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The input-output relation

At the heart of the migration process is the operation of downward continuing data. Given the input data on the plane of the earth's surface z=0, we must manufacture the data that could be recorded at depth z. This is most easily done in the Fourier domain. Mathematically shifting on the z-axis is like shifting on the time t-axis. To shift on time by an amount t0 we multiply by $\exp(i\omega t_0)$ and to shift on depth by an amount z0 we multiply by $\exp(i k_z z_0)$.It is not quite this trivial however: the data field is not evidently a function of kz because we do not have data given on a z-axis so we cannot simply Fourier transform it to kz. Instead we have  
P( \omega , k_x , z ) \ \ =\ \ P( \omega , k_x , 0 )
\ e^{ i\,k_z ( \omega , k_x ) z }\end{displaymath} (9)
Since the operation is a multiplication in the Fourier domain, it may be described as the engineering diagram in Figure 5.

Figure 5


Downward continuation is a product relationship in both the $\omega$-domain and the kx-domain. What does the filter look like in the time and space domain? It turns out like a cone, that is, it is roughly an impulse function of $x^2 \,+\,z^2 \,- \ v^2 t^2$.More precisely, it is the Huygens secondary wave source that was exemplified by ocean waves entering a gap through a storm barrier. Adding up the response of multiple gaps in the barrier would be convolution over x. Superposing many incident ocean waves would be convolution over t.

Now let us see why the downward continuation filter has the mathematical form stated. Every point in the $( \omega , k_x )$-plane refers to a sinusoidal plane wave. The variation with depth will also be sinusoidal, namely $\exp ( i k_z z )$.The value of kz for the plane wave is found simply by solving equation (8):
k_z \ \ & = & \ \ \pm
\sqrt{ { \omega^2 \over v^2 } \ -\ k_x^2 ...
 ...mega^2 } \ }
\\ & = &\ \ \pm \ {\omega \over v }\,\ \cos \, \theta\end{eqnarray} (10)
Choice of the plus sign means that $ \exp ( -i \omega t \ + \ i\,k_z z ) $ is a down going wave (because the phase will stay constant if z increases as t increases). Choice of a minus sign makes the wave up coming. The exploding-reflector concept requires upcoming waves, so we nearly always use the minus sign, whether we are migrating or modeling.

Mathematically we can imagine an arbitrary function of t, x, and z. Its three-dimensional Fourier transform fills a volume in the space of $\omega$, kx, and kz. This arbitrary function is not a wavefield unless it vanishes everywhere except the cone-shaped surface $k_x^2 + k_z^2=\omega^2/v^2$ given by the wave equation. Thus equation (9) makes sense even though we cannot Fourier transform data over the z-axis because we don't observe data along the z-axis. It makes sense because if we know the mathematical function is a wavefield and if we know its Fourier transform over $\omega$ and kx, then we know its Fourier transform at any kz and we can ``delay'' it in z (downward shift) by multiplying by $\exp ( i k_z z )$.

The input-output filter, being of the form $e^{ i \phi }$, appears to be a phase-shifting filter with no amplitude scaling. This bodes well for our plans to deconvolve. It means that signal-to-noise power considerations will be much less relevant for migration than for ordinary filtering.


  1. Suppose that you are able to observe some shear waves at ordinary seismic frequencies. Is the spatial resolution better, equal, or worse than usual? Why?
  2. Scan this book for hyperbolic arrivals on field data and measure the Fresnel zone width. Where zero offset recordings are not made, a valid approximation is to measure $\Delta x_2$ along a tilted line.
  3. Evolution of a wavefield with time is described by
p(x , z , t) \ =\ 
\int \int \ \left[\,P( k_x , k_z , t=0 )
 ... k_x , k_z ) t } \right] \ 
e^{ ik_x x + ik_z z } dk_x \, dk_z \end{displaymath} (13)
    Let P(kx , kz , 0 ) be constant, signifying a point source at the origin in (x , z)-space. Let t be very large, meaning that phase = $ \phi = [ - \omega ( k_x , k_z ) \ +$$ k_x \ (x / t) \ +\ k_z \ (z /t) ] t $ in the integration is rapidly alternating with changes in kx and kz. Assume that the only significant contribution to 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?
  4. Downward continuation of a wave is expressed by
p(x , z , t) \ =\ \int \int \ \left[\,P( k_x , z=0 , \omega ...
 ...) z } \right] \ 
e^{{-i} \omega t + ik_x x } \ d \omega \, dk_x\end{displaymath} (14)
    Let $ P(k_x , 0 , \omega ) $ be constant, signifying a point source at the origin in (x , t)-space. Where is the energy in (x , z , t)-space?

previous up next print clean
Next: Travel-time depth Up: PLANE-WAVE SUPERPOSITION Previous: Waves in Fourier space
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