Download The general solution for plane waves incident on a layered half

5.3.3 The general solution for plane waves incident on a layered halfspace
 The general solution to the Helmholz equation in rectangular
coordinates
 The vector propagation constant
 Vector relationships between vectors k, E, and H
 Properties of uniform plane waves, characteristic impedance
 Reflection and transmission of plane waves at a plane interface
 TE mode reflection and transmission at a plane interface
 TM mode reflection and transmission at a plane interface
 Reflection and transmission of plane waves incident at real angles of
incidence
 Surface impedance formulation of reflection and transmission
 The general impedance of an n-layered half-space
The general solution to the Helmholz equation in rectangular coordinates
In section 5.2 we derived the Helmholtz equation in either magnetic or
electric fields. In the simple description of the MT method in section 5.3.2
we looked at a simple solution for the Helmholtz equation in H when the
horizontal derivatives were zero and there was only one horizontal
component of the field Hy. Here we will look at the general solution to the
vector Helmholz equation where there are no such simplifying assumptions.
The behavior of magnetic and electric fields at the surface of a layered half1
space of arbitrary conductivity and dielectric in the layers as a function of
frequency and the angle of incidence of the field on the surface. This general
treatment will consider the orientation of the electric and magnetic fields
with respect to the interface and the impedance and reflection coefficient for
such fields. We will use a right handed rectangular coordinate system with z
positive down.
For any rectangular component of the field, say Hx, the Helmholz equation
is:
 2 H x    2 H x    2 H x  


 k 2 H x    0
2
2
2
x
y
z
where we have used an eit time dependence. [from here on the 
dependence is implied and will not be denoted explicitly].
A classic method of solution is separation of variables: assume that the
solution can be written as the product of three independent solutions in each
coordinate viz:
Hx  X  x Y  y  Z  z 
Substituting this solution in the Helmholz equation yields three
independent ordinary equations:
1 d2X
 k x 2
2
X dx
1 d2X
 k y 2
2
Y dy
1 d2X
 k z 2
2
Z dz
2
and the condition that
k x 2  k y 2  k z 2  k 2  0
implies that there are only two independent equations since
1 2 X
 k z 2   k 2  k x 2  k y 2
2
Z z


Particular solutions to these ordinary equations are:
X  Ae
 ikx x
, Y  Be
 ik y y
, and Z  Ce
 i k 2 k x 2 k y 2  z
so the general solution is:
H x  x, y , z  
 
  H  kx , k y  e
'
2
2
2
ik x x ik y y i k k x k y z
e
e
k xk y
 
This is a Fourier transform and in fact we get to this solution more elegantly
by simply starting all over and taking the Fourier transform of the Helmholz
equation.
Using the following Fourier transform pair:
1
H  x 
2
H '  x 

 H  kx  e
'
ik x x
dk x


 H  xe
 ik x x
dx

and applying the first directly to the Helmholz equation, in other words
taking the spatial transform, yields an algebraic equation in transform space:






k x 2  H x' k x , k y , k z  k y 2  H x' k x , k y , k z  k z 2  H x' k x , k y , k z  k 2  0
or
k 2  kx2  k y 2  kz 2  0 .
The ki are not independent so the inverse transform back to x,y,z space
becomes:
3
1
H x  x, y , z  
4
 
 

