 Welcome to the 13th lecture in the course Engineering Electromagnetics. We introduced the concept of wave propagation last time. So today we continue with wave propagation and the topics for discussion today are properties of uniform plane waves and the case of sinusoidal time variation. You would recall that we considered the Maxwell's equations last time and wrote these for certain simple situations. The Maxwell's equations were written as del cross H equal to epsilon del E by del t, del cross E equal to minus mu del H by del t, del dot E equal to 0 and del dot H also equal to 0. You would recall that we obtained this form of Maxwell's equations when we considered a source free region alright. We also considered that the medium is homogeneous as well as isotropic. For such a medium which is homogeneous and isotropic the constitutive relations take a very simple form epsilon mu and sigma become scalar constants and when these are incorporated in the Maxwell's equations we get this form of Maxwell's equations under the source free condition. This kind of conditions are satisfied by free space or an infinite perfect dielectric medium. To obtain a law obeyed by just the electric field intensity or just the magnetic field intensity we combined these equations suitably and obtained what we called the wave equation one of the wave equations read as del 2 E equal to mu epsilon del 2 E by del t square which under further simplifications that converted into the following form we had del 2 E by del x square equal to mu epsilon del 2 E by del t square. This was the situation when we assumed that there are no variations with respect to y and z and we considered the case when there are variations only with respect to x present. We further focus attention on one of the component equations of this vector equation for example, the y component and wrote del 2 E y by del x square equal to mu epsilon del 2 E y by del t square. The solutions of which were written as E y equal to f 1 of x minus v naught t and f 2 of x plus v naught t v naught being equal to 1 by square root of mu epsilon where mu and epsilon are the permeability and the permittivity of the medium that we are considering. Since we are considering the medium to be either free space or a perfect dielectric so the conductivity is not appearing in this expression. And then we identified the solutions of E y as representing propagating waves propagating in the positive x direction propagating in the negative x direction their proportion depending on the boundary conditions. This case where there are no variations with respect to y or z direction is known as the case of uniform plane waves. And as I mentioned last time this is an important case although it appears highly simplified and therefore, we will consider the properties of this kind of uniform plane waves. We mentioned that far away from sources most waves can be represented at least locally as uniform plane waves. To consider the properties we focus attention on one of these equations. Let us number these say 1, 2, 3, 4, 5 and let us write 6 here. We start with equation 6 and write the various component equations corresponding to this vector equation the x, the y and the z components giving us del 2 E x by del x squared equal to mu epsilon del 2 E x by del t squared then del 2 E y by del x squared equal to mu epsilon del 2 E y by del t squared. And finally, del 2 E z by del x squared equal to mu epsilon del 2 E z by del t squared. The spatial derivatives with respect to y and z are not appearing because we made this simplifying assumption that there are no variations with respect to y and z direction which is a plane normal to the direction of propagation. And therefore, in planes normal to the direction of propagation it is a completely uniform wave there is no variation and hence the name uniform plane wave. Next we consider equation 3 and write out that also in Cartesian coordinates expand the divergence function, divergence of a vector E and that will give us del E x by del x plus del E y by del y plus del E z by del z equal to 0 under the present conditions that there is no source that we are considering. Now, since y and z variations have been assumed to be 0 this gives us del E x by del x equal to 0 which when used in this first equation tells us that del 2 E x by del t squared must also be 0. So here we have a quantity E x whose x derivative is 0 and the second order time derivative is also 0. So, the possible solutions for E x are it is either 0 or it is some constant let us say k constant with respect to x as well as time or else it is increasing linearly with time mathematically these are the possibilities that come up. However, physically and considering the properties of the characteristics of a wave propagating wave we say that something which is existing for all space and for all time or which is increasing linearly with time for all space is not going to be part of a propagating wave. And therefore, on that basis we reject these two mathematical possibilities and say that E x must be equal to 0 for such a wave. Well, we have considered the set of equations and we see that del E x by del x should be 0 and del 2 E x by del t squared must also be 0. And therefore, the various mathematical solutions for E x would be 0 or constant or something which is linearly dependent on time. The second two solutions would indicate that there is something which is existing at all points in space at all time or else at all points in space it is increasing with time situations which are physically quite impossible. In any case they do not fit in the description of a propagating wave which we described as some disturbance which occurs at one place at some instant of time and then is repeated at other points of space at later instance of time. And the time delay is proportional to the spatial separation between these points. So that aspect is completely missing from these two possibilities and therefore, we reject these and say that for propagating waves E x must be equal to 0 if we are dealing with uniform plane waves. Since the equations that we are dealing with now are completely symmetric in E and H, one could have considered the corresponding magnetic field wave equation and combine it with equation number 4. And arrived at the final result that H x is also equal to 0 running through similar arguments which is an important result. It says that a uniform plane wave propagating in the x direction