Lecture 5c -- Rectangular waveguide - EMPossible

1 10/30/20171 Lecture 5cSlide 1EE 4347 Applied ElectromagneticsTopic 5cRectangular waveguide These notes may contain copyrighted material obtained under fair use rules. Distribution of these materials is strictly prohibited Lecture OutlineLecture 5cSlide 2 What is a Rectangular waveguide ? TEM Analysis TM Analysis TE Analysis Visualization of Modes Conclusions10/30/20172 Lecture 5cSlide 3 What is a Rectangular waveguide ? Lecture 5cSlide 4 Geometry of Rectangular WaveguideStandard convention:ab 10/30/20173 Lecture 5cSlide 5 Analysis of Rectangular WaveguideRectangular waveguides are analyzed along each axis like a parallel plate on the Rectangular waveguide Most classic waveguide example Some of the first waveguides used for microwaves Not a transmission because only one conductor Does not support a TEM mode Exhibits a low frequency cutoff below which no waves will propagateLecture 5cSlide 610/30/20174 Lecture 5cSlide 7TE AnalysisLecture 5cSlide 8 Recall TE AnalysisThe governing equation for TE analysis isAfter a solution is obtained, the remaining field components are calculated according to220,0,20,220zzczHHkHxy 222ckk 0,0,20,0,2zxczycHjHkxHjHky 0,0,20,0,20,0zxczyczHjEkyHjEkxE 10/30/20175 Lecture 5cSlide 9 General Form of the SolutionFrom the geometry of the waveguide .

2 We can immediate write the form of the solution asViewing the Rectangular waveguide as the combination of two parallel plate waveguides, we can apply separation of variables to write H0,z(x,y)as the product of two functions. 0,,,,jzzzHxyzHxye 0,,zHxyXxYy Lecture 5cSlide 10 Separation of Variables (1 of 3)We write our solution as the product of two 1D functions and substitute that back into our differential ,0,20,220zzczHHkHxy 0,,zHxyXxYy 22222222222222200110cccXYXYkXYxyXYYXkXYx ydXdYkXdxYdy We have ordinary derivatives because X(x)and Y(y)have only one independent variable 5cSlide 11 Separation of Variables (2 of 3)First, we focus our attention on the x dependence in our differential equation. 22222110cdXdYkXdxYdy 2xk This definition will let us write the differential equation as a wave Second, we focus our attention on the y dependence in our differential equation.

3 22222110cdXdYkXdxYdy 2yk This definition will let us write the differential equation as a wave Lecture 5cSlide 12 Separation of Variables (3 of 3)We would like to be able to add our two new differential equations together to get the original differential equation. 2222221010xydXkXdxdYkYdy +222222110xydXdYkkXdxYdy 22222200xydXkXdxdYkYdy We get our original differential equation back if222cxykkk 10/30/20177 Lecture 5cSlide 13 General SolutionWe now have two differential equations to 2220ydYkYdy These are essentially the same differential equation so their solution has the same general form. 2220 cossinxxxdXkXXxAkxBkxdx 2220 cossinyyydYkYYyCkyDkydy Slab waveguide along xSlab waveguide along yThe overall solution is the product of X(x)and Y(y). 0,,cossincossinzxxyyHxyXxYyAkxBkxCkyDky Lecture 5cSlide 14 Electromagnetic Boundary Conditions 0,,0 0xEx 0,,0xExb 0,0, 0yEy 0,,0yEay Boundary conditions required that the tangential component of the electric field be zero at the boundary with a perfect 5cSlide 15E0,xand E0,yIn order to apply the boundary conditions, we must derive the electric field components E0,xand E0,yfrom our expression for H0,z.

