 Welcome to the 34th lecture in the course engineering electromagnetism we continue with our discussion on resonators and today we take up waveguide resonators. You would recall that in the last lecture we considered the transmission line resonators and we mentioned that the transmission line resonators can be used over a frequency range which extends from 100 megahertz to roughly 1000 megahertz. As we try to increase the frequency of operation beyond this range the quality factor offered by the transmission line based resonators becomes somewhat low. In fact one can achieve better quality factors by using waveguide resonators, resonators which are derived from waveguides. The waveguide resonators are essentially a volume in a metallic enclosure. These are variously known as waveguide resonators, cavity resonators or resonant cavities. In principle any metallic enclosure could be used as a cavity resonator, but just as we saw that waveguides of arbitrary shape do not offer any particular advantage similarly resonators of arbitrary shape do not offer any particular advantage and in fact it will be difficult to fabricate and therefore resonators which are derived from waveguides are the ones which are utilized in practice. Therefore the resonators derived from cylindrical waveguides, waveguides of rectangular cross section or circular cross section are the ones that we will be talking about here and in fact we will be talking about the resonators derived from rectangular waveguides. Let me show you the resonator that we have in mind the rectangular cavity resonator where the axes are marked as follows this is the x direction, this is the z direction and this is the y direction consistent with the notation that we used for the rectangular waveguide you can notice that the coordinate system is still the same and also one can use the dimensions A for the wider dimension along the cross section and B for the narrow dimension along the cross section and while the waveguide would continue in the z direction depending on the length of the waveguide here we have only a certain length of the waveguide which let us say is D alright this becomes our rectangular cavity resonator to consider the behaviour and when we say that we want to consider the behaviour what are the things that we have in mind we would like to find out we would like to develop an expression for the resonant frequency for a signal in such a resonator. Also as we have seen for the waveguides the field configuration plays a very important role in various aspects. So here also we like to work out the field configuration and finally of course for the resonator the quality factor must be estimated. So these are the three things that we like to work out today we consider that the resonator is obtained by putting short circuiting plates called one and two at z equal to zero and z equal to D that is how it will become a metallic enclosure. The region inside is usually free space or as we may talk about later it may be filled with some dielectric material alright. Now to begin the analysis let us consider that through some mechanism the fields are launched into this resonator fields are launched into let us say the desired mode in this waveguide the launching schemes the schemes of excitation for various modes we have discussed earlier already. In those schemes our objective was to be able to transfer as large in amount of power from the source to the waveguide in the desired mode as possible but resonators usually are lightly coupled to any external agency. So that the loading effect of this external agency is minimal otherwise as you would recall this loading is going to degrade the quality factor that one can achieve from the resonator. So keeping this light coupling aspect in mind through some mechanism through some means we can feed some power to this waveguide. Let us say that that is done somewhere close to the plate one as a result of which wave propagating in the z direction would start right and when this wave comes across a short circuiting plate at z equal to D a reflected wave will start and therefore we expect that standing waves will be formed within this volume. We consider that the fields that have been launched correspond to the TE MN mode okay you would recall that this is the way we obtained the field expressions five field expressions for the TE MN mode in a rectangular waveguide. These symbols here can be recalled to have the following meaning for example B you would recall is N pi by B and similarly A is M pi by A and then I think it is alright you please look up your notes. In any case it does not make too much of a difference and then H squared is A squared plus B squared okay what you are saying is that this is B and this is A alright okay that is a minor thing let us not worry too much about it but it should be consistent with whatever we have written earlier thank you. So let us say that this is the mode that has been launched close to plate one in the resonator now since we are going to talk about wave which is going to be reflected back and therefore it will travel in the negative z direction let us identify this kind of fields with a wave which is travelling in the positive z direction and therefore if we want to find out write down the fields in the cavity resonator then let us write this as H z plus equal to C plus cosine of M pi by A x and cosine of N pi by B y and then we consider these fields in the frequency range