Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

Possible projects in Geoelectrics and Geoelectromagnetics

1 One-dimensional inversion of Schlumberger sounding data

The Matlab function rhoa Schlumberger.m (available at ftp.space.dtu.dk/pub/nio/PAG/) solves theforward problem of calculation ρa(L) from a given 1D resistivity structure ρ(z). The relationship betweenρa(L) and ρ(z) is non-linear, and therefore one has to solve a non-linear inverse problem in order to obtainρ(z) from measured values of ρa(L).

Choose a given geometry of the model (i.e. fixed layer thicknesses dm) and optimize the correspondinglayer resistivities ρm by minimizing the least-squares problem

N∑n=1

(log10(ρobsa /δρa)− log10(ρmoda /δρa)

)2with N as the number of data points.

Linearizing the forward problemy = G(m)

(where y is the data vector, consisting of the measured values log ρa,n, n = 1, . . . , N corresponding tothe electrode separations Ln), m is the model vector, consisting of the unknowns (model parameters)log ρm, and G is the (in this case non-linear) data kernel function. (It is an advantage to work with thelogarithms of the resistivity to make sure that the resistivities are non-negative.)

This non-linear inverse problem can be solved iteratively in Matlab after linearizing, which results in

y = G ·m

The aim of the project is to write a Matlab program that performs a 1D inversion of Schlumbergersounding data, with the option of performing a smooth inversion by regularization of the model (tomake sure that resistivity does not varies too much between neighboring layers: “smooth inversion” or“Occam inversion”).

2 An Introduction to Finite Difference (FD) Calculations

For a 1D earth of σ(z) and harmonic excitation of frequency ω, the complex vertical electric field obeys

d2E

dz2= iωµ0σ(z)E(z) (1)

If we discretize σ(z) and E(z) over evenly spaced intervals ∆z of z such that

z = m∆z, m = 1,M (2)

then to a first approximation for any m

d2E

dz2≈ Em+1 − 2Em + Em−1

∆z2(3)

If you substitute (3) into (1), with a little care you can cast the problem into the linear form

Ax = b (4)

where x is a vector of Em, and which, conveniently, can be solved in MATLAB by typing

x = A\b (5)

1

Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

To get a unique solution you will have to add some boundary conditions to (4), specifying E at the topand bottom of the model. Choose some sensible boundary conditions, and set up the system of linearequations in MATLAB. Solve for a 3 Hz excitation of a 0.5 S/m halfspace. Check your results againstthe known skin depth relationships. Compute C and σa at the surface of the model, as well as the phaseangle (which is the same for both).

Now solve for the excitation of a model in which σ increases linearly with depth from 1 S/m at thesurface to 10 S/m at 1,000 m depth, then is terminated with a 10 S/m halfspace. Compute the responseat 1 Hz, but then extend this to create sounding curves of apparent resistivity and phase and componentsof C.

Think of tricks to improve numerical accuracy for a given M , and ways to assess whether M is largeenough and ∆z is small enough.

3 Magnetotelluric Investigations at Brorfelde

Goal: use simultaneous measurements of the magnetic and electric field at Brorfelde (near Roskilde) toobtain an estimate of the conductivity of the Earth in the depth range down to, say 100 km.

Note: the following text is not meant as a “to-do-list” but as suggestion on topics that could be investi-gated.

The basis material of the investigations are time-series of the (horizontal) magnetic and (Earth-)electricfield obtained at Brorfelde (BFE). We are interested in the period range between few seconds and fewhours (the electric field observations for periods longer than, say, 2 hours are probably contaminated bydrift of the sensors and/or distortion of the electric field).

As an example, the data set ftp.space.dtu.dk/pub/nio/PAG/BFE 02 19-Dec-2007 112500.mat containsmagnetic and electric field observations (sampling rate δt = 2 secs) for the period 19-Dec-2007 11:25:00to 05-Jan-2008 09:51:00.

