 Welcome to the 31st lecture in the course engineering electromagnetics. We continue our discussion on rectangular wave kites. Today first we discuss the answer to a question that was raised in the last lecture and then we go on to discuss the wave impedance and attenuation constant for rectangular wave kites. You would recall that in the last lecture for the rectangular wave kite that we have been dealing with. We considered the following expressions for the Te10 mode which is the dominant mode or the lowest order mode for the rectangular wave kite and we considered the field expressions incorporating explicitly the time variation. Based on this we obtained the field expression, the field configuration for the wave kite and this was plotted in the following manner. I think that is more or less alright and by looking at the field expressions we were able to draw the magnetic field lines shown in the red and the electric field lines shown by blue dots and crosses. The electric field is y oriented plus y or minus y depending on the location and the time instant we choose and the magnetic field is entirely in the XZ plane. The question that was raised was that since we have some circulation of the magnetic field there should be a corresponding displacement current so that right say at the center of these magnetic field line loops there should be some value of the displacement current density. Why do we say this? We say this from the Maxwell's equation which reads as del cross H equal to del D by del t. In the phasor notation this would become del cross H equal to j omega and then epsilon times e and apparently the confusion was raised because at the center of these loops the electric field is 0. However if we consider the derivative of the displacement density which is the displacement current which is epsilon times e and then it is time derivative then we will find that consistent with the maximum value of the magnetic field the curl of the magnetic field here the displacement current density will be maximum at these locations and it will have the appropriate sign. If we consider the first form of the Maxwell's equation if we consider the phasor form del cross H equal to j omega epsilon e then since the factor j represents a phase shift by 90 degrees we will find that if we change the value of this argument say by changing the time by 90 degrees then the maximum value of the displacement current density will appear at these locations with 90 degree phase difference from the magnetic field and therefore as expected the field lines are quite consistent with respect to each other. Then we take up the first topic listed for the day and that is the wave impedance you would recall that we said that the characteristic impedance for the transmission lines plays a very important role and is related to the real power transfer along the transmission line. We would like to evolve a similar quantity for the wave guides for the transmission lines the characteristic impedance is simply the ratio of the voltage and current say for the forward wave but on the wave guides we can have a large number of field components and therefore in the context of the parallel plane guide we had evolved fairly general notion for the wave impedance and you would recall that we wrote expressions of the form z x y plus which indicates the wave impedance involving the x component of the electric field and the y component of the magnetic field looking in the positive z direction and that is given as the ratio of e x and h y. Similarly one could also write equally well z y x again looking in the positive direction of the remaining coordinate that is the z coordinate here which will be minus e y by h x all right and since we have considered waves propagating along the z direction in the rectangular wave guide it is these two quantities z x y plus or z y x plus which will be relevant for us knowing the field expressions for the various modes say for the transverse magnetic modes and similarly for the transverse electric modes it should now be possible for us to obtain these wave impedances okay. Keeping these expressions at the back of our mind and realizing that for the frequency range where omega is greater than the cut off radiant frequency omega c we have gamma bar going to j beta bar where beta bar for the rectangular wave guide is omega squared mu epsilon minus h squared in the general case where h is further available in terms of the dimensions of the wave guide and the indices of the various modes this is whole square root which can be further expressed as omega square root of mu epsilon and then recalling the relationship between h and omega c it can be written in a fairly simple and general form as follows it is 1 minus omega c by omega whole square okay. So for this frequency range where the propagation constant is equal to the is equal to j times the phase shift constant and there is no real part of the propagation constant in this idealized situation we can look at the fields for the transverse magnetic modes okay and concentrate on e x and h y on one hand and e y and h x on the other hand and we find that the ratio is the same for both of these and it is simply gamma bar upon omega epsilon and then upon j omega epsilon and then gamma bar is going to be equal to j beta bar. So making these considerations we obtain the following expression for the wave impedance we can write this as z t m to distinguish it from the wave impedance that we will obtain for the transverse electric modes and we can drop the other subscripts and the superscripts keeping in mind that this is the wave impedance calculated using the x or the y components and looking in the positive z direction then this is equal to beta by omega epsilon whether we use e x and h y or whether we use e y and h x okay if you look at the field components for the transverse magnetic modes it will be very clear that whether we calculate the wave impedance using this expression or this expression the value turns out to be the same. In fact we could combine this in a more general way and write this as equal to e transverse magnitude upon h transverse magnitude combining both x and y field components