 Welcome to the 28th lecture in the course Engineering Electromagnetics. For the last few lectures we have been discussing the parallel plane guide. This guide consisted of conducting planes which were infinite in dimensions. Obviously such a wave guide cannot be realized in practice. A more realizable shape of guide is what is called a rectangular wave guide and that is what we start with today and we shall go on to consider the transverse magnetic modes in such a wave guide. First let me show you what is it that we have in mind when we say rectangular wave guide. The rectangular wave guide can be represented by the following schematic figure. It consists of perfectly conducting metal walls at x equal to 0 and x equal to a and also at y equal to 0 and y equal to b and the region inside this enclosure is completely hollow. It can be evacuated, it can be filled with some inert gas or it may be just simple air. Such a wave guide is called a rectangular wave guide of course it will continue depending on the length of the wave guide required for some distance along the z direction. The rectangular wave guide belongs to the general class of cylindrical wave guides where the cross section could be circular or rectangular. So instead of saying a rectangular cross section cylindrical wave guide for brief we simply say rectangular wave guide. Similarly, a circular wave guide would be one which has a circular cross section. Apart from these straight forward cross sections one could have any cross section in general and it could work as a wave guide. However these other cross sections offer really no advantage, they are not easier to manufacture and they are certainly more difficult to analyze and therefore rectangular wave guides or circular wave guides are by far the most popular types of wave guides. Having introduced what a rectangular wave guide looks like we should also try to explain the significance of going to such a form. Compared to parallel plane wave guide we have already said that that structure is not practically realizable. The cylindrical wave guide or the rectangular wave guide in particular is more realizable but otherwise considering the various types of transmission lines and the other guides that we have considered so far. What is the place of the cylindrical wave guides that we will like to clarify. We have considered different types of transmission lines and guides and the general purpose for these structures was mentioned in the beginning of the course to transport the electromagnetic signal or energy efficiently and without any distortion. These are the requirements for a good transmission structure. Now let us evaluate the various structures we have considered so far from these points of view. Some of the structures for example the parallel wire transmission line and also the parallel plane guide are likely to radiate if they are operated at very high frequencies. Since they are open, radiation is going to be considered in more detail later on but just the fact that they are open can result in radiation if the frequency of operation is raised sufficiently high. Also they are subject to interference because interfering signals can enter the transmission structure. So therefore there are types of structures which are open in nature and they will cause problems sooner or later. One could argue that we consider the coaxial transmission line which is completely enclosed and therefore is not likely to radiate. However the coaxial transmission line as also some other types of transmission lines require dielectric material for support. And therefore in addition to conductor loss, dielectric loss is always present. Although we now have available dielectric materials which have very low loss up to fairly high microwave frequencies like 30, 40 gigahertz, still dielectric loss is going to be present and therefore they are going to be lossy compared to this kind of wave guides. Therefore the advantages of this kind of cylindrical wave guides can be put down. They will not radiate and therefore they are not subject to radiation or interference and also as we have just tried to explain they have very low attenuation. The wave guide will be fabricated out of very good conducting material and it can even be silver plated to reduce the attenuation further and also there is no requirement for using a dielectric material. So dielectric loss is all together eliminated. There is another advantage to using this kind of wave guides and that becomes clear when we consider the power handling capability. The materials that use some sort of dielectric will have limited power handling capability because of the dielectric break down. If the electric field increases beyond a certain value the dielectric material will undergo break down and the guide will not be able to propagate power levels higher than that particular value. The wave guide since it uses no dielectric material can transport much higher power levels. Wave guides can handle megawatts of power level if they are appropriately designed and therefore apart from freedom from radiation or interference and very low attenuation they can handle very high power levels. And while we have other relatively newer transmission structures evolved for example strip lines and micro strip lines which we are not discussing in this course but will be discussed in a subsequent course. When we go to very high microwave frequencies, millimeter wave frequencies for example 100 Gigahertz those alternative newer structures do not offer performance which is comparable to these wave guides. They suffer from higher loss and radiation and therefore at very high frequencies let say millimeter wave frequencies one will still use this kind of