 Welcome to the 22nd lecture in the course Engineering Electromagnetics. For the last few lectures we have been discussing the propagation of uniform plane waves in homogeneous media and then we have considered their reflection, refraction etc. In this work we have found that the behaviour of the plane waves is quite similar to that of the waves on a transmission line and at various points during these last few lectures this similarity has been pointed out to you. The topic for today therefore is the transmission line analogy for plane waves where we will try to formalize the similarity between the behaviour of the plane waves and that of the waves on a transmission line. The similarity between these two types of waves is quite striking particularly if we consider the reflection coefficient on a transmission line or the reflection coefficient related to plane waves travelling from one medium to the other and we find that the expressions for the reflection coefficient v- by v plus for the transmission line and E r by E i for the plane waves have a striking resemblance. Also the formation of standing waves etc are quite similar in the two cases. Now actually this similarity is not a coincidence and if one looks at the evolution of the waves on the two media one finds that there is a basis for fairly rigorous formal analogy or correspondence between these two situations. So we try to recapitulate the various expressions, mathematical expressions that lead us to the waves in these two cases and we consider a uniform transmission line on one hand and we consider a wave propagation or a uniform plane wave in a homogeneous medium and we consider what are the various equations and expressions that hold good in the two cases. For example on the transmission line considering the distributed parameter equivalent circuit for a small section of the transmission line we wrote del v by del z equal to minus l del i by del t minus r i where symbols carry their usual mean. Also del i by del z is equal to minus c del v by del t minus g times v. These are the equations governing the behavior of v and i in terms of the primary constants r l c g of the transmission line. On the other hand what equations do we start with for the plane waves we start from the Maxwell's equations and we write del cross e for example equal to minus mu del h by del t and the second equation is del cross h equal to epsilon del e by del t plus sigma e in the case of conducting media. And we are making the usual assumptions that we are dealing with a homogeneous medium so that these are constants the constitutive parameters mu epsilon sigma are constants and we are not restricting it to a dielectric medium or a perfect dielectric medium. Also we are considering that the medium is source free so that the source term does not complicate the situation. Now these we specialize so that they become simple and it is possible to see the correspondence between the two situations separately. For example if we consider a plane wave propagating in z direction and now recalling the properties of the uniform plane waves the field vectors will be normal to this direction of propagation so that let us say that we have an x component of the electric field somebody please close the door and there is a y component of the magnetic field. And then looking at these two components and considering that there is only spatial variation with respect to the z direction we are going to get one can write out the cross product in terms of the Cartesian coordinates and see that we will get del e x by del z equal to minus mu del h y by del t. On one hand corresponding to the first equation and similarly del h y by del z equal to minus epsilon del e x by del t plus sigma e x on the other hand. And the analogy or the correspondence between these equations representing the plane wave in a homogeneous medium and these equations representing the wave propagation on a transmission line is quite clear. For example one can identify the various quantities which behave similarly or which are analogous on the two sides of the white board. We can see that we have a correspondence between v and the electric field vector say the x component one has units of volts the other has units of volts per meter. Then we have l which has the same role or the same behavior as mu l is the inductance per unit length so has units of henry per meter mu is the permeability of the medium. This one no please correct this should be a negative sign here this one is alright. Thanks for pointing it out anyway. Now so mu has the units of henry per meter and therefore it is it has correspondence with the parameter l and then we do not have something corresponding to the resistance per unit length which exists in the case of the transmission line but right now is absent from the right side of the board. We can consider the reasons for this a little later. We go on to find out the analogous quantities and we see that i has the same role as h y i has the units of amperes h y has the units of amperes per meter then c and epsilon are analogous they both have units of farads per meter and then finally we have the conductance g which is analogous to the conductivity sigma both having units of most per meter therefore one can see there is a almost one to one correspondence between the various quantities on the two sides and the equations are almost identical in nature. We can consider the equations now written for this special case of sinusoidal time variation and assume that these equations were written in phasor notation and then we get del v by del z equal to minus r plus j omega l into i here and del i by del z equal to minus g plus j omega c into v here and what we get here is similarly del e x by del z equal to minus j omega mu h y and del h y by del z equal to minus j omega epsilon minus or plus sigma into e x one can combine these couple partial differential equations using the standard procedure to get let us say an equation in voltage