which is above the cutoff frequency for this particular mode so that the complex propagation constant gamma bar becomes the phase shift constant beta bar and therefore we write this as e to the power minus j beta bar z essentially the same field expressions are going to be written and we will use these with a superscript plus so that we are able to say that this is the field these are the field expressions corresponding to the wave propagating in the positive z direction. So modifying the other field expressions also in a similar manner we write down the other field components as H x plus equal to j beta bar by h square C plus M pi by A so this is the value of B so what we wrote finally was alright then sine of M pi by A x cosine of N pi by B y e to the power minus j beta bar z H y plus becomes j beta bar by h square C plus and here we have N pi by B and then cosine of M pi by A x and then sine of N pi by B y e to the power minus j beta bar z then as far as the electric field components are concerned we can write these from the magnetic field components using the wave impedance relationship so that e x plus is going to be equal to z t e or z z times H y plus where z t e you would recall is equal to omega mu by beta bar so taking care of this and then looking at the H y plus field expression e x plus becomes let us say an amplitude constant which we call e x naught plus and then cosine of M pi by A x sine of N pi by B y and then the exponential e to the power minus j beta bar z here one can make out that the amplitude constant that we have put for the field expression e x naught plus is going to be equal to j omega mu C plus by H square and in addition we have the factor N pi by B so that is this amplitude constant. Similarly we are going to require the y component for the forward travelling wave e y plus which can be written as minus z t e H x plus recalling the definition of the wave impedance in terms of the various field components and this becomes e y naught plus and then the variations along the x and y directions are the same as that of H x plus. So, that it is sine M pi by A x cosine N pi by B y and then we have exponential of minus j beta bar z here the amplitude factor that we have introduced e y naught plus is going to be equal to minus j omega mu C plus by H square multiplied by M pi by A alright. So, this becomes one of the component waves in the standing wave within the resonator volume. Similarly now when this forward travelling wave strikes the short circuiting plate at z equal to d the short circuiting plate is going to impose the boundary conditions that the tangential electric field component must be 0 assuming that the short circuiting plate is a perfect conductor. And therefore, a reflected wave is going to be initiated so that this boundary condition on the field components electric field components which are tangential at that plate are satisfied. From this point of view so that the boundary conditions are satisfied on the entire short circuiting plate at z equal to d that is for all values of x and y. It is reasonable to assume that the reflected wave will have similar x and y variations for the corresponding field components in the reflected wave. And the amplitude factor should be such that the boundary condition that the total tangential electric field is 0 is satisfied. We therefore proceed in the following manner. Let us say that E x minus the x component of the electric field in the wave which is travelling in the negative z direction is equal to E x not minus the amplitude factor of the reflected wave for the x component of the electric field. And then cosine of m pi by A x sin of n pi by B y and then e to the power plus j beta bar z. You would recall that the phase shift constant beta bar is available in terms of frequency the modal indices m n and the waveguide dimensions. Since those things have not changed the reflected wave phase shift constant should also remain unchanged in magnitude. The direction has been taken into account and therefore if we compare this x component of the electric field with the previous one you will find that the only difference we have made is in the amplitude factor and in the direction of the in the sin of the argument of the exponential function. Similarly, we can build up the y component of the electric field in the reflected wave which becomes E y minus equal to and once again we write this as E y not minus sin of m pi by A x and then cosine of n pi by B y e to the power plus j beta bar z and from an alternative point of view these variations along the x and y directions for the reflected wave field components continue to satisfy the boundary conditions which were already existing by the walls of the waveguide at x equal to 0 A y equal to 0 B etc. And therefore this form for the reflected field components is quite reasonable. Therefore what is the total x component of the electric field it will be a summation of the field components for the two waves and therefore it becomes E x not plus e to the power minus j beta bar z and plus E x not minus e to the power j beta bar z. As far as the z variation is concerned the other variations x and y variations remain unchanged because they are common to both field components and therefore we have cosine m pi by A x and sin n pi by B y. Similarly one can construct the total y component of the electric field which will be E y not plus e to the power minus j beta bar z plus E y not minus e to the power