cylindrical or rectangular wave guides. And that clearly brings out the areas where this kind of wave guides are going to be utilized wherever the application requires very low attenuation or high power handling capability or we need to use it at very high frequencies like hundreds of Gigahertz we would utilize this structure. And therefore these wave guides have their own very special place in this area. The objective for us today and the subsequent few lectures is to understand the type of waves that the rectangular wave guide can support and what are the associated modes in these waves and what are the characteristics of these modes that is the propagation characteristics and the field configuration etc. The method that we are going to follow will be very similar to the method that we followed for the parallel plane wave guide that is we shall write down the Maxwell's equations for the region inside the wave guide which will be assumed to be a perfect dielectric at least to begin with. And then these Maxwell's equations will be solved subject to the boundary conditions imposed by this perfectly conducting enclosure constituted by the wave guide alright and therefore from that point of view the concept should be quite familiar. We write down the Maxwell's equations for the region inside the wave guide del cross wave guide del cross h equal to j omega epsilon e and del cross e equal to minus j omega mu h. The way these equations are written a few things are apparent we are assuming that the medium filling the wave guide is isotropic homogeneous time invariant etc. And also we are assuming that the fields are having a time variation which is sinusoidal and we have written these equations using phasor notation. So that each del by del t factor is replaced by the factor j omega. The wave equation satisfied by the magnetic field intensity vector and the electric field intensity vector is also written here del square h equal to minus omega square mu epsilon h which can be substituted in terms of further symbols and a similar equation is satisfied by the electric field intensity vector. Next we need to expand these vector equations into their corresponding components. In a suitable coordinate system a suitable coordinate system will be one where the boundary conditions applicable to the structure at hand can be applied in a simple manner. Since the boundaries are described very easily in terms of x equal to constant or y equal to constant kind of functions therefore the Cartesian coordinate system will be the best one to expand these Maxwell's equations and that is what is done in the following equations. These are the three component equations corresponding to this Maxwell's equation and these are the corresponding three component scalar equations for this Maxwell's equation. Also the wave equation can be written out in terms of the x y and z derivatives for the magnetic field and a similar equation will hold good for the electric field. For example we will have del 2 e by del x square plus del 2 e by del y square plus del 2 e by del y square plus del 2 e by del x square plus del 2 e by del z square equal to minus omega square mu epsilon alright. So these are proceeding in a very general manner the various equations. The next step as you would recall from your experience with the parallel plane guide was to try and put down the nature of these variations with respect to x y and z to the extent possible and there we can easily identify at least the nature of the z variation if we assume that the wave is propagating in the z direction then the variation in the z direction can be written as e to the power minus gamma bar z where gamma bar becomes the propagation constant and in general it is alpha plus j beta that is it may have a real part and an imaginary part. So that as far as del by del z is concerned the partial derivative with respect to the z direction that can be replaced by minus gamma bar where not in a position to put down the variations with respect to x or y directions for the moment those will emerge from the boundary conditions. But the z derivative in these various equations can be replaced by minus gamma bar and that is what we undertake next taking up this equation we will have del h z by del y plus gamma bar h y equal to j omega epsilon e x then minus gamma bar h x minus del h z by del x will be equal to j omega epsilon z by del x. E y and the last component equation will be del h y by del x minus del h x by del y equal to j omega epsilon E z since there is no z derivative in this equation it remains as it was. In a similar manner we write down the three component equations corresponding to the second Maxwell's equation and completing this gives us the following equations del e z by del y plus gamma bar E y as you can notice we can get this set by simply interchaining E and h and epsilon and minus mu. So that is what we are doing this is equal to minus j omega mu and h x. So just by inspection the other set can be written the second equation is minus gamma bar E x minus del e z by del x which should be equal to minus gamma x j omega mu h y and finally del e y by del x minus del e x by del y should be equal to minus j omega mu and h z. The wave equations for the h and e further give us del x minus del 2 h z and we write this for the z components del 2 h z by del x square plus del 2 h z by del y square and then the third term becomes plus gamma bar square h z which should be equal to minus omega square mu epsilon h z. In a similar manner the z component for the wave equations satisfied by the electric field will come out to be del 2 E z by del x square plus del 2 E z by del y square plus gamma bar square E z which should be minus omega square mu epsilon E