 Welcome to the 25th lecture in the course engineering electromagnetics. In this lecture we continue with our discussion on parallel plane guide. The specific topics we take up today are transverse magnetic waves, the field distribution for the transverse magnetic waves and the expression of these fields in terms of plane waves. The parallel plane guide structure that we have been considering is the following. It consists of infinite conducting planes placed at x equal to 0 and at x equal to a. For the perfect electric region between these two conductors we have written down Maxwell's equations earlier and process these so that we obtain expressions for the transverse field components H x, H y and E x, E y in terms of the longitudinal field components E z and H z. From these expressions we make out that in general one of these E z or H z must be non-zero. And we considered earlier the case of transverse electric waves where E z is zero and today we go on to consider the case when H z is zero corresponding to the case of transverse magnetic waves. There is no magnetic field component along the direction of propagation hence the name transverse magnetic. We start with the let us identify which field components exist in this case. H z is zero E z is non-zero and therefore we have E x, H y and of course E z as the non-zero field components. Let us note down this for the T m waves the field components that are non-zero are the following. We have H z equal to zero and of course we have E z as a non-zero field component and then as we identified earlier we have E x and also H y. Out of six possible field components we have three field components which are non-zero just as in the case of the transverse electric waves. We consider the wave equation for the electric field and we write down the component of this wave equation for the z component of the electric field. Because that is what is going to be tangential to the perfectly conducting planes and therefore it will be possible for us to apply the boundary conditions in a simple manner. And therefore we write down the wave equation for the z component of the electric field as follows. Equal to E z by del x squared plus gamma bar squared E z which should be equal to minus omega squared mu epsilon E z. From the vector wave equation for the electric field we pick up the z component for the reason already explained. Then we consider the nature of the z component of the electric field and realizing that E z is equal to a part which varies with respect to x which we call E z naught x and a part which varies with z in this manner by virtue of our assumption that it propagates in the z direction with the propagation constant gamma bar. We realize that this second order partial differential equation can be written as a second order ordinary differential equation for this part of E z. Processing this equation therefore we get del 2 E z by del x squared equal to minus h squared E z where the symbol h was introduced earlier also. And it stands for omega squared mu epsilon plus gamma bar squared and then by substituting for E z this expression we can see easily that it reduces to d 2 E z naught by dx squared which should be equal to minus h squared E z naught. As far as the part which varies with respect to x is concerned it satisfies this simple second order differential equation. The solution for which can be written by inspection we obtain E z naught equal to let us say C 3 sin of h x h is the parameter we have written down earlier plus C 4 cosine of h x. As already noted E z is a field component which is tangential to the perfectly conducting surfaces constituting the parallel plane guide. And therefore it should be 0 at x equal to 0 implying that C 4 must be 0. And since it must be 0 also at x equal to a we get the value of the parameter h as before as h equal to m pi by a m taking on values in general 1, 2, 3 etc. While in the case of Te waves m equal to 0 would have led to a trivial situation all field components becoming 0 in the T m case it does not happen this we shall discuss in detail later on. So, while m equal to 0 is not ruled out for T m waves will restrict ourselves to the choice that m varies from 1, 2, 3 etc. And therefore the complete expression for E z becomes E z naught x which is C 3 sin of m pi by a x e to the power minus gamma bar z. Once we have obtained the expression for E z the expressions for the other field components which are non zero in this case that is E x and H y can be obtained from the equations shown earlier. Since all field components are related through the Maxwell's equations. So, taking up that we can write down the complete expressions for the T m m 0 modes corresponding to different values of the index m the role of which was explained last time we obtain the following field expressions E z of course as we had already written down is C 3 sin of m pi by a x e to the power minus gamma bar z. The other field components which are non zero are E x which is minus gamma bar a by m pi times C 3 cosine of m pi by a x e to the power minus gamma bar z. And finally we have H y which is minus j omega epsilon a by m pi and C 3 times cosine of m pi by a x e to the power minus gamma bar z. As we discussed last time in this case when we are not considering the presence of any kind of losses gamma bar the propagation constant which in general is complex in this case is either completely real or completely imaginary. We are interested in the situation in the range of frequencies where gamma bar is completely imaginary usually. And therefore considering that and considering the expression for the fields for m equal to 1 that is T m 1 0 mode under the situation that gamma bar goes to j beta bar which happens above a certain frequency. We can rewrite these expressions as follows we have E z equal to C 3 sin of m pi by a x and then followed by e to the power minus j beta bar z indicating wave propagation in the z direction with the phase shift constant beta bar with no attendant attenuation. Similarly the other field components become E x equal to since this is for m equal to 1 we need not have written this. Similarly E x becomes minus j beta bar a by pi and then C 3 cosine of pi by a x e to the power minus j beta bar z. And similarly H y becomes minus j omega epsilon a by pi