Rectangular Waveguide

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1 0/30/07 EE 4347 Applied Eletromagnetis Topi 5 Retangular Waveguide Leture 5 These notes ma ontain oprighted material obtained under air use rules. Distribution o these materials is stritl prohibited Slide Leture Outline What is a retangular waveguide? TEM Analsis TM Analsis TE Analsis Visualiation o Modes Conlusions Leture 5 Slide

2 0/30/07 What is a Retangular Waveguide? Leture 5 Slide 3 Geometr o Retangular Waveguide Standard onvention: a b Leture 5 Slide 4

3 0/30/07 Analsis o Retangular Waveguide Retangular waveguides are analed along eah axis like a parallel plate waveguide. Leture 5 Slide 5 Notes on the Retangular Waveguide Most lassi waveguide example Some o the irst waveguides used or mirowaves Not a transmission beause onl one ondutor Does not support a TEM mode Exhibits a low requen uto below whih no waves will propagate Leture 5 Slide 6 3

4 0/30/07 TE Analsis Leture 5 Slide 7 Reall TE Analsis The governing equation or TE analsis is H H 0, 0, kh 0, 0 x k k Ater a solution is obtained, the remaining ield omponents are alulated aording to H H j H x 0, x k 0, k 0, j H 0, E E E j H 0, x k j H x 0, k 0, 0 0, 0, Leture 5 Slide 8 4

5 0/30/07 General Form o the Solution From the geometr o the waveguide, we an immediate write the orm o the solution as,,, j H x H0, xe Viewing the retangular waveguide as the ombination o two parallel plate waveguides, we an appl separation o variables to write H 0, (x,) as the produt o two untions. H x X xy 0,, Leture 5 Slide 9 Separation o Variables ( o 3) We write our solution as the produt o two D untions and substitute that bak into our dierential equation. H H 0, 0, kh 0, 0 x H x X xy 0,, XY XY k XY 0 x X Y Y X k 0 XY x d X d Y 0 k We have ordinar derivatives beause X(x) and Y() X dx Y d have onl one independent variable eah. Leture 5 Slide 0 5

6 0/30/07 Separation o Variables ( o 3) First, we ous our attention on the x dependene in our dierential equation. d X d Y k 0 X dx Y d k x This deinition will let us write the dierential equation as a wave equation. d X 0 k x X dx Seond, we ous our attention on the dependene in our dierential equation. d X d Y k 0 X dx Y d k This deinition will let us write the dierential equation as a wave equation. dy 0 ky d Leture 5 Slide Separation o Variables (3 o 3) We would like to be able to add our two new dierential equations together to get the original dierential equation. d X dx dy d k X x ky X dx Y d d X dy k k x 0 0 d X d Y 0 k x k X dx Y d We get our original dierential equation bak i k k k x Leture 5 Slide 6

7 0/30/07 General Solution We now have two dierential equations to solve. d X dy 0 k x X 0 ky dx d These are essentiall the same dierential equation so their solution has the same general orm. d X k X x X x A k x B k x x x 0 os sin dx dy ky Y C k D k 0 os sin d Slab waveguide along x Slab waveguide along The overall solution is the produt o X(x) and Y(). H0, x, X xy Aos kx x Bsin kx x Cos k Dsin k Leture 5 Slide 3 Eletromagneti Boundar Conditions Boundar onditions required that the tangential omponent o the eletri ield be ero at the boundar with a peret ondutor. E0, x x, b 0 E E 0, 0, 0 0, a, 0 E0, x x,0 0 Leture 5 Slide 4 7

8 0/30/07 E 0,x and E 0, In order to appl the boundar onditions, we must derive the eletri ield omponents E 0,x and E 0, rom our expression or H 0,. E E x, 0, x k x, 0, k j H 0, j Aos kx x Bsin kx x Cosk Dsin k k j k AoskxxBsin kxxcsin kdosk k j H 0, x j Aos kx x Bsin kx x Cos k Dsin k k x j kx Asin kxxboskxxcoskdsin k k Leture 5 Slide 5 Appl Boundar Conditions ( o ) At the x = 0 boundar, we have 0, 0 E 0, j k kx Asin 0 Bos0 CoskDsin k j kxbcoskdsin k k B 0 At the x = a boundar, we have 0, 0 E a, j kx Asin kxacoskdsin k k A = 0 leads to a trivial solution. It must be the sin(k x a) term that enores the BC. 0 sin k a kam m 0,,,... x x Leture 5 Slide 6 8