z. That is as far as one can proceed by identifying the nature of the z variation. The next step would be to use these equations in an appropriate manner and obtain expressions for the transverse field components in terms of the longitudinal field components. The longitudinal field components are the ones which are in the z direction and the ones transverse to this direction x or y components are the transverse field components and those we can express in terms of the longitudinal field components. So, since the longitudinal field components are going to play a somewhat special role we have written the wave equations only for those field components. So next we attempt to write transverse field components in terms of the transverse longitudinal field components. For example, one could pick up this first equation in this lot which involves E x and H y and of course, H z del H z by del y and similarly locate another equation which involves same two transverse field components which happens to be the one here. It involves E x and H y with other multiplying factors and then del E z by del x. So, if we combine these two equations we should be able to get E x and H y in terms of derivatives of E z and H z. For example, the steps will be as follows from here from this equation we will get H y equal to 1. 1 by j omega mu and then gamma bar E x plus del E z by del x which expression for H y can be substituted here. So, that we get j omega epsilon E x equal to del H z by del x del y plus gamma bar by j omega mu into gamma bar E x plus del E z by del x. So, we have got now two terms containing E x which can be combined and then E x can be expressed in terms of del H z by del y and del E z by del x. Upon simplification this gives us the following result. We have E x equal to minus j omega mu by H square del H z by del y minus gamma bar by del x minus gamma bar by del x minus H square del E z by del x where H square is a new symbol substituted for gamma bar square plus omega square mu epsilon. So, this is one illustration of how we can get the various transverse field components in terms of the longitudinal field components. By continuing with the same two equations in that pair this one and this one we will be able to get the expression for H y in terms of the longitudinal field components. And then by picking up these two equations we will be able to get the remaining transverse field components. So, having explained the procedure I think now we can put down the final expressions which read as follows E y equal to j omega mu by H square del H z by del x minus gamma bar H square del H z by del x minus gamma bar H square del E z by del y. And similarly the magnetic field components H x comes out to be minus gamma bar by H square del H z by del x minus gamma del x plus j omega epsilon by H square del E z by del y. And finally H y comes out to be minus gamma bar by H square del H z by del y minus gamma bar by H square del H z j omega epsilon by H square and del E z by del x which completes the expressions for the transverse field components in terms of longitudinal field components. What is the significance of this manner of expressing the transverse field components in terms of the longitudinal field components? Now we can say that one can break up the general solution into two parts. One part would be where let us say E z is 0, but H z is not equal to 0. And the other part would be H z equal to 0 and E z not equal to 0. And of course we said in case both longitudinal field components are non-zero if they exist in a problem we can superpose the particular solutions of these two kinds and obtain the final general solution. As you would recall this solution leads to T E type of solution transverse electric waves because there is no longitudinal electric field component in such a solution. And this solution leads to transverse magnetic type of waves T m type of solutions. And if both E z and H z become 0 then the entire frame work crumbles down there are no fields which exist in that case. One can see that from the basic Maxwell's equations as well. And therefore we say that for the cylindrical waveguides in general and for the rectangular waveguide in particular there is no T E m type of solution which is possible. A transverse electric and magnetic wave does not exist for the cylindrical waveguides. Now it is fine that it is coming out mathematically from this. Is it possible to see in some other simple manner that for cylindrical waveguides the T E m solution is ruled out. It cannot exist. Now fortunately such an argument can be made if we consider the rectangular waveguide like this and make an assumption for a moment that it is a T E m type of wave which is existing on the structure inside the rectangular waveguide. So let us assume a T E m type of wave inside the rectangular waveguide meaning there is no E z and there is no H z then it would be a transverse electric and magnetic wave. Which means any field components that do exist transverse to the direction of propagation to the z direction. Let us focus attention on the magnetic field. It is completely transverse to the direction of propagation it exists in the x y plane. Now the magnetic field lines must exist as closed lines closed loop lines because there are no free magnetic poles. And therefore if we start from the first Maxwell's equation that is say del cross H equal to j omega epsilon E for the sinusoidally time varying fields. And for the situation that there is no source within the waveguide. And if we take the surface integral over the cross section of the waveguide for this then using Stokes theorem it will change to H dot dl equal to j omega epsilon E dot da. And we may consider the z directed component of both