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Biondi, B. and G. Shan, 2002, Prestack imaging of overturned reflections by reverse time migration: 72nd Ann. Internat. Mtg., Soc. of Expl. Geophys., Expanded Abstracts, 1284-1287.

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A

This appendix analytically derives SODCIGs for one shot and well-sampled receiver wavefields in locally constant-velocity media. Let's consider the SODCIG for a dipping reflector at image point (x_m, z_m), as shown in Figure (). The dipping angle is $-\alpha$, while the reflection angle at (x_m, z_m) is $-\gamma$ (assuming positive sign for angles measured in a clockwise direction). The red circle shows the source wavefront at some time t, which intersects the dipping reflector at (x_m, z_m). If we put receivers for every location on the surface, i.e. the receiver wavefield is sufficiently densely sampled, when we downward continue or backward propagate the receiver wavefield, the circular wavefront, which is in blue, can be well reconstructed. It also intersets the dipping reflector at (x_m, z_m) at the same time t. Therefore, cross-correlation of the source wavefield and receiver wavefield will generate the image at (x_m, z_m).

 

dip.full.rec Figure 14 When the receiver wavefield is sufficiently densely sampled, the actual circular wavefront in blue can be well reconstructed.

The equation for the source wavefront in red is

$$\left( x-s \right)^2 + z^2 = \left[\frac{(x_m-s)}{\sin(-\gamma+\alpha)}\right]^2,$$ (8)

while the equation for the receiver wavefront in blue is

$$\left[x-s\cos(-2\alpha)\right]^2 + \left[z-s\sin(-2\alpha)\right]^2 = \left[\frac{(x_m-s)}{\sin(-\gamma+\alpha)}\right]^2.$$ (9)

To get the local offset gather at x_m, we have to apply the multi-offset imaging condition described in equation (1) as follows, where h is the subsurface half-offset.

$$\left\{\begin{array} {l} \left( x-h-s \right)^2 + z^2 = \le... ...rac{(x_m-s)}{\sin(-\gamma+\alpha)}\right]^2 \end{array}\right.$$ (10)

Therefore, by letting x=x_m and subtracting the second equation from the first equation, we get

$$\left( x_m-h-s \right)^2 + z^2 = \left[x_m+h-s\cos(-2\alpha)\right]^2 + \left[z-s\sin(-2\alpha)\right]^2.$$ (11)

After some algebra, we get

$$\left[2x_m-s(1+\cos(-2\alpha))\right]\left[s(\cos(-2\alpha)-1)-2h\right] + (2z-s\sin(-2\alpha))s\sin(-2\alpha) = 0.$$ (12)

Rearranging the above equation:

$$\begin{eqnarray} z &=& \frac{x_ms\sin^2(-\alpha) + x_mh - hs\cos^2(-\alpha)}{s\c... ...rac{\frac{x_m}{s} - \cos^2(-\alpha)}{\cos(-\alpha)\sin(-\alpha)}h.\end{eqnarray}$$ (13) (14) (15)

From Figure (), point (x_m,z_m) on the dipping reflector satisfies:

$$\begin{eqnarray} \tan(-\gamma+\alpha) &=& \frac{x_m - s}{z_m} \\ \tan(-\alpha) &=& \frac{z_m}{x_m}.\end{eqnarray}$$ (16) (17)

Multiplying them together yields

$$\tan(-\alpha)\tan(-\gamma + \alpha) = \frac{x_m-s}{x_m} = \frac{\frac{x_m}{s}-1}{\frac{x_m}{s}},$$ (18)

so  

$$\frac{x_m}{s} = \frac{1}{1-\tan(-\alpha)\tan(-\gamma + \alpha)}.$$ (19)

Substituting equation(19) into equation (15), we get:

$$\begin{eqnarray} z &=& -\tan\alpha x_m + \frac{\frac{x_m}{s} - \cos^2(-\alpha)}{... ...an(-\alpha-\gamma+\alpha)h \\ &=& -\tan\alpha x_m - \tan\gamma h.\end{eqnarray}$$ (20) (21) (22) (23) (24) (25) (26) (27)

Actually the above equation is true for each point (x,z) with dipping angle $\alpha$ and opening angle $\gamma$, so it can be rewritten as follows:

$$z = -\tan \alpha x - \tan \gamma h.$$ (28)

Therefore, when the receiver wavefield is sufficiently densely sampled, the SODCIG is linearly related to the angle gather.

