 Welcome to the 23rd lecture in the course Engineering Electromagnetics. We start today the topic of wave guides and go on to consider perhaps the simplest wave guide that is the parallel plane guide and the boundary conditions that are applicable for this particular guide and also obtain expressions for the transverse field components in terms of the longitudinal field components. For some time now for the past few lectures we have been discussing wave propagation in infinite media or interfaces between infinite media. While this may be a practical situation perhaps for broadcast signals many times waves have to be guided along what are called wave guides, otherwise it is difficult to give a particular direction to the signal. Waves may be guided along conducting surfaces along dielectric surfaces or a combination of conducting and dielectric materials and therefore there could be wave guides of many different types. Even the surface of the earth could act as a wave guide and waves could be guided along this surface that is what happens during the medium wave broadcast of radio signals. We consider today perhaps the simplest type of wave guide constructed out of two parallel conducting planes. This will form a bridge between the waves we considered on transmission lines and waves that exist in general on wave guides. The structure that we consider is as follows. This is called a parallel plane guide. It consists of two infinite perfectly conducting planes. The planes are infinite along the z and the y directions and along the x direction they have a spacing shown as A. The medium between the conductors could be considered to be say a perfect dielectric or may be free space. While such a structure may not appear very practical but in some situations the results that we derived on the basis of the structure are going to be quite useful and otherwise as I said this structure is perhaps the simplest wave guide and it will form a bridge between the transmission line considerations on one hand and the wave guides on the other hand and therefore we consider this structure. First of all one has to consider the boundary conditions that this structure imposes on any wave that may be propagated along this structure and essentially the solution to this problem what kind of waves would propagate on this structure would be found by solving the Maxwell's equations subject to these boundary conditions and therefore it is very important to consider the boundary conditions. You are familiar with the boundary conditions that a perfect conductor imposes what would be these boundary conditions. We say that on the surface of the perfect conductor the electric field should be 0 or the tangential component of the electric field on the perfect conductors that is at x equal to 0 and at x equal to a should be 0. One can show easily from Maxwell's equations that this automatically corresponds to or is the same thing as the time varying normal component of the magnetic field being 0 on the same surfaces that is at x equal to 0 and at x equal to a. This can be done by considering the Maxwell's equation del cross E equal to minus mu del H by del t which is the form in which this equation is written in many cases. Now one can open out the curl operation on this vector using the standard matrix notation so that we write del cross E equal to x cap y cap z cap del by del x del by del y and del by del z and here we have E x E y and E z. Now let's consider the x component of this equation this vector equation has three scalar component equations we consider the x component of this equation and then we get del E z by del y minus del E y by del z which should be equal to minus mu del H x by del t. There will be other two component equations also right now we are concerned with the x component. Now what is the value of the terms on the left hand side at x equal to 0 and x equal to a. We have said that the tangential electric field components must be 0 on a perfect conductor so E y and E z are 0 at x equal to 0 and x equal to a. Can their derivatives be non-zero if E y and E z are 0 everywhere on such a surface the derivatives also must be 0 and therefore we find that the time derivative of the H x which is the component of the magnetic field which is normal to these conductors must also be 0 and therefore H x could at the most have a constant value with respect to time and as far as the time varying fields are concerned which could form part of propagating waves. We say that the normal component of the magnetic field at these two perfect conducting surfaces must also be 0 and we say that applying the boundary condition that the tangential component of the electric field is 0 on a perfect conductor is the same thing as the normal component of the magnetic field being 0 on the same surface. With this boundary condition we now proceed to solve this problem of parallel plane guy and how do we start we write down the Maxwell's equations which will hold good for this structure. Now for which region do we write the Maxwell's equations there are two regions that are appearing here the perfect conductors and the region between these perfect conductors we write the Maxwell's equations for the region between the perfect conductors and then apply the boundary conditions as we just discussed. So we assume that the medium between the perfect conductors is a perfect electric for the sake of simplicity and we know that if the medium happens to be conducting in some situations it will be possible to take care of it by letting the permittivity become complex and therefore we write del cross H equal to j omega epsilon E okay and when we write this we are making the usual simplifying assumptions that we are dealing with the homogeneous medium okay and the medium is isotropic right and we are using phasor notation. So with those usual assumptions we can get Maxwell's equations in this simple form and similarly we have del cross E equal to minus j omega mu H okay now you would recall that these two equations can be combined utilizing the other Maxwell's equations also