Lectures 8 and 9 1 Rectangular waveguides


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1 1 Lectures 8 nd 9 1 Rectngulr wveguides y b x z Consider rectngulr wveguide with 0 < x <, 0 < y < b nd > b. There re two types of wves in hollow wveguide with only one conductor; Trnsverse electric wves (TEwves. E = (E x, E y, 0 nd H = (H x, H y, H z. Trnsverse mgnetic wves (TMwves. E = (E x, E y, E z nd H = (H x, H y, 0. They need to stisfy the Mxwells equtions nd the boundry conditions. The boundry conditions re tht the tngentil components of the electric field nd the norml derivtive of the tngentil components of the mgnetic field re zero t the boundries. 1.1 TEwves We now try to find the electromgnetic fields for TEwves, when E z is zero. The electromgnetic fields re obtined from H z. The eqution to be solved is 2 H z + k 2 H z = 0 H z x (0, y, z = H z x (, y, z = H z y (x,, z = H z y (x, b, z = 0 (1.1 where k = ω/c is the wve number. There re infinitely mny solutions to this eqution ( nπy H zmn (x, y, z = h mn cos cos e jkzz The m, n vlues cn tke the vlues m = 0, 1, 2... nd n = 0, 1, 2..., but (m, n (0, 0. The corresponding trnsverse electric nd mgnetic fields re obtined from
2 2 Mxwell s equtions. The sptil dependence of these components re ( nπy E x cos sin e jkzz ( ( nπy E y sin cos e jkzz ( ( nπy H x sin cos e jkzz ( ( nπy H y cos sin e jkzz Ech of these components stisfies the Helmholtz eqution nd the boundry conditions. The electromgnetic field corresponding to (m, n is clled TE mn mode. Thus there re infinitely mny TE mn modes. The k z is the z component of the wve vector. For given frequency it is given by 2 ( nπ 2 k z = k 2 (1.2 2 ( nπ 2 This mens tht for m nd n vlues such tht k 2 > 0, or, equivlently f > c (mπ 2 ( nπ 2 + then kz is rel nd the TE mn mode is 2π propgting. One cn lso introduce the wvelength in the z direction s λ z = 2π. k z If we move distnce λ z in the zdirection then e jkz(z+λz = e jkzz j2π = e jkzz nd we re bck to the sme phse s we strted ( from. mπ 2 ( nπ 2 For m nd n vlues such tht k 2 < 0, or, equivlently f < c (mπ 2 ( nπ 2 + then kz is imginry nd the TE mn mode is nonpropgting mode. The z dependence of the nonpropgting mode is e αz where 2π α = k zmn Cutoff frequency For TE mn mode the cutoff frequency is the frequency for which k z = 0. This mens tht the mode is in between its propgting nd nonpropgting stges. The cut off frequency for the TE mn mode is f cmn = c (mπ 2 ( nπ π The fundmentl mode TE 10 The fundmentl mode of wveguide is the mode tht hs the lowest cutoff frequency. For rectngulr wveguide it is the TE 10 mode tht is the fundmentl mode. It hs f c10 = c 2. The electric field of the fundmentl mode is ( πx E = E 0 sin e jkzz e y. It is lmost lwys the fundmentl mode tht is used
3 3 in the wveguide. It is then crucil to mke sure tht the frequency is low enough such tht only the fundmentl mode cn propgte. Otherwise there will be more thn one mode in the wveguide nd since the modes trvel with different speeds, s will be seen below, one cnnot control the phse of the wve. Exmple: Consider = 0.3 m nd b = 0.15 m. Then f c10 = 500 MHz, f c01 = f c20 = 1000 MHz nd f c11 = 1118 MHz. It mens tht for frequencies lower thn 500 MHz there re no wves tht cn propgte through the wveguide. In the intervl 500 MHz< f < 1 GHz only the TE 10 mode cn propgte. One hs to be in this frequency spn in order to trnsfer well defined signl. 1.2 TMwves The electromgnetic fields re obtined from 2 E z + k 2 E z = 0 E z (0, y, z = E z (, y, z = E z (x, 0, z = E z (x, b, z = 0 (1.3 where k = ω/c is the wve number. There re infinitely mny solutions top this eqution ( nπy E zmn (x, y, z = e mn sin sin e jkzz The m, n vlues cn tke the vlues m = 1, 2... nd n = 1, The corresponding trnsverse electric nd mgnetic fields re obtined from Mxwell s equtions. The sptil dependence of these components re the sme s for the TEwves ( nπy E x cos sin e jkzz ( ( nπy E y sin cos e jkzz ( ( nπy H x sin cos e jkzz ( ( nπy H y cos sin e jkzz The electromgnetic field corresponding to (m, n is clled TE mn mode. Thus there re infinitely mny TE mn modes. For given frequency k z for the TEmodes is the sme s for the TMmodes 2 ( nπ 2 k z = k Cutoff frequency For TM mn mode the cutoff frequencies re the sme s for the TE mn modes, i.e., f cmn = c (mπ 2 ( nπ 2. + The only difference is tht one cnnot hve m = 0 2π or n = 0. Exmple: When = 0.3 m nd b = 0.15 m, the lowest cutoff frequency for TMmodes is f c11 = 1118 MHz.
