Iterative deblending of simultaneous-source seismic data using seislet-domain shaping regularization

Backward operator

In equation 5, $\mathbf{B}$ is an approximate inverse of $\mathbf{F}$ . Recall that the dithering operator applies a time-domain random time shift to the input seismic data. Taken in the frequency domain, the dithering operator takes the following form:

$$\mathbf{T} = \mathbf{\mathcal{F}}^{-1}\mathbf{P}\mathbf{\mathcal{F}},$$ (9)

where $\mathbf{\mathcal{F}}$ and $\mathbf{\mathcal{F}}^{-1}$ are forward and inverse Fourier transforms, respectively, and $\mathbf{P}$ is a $N\times N$ diagonal block phase-shift operator given by

$$\mathbf{P}=diag(\mathbf{P}_1, \mathbf{P}_2, \mathbf{P}_3, \cdots, \mathbf{P}_N),$$ (10)

where $\mathbf{P}_n$ denotes the individual phase shift operator for $n$ th trace and can be expressed as a $M\times M$ diagonal matrix:

$$\mathbf{P}_n = diag(\overbrace{1,1,1, \cdots, 1}^{M})*\exp(-i\omega\delta t_n).$$ (11)

Here, $w$ denotes the angular frequency, $\delta t_n$ denotes the random dithering time of $n$ th trace, and $M$ and $N$ in equations 10 and 11 denote the number of temporal samples and number of traces, respectively. $diag(\cdot)$ denotes a diagonal matrix.

Considering that the Fourier operator (with symmetric normalization) and the phase shift operator are both unitary, which means $\mathbf{\mathcal{F}}^{-1}=\mathbf{\mathcal{F}}^{T}$ and $\mathbf{P}^{-1}=\mathbf{P}^T$ , it is easy to see that

$$\begin{split}\mathbf{T}^T &=(\mathbf{\mathcal{F}}^{-1}\mathbf... ...thbf{P}\mathbf{\mathcal{F}})^{-1} =\mathbf{T}^{-1}. \end{split}$$ (12)

Thus, we conclude that the dithering operator is also a unitary operator.
Furthermore we notice that

$$\mathbf{F}^T= \left[\begin{array}{cc} \mathbf{I} & \mathbf{T} \... ...{I} & \mathbf{T} \mathbf{T}^{-1} & \mathbf{I} \end{array}\right]= \mathbf{F},$$

and:

$$\mathbf{F}^T\mathbf{F}=\left[\begin{array}{cc} \mathbf{I} & \math... ...I} & \mathbf{T} \mathbf{T}^{-1} & \mathbf{I} \end{array}\right]= 2\mathbf{F}.$$ (13)

The least-squares solution of equation 3 is therefore:

$$\mathbf{\hat{m}}=(\mathbf{F}^T\mathbf{F})^{-1}\mathbf{F}^{T}\math... ...{2}\mathbf{F}^{-1}\mathbf{F}^T\mathbf{\tilde{d}}=\frac{1}{2}\mathbf{\tilde{d}}.$$ (14)

According to equation 14, an appropriate choice for $\mathbf{B}$ is simply $\mathbf{B}=\frac{1}{2}[\mathbf{I}\quad\mathbf{0};\mathbf{0}\quad\mathbf{I}]$ . This form of the derived backward operator is also referred to as scaled pseudodeblending (Mahdad, 2012).

The dithering operator is unitary only if the time-shift range is small and so the constructive summation of the useful components (events) between different traces can be ignored. Even if this condition isn't fully met, we can use the concept of interference to generalize the meaning of the dithering operator $\mathbf{T}$ . Although, in this case, the backward operator might not be most appropriately chosen as half of the identity operator, we can still use it as an approximation.

Iterative deblending of simultaneous-source seismic data using seislet-domain shaping regularization