sides. In this case with this particular assumption the entire result on both sides should be z directed only. So if we consider the z directed component that it should be j omega epsilon E z da right. That is the line integral of the magnetic field in the x y plane should be related to some z directed electric field in this manner. However by virtue of our assumption that it is a TEM wave which is propagating there is no E z everywhere. And therefore this is equal to 0 which means this line integral is 0. Which means that there is no magnetic field which is existing. If it did it must have some non-zero line integral value. For example in a coaxial cable the magnetic field lines are concentric with the axis of the coaxial cable and H dot dl is equal to the current flowing in the inner conductor right. Similar thing should be possible here. It is required if there are closed loop magnetic field lines to exist in the x y plane. However there is no conductor here inner conductor here to support a conduction current. The electric field we are assuming to be 0 and therefore the magnetic field must also be 0 everywhere. And therefore such an assumption is not consistent with a propagating wave. And we say that a TEM wave cannot exist in a rectangular wave guide or in a cylindrical wave guide. The argument is quite general can be extended to all cylindrical wave guides. And therefore we say that no TEM wave on we can generalize this on cylindrical wave guides. So therefore we have to talk of either a TEM mode or a TEM mode on the rectangular wave guide. We can pick up anyone let us pick up the transverse magnetic type of solution which says that H z should be 0 and E z should be non-zero. So we try to work out a solution for the TEM type of wave for which as identified here H z is 0 but E z is non-zero. Now we have to pick up a suitable equation from this large number where we can make out the behavior of the z component of the electric field. And that is the z component of the wave equation for the electric field. E z is also a field component which is tangential to the perfectly conducting magnetic field. And therefore boundary conditions can be applied on this component in a straight forward manner. So from these considerations we consider the equation to be satisfied by the z component of the electric field. And that is del 2 E z by del x squared plus del 2 E z by del y squared plus gamma bar squared E z which should be equal to minus omega squared mu epsilon E z. We essentially have reached a stage where the solution for the rectangular wave guide boils down to solving this equation subject to the boundary conditions. Now here we try to make out what is the nature of E z. E z in general is to be a function of the three coordinates x, y and z. However the z variation has been identified to be corresponding to that of a propagating wave along the z direction. And therefore we can write this as E z naught that is some sort of an amplitude factor which is a function of x and y times e to the power minus gamma bar z. Which can be substituted in this parent equation and a similar equation is going to hold good for this amplitude factor. Then we use a method which is very frequently used in such situations and that is the method of separation of variables or the method of variable separation. What does this method consist of? It amounts to making a very important assumption. The assumption is as follows we say that let E z naught which is now a function of x and y alone be expressible as a product of two functions. One we call x and the other we call y. Where x is a function of x alone and y is a function of y alone. The simplification that this causes will be seen as we proceed. However this kind of a procedure that is assuming that a term which is a function of two coordinates can be expressed as a product of two functions and each function depending on only one direction at a time has been put on a fairly rigorous mathematical footing. And wave equations have been shown to be variable separable in I think 11 coordinate systems including of course the rectangular coordinate system. Therefore once you take it that this step is a regular as mathematically proven step. However for us it is very important because it simplifies the considerations very much. So these expressions with this assumption these are substituted back in the parent equation. And when we do that we get y times d 2 x by d x squared e to the power minus gamma bar z plus x times d 2 y by d y squared e to the power minus gamma bar z plus gamma bar squared x y e to the power minus gamma bar z which should be equal to minus omega squared mu epsilon and x y e to the power minus gamma bar z. We have just made the substitution for e z in the parent equation. Now since e to the power minus gamma bar z is a factor which is common to all terms. It can be dropped and we do not expect it to be 0 in any case. And we can combine the omega squared mu epsilon and the gamma bar squared terms into x squared. Therefore what we are left with is a fairly simple expression which reads as 1 by x times e to the power minus. There is another step that we are squeezing between. We are dividing through by e z naught. That is the product x y is being used to divide through these equations. And once that is done we get 1 by x d 2 x by d x squared plus 1 by y d 2 y by d x squared plus 1 by x d y squared plus h squared which should be equal to 0. One could say that we have divided through by e z. And for a non-trivial solution e z should be non-zero. Now we can write this in a slightly different form and then one can use a very interesting argument. We write this as 1 by x d 2 x by d x squared plus 1 by x d plus h squared which should be