9 0/30/07 Appl Boundar Conditions ( o ) At the = 0 boundar, we have 0, x 0 E x,0 j k A kxx B kxx C k os sin sin 0 D os0 j k AoskxxBsin kxxd k D 0 At the = b boundar, we have 0, x 0 E x, b j k AoskxxBsin kxxcsin kb k C = 0 leads to a trivial solution. It must be the sin(k b) term that enores the BC. 0 sin k b kbn n 0,,,... Leture 5 Slide 7 Revised Solution or H 0, We have determined that B = D = 0 so our expression or H 0, beomes H0, x, ACos kxx os k We write the produt AC as a single onstant A mn. H0, x, Amnos kxx os k Also, reall our onditions or k x and k. m ka x m kx n kb n k a b m x n H0, x, Amnos os a b Leture 5 Slide 8 9

10 0/30/07 Entire Solution ( o ) The inal expression or H 0, is m x n H0, x, Amnos os a b From this, the other ield omponents are jn m x n E x, Amnos sin kb a b 0, x jm mx n E x, Amnsin os ka a b 0, jm mx n H x, Amnsin os ka a b 0, x jn mx n H x, Amnos sin kb a b 0, Leture 5 Slide 9 E0, x, 0 Entire Solution ( o ) The overall eletri and magneti ields at an position are jn m x n Exx,, Amnos sin e kb a b jm m x n Ex,, Amnsin os e ka a b E x,, 0 jm mx n Hxx,, Amnsin os e ka a b jn mx n Hx,, Amnos sin e kb a b mx n Hx,, Amnos os e a b j j j j j Leture 5 Slide 0 0

11 0/30/07 Phase Constant, Reall the uto wave number k k k x Ater analing the boundar onditions, this expression an be written as m n k a b The phase onstant is thereore k k k k mn m n k k k a b Leture 5 Slide Cuto Frequen, Reall our expression or the phase onstant k k The phase onstant must be a real number or a guided mode. This requires k k An time k < k, our mode is uto and not supported b the waveguide. From this, we an derive the uto requen. k k k k mn, k m n a b Leture 5 Slide

12 0/30/07 Charateristi Impedane, Z TE The harateristi impedane is Z TE jn m x n j A os sin mn e Ex kb a b k H jn mx n j Amn os sin e kb a b Leture 5 Slide 3 Cuto or First Order TE Mode ( o ) The uto requen or the TE mn mode was ound to be, mn m n a b What about the TE 00 mode? TE mn0 00, a b The TE 00 mode does not exist. Leture 5 Slide 4

13 0/30/07 Cuto or First Order TE Mode ( o ) What about the TE 0 mode? TE m0, n,0 0 0 a b b What about the TE 0 mode? TE m, n0,0 0 0 a b a Sine a > b, we onlude that the irst order mode is TE 0 beause it has the lowest uto requen. CAUTION: We annot et sa that the TE 0 is the undamental mode beause we have not heked the uto requen o the TM modes. Leture 5 Slide 5 Single Mode Operation ( o ) We wish to determine over what range o requenies the waveguide supports onl a single TE mode. Low Frequen Cuto We just ound the lower requen uto. High Frequen Cuto,0 0 b TE 0 :,0 a b b a,0 a b Leture 5 Slide 6 a The high requen uto is the requen where the seond order TE mode us supported. This ould be the TE 0, TE or TE 0 mode. We must onsider all. TE : 0,0 0 a b b b TE :, a b b a TE will alwas have a higher uto requen than TE 0. The seond order mode depends on our hoie o a and b. a b 3

14 0/30/07 Single Mode Operation ( o ) Bandwidth Tpial retangular waveguides will have a > b, so a a a a a Frational Bandwidth Continuing the assumption that we have a > b, the rational bandwidth an be alulated rom and above as ollows a a FBW 66.7% 3 a Leture 5 Slide 7 a Example # TE Mode Analsis ( o 4) Suppose we have an air illed retangular waveguide with a = 3 m and b = m. What is the uto requen o the waveguide? m s 0 a a r r 0.03 m GH Over what range o requenies is the waveguide single mode? We see that a < b, so the seond order mode is TE m s 7.5 GH b b 0.0 m GH 7.5 GH r r Leture 5 Slide 8 4

