 Welcome to the 29th lecture in the course engineering electromagnetics. In this lecture we continue our discussion on rectangular wave guides which was initiated yesterday and we talk about the TM MN modes and then go on to discuss the TE MN modes. First we have a quick recapitulation of what we did in the last lecture. We are considering the rectangular wave guide shown here consisting of perfectly conducting walls located at x equal to 0, x equal to a and y equal to 0 and y equal to b. To find out the properties of the wave types supported by such a structure we first wrote for the interior of this wave guide the Maxwell's equations expanded these Maxwell's equations in the Cartesian coordinate system and then assuming that the wave propagates along the z direction we simplified these by substituting for del by del z the factor minus gamma bar. Then these equations were further ordered by expressing the transverse field components in terms of the longitudinal field components that is x y components were expressed in terms of the z components and then we were able to make out that we may either have TE type of solution or TM type of solution. We considered first the transverse magnetic type of solution where hz is 0 and ez is non-zero and we obtained the following solution for the z component of the magnetic z component of the electric field that is C sin of m pi by a times x sin of n pi by b y e to the power minus gamma bar z. The indices m and n decide the order of the particular mode and therefore this becomes one of the field components for the transverse magnetic modes and we call this TM, Mn modes. The first subscript m indicates the order of variation along the x direction and the second subscript n indicates the order of variation along the y direction. We can now complete the field expressions for the TM, Mn modes and for the sake of brevity we rewrite ez in terms of the constants a and b so that it becomes C times sin of Ax sin of Bx sorry times sin of Ay e to the power minus gamma bar z. Then using the expressions for the transverse field components which are available in terms of the z components here hz is 0 and we have expression for ez available and therefore we can find out the expressions for all the remaining four field components. The steps are fairly straight forward and therefore we can write down the field expressions straight away. We get Ex equal to minus gamma bar C by H squared then B cos Bx and sin Ay e to the power minus gamma bar z. Then Ey comes out to be minus gamma bar C by H squared A times sin of Bx and cos of Ay e to the power minus gamma bar z. Hx comes out to be j omega epsilon by H squared times C and then A sin Bx cosine Ay e to the power minus gamma bar z and finally Hy comes out to be minus j omega epsilon by H squared times C and then B times cosine of Bx sin of Ay e to the power minus gamma bar z. This is the complete set of fields for the transverse magnetic modes with order m and n. One can make out from these expressions and recalling that B is equal to m pi by A and A is equal to n pi by B that the lowest value of m or n which is required so that these field components do not all become 0 is n equal to 1 and n equal to 1. If either of these becomes 0 the way A and B appear in these expressions the entire set of fields becomes 0 it vanishes. So from this point of view we make this observation and it will play an important role later on that the lowest order Tm mode is Tm 11 where m is equal to 1 and n is also equal to 1. Neither of these can become 0 if that happens then the entire set of fields vanishes that becomes a trivial situation and therefore for a non-trivial situation we must have both m and n non-zero and then the lowest values possible are 1 1 each. So Tm 11 is the lowest order Tm transverse magnetic mode. Before we discuss the properties of the transverse magnetic waves let us also complete the similar exercise for the transverse electric modes. And we proceed in a manner similar to what we did for the transverse magnetic modes for the transverse electric modes the electric field is completely transverse to the direction of propagation and therefore we have E z equal to 0 and of course H z not equal to 0. Once again if both become 0 then the entire set of fields vanishes which is not a situation of interest. So we will consider the component of the wave equation for the magnetic field which involves the z component the H z component and then using the method of variable separation we will obtain a solution for H z alright. I think we can write down the solution straight away since the steps are completely in parallel and therefore H z not the part that varies with respect to x and y turns out to be c 1 cosine of b x plus c 2 sin of b x into the second part is c 3 cosine of ay plus c 4 sin of ay by using the method of variable separation we can reach this stage by following the steps previously described for the Tm modes. Now the question comes how do we apply the boundary conditions because H z is not a field component is not a tangential field component to the perfectly conducting volts and therefore we calculate the field components the electric field components in terms of H z which are tangential to the perfectly conducting volts. For example we can calculate using the expressions for the transverse fields in terms of the longitudinal fields E x and E y E z is 0 and in terms of H z we can write their expressions and E x and E y are going to be tangential to the corresponding perfectly conducting volts. So on E x and E y we can apply the boundary conditions that the tangential electric field component on a perfect conductor must become 0. So in this manner we apply the boundary conditions and we write for E x or to be more precise for E x not we write minus j omega mu by H squared times del H z not by del y and as you can make out the superscript not stands for the fact that we have removed the factor into the power minus gamma bar z because that is common on both sides all right. And similarly we have