-\bar Z_0 & 1        &     .    &  .  \\                .         &-\bar Z_1 &     1    &  .  \\                .         & .        &-\bar Z_2 &  1        \end{array}        \right]\left[        \begin{array}{c}                \tilde s_0 \\                \tilde s_1 \\                \tilde s_2 \\                \tilde s_3                \end{array}        \right]\eq\left[        \begin{array}{c}                u_0 \\                u_1 \\                u_2 \\                u_3                \end{array}        \right] \ \eq \bold u\label{eqn:toomuch}\end{equation}where $\tilde s(z)$ (summed over frequency) is the migrated image.The adjointness of equation~(\ref{eqn:matrecur}) and (\ref{eqn:toomuch})seems obvious,but it is not the elementary form we are familiar withbecause the matrix multiplies the {\em  output}(instead of multiplying the usual {\em  input}).To prove the adjointness, notice thatequation~(\ref{eqn:matrecur}) is equivalent to $\bold u = {\bold M}^{-1}\bold s$ whoseadjoint, by definition,is $\tilde{\bold s} = ({\bold M}^{-1})' \bold u$which is $\tilde{\bold s} = {(\bold M')}^{-1} \bold u$(because of the basic mathematical factthat the adjoint of an inverse is the inverse of the adjoint)which gives $\bold M' \tilde\bold s = \bold u$which is equation~(\ref{eqn:toomuch}).\parWe observe the wavefield only on the surface $z=0$,so the adjointness ofequations~(\ref{eqn:matrecur})and (\ref{eqn:toomuch})is academic because it relates the wavefield at all depthswith the source at all depths.We need to truncate $\bold u$ to its first coefficient $u_0$since the upcoming wave is known only at the surface.This truncation changes the adjoint in a curious way.We rewrite equation~(\ref{eqn:matrecur})using a truncation operator $\bold T$ that isthe row matrix $\bold T = [1,0,0,\cdots ]$ getting$\bold u_0 = \bold T\bold u = \bold T \bold M^{-1} \bold s$.Its adjoint is$\hat {\bold s} = (\bold M^{-1})' \bold T' \bold u_0'                = (\bold M')^{-1} \bold T' \bold u_0' $or$\bold M' \hat {\bold s} = \bold T'\bold u_0$which looks like\begin{equation}\bold M' \, \tilde {\bold s}  \eq\left[        \begin{array}{cccc}                1         & .        &     .    &  .  \\                -\bar Z_0 & 1        &     .    &  .  \\                .         &-\bar Z_1 &     1    &  .  \\                .         & .        &-\bar Z_2 &  1        \end{array}        \right]\left[        \begin{array}{c}                \tilde s_0 \\                \tilde s_1 \\                \tilde s_2 \\                \tilde s_3                \end{array}        \right]\eq\left[        \begin{array}{c}                u_0 \\                0 \\                0 \\                0                \end{array}        \right] \ \label{eqn:conjrecur}\end{equation}The operator \ref{eqn:conjrecur} is a recursion\sx{recursion, downward continuation}beginning from $\tilde s_0 = u_0$and continuing downward with\begin{equation}\tilde s_k \eq \bar Z_{k-1} \  \tilde s_{k-1}\label{eqn:imaging}\end{equation}\parA final feature of the migration applicationis that the image is formed from $\tilde \bold s$by summing over all frequencies.Although I believe the mathematics above%and the code in subroutine \texttt{gazadj()} \vpageref{/prog:gazadj},I ran the dot product test to be sure!%\progdex{gazadj}{phase shift mig.}\noindentFinally, a few small details about the code.The loop on spatial frequency {\tt ikx}begins at {\tt ikx=2}.The reason for the 2, instead of a 1,is to omit the Nyquist frequency.If the Nyquist frequency were to be included,it should be divided into one half at positive Nyquist and one halfat negative Nyquist, which would clutter the codewithout adding practical value.Another small detail is that the loopon temporal frequency {\tt iw}begins at {\tt iw=1+nt/2}which effectly omits negative frequencies.This is purely an economy measure.Including the negative frequencies would assure that thefinal image be real, no imaginary part.Omitting negative frequencies simply gives an imaginary partthat can be thrown away,and gives the same real image, scaled by a half.The factor of two speed up makes these tiny compromises well worthwhile.