and a similar expression as E x. So, C 3 times cosine of pi by a x e to the power minus j beta bar z. This is the nature this is the exact expression for the fields for T m 1 0 mode in the frequency range when the propagation constant gamma bar is completely imaginary. And therefore, it is magnitude is equal to the phase shift constant for this type of wave. Now, to understand the nature of these fields better we consider the corresponding time varying expressions by multiplying these field components by e to the power j omega t and taking the real part. Carrying out that procedure leads to the following time varying field expressions for this particular mode. And of course, they can be obtained for any other mode in general. Then we get E z equal to C 3 sin of pi a by x and then cosine of omega t minus beta z. As actually the field components are a function of both time and z both are sinusoid. Similarly the other field expressions turn out to be E x equal to beta bar a by pi C 3 cosine of pi by a x. And then here we are going to have sin of omega t minus beta bar z. And H y becomes omega epsilon a by pi C 3 cosine of pi by a x sin of omega t minus beta bar z. Which are the time varying field expressions for the TM10 wave. And as we did for the case of the transverse electric waves for the transverse magnetic waves also we can now consider the nature of the fields in a similar manner. We have got the skeleton of the parallel plane waveguide marked here. As far as the magnetic field is concerned now in this case that is the one which has a single component. So let us deal with that first that is a y component and therefore we consider an x y plane kind of cross section of the parallel plane guide that is at some z equal to constant plane we consider the cross section. Of course the actual value of H y will depend upon the actual value of z and the actual time instant chosen. Let us say that we choose omega t minus beta bar z such that it is pi by 2. So that it drops out of consideration and in that case H y can be plotted and we can continue with our convention that values with positive sign are represented to the right of the reference line. And the values with negative sign are represented to the left of the reference plane. So following that convention at x equal to a by 2 these are the locations of the planes. The field is 0 it is maximum at the planes and it may be represented by a cosine function like this. At x equal to a it is a negative value so it is represented to the left of the reference plane. And as we have been saying we may visualize this field in a plane normal to the plane that is being shown here. And positive values may be coming out of this plane negative values may be going down this plane. So this is the representation in magnitude and as we know it can be represented in an alternative manner in magnitude and direction by using the directed line segment kind of representation. And that is going to look like this in the same plane that is x y plane cross section x is 0 and a here and a by 2 here. And we must draw line segments which are oriented in the negative y direction taking care of this sign and are closely spaced here towards the conducting plane. And then as we go down to towards the centre of the wave kite the spacing increases considerably actually right at the centre of the field is 0. There must not be any line drawn there. Similarly the other side can be completed in a symmetric manner. And the arrow sign must be reversed remaining consistent with our sign convention. In this case the magnetic field representation is relatively simpler since the magnetic field is now composed of a single component. It is a y directed magnetic field and it can be represented in these two alternative fashions. Taking up the electric field next which has two field components and therefore is going to require a little more care. We have the components of the electric field which are x and z. And therefore we choose to show this in the x z planes. And the locations of these planes remain as already indicated here at x equal to 0 and at x equal to 8. Then we mark a number of reference planes which are essentially the values of omega t minus beta bar z. And we can choose these values as pi by 2 0 and minus pi by 2 etc. This argument will decrease as the values of z increase. Now at omega t minus beta bar z equal to pi out of the two components E x and E z, E x is going to be 0. Only E z is going to be non-zero. And also this part of this expression is going to have a value minus 1. Otherwise it will have a sinusoidal variation from x equal to 0 to x equal to 8. So with these observations we can now plot E z in this manner. This becomes E z. And with similar observations if we were to plot the fields at this plane where omega t minus beta bar z is equal to 0. Once again only E z will be present and the sign will be reversed. And therefore it can be shown by a curve which looks like this. At the other planes where the argument is pi by 2 or minus pi by 2 only E x will have a value non-zero value. E z will be 0 as far as these exact values are concerned. And keeping our sign convention E x is going to have a cosine sinusoidal variation like this at this plane. And the sign is going to change at the other plane so that it can be represented in this manner. So this is a representation in magnitude. The direction does not come out very clearly in this kind of representation. This limitation can be overcome by going in for the representation in terms of directed line segments. As we did for the t case also and we have this as the x z plane. And at omega t minus beta bar z equal to pi only E z is non-zero. And as you move away from this reference plane the value decreases. So keeping these things in mind we can show this as follows this kind of line segments. And what arrow direction will be put it is a negative sign here following our convention. And therefore it should be directed in this manner. As we move away from this middle plane the value decreases. And therefore as we move in either direction the spacing of the arrows should increase. Also as we move away from this reference plane along the z direction. Once again this field component is going to decrease in value. And while at exactly this reference plane E x was 0 