equal to minus 1 by y d 2 y by d y squared. And recall here that h squared is gamma bar squared plus omega squared mu epsilon. What is the nature of h squared? Given a certain waveguide with certain parameters of the medium filling the waveguide and an operating frequency, the wave will propagate with a certain propagation constant. And therefore h squared is essentially a constant. Let us say at a given for a given structure and for a given frequency. Fine. And therefore if you look at the left hand side, what is the left hand side? It is essentially a function of x alone. It involves terms which are only a function of the x direction x variable d 2 x by d x squared and 1 by x. And by virtue of our assumption that x is a function of x alone entire left hand side is a function of the variable x only. By the same argument the right hand side is a function of y variable alone which equality should hold good over the domain over which we are seeking the solution. That is the interior of the waveguide. That is for different values of x and y x ranging from 0 to a y ranging from 0 to b. This equality should be maintained between 2 functions. One of which is a function of x alone and the other is a function of y alone. This is possible only if either side is a constant. Both left hand side and right hand side must be constant. They should not remain a function of x or y. Then only for varying values of x and y the equality will be maintained. And therefore, we say that both must be equal to a constant. That is they should not remain a function of either x or y. And therefore, we are now in a position to say that let 1 by x d 2 x by d x square plus h square be equal to another constant say a square. By virtue of the arguments we have just gone through. Similarly, the other side 1 by y d 2 y by d y square should be equal to minus a square. What have we achieved? We were dealing with partial differential equation, second order partial differential equation. Now, we have to deal with only second order ordinary differential equations. And they have come out in a form where the solution can be written by inspection. Very familiar form of the second order differential equation. So, that is the purpose of using the method of variable separation. We continue this process further. And here it is quite straight forward. We can write y equal to let us say some amplitude some constant of integration C 3 cosine of A y plus C 4 sin of A y. The other side needs some more manipulations. We rewrite this as 1 by x d 2 x by d x square equal to a square minus h square. And since we prefer to have solution in terms of sin and cosine functions. And that is the kind of boundary conditions. Multiple zeros are expected in the solution. And therefore, we again rewrite this as minus b square. Where quite obviously, h square is equal to a square plus b square which relation we shall draw upon a little later. And therefore, we now have x also equal to some constant of integration C 1 times cosine b x plus C 2 sin of b x. And since we had E z naught equal to x times y. And entire E z is equal to x times y. And E z required another factor e to the power minus gamma bar z. We can now constitute the entire z component of the electric field all variations x y z now coming in place. So, that we can have E z equal to C 1 cosine of b x plus C 2 sin of b x multiplied by C 3 cosine of a y plus C 4 sin of a y times of course, e to the power minus gamma bar z. All right. What remains to be evaluated are these constants of integration. So, this is and if possible a and b. How will these be done? These will be done by ensuring that the boundary condition is satisfied by the z component of the electric field. As we saw earlier the z component of the electric field is a component which is tangential to the walls which are assumed perfectly conducting. And therefore, at the walls x equal to 0 x equal to a and at y equal to 0 y equal to b it must drop to 0. So, enforcing that kind of condition tells us that for E z to be 0 at x equal to 0 would require that C 1 is 0 and C 2 is 0. And for it to be 0 at y equal to 0 would require that C 3 is 0 further for it to be 0 at x equal to a would require that b a is equal to let us say m pi an integral multiple of pi where m can take on values 1, 2, 3 etcetera. It can be seen that m equal to 0 would lead to all field components going to 0. And similarly at y equal to b for E z to be 0 we must have a times b equal to say n pi where n is another integer. So, that we see that b is m pi by a and a is n pi by b. So, we find that these constants capital A and capital B are now coming out in terms of the waveguide dimensions small a and small b. And of course, the indices m and n which as you can make out now is will constitute the various orders of variation along the x and the y direction. And therefore, we are now in a position to complete the expression for E z it turns out to be E z equal to we can combine the constants C 2 and C 4 into a new constant say capital C. And then we have sin m pi by a x sin n pi by b y and of course, e to the power minus comma bar which satisfies the Maxwell's equations and the boundary conditions for the T m type of waves. This is one field component for the T m m n modes belonging to the T m type of waves. The other field components can be obtained by going back to the expression for the transverse field components in terms of the longitudinal field components. This is where we stop today we have introduced the rectangular waveguides. And we have seen how we can work out a solution for propagating waves in the rectangular waveguide. And in particular we are on our way to obtain the field components for the transverse magnetic type of waves. Thank you.