• Calculate transfer functions (Z-response, and from that C, ρa and phase φ) from the time series.You may start with the univariate expression

Ex(ω) = Zxy(ω)By(ω) (6)

Ey(ω) = Zyx(ω)Bx(ω) (7)

– remove a trend from E and B

– calculate auto-and cross spectra, e.g. < Ex · By > is the cross spectrum between Ex(t) andBy(t). You may use the matlab function cspd.m for this purpose. For instance: [EB, f]

= cpsd(E(:,1),B(:,2), [], [], [], 1/2); calculates cross spectrum between Ex and Byand the corresponding frequencies f (in Hz) if the input time series have a sampling rate of2 secs = 1/2 Hz (6th argument of cspd).

– calculate smoothed spectra, by (weighted) average over several neighboring frequencies. Abox-car filter (moving average) is possible, but a more sophisticated filter (for instance acosine bell) is better (less side-lobes!)

– calculate the transfer functions, for instance

Zxy(ω) =< Ex ·By >< By ·By >

,

where < · · · > stands for smoothed spectra. (Note: never smooth transfer functions, alwayssmooth the spectra before calculating transfer functions!)

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

• Calculate a mean impedance, Z = (Zxy − Zyx)/2 (the minus sign is due to the fact that for a 1Dconductivity (Zxy = −Zyx) and from that the C-response, ρa and phase φ. Try a first interpretationof ρa and phase by visual inspection of the curves. Is there evidence for a 1D conductivity structure(in that case there is a relationship between ρa and phase φ)? At what periods do you trust thedetermined transfer functions?

• Derive simple conductivity models that explain the observations.

3

Chapter 1

Electrical Methods

1.1 Overview

Three methods for electrical surveying

1. Self potential: passive method

2. Resistivity method: active method, but DC

3. Induced Polarization (IP): active method (no electromagnetic induction!)

Maxwell’s equations:

∇×E = − ∂

∂tB (1.1a)

∇×B = µj + µ∂

∂tεE (1.1b)

∇ · εE = qc (1.1c)

∇ ·B = 0 (1.1d)

where E is the electric field vector (units [V/m]), j is the electric current density (units [A/m2]), µ = µrµ0

is the permeability of the material with µ0 = 4π10−7 Vs/(Am) as the free space permeability (we assumeµr = 1), ε = εrε0 is the permittivity of the material with ε0 = 1

c2µ0= 8.8541878176 × 10−12 As/(Vm)

as the free space permittivity and εr as the dielectric constant of the material, c ≈ 300, 000 km/s is thespeed of light in vacuum, and qc is the density of electric charges (of unit [As/m3]).

Ohm’s law connects the electric field and the current density:

j = σE (1.2)

where the electrical conductivity σ (units: [S/m]) is a scalar (isotropic conductivity). In general, con-ductivity can be a tensor (in which case the directions of E and j differ).

In geoelectrics: stationary fields, i.e. ∂/∂t = 0. As a consequence of ∇×E = 0:

E = −∇U (1.3)

i.e. the electric field is a potential field, and the potential U (in regions of constant εr) is a solution ofthe Poisson equation

∇2U = −qcε

(1.4)

4

Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

0 .0 , ,. ,

100 000 10 0 00

(IGNEOUS

SAPROLifE

1 000

RES ISTlVI1Y ( Cl m)

" '"

'" " CONDUCT IVITY (mS / m)

, 000 10 0 00 100 000

SHIELC

UNWEATHEREO ROCKS

W'EATHEREO LAYER

GLACIAL SEOIMENTS

SEOIIolENfAR'I' ROCKS

WATER . AQUIFERS

,. , ' 00

Figure 1.1: Electric conductivity of some geomaterials

In regions of constant conductivity σ: ∇ ·E = 0 qc = 0 i.e. no accumulation of charges:

∇2U = 0 (1.5)

i.e. the potential U is a solution of the Laplace equation in regions of constant conductivity.

1.2 Principles of geoelectric sounding

Uniform halfspace:

E =ρI

2πR2= −∂U

∂R(1.6)

U =ρI

2πR(1.7)

Current electrodes A,B at distance rABMeasurement electrodes C,D at distance rCD

Measured: current I (in [A]) and potential difference ∆U = UC − UD > 0 (in [V]).