as we shift from this reference plane E x will assume increasingly non-zero values. So keeping these things in mind we increase the spacing of the arrows in this manner. And we also do not draw these with the same length. Actually this is also an idealization. As far as the z directed field component is concerned it exists alone only at this reference plane. As we shift from this x component does come in with increasing value. And therefore there will be an appropriate curvature of these field lines. So keeping that in mind we do not draw these with the same length. And of course the spacing is increasing. Similarly one may complete this part here since this is already been discussed and is going to follow exactly the same arguments. The only change will be the reversal of the arrow signs. Otherwise it is an identical pattern. Here we had negatively z directed field lines. And here we have field lines directed in the positive z direction. Now the same procedure is applied to the x component of the electric field. At these planes that is pi by 2 and minus pi by 2 E z is completely 0. And E x has the strongest value close to the planes. At the midpoint E x also is 0. And therefore we can draw x directed lines here which are closely spaced. This is in the negative x direction recognizing our sign convention. And this is in the positive x direction. And then the spacing must increase for these lines as we shift from this reference plane. And we also reduce the length recognizing that z component is going to appear as we shift from this reference plane. And therefore we may draw the field lines for E x in this manner here. And similarly in this manner at these locations. As far as drawing these lines at the corresponding plane minus pi by 2 is concerned. It is now a simple matter. We repeat the pattern and change the arrow signs. And therefore we get a pattern which looks like this. With the arrow signs reversed. And similarly here and here this is the kind of skeleton of the complete field lines that we have drawn. And once we have reached this stage it should be fairly clear that actually the field lines have this kind of shape. And if we have proceeded correctly following our convention of sign consistently the arrow directions will come out compatible. And if there is a clash that is an indication that a mistake has been made somewhere. Either in converting the converting from the phasor notation to the time varying notation or in translating the time varying field components into this kind of curves. If we proceed correctly carefully we will get this kind of field patterns consistently. There are text books where this kind of field plots are given for a number of modes. Here we have done it only for T M 1 0 mode and earlier for the T E 1 0 mode. Now, when we look at these field components and earlier also we made comments that the nature of the propagation constant is changing as a function of frequency. It would appear that the waves that we are dealing with now that are supported on the parallel plane guide are different in nature from the uniform plane waves the T E M waves that we considered earlier. In some respects these are different, but one can look at it in an alternative manner and see the link between the uniform plane waves and the T E or T M waves that we are dealing with now. Let us consider the parallel plane guide and try and recall the fields for the T E 1 0 mode. Specifically we are now taking up the topic of expressing these field components as a super position of plane waves. Alternatively one could also call it plane wave decomposition of the wave guide fields. We consider the fields for the T E 1 0 mode which had field components E y H x H z. Let us consider the expression for the E y field component it was in terms of an amplitude constant say C 1 sign of let us make it general let us not restrict it to M equal to 1 mode we make it T E M 0 mode. So, that we write here sign of M pi by A x and then we will write here e to the power minus gamma bar z in general. But let us consider this in the frequency range when gamma bar goes to j beta bar. So, that the exponential factor that we get here is e to the power minus j beta bar z where if you look at the expressions beta bar will come out to be omega squared mu epsilon minus M pi by A whole square and the square root of the entire expression. And for beta bar to be real omega should be greater than a certain value which we shall comment upon a little later. Now, if you look at the two parts of the field expression one giving the variation with respect to z that is the quite similar to the wave propagation that we considered earlier except for the fact that the phase shift constant is now beta bar which is different from the phase shift constant expression that we had for the propagation in an infinite medium. As far as the variation with respect to the x direction is concerned it appears to be similar to that of a standing wave being sinusoidal in nature. So, let us develop this expression further and see what kind of interpretations are possible and how we link it to the standing wave kind of situation along the x direction. We write this as C 1 and the sign expression is written in terms of summation or difference of exponentials. So, that it is 1 by 2 j and then we have e to the power j m pi by a x minus e to the power minus j m pi by a x multiplied by e to the power minus j beta bar z. We multiply through and get C 1 by 2 j and then e to the power minus j beta bar z minus m pi by a x as far as the first term is concerned and in the second term we get e to the power minus j beta bar z plus m pi by a x. Now, we introduce a parameter theta the interpretation of it will become clear as we proceed that is introduced in the following manner. We have beta bar equal to beta cos theta beta bar is less than beta and let us write beta bar equal to beta cos theta and m pi by a equal to beta times sin theta. The basis for this is not clear to us right now as we proceed a very simple interpretation will become available. So, with this we modify this to read as C 1 by 2 j and then we have e to the power minus j beta coming out as common and then we have cos theta z minus sin theta x as far as the first term is concerned and the second term is quite symmetric except for a