15 0/30/07 Example # TE Mode Analsis ( o 4) What is the rational bandwidth o the waveguide? FBW 00% 00% 00% 40% Plot the phase onstant and eetive rerative index or the irstorder and seond order modes rom DC up to 5 GH. The phase onstant is alulated as: m n mn k a b 0 a 0 b The eetive rerative index is alulated as: 0 kn 0 e ne Leture 5 Slide 9 0 Example # TE Mode Analsis (3 o 4) Plot the phase onstant and eetive rerative index or the irstorder and seond order modes rom DC up to 5 GH. 0 a 0 b n 0 e, n 0 e, Leture 5 Slide 30 5

16 0/30/07 Example # TE Mode Analsis (4 o 4) Plot the veloit o the modes as a untion o requen. v 0 0 v ne, ne, Are our modes travelling aster than the speed o light? Leture 5 Slide 3 Summar o TE Analsis Field Solution jn m x n Exx,, Amnos sin e kb a b jm m x n Ex,, Amnsin os e ka a b E x,, 0 jm mx n Hxx,, Amnsin os e ka a b jn mx n H x,, Amn os sin e kb a b mx n Hx,, Amnos os e a b j j j j j, TE 00 mode does not exist TE 0 is the lowest order TE mode a b Phase Constant m n mn k a b Same equation as or TM Cuto Frequen, mn m n a b Same equation as or TM Charateristi Impedane Z TE,mn k mn Leture 5 Slide 3 6

17 0/30/07 TM Analsis Leture 5 Slide 33 Reall TM Analsis The governing equation or TM analsis is E E 0, 0, ke 0, 0 x k k Ater a solution is obtained, the remaining ield omponents are alulated aording to H 0, 0 H H j E 0, x k 0, k 0, j E x 0, E E j E x 0, x k 0, k 0, j E 0, Leture 5 Slide 34 7

18 0/30/07 General Form o the Solution From the geometr o the waveguide, we an immediate write the orm o the solution as,,, j E x E0, x e Viewing the retangular waveguide as the ombination o two parallel plate waveguides, we an appl separation o variables to write E 0, (x,) as the produt o two untions. E x X xy 0,, Leture 5 Slide 35 Separation o Variables ( o 3) We write our solution as the produt o two D untions and substitute that bak into our dierential equation. E E 0, 0, ke 0, 0 x E x X x Y 0,, XY XY k XY 0 x X Y Y X k 0 XY x d X d Y 0 k We have ordinar derivatives beause X(x) and Y() X dx Y d have onl one independent variable eah. Leture 5 Slide 36 8

19 0/30/07 Separation o Variables ( o 3) First, we ous our attention on the x dependene in our dierential equation. d X d Y k 0 X dx Y d k x This deinition will let us write the dierential equation as a wave equation. d X 0 k x X dx Seond, we ous our attention on the dependene in our dierential equation. d X d Y k 0 X dx Y d k This deinition will let us write the dierential equation as a wave equation. dy 0 ky d Leture 5 Slide 37 Separation o Variables (3 o 3) We would like to be able to add our two new dierential equations together to get the original dierential equation. d X dx dy d k X x ky X dx Y d d X dy k k x 0 0 d X d Y 0 k x k X dx Y d We get our original dierential equation bak i k k k x Leture 5 Slide 38 9

20 0/30/07 General Solution We now have two dierential equations to solve. d X dy 0 k x X 0 ky dx d These are essentiall the same dierential equation so their solution has the same general orm. d X k X x X x A k x B k x x x 0 os sin dx dy ky Y C k D k 0 os sin d Slab waveguide along x Slab waveguide along The overall solution is the produt o X(x) and Y(). E0, x, X x Y Aos kxx Bsin kxx Cos k Dsin k Leture 5 Slide 39 Eletromagneti Boundar Conditions Boundar onditions required that the tangential omponent o the eletri ield be ero at the boundar with a peret ondutor. E 0, is tangential to all interaes so we just use it. There is no need to alulate E 0,x or E 0,. E0, x, b 0 E E 0, 0, 0 0, a, 0 E0, x,0 0 Leture 5 Slide 40 0

