equation~(\ref{eqn:huygens}),and in three dimensions,a hyperbola of revolution (umbrella?)carrying a time-derivative waveform.\subsection{Hankel tail}\inputdir{hankel}\parThe waveform in equation~(\ref{eqn:huygens}) often arises in practice(as the 2-D Huygens wavelet).Because of the discontinuities on the left side of equation~(\ref{eqn:huygens}),it is not easy to visualize.Thinking again of the time derivativeas a convolution with the doublet $(1,-1)/\Delta t$,we imagine the 2-D Huygen's wavelet as a positive impulse followedby negative signal decaying as $-t^{-3/2}$.This decaying signal is sometimes called the ``\bx{Hankel tail}.''In the frequency domain$-i\omega= |\omega |e ^ {-i90^\circ}$has a 90 degree phase angle and $\sqrt{-i\omega}= |\omega |^{1/2} e ^ {-i45^\circ}$has a \bx{45 degree phase angle}.\opdex{halfint}{half derivative}{52}{68}{filt/imag}%\newslide\parIn practice, it is easiest to representand to apply the 2-D Huygen's wavelet in the frequency domain.Subroutine \texttt{halfint()} \vpageref{/prog:halfint} is provided for that purpose.Instead of using $\sqrt{-i\omega}$ whichhas a discontinuity at the Nyquist frequencyand a noncausal time function,I use the square root of a causal representationof a finite difference, i.e.~$\sqrt{1-Z}$,which is well behaved at the Nyquist frequencyand has the advantage that the modeling operator is causal(vanishes when $t<t_0$).Fourier transform is done using subroutine \texttt{ftu()} \vpageref{/prog:ftu}.Passing an impulse function into subroutine {\tt halfint()}gives the response seen in Figure~\ref{fig:hankel}.\sideplot{hankel}{width=2.0in}{  Impulse response (delayed) of finite difference operator of half order.  Twice applying this filter is equivalent to once applying $(1,-1)$.}\HideThis{\subsection{Huygen's secondary source}\par\bx{Huygen's secondary source} is the conceptthat a plane wavecan be regarded as broken into many point sources.Each point makes a circular wave in $(x,z)$-spaceor equivalently a hyperbola in $(x,t)$-space.The hyperbolas shown in Figure~\ref{fig:hyplay}are all tangent to the horizontal line.Observe that the line density on the plot is greatnear the line but weaker away from it.We begin analysis from the assumption thateach hyperbola carries a simple impulse function of time.We will add these hyperbolas togetherand discover that the sum is a plane wave,but the waveform is not an impulse function of time,but another function with a tail in the time domain andan $\omega^{-1/2}$ amplitude spectrum.From this we conclude that if the original waveformon the hyperbolahad an $\omega^{1/2}$ amplitude spectrum(instead of being an impulse with a constant spectrum)then we would have an improvedrepresentation of the plane wave as a sum of points.The phase of this $\sqrt{\omega}$spectrum should be such that the final waveform is causalas it appears in Figure~\ref{fig:hankel},indicating the validity of representationof the numerical representation of $\sqrt{-i\omega}$found in subroutine \texttt{halfint()} \vpageref{/prog:halfint}.More theoretical details are found in IEI.\subsection{Huygen's secondary source in two dimensions}First we calculate the density of curvesas a function of time.The density goes to infinity as the hyperbola separation$\Delta x $ tends to zeroin Figure~\ref{fig:hyplay}.Then the hyperbolas sum to a plane wavewhich carries the waveform that we want to know.\activesideplot{hyplay}{width=2.0in}{ER}{	Many hyperbolas tangent to a line.	}\parWhen a hyperbolic event carries a wavelet,the wavelet covers an area in the $(t,x)$-plane.Define this area as that between the two hyperbolasshown in Figure~\ref{fig:hyparea}.These two curves, $x_1(t)$ and $x_2(t)$ are\begin{eqnarray}t &=& v^{-1}\ \sqrt{z^2+x_1^2}		\\t &=& v^{-1}\ \sqrt{z^2+x_2^2} +\Delta t\end{eqnarray}\activesideplot{hyparea}{width=2.0in}{ER}{	Detailed view of one of the many hyperbolas	in figure~\protect\ref{fig:hyplay}.	