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# Class Note for ECE 3317 with Professor Jackson at UH

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This 39 page Class Notes was uploaded by an elite notetaker on Friday February 6, 2015. The Class Notes belongs to a course at University of Houston taught by a professor in Fall. Since its upload, it has received 23 views.

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Date Created: 02/06/15
ECE 3317 Prof David R Jackson Chapter 5 Rectangular Waveguide Ty b 8u a cross section We assume that the boundary is a perfect electric conductor PEC We analyze the problem to solve for EZ or HZ all other fields come from these TMZ EZ only TEZ HZ only HZO EZ i O VZEZ kZEZ O Helmholtz equatior I EZ 0 Oh bOUhdaI y PEG walls Guided wave assumption E Z x 322 E20 xyeijkzz 2 2 2 5 E2 I E z I E2 I kZEZ 0 8x 32 52 I aZEZ 62E 2 I 22 kaZj kZEZ o 63x 6y 62E 62E 22 k2 kEZ 0 6x 6y Define 2 2 We then have 6 E2 6 E2 k2E 0 Note that kc isan I 6x2 632 C Z unknown at this pomt Dividing by the eXpjkzz term we have We solve the above equation by using the method of separation of variables We assume E20 xy XX YO where E20 xy Xquot YO Hence XquotYXYquot kfXY I I Xquot Yquot 2 DIVIde beY kc X Y Hence X k02 Y X Y Both sides of the ThlS has the form F x 2 C31 equation must be a constant Xquot Yquot 2 k3 constant X Y Xquot Set kf constant X General solution Xx A Sinkxx B COSkxx Boundary conditions X0 0 1 Xa 0 2 1 B 0 Xx Asinkxx 2 q sinkxa 0 The second boundary condition is sinkxa 0 This gives us the following result kxazm m12 k a Hence Xx Asinm x 61 Now we turn our attention to the Yy function Now we turn our attention to the Y function Xquotk2Yquotk2 X C Y x Yquot Hence k k Y Define Yquot Then we have k2 Y y General solution Yy Csinkyy D COSkyy Y y C sinky y D cosky y Boundary conditions Y0 0 3 Ya 0 4 3 D0 4 sinkyb0 Equation 4 gives us the following result kyb 1172quot9 n 19 2 mt kyb Hence mry Y y C sin Therefore we have E20 xay Xx Yy ACsinm xjsin a New notation E20 x9 y Amn Sinm xjsin Cl The TMZ field inside the waveguide thus has the following form 2 2 2 Recall that ky kc kx Hence k k k Therefore the solution for kc is given by Next recall that k k2 k Hence Summary of TM Solution for mn Mode HZ O Cl E2 y Z Am Sin mm sin Ly eJk m gtz nl2 2 We start with kW 2 k2kman Z where Note The number kc is the value of k for which the wavenumber k2 is zero Hence 27239f lug 5m 2 whichgivesus 2f a The cutoff frequency fc of the TMWI mode is then th 2 E b Tan 1 fc gt 27 ye This may be written as fTMm Cd I C 27 Z We now start with E 2 H272 2 2 aHZaHZk2 k2H 0 6x2 6y2 Z Z Using the separation of variables method again we have H20 w X x Yv where X x A sinkxx B coskxx Y y C sinky y D cosky y and k3 k kj k3 k2 k Boundary conditions Ty Exx0 0 Ey0y O b Exxb 0 Eyay O The result is m7rx mry H A 20 xy m cos a jcos b This can be shown by using the following equations an a cross section jau 6H jk 6E Z E z z 2 q 07 x kl k3 8y kl k3 ax 6y E Jaw 5H2 sz 5E2 6H2 0 y 2 2 2 2 ax k k2 6x k k2 6y Summary of TEZ Solution for mn Mode E2 O Cl Hz 6 Z Z COS m x COS 6 1kgm nz 2 2 kmn k2 m7 mz Same formula for cutoff z frequency as the TEZ case mQL2H no12 m n 0 0 Note the 00 TEZ mode is not valid since it violates the magnetic Gauss law HZxyzAOOeij VHxyz 0 m7z x 7272 7 kmn A n 8111 8111 y e J Z Z a b myz x 7272 quot2321 A n COS COS y e sz Z a b 2 17252 2 7272 2 k 7 7 I Same formula for both cases a b 2 2 272 a b m O 1 2 m 7 1 2 Z n712 I IE2 74770912 m n i O O Gshsr sl formula for the wsvshumbsr 