AMO impulse response

To derive a time-space representation of the AMO impulse response from its frequency-wavenumber representation, we evaluate the stationary-phase approximation of the inverse Fourier transform along the midpoint coordinates.

The $\rm DMO$ operator and its inverse, ${\rm DMO^{-1}}$, can be defined in the zero-offset frequency $\omega_{0}$ and the midpoint wavenumber ${\bf k}$ as

$$\EQNLABEL{dmo.eq} \rm DMO= \int d{t}_{1}J_1e^{-i\omega_ot_1\... ...ft(\frac{{{\bf k}}\cdot{{\bf h}_{1}}}{\omega_ot_1}\right)}^2}}}$$ (63)

$$\rm DMO^{-1}= \int d\omega_o J_2e^{+i\omega_ot_2\sqrt{1+{{\l... ...t{{\bf h}_{2}}}{\omega_ot_2}\right)}^2}}}. \EQNLABEL{dmoinv.eq}$$ (64)

The $\rm AMO$ operator is given by the cascades of $\rm DMO$ and ${\rm DMO^{-1}}$ and its impulse response can be written as,

$$\rm AMO= \frac{1}{4\pi^2}\int d{{\bf k}} e^{-i{{\bf k}}\cdot... ...}}}{\omega_ot_2}\right)}^2}}} \right)}. \EQNLABEL{amo_freq.eq}$$ (65)

The derivation of the stationary-phase approximation of the integral in $d{{\bf k}}$ is similar to the one presented in Black et al. (1993b) for deriving a time-space formulation for the conventional DMO impulse response. We begin by changing the order of the integrals and rewriting amo_freq.eq as

$$\begin{eqnarray} \rm AMO& =&\frac{1}{4\pi^2} \int dt_1 \int d\omega_o \int d{{\b... ... \Delta m}} \right]}\;. \nonumber \\ \EQNLABEL{amo_freq_change.eq}\end{eqnarray}$$ (66)

The phase of this integral is,

$$\Phi\equiv \omega_o(t_1\eta_1-t_2\eta_2)-{{\bf k}}\cdot{{\bf \Delta m}}, \EQNLABEL{phase_app.eq}$$ (67)

where,

$$\eta_1=\sqrt{1+{{\left(\frac{{\bf k}.{\bf h}_{1}}{\omega_ot_... ...{{\bf k}.{\bf h}_{2}}{\omega_ot_2}\right)}^2}}. \EQNLABEL{eq12}$$ (68)

Next we make the following change of variables and let

$$\beta_1=\frac{{\bf h}_{1}.{\bf k}}{\omega_ot_1} \hspace{.5 i... ...ta_2=\frac{{\bf h}_{2}.{\bf k}}{\omega_ot_2}. \ \EQNLABEL{eq16}$$ (69)

Therefore, $\eta_1$ and $\eta_2$ become

$$\eta_1=\sqrt{1+\beta_1^2}\hspace{.5 in} {\rm and} \hspace{.5 in} \eta_2=\sqrt{1+\beta_2^2}. \EQNLABEL{eq20}$$ (70)

The derivatives of $\eta_1$ and $\eta_2$ with respect to the in-line component of the wavenumber k_x and the cross-line component k_y can be written as

$$\begin{eqnarray} \frac{\partial{\eta_1}}{\partial{k_x}}=\frac{h_{1x}}{w_ot_1}\fr... ...h_{2y}}{w_ot_1}\frac{\beta_2}{\sqrt{1+\beta_2^2}}. \EQNLABEL{eq28}\end{eqnarray}$$ (71)

Making one more change of variables, we let

$$\nu_1=\frac {\beta_1} {\sqrt{1+\beta_1^2}}\hspace{.5 in} {\r... ... in} \nu_2=\frac{\beta_2} {\sqrt{1+\beta_2^2}}. \EQNLABEL{eq32}$$ (72)

Setting the derivative of the phase $\Phi$ to zero yields the system of equations:

$$\left\{ \begin{array} {ll} h_{1x}\nu_1 - h_{2x}\nu_2 = \Del... ...h_{2y}\nu_2 = \Delta m_y \end{array} \right. \ \EQNLABEL{eq36}$$ (73)

which we solve for $\nu_1$ and $\nu_2$ (i.e., $\eta_1$ and $\eta_2$) at the stationary path ${\bf k}_0$. The determinant of the system is given by

$$\Delta=h_{2x}h_{1y}-h_{1x}h_{2y} = h_{1}h_{2}\sin \Delta \theta, \EQNLABEL{eq40}$$ (74)

and the solutions for $\nu_1$ and $\nu_2$ are

$$\nu_{01}=\frac {\Delta m\sin (\theta_{2}-\Delta \varphi)} {h_1 \sin \Delta \theta}, \EQNLABEL{eq44}$$ (75)

and

$$\nu_{02}=\frac {\Delta m\sin (\theta_{1}-\Delta \varphi)} {h_2 \sin \Delta \theta}. \EQNLABEL{eq48}$$ (76)

Now we need to evaluate the phase function $\Phi$ along the stationary path ${\bf k}_0$.By respectively multiplying the equations in eq36 by k_0x and k_0y and summing them together we obtain,

$${{\bf k}_0}\cdot{{\bf \Delta m}} = \frac{{\omega_o t_1 \beta... ... t_2 \beta_{02}}^2}{\sqrt{1+\beta_{02}^2}}. \EQNLABEL{kdotx.eq}$$ (77)

Substituting this relationship into the expression of the phase function [equation phase_app.eq] we obtain

$$\Phi_{0}= \omega_o\left(\frac{t_1}{\sqrt{1+\beta_{01}^2}}- \... ...mega_o\left(\frac{t_1}{\eta_{01}}-\frac{t_2}{\eta_{02}}\right).$$ (78)

The phase function along the stationary path is thus peaked for

$$t_2=t_1\frac{\eta_{02}}{\eta_{01}}=t_1\frac{\sqrt{1-\nu_{01}^2}}{\sqrt{1-\nu_{02}^2}} \EQNLABEL{ratio.eq}$$ (79)

Substituting equations eq44 and eq48 into ratio.eq we obtain amo_surf.eq of the main text:

$${t}_{2}= {t}_{1}\frac{h_{2}}{h_{1}}\sqrt{ \frac{h_{1}^2\sin^... ...^2\sin^2(\theta_1-\Delta \varphi)}}. \EQNLABEL{amo_surf_app.eq}$$ (80)