E y not equal to j omega mu by H squared and del H z not by del x all right del H z not we have just written and therefore these expressions for E y not and E x not can be obtained carrying out this procedure E y not will come out to be j omega mu by H squared and then B times the derivative of this first part and the second part is not going to be affected because it is the x derivative which is involved and therefore we write here minus C 1 sin of B x plus C 2 cosine of B x and this part we write as capital y because before reaching this step one would have written H z not equal to x times y and as a solution of the wave equation x comes out this and y comes out this and y is not getting affected in this operation therefore we retain it as it is which should be 0 wherever it is tangential to the perfectly conducting boundary walls of the waveguide that is at x equal to 0 which implies that C 2 should be 0 similarly it should be 0 at x equal to A and as before that implies a particular value of B that is B equal to M pi by A in this manner we have evaluated C 2 and we have got definite value for the constant capital B similarly taking care of the other part that is ex0 we get ex0 equal to minus j omega mu H squared and here the first part that is capital X is not getting affected and therefore we retain that as such and the second part gives us A times minus C 3 sin of A y plus C 4 cosine of A y which is tangential to the other two walls of the four walls of the rectangular waveguide and therefore we can get some more information about the remaining parameters it should be 0 at y equal to 0 implying that C 4 should be 0 and similarly it should be 0 at y equal to B giving us a certain value of capital A which is N pi by B where M and N are integers taking on values 1, 2, 3 of course whether they can take on the value 0 or not we shall discuss in a moment alright so based on this we now have an expression for H z0 which can be substituted in the expression in the expressions for the transverse field components and then one will get the complete set of fields for the TM modes so carrying out this procedure we get H z the complete expression for H z as equal to a constant C where we have clubbed C 1 and C 3 together C times cosine of B x and cosine of A y and of course e to the power minus gamma bar z which becomes the starting point for the field expressions for the TE modes and depending on the values M and N it becomes the TEMN so for a TEMN mode the z component of the magnetic field comes out like this and then using the expressions for the transverse field components in terms of the longitudinal field components E z is 0 and H z is now available so we can get the remaining four transverse field components and we write down these field expressions as follows we have E x equal to j omega mu C by H squared A times cosine B x sin of A y e to the power minus gamma bar z E y comes out to be minus j omega mu C by H squared B times sin B x and cosine A y e to the power minus gamma bar z H x comes out to be gamma bar C by H squared B sin B x cosine A y e to the power minus gamma bar z and finally H y comes out to be gamma bar C by H squared A times cosine B x sin A y e to the power minus gamma bar z a very high degree of symmetry is obvious in these field expressions and H x and E y behave in a similar manner as far as the x and y variations are concerned and E x and H y behave in a similar manner which will have a bearing on the wave impedance which we shall calculate later on and of course all field components vary along the z direction as e to the power minus gamma bar z corresponding to our assumption that the wave is propagating with the propagation constant gamma bar in this z direction one can look at these field expressions more closely and make sure that each field components satisfies any boundary conditions that are applicable on that particular field. So it becomes a consistent set of field equations which satisfy of course Maxwell's equations and also the boundary conditions which are applicable in this case and since E z is 0 it is a TEMN mode field same. Like we identified for the TEMN modes what is the lowest value the indices M or N can take and then it turns out that any one of M or N can take on value 0 and as long as the other one is non-zero the entire field set does not become 0. We will still have sufficient field components remaining to satisfy the Maxwell's equations even if one of M or N index becomes 0 and therefore we identify this fact that the lowest order TE mode is and right now we are not very sure but what we can say is that it is either TE 10 it is not TE 10 it is TE 10 or TE 01 and if you recall the interpretation of these subscripts this would mean that the fields are constant with respect to y the order of variation is 0 and they have the first order of variation along the x direction one maximum between the x equal to 0 to x equal to a conducting volts and similarly one can interpret these two indices alright. With this observation now we are ready to consider the properties which are common to both transverse magnetic and transverse electric modes. Let us go on to do that next TEM and TMM modes and what about the TEM mode transverse electric and magnetic mode? We identified very clearly that for the cylindrical wave guides that we are considering or the hollow metal wave guides that we are considering the transverse electric and magnetic mode cannot be supported. In fact that becomes the distinguishing feature between the transmission lines and the other wave guides in general. We say that a structure which can support a transverse electric and magnetic mode of wave propagation is a transmission line and structures which do not support this are called wave guides although all structures that we have been considering will support propagating waves but we say that the transmission lines support TEM mode of propagation and wave guides do not. Here we have considered one particular type of wave guides the cylindrical wave guides or the