Dr. K. Pradeep Kumar: Hello and welcome to NPTEL MOOC on electromagnetic waves in guided and wireless medium. So this I module 17 of the course and we have already seen uniform plane waves, the wave equation, and we have seen different type of polarization. In the last module, I gave you an exercise. Let me first solve that exercise to show you that we don’t only have linear polarization, but we have polarizations of different nature as well So we took the electric field to be of the form x hat A at z=0 plane of course. We took this as (x hat A + y hat A e-jpi/2). Now I will first convert this phasor at z=0 plane into the real time dependent expression. So I will now have x hat A cos omega t + y hat A cos (omega t-pi/2). Now cos (omega t-pi/2) would actually be equal to sin pi/2. So I will have, instead of writing this as cos (omega t-pi/2), I can write this as sin pi/2 So then I write that what I will have as A — sorry I wrote cos again — it’s actually sin omega t, so cos (omega t-pi/2) is sine omega t. So now this is what we have in terms of the time dependent electric field. As before, we will draw the axes, mark x and y, and now look at what happens at omega t=0. So at omega t=0 cos0 is 1, sin 0 is 0, so the electric filed have to lie entirely along the x axis, with am amplitude of A. So this is a situation at omega t=0 Now what would be the situation at omega t = pi/4? At omega t = pi/4, both cosine and sin have equal amplitudes and because you have x hat A + y hat A divided root 2 on both sides, the angle with which the resultant vector will be at 45 degrees, okay, so that would be 45-degree angle. However, the amplitude of this one would be at A/root 2 + A/root 2. That would be square root 2 A and you can plot that. So you have A here and then you have A here and then A/root 2, so what you’ll actually be able to find is to find the magnitude of this one and then write the vector at omega t = pi/4, okay Now what is the situation of omega t = pi/2? In which case cos pi/2 will be 0. So the x component will contribute to 0 length vector whereas the y component will contribute to the full A magnitude length vector, okay So you actually have A here at omega t = pi/2 Now you can see a pattern emerging, right So as I start increasing omega t further, I will note that the field vector seems to be going along a particular circle and they would be moving in this particular direction, which is basically counterclockwise. So because you move in the counterclockwise direction, so you can actually start with x and y, and then when you move in the counterclockwise direction, your thumb will be pointing along the z axis So you start off with x axis, which is where at omega t=0 the electric field vector was, and then you move it, move your fingers along or curl it along the direction of the increased changing electric field vector, and you will see that your thumb is pointing along the z axis, and you’ve actually used your right hand in order to get to the z axis, and this is therefore called as right circularly polarized wave, okay, or sometimes called as right handed polarization, right hand circularly polarized Why it is called circularly polarized? Because these electric field vectors are moving along a particular circle of radius A here, correct So this is right hand circularly polarized wave or sometimes simply called as RHCP vector You can show I will leave this as an exercise that if you change the phase from -pi/2 to pi/2, you will end up with a left hand circularly polarized vector, which would actually move in this direction. So it would move in the direction that is opposite to that of the right hand circularly polarized wave. So you do have different types of polarization, linear polarization is one where the phase difference phix – phiy = 0 and you have, let’s say, the circularly polarized version. You have two types of circularly polarized cases. One

you have the left hand circularly polarized wave and then you have the right hand circularly polarized wave, depending on whether the phase phix – phiy is pi/2 or – pi/2, and then the most general case is when you have what is called elliptical polarization. In the elliptical polarization, you can still have it along — I mean the movement of the electrical field vectors can be either left or to the right However, they would not move along the circle, but they would actually move along the ellipse that you can see here. So they would actually be moving along the ellipse, both kinds of ellipses are possible, left handed ellipse and the right handed ellipse. So these are the different types of polarization of a uniform plane wave Now let us go back to the type of media that we were considering. We considered a linear medium, which is fine, we will consider the same medium. We consider isotropic medium, meaning that your direction of the propagation would not matter. The properties of the matter would remain the same whether you were propagating along the z direction, oriented along x or oriented along y Then we have a homogenous medium, meaning that the material properties, say, the permittivity and permeability would remain constant and it would remain the same at different points in the space, and finally, we had considered the lossless materials. In the lossless material scenario, what we had was the material made out of pure dielectric, meaning