∆U =ρI

2π

[(1

rAC− 1

rCB

)−(

1

rAD− 1

rDB

)](1.8)

Apparent resistivity ρa is the resistivity of a fictitious homogeneous halfspace that would yield the samepotential difference ∆U as the Earth over which the measurements actually were made.

1. Wenner configuration

ρa = 2πa∆U

I(1.9)

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

Figure 1.2: Four-electrode configuration

2. Schlumberger configuration

ρa =πL2

2l

∆U

I(1.10)

3. Dipole-Dipole configuration

ρa = πn(n+ 1)(n+ 2)2l∆U

I(1.11)

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

1.3 Forward modeling: ρa(L)

Let us consider a layered halfspace, consisting of M layers each of which has constant resistivity ρm:

z =01

z2

z3

zM-1

zM

d r1 1

d r2 2

d rM-1 M-1

rM

In each layer the electric potential U has to obey Laplace’s equation, ∇2U = 0. At layer boundaries thetangential component of the electric field E is continues, which results in continuity of U across layerboundaries. In addition, the normal (vertical) component jz = − 1

ρ∂U∂z of the current density j has to be

continuous.

In cylindrical coordinates (assuming ∂/∂φ = 0)

0 = ∇2U =∂2U

∂r2+

1

r

∂U

∂r+∂2U

∂z2(1.12)

Separation of variables U(r, z) = R(r) · Z(z) results in

Zd2R

dr2+ Z

1

r

dR

dr+R

d2Z

dz2= 0

which separates into two ordinary differential equation:

d2Z

dz2− λ2Z = 0

has the solution

Z(z) = Ae−λz +Be+λz

and

d2R

dr2+

1

r

dR

dr+ λ2R = 0

This is a Bessel differential equation (of order zero) with the two solutions R(r) = J0(λr) and R(r) =K0(λr). Only the first solution is physically meaningful since K0(λr)→∞ for r → 0.

The general solution for the potential U in the mth layer (zm ≤ z ≤ zm+1,m = 1 . . .M) is therefore

Um(r, z) =

∫ ∞0

(Am(λ)e−λz +Bm(λ)e+λz

)J0(λr)dλ (1.13)

Introducing f(λ) as the Hankel Transform (of order n = 0) of f(r):

f(λ) =

∫ ∞0

f(r)J0(λr)rdr

f(r) =

∫ ∞0

f(λ)J0(λr)λdλ

leads to the general solution in each layer

Um(λ, z) =1

λ

(Am(λ)e−λz +Bm(λ)e+λz

), zm ≤ z ≤ zm+1,m = 2, . . . ,M − 1 (1.14a)

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

The solution in the bottom layer (which is a uniform halfspace extending to z =∞) is

UM (λ, z) =1

λAM (λ)e−λz, zM ≤ z (1.14b)

From the boundary condition at the surface,

0 = jz|z=0 = − 1

ρ1

∂U

∂z=A1 −B1

ρ1

follows A1 = B1 which leads to

U1(λ, z) =1

λA1(λ)

(e−λz + e+λz

), 0 = z1 ≤ z ≤ z2 (1.14c)

for the top layer.

The set of equations 1.14a to 1.14c contains 2(M − 2) + 1 + 1 = 2(M − 1) unknowns Am, Bm,m =2 . . .M − 1, AM , A1. At each of the M − 1 layer boundaries the electric fields as well as the verticalcurrents have to be continuous, which yields 2(M − 1) conditions. Starting from the bottom “layer”,m = M , it is possible to solve for all Am, Bm in order to get A1(λ) at the surface(z = 0) where

U1(λ)∣∣∣z=0

=1

λA1(λ)

or in the spatial domain

U1(r)|z=0 =

∫ ∞0

A1(λ)J0(λr)dλ

=I

2π

∫ ∞0

T (λ)J0(λr)dλ

=I

2πH0T (λ)

where H0T (λ) is the Hankel transform (of order n = 0) of the “resistivity function” T (λ), in units of[Ωm]. Its surface value is T1(λ).

For a uniform halfspace of resistivity ρ: H0T (λ) = ρ/r and thus

U1(r)|z=0 =I

2π

ρ

r

in agreement with eq. 1.7.