H ' k x , k y eikx xe
2
2
2
ik y y i k k x k y z
e
k xk y
 
Before going on to apply the general solution to particular problems, we will
examine the particular solutions and find out the physical significance of the
ki parameters.
The Vector propagation constant
First, to simplify the ensuing algebra, we will consider a two
dimensional model of the Earth-air interface and a field above the interface
that has no variation in y, so k y is zero. [Over a uniform half-space the
coordinates can always be rotated such that y axis coincides with the
direction in which the field is invariant]. Consider the particular solution for
a component of the electric field in the y direction:
E y  E y 0eikx xeikz z  E y 0e 
i kx xkz z 
The bracketed term in the exponent has the form of a vector dot product of a
position
vector
r   ix   z 
and
a
vector
propagation
constant
k   ik x   k z  where i , j and  are unit vectors in the x, y and z axes
respectively; this solution has the form
E y  E y 0eik
r
The following sketch illustrates these vectors in two dimensions:
4
Figure 5.3.3.1
Here we can identify k x and k z as the rectangular components of a vector k
making an angle  with the vertical, z, axis. In this illustration k x and k z can
be
interpreted
as
the
rectangular
components
of
k
,
kx  k sin  and k z  k cos implying values limited to k. But be careful
because in the general solution k x or k z can have values far greater than k
suggesting meaning for  that is not intuitive. We return to these solutions
later.
A small digression at this point will help explain why, in later sections,
we will study things like reflection coefficient and surface impedance in
terms of k x . If we consider the full solution for a field, eiwt ikx x , and consider
where a field of constant phase moves in space and time we have:
t  kx x  a constant . And so
dx

   Vx , the phase velocity is in the x
dt
kx
direction. From the relation that wavelength,  ,
5
V
, we find that such a
f
wave has a spatial wavelength of x 

kx f

2
. So a wave propagating in
kx
the x direction has an implicit horizontal wavelength. The general solution,
which is an integral over all k x is made up of the superposition of waves of
different horizontal wavelength. We will see how the reflection and
refraction, and surface impedance, of such a wave depends strongly on x .
Vector relationships between k , E and H
We can now show the vector relationships between the fields and the
vector propagation constant by taking the Fourier transform of the individual
Maxwell equations. For example the   E equation can be written out in
full as:
Ez E y

 i H x
y
z
Ex Ez

 i H y
z
x
E y Ex

 i H z
x
y
Taking the Fourier transform of each of these yields:






ik z Ex  k x , k y , k z   ik x Ez  k x , k y , k z   i H y  k x , k y , k z 
ik x E y  k x , k y , k z   ik y Ex  k x , k y , k z   i H z  k x , k y , k z 
ik y Ez k x , k y , k z  ik z E y k x , k y , k z  i H x k x , k y , k z
6
Since k x , k y and k z and kz are now known as components of a vector k this
set of equations can be written in a much more compact form:
k  E   H
and the same process yields the second curl equation
k  H    i  E
Finally the divergence equations   E  0 and   H  0 and Ñ × H = 0
simply transform to:
k E 0
k H 0
So for this propagating plane wave we can see that E and H must always
be perpendicular to one another and that both are perpendicular to the vector
direction of the propagation constant. These simple rules show that there can
never be component of the field in the direction of propagation of a uniform
plane wave field.
Properties of uniform plane waves
To simplify the algebra involved let’s consider the nature of a wave
propagating in the z direction. Since solutions for each component are
simply additive we can illustrate properties by choosing any component.
First lets consider the solution for E y i.e. a solution to
2Ey
z
2
 k 2Ey  0
7
is E y  E y 0eikz it and from Faraday’s law 
E y
z
 i H x  ikE y so the
ratio of the electric to magnetic field for a plane wave propagating in the
positive z direction is the characteristic or intrinsic impedance, denoted here
by  and is equal to:
Ey
Hx
  Ey  

k
Note a couple of things about the characteristic impedance. We could have
started this process from the Helmholz equation in E x viz.
 2 Ex
 k 2 Ex  0
2
z
The solution is for a wave travelling in the +z direction is Ex  Ex 0eikz it
and now from Faraday’s law
Ex
 i H y  ikEx , so
z
Ex 

