E : Ground-penetrating radar (GPR)

Transcription

1 Geophysics 3 March 009 E : Ground-penetrating radar (GPR) The EM methods in section D use low frequency signals that trael in the Earth by diffusion. These methods can image resistiity of the Earth on arious scales. Because they are diffusie, they will not be reflected from sharp interfaces between layers with different resistiities. Seismic exploration is the most widely used exploration method in geophysics. It uses the propagation of elastic waes in the Earth and can gie detailed images of the subsurface. Can show that the best ertical resolution that can be obtained in one quarter of a waelength. Howeer it can be difficult to apply seismics to the study of shallow structure on the scale of 0 s of metres as high frequency signals are needed. In addition, seismic sources can be awkward to operate repeatedly. RADAR (Radio detection and ranging) was deeloped during WW to measure the distance and elocity of an aircraft. From In the 960 s and 970 s radar was applied to imaging subsurface structure. Frequency must be chosen carefully to ensure that (a) signals trael as waes and (b) that the depth of penetration is not limited by the skin depth phenomena. Ground penetrating radar (GPR) is now a widely used technique for imaging near surface structure. The radio wae pulses can be transmitted much more easily than in seismic exploration. Most seismic data processing techniques can be easily applied to GPR data. Calculation What will be the round-trip trael time for a reflection from a surface 5 m from the TX-RX? Assume speed of light is 3 x 0 8 m/s. Compare this with sound waes traelling at 300 m/s

2 Geophysics 3 March 009 E : Wae propagation In section D we considered how to determine if an EM signal will trael in the Earth as a wae or by diffusion. To determine if wae propagation or diffusion will dominate, can consider the σ ratio defined as r = πfε, where f is the frequency, ε is the permittiity of the subsurface and σ is the electrical conductiity. Assume that the permittiity has the free space alue ε = ε 0 = F/m If r is large, then EM diffusion will occur. If r is small then wae propagation occurs E. : Velocity of propagation In free space radio waes trael at the speed of light, c, where c = μ0ε 0 and μ 0 is the permeability of free space and ε 0 is the permittiity of free space μ 0 = 4π x 0-7 H/m and ε 0 = 8.85 x 0 - F/m giing c = 3 x0 8 m/s In the Earth the permeability (μ) and permittiity (ε) do not hae their free space alues. This is because the atoms/molecules behae as magnetic and/or electric dipoles. The magnetic behaiour of the atoms/molecules can be expressed in terms of relatie permeability (μ r ) as μ = μ r μ0 and section C describes this topic. The relatie permeability is usually close to, except in deposits of magnetic minerals. The relatie permittiity (ε r ) is defined as ε = ε rε 0. This quantity is also called the dielectric constant (K). The permittiity will be greater than if the molecules can act as electric dipoles. Water is one of the most polar molecules since the hydrogen atoms deelop a positie charge and the oxygen deelops a negatie charge.

3 Geophysics 3 March 009 Can write the elocity of a radar wae in the Earth () in terms of the relatie permittiity and permeability as c = μ ε r Some typical alues are listed below (from Dais and Annan, 989) r Material ε r σ(ms/m) (m/ns) δ (f = 00 MHz) Air Distilled water m Fresh water m Seawater m Dry sand m Saturated sand m Limestone m Clay m Granite m Dry salt m Ice m Most materials are mixtures so need to consider a combination of properties. Topp et al., 980 deried an empirical relationship between the dielectric constant (K a ) and the water content (θ ). Figure 5 from this paper is illustrated below. 3

4 Geophysics 3 March 009 K a = θ = θ + 46θ 76.7θ x0 +.9x0 K a 5.5x0 K a + 4.3x0 K a Calculation A 00 MHz GPR signal traels in the Earth with = 0.06 m/ns. What is the waelength? E. Reflection at interfaces When a radar pulse reaches an interface between two layers, some energy will be reflected and some will be transmitted. Consider the case when the signal is normally incident on a horizontal interface In the upper layer the elocity is and in the lower layer it is If the incident wae has an amplitude A =, then can show that the amplitude of the reflected and transmitted wae is gien by the coefficients R and T. Ar A i = + T = A A t i = + If the elocity ariation is solely due to changes in permittiity, then ε ε + ε ε T = ε ε + ε 4

5 Geophysics 3 March 009 Example : Increase in elocity with depth This example has an increase in elocity from 0.06 to 0.5 m/ns which would correspond to water saturated sediments oerlying granite. Note the positie reflection coefficient as predicted by + Confirm that the round trip trael time is correctly computed. 5

6 Geophysics 3 March 009 Example : Decrease in elocity with depth Velocity decreases from 0.5 to 0.06 m/ns corresponding to the water table Note the decrease in elocity that corresponds to the increase in permittiity Reflection has negatie polarity as predicted by + 6

7 Geophysics 3 March 009 Example 3 : Low elocity layer Low elocity layer produced by a zone of water saturated sand. The 3 layers hae elocities, and 3 Note the negatie polarity reflection from the top of the layer which is caused by a decrease in elocity according to + The second reflection comes from the base of the low elocity layer and has a positie polarity. To calculate the amplitude must consider both transmission at the top of the layer and reflection at the base. Amplitude = Note that the middle term can be positie or negatie. Howeer the first and third terms (transmission on downward and upward paths) are both positie. Thus the polarity of the second reflection is determined by the elocity change where it is reflected. 7

8 Geophysics 3 March 009 Can you confirm the relatie magnitude of these two reflections for the numerical alues used in this case? In the animation you can also see reerberation (multiple internal reflections) in the low elocity layer. As the layer gets thinner, the reflections arrie closer in time. Can show that the λ two reflections can only be distinguished if t > where t is the thickness of the 4 layer. Calculation A 00 MHz surey takes place in a region where dry sand has = 0.5 m/ns. What is the thinnest layer that could be detected? E.3 Factors that reduce the amplitude of radar waes The amplitude of radar waes will be reduced by two primary effects (a) Geometric spreading. As the wae spreads out to coer a larger area, conseration of energy requires that the amplitude decreases. This is ery similar to the way that ripples on a pond decrease in amplitude. (b) Attenuation through the skin depth effect. This is the most important effect and you should always compute the skin depth from the frequency and conductiity. At high frequency in a high conductiity medium, the skin depth can be just centimetres (see table aboe). 8