Two hyperbolas are separated by $\Delta t$.	An impulsive signal is defined as one that	is zero everywhere but between the two hyperbolas	where its amplitude is $1/\Delta t$.	}\parIn the limit $\Delta t \rightarrow 0$,the waveform is an impulse functionwhose temporal spectrum is constant.In that limit, the many waveforms in Figure~\ref{fig:hyplay}superpose giving a great concentration at the apex.Since the many hyperbolas are identical with one another,the time dependence of their sum is (within a scale factor)the same time dependence of the shaded area of the single hyperbolain Figure~\ref{fig:hyparea}.\parThe separation between the two curves, $\Delta x$ is\begin{equation}\Delta x \eq x(t_1) - x(t_2) \eq   \sqrt{ t          ^2v^2-z^2}  \ -\    \sqrt{(t-\Delta t)^2v^2-z^2}\end{equation}The area in Figure~\ref{fig:hyparea} is a length, integrated over time.Between the two hyperbolas of separation $\Delta t$,we take a constant signal of strength $1/\Delta t$ whichadds up to an amplitude at time $t$ of $A_{\rm hyp}(t)$.\begin{equation}A_{\rm hyp}(t) \eq{\Delta x \over \Delta t} \eq{d \over dt} \ \sqrt{t^2 v^2-z^2} \eq{t v^2\over \sqrt{t^2 v^2-z^2} }\quad \quad {\rm for} ~ t>z/v\label{eqn:Shyp}\end{equation}To avoid much clutter that belongs in a mathematics bookrather than in a seismology book,we concentrate our attention on the discontinuity itself.Mathematically, this amounts to ignoring all but the highest frequencies.Seismologically, we say thatafter a signal like equation~(\ref{eqn:Shyp}) passes through the filtersand gain control processes typical of data collection,the numerator $t$ does not warrant comment,but the power of $t$ at the pole itself is important.\parBy making this high frequency approximation,we are setting aside some calculationsthat we should be able to do for a constant velocity mediumbut which would not apply to a medium with velocity as a function of depth.Divergence considerations, for example,suggest that the amplitude of each hyperbola should not bea constant, but should drop off proportional to $t^{-1/2}$.Another example is that of parabolas insteadof hyperbolas, i.e.~$x(t)=\sqrt{t-t_0}$.(When velocity changes with depth, we no longer have hyperbolas,but the tops of the traveltime curve could be approximated by a parabola.)Then the amplitude is $A_{\rm par}(t)= 1/\sqrt{t-t_0}$.Notice that the divisor in equation~(\ref{eqn:Shyp}) for the hyperbola$\sqrt{t^2 v^2-z^2}=\sqrt{tv-z}\sqrt{tv+z}$has the same half power discontinuity at $z=tv$ as the parabola.Thus the amplitude spectrum at high frequencies for eitherthe parabola or the hyperbola will drop off as the inversehalf power of frequency according to equation~(\ref{eqn:halfint}).To get a plane wave with the constant spectrum of an impulseinstead of this inverse-square-root spectrum,we must use hyperbolas (or parabolas),not carrying an impulse function,but instead carrying a square-root spectrum(to cancel the {\em  inverse}-square-root spectrum that arisesfrom superposing the hyperbolas.)Equation~(\ref{eqn:huygens}) gives us a causal waveform with the desired spectrum.Thus the decomposition of an impulsive 2-D planewave into hyperbolas requires the hyperbolas to carrythe ``half-order differential'' waveform given in equation~(\ref{eqn:huygens}).