uniform plane wave where y and z variations are absent propagating in the x direction therefore, will have no x components of either the electric field intensity or the magnetic field intensity. Now, since the choice of x y z is arbitrary it is in our control we could change this by simply rotating the coordinate system. We generalize the statement by saying that a uniform plane wave has no components along the direction of propagation. We can write this down a uniform plane wave has no field components in the direction of propagation. If it is propagating in the x direction then the x components would be absent and so on. What could exist are the other two components y and z which would be transverse to the direction of propagation putting it in general terms. We have no restriction on those in fact, anticipating such a result we have focus attention on the y component of the electric field to make out what kind of solutions this equation is going to have. So, one of the first properties of the uniform plane wave that we have been able to see is that such a wave has no field components in the direction of propagation. And once we are putting it in this kind of general words there should be no confusion at which component would be absent which could be present. Then we go on to the next property. And for this purpose we consider the curl equations that is equations 1 and 2. And first we obtain the curl let us say of h using the expression like this x cap, y cap, z cap and then del by del x del by del y which we have assumed 0. So, we write 0 here. So, that we do not forget and similarly del by del z is also 0. Then we come to the components of h x h y h z. However, h x as we have just seen for the situation that we have set up is going to be 0. And therefore, we will have only h y and h z. Therefore, del cross h or let us say from equation 1 we get the left hand side is equal to minus del h z by del x this should be del by del x minus del h z by del x y cap plus del h y by del x z cap. On the other hand the RHS can be written as epsilon del E y by del t y cap plus epsilon del E z by del t z cap. Since the x component is not going to be existing E x is 0 as we just made out. Now, comparing these two equations and comparing the corresponding components y and the z components we obtain two more equations reading minus del h z by del x equal to epsilon del E y by del t on one hand. And del h y by del x equal to epsilon del E z by del t on the other hand. One could as well have started from equation 2 obtaining two more relations like this. But the conclusion that we want to draw for that purpose these are enough. Now, we substitute the solution for E y that we have obtained. Let us say the one part the part which represents waves propagating in the positive x direction. What difference would be there if one chose the other part we will mention later on. Let us focus attention on one of the parts and consider a situation where only a forward propagating wave or a wave propagating in the positive x direction exists that is E y is equal to some function with argument x minus v naught t. Let E y be this. So, that del E y by del t is equal to del f 1 of x minus v naught t by del x minus v naught t multiplied by the derivative of the argument with respect to time that is del x minus v naught t by del t. Realizing that this part is v naught and this part minus v naught thank you and this part may be represented as f 1 prime representing the derivative with respect to the argument entire argument. This comes out to be minus v naught f 1 prime. We may number these equations also this one is number 7 and this one is number 8. Now, considering equation 7 we obtain nullifying the negative sign on both sides del h z by del x equal to epsilon v naught times f 1 prime where f 1 is E y. Therefore, what is the expression for h z h z is going to be epsilon v naught f 1 prime integrated with respect to x and for this indefinite integral some constant something which is constant with respect to x. Now, utilizing this kind of mathematics it is easy to see that this part will be simply f 1 and since c will be something which is constant for all space and would not form a part of a propagating wave and therefore, it is ignored or rejected giving us h z equal to epsilon v naught times f 1 and recognizing that v naught is 1 by square root of mu epsilon it is going to be equal to epsilon by mu square root into E y since f 1 is E y. So, that E y by h z is equal to square root of mu by epsilon there exists a very simple relationship between the y component of the electric field and the z component of the magnetic field. One could have considered the other pair say equation 8 where it is E z and h y which are involved and obtain the relation E z by h y equal to minus square root of mu epsilon. The steps are identical there is no point in going through the steps it can be verified very easily. Now, let us see if there is a relation which exists between the overall electric field and the overall magnetic field at the components level this kind of relation we have seen for that purpose therefore, let us write E which is the magnitude of the complete electric field. Now, E x is 0 and therefore, this is simply E y square plus E z square and substituting for E y and E z from these relations what we are going to get is that E is equal to square root of mu epsilon into square root of h z square plus h y square or square root of mu by epsilon magnitude h. So, that we obtain between overall E and overall h also the relation that E by h is equal to square root of mu by epsilon. So, it is a recurring relation it appears whenever we consider appropriate field components. Now, if we consider the units of E and h volts per ampere volts per meter and amperes per meter. So, the units of the left hand side are volts upon amperes that is ohms and therefore, the units of this new expression new quantity on the right hand side should also be ohms. So, on this basis this quantity square root of mu by epsilon is called the characteristic or the intrinsic impedance of the media. It is a property intrinsic property of the medium in which we are considering the uniform plane wave and this is the intrinsic impedance. There is some distinction which is made between the nomenclature for lossless media and for conducting media. We will not go into that here we will call this intrinsic impedance and usually given the symbol eta. For free space eta can be calculated