Hale's DMO

Given the geometry in Figure we have proved in Part 1 that  

$$t_0^2={t_h^2-\left( {{2h \cos \theta} \over v} \right)^2}$$ (10)

where t₀ is the traveltime from CMP to the reflector and back, t_h is the source-receiver traveltime, 2h is the distance between source and receiver and v is the velocity of the medium. Note that in this situation the reflection point R in the nonzero-offset case differs from the actual reflection point S in the zero-offset case.

 

HaleDMO
Figure 12 Geometry for a dipping reflector in a constant velocity medium. Notice that the reflection point for the nonzero-offset ray R is different from the zero-offset reflection point S. The dipping angle is $\theta$.

Hale (1984) uses equation (10) to write  

$$t_0^2={t_h^2-{{4h^2} \over v^2}+ {{4h^2{\sin \theta}^2} \over {v^2}}}$$ (11)

and we observe that the NMO corrected time is  

$$t_{n}^2= t_h^2-{{4h^2} \over v^2}.$$ (12)

Substituting t_n in equation (11) we have  

$$t_0^2={t_{n}^2+{{4h^2{\sin \theta}^2} \over {v^2}}}.$$ (13)

Let us consider a pressure field p(t_h,y,h) recorded as a function of nonzero-offset time t_h, midpoint y and offset h. In a constant-offset section we set the variable h to a constant value, so we have a 2-D field p(t_h,y;h=h₀). For all the values of the offset h we have a 3-D field p(t_h,y;h). We define a new field p_n (t_n,y,h) as  

$${p_{n} (t_{n},y,h)} \equiv { p( {\sqrt{ t_{n}^2+{{4h^2} \over {v^2}}}},y,h) }$$ (14)

obtained by replacing the value of the constant-offset traveltime t_h in p(t_h,y,h) by its expression in equation (12)

$$t_h^2={t_n^2+{{4h^2} \over v^2}}.$$

Note that for a constant value of h this transformation amounts to shifting a value in a trace from t_h to t_n.

Next we define another field p₀(t₀,y,h) as  

$${p_{0}(t_0,y,h)} \equiv {p_{n}(\sqrt{t_0^2-{{4h^2\sin^2 \theta} \over v^2}},y,h)}$$ (15)

obtained by replacing the value of the NMO corrected traveltime t_n in p_n(t_n,y,h) by its expression in equation (13). Equation (15) is dip dependent as it contains the variable $\sin \theta$.The new field p₀(t₀,y,h) so far is unknown and further computations are needed to determine it. However equation (15) formally represents a mapping from a NMO corrected field to a DMO corrected field. Remember again that in this formulation the nonzero-offset reflection point R does not correspond to the zero-offset reflection point S.

So far in equation (15) the only variable that we cannot easily determine is $\sin \theta$ so we will try to find a transformation to express $\sin \theta$ as a function of other variables. We have in a zero-offset section

$${\sin \theta} = { {v \Delta t_0 } \over {2 \Delta y}}$$

as seen in Figure . In equation (9) we proved that for a dipping segment we have  

$${{dt_0} \over {dy}}={{k_y} \over {\omega_0}}={{ 2 \sin \theta} \over v}$$ (16)

Now we need to Fourier transform the pressure field p₀(t₀,y,h) to take advantage of the new variables $k_y,\omega$ that we used in equation (16). We have  

$$P_0(\omega_0,k_y,h)={\int_{t_0}dt_0 \int_{y}dy {e^{i (\omega_{0} t_0- k_y y) }}p_0(t_0,y,h)}.$$ (17)

We can use the mapping we defined in equation (15) and replace p₀(t₀,y,h) by p_n(t_n,y,h) in equation (17). By changing the variable of integration we need to calculate the Jacobian of the transformation (13). We have

$$t_0={\sqrt{t_n^2+ h^2 {k_y^2 \over \omega_0^2} } }$$

and

$$dt_0={t_{n} \over {\sqrt{t_n^2+h^2 {k_y^2 \over \omega_0^2}}}} dt_n.$$ (18)

We can now rewrite equation (17) as  

$$P_0(\omega_0,k_y,h)={\int_{t_n}dt_n\int_{y}d_y {t_{n} \over ... ...mega_{0}{\sqrt{t_n^2+h^2 {k_y^2 \over \omega_0^2}}}- k_y y)}}},$$ (19)

which is Hale's DMO by Fourier transform.