\subsection{Three dimensions}Interestingly,the Huygen's wavelet is different in \bx{three dimensions}than in two dimensions.In three dimensions, there is no Hankel tail.In three dimensions we have the definition$x^2+y^2=r^2$ andfrom $t^2v^2=x^2+y^2+z^2$ we find on a plane of constant $z$,that the equation for a circle expanding with time is$r=\sqrt{t^2v^2-z^2}$.Between time $t$ and $t+\Delta t$ is a ring with an area$2\pi r \Delta r$.Taking the signal amplitude in the ring to be $1/\Delta t$,analogous to equation~(\ref{eqn:Shyp}) the amplitude at time $t$ is\begin{eqnarray}A(t) &=&  {\rm step}(t-z/v) \ 	2\pi r \ {\Delta r \over \Delta t }				\\A(t) &=&  {\rm step}(t-z/v)  \ 	2\pi \sqrt{t^2v^2-z^2}\ {d~\over dt}\ \sqrt{t^2v^2-z^2}		\\A(t) &=&  {\rm step}(t-z/v) \ 2\pi v^2 \ t \label{eqn:tstep}\end{eqnarray}As before, in seismologywe are interested in the high frequency behaviorso the scaling $t$ in equation~(\ref{eqn:tstep})is not nearly so important as is the step function.By equation~(\ref{eqn:iterint})the step function causes the spectrum to decay as $\omega^{-1}$.Our original erroneous assumptionthat Huygen's hyperbola of revolutionshould carry a positive impulseleads to the contradictionthat an impulsive plane wavedecomposed into Huygen's sources and added together againdoes not preserve the constant spectrumof the original impulsive waveform.We can get the missing $\omega$ back into the spectrumby having the hyperbola of revolution carrya time-derivative filter instead of an impulse.Thus in three dimensions,a plane wave can be regarded, approximately,as the superposition of many hyperboloidal responses,each carrying a $d/dt$ waveform.\subsection{Amplitude versus offset on Huygen's wave}\sx{amplitude}\sx{AVO}The analysis above did not show that in two dimensionsthe hyperbolas should carry the scaling factor $x/t^{3/2}$and in three dimensions the scaling factor $x/t^2$.These scaling factors can be explainedas combination of two parts.The first part of the scaling factor isa divergence correction dueto expansion of the wavefront.The second part of the scaling factor isa cosine function $(\tau/t)$ called the obliquity functionwhich is explained as the difference between a point sourceand a source like a short line or patch which has theability to radiate preferentially along the directionof the maximum of the cosine.\parYou might fear that analysis of weights and waveshapeswould be a big chore in stratified media $v(z)$.Luckily it is all made easy by Fourier domainmethods in chapters~\ref{ft1/paper:ft1} and~\ref{dwnc/paper:dwnc}.\begin{exer}\item	If we input a one layer model $\delta(t-t_0)\,{\rm const}(x)$	to subroutine \texttt{kirchslow()} \vpageref{/prog:kirchslow},	we find an impulse function followed by a	${\rm step}(t-t_0)/\sqrt{t}$ waveform.	What data would result from	a 3-D version of {\tt kirchslow()}	on the model $\delta(t-t_0)\,{\rm const}(x,y)$?\item   At the bottom of a well (deep hole in the ground) is an explosion.	An air wave propagates up the well and when emerging at the surface	it becomes a spherical wave.	At the mouth of the well	$(x,y,z)=(0,0,0)$ an air wave time function $f(t)$ is observed.	The speed of sound in air is a constant 340 m/s.	\par	Write an equation	for the air wave	expanding spherically from the top of the well.	Assume wavelengths and observation distances are 	are large compared to the well diameter.	Be sure your equation carries the correct time-dependent waveform.\item	Given the signal ${\rm step}(t)/\sqrt{t}$	\begin{enumerate}	\item What is its {\em  energy} spectrum?	\item Would you say the spectral color is red, white, or blue?	\end{enumerate}\end{exer}}% end of HideThis\section{References}\reference{	Special issue on fast Fourier transform, June 1969:	IEEE Trans.~on Audio and Electroacoustics	(now known as IEEE Trans.~on Acoustics, Speech,	and Signal Processing), {\bf AU-17}, entire issue (66-172).	}%\newpage%\showiex%\par%\iex{Exer/Inter2}{inter2}%\iex{Exer/ed1D}{Lab2}%\newpage