4 4 2 Derivtion of the modes In order to solve Eqs. (1.1 nd (1.3 we use the method of seprtion of vribles. Consider Eq. (1.1. We ssume tht ll solutions to the eqution cn be written s product H z (x, y, z = F (xg(yk(z. When we insert this into the eqution we get GK 2 F x + F K 2 G 2 y + F K 2 G 2 z + 2 k2 F GK = 0 We divide by F GK 1 2 F F x G 2 G y + 2 k2 = 1 2 K K z 2 The left hnd side is independent of z nd the right hnd side is independent of x, y. The only wy tht this cn hppen is if both sides re constnt. We cll this constnt k 2 z nd get the eqution for K The solution is 2 K z 2 + k2 zk = 0 K(z = p e jkzz + n e jkzz These solutions correspond to wves trveling in the positive z direction (e jkzz nd wves in the negtive z direction (e jkzz. The eqution for F G cn now be written s 1 2 F F x + 2 k2 kz 2 = 1 2 G G y 2 Also here both sides hve to be constnt nd we cll this constnt k 2 y. Then the eqution for G is the following eigenvlues problem The solution to the eqution is 2 G(y + k y yg(y 2 2 G (0 = G (b = 0 G(y = α 1 cos(k y y + α 2 sin(k y y The solution hs to stisfy the boundry conditions G (0 = G (b = 0, where G (y = α 1 k y sin(k y y + α 2 k y cos(k y y. The condition G (0 = 0 gives α 2 = 0 nd the condition G (b = 0 gives α 1 k y sin(k y b = 0. We must hve α 1 0, otherwise H z = 0. This mens tht k y b = nπ nd k y = nπ. Thus the eigenvlue problem hs ( b nπy ( nπ 2. the eigenfunctions G(y = α 1 cos nd eigenvlues ky 2 = b b The eqution for F is 2 F (x x 2 + (k 2 (nπ/b 2 k 2 zf (x = 0
5 5 By introducing k 2 x = k 2 (nπ/b 2 k 2 z then 2 F (x + k x xf 2 (x 2 F (0 = F ( = 0 This gives the eigenfunctions F (x = β 1 cos nd eigenvlues ky 2 = We hve seen tht the solutions to Eq. (1.1 re ( nπy H zmn (x, y, z = h mn cos cos e jkzz 2. where k z = k 2 (mπ/ 2 (nπ/b 2. There is lso corresponding field trveling in the negtive z direction ( nπy H zmn (x, y, z = h mn cos cos e jkzz It turns out tht when both m nd n re zero, (m, n = (0, 0, then the field contrdicts the Mxwell equtions nd for this reson (m, n (0, 0. The solutions hve the importnt property tht they re orthogonl over the cross section of the wveguide, i.e., 1 b 2 h2 mnδ mm δ nn if m = 0 or n = 0 H zmn (x, y, zhzm n dx dy = h2 mnδ mm δ nn if m > 0, n > 0 where denotes complex conjugte nd δ mm is one if m = m nd zero if m m. In generl the totl field in wveguide is superposition of wveguide modes trveling in both directions. 2.1 Phse nd group speeds The phse speed is the speed tht the pttern of the wve moves. The group speed is the speed tht the energy is trveling with. For plne wve tht trvels in vcuum the phse nd group speeds re the sme nd equls the speed of light c = 1/ ε 0 µ 0 = m/s. Also for trnsmission line the two speeds re the sme nd equls the speed of light in the mteril between the conductors. In wveguide it is different. Then the phse speed turns out to be lrger thn the speed of light nd the group speed is smller thn the speed of light. One cn show tht the phse speed is given by v f = ω k z nd tht the group speed is given by the derivtive v g = dω ( dkz = dk z dω 1