that epsilon r was present and any conductivity of the medium that could normally be present in a typical medium was actually taken to be equal to 0. So the material was pure insulator or pure dielectric. Of course, we have non-magnetic medium which we will continue to use Now what I want to do is to relax this assumption of losslessness. I mean we want to consider after all propagation of waves in real media, and most real media will actually be lossy So if you take a glass slab for example, and then send light in, and then you measure the power of the light that is coming out and compare it to the power of the light that was put into the glass, then you will see that there is some difference. That difference is actually being observed or is the result of absorption of electromagnetic wave, some part of electromagnetic wave by the material itself What it does is to slightly change the temperature of the material? But we will not worry about that one. For us, whatever that has gone in is not the same as that is coming out of the medium, then it means that we are considering a lossy medium. Of course, we could also have the output of the particular medium being having the power which is greater than the input power. In that case, we would be considering what is called as active medium or gain medium We will not consider the gain medium in this course. We will stick with lossy medium Now I’ve already given you a hint as to how to approach modeling of a lossy medium Lossy medium can be modeled by letting sigma be non-zero and having some finite value of that. Now the notable change that would happen in Maxwell’s equation for such a lossy medium is this curl expression of the Maxwell-Faraday expression wherein the previously neglected term, which is j is now brought back into it, so which is given by sigma e plus assuming still that we are working with phasors, I can write the phasor expression as j omega epsilon E This comes from another observation timing, perhaps a different course that conduction current density is actually proportional to the electric field and this proportional to constant is called as the conductivity of the medium, and this conductivity carries units of siemens per meter. The old notation was mhos per meter. Now this is the preferred SI notation of siemens per meter. And of course, it makes sense, because j is current density, which would be measured in ampere per meter square, and this fellow is siemens per meter and there’s voltage, v per meter, so v into s is like voltage times conductivity, which is basically current ampere, so ampere per meter square is what you’re going to get So this is the only change that would happen in your Maxwell’s equation. The other equations are as they are. So you can even simplify this expression by writing this as sigma plus j omega epsilon E and I can write this as j omega epsilon c E, where you can see that epsilon c is a parameter that has simply introduced and this parameter should actually be equal to some sort of — that would actually be complex and that should be related to sigma and epsilon. How should it be written. So we know that j omega epsilon c should be equal to sigma plus j omega epsilon, these two equations has to remain the same, and then I know that epsilon c is a complex permittivity that has introduced, which I can write as epsilon’

–j epsilon”. So I now have j omega epsilon’ – j epsilon” should be equal to sigma + j omega epsilon, where epsilon is your original epsilon that we had considered not the complex permittivity that we have considered. So I can write this as j omega epsilon + j omega — sorry j is not there, because that has gone now — so omega epsilon”. This should be equal to sigma + j omega epsilon. So the complex part can be equated to this one saying that epsilon, which we had taken, is actually the real part of the complex permittivity epsilon’ and then whatever that is left out, omega epsilon’ is equal to sigma. Alternatively, epsilon” is equal to sigma/omega So I can write the equation without changing anything, by simply introducing this complex permittivity epsilon c and working with the entire thing, identifying that the original permittivity of the lossless media can be taken as the real part of the complex permittivity So this is something that I can do. In fact, you would see epsilon’ and epsilon’’ being used in many electromagnetic materials that we considered So what we have seen is this epsilon” can be written as sigma/omega and this can be included to talk about the lossy medium. Let us now see what happens to the wave function when we include this lossy materials or lossy medium. So I have DEL x E = -j omega μ₀ H as before, so there’s no change in this one, and now I will take the curl of this electric field again. So I will get curl of electric field. Sorry, I should take the curl here. So this would be curl of the electric field that should be -j omega μ₀, and then I will end up with DEL x H which as I have seen can be written as sigma + j omega epsilon E or equivalently I can write this as j omega epsilon c E I’ll go with the former, because I want to bring in the transmission line analogy as well. So I am going to use the full expression for complex permittivity. So I will write this as sigma + j omega epsilon only, and in this case epsilon will be a real quantity Sigma