Instead of solving for the unknowns Am, Bm of eq. 1.14a to 1.14c it is more convenient to work withT (λ), starting from the bottom layer, TM = ρM and determine all other values according to

bottom layer: TM = ρM (1.15a)

m = M − 1, . . . 1 : Tm = ρmTm+1 + ρm tanh(hmλ)

ρm + Tm+1 tanh(hmλ)(1.15b)

• One layer model (“uniform halfspace”): T (λ) = ρ = const.

• Two layer model: T (λ) = ρ1ρ1+ρ2 tanh(dλ)ρ1+ρ2 tanh(dλ)

with d as the thickness of the top layer.

• M -layer model: Matlab program rhoa Schlumberger.m to calculate the response for a Schlum-berger sounding geometry.

8

Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

We approximate the potential difference between the electrodes C and D of a Schlumberger soundingby the horizontal derivative at r = L:

−dUdr

= − I

2π

∫ ∞0

T (λ)d

drJ0(λr)︸ ︷︷ ︸

−λJ1(λr)

dλ

=I

2π

∫ ∞0

T (λ)J1(λr)λdλ︸ ︷︷ ︸H1 of order n = 1 of T (λ)

The potential difference in the middle of the two current electrodes A and B (i.e. at r = L) due to twocurrent electrodes introduces a factor of 2:∣∣∣∣dUdr

∣∣∣∣ =I

π

∫ ∞0

T (λ)J1(λr)λdλ

For a Schlumberger sounding

ρa =πL2

I

∆U

2l

∣∣∣∣r=L

≈∣∣∣∣dUdr

∣∣∣∣r=L

which yields the apparent resistivity of a layered halfpace

ρa(L) = ρ1L2

∫ ∞0

T (λ)J1(λL)λdλ (1.16)

with surface resistivity transfer function T (λ) = T1(λ).

Numerical implementation of the Hankel transform:

ρa(L) =

K∑k=1

akT (λk) (1.17)

with filter factors ak and corresponding spacing λk (both depend on the model, i.e. on the ρm, dm). Thisis implemented in the Matlab program rhoa Schlumberger.m using K = 13 filter coefficients.

9

Electrical Methods – Exercises

1.4 Apparent resistivity curves

Let’s have a closer look at the apparent resistivity curves that result from synthetic Schlumberger datasurveyed over a horizontally stratified conductivity model. The Schlumberger configuration is mostcommonly used for vertical electrical sounding (VES). The mid-point of the array is kept fixed while thedistance 2L between the current electrodes A and B is progressively (usually exponentially) increased.This causes the currents to penetrate to ever greater depths and sense the conductivity at those depths.

The MATLAB function rhoa Schlumberger.m (available at ftp.space.dtu.dk/pub/nio/PAG/) calculatesapparent resistivity ρa in dependence of the half distance L of the current electrodes A,B for a Schlum-berger survey configuration.

1. Start with a three-layer resistivity model, rho = [10 1000 5] with layer thicknesses d = [10 1]

(Note that the third, lowermost, layer is a half-space of infinite thickness). Calculate and plot theapparent resistivity ρa as a function of electrode separation L. Interpret the result

2. Try to find combinations of thickness and resistivity pairs which quantitatively correspond to thefour curves shown in Figure 1.3.

3. Resistivity soundings aim at resolving a layered resistivity structure varying with depth. Is itpossible to resolve the resistivity structure below (i) a perfect conductor or (ii) a perfect insulator?

4. Try also some other combinations: how does, for example, the apparent resistivity curve look likefor a thin resistive or thin conductive upper or middle layer?

5. What is the thinnest layer you can resolve? Summarize your experience in a few words.

6. Set your model to rho = [100 1000 100] and d = [2 10]. Keep the same resistivity value forthe first and the last layer, but change these values systematically over one or two magnitudes andcompare the resultant curves. Explain your findings.

7. Consider the model rho = [100 1000 100] and d = [3 1]. Change the thickness and resistivityof the middle layer so that the product resistivity × thickness (the so called resistance) remainsconstant. Try different models and document your experience. Keep the thickness of the middlelayer smaller than 2 m.