2 2 llIZ or TEZ mods kz k2 7 k with kc 72 2 j I 2 2 j a b Note The 77171 hotstior I is suppressed her s k2 k1ikc IC2 Recall kc wasL s k1 7 L7 f2 Hence k2 le f f2 f 39 Wheh we are below cutoff it is COhvehieht to write k2 k2ikc2ij kfik2 Hehce kz 7j 16627162 7jkc17kkc2 7jkC17fLf 2 Recal that Hence we have k 1 fCf2 fgtJi akc 1 ff2 flt l CWT DZC T TI General behavior of the wavenumber k1fCf2 fgt fez d akc 1 f222 fltJZ 2 Lowest TMZ mode TM11 Lowest TEZ mode TE1O 1 I b an a cross section The dominant mode is the TE1O mode TEZ TMZ m012 quot1213quotquot n012 n12 man 070 Formulas for the dominant TE1O mode Hz 90 y Z A10 cos e jkzz a At the cutoff frequency k1 fcf2 fgt12 0kcx1ffc2 f lt2 What is the mode with the next highest cutoff frequency 6 m7 2 117239 2 27 a b The next highest is the TE20 mode f20 2f10 useful operating region lt gt i l TE10 TE20 f6 Fields of the dominant TE10 mode Hz 36 y Z A10 0036 1792 or Find the other fields from these equations a2 kZ kf y k2 kf ax y kZ k ax kZ k ay H jwa 0E1 jkz 6H2 x CZ k ay CZ k ax H jag at jkz 632 y kZ k ax kZ k y From these we find the other fields to be 1xa yaz JA Elo Sin 6 1722 a A k Hx9 yDZ E10 sln szz W a jam 7 Where E10 2 I Stahdar d X bahd waveguide air filled a 5 0900 ir ICheS 2288 cm 0400 ir ICheS I 0 I8 cm Fir Id the operatihg frequehcy regior t i 2a Use f1 0 X bar Id from 80 to I2 GHZ Standard Xband waveguide airfilled a 0900 inches 2286 cm I 0400 inches 1016 cm Find the phase constant of the TE1O mode at 900 GHz Find the attenuation in dBm at 500 GHz Recall 10 656 GHz Zk1fCf27 fgtfc kauoao27rfc27ri0 2 At 900 GHZ k18862 radm 0 ch ffey fltfc kc 7ra13743radm At 500 GHZ a 8891 nepersm Ey x y z 2 A10 sin 6 Therefore dBm 2010g10 11 A very rapid attenuation Note We could have also used dBm 868589 nepersm 86858905 The guide wavelength is is the distance Z that it takes for the wave to repeat itself This assumes that we are above the cutoff frequency From this we have L 27239 27239 g k1 ff2 1ff 2 Hence we have the result Hence a a 1 P k17f f2 a ng17J f2 ug 17J f2 Cd We then have VP 2 xl J f For a hollow waveguide Cd c VP 2 ll J f Hence v gt C This does not violate relativity V The group velocity is the velocity at which a pulse travels on a structure The group velocity is given by The derivation of this is omitted V l A pulse consists of a quotgroupquot of frequencies according to the Fourier transform I ll 0 V t 1 waveguiding system If the phase velocity is a function of frequency the pulse will be distorting as it travels down the system v i gp A I V t at waveguiding system I A pulse will get distorted in a rectangular waveguide I o calculate the group velocity for a waveguide we use Hence we have d5 1 2 2 12 eye 8 a 8 e kc Zeye r 2 We then have the fOIIOWIr Ig final result vg Cd 7 For a hollow waveguide For a lossless transmission line or a plane wave TEMZ waves kw We then have v22 0 1 c 1 3 WE J78 d vg E JE cd da For a lossless transmission line there is no distortion Consider the electric field of the dominant TE10 mode 1xyz ElO sin e jkzz a ejkxx e jkxx k E10 6122 ka a 2 where ltgt Collecting terms we have E E E x Z A 10 ekaxe kzZ A 10 e kaxe szZ y 42 4 2 This form is the sum of two plane waves l Clz kz ikx E22162 At the cutoff frequency the angle 6 is 90 At high frequencies well above cutoff the angle 6 approaches zero Picture of two plane waves 1 crests of waves 121 1 441 g The two plane waves add to give an electric field that is zero on the side walls of the waveguide x 0 and x a

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