\subsection{Vertical exaggeration example}\inputdir{vofz}To examine questions of \bx{vertical exaggeration}and spatial \bx{resolution}we consider a line of point scatters alonga $45^\circ$ dipping line in $(x,z)$-space.We impose a linear velocity gradient such asthat typically found in the Gulf of Mexico, i.e.~$v(z)=v_0+\alpha z$with $\alpha=1/2 s^{-1}$.Viewing our point scatterers as a function of traveltime depth,$\tau = 2\int_0^z dz/v(z)$in Figure~\ref{fig:sagmod}we see, as expected,that the points,although separated by equal intervals in $x$,are separated by shorter time intervals with increasing depth.The points are uniformly separatedalong a straight line in $(x,z)$-space,but they are nonuniformly separated along a {\em  curved} linein $(x,\tau)$-space.The curve is steeper near the earth's surfacewhere $v(z)$ yields the greatest vertical exaggeration.Here the vertical exaggeration is about unity (no exageration)but deeper the vertical exaggeration is less than unity(horizontal exaggeration).\sideplot{sagmod}{width=3.00in}{        Points along a 45 degree slope as seen        as a function of traveltime depth.        }%\newslideApplying %subroutine \texttt{gazadj()} \vpageref{/prog:gazadj}the points spray out into hyperboloids (like hyperbolas, but not exactly)shown in Figure~\ref{fig:sagdat}.The obvious feature of this synthetic data is that the hyperboloidsappear to have different asymptotes.\sideplot{sagdat}{width=3.00in}{        The points of Figure~\protect\ref{fig:sagmod} diffracted        into hyperboloids.        }%\newslideIn fact, there are no asymptotes because an asymptoteis a ray going horizontal at a more-or-less constant depth,which will not happen in this modelbecause the velocity increases steadily with depth.\par(I should get energetic and overlay these hyperboloidson top of the exact hyperbolas of the Kirchhoff method,to see if there are perceptible traveltime differences.)\subsection{Vertical and horizontal resolution}In principle, migration converts hyperbolas to points.In practice, a hyperbola does not collapse to a point,it collapses to a {\em  focus.}A focus has measurable dimensions.Vertical resolution is easily understood.For a given frequency, higher velocity gives longer vertical wavelengthand thus less resolution.When the result of migration is plotted as a function oftraveltime depth $\tau$ instead of true depth $z$, however,enlargement of focus with depth is not visible.\parHorizontal resolution works a little differently.Migration is said to be ``good'' because it increases spatial \bx{resolution}.It squeezes a large hyperbola down to a tiny focus.Study the focal widths in Figure~\ref{fig:sagres}. %Notice the water-velocity focuses hardly broaden with depth.We expect some broadening with depth because the latehyperbolas are cut off at their sides and bottom(an aperture effect),but there is no broadening herebecause the periodicity of the Fourier domainmeans that events are not truncated but wrapped around.\plot{sagres}{width=6.00in,height=2.25in}{        Left is migration back to a focus        with a constant, water-velocity model.        Right is the same,        but with a Gulf of Mexico velocity,        i.e.~the hyperboloids of Figure~\protect\ref{fig:sagdat}        migrated back to focuses.	Observe focus broadening with depth.        }%%\newslide\parWhen the velocity increases with depth,wider focuses are found at increasing depth.Why is this?Consider each hyperbola to be made of many short plane wave segments.Migration moves all the little segments on top of each other.The sharpness of a focus cannot be narrowerthan the width of each of the many plane-wave segmentsthat superpose to make the focus.The narrowest of these plane-wave segmentsis at the steepest part of a hyperbola asymptote.Deeper reflectors (which have later tops)have less steep asymptotes becauseof the increasing velocity.Thus deeper reflectors with faster RMS velocities have wider focusesso the deeper part of the image is more blurred.A second way to understand increased blurring with depthis from equation~(\ref{eqn:kxofvp}),that the horizontal frequency $k_x=\omega p = \omega v^{-1}\sin\theta$is independent of depth.The steepest possible angle occurs when $|\sin\theta| = 1$.Thus, considering all possible angles,the largest $|k_x|$ is $|k_x|=|\omega|/v(z)$.Larger values of horizontal frequency $|k_x|$could help us get narrower focuses,but the deeper we go (faster velocity we encounter),the more these high frequencies are lostbecause of the evanescent limit $|k_x|\le |\omega/v(z)|$.The limit is where the ray goes no deeper but bends around andcomes back up again without ever reflecting.Any ray that does this many times is said to be a surface-trapped wave.It cannot sharpen a deep focus.