for both the electric field magnitude and the magnetic field magnitude okay now this turns out to be recalling that beta bar can be written like this it comes out to be equal to eta times square root of 1 minus omega c by omega whole square root where eta is the familiar expression for the intrinsic impedance of the medium filling the wave guide and it is equal to square root of mu upon epsilon okay you may be able to recall that this is the same expression that we got for the transverse magnetic modes in parallel plane guide okay and we said that these expressions are fairly general they hold good for various wave guides in general in a similar manner if we were to consider the TE MN modes and we have to calculate the wave impedance using the x and y electric and magnetic field components then you will see that in the same frequency range if you consider e x upon h y we are going to get omega mu upon beta bar or if we consider e y and h x with the negative sign in front once again we are going to get omega mu by beta bar okay this expression is different from that for the TE MN modes but otherwise it is the same whether we use one definition or the other noting down this here gives us the wave impedance for the TE modes as omega mu upon beta bar this is beta bar and so is this okay and this also can be rewritten in a simpler form as eta times 1 minus omega c upon omega whole squared and whole square root of the denominator. Now as far as the expression in terms of the frequency and the cutoff frequency is concerned these expressions as I have been saying are quite general and therefore the variation of the wave impedance as a function of frequency that we considered for the parallel plane guide is going to be maintained here also the only difference is that for the rectangular wave guide there is no TE M mode that can propagate that we have seen earlier and as far as the variation of the wave impedance as a function of frequency for TE M and TE modes is concerned that remains qualitatively the same and in fact in terms of the cutoff frequency the behavior of various TE M and TE modes is going to be given by these curves what is going to happen below the cutoff frequency below the cutoff frequency the wave impedance will not remain real it will become imaginary and it will not correspond to any average power transfer along the axis of the wave guide we should also notice that close to the cutoff frequency the impedance varies rather sharply and therefore from the point of view of systems application one would like to avoid a frequency which is too close to the cutoff frequency okay this point will come up later also again in the context of the attenuation taking up the calculation of the attenuation constant next we had listed a general procedure for calculating the attenuation constant we said that a good engineering approximation is to calculate the fields ignoring losses and then make an assumption that the fields remain unchanged even in the presence of a small amount of losses which is going to hold good when we use a low loss conducting or dielectric materials for the construction of the wave guide and then we calculate the power that is transmitted estimate the power loss in the lossy conductor and or the dielectric as the case may be and then calculate the attenuation constant using this definition which was evolved by using the transmission line analogy so that the attenuation constant alpha is given in terms of WL which is the power loss per unit length divided by twice the total power transmitted this method we have applied to the parallel plane guide now let us apply it to the rectangular wave kind continuing as general continuing to be as general as possible as far as the calculation of the power transmitted is concerned we will be interested in the pointing vector in the Z direction okay let us call it the axial pointing vector and we call it PZ okay and as was mentioned earlier as far as the quantities involved in the attenuation constant are concerned we could either use both numerator and denominator as average quantities or peak quantities but both should be of the same kind and usually it is safer to calculate average values so that one does not make a mistake so what we are trying to calculate here is the average value of the axial pointing vector which has the standard expression that is half real of E cross H star and we are interested in the Z component of this vector product which can be written as half real of now which components of these fields will be involved in the Z component it will be the transverse components of the electric and magnetic field and therefore we can write this equivalently as E transverse cross H transverse okay this can save some work because otherwise there may be Z components also existing in a certain problem but this clearly shows that those are of no interest in this context further as we have just seen in the calculation of the wave impedance the transverse field components are all in phase and therefore we can write this simply as half of magnitude E transverse times magnitude of H transverse okay. However their ratio E transverse upon H transverse is the wave impedance and therefore we can simplify it further to read as half and either Z Z times magnitude H transverse squared or equivalently 1 by 2 Z Z times magnitude squared of the transverse electric field okay one can use either definition. However what we are interested in is the total power transmitted what have we got right now is the power transmitted per unit area and therefore as far as the total power transmitted is concerned denoting this by WT this is equal to PZ integrated over the cross section of the waveguide okay which in terms of one of these expressions turns out to be half Z Z which is a constant for the given waveguide that will not be involved in the integration and then integral of magnitude H transverse squared times DA where as noted earlier the integration is over the cross section of the waveguide in a similar manner as far as the power loss is concerned and we could first consider the power loss in the conductor we can write this as half RS and then magnitude J squared this will be the