4 0,0,222,cossincossincossinsincoszxcxxyyc yxxyycHjExykyjAkxBkxCkyDkykyjkAkxBkxCkyD kyk 0,0,222,cossincossinsincoscossinzycxxyyc xxxyycHjExykxjAkxBkxCkyDkykxjkAkxBkxCkyD kyk Lecture 5cSlide 16 Apply Boundary Conditions (1 of 2)At the x= 0 boundary, we have 0,200,sin 0yxcEyjkAk 2cos 0 cossincossinyyxyycBCkyDkyjkBCkyDkyk 0B At the x= aboundary, we have 0,20,sincossinyxxyycEayjkAkaCkyDkyk A= 0 leads to a trivial solution. It must be the sin(kxa) term that enforces the BC. 0sin 0,1,2,.. xxkaamkm 10/30/20179 Lecture 5cSlide 17 Apply Boundary Conditions (2 of 2)At the y= 0 boundary, we have 0,20,0cossinsin 0xyxxcExjkAkxBkxCk 2cos 0cossinyxxcDjkAkxBkxDk 0D At the y= bboundary, we have 0,20,cossinsinxyxxycExbjkAkxBkxCkbk C= 0 leads to a trivial solution. It must be the sin(kyb) term that enforces the BC.

5 0sin 0,1,2,.. yykbbnkn Lecture 5cSlide 18 Revised Solution for H0,z 0,,coscoszxyHxyACkxky We have determined that B= D= 0 so our expression for H0,zbecomesWe write the product ACas a single constant Amn. 0,,coscoszmnxyHxyAkxky Also, recall our conditions for kxand ky. xxmkamka yynkbnkb 0,,coscoszmnmxnyHxyAab 10/30/201710 Lecture 5cSlide 19 Entire Solution (1 of 2)The final expression for H0,zisFrom this, the other field components are 0,,coscoszmnmxnyHxyAab 0,20,20,20,2,cossin,sincos,sincos,cossin xmncymncxmncymncjnmxnyExyAkbabjmmxnyExyA kaabjmmxnyHxyAkaabjnmxnyHxyAkbab 0,,0zExy Lecture 5cSlide 20 Entire Solution (2 of 2)The overall electric and magnetic fields at any position are 2222,,cossin,,sincos,, 0,,sincos,,cossinjzxmncjzymnczjzxmncymnc jnmxnyExyzAekbabjmmxnyExyzAekaabExyzjmmx nyHxyzAekaabjnmxnyHxyzAkba ,,coscosjzjzzmnebmxnyHxyzAeab 10/30/201711 Lecture 5cSlide 21 Phase Constant, Recall the cutoff wave number222cxykkk After analyzing the boundary conditions, this expression can be written as222cmnkab The phase constant is therefore2222222222 2ccmnckkkkmnkkkab Lecture 5cSlide 22 Cutoff Frequency, fcRecall our expression for the phase constant2cccckkkfk 22ckk The phase constant must be a real number for a guided mode.

6 This requiresckk Any time k< kc, our mode is cutoff and not supported by the waveguide . From this, we can derive the cutoff frequency ,122ccmnkmnfab 10/30/201712 Lecture 5cSlide 23 Characteristic Impedance, ZTEThe characteristic impedance is2TE2cossincossinjzmnxcjzymncjnmxnyAeEkbabkZjnmxnyHAekbab Lecture 5cSlide 24 Cutoff for First Order TE Mode (1 of 2)The cutoff frequency for the TEmnmode was found to be22c,12mnmnfab What about the TE00mode?00TE0mn 22c,0010002fab The TE00mode does not 5cSlide 25 Cutoff for First Order TE Mode (2 of 2)What about the TE01mode?01TE0, 1mn 22c,01101 122fabb What about the TE10mode?10TE1, 0mn 22c,1011 0 122faba CAUTION: We cannot yet say that the TE10is the fundamental mode because we have not checked the cutoff frequency of the TM a> b, we conclude that the first order mode is TE10because it has the lowest cutoff 5cSlide 26 Single Mode Operation (1 of 2)We wish to determine over what range of frequencies the waveguide supports only a single TE Low Frequency CutoffWe just found the lower frequency High Frequency CutoffThe high frequency cutoff is the frequency where the second order TE mode us supported.

