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\markright{Converting slant stacks --- W.S. Harlan}
\title{Tieman's conversion of common-midpoint slant stacks to common-source}

\author{William S. Harlan}

\date{1996}

\maketitle

% \section{Objectives}
% \begin{enumerate}
% \item 
% \item  
% \item  
% \end{enumerate}
% 
%\section{Summary}


%\section{Conclusions}

%\section{Introduction}


%\section{Recommendations}

%\section{Plans for Future Work}

%\section{Acknowledgments}

%\section{Methods}
Hans Tieman spoke to the Stanford Exploration Project on Jan 23, 1996
about depth imaging with slant slacks.  Among his techniques
is a clever method of converting slant stacks of midpoint gathers
into equivalent slant stacks of source gathers.
A source gather best represents a physical experiment
that can be modeled easily by wave-equation methods.
Midpoint gathers, however, better include the coherence of steep
reflections and better avoid aliasing.
A conversion takes advantage of the best of both domains.

\subsection{The data}
Seismic amplitudes $d(s,r,t)$ are
recorded over time $t$ as a function of surface horizontal 
positions for source $s$ 
and receiver $r$. Although these positions are one-dimensional we must also
be prepared to think of them as vectors indicating a surface
position.  The time coordinate is sampled evenly and densely enough that
we can think of it as continuous.  

For a given source, we have a limited range of receivers (perhaps
3--5 kilometers), and vice versa.
Receiver positions are often sampled
two or four times as densely as source positions.  In marine
data, both are relatively evenly sampled, but a spatial Fourier transform
must pay attention to aliasing or
edge effects from the short span.  Land data will be much more
arbitrarily sampled.

Define the coordinates of full offset $h \equiv r-s$ 
and midpoint $y \equiv  (s+r)/2$.  Resorted data can be written as
$d(s=y-h/2, r=y+h/2, t)$. The well-sampled midpoint coordinate covers
the entire span of the survey.  

\subsection{A conventional slant stack}
Slant stacks are commonly applied to unsorted data, one
shot at a time.  This form is well suited to deconvolution
of multiple reflections from flat reflectors.  Such multiple
reflections are periodic at zero-offset, but not at a single
finite offset $h$.  

A slant stack attempts to describe our recorded data as
a sum of dipping lines.
A dip $p_s$ will measure the slope of
time with offset holding a source position constant.
\begin{equation}
p_s  \equiv \left.{ \partial t \over \partial h }\right|_{s}
\label{ps}
\end{equation}
With ideal sampling and infinite offsets, 
this equation would describe a plane-wave
source on the surface.  A plane wave reflecting from
flat reflectors would produce periodic multiples at any $p_s$.
Predictive deconvolutions can detect this periodicity and
remove multiple reflections.

The simplest slant-stack sums data over all lines within
a feasible range of dips.
Let $\tau_s$ be the intersection at zero offset of our imaginary
plane wave in the shot gather.
\begin{equation}
S(s,p_s , \tau_s ) \equiv \int d(s,r=s+h,t=\tau_s + p_s h ) dh
\label{ss}
\end{equation}
In practice the integral over offset $h$ must be a discrete
sum with a limited range of offsets.

The inverse of this transform looks much like another slant
stack, with some adjustments of the spectrum.  Papers
are readily available to explain this inverse.  I will concentrate
instead on the conversion of one type of slant stack to another.

\subsection{The Fourier version}
Because the time axis is well sampled and unaliased, we can safely
Fourier transform the data between time $t$ and frequency $f$:  
\begin{equation}
d (s,r,t) \equiv \int \exp( i 2 \pi f t ) \tilde d(s,r,f) df .
\label{fd}
\end{equation}
Tildes will indicate Fourier transforms.
Transform the slant stack from $\tau_s$ to its frequency $f_s$:
\begin{equation}
\tilde S (s,p_s,f_s) = 
\int \exp( -i 2 \pi f_s \tau_s ) S (s,p_s,\tau_s) d \tau_s .
\label{fss}
\end{equation}
The slant stack simplifies numerically.
Substitute the transform (\ref{fd}) into the slant stack (\ref{ss}),
then take the transform (\ref{fss}) of both sides of the equation:
\begin{equation}
\tilde S (s,p_s ,f_s ) = 
   \int \int \int \exp ( -i 2 \pi f_s \tau_s ) 
    \exp( i 2 \pi f (\tau_s + p_s h)  ) 
	\tilde d (s,r=s+h,f) dh df d \tau_s.
\label{a}
\end{equation}
Rearranging terms, we reduce integrals
\begin{eqnarray}
\tilde S (s,p_s ,f_s ) &=&
   \int \int \{ \int \exp[ -i 2 \pi (f_s - f) \tau_s ] d \tau_s \} 
    \exp( i 2 \pi f  p_s h ) \tilde d (s,r=s+h,f) dh df \nonumber\\
&=&
   \int \int \{ \delta (f_s - f) \} 
    \exp( i 2 \pi f  p_s h ) \tilde d (s,r=s+h,f) dh df \nonumber\\
&=&
   \int \exp( i 2 \pi f_s  p_s h ) \tilde d (s,r=s+h,f=f_s) dh .
\label{b}
\end{eqnarray}
The second step uses the Fourier transform of a delta function
$\delta(f) = \int \exp ( -i 2 \pi f t ) dt$.
The third uses
the behavior of a delta function in an integral
$ \int g(f) \delta (f-f_0 ) df = g(f_0)$.