power loss in the conductor per unit area and for the total power loss in the conductors we will need to integrate this over a surface over the conducting surface of unit length along the direction of propagation okay which could be noted down here a surface of unit length however we have identified earlier that the linear surface current density in magnitude is equal to the tangential component of the magnetic field therefore this also can be written in terms of the magnetic field again giving us half RS magnitude H tan squared DA integrated over such a surface okay and in the form in which WT the power transmitted and the power loss in the conductor are written we only require the magnetic field components okay so to that extent it is somewhat simpler and if one were to estimate the power loss in the dielectric WLD then one would write 1 by 2 sigma magnitude E squared which becomes the power loss in the dielectric per unit volume and therefore is to be integrated over a volume which is of unit length along the direction propagation okay usually the dielectric material if at all there is any involved in the wave guide is of a much lower loss than the conductor and even if the dielectric is not utilized conductor has to be present and therefore conductor loss is always present therefore usually in calculating the attenuation constant we concentrate on the loss due to the conductor now this calculation can be carried out therefore for any particular mode that one is interested in since it is the dominant mode T10 mode which is of greatest practical utility we will carry out this calculation for that mode only the starting point will be the fields for the T10 mode okay ignoring any losses assuming that the dielectric is perfect and the conductors are perfect and then for the T10 mode we have these three field components so starting with these we can now calculate the various quantities required for the attenuation constant for example if we consider the calculation of the power loss in the conductor let us call it WLC okay this is going to be given by half and then RS and magnitude H10 squared DA the tenetial component of the magnetic field is related to the linear surface current density and we have been able to relate JS and H10 and therefore if we consider the tenetial component of the magnetic field to the conductors depending on the particular wall that one is considering the tenetial field component of the magnetic field will be different for example it is the it is both these magnetic field components HZ and HX which are tenetial to the broad walls of the wave kite the wave kite you would recall is like this the broad walls are in the XZ plane and the narrow walls are in the YZ plane and the narrow walls are located at X equal to 0 and X equal to A as far as the broad walls are concerned both HX and HZ field components are tenetial to these walls and are going to cause corresponding current flow in these walls and which will result in some ohmic loss as far as the narrow walls are concerned it is only the Z component of the magnetic field which is tenetial to the narrow wall and also the value of the Z component that needs to be utilized in the calculation is the value at X equal to 0 and X equal to A okay so keeping these things in mind and looking at these field components we can write the expression for the conductor loss in the following manner we will have integral from 0 to A of beta C by pi by A and then sine of pi X by A whole square DX which corresponds to magnitude HX okay similarly we will write plus integral from 0 to A for C times cosine of pi X by A whole square DX which corresponds to the magnitude of HZ on the broad walls and as far as the narrow walls are concerned we have a constant value HZ of magnitude C so C squared and then multiplied by the height or the width of the wave guide at the narrow wall at X equal to 0 or A the field is constant so an integration is not required and along the length we are using a unit length so that completes the expression for the loss in the wave guides for the conductor loss in the wave guide there is something that we have forgotten we have to consider two walls of each type and therefore a factor of two can be incorporated here the integrals that are involved are fairly straight forward both sine squared pi X by A or cos squared pi X by A when integrated from 0 to A give us a value A by 2 so recognizing this we get a value for this quantity which is RS C squared and then beta by pi by A whole squared into A by 2 for the first integral plus A by 2 for the second integral and for the third one third term we get simply plus which takes into account the power loss in the conductor as ohmic loss similarly we have already written down the expression for the power transmitted WT as half ZZ and then integral of H transfers magnitude squared over the cross section of the wave kind alright and the transfers component of the magnetic field for this mode is only HX the component which is transfers to the direction of propagation that is a direction and therefore we need to consider only the X component of the magnetic field and using that and recalling that here ZZ is simply eta by 1 minus omega C by omega whole squared whole square root which for the sake of brevity is written as eta by K 1 0 it is just a new symbol that we have introduced for the denominator and we write K 1 0 because we intend to use this for the TE 1 0 mode so using this and the integral is fairly straight forward what we get is the value equal to half eta by K 1 0 and then beta C by pi by A whole square of the whole quantity times A and B by 2 while we are at this point it can be pointed out that this is the expression that one will utilize for calculating the total power transmitted in the waveguide and from this one can make out what will be the maximum power handling capability of the waveguide it of course depends upon these various other quantities the frequency of operation cut off frequency etc the waveguide cross sectional area and the value of the amplitude constant C which value is limited by the breakdown electric field value for the medium inside the waveguide. So many times to improve the