Matrix expression of stacking operators

Stacking along hyperbolas is a linear operator mapping a two-dimensional (t,x)-space into two-dimensional $(\tau,m)$-space. The operator may therefore be expressed as a matrix in a four-dimensional space. This is not convenient for our imagination. The representation of the operator in two dimensions is preferable. This can be achieved by ordering all the columns of (t,x)-space into one vector, and the same for $(\tau,m)$-space. To get an idea of what the matrix looks like, let us write the equation transforming (t,x)-space into $(\tau,m)$-space when the number of samples is six, two offsets and two sloths. The nearest neighbor interpolation was used.

$$\pmatrix{ \pmatrix{ u_{11} \cr u_{21} \cr u_{31} \cr u_{41} ... ...cr d_{22} \cr d_{32} \cr d_{42} \cr d_{52} \cr d_{62} \cr} \cr}$$ (14)

Each submatrix in the matrix represents ${\bf NMO}$ transformation (Claerbout, 1989). The transpose matrix applied on u gives d:

$$\pmatrix{ \pmatrix{ \tilde{d}_{11} \cr \tilde{d}_{21} \cr \t... ...r u_{22} \cr u_{32} \cr u_{42} \cr u_{52} \cr u_{62} \cr} \cr}$$ (15)

Generally, if we denote $\bold d_{i}$ as the i-th trace of a gather, $\bold u_{i}$ as the i-th trace of a velocity analysis panel and $\bold N_{ij}$ as an NMO matrix transforming i-th offset with j-th sloth, then the velocity stacking transformation may be expressed by the equation

$$\pmatrix{ \bold u_1 \cr \bold u_2 \cr \vdots \cr \bold u_k \... ...pmatrix{ \bold d_1 \cr \bold d_2 \cr \vdots \cr \bold d_n \cr}$$ (16)

If we want to compute $\bold d$ from $\bold u$ by least squares, then the matrix $\bold N^{\bold T} \bold N$

$${\bf N^T N}= \pmatrix{ \sum \bold N^{\bold T}_{i1}\bold N_{i... ... N_{i2} & \ldots & \sum \bold N^{\bold T}_{in}\bold N_{in} \cr}$$ (17)

should be inverted. This matrix is not even approximately unitary, as it is for ${\bf NMO}$ (Claerbout, 1989). For our example we have  

$$\bold N^{\bold T}\bold N = \pmatrix{ . & . & . & . & . & . &... ... & 3 & . \cr . & . & . & . & 1 & 1 & . & . & . & . & . & 2 \cr}$$ (18)

Hence just a transpose operation to stacking cannot be used for inverting the velocity analysis panel.

The rank of the matrix ${\bf N^T N}$ cannot be higher than max(n.nt,k.nt), where nt is the number of samples of a trace. From this it follows that the matrix ${\bf N^T N}$ is singular if the number of traces in the $(\tau,m)$-space is less than the number of traces in (t,x)-space. Generally, this matrix may be expected to be singular even for the opposite case, as we can see from equation (18).