CHAPTER 14 14.1. A parallel-plate waveguide is kow to have a utoff wavelegth for the m 1 TE ad TM modes of λ 1 0.4 m. The guide is operated at wavelegth λ 1 mm. How may modes propagate? The utoff wavelegth for mode m is λ m 2d/m, where is the refrative idex of the guide iterior. For the first mode, we are give λ 1 2d 1 Now, for mode m to propagate, we require 0.4 m d 0.4 2 0.2 m λ 2d m 0.4 m m 0.4 λ 0.4 0.1 4 So, aoutig for 2 modes TE ad TM) for eah value of m, ad the sigle TEM mode, we will have a total of 9 modes. 14.2. A parallel-plate guide is to be ostruted for operatio i the TEM mode oly over the frequey rage 0 <f <3 GHz. The dieletri betwee plates is to be teflo ɛ R 2.1). Determie the maximum allowable plate separatio, d: We require that f<f 1, whih, usig 7), beomes f< 2d d max 2f max 3 10 8 2 2.1 3 10 9 ) 3.45 m 14.3. A lossless parallel-plate waveguide is kow to propagate the m 2 TE ad TM modes at frequeies as low as 10GHz. If the plate separatio is 1 m, determie the dieletri ostat of the medium betwee plates: Use f 2 d 3 1010 1) 10 10 3orɛ R 9 14.4. A d 1 m parallel-plate guide is made with glass 1.45) betwee plates. If the operatig frequey is 32 GHz, whih modes will propagate? For a propagatig mode, we require f > f m Usig 7) ad the give values, we write f> m 2d m<2fd 232 109 )1.45).01) 3 10 8 3.09 The maximum allowed m i this ase is thus 3, ad the propagatig modes will be TM 1,TE 1,TM 2, TE 2,TM 3, ad TE 3. 14.5. For the guide of Problem 14.4, ad at the 32 GHz frequey, determie the differee betwee the group delays of the highest order mode TE or TM) ad the TEM mode. Assume a propagatio distae of 10 m: From Problem 14.4, we foud m max 3. The group veloity of a TE or TM mode for m 3is v g3 f3 f where f 3 33 1010 ) 21.45)1) 3.1 1010 31 GHz 249
14.5. otiued) Thus v g3 3 1010 1.45 ) 31 2 5.13 10 9 m/s 32 For the TEM mode assumig o material dispersio) v g,t EM / 3 10 10 /1.45 2.07 10 10 m/s. The group delay differee is ow 1 t g z v g3 1 v g,t EM ) ) 1 10 5.13 10 9 1 2.07 10 10 1.5 s 14.6. The utoff frequey of the m 1 TE ad TM modes i a parallel-plate guide is kow to be f 1 7.5 GHz. The guide is used at wavelegth λ 1.5 m. Fid the group veloity of the m 2 TE ad TM modes. First we kow that f 2 2f 1 15 GHz. The f /λ 3 10 8 /.015 20 GHz. Now, usig 23), v g2 was ot speified i the problem. f2 f ) 15 2 2 10 8 / m/s 20 14.7. A parallel-plate guide is partially filled with two lossless dieletris Fig. 14.23) where ɛ R1 4.0, ɛ R2 2.1, ad d 1 m. At a ertai frequey, it is foud that the TM 1 mode propagates through the guide without sufferig ay refletive loss at the dieletri iterfae. a) Fid this frequey: The ray agle is suh that the wave is iidet o the iterfae at Brewster s agle. I this ase θ B ta 1 2.1 4.0 35.9 The ray agle is thus θ 90 35.9 54.1. The utoff frequey for the m 1 mode is f 1 2d ɛ R1 3 1010 21)2) 7.5 GHz The frequey is thus f f 1 / os θ 7.5/ os54.1 ) 12.8 GHz. b) Is the guide operatig at a sigle TM mode at the frequey foud i part a? The utoff frequey for the ext higher mode, TM 2 is f 2 2f 1 15 GHz. The 12.8 GHz operatig frequey is below this, so TM 2 will ot propagate. So the aswer is yes. 14.8. I the guide of Problem 14.7, it is foud that m 1 modes propagatig from left to right totally reflet at the iterfae, so that o power is trasmitted ito the regio of dieletri ostat ɛ R2. a) Determie the rage of frequeies over whih this will our: For total refletio, the ray agle measured from the ormal to the iterfae must be greater tha or equal to the ritial agle, θ, where si θ ɛ R2 /ɛ R1 )1/2. The miimum mode ray agle is the θ 1 mi 90 θ. Now, usig 5), we write ) ) π 90 θ os 1 os 1 π k mi d 2πf mi d 4 ) os 1 4df mi 250