8. Consider the general 4-point electrode configuration consisting of a pair of current electrodes anda pair of potential electrodes. How does an exchange of current and potential electrodes, meaningwe use A and B as detection electrodes and C and D as current electrodes, change your result?

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

10−1

100

101

102

103

101

102

103

104

Electrode separation AB/2 [m]

App

aren

t res

istiv

ity ρ

a [Ω m

]

10−1

100

101

102

103

101

102

103

104

Electrode separation AB/2 [m]

App

aren

t res

istiv

ity ρ

a [Ω m

]

10−1

100

101

102

103

101

102

103

104

Electrode separation AB/2 [m]

App

aren

t res

istiv

ity ρ

a [Ω m

]

10−1

100

101

102

103

101

102

103

104

Electrode separation AB/2 [m]

App

aren

t res

istiv

ity ρ

a [Ω m

]

Figure 1.3: Four typical Schlumberger sounding curves.

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

100

101

102

101

102

103

Electrode separation AB/2 [m]

App

aren

t res

istiv

ity ρ

a [Ω m

]

Figure 1.4: Observed apparent resistivity values at the waste pit site Stetten in Switzerland

1.5 Field data: Determine the thickness of a waste pit

We will now analyze a field data set measured with a Schlumberger configuration over the Stetten landfillsite in Switzerland (for more information, see Green et al., 1999, A template for geophysical investigationsof small landfills, The Leading Edge, p. 248-254). Here, industrial and household waste was dumped ina former gravel pit. As the dumped waste could possibly contaminate the ground water, the landfill sitehas been intensively studied, also with the goal to show the potential of geophysical methods in locatingand characterizing the in-fill and the geology of the surrounding.

A Schlumberger-sounding gave the apparent resistivities shown in Figure 1.4 (data available atftp.space.dtu.dk/pub/nio/PAG/Stetten Schlumberger.txt)

1. Determine the depth to the top of the in-fill and the thickness of the waste pit.

Try to fit a model curve to the observed data points by try-and-error. Change the model parametervalues only by small amounts and in a systematic manner. Interpret your model curve.

Try two- and three-layer models. What is the minimum rms (root-mean-squared) misfit (rms

=√

(∑Nn=1 (log10(ρobsa /δρa)− log10(ρmoda /δρa))

2)/N with N as the number of data points and

δρa = 1Ωm as the measurement error) that you can achieve?

2. Use the res1d.exe inversion program (available at ftp.space.dtu.dk/pub/nio/PAG/res1d.exe)to do a 1D inversion of the data for models with different numbers of layers.For this program you need the data in a slightly different format, available atftp.space.dtu.dk/pub/nio/PAG/Stetten Schlumberger.dat.

Start with the following steps:

(a) start the program res1d.exe

(b) Open the data file (“File” → “Read 1D resistivity file” → Stetten Schlumberger.dat)

(c) Do an automatic inversion (“Inversion” → “Automatic Inversion”)

(d) Try different number of layers (“Inversion” → “Invert with user model”)

12

Chapter 2

Electromagnetic Methods

Combining the two Maxwell equations

∇×E = − ∂

∂tB

∇×B = µ0j + µ0∂

∂tεE

and Ohm’s lawj = σE

yields

∇×∇×E = −∂∇×B

∂t= −µ0σ

∂E

∂t− µ0ε

∂2E

∂t2

In regions of constant conductivity σ (which means ∇ ·E = 0) this results in the telegraph equation (forE)

∇2E = µ0σ∂E

∂t+ µ0ε

∂2E

∂t2(2.1)

A similar equations can be derived for B (valid in regions of constant ε and σ):

∇2B = µ0σ∂B

∂t+ µ0ε

∂2B

∂t2(2.2)

Assuming a harmonic time dependence F ∝ eiωt (where F is either E or B) results in

∇2F = (iωµ0σ − ω2µ0ε)F = k2F (2.3)

The term ∝ ω2µ0ε is due to displacement currents, while the term ∝ iωµ0σ is due to conduction. Theirratio is

α =|displacement current||conduction current|

=ωε

σ(2.4)

Limiting cases:

• α 1: conduction dominates. This leads to the diffusion equation

∇2F = µ0σ∂F

∂t(2.5)