variable as del 2 v by del z squared equal to r plus j omega l into g plus j omega c into v or simply gamma squared v with gamma squared having the expression r plus j omega l into g plus j omega c and similarly on the other side we are going to get del 2 e x by del z squared equal to j omega mu into sigma plus j omega epsilon e x and this also can be written say as gamma squared e x where gamma squared here is j omega mu into sigma plus j omega epsilon a similar equation will be obtained in h y here and in i here and considering this final result that we have got we can see that we can write either v or i as a result which reads as let us say a times e to the power minus gamma z plus b times e to the power plus gamma z here and similarly we can write either e x or h y as a e to the power minus gamma z plus b e to the power plus gamma z and therefore we find that it was not just a coincidence that we got some two isolated results which was similar in these two cases the entire mathematics is quite parallel and there is a one to one correspondence in these two situations. One also has the expression for the impedance for example the characteristic impedance here is going to be r plus j omega l upon g plus j omega c and similarly we have the expression for the intrinsic impedance here eta which is j omega mu upon sigma plus j omega epsilon whole square root z naught is the ratio of the voltage and current in one particular way and the sign is chosen appropriately depending on the direction of propagation. Same is the case with eta the intrinsic impedance it is the ratio of the fields electric field and the magnetic field in one particular wave with an appropriate sign and we have also used the specialized results for the lossless case where we have r and g as equal to 0. So, that z naught becomes simply root l by c here and beta becomes omega times square root of l c and here for the lossless case we will have sigma equal to 0 and then eta will specialize to square root of mu by epsilon and beta will become simplified to omega times square root of mu epsilon. Now one can comment for a short while on something on this side missing corresponding to the resistance per unit length in the transmission line. One could look at it from two points of view. First point of view would be that here the transmission line is constructed out of two materials one is the conductors constituting the conductors of the transmission line a good conductor the other is the dielectric medium filling the region between the two conductors or supporting the conductors. In practice both are going to be lossy and therefore r and g cater to the loss caused by the conductor and the dielectric material respectively. For plane waves we do not really need two separate media they can propagate in free space or in one single homogeneous medium and therefore only one loss mechanism is present and therefore we have a term corresponding to the r missing here that is one way of looking at it. Alternatively we have seen that particularly when we use the phasor notation we can incorporate sigma in the complex permittivity of the dielectric medium. Similarly if in addition to dielectric losses what does the complex permittivity actually signify signifies that there are losses present in the dielectric medium. Similarly if it is not a dielectric medium if it is a magnetic medium magnetic material which also has some losses the losses can be accounted for at least mathematically by replacing the real mu real permeability by a complex permeability in which case we will find immediately that a term corresponding to r will appear in the simplified Maxwell's equations. And therefore we do not pay much attention to this fact that for this simple case there is no term corresponding to r present in the case of the plane waves. This homework should put the analogy between the transmission lines or the wave phenomena on transmission lines and the wave phenomena in homogeneous media on a very formal and rigorous footing. It is not a coincidental similarity between one or two results. The entire body of results that we have derived for the transmission lines or that are available for the transmission lines can now be so to say lifted and applied to these situations which can cause a considerable amount of simplification in the work that one has to do. This formal correspondence or analogy was first noticed and brought to light by S. A. Shalkunov. Now what is it that we want to really gain out of this formal analogy or correspondence between the two situations? We can gain in two ways. One is that we gain a better understanding of the plane wave phenomena. Plane wave phenomena are usually more complex because the three dimensional media are involved. On the other hand transmission lines are a single dimensional entity. On the other hand there are certain topics for example impedance matching where large amount of work has been done in the context of the transmission lines, switch chart etc. have been developed they can be utilized. That entire work can be utilized for work with plane waves. A single example should suffice to convince us of the utility of this analogy. In the last lecture you would recall that we considered the behavior of a plane slab. A plane slab inserted between two different media and we analyzed the behavior by considering multiple reflections at the two interfaces of the plane slab. And you found that the expressions were fairly tedious. However if we consider the transmission line equivalent of the problem you can visualize already that the work can be considerably simplified. So it is this that we shall attempt to do in the rest of the time. Before we do that we are going to require the definition of impedance. This is the characteristic impedance. This is the ratio of the voltage in current for one particular wave. If it is a uncorrupted propagating wave then this is the ratio of the voltage in current. However if it is a combination of an incident wave and a