  Ex  
.
Hy
k
The component chosen affects the sign of the intrinsic impedance.
If we had chosen a wave travelling in the negative z direction the signs of
the z derivative would have reversed and so  Ey    Ey and Ex    Ex
Finally some sharp eyed reader may have tried to get to these relationships
by starting with the Helmholz equation in, say, H x viz.
2H x
 k 2H x  0
2
z
The solution is H x  H x 0eikz it and now from Faraday’s law
8
H x
   i  E y  ikH x
z
And so  Ey 
Ey
Hx

ik

which doesn’t look much like 
.
  i
k
But wait, with a little multiplying top and bottom and factoring watch what
happens:
  2   i
 2   i
ik




2
k
  i
  i
   i
 2   i


Either definition of the intrinsic impedance is ok.
Now let’s look at the propagation characteristics for various values of
,  , and  .
In general k is complex, that is k  kRe  ikIm   2   i .
Squaring both sides and equating real and imaginary terms yields the
following relations:
2
2
kRe
 kIm
  2 
and
2kRekIm   .
After a tedious bit of algebra we solve for kRe and kIm :
After a tedious bit of algebra we find the following equations for k R and k I :
k Re
1


2 2
     
1  1      

  
    


2






9
and
kIm
1


2 2




 
 1  1      
  
 
   


2






There are three limiting cases to be considered.
1) If   0 (free space) then kIm  0 and kRe   
2) If 
 then kRe  kIm 
3) If   , then

2

is small and ,

1
2 2

2
  
1  
1  
 1 

    
2   


1
[based on 1    2  1 

2
when 
1]
In this high frequency limit kRe    , the same value as in a nonconductive medium and kIm 
The skin depth  is therefore

2


.
2 


 2
.


This last result is very important because it means that in a conductive
medium at very high frequencies the skin depth becomes independent of
frequency. The skin depth may be very small but it depends only on the
10
conductivity, dielectric permittivity and magnetic permeability. The phase
velocity on the other hand is independent of the conductivity. We will return
to these results in section 5.11 on Radar.
There are a few details left to discuss about propagation in free space. Again
we will consider an E y field propagating in the positive z direction the form
for which is
E y  E y 0eikz it
Now let’s consider this solution for fields in free space where   0 and so k
is real and equal to  0 0 . The complete solution is:
E y  E y 0e

i  0 0  z t

This is a true propagating wave. A point of constant phase is defined by
t  0 0  z = constant
and
dz
1

 V ph the phase velocity ( in + z).
dt
0 0
The constants 0  4 107 and  0  8.85 1012 yield a phase velocity of
2.998 108 which is the velocity of light in free space. This remarkable
derivation is due to Maxwell who predicted it from first principles far ahead
of its accurate experimental determination.
The characteristic impedance of free space is:
 Ey  
0
k

0
 377 Ohms.
0
11
For completeness the basic solution we derived in section 5.3.2 for an H y
component ‘propagating’ in a good conductor where 
 is repeated
here.
2H y
z
2
 k 2H y  0
the solution for which is (including the time dependence):
H y  H y 0eit ikz
The sign must be chosen so that the fields do not grow exponentially with z.
In this case we choose e-ikz so that on expanding k we get the solution:
H y  H y 0e