= iωµ0σF (2.6)

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

with solutionF = F0e

i(ωt−z/d)e−z/d = F0eiωte−

zd (1+i)

EM methods in quasi-stationary regime: diffusion of EM wave in the Earth with skin depth

d =√

2ωµ0σ

• α 1: displacement currents dominate. This leads to the wave equation

∇2F =1

c2∂2F

∂t2(2.7)

= −ω2µ0εF (2.8)

with solutionF = F0e

i(k·r±ωt)

GPR (Ground Penetrating Radar, Georadar), velocity of wave: c = ω|k| = 1√

µ0ε

Classification of EM methods:

∇×B ∇×E α d/L

Wave propagation µ0(j + ∂(εE)/∂t) −∂B/∂t 1 1 radio waves, GPR, HF (f > 10 MHz)Diffusion µ0j −∂B/∂t 1 1 induction methods, FEM, TEM, LF (f < 1 MHz)Stationary µ0j 0 1 ' 1 geoelectrics, “DC”Magnetostatic 0 0 potential field methods

2.1 Principles of Electromagnetic Induction

Consider a magnetic field variation with time dependence eiωt Let z be positive downward, with z = 0as the Earth’s surface.

For a uniform halfspace (σ is constant), the electromagnetic field decays in the region below the surface(z > 0): F ∝ ez/C with

C =

√1

iωµ0σ=

√2

µ0ωσ︸ ︷︷ ︸skin-depth d

1

1 + i=d

2(1− i)

C = CR − iCI , CR = CI = d/2

For a general 1D conductivity (σ(z) depends on depth z):

CR > 0

CI ≥ 0

Estimation of the conductivity-depth profile σ (z) is possible from knowledge of the frequency dependenceof C (ω).

C (ω) can be derived from measurements of the electric and magnetic field at the surface:

Every divergence-free vector field (e.g. B and, in regions of constant σ, also E) can be expressed as asum of a toroidal and a poloidal part:

E = Etor + Epol (2.9)

= ∇× zΦ +∇×∇× zΨ (2.10)

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

The toroidal part has no vertical component, while the poloidal part in general contains of vertical as wellas horizontal components. The toroidal electric field is therefore also called TE (“Transversal Electric”);the toroidal part of the magnetic field is called TM (“Transversal Magnetic”).

Induction equations:

iωµ0σB = ∇2B (2.11)

iωµ0σE = ∇2E (2.12)

Below surface (z ≥ 0):Only horizontal currents if 1D conductivity. Hence the electric field E = 1

σJ is horizontal, too, whichmeans that there is only a TE field (“TE mode”).

E = ∇× (zΦ) is toroidal

−iωB = ∇×E = ∇×∇× (zΦ) is poloidal

The toroidal potential Φ has to fulfill the induction equation

∇2Φ = iωµ0σΦ

In following we assume an incident (primary) wave in the z−x-plane (no dependence on the y-direction).In regions of constant σ:

Φ (x, z) =ω

k

(Ae−z/C +Bez/C

)ei(ωt+kx), ∇2Φ 6= 0

Φ (z →∞) ≤ ∞,→ B = 0

E = ∇× (zΦ) =

0∂xΦ

0

→ Ey = iωei(ωt+kx)Ae−z/C

z=0= iωAei(ωt+kx) (2.13)

B = − 1

iω∇×E =− 1

iω

−∂zEy0∂xEy

Bx = − 1

CAei(ωt+kx)e−z/C

z=0= − 1

CAei(ωt+kx) (2.14)

By = 0

Bz = −ikAei(ωt+kx) z=0= −ikAei(ωt+kx) (2.15)

Measuring the electric and magnetic field at the Earth’s surface (z = 0) allows to eliminate the unknownvalue A and to determine the transfer function C(ω):

Ey = i iωC︸︷︷︸impedance Z=−Ey/Bx

Bx (2.16)

= −C ∂Bx∂t

in time domain (2.17)

In general: (Ex (ω)Ey (ω)

)=

(Zxx (ω) Zxy (ω)Zyx (ω) Zyy (ω)

)(Bx (ω)By (ω)

)(2.18)

15

Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

and for a 1D conductivity: Zxx = Zyy = 0; Zxy = −Zyx = iωC.

