noc18-ee28-Lecture 15-Systematic analysis of parallel plate metallic waveguide

[Music] fiber-optic communication systems and techniques in the previous module because of the time constraint we had to stop our six step procedure of systematically analyzing a waveguide structure to just about step one where we wrote down the Maxwell's equations and beyond that we did not do anything so in this module what I want to do is to go through the remaining steps and show that these steps lead to the same kind of expression that we have already seen for the rectangular or for a parallel plate waveguide which is made out of perfect electric conductors the reason why we are doing this I have already told you is because just a simple you know a picture kind of an approach to analyzing wave guides does not give us fully information about that so having said that we have covered first step which is basically Maxwell's equation so now we are going to write down those Maxwell's equation in the coordinate system the first step was to choose the coordinate system which we chose as the rectangular coordinate system now we are going to write down Maxwell's equation in rectangular coordinates okay so this step should be familiar to you so you have to curl equations of Maxwell's that are important in this context so I am writing the Maxwell's equations in the region between the waveguides okay that is the region where we have the waves propagating or the modes propagating and outside the region it's not really of concern at this point but you can actually show that once you have a perfect electric conducting walls then those fields outside would not really be present okay at least we can not worry about whatever the radiation fields that might exist outside because we do assume that those radiation fields actually quickly go to zero okay anyway these are propagating structures they are not meant for radiating we have the first equation Carlos electric field being equal to minus J Omega mu not hedge and curl of H equals J Omega epsilon knot e now I am going to write down component twice the equations you know the best way to write down this set of equations is to express this curl operation in terms of a determinant which I have shown you in the previous module so pulling those equations from the X component will be something like this so please note that each of those components themselves can be a function of XY and Z okay so it's easy it can be a function of XY and z ey can be a function of XY and z okay so this would be equal to say minus J Omega mu not H X similarly you have then e X by Del Z minus del H by Del X this is the Y component of your feed ramp I think so this will be minus J Omega mu not H Y similarly you have del e Y by Del X Y and X component so del e Y by Del X minus del X by Del Y okay so this should be equal to minus J Omega mu not H set so I have this set of equations as I have obtained from the first curl equation now I have another curl equation which will allow me to write down the spatial derivatives of hedge in terms of the components x ey and E Z or rather equate the spatial derivatives of hedge with the electric field components on the right hand side okay so in the left hand side you had curl of e and the right hand side you have curl of H you simply have to replace Y by hedge to obtain the corresponding expressions and I'm going to do that one now okay so I am writing this in parallel because I want to show you we know how to combine these equations and express exp y HX and hy in terms of Ease add and headset and their derivatives okay so I have del H Z by Del Y minus del H Y by Del Z is equal to plus J Omega epsilon not px similarly I have Dell H X by Del Z minus del headset by del X equal to J Omega epsilon not V Y then you have del H Y by Del X minus del H X by Del y equals J Omega epsilon not okay so I have these two equations now what I'm going to do is so let me just go back to the equation we are still in the step two where we had these transverse components that is e ex ey HX and hy as and we need to express them as functions of either then headset okay so we are still in the second procedure itself or second step itself and for which we actually have this set of equations written already okay so in the same step right what we need to now do is to combine this equation now before doing that I already know that no matter what electric field component I am going to consider whether I am going to consider e X dy or is it okay and similarly HX hy or H that all these components in terms of said dependence would show a dependence in the form of e power minus J beta times and said okay that is because they are actually going to be modes which are propagating along z axis okay so these are the waves which would be propagating along the z axis so clearly if I take you know if I differentiate any of these components on the left with respect to said what I would have is you know we already know that the way in which we will have our modes will be of the form f of X Y e power minus J beta said of course this F will be different for different components so if you actually differentiate this mode with respect towards the mode function the total mode function with respect to said clearly what you are going to get is this f of X Y is independent of Z therefore it won't be differentiated it could be treated as constant so it will come out of the differential and then you have e power minus J beta Z when you differentiated you're going to pull - J beta and because it's an exponential function this will be e power minus J beta as it is times f of X Y so this was the original field and this is minus J beta so the operation del by Del Z it can actually be replaced by multiplying that corresponding component by minus J beta this is very very important ok now I'm going to put in all these equations that we have written this minus JB I'm going to substitute for example I have a minus del by Del Z in the first equation so I'm going to substitute with minus J beta so that the first part can be rewritten as plus J beta okay the reason is plus J beta because there is a minus sign out there similarly I will go ahead and make all the other substitutions I request you to also do these substitutions in your note in case you are looking at them ok and you know you would actually be able to follow