2
z
e

 
i t 
z 
2


And we defined the skin depth,  , as

2


2

 500
7
f
2 f 4  10
We now have all the basic definitions, formulations and propagation
characteristics to address what happens when a propagating field in one
medium is incident on another medium. Our major interest in in plane waves
incident on a layered earth.
The schematic diagram below shows an incident field approaching a
horizontal interface between medium 1 and medium 2 with a propagation
vector. The position vector lies in the interface and the unit vector is normal
to the interface and is by our coordinate convention positive down. We will
assume from experimental results that there is energy reflected from the
12
interface and some energy may be transmitted across the interface into the
medium 2.
Figure 5.3.3.2
We have drawn this diagram in the conventional way used in most texts to
show the incident field as a ray approaching the interface at an angle 1 .
Remember though that k1 is a vector composed of an x component and a z
component where k x and k z are can have values far higher than the
geometric interpretation k1 cos1 or k1 sin 1 . As drawn the diagram
represents a field incident at what is termed a real angle of incidence. In the
derivations that follow we will consider the general case of arbitrary k x and
k z and highlight the case of real angles of incidence as we go. We define the
plane of incidence as the plane containing k1 and the unit normal to the
surface n .
Before going on to derive the amplitude relations of the reflected and
transmitted fields we can derive some important properties of the fields at an
interface from their vector representation. We have prescribed and incident,
13
reflected
and
E1eik1 r , E1'eik1
'
transmitted
r
field,
say
vector
electric
field,
by
and E2eik2 r respectively. Exactly the same form applies to
the vector magnetic field.
The boundary conditions are that the tangential components of the field must
be continuous across the interface. Thus, on the interface the sum of the
tangential incident and reflected fields must equal the tangential transmitted
field. First consider the incident field to have only a y component, i.e. only a
tangential component. Then the boundary condition yields:
E y1eik1 r  E y' 1eik1 r  E y 2eik2
'
r
This condition must hold for any position, r , in the interface and this can
'
only be assured if ; k1 r  k1 r  k2 r .
Now suppose that r locates an arbitrary point on the interface, it is on the
horizontal xy plane in the above sketch. Now choose another vector t also
lying in the interface so that r can be written as r  t  n . Substituting this
form for r in the above equation and noting the vector identity


k  t  n   t n  k we arrive at:
n  k1  n  k1'  n  k2
This is the general form of Snell’s Law.
An immediate consequence of this relationship is that all the propagation
vectors must lie in the same plane, the plane of incidence. Also, in the plane
of the interface, z is zero so
14
k r  kx x  kz z  kx x
so again if k1 r  k1' r  k2 r and this holds for any point in the interface,
any x, then we have the important result that
k1x  k2 x .
To put this general form of Snell’s Law in its familiar form we return to the
representation of the incident fields in the above sketch where the
propagation vectors of the incident reflected and transmitted fields lie at
'
angles 1,1 and 2 with respect to the normal to the interface n . Because
n  k1  n  k1'  n  k2 can now be written
k1 sin 1  k '1 sin '1  k2 sin2
'
'
Since k1  k1 then 1  1 so the angle of reflection is equal to the angle of
incidence.
Further in non conducting media k   
and the phase velocity ,V, is

so Snell’s Law can be written in the familiar form:
k
sin 1 sin  2

V1
V2
Another important result for plane waves in free space incident at a real
angle to a conducting ground can be found from
k1 sin 1  k2 sin  2
or
sin 2 
k1 sin 1
k2
If k2 >> k1 then sin 2  0 and 2  90 no matter what the incident real
angle. The refracted wave propagates vertically even for incident fields
propagating horizontally. This condition applies for a range of k1x beyond k1.
15
For the wave in medium 2 to propagate vertically k2 x
k2 z and since
k1x  k2 x then in terms of horizontal wavelength in the incident medium
1x
2
k2 . In a practical case where the ground has a resistivity of 100 Ohm-
m and at 1,000 Hz we find that the horizontal wavelength must only be
greater than ~700 m for the wave to be refracted vertically. This implies that
if the incident field is actually produced by a finite nearby source then it is
simply necessary to ensure that the bulk of the integral for the fields comes
from horizontal wave lengths greater than 700 m. This is a fundamental
statement about using controlled sources to simulate MT results in systems
like CSAMT and Stratagem.
Reflection and refraction of plane waves at a plane boundary
We now have some basic properties of incident, reflected and refracted
fields and we will now derive the amplitudes of the reflected and refracted
fields, and in the process determine the surface impedance of a layered
media.
Any incident plane wave field has the electric and magnetic field vectors
orthogonal in the plane that is perpendicular to the direction of propagation.
The field vectors can always be broken into two component one of which is
perpendicular to the plane of incidence and consequently parallel or
tangential to the interface. Since the boundary conditions are on the
tangential components it is convenient to set up the reflection refraction
16
problem in terms of electric or magnetic field components that are
perpendicular to the plane of incidence. This field decomposition has two
important definitions:
(1) An incident wave with electric field normal to the plane of incidence is
called the TE mode. The incident wave has an electric field transverse to the
plane of incidence .
(2) An incident wave with magnetic field normal to the plane of incidence is
called the TM mode. The incident wave has an magnetic field transverse to
the plane of incidence .
We will see that the amount of incident field that is reflected or transmitted
depends on the angle of incidence and on the mode.
TE mode reflection and refraction at a plane boundary
We will assume that the coordinate system can always be rotated such that
the xz plane is the plane of incidence. The TE mode then has only a y
component of E and the TM mode has only a y component of H. [Note that
an incident field with E and H in arbitrary directions can always be broken
down into a field with an E component and an H component perpendicular
to the plane of incidence, solved for each mode separately, and combined for
the total field in the reflected and transmitted fields.]
We will set up the problem for the y component of E, Ey,
The incident, reflected and transmitted fields are
E y1eik1 r , E y' 1eik1
'
r
and E y 2eik2
r
The tangential E continuous boundary condition simply yields:
17
E y1  E ' y1  E y 2
From   E  ki  E  
B
 i H and that the tangential component of
t
H n
we find that (dropping the y subscript):
k  E   n  k  E   n  k
'
1
'
1
1
'
1
2