Similar to the apparent resistivity in geoeletrics (where ρa(L) is measured in dependence on electrodeseparation 2L), apparent resistivity can also be defined for electromagnetics (in dependence on frequencyω):

ρa(ω) = ωµ0|C|2 (2.19)

φ(ω) = 90 + argC (2.20)

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Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

10−6

10−4

10−2

0

20

40

60

80

100

T [secs]

C [m

]

uniform halfspace of ρ = 100 Ω m

10−6

10−4

10−2

102

103

104

T [secs]

ρ a [Ω m

]

10−6

10−4

10−2

0

15

30

45

60

75

90

T [secs]

φ [d

eg]

10−6

10−4

10−2

0

20

40

60

80

100

T [secs]

C [m

]

uniform halfspace, overlaid by 20 m thick insulator

10−6

10−4

10−2

102

103

104

T [secs]

ρ a [Ω m

]

10−6

10−4

10−2

0

15

30

45

60

75

90

T [secs]

φ [d

eg]

ReC−ImC

ReC−ImC

Figure 2.1: C-response, apparent resistivity ρa(ω) = ωµ0|C|2 and phase φ(ω) = 90 + argC for auniform halfspace (left) and a uniform halfspace overlaid by a 20 m thick insulator.

17

Electromagnetic Methods –Exercises

1. You are using a VLF instrument receiving a 16 kHz signal to investigate an area with an averageresistivity of 1000 Ωm.

• Compute the skin depth d.

• To what fraction of the original amplitude is the amplitude decayed at that depth?

• How could you increase the investigation depth of this method?

2. Calculate the skin depths for electromagnetic fields with 10 kHz, 500 kHz, and 2000 kHz in

• Wet sandstone with a conductivity of 10−1 S/m;

• Massive limestone with a conductivity of 2.5× 10−4 S/m;

• Granite with a conductivity of 10−6 S/m.

3. You want to perform a Schlumberger sounding. AC current is used, to prevent polarization of thecurrent electrodes. What is the maximum frequency of the AC current that you may use? Choserealistic values for L and ρ.

4. Comparison DC and MT sounding. The MATLAB function C wait.m (available atftp.space.dtu.dk/pub/nio/PAG/) calculates apparent resistivity ρa (as well as C and phase φ)in dependence on the period T of the primary signal for a MT survey.

• Compare DC and MT sounding curves. Use for instance a 3-layer model rho = [110 30 65]Ωm,d = [20 120] m. How does the phase φ of the MT sounding curve change? (Hint: look atperiods ranges for which φ > 45, resp. φ < 45).

• Introduce a very poor conducting thin layer (ρ ≈ 10−8Ωm) at 10 m depths. How does thisinfluence the DC, resp. MT sounding curves?

Figure 2.2 shows ρa and phase φ for the four conductivity models used to produce Figure 1.3.

18

Introduction to Geophysical Exploration Klaus Mosegaard and Nils Olsen January 2012

10−10

10−8

10−6

10−4

10−2

100

101

102

103

104

Period

App

aren

t res

istiv

ity ρ

a [Ω m

]

10−10

10−8

10−6

10−4

10−2

100

101

102

103

104

Period

App

aren

t res

istiv

ity ρ

a [Ω m

]

10−10

10−8

10−6

10−4

10−2

100

101

102

103

104

Period

App

aren

t res

istiv

ity ρ

a [Ω m

]

10−10

10−8

10−6

10−4

10−2

100

101

102

103

104

Period

App

aren

t res

istiv

ity ρ

a [Ω m

]

10−10

10−8

10−6

10−4

10−2

100

0

45

90P

hase

[deg

]]

10−10

10−8

10−6

10−4

10−2

100

0

45

90

Pha

se [d

eg]]

10−10

10−8

10−6

10−4

10−2

100

0

45

90

Pha

se [d

eg]]

10−10

10−8

10−6

10−4

10−2

100

0

45

90

Pha

se [d

eg]]

Figure 2.2: MT sounding curves

19