paper.tex

国外免费地震资料处理软件包

%hyperbola case all over again!
%Figure~\FIG{ottohyp} shows a three-dimensional sketch
%of the hidden hyperbola.
%\activeplot{ottohyp}{height=2.8in}{}{
%	An unexpected hyperbola in Cheops' pyramid is the
%	diffraction hyperbola on a radial-trace section.
%	(Harlan)
%	}
%\subsection{Anti-alias characteristic of dip moveout}
%\par
%You might think that if $(y,h,t)$-space is sampled
%along the $y$-axis at a sample interval  $\Delta y$,
%then any final migrated section  $P(y,z)$  would
%have a spatial resolution no better than  $\Delta y$.
%This is not the case.
%\par
%The basic principle 
%at work here has been known since 
%the time of Shannon.
%If a time function and its derivative are sampled
%at a time interval $2 \Delta t$,
%they can both be fully reconstructed provided that
%the original bandwidth of the signal is lower than  $1/(2 \Delta t)$.
%More generally, if a signal is filtered with  $m$  independent filters,
%and these  $m$  signals are sampled at an interval  $m \Delta t$,
%then the signal can be recovered.
%\par
%Here is how this concept applies to seismic data.
%The basic signal is the earth model.
%The various filtered versions of it are the constant-offset sections.
%Recall that the CDP reflection point moves up dip as the offset is increased.
%Further details can be found in a paper by
%Bolondi, Loinger, and Rocca [1982],
%who first pointed out the anti-alias properties of dip moveout.
%At a time of increasing interest in 3-D seismic data,
%special attention should be paid to the anti-alias character of dip moveout.
%\section{DIPPING-REFLECTOR RAY GEOMETRY (Black)}
%\par
%While the finite-offset traveltime curves
%and raypath positions for a reflector dipping at angle $\theta$
%from the horizontal are simple,
%they are not simple to derive.
%Before presenting the derivations,
%we will state the two principal results.  
%The first result is that 
%the traveltime curve seen at midpoint location $y$ is 
%\begin{equation}
%t^2  \eq 
%t_m^2 
%\ +\ 
%{4 \, h^2 \, \cos^2\theta \over v^2 }
%\EQNLABEL{levin1}
%\end{equation}
%where $t_m$ is the traveltime at $h=0$ at the same midpoint location.
%\par
%For a common-midpoint gather at midpoint $y$, 
%equation \EQN{levin1} looks
%like  $t^2 \,=\, t_m^2 \,+$
%$4h^2 / v_{\rm apparent}^2$.
%Thus the common-midpoint gather contains an
%{\it exact}
%hyperbola, regardless of the earth dip angle  $\theta$.
%The effect of dip is to change the asymptote of the hyperbola,
%thus changing the apparent velocity.
%The result has great significance in practical seismic processing and is
%known as Levin's dip correction [1971]:
%\begin{equation}
%v_{\rm apparent}  \eq  {v_{\rm earth}  \over  \cos\theta }
%\EQNLABEL{levin2}
%\end{equation}
%(See also Slotnick [1959]).
%This increase of stacking velocity with dip is the first complication of
%the combination of offset and dip.
%\par
%The second complication is more subtle, but equally important.
%Figure~\FIG{conflict} depicts some rays from a common-midpoint gather.
%\activesideplot{conflict}{width=3.5in}{}{
%	Finite-offset rays for dipping and flat reflectors
%	at a common midpoint M.
%	As the offset increases,
%	the shot S and receiver G move symmetrically away from M.
%	Note how the reflection point ``creeps up''
%	the dipping reflector.
%	Press button for \bx{movie}.
%	}
%Notice that each ray strikes the dipping bed at a different reflection point.
%In fact, there is a simple formula for the position of the reflection point
%along the dipping reflector:
%\begin{equation}
%\Delta L \eq {h^2 \over d} \cos\theta \ \sin\theta \ \ ,
%\EQNLABEL{cdpsmear}
%\end{equation}
%where $d$ is the perpendicular distance from the midpoint to the reflector.