 E2  n
1



2
 

Using the vector identity A  B  C  B C A  A C B and noting that
n E1  n E1'  n E2  0 because E is perpendicular to the plane of incidence
and therefore perpendicular to n , this boundary condition reduces to:
 E  n k   E  n k   E  n k 
1
'
1
1
'
1
2
1
2
2
Using both boundary condition equations and solving for E1' and E2 (and
noting that n k1'  n ki ), we find that:
E1'








 2 n k1'  1 n k2 

 E1 
 2 n k1'  1 n k2 


and
E2  E1
 n k  2
 n k    n k 
1
1
2
2
2
1
These are the Fresnel equations in vector form for the reflected and
transmitted E field amplitudes for the TE Mode. The transmitted field is
18
E1'
E
and 2 are the
called the refracted field in studies of light. The ratios of
E1
E1
reflection coefficient and transmission coefficient  and t respectively.
Noting that n k  k z and that k 2  k x2  k z2 and k1x  k2 x we can expand the
above Fresnel equations into:
E1'
  k2  k2   k2  k2
 2k1z  1k2 z 
2
1
1x
1
2
1x
 E1 
  E1 
2
2
2
  k  k   k  k2
 2k1z  1k2 z 
1x
1
2
1x
 2 1




and similarly

22 k12  k12x

E2  E1
  k2  k2   k2  k2
1x
1
2
1x
 2 1




TM mode reflection and refraction at a plane boundary
Following the same derivation used above for the TE mode we can obtain
the Fresnel equations for the TM mode: now H is perpendicular the plane of
incidence and E is in the plane of incidence.
  k k2  k2   k k2  k2
2 1
2
1x
 H1  1 2 2 1x
  k k2  k2   k k2  k2
2 1
2
1x
 1 2 2 1x