7 This could be the TE01, TE11or TE20mode. We must consider all.,012,20 2 2cccfabffab 2201c,0122211c,112220c,20101 1TE : 2211TE : 12212 0 12TE : 22fabbbfababbfabab TE11will always have a higher cutoff frequency than second order mode depends on our choice of aand 5cSlide 27 Single Mode Operation (2 of 2)Typical Rectangular waveguides will have a> 2b, soBandwidthc112fa 21cfa c2c111 122fffaaa Fractional Bandwidth Continuing the assumption that we have a> 2b, the fractional bandwidth can be calculated from fc1and fc2above as followsLecture 5cSlide 28 Example #1 TE Mode Analysis (1 of 4)Suppose we have an air filled Rectangular waveguide with a= 3 cm and b= 2 is the cutoff frequency of the waveguide ? 011299792458 m GHz222 m Over what range of frequencies is the waveguide single mode?

8 We see that a< 2b, so the second order mode is TE01. 021299792458 m GHz222 m GHzf 10/30/201715 Lecture 5cSlide 29 Example #1 TE Mode Analysis (2 of 4)What is the fractional bandwidth of the waveguide ? 100%200%200%40% Plot the phase constant and effective refractive index for the first order and second order modes from DC up to 15 2mnfcamnkabfcb The phase constant is calculated as:00effeff0 22cknnffc The effective refractive index is calculated as: Lecture 5cSlide 30 Example #1 TE Mode Analysis (3 of 4)Plot the phase constant and effective refractive index for the first order and second order modes from DC up to 15 0eff,112cnf 22202fcb 0eff,222cnf 10/30/201716 Lecture 5cSlide 31 Example #1 TE Mode Analysis (4 of 4)Plot the velocity of the modes as a function of ,1cvn 02eff,2cvn Are our modes travelling faster than the speed of light?

9 Lecture 5cSlide 32 Summary of TE AnalysisField Solution 2222,,cossin,,sincos,,sincos,,cossi,, 0njzxmncjzymncjzxmnmnczcyjnmxnyExyzAekba bjmmxnyExyzAekaabjmmxnyHxyzAekaabjnmxnyH xyzAkbExyza ,,coscosjzjzzmnmxnyHxyzAeabeb Phase Constant222mnmnkab Cutoff Frequency22c,12mnmnfab Characteristic ImpedanceTE,mnmnkZ TE00mode does not exist TE10is the lowest order TE mode, ab Same equation as for TMSame equation as for TM10/30/201717 Lecture 5cSlide 33TM AnalysisLecture 5cSlide 34 Recall TM AnalysisThe governing equation for TM analysis isAfter a solution is obtained, the remaining field components are calculated according to220,0,20,220zzczEEkExy 222ckk 0,0,20,0,2zxczycEjHkyEjHkx 0,0,20,0,2zxczycEjEkxEjEky 0,0zH 10/30/201718 Lecture 5cSlide 35 General Form of the SolutionFrom the geometry of the waveguide , we can immediate write the form of the solution asViewing the Rectangular waveguide as the combination of two parallel plate waveguides, we can apply separation of variables to write E0,z(x,y)as the product of two functions.

10 0,,,,jzzzExyzExye 0,,zExyXxYy Lecture 5cSlide 36 Separation of Variables (1 of 3)We write our solution as the product of two 1D functions and substitute that back into our differential ,0,20,220zzczEEkExy 0,,zExyXxYy 22222222222222200110cccXYXYkXYxyXYYXkXYx ydXdYkXdxYdy We have ordinary derivatives because X(x)and Y(y)have only one independent variable 5cSlide 37 Separation of Variables (2 of 3)First, we focus our attention on the x dependence in our differential equation. 22222110cdXdYkXdxYdy 2xk This definition will let us write the differential equation as a wave Second, we focus our attention on the y dependence in our differential equation. 22222110cdXdYkXdxYdy 2yk This definition will let us write the differential equation as a wave Lecture 5cSlide 38 Separation of Variables (3 of 3)We would like to be able to add our two new differential equations together to get the original differential equation.