to read as del 2 H equal to minus omega squared mu epsilon H which is the wave equation governing the behavior of the magnetic field intensity and similarly of course we will have del 2 E equal to minus omega squared mu epsilon E which can be written as say minus beta squared H or plus gamma squared H okay generally when the medium is conducting we use the symbol gamma for the propagation constant for a perfect dielectric beta in magnitude is equal to the propagation constant itself otherwise in general beta is the real part of the propagation constant gamma okay the beta is the imaginary part of the propagation constant gamma. Now we expand these curl equations in their component equations what a coordinate system should be used the coordinate system used should be such that the problem at hand can be easily described in terms of that coordinate system for example the structure we are dealing with has conducting surfaces at x equal to 0 and at x equal to a and therefore the Cartesian coordinate system should be the most convenient one to work with and it will be very easy to apply the boundary conditions in the Cartesian coordinate system and doing that and using the determinant for expanding the curl equations we get del H z by del y minus del H y by del z equal to j omega epsilon E x which is the x component of this vector equation alright we can get the other two equations as well by looking at the determinant or there is a simpler way out one can change the subscripts in the cyclic order then also one will get the correct component equations which is fairly straight forward and doing that we get del H x by del z minus del H z by del x which should be equal to j omega epsilon E y and the third equation will be del H y by del x and of course on the left hand side you can see that the field component and the derivative are appearing in pairs so for example it is del H y by del x and then the next one should be del H x by del y so this kind of symmetry can be utilized to check for any mistakes so this becomes j omega epsilon E z and similarly we need to write the component scalar equations for the other Maxwell's curl equation and we get we just need to substitute H by E and E by H taking care of the negative sign and also replace epsilon by mu and therefore the procedure is fairly straight forward we get del E z by del y minus del E y by del z which should be equal to minus j omega mu H x and similarly we get del E x by del z minus del E z by del x which should be minus j omega mu H y and del E y by del x minus del E x by del y equal to minus j omega mu H z so these are the expanded versions of these vector curl equations what will be the corresponding expanded version of the wave equations those also can be written down we are going to get del 2 H by del x squared plus del 2 H by del y squared plus del 2 H by del z squared which is the Laplacian which should be equal to minus omega squared mu epsilon H and an equation in identical in form is going to be obtained corresponding to this equation that is in the electric field right now there are many field components there are many derivatives and we need to put some order in this system and then we consider the variations in the different directions separately for example if the guide that we are dealing with is like this and we assume that the wave propagation takes place along the z direction or alternatively put we orient the z axis along the direction of wave propagation okay so assuming that the propagation is in z direction what is the implication of this assumption we have seen that propagation in a certain direction implies a certain a particular mathematical form only when the mathematical form is consistent with that will the expression correspond to a propagating way and therefore we are able to assume a certain form of the fields as far as the z direction is concerned that is the implication of this assumption for example if we consider say the y component of the magnetic field just to take some example alright in general it is a function of the three coordinates x y and z and as of now we do not know the variation with respect to these three coordinates but by virtue of this assumption we are now in a position to write that this is equal to h y and let us say not which becomes the amplitude factor as a function of x and y and as far as the z direction is concerned we can write this as e to the power minus gamma bar z okay our experience with the transmission lines and with waves propagating in infinite media tells us that for wave propagation in z direction the mathematical form of the field components should be of this form e to the power minus gamma bar z where gamma bar may have a real part and an imaginary part in general but in specific cases may have only a real part or only an imaginary part okay also you will notice that we are using a different symbol gamma bar from gamma gamma was the propagation constant for the infinite media now we are dealing with a situation which is not the same thing as an infinite medium and therefore in general gamma bar the propagation constant in this case is going to be different and therefore we use a different symbol but the moment we write this the z derivative becomes extremely simple to handle for example we have del h y by del z what will it be it will be equal to minus gamma bar and then h y itself right and therefore in the previous system of equations 8 equations or so wherever we encounter derivative with respect to z we can replace that by minus gamma bar okay which will make the equations look much more simple now what is the situation regarding the y direction in the y direction as in the z direction the structure is infinite and there is no particular constraint imposed on the wave and the corresponding field components there is no boundary condition along the y direction okay and therefore the simplest possible situation is that the wave the fields are completely uniform along the y direction there is no variation along the y direction of course the moment this structure does not remain infinite