The Rectangular Waveguide - EMPossible

1 8/15/20211 Electromagnetics:Electromagnetic Field TheoryThe Rectangular WaveguideLecture Outline What is a Rectangular Waveguide ? TM Analysis TE Analysis Visualization of Modes ConclusionsSlide 28/15/20212 Slide 3 What is a Rectangular Waveguide ?Geometry of Rectangular WaveguideSlide 4 Standard size convention:ab bazyx , 8/15/20213 Analysis of Rectangular WaveguideSlide 5 Rectangular waveguides are analyzed a bit like each axis were its own parallel plate on the Rectangular Waveguide Most classic Waveguide example Some of the first waveguides used for microwaves Not a transmission line because it has only one conductor Does not support a TEM mode Exhibits a low frequency cutoff below which no waves will propagateSlide 68/15/20214 Slide 7TE AnalysisRecall TE AnalysisSlide 8 The governing equation for TE analysis isAfter a solution is obtained, the remaining field components are calculated according to220,0,2c0,220zzzHHkHxy 22 2ckk 0,0,2c0,0,2czxzyHjHkxHjHky 0,0,2c0,0,2c0,0zxzyzHjEkyHjEkxE 8/15/20215 General Form of the SolutionSlide 9 From the geometry of the Waveguide .

2 The general form of the solution can be immediately written asViewing the Rectangular Waveguide as the combination of two parallel plate waveguides, apply separation of variables to write H0,z(x,y)as the product of two functions. 0,,,,jzzzHxyzHxye 0,,zHxyXxYy 222c22110dXdYkXdxYdy Separation of Variables (1 of 3)Slide 10 The solution is written as the product of two 1D functions, X(x)and Y(y). Substitute this solution back into the differential ,0,2c0,220zzzHHkHxy 0,,zHxyXxYy The derivatives become ordinary because X(x)and Y(y)have only one independent variable 222c220 XYYXkXYxy To be compact, drop the and out of the operation and out of the of Variables (2 of 3)Slide 11 First, attention is focused on the x dependence in the differential equation. 222c22110dXdYkXdxYdy 2xk This definition of lets the differential equation be written as a wave Second, attention is focused on the y dependence in the differential equation.

3 222c22110dXdYkXdxYdy 2yk This definition of lets the differential equation be written as a wave Separation of Variables (3 of 3)Slide 12If all of this is correct, then it should be possible to add the two new differential equations together to get the original differential equation. 2222221010xydXkXdxdYkYdy +222222110xydXdYkkXdxYdy 22222200xydXkXdxdYkYdy The original differential equation is obtained if222cxykkk 222c22110dXdYkXdxYdy Original differential equation8/15/20217 General SolutionSlide 13 There are now two differential equations to 2220ydYkYdy These are essentially the same differential equation so their solution has the same general form. 2220 cossinxxxdXkXXxAkxBkxdx 2220 cossinyyydYkYYyCkyDkydy PP Waveguide along xPP Waveguide along yThe overall solution is the product of X(x)and Y(y).

4 0,,cossincossinzxxyyHxyXxYyAkxBkxCkyDky Electromagnetic Boundary ConditionsSlide 14 Boundary conditions require that the tangential component of the electric field be zero at the boundary with a perfect , 0,,00xEx 0,,0xExb 0,0,0yEy 0,,0yEay 8/15/20218E0,xand E0,ySlide 15In order to apply the boundary conditions, the electric field components E0,xand E0,ymust be derived from the expression for H0,z. 0,0,2c2c2c,cossincossincossinsincoszxxxy yyxxyyHjExykyjAkxBkxCkyDkykyjkAkxBkxCkyD kyk 0,0,2c2c2c,cossincossinsincoscossinzyxxy yxxxyyHjExykxjAkxBkxCkyDkykxjkAkxBkxCkyD kyk Apply Boundary Conditions (1 of 2)Slide 16At the x= 0 boundary, 0,2c00,sin 0yxEyjkAk 2ccos 0cossincossinyyxyyBCkyDkyjkBCkyDkyk 0B At the x= aboundary, 0,2c0,sincossinyxxyyEayjkAkaCkyDkyk A= 0 leads to a trivial solution.

5 It must be the sin(kxa) term that enforces the BC. 0sin 0,1, 2, .. xxkaamkm The specific values mcan be will be considered Boundary Conditions (2 of 2)Slide 17At the y= 0 boundary, 0,2c0,0cossinsin 0xyxxExjkAkxBkxCk 2ccos 0cossinyxxDjkAkxBkxDk 0D At the y= bboundary, 0,2c0,cossinsinxyxxyExbjkAkxBkxCkbk C= 0 leads to a trivial solution. It must be the sin(kyb) term that enforces the BC. 0sin 0,1, 2, .. yykbbnkn The specific values ncan be will be considered Solution for H0,zSlide 18 0,,coscoszxyHxyACkxky It was determined that B= D= 0 so the expression for H0,zbecomesThe product ACis written as a single constant Amn. 0,,coscoszmnxyHxyAkxky Also, recall the conditions for kxand ky. xxmkamka yynkbnkb 0,,coscoszmnmxnyHxyAab 8/15/202110 Entire Solution (1 of 2)Slide 19 The final expression for H0,zisFrom this, the other field components are 0,,coscoszmnmxnyHxyAab 0,2c0,2c0,2c0,2c,cossin,sincos,sincos,co ssinxmnymnmnxmnmnymnjnmxnyExyAkbabjmmxny ExyAkaabjmmxnyHxyAkaabjnmxnyHxyAkbab 0,,0zExy Entire Solution (2 of 2)