what we have been writing over here okay so I have del by Del Z being replaced by minus J beta so I have a minus J beta at this point the third equation has no del by Del Z component so we will just leave it as it is whereas on this side on the right hand side that I have again I'm going to make the same substitution so here I have del by Del Z being substituted by plus J beta okay so I'm removing this under things so this is H y and then I have minus del by Del Z on this side so I'm going to replace that one by minus J beta now I look at these equations okay so look at these equations there is a minus sign throughout so you can remove all the minus signs and then convert everything into plus in this case you remove the plus I mean minus signs here and then put a minus sign on to the right-hand side okay this is just to kind of you know simplify the equations a little bit all right so wherever possible I will try to simplify the equations in this manner and these are the equations that we have simplified so I hope that you all have understood how we have come up to step two so we started off by choosing the coordinate system and second step was to write down Maxwell's equations individually the curly essentially what we are writing them down the divergence equations don't really help us in this particular case in the curl expressions you can see that wherever del by Del Z term appears because you have assumed a propagation along z axis so you can replace del by del read with minus J beta okay so anyway after doing all that this is the set of equations that we obtain now we get to the interesting part what we do is I look at this equation okay and I pair up with this second equation that I have what is the meaning of pairing up what I want to do is I look at this expression you know on the left hand side and express H X in terms of the left hand side expression that is I am going to invert this left hand and the right hand convention and when I do that I can write H X as 1 by or minus J Omega mu naught times del is that by Del Y plus J beta even correct so this is all I have done is to simply rearrange the first equation and now what I will do is I will after rearrange that equation I will put that one into the equation on the right hand side which is the second equation on to the right hand side equally I could have treated this Y component express ie Y in terms of the others and then substituted into the second expression I could have done that one and in fact I would like you to do that one because doing that will allow you to express Eve I in terms of the other components okay or rather HX in terms of the other components okay so now let's go forward I have written down HX in terms of del Z del Y and dy now I'm going to put that one into the second equation here to obtain an equation now I'm going to erase this part I hope you understood this one the reason I'm doing it on this slide itself is because I want you to keep looking at the equations okay so where I am starting and where is the equation at which I am substituting okay so put HX in this place so what you get for the second equation on the right hand side is that you have J beta there is a minus J Omega mu naught del e Z by Del y plus J beta ey which covers the first part plus del H Z by Del X equals minus J Omega epsilon not ey okay so if you want you can if you wish you can cancel this J from denominator and numerator and then you take this term and multiply individually to each term in the bracket here so you get del ez by Del Y or rather that would be minus beta divided by Omega mu knot then ease it by Del Y and then you have minus J beta divided by Omega mu knot sorry that would be beta square because there is already a beta here then there is one more beta let me remove the minus J here okay so that now you are alright okay so I've written this minus J beta square divided by Omega mu knot this would be e y plus del H Z by Del X equals minus J Omega epsilon not ey correct so now what you have you have one ey component here another rebuy component here you can move this one onto this side and then interchange the equations and then write down what would be e Y in terms of easy or other the derivative of easy and derivative of headset okay so when you rearrange and write it I will not write the intermediate step I will leave that as an exercise for you you can just look at this one and then see that you are going to get the same expressions that I am getting over here of course if I have made a mistake here you would also be able to figure that out but most importantly what you have is that you have kind of expressed everything in terms of you know easy and H said right or their derivatives now the reason why we are doing this is because remember we have to express all our components x ey HX and hy in terms of easier and headset so that when we solve a wave equation for easy or hedge there depending on which mode we are considering once we have solved that equation for either or headset then it is very easy to go back and find out the mode functions or how the modes I mean other components of the mode function behave because I have all the information in terms of their expressions of exe by HX and hy in terms of easier than H then again it is not necessary that you choose either and had said as the coordinates in which you I mean components in which you express other components okay you could have equally chose an ey and H Y but this is a very systematic procedure which in almost all cases is going to give you solutions which you are looking for in a very least possible suffering from your site if I were to say that okay so you have Eva expressed and please take this exercise very seriously once in lifetime you are you know you can do this one what you have to do is to express other components that is e x HX and hy in terms of derivatives of either dent and said I have shown you one way to do this one if you go back to this set of expressions that we have so you had this expression where we you know you we substituted for hey chicks and then we expressed the component ey if you substitute for e Y then you can obtain HX from these two equations and the remaining two equations also can be similarly manipulated to obtain H Y you know H Y and X components accordingly so please do that and once you also look at