%So a common-%
%\it midpoint %
%\rm gather %
%is not a 
%\it common-reflection-point
%\rm gather.
%The reflection point moves
%{\it up}
%dip with increasing offset.
%This effect is usually referred to as 
%{\it CDP smear} or {\it reflection-point smear}.
%It means that stacking the traces in a common-midpoint 
%gather does not generally create an accurate estimate of the zero-offset 
%section when there are dipping reflectors.
%\par
%\subsection{Moveout derivation}
%\par
%We now derive the two equations we have just presented, 
%starting  with a geometrical derivation of equation~\EQN{levin1}.
%Figure~\FIG{clayer} shows the raypath geometry for that go from S to R to G.
%Inspection of the figure shows
%that the traveltime along SRG is the same as the traveltime along S'RG.  
%The point S' is the ``image source,''
%the mirror-image of S in the reflector.
%\activesideplot{clayer}{height=2.7in}{}{
%	Traveltime from source  $S$ to reflection point  $R$
%	to receiver $G$ is the same as the straight-line traveltime 
%	from image
%	source  $S'$ to  $R$ to  $G$.
%	}
%\par
%Triangle S'XG is a right triangle,
%whose hypotenuse length $S'G$ gives the desired traveltime.
%Once the lengths $XG$ and $S'X$ are known,
%simple application of the Pythagorean theorem
%will yield equation~\EQN{levin1}.
%The expression for $XG$ is very easy to derive.
%Since $SG = 2h$, 
%the trigonometry of the right triangle XSG yields
%\begin{equation}
%XG \eq 2\, h\, \cos\theta \ \ \ . 
%\EQNLABEL{XG}
%\end{equation}
%The length $XS'$ is only slightly more complicated,
%being the sum
%\begin{displaymath}
%XS' \eq XS \ + \ SA \ + \ AS' \ \ . 
%\end{displaymath}
%The first term in this equation, $XS$, is again related to $SG$
%by simple trigonometry on triangle XSG:
%\begin{displaymath}
%XS \eq 2\, h\, \sin\theta   \ \ \ . 
%\end{displaymath}
%Getting the lengths $SA$ and $AS'$ uses 
%$SA = BA - BS$, where $BA$ is simply $d$, 
%the perpendicular distance from reflector to midpoint.
%The trigonometry of the right triangle BSM shows that
%$BS = h \sin\theta$ so that 
%%
%\begin{equation}
%SA \eq AS'  \eq d \ - \ h \ \sin\theta \ \ \ . 
%\EQNLABEL{ASp}
%\end{equation}
%%
%where the first equality comes from the fact that S' is the mirror image of
%S through point A.
%Thus the final expression for $XS'$ is
%\begin{equation}
%XS' \eq 2\,h\,\sin\theta \,+\, 2\,(d-h\sin\theta) \eq 2\,d
%\EQNLABEL{XSp}
%\end{equation}
%Finally, applying the Pythagorean theorem yields the desired traveltime:
%\begin{displaymath}
%v^2t^2 \eq 4\,d^2 + 4\,h^2\,\cos^2\theta \ \ , 
%\end{displaymath}
%which is equation~\EQN{levin1}
%with the definition of two-way zero-offset traveltime
%at the midpoint as
%\begin{equation}
%t_m \eq {2\,d \over v}
%\EQNLABEL{TM}
%\end{equation}
%\subsection{Smear derivation}
%\par
%Now turning to \EQN{cdpsmear},
%we want to compute the distance $RR'$ in Figure~\FIG{clayer},
%which  is the quantity $\Delta L$.
%We begin by noting that $RR'=AR'-AR$ and that 
%\begin{equation}
%AR' \eq BM \eq h \, \cos\theta \ \ \ .
%\EQNLABEL{ARp}
%\end{equation}
%Thus all we need to find out is the distance $AR$.
%From the figure it is apparent that triangle $AS'R$
%is similar to triangle $XS'G$.
