Problems in Exploration Seismology and their Solutions Lloyd P. Geldart and Robert E. Sheriff 3 47 - 77 http://dx.doi.org/10.1190/1.9781560801733 ISBN 9781560801153 SEG Online Store

## Problem 3.7a

Calculate the reflection and transmission coefficients, ${\displaystyle R}$ and ${\displaystyle T}$ of equations (3.6a,b), for a sandstone/shale interface (for a wave incident from the sandstone) for the following:

1. ${\displaystyle V_{ss}=2.43}$ km/s, ${\displaystyle V_{sh}=2.02}$ km/s, ${\displaystyle \rho _{ss}=2.08}$ g/cm${\displaystyle ^{3}}$, and ${\displaystyle \rho _{sh}=2.23}$ g/cm${\displaystyle ^{3}}$;
2. ${\displaystyle V_{ss}=3.35}$ km/s, ${\displaystyle V_{sh}=3.14}$ km/s,
 ${\displaystyle \rho _{ss}=2.21}$ g/cm${\displaystyle ^{3}}$, and ${\displaystyle \rho _{sh}=2.52}$ g/cm${\displaystyle ^{3}}$.

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 ${\displaystyle \psi =A\sin \omega t}$, the particle velocity is ${\displaystyle \partial \psi /\partial t=\omega A\cos \omega t}$ and the kinetic energy/unit volume is ${\displaystyle {\frac {1}{2}}\rho \omega ^{2}A^{2}\cos ^{2}\omega t}$, ${\displaystyle \rho }$ 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 ${\displaystyle \cos ^{2}\omega t=+1}$, so the total energy/unit volume${\displaystyle {}=E={\frac {1}{2}}\rho \omega ^{2}A^{2}}$, where ${\displaystyle E}$ is the energy density (problem 2.3).

The coefficients ${\displaystyle R}$ and ${\displaystyle T}$ 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 ${\displaystyle E_{R}}$ and ${\displaystyle E_{T}}$; “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, ${\displaystyle \left({\frac {1}{2}}\rho \omega ^{2}A^{2}\right)\alpha }$ for a P-wave. Therefore

 {\displaystyle {\begin{aligned}E_{R}=R^{2},\qquad E_{T}={\frac {\rho _{2}\alpha _{2}\omega ^{2}A_{2}^{2}}{\rho _{1}\alpha _{1}\omega ^{2}A_{0}^{2}}}=\left({\frac {Z_{2}}{Z_{1}}}\right)T^{2}={\frac {4Z_{1}Z_{2}}{(Z_{1}+Z_{2})^{2}}}.\end{aligned}}} (3.7a)

The coefficients ${\displaystyle E_{R}}$ and ${\displaystyle E_{T}}$ are sometimes referred to as reflection and transmission energy coefficients to distinguish them from ${\displaystyle R}$ and ${\displaystyle T}$. Note that both ${\displaystyle E_{R}}$ and ${\displaystyle E_{T}}$ are independent of the direction of travel through the interface.

### Solution

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

{\displaystyle {\begin{aligned}Z_{ss}=2.08\times 2.43=5.05,\qquad Z_{sh}=2.23\times 2.02=4.50\end{aligned}}}

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

{\displaystyle {\begin{aligned}R=\left(4.50-5.05\right)/\left(4.50+5.05\right)=-0.55/9.55=-0.058.\end{aligned}}}

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

{\displaystyle {\begin{aligned}T=2\times 5.05/9.55=1.06.\end{aligned}}}

Note that the amplitude of the transmitted wave is larger than that of the incident wave when ${\displaystyle R}$ is negative (see problem 3.6c).

ii) ${\displaystyle Z_{ss}=2.21\times 3.35=7.40,\ Z_{sh}=2.52\times 3.14=7.91}$.

Then,

{\displaystyle {\begin{aligned}R&=\left(7.91-7.40\right)/\left(7.91+7.40\right)=0.51/15.3=0.033,\\T&=2\times 7.40/15.3=0.967.\end{aligned}}}

iii) From problem 2.17, we have nepers ln(amplitude ratio) and 1 neper ${\displaystyle {}=8.686}$ dB. Therefore,

for (i),

{\displaystyle {\begin{aligned}R&=\ln \left(0.058\right)=-2.8\ \mathrm {nepers} =-24\ \mathrm {dB} ,\\T&=\ln \left(1.06\right)=0.058\ \mathrm {nepers} \ =0.51\ \mathrm {dB} .\end{aligned}}}

For (ii),

{\displaystyle {\begin{aligned}R&=\ln \left(0.033\right)=-3.4\ \mathrm {nepers} \ =-30\ \mathrm {dB} ,\\T&=\ln \left(0.967\right)=-0.034\ \mathrm {nepers} \ =-0.29\ \mathrm {dB} .\end{aligned}}}

Negative values of nepers and ${\displaystyle dB}$ merely mean that the values are less than unity.

## Problem 3.7b

Calculate the energy coefficients ${\displaystyle E_{R}}$ and ${\displaystyle E_{T}}$ for cases (i) and (ii) in part (a).

### Solution

We use equation (3.7a) to get,

for (i),

{\displaystyle {\begin{aligned}E_{R}&=R^{2}=(0.058)^{2}=0.0034,\\E_{T}&=\left(Z_{2}/Z_{1}\right)T^{2}=\left(4.50/5.05\right)\times 1.06^{2}=1.00.\end{aligned}}}

For (ii),

{\displaystyle {\begin{aligned}E_{R}&=0.033^{2}=0.001,\\E_{T}&=(\left(7.91/7.40\right)\times 0.967^{2}=1.00.\end{aligned}}}

Note that ${\displaystyle E_{R}+E_{T}=1}$ (within the accuracy of the calculations).