%\tilde z_2' z_2%\end{equation}%Here you have two tests,%one on the real part%and one on the imaginary part.%%%%Let $x$ denote a vector of reals and $z$ denote a vector of complex values.%%Suppose one is derived from the other by what appears to be%%a linear operator $A$.%%The question is whether the operator $A$%%will/can pass the dot product test,%%and the answer is surprising.%%Given a random real vector $x$ and a random complex vector $z$,%%we define two new vectors, $\tilde x$ and $\tilde z$.%%\begin{eqnarray}%%\tilde z &=& A x	\\%%\tilde x &=& A' z%%\end{eqnarray}%%where $A'$ is the adjoint of $A$.%%The dot-product question is whether %%\begin{equation}%%\tilde x'x \eq \tilde z' z \quad ?%%\end{equation}%%The answer is no.%%The left side is necessarily real%%whereas the right side is necessarily complex.%%Since $z$ and $\tilde z$ come from different sets of random numbers,%%there is no reason for the imaginary part $\Im \tilde z' z$ to vanish.%%The paradox above arises because%%you cannot make a linear operator $A'$%%whose function is to truncate an imaginary part of a complex number,%%or a linear operator $A$ whose function is to append a zero imaginary part%%to a real number to make a complex number.%%\par%Now for the second surprise.%Consider real to complex Fourier transform,%the operator that takes real valued signals and%converts them to complex valued frequency functions.%The real part of the frequency function is an even function of frequency,%and the imaginary part of the frequency function is odd.%Thus $\Im \tilde z' z$ is a product of an even and an odd function,%so it is odd.%The integral of an odd function vanishes,%and it is this integral that is the imaginary part of the dot product.%Thus, the real to complex FT operator passes the dot product test.\begin{exer}\itemMost time functions are real.Their imaginary part is zero.Show that this means that  $F( \omega , k )$  canbe determined from  $F( - \omega , - k )$.\itemWhat would change in Figure~\ref{fig:ft2dofpulse}if the pulse were moved(a) earlier on the $t$-axis,  and(b) further on the $x$-axis?What would change in Figure~\ref{fig:ft2dofpulse}if insteadthe time axis were smoothed with (1,4,6,4,1)and the space axis with (1,1)?\itemWhat would Figure~\ref{fig:airwave}look like on an earth with half the earth velocity?\itemNumerically (or theoretically)compute the two-dimensional spectrumof a plane wave [$\delta (t-px)$], wherethe plane wave has a randomly fluctuating amplitude:say, rand$(x)$ is a random number between $\pm 1$,and the randomly modulated plane wave is[$(1 \ +\ .2\,{\rm rand}(x)) \, \delta (t-px)$].\itemExplain the horizontal ``layering'' in Figure~\ref{fig:plane4}in the plot of  $P( \omega  , x)$.What determines the ``layer'' separation?What determines the ``layer'' slope?\end{exer}\subsection{Magic with 2-D Fourier transforms}\inputdir{brad}We have struggled through some technical detailsto learn how to perform a 2-D Fourier transformation.An immediate reward next is a few "magical" results on data.\parIn this book waves go down into the earth; they reflect;they come back up; and then they disappear.In reality after they come back up they reflectfrom the earth surface and go back down for another episode.Such waves, called multiple reflections,in real life are in some places negligiblewhile in other places they overwhelm.Some places these multiply reflected waves can be suppressedbecause their RMS velocity tends to be slowerbecause they spend more time in shallower regions.In other places this is not so.We can always think of making an earth model,using it to predict the multiply reflected waveforms,and subtracting the multiples from the data.But a serious pitfall is that we would need to have the earth modelin order to find the earth model.\parFortunately, a little Fourier transform magic goes a longway towards solving the problem.Take a shot profile $d(t,x)$.Fourier transform it to$D(\omega, k_x)$.For every $\omega$ and $k_x$, square this value $D(\omega, k_x)^2$.Inverse Fourier transform.InFigure~\ref{fig:brad1}we inspect the result.For the squared part the $x$-axis is reversedto facilitate comparison at zero offset.A great many reflections on the raw data (right) carry over into thepredicted multiples (left).If not, they are almost certainly primary reflections.This data shows more multiples than primaries.