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

{\displaystyle {\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 ${\displaystyle (a^{2}+b^{2})^{1/2}}$ and an advance in phase by ${\displaystyle \tan ^{-1}\left(b/a\right)}$.

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

\displaystyle \begin{align} A_{1} + A_{2} &= A_{0}, \\ Z_{1} A_{1} -Z_{2} A_{2} &=-Z_{1} A_{0}, \end{align}

${\displaystyle Z_{1}}$, ${\displaystyle Z_{2}}$, being impedances (see problem 3.2). The solution of these equations is

 \displaystyle \begin{align} 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{align} (3.6a)

 \displaystyle \begin{align} 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{align} (3.6b)

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

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

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

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

 {\displaystyle {\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

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

{\displaystyle {\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 ${\displaystyle \left(a+\mathrm {j} b\right)=(a^{2}+b^{2})^{1/2}e^{\mathrm {j} \phi }}$ where ${\displaystyle \tan \phi =b/a}$ (see Sheriff and Geldart, 1995, section 15.1.5), so

{\displaystyle {\begin{aligned}\psi _{1}=(a^{2}+b^{2})^{1/2}A_{0}e^{\mathrm {j} (kr-\omega t+\phi )}.\end{aligned}}}

Since ${\displaystyle (a^{2}+b^{2})^{1/2}<1}$, the amplitude is reduced by this factor and the phase is advanced by ${\displaystyle \phi }$

## Problem 3.6b

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

### Solution

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

{\displaystyle {\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 ${\displaystyle R}$ and ${\displaystyle T}$ will in general be complex, so phase shifts will occur.

## Problem 3.6c

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

### Solution

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