6 Slide 20 The overall electric and magnetic fields at any position are 2c2c2c2c,,cossin,,sincos,,0,,sincos,,cos mnmnmnjzxmnjzymnzjzmnxmnmnymnjnmxnyExyzA ekbabjmmxnyExyzAekaabExyzjmmxnyHxyzAekaa bjnmxHxyzAkba sin,,coscosmnmnjzjzzmnnyebmxnyHxyzAeab 8/15/202111 Phase Constant, Slide 21 Recall the cutoff wave number222cxykkk After analyzing the boundary conditions, this expression can be written as222cmnkab The phase constant is therefore22 2c222c22222cmnkkkkmnkkkab Cutoff Frequency, fcSlide 22 Recall the expression for the phase constantcccc2kkkfk 22cmnkk The phase constant must be a real number for a guided mode. This requiresckk Any time k< kc, the mode is cutoff and not supported by the Waveguide . From this, the cutoff frequency fcis derived to be22cc,122mnkmnfab 8/15/202112 Characteristic Impedance, ZTES lide 23 The characteristic impedance of the TE mode is2cTE2ccossincossinjzmnxjzmnymnmnmnjnmx nyAeEkbabkZjnmxnyHAekbab Cutoff for First Order TE Mode (1 of 2)Slide 24 The cutoff frequency for the TEmnmode was found to be22c,12mnmnfab What about the TE00mode?

7 00TE0mn 22c,0010002fab The TE00mode does not exist because it is impossible for a mode in this Waveguide to have a cutoff frequency of 0 for First Order TE Mode (2 of 2)Slide 25 What about the TE01mode?01TE0,1mn 22c,01101122fabb What about the TE10mode?10TE1,0mn 22c,1011 0 122faba CAUTION: It cannot yet be said that the TE10is the fundamental mode because the cutoff frequency of the tm modes has not yet been a> b, it is concluded that the first order TE mode is TE10because it has the lowest cutoff Mode Operation (1 of 2)Slide 26 Over what range of frequencies does a Rectangular Waveguide supports only a single TE mode?c1c2fff Low Frequency CutoffThe lower frequency cutoff was just High Frequency CutoffThe high frequency cutoff is the frequency where the second order TE mode is supported.

8 This could be the TE01, TE11or TE20mode. All must be considered.,012,20 2 y 2 (t pical)cccfabffab 2201c,0122211c,112220c,201011TE : 2211TE : 12212 0 12TE : 22fabbbfababbfabab TE11will always have a higher cutoff frequency than second order mode depends on choice of aand Mode Operation (2 of 2)Slide 27 Typical Rectangular waveguides will have a> 2b, soBandwidthc112fa 21cfa c2c111 122fffaaa Fractional Bandwidth Continuing the assumption that a> 2b, the fractional bandwidth can be calculated from fc1and fc2above as followsExample #1 TE Mode Analysis (1 of 4)Slide 28 Suppose there exists an air filled Rectangular Waveguide with a= 3 cm and b= 2 is the cutoff frequency of the Waveguide ? 011299792458 m GHz222 Over what range of frequencies is the Waveguide single mode?

9 Observing that a< 2b, so the second order mode is TE01. 021299792458 m GHz222 GHzf 8/15/202115 Example #1 TE Mode Analysis (2 of 4)Slide 29 What is the fractional bandwidth of the Waveguide ? Plot the phase constant and effective refractive index for the first order and second order modes from DC up to 15 : 2TE : mnfcamnkabfcb The phase constant is calculated as:00effeff0 22cknnffc The effective refractive index is calculated as:Example #1 TE Mode Analysis (3 of 4)Slide 30 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 8/15/202116 Example #1 TE Mode Analysis (4 of 4)Slide 31 Plot the velocity of the modes as a function of ,1cvn 02eff,2cvn Are the modes travelling faster than the speed of light?

10 Summary of TE AnalysisSlide 32 Field Solution 2c2c2c2c,,cossin,,sincos,,sincos,,co0s,, mnmnmnjzxmnjzymnjzmnxmnmnymnzjnmxnyExyzA ekbabjmmxnyExyzAekaabjmmxnyHxyzAekaabjnm xHxyzAkbExzay osin,,cscosmmnnjzzmnjzmxnyHxyzAeyanbeb 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 TM8/15/202117 Slide 33TM AnalysisRecall TM AnalysisSlide 34 The governing equation for TM analysis isAfter a solution is obtained, the remaining field components are calculated according to220,0,2c0,220zzzEEkExy 22 2ckk 0,0,2c0,0,2czxzyEjHkyEjHkx 0,0,2c0,0,2czxzyEjEkxEjEky 0,0zH 8/15/202118 General Form of the SolutionSlide 35 From the geometry of the Waveguide , the general form of the solution can be immediately written asViewing the Rectangular Waveguide as the combination of two parallel plate waveguides, apply separation of variables to write E0,z(x,y)as the product of two functions.