THE NORMAL MOVEOUT MAPPING

Recall the traveltime equation ().

$$\begin{eqnarray} v^2 \, t^2 &=& z^2 \ +\ x^2 \\ t^2 &=& \tau^2 \ +\ { x^2 \over v^2 } \end{eqnarray}$$ (8) (9)

where $\tau$ is traveltime depth. This equation gives either time from a surface source to a receiver at depth $\tau$,or it gives time to a surface receiver from an image source at depth $\tau$.

A seismic trace is a signal d(t) recorded at some constant x. We can convert the trace to a ``vertical propagation'' signal $m(\tau)=d(t)$by stretching t to $\tau$.This process is called ``normal moveout correction'' (NMO). Typically we have many traces at different x distances each of which theoretically produces the same hypothetical zero-offset trace. Figure 1 shows a marine shot profile before and after NMO correction at the water velocity. You can notice that the wave packet reflected from the ocean bottom is approximately a constant width on the raw data. After NMO, however, this waveform broadens considerably--a phenomenon known as ``NMO stretch."

 

stretch Figure 1 Marine data moved out with water velocity. Input on the left, output on the right. Press button for movie sweeping through velocity (actually through slowness squared).

The NMO transformation ${\bf N}$ is representable as a square matrix. The matrix ${\bf N}$ is a $(\tau,t)$-plane containing all zeros except an interpolation operator centered along the hyperbola. The dots in the matrix below are zeros. The input signal d_t is put into the vector $\bold d$.The output vector $\bold m$--i.e., the NMO'ed signal--is simply (d₆,d₆,d₆, d₇,d₇, d₈,d₈, d₉, d₁₀, 0). In real life examples such as Figure 1 the subscript goes up to about one thousand instead of merely to ten.

 

$${\bf m\eq Nd} \eq \left[ \begin{array} {c} m_1 \\ m_2... ...\ d_6 \\ d_7 \\ d_8 \\ d_9 \\ d_{10} \end{array} \right]$$ (10)

You can think of the matrix as having a horizontal t-axis and a vertical $\tau$-axis. The 1's in the matrix are arranged on the hyperbola $t^2=\tau^2+x_0^2/v^2$.The transpose matrix defining some ${\bf \tilde d}$from $\bold m$ gives synthetic data ${\bf \tilde d}$ from the zero-offset (or stack) model $\bold m$, namely,

 

$${\bf \tilde d \eq N' m } \eq \left[ \begin{array} {c} \t... ...\\ m_6 \\ m_7 \\ m_8 \\ m_9 \\ m_{10} \end{array} \right]$$ (11)

A program for nearest-neighbor normal moveout as defined by

equations (10) and (11) is nmo0(). Because of the limited alphabet of programming languages, I used the keystroke z to denote $\tau$. 

subroutine nmo0( adj, add, slow,    x, t0, dt, n,zz,  tt )
integer  it, iz, adj, add,                        n
real  xs, t , z,           slow(n), x, t0, dt, zz(n), tt(n)
call adjnull(    adj, add,                     zz,n,  tt,n)
do iz= 1, n {   z = t0 + dt*(iz-1)              # Travel-time depth
                xs= x * slow(iz)
                t = sqrt ( z * z + xs * xs)
                it= 1 + .5 + (t - t0) / dt      # Round to nearest neighbor.
                if( it <= n )
                        if( adj == 0 )
                                tt(it) = tt(it) + zz(iz)
                        else
                                zz(iz) = zz(iz) + tt(it)
        }
return; end

A program is a ``pull'' program if the loop creating the output covers each location in the output and gathers the input from wherever it may be. A program is a ``push'' program if it takes each input and pushes it to wherever it belongs. Thus this NMO program is a ``pull'' program for doing the model building (data processing), and it is a ``push'' program for the data building. You could write a program that worked the other way around, namely, a loop over t with z found by calculation $z=\sqrt{t^2/v^2-x^2}$.What is annoying is that if you want a push program going both ways, those two ways cannot be adjoint to one another.

Normal moveout is a linear operation. This means that data can be decomposed into any two parts, early and late, high frequency and low, smooth and rough, steep and shallow dip, etc.; and whether the two parts are NMO'ed either separately or together, the result is the same. The reason normal moveout is a linear operation is that we have shown it is effectively a matrix multiply operation and that operation fulfills ${\bf N(d_1+d_2) = Nd_1+Nd_2}$.