\subsection{The midpoint gather}
We should prefer a transform of a single source gather
because these gathers correspond to a physical experiment
that can be modeled easily by wave-equation methods.
Unfortunately,
reflections from dipping layers and point scatters may have
a very complicated expression in a source gather. 
We may be obliged to use many dips to capture their coherence.
Worse, many reflections will have minimum times at finite offset,
and a slant stack will alias some of their energy.

If the data are first sorted by midpoint $y$ and half-offset $h$,
then reflections from dipping lines and from points will
still remain symmetric about zero offset.  A slant
stack of a midpoint gather will better capture the
coherence of the reflections: 
\begin{equation}
Y(y,p_y, \tau_y ) \equiv \int d ( s=y-h/2, r=y+h/2, t=\tau_y + p_y h ) dh
\label{c}
\end{equation}
where
\begin{equation}
p_y = \left. {\partial t \over \partial h } \right|_y .
\label{d}
\end{equation}
The Fourier version of a common-midpoint slant stack 
can be derived exactly as before.  
Let $f_y$ be the Fourier frequency of $\tau_y$:
\begin{equation}
\tilde Y (y,p_y ,f_y ) =
   \int \exp( i 2 \pi f_y  p_y h ) \tilde d (s=y-h/2,r=y+h/2,f=f_y ) dh .
\label{e}
\end{equation}
Unfortunately, this slant stack does not correspond to any
single seismic experiment, and wave-equation modeling is much
more awkward. 

\subsection{Conversion of midpoint to source gather}
Fortunately, we can convert this common-midpoint transform (\ref{e}) into 
an equivalent common-source transform (\ref{b}).
Let us make two additional Fourier transforms over spatial dimensions
of $s$ and $y$ for the spatial frequencies $k_s$ and $k_y$:
\begin{eqnarray}
\tilde {\tilde S} (k_s, p_s, f_s) &=& 
\int \exp( -i 2 \pi k_s s ) \tilde S (s, p_s , f_s) ds  \nonumber\\
&=& \int \int \exp( -i 2 \pi k_s s ) 
   \exp( i 2 \pi f_s  p_s h ) \tilde d (s,r=s+h,f=f_s) dh ds
\label{f}
\end{eqnarray}
and
\begin{eqnarray}
&\tilde {\tilde Y}& (k_y, p_y, f_y) 
= \int \exp( -i 2 \pi k_y y ) \tilde Y (y, p_y , f_y) dy  \nonumber\\
&=& \int \int \exp( -i 2 \pi k_y y ) 
   \exp( i 2 \pi f_y  p_y h ) \tilde d (s=y-h/2,r=y+h/2,f=f_y ) dh dy .
\label{g}
\end{eqnarray}
To place the second integral (\ref{g}) in the form of the
first (\ref{f}), we should change the variables of integration
from $h$ and $y$ to $h$ and $s$.  (The Jacobian of this transformation
is $\partial(h,y)/\partial(h,s)=1$.)  Substituting $y=s+h/2$ we get
\begin{eqnarray}
&\tilde {\tilde Y}& (k_y, p_y, f_y)  \nonumber\\
&=& \int \int \exp[ -i 2 \pi k_y (s+h/2 ) ]
   \exp( i 2 \pi f_y  p_y h ) \tilde d (s,r=s+h,f=f_y ) ds dy \nonumber\\
&=& \int \int \exp( -i 2 \pi k_y s )
   \exp[ i 2 \pi f_y h (p_y - k_y/2f_y ) ] \tilde d (s,r=s+h,f=f_y ) ds dy 
\nonumber\\
&=& \tilde {\tilde S} ( k_s = k_y , p_s = p_y - k_y/2f_y , f_s = f_y) .
\label{h}
\end{eqnarray}
Thus, a two-dimensional stretch of the midpoint-gather transform
becomes equivalent to the source-gather transform.
For a given dip over offset in a midpoint gather
$p_y$, we can identify a dip over midpoint
\begin{equation}
- k_y/f_y =
\left. {\partial \tau_y \over \partial y}\right|_{p_y} =
\left. {\partial t \over \partial y}\right|_h .
\end{equation}
The adjustment of $p_s = p_y - k_y/2f_y$ subtracts
half of this midpoint dip from the offset dip.
With a careful application of the chain rule, and carefully
distinguishing partial derivatives, we could arrive at the
same result
\begin{equation}
\left. {\partial t \over \partial h}\right|_s =  
\left. {\partial t \over \partial h}\right|_y + {1 \over 2}
\left. {\partial t \over \partial y}\right|_h .
\end{equation}

\end{document}