will of course also be real. The left hand side we have already seen can be reduced down to –DEL square E. the presence of conduction current density does not alter this equation, DEL.D = 0. This equation will still remain the same and consequently DEL.E will also be equal to 0, and in this expression of the vector Laplacian, you can see that when you expand this out, that DEL(DEL.E) term will go to 0. So I am having on the left hand side a –DEL square E, which of course would be onto the right hand side given by -j omega μ₀ (sigma + j omega epsilon) E Minus sign can be cancelled off and also considering the fact that we’re not looking at waves in arbitrary direction, we’re looking at waves which are propagating along the z direction, and x polarized waves is what we have been considered, x polarized linear waves that we have considered. I can replace this DEL square simply by DEL square /DEL z square, which is what I am actually interested in When I do that, I’ll have DEL square Ex/DEL z square to be equal to j omega μ₀ (sigma + j omega epsilon) Ex. So this would be the expression that I would have. Of course, instead of DEL square / DEL z square, because we are dealing with a phasor. I can simply write it as d square Ex/dz square. I can now write this entire thing as say gamma square Ex and my solutions with respect to z, the phasor solutions with respect to z will be some amplitude, which we will call as some Ex0, and because we are considering z direction wave, so we can write this as e-gamma z, where gamma square is equal to j omega μ₀ (sigma + j omega epsilon), or gamma itself is equal to root j omega μ₀ sigma + j omega epsilon, which can be further written as alpha + jβ, and alpha is the real part of gamma, which is called as attenuation constant or attenuation coefficient, and β is the imaginary part

of gamma , which is called as the propagation constant or the propagation coefficient This has given by square root of j omega μ₀ (sigma + j omega epsilon) that gamma. Now if you recall the transmission line where we have taken, say, let’s assume that we’ll take R=0, we’ll take G not equal to 0, we will take L not equal to 0 and not equal to 0, c also not equal to 0, then this gamma expression in the transmission line case would actually be given by (j omega L) (G+j omega C). Is that correct? Yes, because I’ve taken only R=0, these, I am neglecting the series resistance of the wires. So I am assuming that the wires actually have some amount of or the material that I have considered is lossy material, which is modeled by having this conductance G there. So this is what your gamma in the transmission line expression would look like, and now you can see that it would essentially be similar to the expression that we have written in a lossy medium Here also the transmission line was lossy because of the presence of this G, and clearly because G is kind of similar to sigma, this corresponding medium is now lossy as a result of this. So the expressions that we have written for gamma is essentially similar to the expression that we have written or used in the transmission line case as well. Now you may also find out that the impedance that we are going to write, earlier the characteristic impedance of a transmission line when it was lossy was given as R + j omega L/G + j omega c under root So by taking this analogy between the transmission line and the lossy medium that we are considering the characteristic impedance or the wave impedance or the intrinsic impedance of the medium when the medium is lossy would be correspondingly given by j omega μ₀/sigma + j omega epsilon This expression equal eta₀, that is eta = eta₀ when sigma = 0 and epsilon is actually equal to epsilon₀. So this expression that we have written, the medium impedance or the wave impedance or the intrinsic impedance, this would be equal to the — it would be similar to the expression for a lossless or lossy transmission line. So this is the propagation of wave in a general loss media There are some situations where the sigma is very, very small compared to omega epsilon In this case, the medium is mostly a dielectric medium. However, at different frequencies it may not remain a dielectric material. When sigma is much greater than omega epsilon, then you have a pure conductive medium which is very, very lossy. That is the reason why most metals actually are very lossy at very high frequencies. So basically their sigma value becomes very high omega epsilon. Of course, you can bring them to be kind of a dielectric, but you have to actually increase the value of omega to very large extent there So it’s also interesting to look at what would be the expression for gamma when it is pure dielectric. When it dielectric, we know that gamma is given by pure j omega root μ₀ epsilon, epsilon = epsilon₀ will give you the propagation constant in free space, epsilon – epsilon₀ epsilon r will give propagation in a dielectric medium where this condition holds. So this sigma is so small that I can neglect that one. The corresponding impedance is also going to be quite real. It would be root μ₀/epsilon. This is for the dielectric scenario On the other hand, for the conductive scenario, gamma will be equal to root j omega μ₀ sigma, because j omega epsilon is very small compared to this one. So gamma will be equal to root j omega μ₀ sigma, which of course will be equal to alpha + jβ, correct. Any complex number is equal to real and imaginary party. So what is alpha here? In fact, the magnitude of alpha and β both are equal, and they are given by root omega μ₀ sigma/2 under root. Why is this? Because I can write this square root j omega μ₀ sigma as square root e-jpi/2 omega μ₀ sigma. This can be written as root ejpi/2 square root omega μ₀ sigma, and the square root ejpi/2 is basically eipi/4, which is actually 1 + j/root 2. So now you can see that alpha and β both should be equal to root omega μ₀ sigma/2. So this actually means that the medium is lossy, but also has a propagation constant, lossy, but