plus j beta bar z and then we have sin term that is sin n pi by A x and a cosine term cosine of n pi by B. So that now one is in a position to apply the boundary conditions that we have been mentioning both these x and y electric field components are tangential to the new walls that we have placed at z equal to 0 and z equal to D and therefore these x and y electric field components must become 0 at these new walls and therefore we say that for E x and E y to be 0 at z equal to 0. What we require is that E x not minus should be exactly equal to minus E x not plus then only this condition is going to be satisfied which let us say is equal to E x not so that we can have more compact notation by the same argument E y not minus is going to be equal to minus E y not plus and therefore let us call it E y not so that the total x and y electric field components can now be simplified by making this substitution and they now become as follows we have E x equal to twice j E x not sin beta z and then cosine of n pi by A x sin of n pi by B y and one thing is clear right now that is along the z direction it is no more a travelling way it varies as a sinusoidal function along the z direction therefore it is now a standing way at z equal to 0 this field component is certainly 0 we still have to apply the boundary condition at z equal to D in a similar manner one obtains the expression for the y component which is 2 j E y not where E x not and E y not expressions have been written earlier so those same expressions can be fitted here so that this is sin beta z once again and then we have sin m pi by A x cosine n pi by B y where now one can apply the remaining boundary condition which requires that E x and E y should be 0 at z equal to D also which gives us the result that beta D should be equal to an integral multiple of pi and this integer we now choose to be different from m and n which are already in use and therefore we choose the integer l so that beta D is equal to l pi or in other words beta is equal to l pi by D where l takes on values normally 1 2 3 etc but once the complete set of field expressions has been worked out it will be possible to identify whether any of these indices can assume the value 0 or not right and therefore now we have the complete information one thing that is once again clearly noticeable is that whereas earlier for example for the infinite rectangular waveguide beta was a continuous function now it can assume only some discrete values and therefore the waveguide is going to support fields only for some discrete frequencies which will be the resonant frequencies for the resonator and now we make the substitution for beta since now beta is available in terms of the index l and the dimension D and E x becomes 2 j E x not cosine of m pi by a x sin of n pi by b y and finally sin of l pi by D z that is along all the three dimensions along all the three coordinates x y z now it is a sin or a cosine kind of variation as far as the field is concerned which will be true for all field components not just for E x but the actual variation whether it is sinusoidal or cosine sinusoidal will depend upon the field component that is being considered E y becomes 2 j E y not and then sin of n pi by a x cosine of n pi by b y and the z variation remains identical now in this manner one can complete the expressions for the other field components also proceeding in a similar manner or else working out the field components magnetic field components for the reflected wave in terms of the wave impedance that is applicable to the wave traveling in the negative z direction and then super posing the magnetic field components for the two waves whichever wave one proceeds the field expressions are as follows since these are going to be required when we calculate the quality factor let me put down these field expressions we obtain h z as twice j and then either c plus or simply c this will be equal to c plus that we used earlier and then cosine of m pi by a x cosine of n pi by b y and sin of l pi by d z you can notice that the z component of the magnetic field which is a normal field component to the short circuiting plates 1 and 2 that we have put normal to the z axis is behaving just the same way as the x and y electric field components that we have we have noticed earlier that the tangential field components at a perfect conductor and the normal magnetic field component at a perfect conductor have the same behavior. So, that is coming out quite neatly as far as the other magnetic field components are concerned they turn out to be twice j beta bar by h squared c m pi by a sin of m pi by a x cosine of n pi by b y and then this x component of the magnetic field is not normal to the new short circuiting plates this has a variation which is cosine of l pi by d z and similarly the remaining field component which is h y is x 2 j beta bar by h squared c n pi by b and then we have cosine of m pi by a x sin of n pi by b y and once again cosine of l pi by d z this is the complete set of field expressions for which mode it was derived from the TE m n mode of the parent waveguide, but now there is a sinusoidal or cosine sinusoidal variation along the z direction also and the third index also appears which is l. So, these are the field expressions for the TE m n l modes for the rectangular cavity resonator in a similar manner one can obtain complete field expressions for the resonator modes which are derived from the transverse