%Thus it follows that
%\begin{displaymath}
%{AR \over AS'} \eq {XG \over XS'}  \ \ . 
%\end{displaymath}
%But we already know $AS'$, $XG$, and $XS'$ from equations~\EQN{ASp},
%\EQN{XG}, and \EQN{XSp} above.
%Putting these all together yields
%\begin{displaymath}
%AR \eq (d\,-\,h\,\sin\theta)\ {h\,\cos\theta \over d} \ \ \ . 
%\end{displaymath}
%Combining this result with equation~\EQN{ARp} and $RR'=AR'-AR$ yields
%\begin{displaymath}
%\Delta L \eq h\,\cos\theta - (d\,-\,h\,\sin\theta) %
%	{h\,\cos\theta \over d} \ , 
%\end{displaymath}
%which simplifies to equation~\EQN{cdpsmear}.
%\subsection{Transforming to zero offset correctly}
%\par
%The above derivations show that applying NMO and stack
%to the traces in a common-midpoint gather
%causes two problems when the reflector is dipping:
%\begin{itemize}
%\item The NMO velocity depends on the dip angle $\theta$ according to
%equation~\EQN{levin2}.
%\item  Energy from different points along the reflector are being
%stacked together as though they were from the same point.
%This happens because the finite-offset reflection point $R$ creeps 
%up the reflector away from the zero-offset reflection point $R'$ as 
%offset increases in Figure~\FIG{clayer}.
%\end{itemize}
%These facts often prevent the resulting stacked trace from being
%a good estimate of a zero-offset trace 
%recorded at midpoint location $M$ in Figure~\FIG{clayer}.
%We will now show {\em what should have happened}
%in converting the finite-offset reflections
%in Figures~\FIG{conflict} and \FIG{clayer} to zero-offset reflections.  
%%In a later section in this chapter, we will show {\em how}
%%to make this happen with a process called {\em dip-moveout}.
%\par
%Figure~\FIG{slayer} is a simplified version of Figure~\FIG{clayer},
%showing how a finite-offset dipping event
%should be converted to zero-offset.
%Recall that the finite-offset event
%in this figure corresponds to raypaths $SR$ and $RG$.
%We have drawn an additional line segment $ZR$ in this figure.
%This segment is called a {\em normal ray} because it strikes the reflector 
%at normal incidence.
%Thus the normal ray from reflection point $R$
%corresponds to what a zero-offset seismic survey
%above this reflector would record.
%Note that the zero-offset survey would ``see'' the reflection energy 
%from $R$ not at point $M$ but rather at point $Z$,
%where the normal ray hits the earth's surface.
%Thus the proper conversion from finite offset to zero offset
%must somehow ``move'' this reflection energy
%to surface location $Z$ and give it a two-way traveltime
%corresponding to the zero-offset raypath $ZR$.
%This movement to the correct zero-offset surface location
%and traveltime is a type of prestack migration called ``dip-moveout'' or
%``prestack partial migration.''
%\activesideplot{slayer}{height=2.4in}{}{
%	Reflection event from source {\bf S} to reflection point {\bf R}
%	to receiver {\bf G} needs to be ``migrated'' to zero-offset, 
%	corresponding to {\bf ZR}.
%	}
%\par
%To see what kind of movement is required,
%we first must clarify where we are moving {\em from}.
%In midpoint-offset coordinates,
%the reflection energy in Figure~\FIG{slayer} is initially
%located at midpoint $M$ with offset $2\ h$ and traveltime given by 
%equation~\EQN{levin1}.
%The distance along the surface from initial finite-offset position
%$M$ to final zero-offset position $Z$ is easily deduced from the geometry of
%the small triangle $ZPM$ in the figure.
%\begin{equation}
%\Delta y \eq  -\ ZM \eq -\ {\Delta L \over \cos\theta} \eq %
%			-\ {h^2 \over d} \sin\theta  \ \ ,
%\EQNLABEL{dmody}
%\end{equation}
%where we have used equation~\EQN{cdpsmear} and the fact that $ZP = \Delta L$.