along the y direction this simple situation is going to change but right now since the structure is assumed infinite along the y direction and there is no there are no boundary conditions along this direction we can say that along the y direction fields are uniform and what is the implication on the derivative with respect to y of this observation the implication is that del by del y can be replaced by 0 then comes the x direction along the x direction we do have certain boundary conditions that the fields must satisfy the tangential field and the normal tangential electric field and the normal magnetic field they must be 0 at the perfect conductors and therefore we are not in a position to say at this moment anything specific about the x derivative and that we retain as it is. So therefore the rule that we form is that del by del z is to be replaced by minus gamma bar and del by del y is to be replaced by 0 and with this we now go back to the system that we had developed starting from the Maxwell's equations and make the substitutions that we have just discussed for example here in this equation we will get gamma bar h y equal to j omega epsilon e x corresponding to this taking care of the y and z variations as we have discussed this is the simplified corresponding equation. Similarly one can get the simplified versions of the other five components here and also for these equations. So let us write out these we will have minus gamma bar h x minus del h z by del x which should be equal to j omega epsilon e y and then del h y by del x should be equal to j omega epsilon e z. Similarly for the other set of equations we are going to get gamma bar e y equal to minus j omega mu h x and then minus gamma bar e x minus del e z by del x which should be minus j omega mu h y and then del e y by del x should be equal to minus j omega mu h z. We have been able to replace the y and z derivatives the x derivative remains. Similarly we take care of the wave equation and the wave equation will also get simplified it will be written as del 2 h by del x squared plus gamma bar squared h equal to minus omega squared mu epsilon h and similarly here we will get del 2 e by del x squared plus gamma bar squared e equal to minus omega squared mu epsilon e. Now while the laws satisfy the magnetic field intensity and the electric field intensity appear identical in form does it mean that they will have identical solutions it does not mean that because the boundary conditions on the magnetic field and the electric field are different. So same equations will yield different solutions for a particular structure in general. Now we focus attention on this system of equations. There are transverse field components x and y e x e y h x h y and there are longitudinal field components e z and h z and all these field components are governed by these simplified wave equations. A general method of putting these equations in to a more simplified order is to express the transverse field components x and y components in terms of the longitudinal field components x y components express in terms of the z components how that helps should become clear as we proceed. For example if we consider let us say this equation which involves e x and h y find another equation which also involves e x and h y and that is this equation alright and one can combine these solve these simultaneously so to say to obtain expression for both e x and h y in terms of e j. The procedure can be indicated as follows we consider let us say this equation and we are going to get h y equal to 1 by j omega mu into gamma bar e x plus del e z by del x right. This value of h y is substituted in this equation giving us here we have e x equal to gamma bar by j omega epsilon times h y and substituting this expression for h y here we get this as minus gamma bar by omega squared mu epsilon and then gamma bar e x plus del e z by del x. Now one can combine terms so that we get e x into 1 plus gamma bar squared upon omega squared mu epsilon which should be equal to minus gamma bar by omega squared mu epsilon del e z by del x which simplifies to e x equal to minus gamma bar and introducing a new symbol h squared and then del e z by del x. What is this h squared? h squared is simply omega squared mu epsilon plus gamma bar squared. What have we achieved now? We have obtained an expression for one of the transverse field components e x in terms of the longitudinal field component e z alright involving the propagation constant and the other parameters in the problem the frequency and the constitutive parameters of the medium filling the space between the parallel planes. Substituting this expression for e x here we are going to get a similar expression in h y in terms of e z or some derivative of e z. So this is the procedure that one can utilize for obtaining expressions for transverse field components in terms of longitudinal field components. Carrying out this procedure now for example here we had selected this equation and this equation for obtaining expressions for e x and h y if we selected these two equations we would have obtained expressions for h x and e y in terms of h z or its derivative. So carrying out this procedure we get the following results. We have h x equal to minus gamma bar by h squared and del h z by del x. Similarly we have e x equal to minus gamma bar by h squared as we just obtained times del e z by del x. Similarly we have h y equal to minus j omega epsilon by h squared del e z by del x on one hand and e y equal to j omega mu by h squared del h z by del x. The new symbol that we have introduced here in these equations that is h squared has the expression omega squared mu epsilon plus gamma bar square which completes the task of writing or obtaining expressions for the transverse field components e x e y h x h y in terms of the longitudinal field components e z or its derivative and h z. Now what is the significance of doing this? The next step that we make out from here is the following. We find that in general both e z and h z are 0 all field components become 0. Now in