Kurt Peek (NAM-PTU/E/S) 25 April 2014

Seismic Processing Online Assignment 13

Question The apparent velocity of an event is the velocity which one observes along surface. So: if an event originates vertically

below the surface, and falls in perpendicularly, the event has an infinite apparent velocity, and a wavenumber along

surface of exactly 0 (as wavenumber is inversely proportional to velocity).

In the exercise below we will look at three (plane wave) seismic events, and you will have to determine their apparent

velocity and apparent wavenumber. Please list the numerical values and submit these to Moodle. Remember to put your

name in the file name of the documents you submit.

Below you will also find a document that tries to clarify aliasing in 2D space by considering that regular sampling in the

time-space domain translates into periodicity in the frequency wavenumber domain.

Exercise_FK_diagram

2D_Aliasing.pdf

Answer

a)

See Figure 1.

b)

The events are numbered from top to bottom, and for each event a straight line was drawn along a

part of the wave with constant phase (e.g. a trough). The slopes of these lines give the apparent velocity

according to

( 1 )

where is the increment of offset measured at the surface, and is the corresponding

increment of two-way time.

The apparent velocities for the three events were determined at = 753 m/s, = 193 m/s, and

= 197 m/s1.

c)

The frequencies were determined by measuring two periods using a ruler and interpolating to the height of the X-T diagram. I found:

1 The straight lines were extended across the full width of the X-T diagram, so that = = = 140 m. The corresponding time intervals were measured with a ruler and interpolated (c.q. extrapolated) from the height of the X-T diagram, which is 700 ms and was measured at 11.3 cm. Thus I determined TWT1 = (3 cm)/(11.3 cm) 700 ms = 186 ms, TWT2 = (11.7 cm)/(11.3 cm) 700 ms = 725 ms, and TWT3 = (11.45 cm)/(11.3 cm) 700 ms = 709 ms.

https://www.moodle.shell.com/file.php/9459/Roster%20Attachments/Row_114_A.%2014/exercise_fk_diagram_v2.pdfhttps://www.moodle.shell.com/file.php/9459/Roster%20Attachments/Row_114_A.%2014/exercise_fk_diagram_v2.pdfhttps://www.moodle.shell.com/file.php/9459/Roster%20Attachments/Row_114_A.%2014/2D_Aliasing.pdf

T1 = (1.45 cm/2)/(11.3 cm) 700 ms = 44.9 ms f1 = 22.3 Hz T2 = (2.85 cm/2)/(11.3 cm) 700 ms = 88.3 ms f2 = 11.3 Hz T3 = (1.4 cm/2)/(11.3 cm) 700 ms = 43.4 ms f3 = 23.1 Hz

d)

The apparent wavenumbers were calculated using the formula

( 2 )

where is the frequency (in Hz) of the wave train and the apparent velocity as given by Equation

( 1 ). We find 0.030 cycles/m, = 0.059 cycles/m, and = 0.117 cycles/m.

The last apparent wavenumber, , is above the Nyquist wavenumber = 0.1 cycles/m. Hence

with 5 m spatial sampling, its alias within the Nyquist band is = 0.117 0.1 = 0.017

cycles/m.

Figure 1 X-T diagram showing three numbered events.

Figure 2 F-K diagram corresponding to the X-T diagram in Figure 1. Event 3 is aliased in the spatial domain.