6 Exmple The TE mn nd T M mn modes in rectngulr wveguide hve the z component of the wve vector given by Eq. (1.2 The phse speed is given by k z = v f = ω k z = k 2 2 ( nπ ω 2 ( nπ k 2 2 (2.1 Since k = ω/c it mens tht v f < c, when the mode is propgting. When we decrese the frequency such tht f f cmn then v f. When f then v f c. The group speed is given by v g = ( 1 dkz = k z dω k c This mens tht v g < c nd s f f cmn then v g 0. As f then v g c. 2
7 7 3 Lecture: Comsol for wveguides Determintion of losses in wveguide y b x z Consider rectngulr wveguide m 2 mde out of copper nd filled with ir. The fundmentl mode propgtes in the wveguide. Use Comsol to determine the rel nd imginry prt of k z s function of frequency in the intervl 501 MHz1 GHz. Choose 2d, Electromgnetic wves nd Mode nlysis Choose Prmeters under Globl definitions nd define prmeter clled frequency. Write freq in expression since tht is the predefined nme of the frequency. Build the geometry. Choose ir nd copper s mterils nd let the internl domin be ir nd the boundry be copper. Choose Impednce boundry condition by right click on Electromgnetic wves. Choose Prmetric sweep under bf Study 1. Choose frequency s prmeter by clicking on the plus sign nd then click on the digrm symbol to the right of the sve button. Let the frequency strt t 510 MHz, stop t 1001 MHz with step 20 MHz. Choose Step 1 : Mode nlysis nd put 1 in desired number of modes nd 0.7 s effective mode index in Serch for modes round: This mens tht it will find one mode nd it will strt to serch t 0.7 for the effective mode index (n = c/v f.
8 8 Run. Mke sure tht it is the fundmentl mode for ll frequencies. Choose Globl plot under 1D Plot Group under Results. mke sure tht you hve the correct solution in Dt set. Pick ttenution constnt on the yxis nd frequency on the yxis. Under prmeter it should sy expression Propgtion of wves in system of wveguides Consider lossless rectngulr irfilled wveguide with cross section m 2 nd with shpe given by the figure. Determine the propgtion of the TE 10 mode for the frequency 704 MHz. b Determine the propgtion of the TE 10 mode for the frequency 1500 MHz. c Determine the propgtion of the TE 20 mode for the frequency 1500 MHz. d Which other modes couple to the TE 10 nd TE 20 modes in the wveguide. Choose 2d, Electromgnetic wves nd Frequency domin Build the geometry. Remember to do Union under Boolen ; opertions in Geometry. Choose port under Electromgnetic wves nd dd the surfce tht is the entrnce port. Choose rectngulr under Type of port. Click on in Wve excittion. Let the mode be TE nd number 1 (gives TE 10. Choose port gin nd dd the surfce for the exit port. Wve excittion s to be off here but the mode is TE nd number 1. Refine the mesh to finest. Click Frequency domin under Study nd dd the frequency.
9 9 Run. The defult plot is the the norm of the electric field. Using tht it is hrd to see the wves. Choose the zcomponent of the electric field insted nd tke the rel prt of this. It should sy rel(ez under Expression. Mke movie by using nimtion under Export. Use dynmic dt extension under Sweep type.
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