21k2 k22  k12x
H 2  H1 
  k k2  k2   k k2  k2
2 1
2
1x
 1 2 2 1x




H1'
and
19
The amplitude reflection coefficients are defined for each mode as
rTE 
E y' 1
E y1
and rTM 
H y' 1
H y1
.
It is quite clear that these reflection coefficients are different for each mode,
and that the reflected (and for that matter the transmitted) fields can have a
different phase than the incident field. This can be simply illustrated with the
limiting case of normally incident fields, i.e. when k1x  k2 x  0 . For added
simplicity let 1  2 .
Then
rTE 
k1  k2
k k
and rTM  2 1
k1  k2
k1  k2
So in this simple situation we see that the sign of the coefficient is different
for each mode. This has particular significance when the incident field is in
free space and k1    and the field is incident on a conducting halfspace for which k2  i . In this situation, which is typical for most
earth conductivities and frequencies less than a megahertz, k2
H y' 1
H y1
k1 and so
 1 . The reflected field is reflected in phase and is essentially the
same amplitude as the incident field. This means that the total magnetic field
on the ground surface is very close to twice the amplitude of the incident
field and is independent of the ground conductivity.
On the other hand
E y' 1
E y1
 1 so the electric field is reduced almost to zero.
Just how small the net surface electric field is can be found from the
transmission coefficient:
20
22 k12  k12x
E2

E1  k 2  k 2   k 2  k 2
2
1
1x
1
2
1x
For simplicity suppose 2  1 and assume normal incidence, i.e. k1x  0 ,
then
E2
2 k1

and if k2
E1 k1  k2
k1 then

E2 2 
 i 4

2
e

E1
i
The net surface field has a phase shift of 45 degrees and for typical ground
conductivity of 0.01 S/m and for an angular frequency of 1,000 (f = 169 Hz)
the net field is reduced by 1.88 *10-3.
Neither result should be surprising since these are exactly the relationships
that hold in the limit at the boundary of a perfect conductor. What is
important to note about this result is that the small residual electric field on
the boundary is very sensitive to conductivity. In all geophysical
measurements employing incident plane wave fields it is the electric field
that has information about the conductivity of the ground.
The graphical window associated with this section has an option to calculate
the TE and TM reflection coefficients for arbitrary kx for any values of
.
21
Reflection and refraction of plane waves at incident on the ground at real
angles of incidence
When the plane wave is incident at a real angle of incidence the Fresnel
equations can all be rewritten in terms of sines and cosines of the angle of
incidence 1 . For the TE mode the reflection and refraction coefficients
become:
E1' 2k1 cos1  1

rTE 
E1  k cos  
2 1
1
1
tTE 



 k sin  
k22  k12 sin 2 1
k22
2
1
2
1
E2
22k1 cos1

E1  k cos   k 2  k 2 sin 2 
2 1
1
1
2
1
1


and for the TM mode they are:
rTM
2
H1' 1k2 cos1  2k1


H1  k 2 cos   k
1 2
1
2 1



 k sin  
k22  k12 sin 2 1
k22
2
1
2
1
and
tTM 
H2
21k22 cos1

H1  k 2 cos   k k 2  k 2 sin 2 
1 2
1
2 1
2
1
1


The graphical window associated with this section has an option to calculate
the TE and TM reflection coefficients for arbitrary 1 for any values of
or.
22
Surface impedance of a layered ground
The practical problem with plane waves from distant sources is that we have
no information about the incident electric and magnetic fields themselves-we
only have the total fields on the interface. We have seen that the observed
magnetic field is twice the incident field and that it is the electric field, while
much reduced, that is sensitive to the ground conductivity. It would seem
that measuring the electric field would provide the information needed. But
the incident fields are essentially random in amplitude so the measured
amplitudes vary widely in time and cannot be related directly to ground
conductivity. The solution is to reference the electric field to the magnetic
field by measuring the ratio of orthogonal tangential E and H fields on the
ground surface. We defined this ratio earlier as the surface impedance and
we will briefly re-derive the reflection and transmission coefficients in terms
of the characteristic and surface impedances, and then derive expressions for
the surface impedance of a layered medium. We will adopt a revised
notation to be compatible with the n-layered programs in the graphical
windows of this section.
First consider again a TE mode solution for an arbitrary medium i:
E yi  E yi eikiz z eikix x
for a wave travelling in the positive z and x direction.
For a wave travelling in the negative z direction a solution is:
E yi  E yi eikiz z eikix x
The x component of the accompanying magnetic field is obtained from
  E  i H so
23
H xi
The ratio ,
E yi
H xi