general both could be non-zero and then all field components would exist but we can divide the problem into two simpler parts. One part would be when say e z is 0 and correspondingly h z is not equal to 0 because if both are 0 then as we observe in general there is an important exception which we shall deal with later. In general if both are 0 all field components become 0 which is a trivial situation. So we could consider the simpler case that one of these is 0 the other is non-zero. In which case not all transverse field components would exist for example here since h z is non-zero we will have h x and e y which are non-zero but e x and h y will become 0. This will become a particular solution for the problem that we have started to solve. There is a name given to this particular solution. It can be derived either stressing the fact that e z is 0. So that if any electric field components exist they are transverse to the direction of propagation. So if we emphasize this then this becomes a transverse electric field case or for brevity the Te case. The wave solution that we will obtain this way will be called a Te wave. And as we shall see there can be many types of Te wave solutions also giving us various types of Te modes. As we proceed the meaning of these terms will become clearer. On the other hand if we want to emphasize the fact that h z is non-zero we can call it an h wave. This later designation is older. This kind of waves were earlier called h waves but now they are called Te waves transverse electric waves meaning that there is no longitudinal component of the electric field or put it the other way round any field components for the electric field that exist are transverse to the direction of propagation. Similarly one can consider the corresponding case when h z is 0 that is we will have only e x and h y components h x and e y will become 0. And e z is not equal to 0 giving us what can be called as the transverse magnetic field case or transverse magnetic field waves or in short TM waves. And in parallel with the nomenclature we used for the first case this can also be called an e wave. As I mentioned it is the Te and TM waves designation which is more commonly used now. Now that is the advantage of putting the earlier equations in this form that is transverse field components expressed in terms of the longitudinal field components and now we can get the solution for these two simpler particular situations. Of course in general both e z and h z could exist and that general solution can be obtained as a superposition of these two solutions. As far as the mathematical solution is concerned as we shall discuss later it is possible to select selectively excite one of these two types by choosing an appropriate field configuration of the excitation source. Therefore this segregation into two simpler solutions is not too artificial and of course mathematically it is very convenient. And therefore we now take up the solution for the Te waves and as we have just noticed for this case we have e z equal to 0 and h z is non-zero. For this purpose we focus attention on this equation the wave equation for the electric field. For the Te waves which field components are going to exist since e z is 0 we will have h x and e y and of course h z. So the field components that exist in this case are h z and h x e y. Now we can pick any one of these components and consider the component equation for that particular component from these wave equations. And if we choose a field component or the corresponding component equation on which the boundary conditions are easily applied it will be easier for us and from this point of view we pick the e y field component and write the y component of this equation. Which we will read as follows we will have del 2 e y by del x squared plus gamma bar squared e y which should be equal to minus omega squared mu epsilon e y choosing the y component on both sides of that vector equation. Or in other words we have del 2 e y by del x squared equal to minus h squared since h squared is the sum of omega squared mu epsilon and gamma bar squared minus h squared e y which becomes the second order partial differential equation. Involving only the x variable we have been able to considerably simplify the previous maze of equations. We can simplify this even further for example we have e y equal to e y naught which is now a function of x alone and of course of z but the z variation is e to the power minus gamma bar z. And when this form of the y component of the electric field is substituted here it is easy to see that we get an ordinary differential equation instead of a partial differential equation that is d 2 e y naught by d x squared equal to minus h squared e y naught since the z the z part drops out from both sides. And since e y naught is a function of x alone it becomes an ordinary differential equation. We take one more step and that is the solution can now be written very simply it is e y naught which is a function of x is equal to c 1 sin of h x it is not sin hyperbolic x sin of h x h standing for the symbol h standing for omega squared mu epsilon plus gamma bar squared plus of course c 2 cosine of h x where the constants of integration c 1 and c 2 should be evaluated using the boundary conditions. If you have any questions we can try those out now yes please won't h z and no that we have seen for the t e waves since e z is 0 h z is non-zero wherever e z is involved those field components would be 0 and we will have only h x and e y and of course h z. So, those are the three components we have recognized as non-zero for the t e case all right. So, in this lecture we have started our consideration of the parallel plane guide which is the simplest type of wave guide. And we have considered the boundary conditions that are applicable for this case and we have obtained expressions for the transverse field components in terms of the longitudinal field components and we have introduced the possibility of obtaining particular solutions called t e wave solution and t m wave solution. Thank you.