has a propagation constant β whose magnitude is actually equal to alpha. So this is a very interesting, this one In fact, this alpha which is given by root omega μ₀ sigma and writing omega as 2 pi f and substituting omega as 2 pi f and canceling this 2 will give square root pi f μ₀ sigma The expression that you have e-alpha z as the z propagating wave or the attenuation along z can be re-written as e power -z/delta where delta is equal to 1/alpha, which is equal to 1/root pi f μ sigma or μ₀ sigma for non-magnetic medium This delta is a very important parameter called skin depth. This is called a skin depth meaning that if we launch an electromagnetic wave into a completely lossy or conductive medium, with an initial amplitude of A, then the amplitude decays exponentially as A e power –z/delta, this is a z axis, so this is the medium, let’s say, which is medium 1 have sent in the wave medium 2. We will have to say more about what happens when medium 1 and medium 2 interact, but for now, we will simply assume that the wave that has been launched into the second medium, which is basically lossy and conductive actually starts off with an amplitude to A at z=0, but as you have seen here, the amplitude decays exponentially and over phase z equal to some four delta or so, the amplitude would have actually decayed almost to 0. The amplitude would be just about 1% of what it started out with So that is why this delta measure the depth over which the wave can actually, the typical width over which there is important distance of the depth over which the wave can go into the lossy medium can penetrate the lossy medium is about 4 delta. So DEL is called as skin depth, and it actually is important not only to know what is the depth, but actually is important in practical scenario when you’re designing shields Suppose I have a source of electromagnetic waves and I want to prevent. So I have some device let’s say, so let’s say, this device is motherboard and these EM waves are being generated externally for whatever reasons, maybe lightning or whatever other reasons, or it could be that EM waves are generated by the motherboard and this device is, let’s say, television or some other kind of a device that I am looking at, and I want to shield the electromagnetic waves between these two The EM waves which are going to impinge on the device, if the device is not shielded would actually simply go and fall onto or incident onto the device However, if I build a metal, I build a metal shield or I kind of build a metal ground around this one of thickness which is approximately 4 delta, the thickness of this metal would be approximately 4 delta, and please remember this delta is now dependent on sigma as well as the frequency. So if the EM waves are coming in at 50 hertz, then frequency omega is very small, delta is actually going to be very small as well. However — sorry, delta is inversely proportional to this. So delta we wrote it as 1/root pi f μ sigma and at low frequencies the skin depth will be larger, meaning that the waves can penetrate more into the metal or the lossy material, whereas when frequency increases, the skin depth reduces So if you know that the worst case electromagnetic waves that would be incident on your device would be at 50 hertz, then if you build a metal house or you house this device inside a metal with a depth of, say, 4 delta as determined by the material conductivity, then that metal or that house is actually sufficient, or the thickness of the device is sufficient to withstand, even when your f is, say, 2.4 gigahertz, which would come from your cell phone So this idea of what should be the thickness which is determined by the material conductivity and the frequency of operation, the worst case operation and best case operation is not so straight-forward as I have written, but this still gives you a basic idea of what to expect when you want to shield certain devices from electromagnetic interference The concept of skin depth of a material is very important in this case in order to determine the requisite thickness of the material So with this, we stop our discussion on uniform plain waves in lossy materials. We will have

to say more about the interfaces of these waves. So as we have seen here that we start off with a wave in one medium, and then we somehow said that the medium two will carry the wave, but exactly what happens at the interface and whether this — I can just take the electromagnetic wave from one medium and then convert or push the electromagnetic wave into the second medium without any changes at the boundary is something that we have to explore, and we’re going to do that in the next module. Thank you very much