reflected wave then we know that the impedance behaves in a different manner. And that impedance can be written as let us say z as a function of z and it is equal to the total voltage at that location divided by the total current at that location and we have got expressions for this impedance at different distances from a certain load impedance on the transmission line. In a similar manner here also we could say that the wave impedance or the field impedance which in the general situation that is when we are considering a superposition of an incident wave and a reflected wave is not the same as the intrinsic impedance can be written as z as a function of z equal to E x upon H y. Whatever is the total E x at a certain distance from an interface and whatever is total H y at that distance we take the ratio and that we use as the field impedance or the wave impedance. Using this concept we first take a very simple situation and then consider the plane slab problem once again. Now while we have tried to establish an exact analogy when there is just one kind of propagating wave for example in the positive z direction or in the negative z direction what happens to this analogy when there is a discontinuity either in the path of the transmission line or in the path of the plane wave. We have seen recently that in the case of the plane waves the reflection, refraction etcetera take place and these are governed by the boundary conditions which have to be satisfied by the field components at the interface. Now how does the analogy hold between these two situations at that time when there is a discontinuity and there is an interface of two different media or there is a joint between two different transmission lines. Our experience with the expressions for the reflection coefficient shows that the analogy should hold good. The reflection coefficient that we have got in the two cases virtually has the same form. So one would for example be curious how this analogy is holding good even when a discontinuity is there. So actually the problem is straight forward. We consider the boundary conditions for the case of the plane waves we have the tangential electric field at the interface continuous and the tangential magnetic field also continuous. Now we just have to realize that is there something corresponding to these boundary conditions or these continuity relations which applies to the transmission line situation as well. And actually it does if there is a joint let us say between two different transmission lines there must be the voltage must be the same on the two transmission lines at the joint. Similarly the current that is coming out of one transmission line at the joint should have the same value as the current that is going into the other transmission line at the joint. And therefore in effect these boundary conditions are identical and therefore it should be of no surprise that the behavior the analogy between the transmission lines and the plane waves is maintained even in the presence of discontinuities. We consider reflection at a perfect conductor for the purpose of demonstrating this analogy even in the presence of discontinuities. Let us say this is the perfectly conducting interface existing at z equal to 0 and let this be the normal. And therefore we are considering wave propagating in the positive z direction and impinging on this interface at z equal to 0. We will have let us say the x directed electric field and the y directed magnetic field. Now with our experience we can write down the total fields in a fairly straight forward manner. We will have e x equal to say e plus into e to the power minus j beta z minus e to the power plus j beta z. Taking care of the direction of propagation of the incident wave and the reflected wave and the fact that the tangential field at the interface must be 0 which is equal to minus 2 j e plus and then sin of beta z. Similarly h y is going to be e plus by eta eta being the intrinsic impedance of the medium into e to the power minus j beta z and plus e to the power plus j beta z. Taking care of the fact that the relationship between the field components changes sin depending on the direction of propagation which simplifies to twice e plus by eta cosine of beta z. Therefore what is going to be the value of the field impedance or the wave impedance at some distance z from the interface? We will have z as a function of z equal to total e x by total h y and it is going to be minus j eta times tan beta z. And for the case of the transmission line what is the input impedance at different distances from the short circuit from a short circuit termination that is z i equal to j z naught tan beta l where l was taken as minus z and therefore if we do that substitution that for l equal to minus z this is going to be simply j eta tan beta l where l is the distance from the interface. So we started with the field description. We arrived at some wave impedance result which is identical to what we would have arrived had we just used the transmission line analogy. Just a simple demonstration of the analogy and therefore one can use the results that we have got for the transmission line for input impedance under different situations straight away for the plane waves situation. We have got the input impedance in the general case as z i equal to z naught times z l cos beta l plus j z naught sin beta l divided by z naught cos beta l plus j z l sin beta l. You can write this in different alternative forms that does not matter so much. Now it is this input impedance expression that we shall try to use for the problem of the plane slab. Let us make space here. We consider a plane slab and we have three regions. Let us say this is medium one with intrinsic impedance eta 1 this is medium two with intrinsic impedance eta 2 and this is medium three with intrinsic impedance eta 3. Let the slab thickness be l or let us just use l. Let this interface between the two media media 1 and 2 be located at z equal to 0. And let