magnetic modes in the parent waveguide. So, this is as far as the field expressions are concerned we said that we will like to calculate the resonant frequency also. Now, where is the information about the resonant frequency in this system that will come from the expression for beta bar and to be consistent with the notation we have been using earlier all this should be beta bar here as well as here alright. So, now let us recall the expression for the phase shift constant beta bar in a frequency range above the cutoff frequency and then we see that beta bar squared was omega squared mu epsilon minus h squared or which is minus m pi by a whole squared minus n pi by b whole squared fine this relationship will continue. However, now beta is allowed only some specific values. So, that the new boundary conditions are satisfied. So, that this becomes l pi by d whole squared which will be satisfied only at some particular frequencies the resonant frequencies and therefore, we may put the symbol omega naught squared here and therefore, we find that the expression for the resonant frequency or radiant frequency omega naught can be obtained from the relation omega naught squared mu epsilon should be equal to m pi by a whole squared plus n pi by b whole squared plus l pi by d whole squared which can be manipulated in a simple way to obtain the expression for the resonant frequency giving us the resonant frequency f naught equal to c by 2 this c is the velocity of light in the medium filling the waveguide and then whole square root of m by a whole squared plus n by b whole squared plus l by d whole squared whole squared root and therefore, we find that the same waveguide with given dimensions a b and d can resonate at different frequencies depending on the order of these indices m n l these are the various resonant frequencies of the resonant cavity. Now, just as we have been making some effort in trying to identify what is the lowest order mode T e mode or T m mode which can propagate in the rectangular waveguide because the lowest order mode has some particular significance similarly, here also it is required to be able to identify what is the lowest order mode for example, the T e mode that can be supported in the waveguide and when we look at the field expressions for the T e m n l modes it turns out that l cannot be 0. However, either m or n could be 0 and therefore, using the convention that the larger dimension of the waveguide along the cross section is oriented along the x axis we can say that the lowest order T e m n l mode is T e 1 0 1 mode. It is not that 0 value of the first index is not allowed, but that will not be an order which is lower than this particular mode using the convention that we have been following. And therefore, this becomes the lowest order mode it will be possible to arrange resonance of this single mode in a cavity some other mode cannot does not enjoy the same privilege that is the advantage of this identification. Similarly, as I said one can work out the total fields for a resonant mode starting from the transverse magnetic modes and it is of some interest to identify the lowest order mode there also. And that comes out to be there also we will get modes which would be in general T m m n l modes and for the T m m n l modes the lowest order mode turns out to be 1 1 0. We had earlier seen that for the parent waveguide the lowest order T m mode is T m 1 1. And for this set of fields l can be 0 it does not make too many field components 0. And therefore, this is the lowest order mode for the T m modes. So, by now we have worked out the field expressions for the T e m n l modes and the resonant frequency. The third thing that remains is the calculation of the quality factor quality factor or the selectivity is a very important performance parameter for any resonator. And quality factor definition we have been writing now for some time. So, it should be quite familiar q is omega naught w by p l where the interpretation of the symbols should be fairly clear. Omega naught is the resonant frequency and then w is the average or time averaged energy stored in the cavity. So, for brevity we write this as average stored energy and the denominator p l is the average power loss in the resonator. And by average we mean the time averaged power loss. What is the mechanism of this power loss? The power loss will take place because of the finite conductivity of the walls constituting the metallic enclosure the cavity or in addition if the dielectric that is filling the cavity has some non-zero conductivity. So, once again the power loss could have an origin in conductor loss or in dielectric loss or both. As far as the calculation of the average stored energy is concerned that is w. We can write this as a summation of the average stored energy in the electric field plus the average energy stored in the magnetic field which we have already identified at a general level for all resonators that both these terms are equal. And therefore, we may write this as either twice w e or twice w m. Since the magnetic field components are required anyway for the conductor loss calculation we can work out the average stored energy in terms of w m. Of course, once you have all the field components available it makes very little difference which kind of field you are using for this calculation. We can write both perhaps w e is going to be the energy stored in the electric field per unit volume integrated over the