hollow metal wave guides. There are possibilities for other types of wave guides as well for example dielectric slab or dielectric rod. In fact an optical fiber is also a wave guide operating at very high frequencies. All these other wave guides other than transmission lines do not support TEM mode of propagation. So that becomes a distinguishing feature between transmission lines and other wave guiding structures. And now let us see what is the significance of this observation or what are the properties associated with non TEM that is TE or TM modes. For this purpose you would recall that we wrote h squared equal to gamma bar squared plus omega squared mu epsilon on one hand and also in terms of the constants a and b we had h squared equal to a squared plus b squared which was further equal to n pi by b whole squared plus m pi by a whole squared. So in this manner h squared is related to the order of the mode m or n and the wave guide dimensions. The other parameters are the frequency and the constitutive parameters of the medium filling the interior of the wave guide and therefore all these things are coming in into picture and they will decide the propagation constant gamma bar. We write gamma bar equal to h squared minus omega squared mu epsilon and therefore it is m pi by a whole squared plus n pi by b whole squared minus omega squared mu epsilon whole squared. And while in general we said that gamma bar the propagation constant can be complex with real part alpha bar and an imaginary part beta bar in this idealized situation where we have ignored losses in the conducting walls or in the dielectric medium filling the wave guide gamma bar would either be completely real below a certain frequency or it would be completely imaginary above a certain frequency. We will have for frequencies less than what is called the cutoff frequency gamma bar going to alpha bar it will become completely real and the fields instead of propagating will simply decay exponentially as e to the power minus alpha z. Such a wave is also called an evanescent wave after a certain distance along the z direction in this case the fields will become negligible as the meaning of the word evanescent would suggest. However above this cutoff frequency omega greater than omega c will have gamma bar completely imaginary and we will have wave propagation without any attendant attenuation at least in this idealized situation. This is the range of frequencies we are normally interested in. This cutoff frequency can be identified from this expression these expressions we have omega c equal to 1 by square root of mu epsilon and then square root of n pi by a whole squared plus n pi by b whole squared and the square root of this whole quantity or this is the radiant frequency the frequency in hertz can also be written it will be replacing 1 by square root of mu epsilon by the velocity of light in the medium filling the wave guide it will be c by 2 and then we have m by a whole squared plus n by b whole squared and the square root of the quantities involving m and n. So this is the cutoff frequency which is related to the velocity of light in the medium in the interior of the wave guide the order of the modes m and n and the wave guide dimensions. The higher the values of m and n the higher the corresponding cutoff frequency. So from this point of view now one can appreciate the significance of the fact what is the lowest order mode for the T m or the T e modes. We said that T m mode has as its lowest order T m 1 1 mode and then in the T e modes we had T e 0 1 as a possibility and also T e 1 0 as the other possibility. For T m 1 1 the cutoff frequency will be given by m equal to 1 and n equal to 1. So for a given wave guide the cutoff frequency of this which we may write as f c 1 1 as far as the expressions are concerned they are not different for the T m modes or the T e modes. So we just write f c 1 1 which of course is less than the cutoff frequency corresponding to either of these less than or greater than it is greater than. For T e 0 1 mode the cutoff frequency is f c 0 1 and it is equal to c by 2 b quite simply and the corresponding cutoff frequency for the last one f c 1 0 is going to be c by 2 a. Now by convention a is the dimension along the x direction and b is the dimension along the y direction and a is taken to be the greater dimension. You could put it in other words x x 6 is oriented along the wider dimension of the cross section. This is almost a universal convention 1 or 2 or text books may use the convention other way round but normally x is the direction associated with the longer dimension along the cross section. From this following this convention now we can see which of these two last cutoff frequencies is greater and we can easily put this kind of sign. Since a is greater than b this cutoff frequency is greater than this cutoff frequency and therefore we find based on this consideration that it is this mode which has the lowest cutoff frequency. If we are given a waveguide and we start increasing the operating frequency it is this mode which will start propagating first of all and as the frequency continues to increase first this mode may appear will be supported by the waveguide and then other modes may appear progressively. So, from that point of view these other modes which appear later at higher frequencies are called higher order modes and this becomes the dominant mode the lowest order mode which is an observation of considerable significance. We have already mentioned that normally we like the propagation in a structure to be in a single mode because if the signal is propagating in a number of modes and these different modes are going to have different characteristics different field configurations and as we shall see different phase velocities distortion is inevitable. And also the fact that a signal is distributed over a number of modes affects the efficiency of excitation of say the waveguide and also the efficiency of collection of the power for that particular signal. And