 E yi   kiz



ii  z  i
1

is the characteristic impedance, iTE
=
i
. For a wave
kiz


travelling in the negative z direction iTE
.
 iTE
For the TM mode we have the solution in Hy;
H yi  H yi eikiz z eikix x
and from   H    i i  E we derive the characteristic impedance for

the TM mode to be; iTM

ikiz
.
  i i
We can now set up the interface problem in terms of impedance. Let the free
space above the interface be denoted now by the index 0 and the ground by
index 1. At z=0, our solutions for E and H have to satisfy the boundary
condition that each is continuous at the interface. This leaves the two
equations:
E y0  E y0  E y1
E y 0
0TE

E y 0
0TE

E y1
1TE

E y1
Z1TE
We have introduced an important concept in recognizing that on the
interface the total E and H fields define the surface impedance. For this
simple model of two media the surface impedance is the characteristic
impedance of the second medium. In the multilayer model to follow the
boundary conditions are met with the surface impedance of a layered model
24
so Z1TE will be replaced with the general expression for the impedance of a
layered model.
Combining these two equations we arrive at a ‘new’ expression for the
reflection coefficient:
E y0
E y0
Z1TE  0TE

Z1TE  1TE
Similarly the TM reflection coefficient becomes:
H y0
H y 0
0TM  Z1TM
 
0TM  Z1TM
These are the Fresnel equations rewritten in terms of the characteristic and
surface impedances. The program behind the graphic window uses these
expressions with either a general kx or a real angle of incidence and uses the
general n-layer impedance formula of the next section of Z. Note, as we saw
before in the general derivation of the reflection and transmission
coefficients, these equations are functions of the spatial wave number kz. as
are the characteristic impedances themselves and the surface impedance.
To bring us back to the practical matters of magnetotellurics, it is the surface
impedance that we actually compute from measurements of electric and
magnetic fields on the ground surface. Then, as we saw in the simple
introduction to MT, we assume for purposes of representing the data that the
computed impedance at a given frequency is for a simple half-space model.
This allows us to calculate an apparent resistivity via
A 
1

25
Z
2
The practical impact of a finite horizontal wave number, k0 x on the TE and
TM surface impedances and hence on the apparent resistivities can be
appreciated in the following little analysis. We will assume that the earth is a
good conductor so 
 .
The TM surface impedance of a half-space, medium 1, is:
ZTM 
but since k1x  kox and
ik1
1
ik1z
1

i k12  k12x
1

ik1
1
k12x
1 2
k1
is the surface impedance for normal incidence we
have the interesting result that as k0 x increases beyond k1 the ‘normal’
impedance, and the apparent resistivity calculated from it, goes up.
For incident waves in the TE mode

Z1TE  

k1z


k12  k02x

k1
k02x
1 2
k1
So now as the horizontal wave number goes up the impedance and the
apparent resistivity goes down.
We have spent a lot of effort on the topic of arbitrary horizontal wave
number because the solutions for the fields from finite sources are integrals
over such wavenumbers and the behavior of the solutions can often be
deduced from the behavior at a single wavenumber that is representative of
the main contribution from the wavenumber spectra in the integral for the
26
full solution. This concept is dealt with fully in section 5.4 where, for
example, we derive the solution for a line source in the y direction.
These fields are TE and as we get closer to the source the solution requires
shorter and shorter horizontal wavelengths to represent the large spatial
variations in the field. In that case apparent resistivities calculated in this
zone will be lower than those expected from a normally incident plane wave.
Other source would cause the apparent resistivities to go up. This issue is
important in assessing the actual sources for magnetotelluric fields and
whether apparent resistivities calculated from field data really are source
independent.
The general impedance of an n-layered half-space
We now derive the general form for the impedance of a layered halfspace. We will use the notation shown in Figure 5.3.3.3 below.
The solution for the TM mode in any layer is:


H yi  H yi eikiz z  H yi e  ikiz z e ikix x


Exi  iTM H yi eikiz z  H yi e  ikiz z e ikix x
From the general form of Snell’s Law k0 x  k0 x  k1x ......  kix so in all ratios
of E to H these terms cancel and so will be dropped in the following
derivation.
27
Figure 5.3.3.3
At the surface of layer i, z  zi and the above equations can be written in
matrix form viz:
Exi
H yi

iTM eik
iz zi
e
z  zi

 iTM e ikiz zi H yi
 ikiz zi
e
 ikiz zi
H yi

 A
H yi 
H yi 
and at the bottom of the layer, z  zi 1
Exi
H yi

z  zi 1
iTM eik

 iTM eikiz zi 1 H yi
 ikiz zi 1
 ikiz zi 1
iz zi 1
e
e
H yi 
B
H yi 
H yi 
These two equations can be combined to yield an equation for the fields at
zi in terms of the fields at zi 1 both within layer i viz:
Exi
H yi
 A B
z  zi
28
1

Exi
H yi
z  zi 1
After a horrible algebraic exercise of multiplying this out and noting that
zi1  zi  hi we obtain:
Eix
H iy

z  zi


iTM eik h  eik h
1
e
2iTM
iz i
ikiz hi
iz i
 eikiz hi


e
2
iTM
eik h  eik h
iTM
iz i
ikiz hi
iz i
 eikiz hi


Eix
H iy
z  zi 1
Now multiply this out into two equations and divide one by the other noting
that
Exi
H yi

z  zi 1
Exi 1
H yi 1
z  zi 1
because E and H are continuous at the zi 1 interface:
Eix
H iy
zi
 ZiTM
z  zi





2
iTM eik h  eik h Ei 1x z  iTM
eik h  eik h H i 1 y
e
zi
ikiz hi
iz i
e
iz i
 ikiz hi
E
i 1x zi 1
i 1

 iTM e
ikiz hi
iz i
e
iz i
ikiz hi
H
zi 1
i 1 y z
i 1
Finally if we divide top and bottom of the right hand side by
e
ikiz hi

 eikiz hi H i 1 y
zi 1
we get the disarmingly simple form for the
impedance at the top of layer i in terms of the impedance at the top of the
layer, i+1,below:
ZiTM  iTM
Zi 1TM  iTM tanh(ikiz hi )
i 1TM  Zi 1TM tanh(ikiz hi )
To find the surface impedance of an n-layer medium start by calculating the
surface impedance at the top of the layer that is on the basement, at depth
zn1 , and use the characteristic impedance of the basement nTM for Z nTM .
Then by a recurring operation calculate the surface impedance at the top of
each successive layer all the way to the ground surface.
29
The TE mode impedance is derived the same way. We now start with an
incident H x field:


H xi  H xi eikiz z  H xi eikiz z eikix x
and the accompanying E y field is now written via the TE characteristic
impedance:


E yi  iTE H xi eikiz z  H xi eikiz z eikix x
The same procedure used above leads to the same layer to layer impedance
formula except with the TE characteristic impedances and surface
impedances.
ZiTE  iTE
Zi 1TE  iTE tanh(ikiz hi )
i 1TE  Zi 1TE tanh(ikiz hi )
These expressions are used recursively to find the surface impedance for an
arbitrary horizontal wave number k0 x .[ kiz  ki2  k02x
( ki2  k02 sin 2 1
for a finite angle of incidence)].
The graphical window shows plots of the apparent resistivity derived from
this impedance, and the phase of the impedance, for multilayered models.
The resistivity 1, the permeability i and the susceptibility i for each layer
are entered as input as are the layer thicknesses and desired frequency range.
The default settings assume that the permeability and susceptibility are free
space values. Finally the impedances are calculated for either specified
horizontal wavelength ( = 2/k0x) or real angle of incidence 0. The default
is normal incidence, 0 = 0. For any model the mode, TE or TM, must be
specified on opening the window.
30