power handling capability one will evacuate the waveguide make sure that it is free of moisture or fill it up with some dry inert gas to get the maximum power transferred through the waveguide anyway this is the general expression for the total power transmitted and now as far as the calculation of alpha is concerned which is W L by twice W T one can take the ratio of the expressions already evaluated and what we get is the following we also recognize that for the T 1 0 mode pi by A whole squared a quantity which is appearing in both the numerator and the denominator is equal to omega C squared mu epsilon in fact this is the expression utilized for calculating the cut off frequency and therefore pi by A squared can be replaced by this so that we can express the various quantities in as the ratio of omega C by omega or in terms of this ratio. So using that we get a relatively simple expression which reads as R S upon eta times A B and then k 1 0 the symbol that is been introduced earlier and then we have A plus 2 B and omega C squared by omega square which is the expression for the attenuation constant considering the conductor loss. So we could put a subscript C to make sure that we interpret it appropriately this is the expression for the attenuation constant due to conductor loss for the T 1 0 mode one can calculate the attenuation constant due to conductor loss for TM and TE modes in general also and the procedure is going to be along these lines only for the sake of completeness let me write down the expressions for the attenuation constant due to conductor loss for the rectangular wave kind the attenuation constant due to conductor loss for the TM MN modes turns out to have an expression which is twice RS upon eta AB KMN and then we have N squared AQ plus M squared BQ divided by N squared A squared plus M squared B squared where RS is the skin resistance of the surface resistance of the conducting material constituting the waveguide walls and has an expression omega mu M by 2 sigma M whole square root and the symbol KMN stands for 1 minus omega C by omega whole squared and then the square root of the whole expression in a similar manner the attenuation constant due to conductor loss for the TE MN modes in general has an expression which is twice RS upon eta AB times KMN that is as for the TM MN modes then we have a relatively long expression here which reads as A plus B multiplied by omega C squared by omega squared plus AB times KMN prime squared and then we have M squared B plus N squared A divided by M squared B squared plus N squared A squared here the new symbol that is been introduced KMN prime means delta N by 2 minus omega C by omega whole squared and the square root of this entire expression where delta N is the chronicle delta standing for 1 whenever the subscript N is 0 and otherwise it stands for 2 so that is delta N is equal to 2 for N greater than 0 with this arrangement the same expression can give us the attenuation constant due to conductor loss for the TE 1 0 modes also the variation of the attenuation constant turns out to be like this attenuation for various modes in a rectangular brass waveguide the waveguide dimensions have to be specified before we can talk of numerical values of the attenuation constant the waveguide is shown here it is a 2 inch by 1 inch brass waveguide so the conductivity of the metal walls constituting the waveguide is also specified and this is the variation of the attenuation in decibels per foot okay that is a slight conversion involved which is on the log scale and so is the frequency which is on the x axis on log scale we have the frequency of 10 GHz here and then 15, 20, 30, 40 etc this waveguide that is been selected for illustration is a C band waveguide it is meant for the frequency range 4 to 8 GHz if you notice the dimensions are more or less the double that of the dimensions for the x band waveguide the x band frequency range is 8 to 12 GHz therefore this is meant for C band which is 4 to 8 GHz if you calculate the cutoff frequency it will come out to be roughly 3 GHz and as we approach the cutoff frequency the attenuation is increasing steeply this is the variation of the attenuation for the TE 1 0 mode one can see clearly that after a certain frequency other higher order modes can start propagating but within this frequency range we can have single mode of operation the dominant mode the TE 1 0 mode has the lowest attenuation compared to the other possible modes and there is a certain range even in the single mode frequency range where the attenuation is the lowest and is relatively flat which is somewhat far away from the cutoff frequency for this particular mode when we have shifted so much from the cutoff frequency the variation in the wave impedance also will be fairly small and therefore this range becomes a range where one can utilize the waveguide quite efficiently and effectively without obtaining too much of distortion in the signal what we use more frequently for example in the laboratories are X band waveguides which have slightly different dimensions and they are utilized in the TE 1 0 mode and that attenuation is shown here it is a waveguide which is different from the previous one it is a copper waveguide with dimensions in centimeters as 2.286 and 1.016 centimeters which corresponds to 0.9 inches and 0.4 inches and the material that has been specified is copper for the dominant mode the loss attenuation in decibels per meter varies in this fashion as a function of frequency we did calculate the cutoff frequency for the various modes for this waveguide earlier and you would recall that the cutoff frequency for the TE 1 0 mode was around 6.55 gigahertz so as we would approach that frequency the attenuation would rise steeply but somewhat far away from that frequency the attenuation is not varying so sharp this is where we like to stop unless you have question. So in the lecture today we have considered the wave impedance for the rectangular waveguide both for the TE modes and the TM modes and then we have calculated the attenuation constant due to conductor loss once again for the various modes that the rectangular waveguide can support thank you.