%Similarly, the expression for two-way traveltime $t_0$ along the 
%zero-offset raypath
%$ZR$ is obtained by noting that $ZR = d - PM$, which yields
%\begin{equation}
%v\,t_0\, /\,2 \eq d            \ -\ \Delta L\  \tan\theta \eq %
%                  v\,t_m\, /\,2\ -\ {h^2 \over d} \sin^2\theta \ \ ,
%\EQNLABEL{dmot0}
%\end{equation}
%where we have used equations~\EQN{TM} and \EQN{cdpsmear}
%to get the second equality.
%Notice that both the spatial shift in equation~\EQN{dmody} and the
%temporal shift in equation~\EQN{dmot0} vanish when either dip is zero
%or offset is zero,
%so that they are definitely consequences
%of the occurrence of dip and offset together.
%	\subsection{
%	OTHER THINGS TO POSSIBLY PUT IN THIS SECTION:}
%	\begin{itemize}
%	\item Illustration of CDP smear by comparing two common-offset section
%	\begin{itemize}
%	\item A near-offset section with NMO appropriate for the steeply-dipping
%	reflector applied.
%	\item A far-offset section with NMO appropriate for the steeply-dipping
%	reflector applied.
%	\end{itemize}
%	\end{itemize}
%\section{DMO VIA DIPPING-BED ANALYSIS (BLACK)}
%\par
%In this section we investigate a two-step process for carrying out the 
%prestack migration described in the previous section.
%The first step is to convert the finite-offset data to 
%equivalent zero-offset data by a procedure that is closely related to normal
%moveout.  This procedure is most commonly called ``dip moveout''.  It is also
%known by the names ``prestack partial migration'' and ``offset continuation''
%and ``migration to zero offset.''  The second step is to perform 
%zero-offset migration on the data that has been converted to zero offset.  This
%two-step approach has considerable practical advantages over full prestack 
%migration, especially in processing 3--D seismic data.  
%%
%\subsection{From prestack migration to DMO}
%In earlier sections, we showed that the difficulty with 
%NMO and stack on common-midpoint gathers is that they do
%not produce the correct zero-offset sections when there are dipping reflectors.
%We also derived  expressions for what kind of spatial and temporal shifts
%would be required to fix things up.
%In this section, we are finally ready to derive the migration operator that
%carries out the correct transformation from finite offset to zero offset.  
%This transformation is called ``{\em dip moveout}'' or ``{\em DMO}'' for short.
%Instead of
%relying on the dipping reflector for this derivation
%we begin by asking the question ``What
%is the impulse response of the DMO operator?''  
%This is very similar to the
%question that we asked about zero-offset migration in chapter~\CHAP{krch}
%and about prestack migration in the previous section of the current chapter.
%In fact we can directly employ the prestack migration ellipse to answer our
%question.  The logic, first invented by Deregowski and Rocca[???] goes 
%like this:
%\activesideplot{ellipse2}{width=3.5in}{}{
%	The zero-offset ray, which is normal to  the prestack migration 
%	ellipse at the reflection point {\bf R}.
%	}
%\begin{enumerate}
%\item A spike in the finite-offset input data implies the
%elliptical reflector shown in Figure~\FIG{ellipse2}.
%\item Imagine collecting a zero-offset seismic section by moving
%a source and receiver along the earth's 
%surface above this elliptical reflector.  (The source and receiver in this
%thought experiment are, of course, on top of each other at point $Z$.)
%\item The resulting section is thus the zero-offset response to a spike in
%the finite-offset input data.  Thus this is the desired DMO impulse response.
%\end{enumerate}
%To make progress, we need to compute the zero-offset raypaths for the
%elliptical reflector of Figure~\FIG{ellipse2}.  To do this it is necessary
%to find the normal ray at each point along the ellipse.  Such a normal
%ray is shown as line segment $RZ$, 
%which is perpendicular to the ellipse at point $R$ in Figure~\FIG{ellipse2}.  
%The traveltime
%along $RZ$
%and the po