what we have on ey expression the denominator that we have is Omega square mu naught epsilon naught which you recognize from your earlier classes that this is equal to free space wave number that is square of free space wave number which I am going to write this this k0 square if the wave guide were to be filled with a certain material of epsilon R then instead of K 0 square you would have obtained k 0 square epsilon R and since epsilon R is the refractive index so you could have written this as K 0 square n Square where n is the refractive index of the material that fills up this perfect electric conducting parallel plane waveguide ok so I hope that is kind of understood from your side so you can simplify these expressions by removing Omega square mu naught epsilon naught and then writing this as K 0 squared ok now the step 3 that we are looking for is that we need a wave equation for even and very equation we have already derived earlier in one of the introductory classes where we have seen that curl óscar loss electric field e will be equal to or rather let me just write down the final expression which you are already familiar with so I have derived this equation which says del square e plus k0 square e equals 0 right this is the so-called Helmholtz equation that we wrote down and here this del square is the scalar laplacian which is individually given by del square by Del X square plus del square by Del Y square plus del square by Del Z square luckily in the coordinate system that we have chosen this equation can be separately applied to e^x separately applied to ey separate applied to easy so in all that cases what you get is this is the wave equation that you are going to get right and since we have expressed all field components in terms of easy this is the equation that I'm actually looking for I don't want to look at these equations which correspond to X and D Y wave equations I just want to know how easy wave equation would be looking like and this is exactly the equation that I am going to get and in this equation I can further make the substitution of del square by Del Z square by minus beta square okay why is this so go back to the start of this model and you will understand what why we have written this which works out this way okay so now quickly so I have del square by Del X square plus del square by Del Y square okay and since my parallel plane waveguide is assumed with a coordinate system such that this is X and this is the plate X equal to 0 this is the plane X equal to H and this direction is the z axis and along y direction we have assumed it to be infinity okay so you have the y axis direction in which the parallel plate wave weight actually extends to infinity so all of those del by Del Y terms can be made equal to 0 that is the modes are not changing as a function of Y okay so the only thing that is they are changing as a function of X and now there is no need to write a partial derivative you'll simply have a total derivative of the component that you are considering so you have d square ez by d x square plus d square ez by DX square plus basically k0 square minus beta square times ez equals 0 now this is the equation that we wrote for easy but now I mean you can write a similar equation for H side as well so replace either by H said in these equations and then you will be able to obtain an equation for head set now let us ask are we going to talk about transverse electric modes or are we going to talk about transverse magnetic mode right the essential difference between transverse electric and transverse magnetic mode we have seen is that transverse electric modes will have a component of electric field along y direction they do not have any component along I'm set direction whereas transverse magnetic waves will have component of the magnetic field along y direction and two in plane components e^x and easy in the X and z plane okay since we find it a little bit easier to talk about transverse electric modes okay just for the mathematical convenience we are going to go with transverse electric mode in literature sometimes you will also you will see that this is written as t is e where this Z is just to indicate that this is the mode which is propagating along z axis okay so I won't write this Z every time at in the superscript that will confuse us a little bit so I am going to consider transverse electric modes which means I can't have easy at all so in fact for transverse electric mode either it is zero so is e X okay the only thing that is nonzero is HX right Hz and ey so these are the nonzero components for transverse electric mode and now the wave equation I should not be writing in terms of easy but I should be writing in terms of H Z ok so this is the equation that I would like to now write down for headset and that simply comes from the Helmholtz equation so I have del d square headset by DX square plus k0 square minus or rather let me write it like this I will take this term to the right hand side I will have k0 square minus beta square times headset so I hope this equation written in this fashion should be on right okay so now I have this equation now what are the solutions of this equation well this is where we go to step four okay where our solutions are to be considered within the waveguide okay so I have X equal to 0 and X equal to H here as usual and said right so along Z I already know what is this e power minus J beta Z so I do not need to worry about the Z dependence but on the X dependence the solution now depends on the sign of the argument in the brackets right if k0 square minus beta square is positive then I get solutions right so since in that case I will get a solution which would be the characteristic equation of this one will be what let's first write down the characteristic equation remember the solutions are assumed of the form e power SX okay so I will have s 1 and 2 square being equal to minus k0 square minus beta square is the characteristic equation and therefore s12 which is the square root of the characteristic equation will be square root of minus k0 square minus beta square okay now I want k0 square minus beta square to be positive because I want a sinusoidal II oscillating solutions in the region between the parallel plate waveguide okay I don't want a decaying solution since between the waveguide so this quantity will