B

This appendix analytically derives SODCIGs for one shot and one receiver in locally constant-velocity media. Let's again consider a dipping reflector with the dip angle $\alpha$, as shown in Figure . The source is located at (s,0), while receiver is located at (r,0). Since there is only one receiver, when we downward continue or backward propagate the receiver wavefield, the receiver wavefront that intersects (x_m, z_m) is the green one shown in Figure instead of the blue one shown in Figure .

 

dip.one.rec Figure 15 In the extreme situation where there is only one receiver, the receiver wavefront is the one shown here in green, instead of the blue one shown in Figure .

The image point at (x_m, z_m) on the dipping reflector certainly satisfies:

$$\begin{eqnarray} \tan(-\gamma + \alpha) &=& \frac{x_m - s}{z_m} \\ \tan(-\gamma - \alpha) &=& \frac{r - x_m}{z_m}.\end{eqnarray}$$ (29) (30)

Thus, we have:

$$z_m = \frac{x_m - s}{\tan(-\gamma+\alpha)} = \frac{r - x_m}{\tan(-\gamma - \alpha)},$$ (31)

which rearranges to  

$$x_m = \frac{s\tan(-\gamma - \alpha) + r\tan(-\gamma + \alpha)}{\tan(-\gamma - \alpha) + \tan(-\gamma + \alpha)}.$$ (32)

Since (x_m, z_m) is the intersecting point of the source wavefront and the receiver wavefront, it should satisfy the source wavefront equation:

$$\left( x-s \right)^2 + z^2 = \left[\frac{x_m-s}{\sin(-\gamma+\alpha)}\right]^2,$$ (33)

and the receiver wavefront equation:

$$\left( x-r \right)^2 + z^2 = \left[\frac{r-x_m}{\sin(-\gamma-\alpha)}\right]^2.$$ (34)

To get the SODCIG, we have to apply the multi-offset imaging conditon described in equation (1) as follows, where h is the subsurface half-offset,

$$\left\{\begin{array} {l} \left( x-h-s \right)^2 + z^2 = \lef... ...frac{(x_m-r)}{\sin(-\gamma-\alpha)}\right]^2.\end{array}\right.$$ (35)

Therefore, by letting x=x_m and summing the above two equations together, we can derive the relationship between z and h for surface location x_m:  

$$(x_m-h-s)^2 + (x_m+h-r)^2 + 2z^2 = \left[\frac{x_m-s}{\sin(-... ...)}\right]^2 + \left[\frac{r-x_m}{\sin(-\gamma-\alpha)}\right]^2$$ (36)

From equation (32), it is easy to get:  

$$x_m - s = \frac{s\tan(-\gamma - \alpha) + r\tan(-\gamma + \a... ...mma + \alpha)}{\tan(-\gamma - \alpha) + \tan(-\gamma + \alpha)}$$ (37)

and  

$$r - x_m = r - \frac{s\tan(-\gamma - \alpha) + r\tan(-\gamma ... ...ma - \alpha)}{\tan(-\gamma - \alpha) + \tan(-\gamma + \alpha)},$$ (38)

Substituting equations (37) and (38) into equation (36), after some algebra, we get

$$z^2 = -h^2 + (r-s)h+\frac{(r-s)^2}{\left[\tan(-\gamma -\alpha) + \tan(-\gamma+\alpha)\right]^2}.$$ (39)

Obviously, in this situation, depth z is not linearly related to subsurface half-offset h. When $\alpha = 0$, i.e. for a flat reflector, the above equation can be simplified as follows:

$$z^2 = -h^2 + (r-s)h + \frac{(r-s)^2}{4\tan^2\gamma}.$$ (40)