21 0/30/07 Appl Boundar Conditions ( o ) At the x = 0 boundar, we have E0, Aos0 Bsin 0 Cosk Dsin k 0 0, ACoskDsin k A 0 At the x = a boundar, we have 0, 0 E a, x Bsin k a Cos k Dsin k B = 0 leads to a trivial solution. It must be the sin(k x a) term that enores the BC. 0 sin kxa kxa m m,,... m = 0 leads to a trivial solution. Leture 5 Slide 4 Appl Boundar Conditions ( o ) At the = 0 boundar, we have 0, x 0 E x,0 Aos kxx Bsin kxx Cos 0 Dsin 0 sin Aos kxx B kxx C C 0 At the = b boundar, we have 0, x 0 E x, b x x A os kx Bsin kx Dsin kb D = 0 leads to a trivial solution. It must be the sin(k b) term that enores the BC. 0 sin kb kb n n,,... n = 0 leads to a trivial solution. Leture 5 Slide 4

22 0/30/07 Revised Solution or H 0, We have determined that A = C = 0 so our expression or E 0, beomes E0, x, BDsin kxx sin k We write the produt BD as a single onstant B mn. E0, x, Bmnsin kxx sin k Also, reall our onditions or k x and k. m ka x m kx n kb n k a b mx n E0, x, Bmnsin sin a b Leture 5 Slide 43 Entire Solution ( o ) The inal expression or E 0, is mx n E0, x, Bmnsin sin a b From this, the other ield omponents are jm mx n E x, Bmnos sin ka a b 0, x jn mx n E x, Bmnsin os kb a b 0, jn m x n H x, Bmnsin os kb a b 0, x jm mx n H x, Bmnos sin ka a b 0, Leture 5 Slide 44 H0, x, 0

23 0/30/07 Entire Solution ( o ) The overall eletri and magneti ields at an position are jm mx n Exx,, Bmnos sin e ka a b jn mx n Ex,, Bmnsin os e kb a b mx n j Ex,, Bmnsin sin e a b jn m x n Hxx,, Bmnsin os e kb a b H H j j j jm m x n x,, B os sin mn e ka a b x,, 0 Leture 5 Slide 45 j Phase Constant, Reall the uto wave number k k k x Ater analing the boundar onditions, this expression an be written as m n k a b The phase onstant is thereore k k k k mn m n k k k a b Leture 5 Slide 46 3

24 0/30/07 Cuto Frequen, Reall our expression or the phase onstant k k The phase onstant must be a real number or a guided mode. This requires k k An time k < k, our mode is uto and not supported b the waveguide. From this, we an derive the uto requen. k k k k, mn k m n a b This is the same equation as or the TE modes. Leture 5 Slide 47 Charateristi Impedane, Z TM The harateristi impedane is Z TM jm mx n j B os sin mn e Ex ka a b H jm m x n j k Bmn os sin e ka a b Leture 5 Slide 48 4

25 0/30/07 Cuto or First Order TM Mode ( o ) The uto requen or the TM mn mode was ound to be, mn m n a b Note, we annot have n = 0 or m = 0 or the TM mode. So The TM 00 mode does not exist. The TM 0 mode does not exist. The TM 0 mode does not exist. The TM 0 mode does not exist. The TM 0 mode does not exist. The TM 03 mode does not exist. The TM 30 mode does not exist. et. Leture 5 Slide 49 Cuto or First Order TM Mode ( o ) What ombination o m and n minimies?, mn Sine a > b, the TM mode will have the lowest uto requen. m, n m n a b a b a b CAUTION: We annot et sa that the TM is the undamental mode beause we have not heked this against the TE modes. Leture 5 Slide 50 5

26 0/30/07 Example # TM Mode Analsis ( o 3) Suppose we have an air illed retangular waveguide with a = 3 m and b = m. What is the uto requen o the waveguide? a b 0 r 0 r a b 0 0 r r a b Reall that r r a b m s 9.0 GH m 0.0 m Leture 5 Slide 5 Example # TM Mode Analsis ( o 3) Plot the phase onstant and eetive rerative index or the irstorder mode rom DC up to 5 GH. The phase onstant is alulated as: The eetive rerative index is alulated as: m n k a b 0 a 0 kn 0 e ne 0 Leture 5 Slide 5 6