be greater than zero accordingly you will have plus or minus J square root of K 0 square minus beta square or instead of writing this plus or minus J kind of a thing you can write down the solution for head set of X in terms of the linear combinations of these complex Exponential's which are nothing but sine and cosine functions so you will have a cos now I need to find out certain you know interesting letter for this one so let me just write down this as s itself so I have cosine of s X plus B sine of s X ok so I have this solution now I know what is e Y right ey as a function of X we have already I mean derived it this would be if you go back and substitute or if you go back and look at the equation you will have J Omega mu naught k0 square minus beta square in the denominator you have del headset by Del X okay so you can see the previous equation out there and of course what I have now is that k0 square minus beta square is actually a positive s quantity right so I I mean I can just write down as it is I don't want to worry about that one so now write down what is Del headset by Del X here so writing del headset by Del X by substituting from the top equation what you get is J Omega mu naught by K 0 square minus beta square is as it is okay so the solution that we have assumed is that this is going to be minus s square anyway so Dell headset by del X will give you differentiating headset will pull s out from the first equation then you have minus a sign a soft X plus B so s is the common thing so I am going to remove this one from the entire thing so I will have s here now B sine of s X derivative will give you cosine of s X and this is the solution for electric field that we are going to obtain okay and what is this k0 square minus beta square well k0 square minus beta square is basically so squaring up on both sides will give you s square equals minus k0 square minus beta square so I can instead of writing k0 square minus beta square in the denominator replace that one by minus s square okay and one s is in the numerator one s is in the denominator so I can remove that s from the numerator and in the denominator I will have a minus s here okay so this is the expression that I have now I have obtained dy since I have did not derive the expression for HX you please derive the expression for H X and then you can obtain the corresponding solution ok so this I will leave it as an exercise for you ok now we are not completely finished let us now go to step five where we are going to apply boundary conditions okay so what is the boundary condition I know that on this surface right my Y which is essentially tangential component must be equal to 0 at X equal to 0 or rather this is X equal to H and similarly Y being a tangential component will be again equal to 0 at X equal to 0 as well ok so when you substitute these equations essentially what you get is B equal to 0 okay so from the expressions that you substitute you will see that B will be equal to 0 and for the second boundary condition that at X equal to H when you apply you'll get a very interesting solution you will get minus J Omega mu naught by s is sine of H equal to zero okay now when can this sine function go to zero when this term right s times H is equal to some integer multiple of pi okay now s seems to be pretty bad way of representing the solution so let me actually rewrite this one in terms of what is called as Kappa F F stands for the wave guide or the film okay or the slab if you wish that would be so what we have the condition is Kappa F H equals nu pi where Kappa s is given by square root of K 0 square minus beta square okay so what we have seen is a very interesting expression out there that you will get electric field component eva of x being proportional to and put all the proportionality constant into some c and then you have sine of kappa f times x clearly this function will be equal to 0 at x equal to 0 as well as at x equal to h because we have written kappa f in terms of some integer nu pi right so this would be C sine nu PI by H times X so this is the expression and new of course can take on different values nu equal to 0 makes no sense because then the field itself will be equal to 0 so nu is not equal to 0 but the first term that you are going to get will be nu equals 1 2 3 and so on and these are all the different orders of mode ok so these are all the different modes and now let me just sketch for you how would the functions look like for let us say nu equal to 1 okay so nu equal to 1 will have sine of PI X by H which of course must start at 0 this is X equal to 0 plane this is the X equal to H plane it must start at 0 and then at H equal to I mean X equal to H it must go to 0 again so this is e Y which I am plotting so you can of course plot H 0 and H by separately for you I mean if you wish and this is the mode that would actually be propagating along the z direction with a beeper - J beta Z function okay when you take new equal to two you will actually see that again the function will start at zero but now it will actually go something like this okay so it will actually exhibit one maxima two Maxima so it will actually exhibit two Maxima which is equal to new in fact the first one actually exhibited one Maxima which was equal to the value of new so in fact new tells you the number of Maxima that your wave is going to go or the mode that is going to present between the two parallel plates okay so I have left out couple of in fact let me just find point out finally that this mode so if you were to draw a line here at the center you would find that this mode is an even function the green one so this is called as the even mode and this mode is called as the odd mode because it is basically in an odd function okay the even and odd modes actually depend on the symmetry of the line that you draw so because of that this becomes even and odd so this is not really a standard convention out there so you have seen that all the steps that we have written except for the last step where you will be writing x ey HX and H bar in terms of the other components okay so you can derive all the other component mode functions and this is how you analyze systematically any given wave guide structure so starting from choosing the coordinate system to writing the final solutions okay thank you very much [Music] [Music]