since we know mu naught and epsilon naught. Let us write it here eta for free space and it comes out to be 120 pi or 377 ohms an interesting result. Just free space also has some intrinsic impedance from this point of view and it has a certain value 120 pi or 377 ohms. If one had a medium with different mu and epsilon this intrinsic impedance value will change according to this expression square root of mu by epsilon. So, another very important property that for a uniform plane wave the electric and the magnetic field components as well as the overall electric and magnetic field intensities are related through the intrinsic impedance of the media. It would have been nice if one did not have to care about this negative sign. If one had a uniform relation for whatever ratios one considered, but there is a very simple way out for remembering when to put the negative sign and when not to. You consider the cyclic order of the coordinates x, y, z. The order in which they are appearing e, y, h, z that is the proper cyclic order y, z, x and the wave is propagating in the x direction and therefore, this has a positive sign. Here the cyclic order is disturbed. So, we get a negative sign. In fact, based on this one can now consider the other part of the general solution for e y which represents waves propagating in the negative x direction and then apply this consideration and obtain appropriate ratios of e y, h, z and e z, h, y. However, the overall e and h will have this magnitude square root of mu by epsilon. So, this is the second important property that for a uniform plane wave the electric field and the magnetic field are related through the intrinsic impedance of the media. What was the first property? The first property was that a uniform plane wave has no components, field components in the direction of propagation which are also many times called the longitudinal field components. So, a uniform plane wave has only transverse field components and they are related through the intrinsic impedance of the media. Let us go on to the next property. For the next property we consider the dot product of e and h. It will give us the relative orientation of the electric and the magnetic field. So, the dot product is going to be e y, h y plus e z, h z. What about the third term? e x, h x? e x and h x are 0 for this way. Now, one can substitute for say electric field in terms of the magnetic field components and vice versa. Let us substitute the electric field components in terms of the magnetic field components. So, that e y is square root of mu by epsilon which can be taken out as common and h z, h y and e z is minus square root of mu by epsilon, h y so that it is minus h y, h z which is equal to 0. Indicating clearly that the electric and the magnetic field intensity vectors are orthogonal. They are perpendicular to each other. As it is, they existed in a plane transverse to the direction of propagation and among themselves they are normal to each other or they are orthogonal. And therefore, this wave can also be called a transverse electric and magnetic wave, a wave which has no longitudinal field components and in such a wave e and h are also orthogonal to each other. And that simplifies many things. Now, if only one of the field components is given to us, we can easily make out through this relation and through this observation the value of the other field component. And this system that appeared such a maze considerably simplifies particularly for the uniform plane waves. Another operation that can be performed is the cross product between the electric field and the magnetic field and that will give us, one can write it here so that no mistake is made x cap, y cap, z cap, e x of course is 0. So, 0 e y e z and similarly 0 h y h z. So, that e cross h is going to be x cap into e y h z minus e z h y with no y or z component, which can be simplified to read as, we can once again substitute for e y and e z in terms of the magnetic field components. So, that we get square root of mu by epsilon here and here we have h z square and here one will get h y square that is x cap eta times h square. It could as well have been put in the form of e square in terms of e square, but the direction would remain x cap, which is what we are aiming at just now. The cross product e n h is in the direction of propagation. This is the let us say the fourth property of uniform plane waves. One can recapitulate these. The first property we observed was that there are no longitudinal field components, no components in the direction of propagation. Next was that e n h are related through the intrinsic impedance. The third was that e n h are mutually perpendicular and as we have seen here also to the direction of propagation. So, e h and the direction of propagation are mutually perpendicular and e cross h is in the direction of propagation. Excuse me sir, does this cross product represent anything like the energy or power? You have gone a little ahead of us, e cross h we will see will be called the pointing vector and it can be related to the flow of power associated with the wave, which is also going to be an important consideration. Because after all when a wave propagates from a source say from a transmitter to the receiver some information is transmitted or perhaps something is actuated. So, there is some transfer of energy on an overall basis and we must have some expression to estimate or calculate this transfer of power and energy. So, this vector will help us at that time, perhaps the next lecture. Any other questions at this point of time? So, then we give a stop to the properties of uniform plane waves go on to the next topic. Now, so far we have not considered any specific or particular time variation, but as you all know the case of sinusoidal time variations is very important. Most generators generate signals which vary with time in a sinusoidal manner even if they do not then we know that all periodic signals can be expressed as some of sinusoidal using Fourier series expansion. And therefore, the case of sinusoidal time variations is very important and that is what we take up next. We know that the sinusoidal time variations can be very easily handled by adopting the phasor notation. So, that although we have quantities say e or h which are say functions of space let us say x, y and z or as in this case we saw as a function of x and time. They could be represented in terms of a phasor which is just a function of x