change of sin in the term within the bracket. So, that is e to the power minus j beta cos theta z plus sin theta x we further write this in the following form. It is C 1 by 2 j and we write this as e to the power minus j beta and let us write the position vector of a general point in the x z plane as x x cap plus z z cap and then if we multiply this take the dot product of this with minus sin theta x cap and plus cos theta z cap we recover the previous expression except that now it is in terms of the dot product of two vectors where the first vector is the position vector of some point in the x z plane and the second vector represents a unit vector. So, it is a certain direction in the x z plane with the direction cosines as minus sin theta and cos theta this can be interpreted as a uniform plane wave propagating with the phase shift constant beta not beta bar beta which is square root of omega square mu epsilon the phase shift constant that one will have if the wave propagated in an infinite medium. But, propagating in a direction which is oblique to the z and the x direction. Now, this is not alone this is just the first term, but the second term can also be written in the same form in a similar form and one can have a similar interpretation also for that it will be minus e to the power minus j beta and then x x cap plus z z cap dot sin theta x cap plus cos theta z cap which two terms going by the interpretation we have just offered represent uniform plane waves with phase shift constant of a uniform plane wave. But, propagating obliquely to the z direction or to the x direction what is this direction one can consider the parallel plane guide and consider the x z plane here this is the x equal to 0 plane and this is the x equal to a plane. Then the first term of the previous expression if we consider its direction of propagation is going to be represented by a ray which looks like this and is inclined to the z direction by an angle theta in this manner and it is propagating in this direction. On the other hand the other term represents a wave which is propagating also inclined to the z direction, but inclined in a symmetrically different direction that is making an angle theta like this. What are the associated fields of these plane waves we can recall the properties of the fields in the uniform plane wave that is a transverse electric and magnetic wave this has a y component of the electric field. Let us say coming out of the plane of the black board and an associated magnetic field which is orthogonal to the electric field and to the direction of propagation it is a magnetic field in the x z plane and as you see we have an x component of the magnetic field and a z component of the magnetic field. The vector may be represented in this manner this is the electric field and this is the magnetic field for the second wave keeping in mind that there is a negative sign in front of this term. We can put down the electric field as something which is going into the plane of the black board and it is propagating in this direction and therefore, the magnetic field vector is going to be in this manner and it is this super position of these two symmetrically propagating these two uniform plane waves propagating at an inclination to the z direction in a symmetric fashion which constitutes the actual fields for the T e m 0 mode. Now, theta here that we have got is related to the mode index m we have sin theta equal to m pi by a times beta and therefore, it is a straight forward to make out that this angle of inclination theta is going to depend upon the index m the mode that we choose the value of m that we choose will have a bearing on this inclination angle theta for theta equal to 90 degrees which is going to happen when we have omega square mu epsilon equal to m pi by a whole square which if you recall is the dividing line between the frequency ranges below which gamma is completely real and above which gamma bar is completely imaginary and we see that at exactly this point theta is 90 degrees and therefore, there is really no wave motion in the positive z z direction or in the z direction and the wave just propagates back and forth with reflections at these two planes. In the x direction the nature is that of a standing wave we can now decompose these two plate waves into a part propagating in the x direction and a part propagating in the z direction and the x directed components are superimposing to form a standing wave and therefore, the spacing a must have a relationship which is quite straight forward in terms of the wave length along the x direction it must be equal to let say m times the wave length in the x direction divided by 2 like any standing wave where lambda x is equal to the wave length in free space divided by sin theta which kind of expressions we have obtained earlier also whenever we have considered waves propagating at an inclination to an interface. And therefore, here one can straight away see the phenomenon of the cut off frequency coming into play for a given spacing if the wave length in free space is so large that for the smallest value of m also this relation is not satisfied the guide cannot support any wave propagation. So, this is where a major difference comes in propagation in infinite media and propagation in a wave guide. Since the wave guide imposes certain boundary conditions those are satisfied only above a certain frequency range which frequency the is called the cut off frequency and is obtained by writing omega equal to omega c and it is going to be m pi by a divided by square root of mu epsilon or the corresponding cut off frequency is going to be m by 2 a divided by square root of mu epsilon 1 by square root of mu epsilon is the velocity of light in the medium filling the parallel plane guide. Therefore, this cut off frequency is related to the cut off frequency is different for modes of different order m this is where I like to stop if there are any questions we can try out these. So, in the lecture today we have considered the transverse magnetic waves for the parallel plane guide we have seen the corresponding field distribution and we considered the very important concept that the waves that we are considering now T e or T m waves can be expressed as a super position of plane waves. Thank you.