volume of the cavity that is wherever the fields are non-zero. And therefore, since we are calculating the average energy stored it will be epsilon by 4 and then integral of the magnitude e squared over the volume of the cavity. Similarly, w m is going to be mu by 4 magnitude h squared integral over the volume of the cavity which for the rectangular cavity what kind of integral will it turn out to be it is going to be epsilon by 4 and then a triple integral from 0 to a 0 to b and 0 to c of magnitude e squared dx dy dz as you have pointed out this is d. Thank you. So, since the functions that are involved are sin and cosine functions the integrals are fairly straight forward. And the at least for the T e modes the complete field expressions are available. So, we can calculate either w e or w m and twice of either of these will become the average stored energy. As far as the power loss is concerned focusing attention for the moment on conductor loss it is going to be r s by 2 and then integral of magnitude h tan squared d a over all the walls of the cavity where as you would remember by now r s is the surface resistance and is equal to omega mu by 2 sigma or 2 sigma m whole square root. As far as this integration is concerned this will have to be carried out for the 3 pairs of walls separately because we need to identify the tangential magnetic field which is different at different pairs of walls. For example, on this wall the one at x equal to 0 what are the tangential field magnetic field components those are h y and h z. So, those are the ones which will be involved in the calculation of the power loss due to the conductor. And similarly for all the other walls the loss in this wall will be identical to the loss in this wall. And therefore, if one tries to expand this integration to clarify this calculation a little more it will come out to be r s recognizing that there are a pair of walls of either of each type. So, that this is integral over say y z of magnitude h y squared plus magnitude h z squared d y d z integrate from 0 to b and 0 to d. This will take care of the loss on the x equal to 0 and x equal to a walls. Similarly, one can write the other two terms say by cyclic rotation of the relevant quantities. So, that it becomes h z magnitude squared plus h x magnitude squared then d z d x or d x d z from 0 to a and 0 to d. And finally, the third term which will be involving h x and h y. So, that the integral also is over d x and d y from 0 to a and 0 to b. All these field expressions are available the evaluation of the integrals is not difficult. And therefore, the quality factor can be calculated by substituting these expressions for w and p l in the definition of the quality factor. For the sake of reference let me give you the quality factors for the T e and T m related resonant modes in the cavity. We have the quality factor for the T e m 0 l mode which will be required for calculating the quality factor for the T e 1 0 1 resonant mode which we identified is the lowest order resonant mode for the rectangular wave kind. And that is pi times eta which is the intrinsic impedance of the medium filling the cavity pi eta b. And then we have a squared l squared plus b squared m squared raise to the power 3 by 2. Then the denominator reads as twice r s and then we have a number of terms here that is a d plus 2 a b a squared l squared in a second term which is symmetric and reads as a d plus 2 b d into d squared m squared m 0 l. I used a slightly different type phase, but that is slightly confusing. So, let us correct that. The n index is not appearing because that is been made 0. Similarly, the quality factor for the T m m n l modes comes out to be pi eta then b squared m squared plus a squared n squared. And then the second term reads as b squared d squared m squared plus a squared d squared n squared plus a squared b squared l squared. And the second term is under the square root sign. The denominator is relatively simpler and it reads as 4 r s into a plus d into b cube m squared plus b plus d into a cube n squared. You would notice that the resonant frequency omega naught is not appearing in these expressions. So, what has been done is that the resonant frequency has been substituted in terms of the modal indices and the dimensions. So that is incorporated in the expressions. Now before we stop this lecture, let me give you a brief illustration of a typical cavity. Let us consider a cavity made of copper with dimensions a equal to b equal to d equal to 3 centimeters. It is made of copper so that the conductivity of copper is 5.8 into 10 to the power 7 simons per meter. The resonant frequency for the TE101 mode for this can be calculated from the formula derived earlier is found to be F naught equal to 7,070 megahertz 7.07 gigahertz. At this frequency the surface resistance is calculated for copper to be 0.022 ohms. And when we use the formula for the quality factor for the TE101 mode, the Q turns out to be 12,700 which is a value which we would not have got if one had used say a transmission line resonator. Q's of the order of 10,000 are quite typical for a waveguide cavity resonators. This is where we will stop today. In this lecture we have considered the waveguide resonators derived from the rectangular waveguides. We have worked out the field expressions for the TEM and L modes. We obtained an expression for the resonant frequency and then we saw how the quality factor can be estimated. Thank you.