and we could put a vector sign which is usually the case of space where the time variation is not there explicitly. However, the time variation can be brought out or incorporated explicitly by multiplying by e to the power j omega t and considering the real power. And using phasor notation you would recall that the time derivative is very easy to handle it is equivalent to multiplication by a factor j omega. The first thing that we need to do to handle sinusoidal time variations is to write the Maxwell's equations for this case for sinusoidal time variations using phasor notation. The process is very simple, but for the sake of record let us write it down we have del cross h equal to j omega epsilon. Similarly, del cross e is equal to minus j omega mu h and del dot e is equal to 0 del dot h is also equal to 0. This of course is not the general form of Maxwell's equations for sinusoidal time varying signals. It contains the various simplifying assumptions that we made in writing this set in particular. These are the equations for the source free regions and for media which are homogeneous and isotropic. So, all those assumptions we continue with for the sake of simplicity. Now, one could either combine these equations once again to obtain a law obeyed by e or h or utilize this simplified thumb rule that each del by del t is actually equivalent to a factor j omega in the phasor notation. And once we do that various equations can be written down in a simple manner for the sinusoidal time variation situation. For example, the wave equation would become del 2 e e equal to minus omega squared mu epsilon e. Equation 5 would convert to such an equation for sinusoidal time variations. Now, it is derived from wave equation, but it is given a separate name and it is called vector Helmholtz equation. The difference between the wave equation and the Helmholtz equation being that the time variation has been taken care of a time variation has been specified. In particular for sinusoidal time variations the wave equation converts to a Helmholtz equation. If you consider the components of this vector equation there will be scalar Helmholtz equations. A second order partial differential equation and the literature is full of the techniques for solving this for various situations. Now, once again we can continue with the simplifications. So, that we reach the uniform plane wave solution the corresponding solution for the sinusoidal time variation. And therefore, keeping the assumption that there are no y or z variations we are going to have del 2 e by del x squared equal to minus omega squared mu epsilon e. And then focusing attention on the y component we will have del 2 e y by del x squared equal to minus omega squared mu epsilon e y. So, that now one can attempt a solution by writing this as minus beta squared e y with beta being equal to omega times square root of mu epsilon. And one is now in a position to write the general solution it will be e y equal to some constant c 1 times e to the power minus j beta x plus c 2 e to the power plus j beta x. And how such a solution corresponds to these general solutions after all this is just a specific time variation. And it should belong to this general set that one can see by realizing or recognizing that this is in phasor notation. And time variation can be brought in explicitly by multiplying by e to the power j omega t and taking the real part. When that is done we get the actual time varying e y which is a function of x as well as t equal to c 1 cosine omega t minus beta x plus c 2 cosine omega t minus beta x plus beta x considering that c 1 and c 2 are real. If they are not they will contribute some phase angle to the argument. And continuing here this can be rewritten as e y as a function of x and t is equal to c 1 cosine of one can take out minus beta x. This common so that one has x minus omega by beta t on one hand and c 2 times cosine of plus beta x plus omega by beta t on the other hand. And now by comparing this form with the earlier general form one is in a position to say that this solution this particular solution belongs to the general solution we proposed earlier. That is it is in the form of functions which are of argument x minus omega by beta t. So, that one can identify omega by beta with the velocity of propagation in the two different directions beta expression is available to us. So, that it is simply one by square root of mu epsilon. So, whether it is the case of general time variation or the specific case of sinusoidal time variation uniform plane waves would propagate with the same velocity v naught equal to one by square root of mu epsilon. Another thing that brings out that is brought out very clearly here is that for a sinusoidal time varying signal the space variation is also sinusoidal. If you considered a snapshot of any one of these functions as a function of x as a function of the space coordinate that also will be a sinusoidal variation which is something that we have seen for the case of the transmission lines also. And therefore, whether we consider a snapshot in time or we consider at a particular value of x time variation they are both sinusoidal. Beta here is called the phase shift constant and one can also introduce the concept of wave length the distance over which the phase changes by 2 pi radians. And therefore, beta lambda should be equal to 2 pi giving us a relation between lambda and beta that is lambda is 2 pi by beta or one could put it in any other way. So, that the wave length is related to the phase shift constant. When we are talking about sinusoidal time varying signals now one can identify a certain phase front and say clearly that uniform plane waves have a plane phase front or plane wave front. If we consider equi phase points they will lie on a plane. This justifying the nomenclature uniform plane waves. If you have any questions in the two topics that we have considered today we can try out those now. In today's lecture we first considered the properties of uniform plane waves. We saw that they have no longitudinal field components. The electric and the magnetic fields are related through the intrinsic impedance of the medium. E and H vectors are perpendicular to each other and also to the direction of propagation. And then we considered the special and quite important case of sinusoidal time variation and identified the expression for the phase shift constant and the wave length. So, we end this lecture here.