therefore we will prefer that the waveguide or the structure at hand operates in a single mode and that is possible only for the dominant mode of propagation. So, what is the lowest order mode what is the dominant mode for a particular structure is therefore is a consideration of great significance. The other characteristics of TE and TM modes can now be written down we have the expression for the phase shift constant in the frequency range greater than omega c beta bar equal to omega squared mu epsilon minus m pi by a whole squared minus n pi by b whole squared and of course whole squared. From this expression itself we can make out that the phase shift constant beta is not a linear function of frequency. In fact this is more clear if you write the expression for the phase velocity v p bar which is going to be omega by beta bar which is now a function of omega unlike the situation for the infinite medium where these other terms do not appear and we have beta bar equal to square root of omega squared mu epsilon in that case the phase shift the phase velocity is going to be constant at all frequencies but here the phase velocity is not constant with frequency. This is directly a consequence of the fact that the waveguide is not supporting a TEM mode of propagation. This means that some dispersion is unavoidable if the phase velocity is a function of frequency it is different and different frequencies that is going to lead to dispersion. That precisely is dispersion and therefore some dispersion is inevitable which is again a universal characteristic of TEM and TEM modes and therefore all structures which are not transmission lines. That is why we have been stressing the fact whether a structure supports TEM mode or not. If it is not a TEM mode of propagation then the phase velocity is a function of frequency and dispersion and therefore related distortion is unavoidable. As a result of this dispersion if there is a signal which has frequency content which spreads too widely it is likely to get distorted or else it should be propagated over rather short distances and therefore this will have an effect on the bandwidth of the signal that the waveguide can handle or the distance over which a signal with a certain bandwidth can be propagated. The guide wavelength can also be obtained from here lambda bar which is going to be 2 pi by beta bar and one can make out that just as this is less than omega square mu epsilon whole square root. This is greater than the velocity of light in the medium filling the interior of the waveguide and this also is greater than lambda or lambda naught that will be applicable to the interior of the waveguide. Now there are two things that I like to point out here as far as the waveguide is concerned. First thing is that as compared to transmission lines in a waveguide some dispersion is always present. That one could say is a disadvantage of using waveguides. We listed a number of advantages of waveguides there are certain disadvantages also this is one of those. What could be the other disadvantage that comes from the fact that for the waveguide to be able to propagate signal at a certain frequency it must have certain minimum dimensions. So that the cut off frequency that the lowest order mode that can be supported on the waveguide is less than the frequency of the signal that we want to transmit. The cut off frequency as we have seen for even for the lowest order mode f c 1 0 is equal to c by 2 a. So if you want to transmit let us say signal of frequency 5 gigahertz this dimension must have a certain minimum value which will turn out in centimeters. We shall consider an example and this will become clearer and therefore the waveguides are rather bulky. They are not compact because they must have certain minimum cross sectional dimensions to be able to propagate signal at a given frequency. So these are two disadvantages of the waveguides and novel structures transmission structures have been proposed like stripline and transmission line which at least remove the bulkiness part of the waveguides but at the cost of something else. Also these are not easily mass produced these in large quantities is not very economical. But the other structures which may remove some of these limitations will suffer from some other limitations like lower power handling capability and higher attenuation etcetera. We consider the example of an expand waveguide and see what kind of cut off frequencies the waveguide has. The expand waveguides are used very frequently in laboratories it has dimensions which are 0.9 by 0.4 inches or 2.286 centimeters by 1.016 centimeters. The expression for the cut off frequency has been written earlier. Let me just write down the cut off frequencies for the modes of various orders. For the T10 mode the cut off frequency is 6.55 gigahertz and for transmitting signals which are greater than this frequency this waveguide can be utilized which is certainly quite large compared to a thin coaxial cable. So that supports the statement that waveguides are not very compact. The next mode in the order of increasing cut off frequencies is TE20 mode which has a cut off frequency of 13.1 gigahertz. Then TE10 mode has a cut off frequency which is 14.0 30 TE01 thank you. The TE01 mode has a cut off frequency of 14.78 gigahertz and then if you go to TE11 or TM11 the cut off frequency is 16.14 gigahertz. And as we are interested in single mode of operation over the frequency band over which the waveguide is to be utilized one can see that if we restrict our signals within the frequency range 8 to let us say 12 gigahertz then we will maintain single mode of operation. The other modes will be above cut off the waveguide will not support this which is the X band frequency X band frequency range. So in the lecture today we have considered the field expressions for the TM MN and the TE MN modes. We have considered the characteristics of these modes and we considered the example of an X band waveguide. So this is where we stop this lecture. Thank you.