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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 eit 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 xk 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 xe 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 xk 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 xkz 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 0eikz it 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 0eikz it 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 0eikz it 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 0eikz it 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 4 107 and 0 8.85 1012 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 0eit 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 cos1 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 E1eik1 r , E1'eik1 ' transmitted r field, say vector electric field, by and E2eik2 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 y1eik1 r E y' 1eik1 r E y 2eik2 ' 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 sin2 ' ' 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 y1eik1 r , E y' 1eik1 ' r and E y 2eik2 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 22 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 21k2 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 22 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 cos1 1 rTE E1 k cos 2 1 1 1 tTE k sin k22 k12 sin 2 1 k22 2 1 2 1 E2 22k1 cos1 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 cos1 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 21k22 cos1 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 eikiz z eikix 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 eikix 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 ii 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 eikix 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 y0 E y0 E y1 E y 0 0TE E y 0 0TE E y1 1TE E y1 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 y0 E y0 Z1TE 0TE Z1TE 1TE Similarly the TM reflection coefficient becomes: H y0 H y 0 0TM 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 eikiz z H yi e ikiz z e ikix x Exi iTM H yi eikiz 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 eik 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 eik iTM eikiz 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 zi1 zi hi we obtain: Eix H iy z zi iTM eik h eik h 1 e 2iTM iz i ikiz hi iz i eikiz hi e 2 iTM eik h eik h iTM iz i ikiz hi iz i eikiz 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 eik h Ei 1x z iTM eik h eik 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 eikiz 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 zn1 , 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 eikiz z H xi eikiz z eikix x and the accompanying E y field is now written via the TE characteristic impedance: E yi iTE H xi eikiz z H xi eikiz z eikix 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