us say that we want to calculate the reflection coefficient for a wave impinging on this plane slab from medium one. Using the transmission line analogy all that we need to do is to consider the input impedance at this point. Let us call it z l 1 which can be calculated if we know let us say the load impedance at this point which we may call z l 2. Now if there is no other discontinuity in medium three in the path of a wave which is transmitted through the plane slab then what would be z l 2 then z l 2 would be simply eta 3. So, making that assumption we can write z l 2 equal to eta 3. The other quantities in this case for example, z naught will go to eta 2. And let this z l go to z l 1 we still keep it z l 2 for the time. Now z l 1 is going to be equal to the characteristic impedance which is eta 2 times the load impedance which is eta 3 cos beta l plus j eta 2 sin beta l divided by eta 2 cos beta l plus j eta 3 sin beta l which is what a wave impinging on this interface will see. And again using the transmission line analogy the reflection coefficient is going to be rho equal to z l 1 minus eta 1 upon z l 1 plus eta 1. Almost in a minute you are in a position to write down the reflection coefficient if we draw upon the transmission line analogy for the plane waves. Now there were two results of general significance that we had derived that we had stated last time. For example, one was that the reflection coefficient can be eliminated if eta 2 is equal to the geometric mean of eta 1 and eta 3. And if beta l is equal to an odd multiple of lambda by 4. Now, to make sure which lambda we have in mind we should put proper subscripts here for example, this is beta 2 it is the phase shift constant of a plane wave in medium 2 that will be involved here. And similarly this will be the wave length in medium 2 which should be utilized for deciding the thickness of the plane slab. Such a slab was given the name a quarter wave plate or a quarter wave coating. Similarly we had results for a half wave plate. Actually it is possible to derive those results in a rigorous mathematical manner which is we will attempt here just a minute. Yes please I think you are right this is how it will be. This is the process result what we need is beta 2 l equal to an odd multiple of pi by 2. Thank you. And when we substitute for beta 2 in terms of 2 pi by lambda 2 then we get this. Thank you. Let us consider the expression for Z l 1 and in terms of this the reflection coefficient. Let us say we require that the reflection coefficient should be 0. That requires that Z l 1 should be equal to eta 1 and this is the condition we use for no reflection. We put the condition that Z l 1 should be equal to eta 1. And therefore, we have eta 2 into eta 3 cos beta 2 l plus j eta 2 sin beta 2 l divided by eta 2 cos beta 2 l plus j eta 3 sin beta 2 l on the left hand side. And let us rationalize the denominator. So, that we multiply by eta 2 cos beta 2 l minus j eta 3 sin beta 2 l divided by the same quantity. So, eta 2 cos beta 2 l minus j eta 3 sin beta 2 l which entire expression should be equal to eta 1. And now we get eta 2 into eta 3 squared sorry eta 2 eta 3 on one hand. The product of these two terms they simplify to this eta 2 plus eta 3 and the other terms give us plus j sin beta 2 l cos beta 2 l into eta 2 squared minus eta 3 squared divided by eta 2 squared cos squared beta l plus eta 3 squared sin squared beta 2 l which should be equal to eta 1. Now, assuming that eta's are all real the imaginary part on the left hand side must be equal to 0. And therefore, we say that sin beta 2 l cos beta 2 l into eta 2 squared minus eta 3 squared should be equal to 0. Now, one can see that if we choose the option that eta 2 is equal to eta 3 one lands into a trivial situation. And then there will be no interface between these so called 3 different media. So, leaving aside that we have two possibilities one is that either sin beta 2 l is equal to 0 or cos sin beta 2 l is equal to 0. Considering the second possibility first the other one will also be considered this tells us that the simplest solution is that beta 2 l is equal to pi by 2 or in general an odd multiple of pi by 2. We make this substitution in the expression and then we find that the imaginary part is already 0. And we are left with eta 2 squared by eta 3 equal to eta 1 you just substitute beta 2 l equal to pi by 2. So, only this term will remain in the denominator and this will be the final result that is eta 2 should be equal to the geometric mean of eta 1 and eta 3. And the other condition that we got is exactly corresponding to the general condition that we had stated earlier. Similarly, one can see that the first choice sin of beta 2 l equal to 0 which will translate into let us say the taking the first solution beta 2 l equal to pi this will lead to the general result that eta 1 is equal to eta 3 and that there is no reflection if beta 2 l is equal to pi. So, this corresponds to lambda by 2 plane slab and the other choice that we have discussed is the lambda by 4 plane slab. Now, supposing we are interested in eliminating reflections over a broad range of frequencies then we are aware of the limitations of a single coating. For example, here it will have a limited bandwidth beyond the design wavelength or the design frequency the reflections are bound to increase and only where such a relation is satisfied the reflection will be eliminated. And as a solution to this problem we had stated that one can use multi section quarter wave coatings or in transmission line terminology multi section quarter wave transformers. And one can now see how one is going to analyze this kind of multi junction or multi interface problems using the multiple reflection point of view that we had adopted earlier. It will be virtually impossible to do that but using the transmission line analogy one can handle that kind of situation without much difficulty. So, that brings the power of the transmission line analogy into focus.