\plot{brad1}{width=6in,height=8.5in}{  Data (right) with its FT squared (left).}\parWhy does this work?  Why does squaring the Fourier Transform of the rawdata give us this good looking estimate of the multiple reflections?Recall $Z$-transforms $Z=e^{i\omega\Delta t}$.A $Z$-transform is really a Fourier transform.Take a signal that is an impulse of amplitude r at time $t=100\Delta t$.Its $Z$-transform is $r Z^{100}$.The square of this $Z$-transform is $r^2 Z^{200}$,just what we expect of a multiple reflection ---squared amplitude and twice the travel time.That explains vertically propagating waves.When a ray has a horizontal component,an additional copy of the ray doubles the horizontal distance traveled.Remember what squaring a Fourier transformation does -- a convolution.Here the convolution is over both $t$ and $x$.Every bit of the echo upon reaching the earth surfaceturns around and pretends it is a new little shot.Mathematically, every point in the upcoming wave $d(t,x)$ launches a replicaof $d(t,x)$ shifted in both time and space -- an autoconvolution.\parIn reality, multiple reflections offer a considerable numberof challenges that I'm not mentioning.The point here is just that FT is a goodtool to have.\subsection{Passive seismology}\parSignals go on and on, practically forever.Sometimes we like to limit our attention to something more limited suchas their spectrum, or equivalently, their autocorrelation.We can compute the autocorrelation in the Fourier domain.We multiply the FT times its complex conjugate$D(\omega, k_x) \overline{D(\omega, k_x)}$.Transforming back to the physical domain we see Figure~\ref{fig:brad2}.We expect a giant burst at zero offset (upper right corner).We do not see it because it is "clipped",i.e. plot values above some threshhold are plotted at that threshhold.I could scale the plot to see the zero-offset  burst,but then the interesting signals shown here would be too weak to be seen.\par\plot{brad2}{width=6in,height=8.5in}{  The 2-D autocorrelation of a shot profile resembles itself.}\parFigure~\ref{fig:brad2} shows us that the 2-D autocorrelationof a shot profile shares a lot in common with the shot profile itself.This is interesting news.If we had a better understanding of thiswe might find some productive applications.We might find a situation where we do not have (or do not want)the data itself but we do wish to build an earth model.For example, suppose we have permanently emplaced geophones.The earth is constantly excited by seismic noise.Some of it is man made;  some results from earthquakeselsewhere in the world; most probably results from naturalsources such as ocean waves, wind in trees, etc.Recall every bit of acoustic energy that arrives at the surface from belowbecomes a little bit of a source for a second reflection seismic experiment.So, by autocorrelating the data of hours and days durationwe convert the chaos of continuing microseismic noiseto something that might be the impulse response of the earth,or something like it.Autocorrelation converts a time axis of length of days to one of seconds.From the autocorrelation  we might be able to draw conclusions in usual ways,alternately,we might learn how to make earth models from autocorrelations.\parNotice from Figure~\ref{fig:brad2}that since the first two seconds of the signal vanishes(travel time to ocean bottom),the last two seconds of the autocorrelation must vanish(longest nonzero lag on the data).\par%The 2-D autocorrelation of 2-D random noise is an impulse%at the origin.%All data has random noise in it; and this data has plenty.%The huge burst in the neighborhood of the origin%of the autocorrelation%is not visible because it%is simply clipped away by the plotting software.\parThere are many issues onFigure~\ref{fig:brad2} to intrigue an interpreter(starting with signal polarity).We also notice that the multiples on the autocorrelationdie off rapidly with increasing offset and wonder why,and whether the same is true of primaries.But today is not the day to start down these paths.\parIn principal an autocorrelation is not comparable to the raw dataor to the ideal shot profile becauseforming a spectrum squares amplitudes.We can overcome this difficulty by use ofmultidimensional spectral factorization ---but that's an advanced mathematical conceptnot defined in this book.See my other book, Image Estimation.