%\par%When the velocity increases with depth,%wider focuses are found at increasing depth.%Why is this?%Deep waves travel fast and so they must have a long wavelength.%As a deeper wave comes up its wavelength shrinks,%but its horizontal component $k_x$%remains constant according to Snell's law%(equation~(\ref{eqn:kxofvp}))%so the horizontal component measured at the earth's surface%is exactly the same as at the bottom of the ray%where it is determined only by the frequency and the material velocity.%The sharpness of a focus cannot be narrower%than the width of each of the many plane-wave segments%that superpose to make the focus,%and the narrowest of these plane-wave segments is at the%the steepest part of a hyperbola asymptote.%The deeper hyperboloids (which have later tops) have less steep asymptotes%which is why deep focuses are wider.%%The depth of a scatterer is indicated by the hyperbola top.%The long wavelengths from great depth show up%as less-steep asymptotes on the deeper hyperboloids.\subsection{Field data migration}Application of subroutine~\texttt{gazadj()} \vpageref{/prog:gazadj}to the Gulf of Mexico data set processed in earlier chaptersyields the result in Figure~\ref{fig:wgphase}.\plot{wgphase}{width=6.20in,height=8.5in}{        Phase shift migration of        Figure~\protect\ref{vela/fig:agcstack}.%        Press button for movie to compare to stack and Kirchhoff migration of%        Figure~\protect\ref{vela/fig:wgstack}.        }%\newslide%\subsection{Downward continuation movie such as jfcmig}%This was Jim's idea and I don't see a way to build it%without doing something artificial.%This fits better in a finite-difference migration chapter.\begin{exer}\itemDevise a mathematical expression for a plane wavethat is an impulse function of time with a propagation angle of 15$^\circ$from the vertical $z$-axis in the plus  $z$  direction.Express the result in the domain of\item[{}] (a) $\ \ (t,x,z)$\item[{}] (b) $\ \ ( \omega ,x,z)$\item[{}] (c) $\ \ ( \omega , k_x , z)$\item[{}] (d) $\ \ ( \omega ,p,z)$\item[{}] (e) $\ \ ( \omega , k_x , k_z)$\item[{}] (f) $\ \ (      t , k_x , k_z)$                \todo{  %remove this temporarily to save two sheets of paper.\itemSuppose that you are able to observe some shear waves atordinary seismic frequencies.Is the spatial \bx{resolution} better, equal, or worse than usual?Why?\itemEvolution of a wavefield with time is described by\begin{equation}p(x , z , t) \ =\ \int  \int \  \left[\,P( k_x , k_z , t=0 )\, e^{ -i \omega ( k_x , k_z ) t } \right] \ e^{ ik_x x + ik_z z }  dk_x \, dk_z \end{equation}Let  $ P(k_x , k_z , 0  ) $  be constant, signifying a pointsource at the origin in  $(x , z)$-space.Let  $t$  be very large, meaning thatphase = $ \phi = [ - \omega ( k_x , k_z ) \ +$$ k_x \  (x / t) \ +\  k_z \  (z /t) ] t $  inthe integration is rapidly alternating withchanges in  $ k_x $  and  $ k_z$.Assume that the only significant contributionto the integral comes when the phase is stationary,that is,where  $ \partial \phi / \partial k_x $  and$ \partial \phi / \partial k_z $  both vanish.Where is the energy in $(x , z , t)$-space?\itemDownward continuation of a wave is expressed by\begin{equation}p(x , z , t) \ =\  \int \int \  \left[\,P( k_x , z=0 , \omega ) \  e^{ik_z ( \omega , k_x ) z } \right] \ e^{{-i} \omega t + ik_x x } \  d \omega \, dk_x\end{equation}Let  $ P(k_x , 0 , \omega  ) $  be constant, signifying a pointsource at the origin in  $(x , t)$-space.Where is the energy in $(x , z , t)$-space?                     } %remove this temporarily to save two sheets of paper.\end{exer}%\iex{Exer/Intro}{assignment}%\iex{Exer/Shift}{assignment}%\iex{Exer/2Dft}{assignment}%\iex{Exer/Phasemod}{assignment}%\iex{Exer/Phasedown}{assignment}%\iex{Exer/Phasemig}{assignment}%\iex{Exer/Stolt1}{assignment}%\iex{Exer/Stolt2}{assignment}