# Complex coefficient of reflection

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 3.6a

Using the expression $\psi =Ae^{\mathrm {j} \omega \left(r/V-t\right)}$ to represent a plane wave incident on a plane interface, show that a complex coefficient of reflection,

{\begin{aligned}R=a+\mathrm {j} b,\quad a^{2}+b^{2}<1,\end{aligned}} R [defined by equation (3.6a) below] corresponds to a reduction in amplitude by the factor $(a^{2}+b^{2})^{1/2}$ and an advance in phase by $\tan ^{-1}\left(b/a\right)$ .

### 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, $\delta _{1}=0=\delta _{2}$ , so the equations reduce to

{\begin{aligned}A_{1}+A_{2}&=A_{0},\\Z_{1}A_{1}-Z_{2}A_{2}&=-Z_{1}A_{0},\end{aligned}} $Z_{1}$ , $Z_{2}$ , being impedances (see problem 3.2). The solution of these equations is

 {\begin{aligned}R&={\frac {A_{1}}{A_{0}}}={\frac {\rho _{2}\alpha _{2}-\rho _{1}\alpha _{1}}{\rho _{2}\alpha _{2}+\rho _{1}\alpha _{1}}}={\frac {Z_{2}-Z_{1}}{Z_{2}+Z_{1}}},\end{aligned}} (3.6a)

 {\begin{aligned}T&={\frac {A_{2}}{A_{0}}}={\frac {2\rho _{1}\alpha _{1}}{\rho _{2}\alpha _{2}+\rho _{1}\alpha _{1}}}={\frac {2Z_{1}}{Z_{2}+Z_{1}}},\end{aligned}} (3.6b)

$R$ and $T$ being the coefficient of reflection and coefficient of transmission, respectively. Although equations (3.6a,b) hold only for normal incidence, the definitions $R=A_{1}/A_{0}$ and $T=A_{2}/A_{0}$ are valid for all angles of incidence. A negative value of $R$ means that $A_{1}$ in equation (3.6a) is opposite in sign to $A_{0}$ . Since $e^{\mathrm {j} \pi }=-1$ , the minus sign is equivalent to adding $\pi$ to the phase of the waveform in part (a), that is, reversing the phase. Note that (except for phase reversal) $R$ is independent of the direction of incidence on the interface; however, the magnitude of $T$ depends upon this direction, and when necessary, we shall write $T\downarrow$ and $T\uparrow$ to distinguish between the two values. Note the following relations:

 {\begin{aligned}R+T\downarrow =1,\quad T\uparrow +T\downarrow =2,\quad T\uparrow T\downarrow =E_{T},\end{aligned}} (3.6c)

where $E_{T}$ is the fraction of energy transmitted as defined in equation (3.7a).

Euler’s formulas (see Sheriff and Geldart, 1995, p. 564) express $\sin x$ and $\cos x$ as

 {\begin{aligned}\sin x=\left(e^{\mathrm {j} x}-e^{-\mathrm {j} x}\right)/2\mathrm {j} ,\quad \cos x=\left(e^{\mathrm {j} x}+e^{-\mathrm {j} x}\right)/2.\end{aligned}} (3.6d)

The hyperbolic sine and cosine are defined by the relations

 {\begin{aligned}\sinh x=\left(e^{x}-e^{-x}\right)/2,\quad \cosh x=\left(e^{x}+e^{-x}\right)/2.\end{aligned}} (3.6e)

### Solution

Writing $\psi 0=A_{0}e^{\mathrm {j} \omega \left(r/V-t\right)}=A_{0}e^{\mathrm {j} \left(\kappa r-\omega t\right)}$ , we have

{\begin{aligned}\psi _{1}=R\psi _{0}=\left(a+\mathrm {j} b\right)A_{0}e^{\left(\kappa r-\omega t\right)}.\end{aligned}} But $\left(a+\mathrm {j} b\right)=(a^{2}+b^{2})^{1/2}e^{\mathrm {j} \phi }$ where $\tan \phi =b/a$ (see Sheriff and Geldart, 1995, section 15.1.5), so

{\begin{aligned}\psi _{1}=(a^{2}+b^{2})^{1/2}A_{0}e^{\mathrm {j} (kr-\omega t+\phi )}.\end{aligned}} Since $(a^{2}+b^{2})^{1/2}<1$ , the amplitude is reduced by this factor and the phase is advanced by $\phi$ ## Problem 3.6b

Show that an imaginary angle of refraction $\theta _{2}$ (see Figure 3.1b) in equations (3.2e,f,h,i) leads to a complex value of $R$ and hence to phase shifts.

### Solution

Let $\theta _{2}=\pi /2-\mathrm {j} \theta$ , where $\theta$ is real. Then, using equations (3.6d,e), we have

{\begin{aligned}\sin \theta _{2}=\cos \left(\mathrm {j} \theta \right)=\cosh \theta ,\qquad \cos \theta _{2}=\sin \left(\mathrm {j} \theta \right)=\mathrm {j} \sinh \theta ,\end{aligned}} hence some of the coefficients in equations (3.2e,f,h,i) are imaginary and $R$ and $T$ will in general be complex, so phase shifts will occur.

## Problem 3.6c

Show that when $R$ is negative, $A_{2}>A_{0}$ .

### Solution

If $R$ is negative, $Z_{1}>Z_{2}$ in equation (3.6a), so $2Z_{1}>\left(Z_{1}+Z_{2}\right)$ . Therefore, from equation (3.6b) we see that $T>1$ , and since $T=A_{2}/A_{0}$ , $A_{2}>A_{0}$ .