27 0/30/07 Example # TM Mode Analsis (3 o 3) Plot the veloit o the modes as a untion o requen. 0 v n e Are our modes travelling aster than the speed o light? Leture 5 Slide 53 Summar o TM Analsis Field Solution jm mx n Exx,, Bmnos sin e ka a b jn mx n Ex,, Bmnsin os e kb a b mx n j Ex,, Bm n sin sin e a b jn m x n Hxx,, Bmnsin os e kb a b H H j j j j m m x n x,, B os sin mn e ka a b x,, 0 j, m 0 and n 0, so TM 00, TM 0, TM 0, TM 0, TM 0, et. are not supported modes. TM is the lowest order TM mode a b Phase Constant m n mn k a b Same equation as or TE Cuto Frequen, mn Same equation as or TE m n a b Charateristi Impedane Z TM, mn k mn Leture 5 Slide 54 7

28 0/30/07 Visualiation o Modes Leture 5 Slide 55 Animation o TE 0 Notie one bright spot along x and ero along. (m =, n = 0) Leture 5 Slide 56 8

29 0/30/07 Animation o TE 0 Notie two bright spots along x and ero along. (m =, n = 0) Leture 5 Slide 57 Animation o TE 0 Notie ero bright spots along x and one along. (m = 0, n = ) Leture 5 Slide 58 9

30 0/30/07 Animation o TE Notie one bright spot along x and one along. (m =, n = ) Leture 5 Slide 59 Animation o TE Notie two bright spots along x and one along. (m =, n = ) Leture 5 Slide 60 30

31 0/30/07 Animation o TM Leture 5 Slide 6 Conlusions Leture 5 Slide 6 3

32 0/30/07 The Fundamental Mode The undamental mode is the mode whih has the lowest uto requen. This is either the TE 0 or the TM mode.,te,tm a We an see that the TE 0 mode will have the lowest uto requen. a b We onlude that the TE 0 mode is the undamental mode o the waveguide. This is also alled the dominant mode. When multiple modes are exited, usuall most o the power ends up in the undamental mode. Leture 5 Slide 63 Example #3 Mode Analsis ( o 3) Suppose we have an air illed retangular waveguide with a = 4 m and b = m. Over what range o requenies is this waveguide single mode? An eas wa to do this is to alulate a table using a desktop omputer. m n TE Cuto TM Cuto GH GH 3 0. GH GH GH 8.37 GH 8.37 GH 0.58 GH 0.58 GH GH 3.49 GH GH 6.73 GH GH 5.43 GH 5.43 GH 6.73 GH 6.73 GH GH 8.7 GH 4.6 GH.6 GH GH 3.76 GH.76 GH GH 3.66 GH GH 5.0 GH GH 6.98 GH GH GH 30.6 GH GH GH GH 3.96 GH GH GH Leture 5 Slide 64 3

33 0/30/07 Example #3 Mode Analsis ( o 3) Suppose we have an air illed retangular waveguide with a = 4 m and b = m. Over what range o requenies is this waveguide single mode? An eas wa to do this is to alulate a table using a desktop omputer. Then sort the table in order o inreasing uto requen. Mode Cuto TE GH TE GH TE GH TE 8.37 GH TM 8.37 GH TE 0.58 GH TM 0.58 GH TE30. GH TE GH TM GH TE GH TE GH TE 5.43 GH TM 5.43 GH TE GH Leture 5 Slide 65 Example #3 Mode Analsis (3 o 3) Suppose we have an air illed retangular waveguide with a = 4 m and b = m. Over what range o requenies is this waveguide single mode? An eas wa to do this is to alulate a table using a desktop omputer. Then sort the table in order o inreasing uto requen. We immediatel see that the TE 0 mode is the undamental mode with the lowest uto requen o 3.74 GH. Mode Cuto TE GH TE GH TE GH TE 8.37 GH TM 8.37 GH TE 0.58 GH TM 0.58 GH TE30. GH TE GH TM GH TE GH TE GH TE 5.43 GH TM 5.43 GH TE GH The seond order mode is taken rom the table to be either the TE 0 or TE 0 mode beause both o these have the same uto requen o 7.48 GH. The overall range o requenies or single mode operation is thereore 3.74 GH 7.48 GH Leture 5 Slide 66 33

34 0/30/07 Ke Points The retangular waveguide is not a transmission line beause it has less than two ondutors. When illed with a homogeneous dieletri, the retangular waveguide supports TE and TM modes, but not TEM. The uto requenies or TE and TM modes are the same. The TE 00 mode does not exist. For TM modes, m 0 and n 0. The TE 0 is the dominant mode beause the TM 0 mode does not exist. It is onl the phase veloit that exeeds the vauum speed o light. Leture 5 Slide 67 Summar rom Text Book Leture 5 Slide 68 34

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