\section{THE HALF-ORDER DERIVATIVE WAVEFORM}%First we will review the calculus of step functions%and power functions and their Fourier transforms%leading up to the half-order differential operator.%With this background we can answer the question,%what is the difference between a line%and the sequence of short dashes that make up the line?%Likewise, what is the difference between a plane%and a dense array of tiny patches that make up the plane?%Is each tiny patch the same as a point source?%These questions will be answered from the viewpoint%of wave propagation where it is conceptually and computationally%convenient to make plane waves by superposing an array%of spherical waves.%We will find that a plane wave with an impulsive waveform%can be made from a superposition of circles (or hyperbolas)%or spheres (or hyperboloids of revolution)%but that in three dimensions the hyperboloid carries%not an impulsive waveform,%but the time derivative $d/dt$,%and in two dimensions the hyperbola carries the half-order derivative waveform.%These waveforms are generally clear on synthetic data%but on field data%they are usually obscured by waveforms from other sources.%%\subsection{Fractional order operators}\parCausal integrationis represented in the time domainby convolution with a step function.In the frequency domain this amounts to multiplication by $1/(-i\omega)$.(There is also delta function behavior at $\omega=0$which may be ignored in practice and sinceat $\omega=0$, wave theory reduces to potential theory).Integrating twice amounts to convolution by a ramp function,$t\, {\rm step}(t)$, which in the Fourier domain is multiplication by$1/(-i\omega)^2$.Integrating a third time is convolution with$t^2\, {\rm step}(t)$ which in the Fourier domain is multiplication by$1/(-i\omega)^3$.In general\def\eq{\quad =\quad}\begin{equation}\label{eqn:iterint}t^{n-1}\ {\rm step}(t) \eq {\rm FT}\ \left( { 1 \over (-i\omega)^n} \right)\end{equation}Proof of the validity of equation~(\ref{eqn:iterint}) for integer values of $n$is by repeated indefinite integration which also indicatesthe need of an $n!$ scaling factor.Proof of the validity of equation~(\ref{eqn:iterint}) for fractional values of $n$would take us far afield mathematically.Fractional values of $n$, however,are exactly what we need to interpret Huygen's secondary wave sources in 2-D.The factorial function of $n$ in the scaling factor becomes a gamma function.The poles suggest that a more thorough mathematical study of convergenceis warranted, but this is not the place for it.\par%We will see that theThe principal artifactof the hyperbola-sum method of 2-D migration is the waveformrepresented by equation~(\ref{eqn:iterint}) when $n=1/2$.For $n=1/2$, ignoring the scale factor,equation~(\ref{eqn:iterint}) becomes\begin{equation}\label{eqn:halfint}{1\over \sqrt{t}} \ {\rm step}(t) \eq{\rm FT}\ \left( { 1 \over \sqrt{-i\omega}} \right)\end{equation}A waveform that should come out to be an impulseactually comes out to be equation~(\ref{eqn:halfint}) because Kirchhoffmigration needs a little more than summing or spreading on a hyperbola.To compensate for the erroneous filter response of equation~(\ref{eqn:halfint})we need its inverse filter.We need $\sqrt{-i\omega}$.To see what $\sqrt{-i\omega}$ is in the time domain,we first recall that\begin{equation}\label{eqn:derivative}{d \ \over dt} \eq{\rm FT}\ \left(  -i\omega \right)\end{equation}A product in the frequency domain correspondsto a convolution in the time domain.A time derivative is like convolution with a doublet $(1,-1)/\Delta t$.Thus, fromequation~(\ref{eqn:halfint}) and equation~(\ref{eqn:derivative})we obtain\begin{equation}\label{eqn:huygens}{d \ \over dt} \ {1\over \sqrt{t}} \ {\rm step}(t) \eq{\rm FT}\ \left(  \sqrt{-i\omega} \,\right) \end{equation}Thus, we will see the way to overcomethe principal artifact of hyperbola summationis to apply the filter of equation~(\ref{eqn:huygens}).In chapter~\ref{dwnc/paper:dwnc}we will learn more exact methods of migration.There we will observe that an impulse in the earthcreates not a hyperbola with an impulsive waveformbut in two dimensions,a hyperbola with the waveform of