Complex coefficient of reflection

Problems in Exploration Seismology and their Solutions

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

Problem 3.6a

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

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

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 $T=A_{2}/A_{0}$, $A_{2}>A_{0}$.

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Table of Contents (book)

Also in this chapter

• General form of Snell’s law
• Reflection/refraction at a solid/solid interface and displacement of a free surface
• Reflection/refraction at a liquid/solid interface
• Zoeppritz’s equations for incident SV- and SH-waves
• Reinforcement depth in marine recording
• Reflection and transmission coefficients
• Amplitude/energy of reflections and multiples
• Reflection/transmission coefficients at small angles and magnitude
• Magnitude
• AVO versus AVA and effect of velocity gradient
• Variation of reflectivity with angle (AVA)

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