14.8a. otiued) Now os90 θ ) si θ ɛ R2 ɛ R1 4df mi Therefore f mi /2 2.1d) 3 10 8 )/2 2.1.01)) 10.35 GHz. The frequey rage is thus f>10.35 GHz. b) Does your part a aswer i ay way relate to the utoff frequey for m 1 modes i ay regio? We ote that f mi /2 2.1d) f 1 i guide 2. To summarize, as frequey is lowered, the ray agle i guide 1 dereases, whih leads to the iidet agle at the iterfae ireasig to evetually reah ad surpass the ritial agle. At the ritial agle, the refrated agle i guide 2 is 90, whih orrespods to a zero degree ray agle i that guide. This defies the utoff oditio i guide 2. So it would make sese that f mi f 1 guide 2). 14.9. A retagular waveguide has dimesios a 6madb 4 m. a) Over what rage of frequeies will the guide operate sigle mode? The utoff frequey for mode mp is, usig Eq. 54): f,m m p + 2 a b where is the refrative idex of the guide iterior. We require that the frequey lie betwee the utoff frequeies of the TE 10 ad TE 01 modes. These will be: f 10 f 01 2a 3 108 2.06) 2b 3 108 2.04) 2.5 109 3.75 109 Thus, the rage of frequeies over whih sigle mode operatio will our is 2.5 GHz <f <3.75 GHz b) Over what frequey rage will the guide support both TE 10 ad TE 01 modes ad o others? We ote first that f must be greater tha f 01 to support both modes, but must be less tha the utoff frequey for the ext higher order mode. This will be f 11, give by The allowed frequey rage is the f 11 ) 1 2 ) 1 2 + 30 2.06.04 2 4.5 109 3.75 GHz <f <4.5 GHz 251
14.10. Two retagular waveguides are joied ed-to-ed. The guides have idetial dimesios, where a 2b. Oe guide is air-filled; the other is filled with a lossless dieletri haraterized by ɛ R. a) Determie the maximum allowable value of ɛ R suh that sigle mode operatio a be simultaeously esured i both guides at some frequey: Sie a 2b, the utoff frequey for ay mode i either guide is writte usig 54): f mp m p + 4b 2b where 1 i guide 1 ad ɛ R i guide 2. We see that, with a 2b, the ext modes havig the ext higher utoff frequey) above TE 10 with be TE 20 ad TE 01. We also see that i geeral, f mp guide 2) <f mp guide 1). To assure sigle mode operatio i both guides, the operatig frequey must be above utoff for TE 10 i both guides, ad below utoff for the ext mode i both guides. The allowed frequey rage is therefore f 10 guide 1) <f <f 20 guide 2). This leads to /2a)<f </a ɛ R ). For this rage to be viable, it is required that ɛ R < 4. b) Write a expressio for the frequey rage over whih sigle mode operatio will our i both guides; your aswer should be i terms of ɛ R, guide dimesios as eeded, ad other kow ostats: This was already foud i part a: 2a <f < ɛ R a where ɛ R < 4. 14.11. A air-filled retagular waveguide is to be ostruted for sigle-mode operatio at 15 GHz. Speify the guide dimesios, a ad b, suh that the desig frequey is 10/while beig 10% lower tha the utoff frequey for the ext higher-order mode: For a air-filled guide, we have f,mp m 2a p + 2b For TE 10 we have f 10 /2a, while for the ext mode TE 01 ), f 01 /2b. Our requiremets state that f 1.1f 10 0.9f 01. So f 10 15/1.1 13.6 GHz ad f 01 15/0.9 16.7 GHz. The guide dimesios will be a 3 1010 2f 10 213.6 10 9 1.1m ad b 3 1010 f 01 216.7 10 9 0.90 m ) 14.12. Usig the relatio P av 1/2)Re{E s Hs }, ad Eqs. 44) through 46), show that the average power desity i the TE 10 mode i a retagular waveguide is give by P av β 10 2ωµ E2 0 si2 κ 10 x)a z W/m 2 ote that the si term is erroeously to the first power i the origial problem statemet). Ispetig 44) through 46), we see that 46) iludes a fator of j, ad so would lead to a imagiary part of the 252
total power whe the ross produt with E y is take. Therefore, the real power i this ase is foud through the ross produt of 44) with the omplex ojugate of 45), or P av 1 2 Re { E ys Hxs } β 10 2ωµ E2 0 si2 κ 10 x)a z W/m 2 14.13. Itegrate the result of Problem 14.12 over the guide ross-setio 0 <x<a,0<y<b, to show that the power i Watts trasmitted dow the guide is give as P β 10ab 4ωµ E2 0 ab 4η E2 0 si θ 10 W where η µ/ɛ ote misprit i problem statemet), ad θ 10 is the wave agle assoiated with the TE 10 mode. Iterpret. First, the itegratio: b a β 10 P 0 0 2ωµ E2 0 si2 κ 10 x)a z a z dx dy β 10ab 4ωµ E2 0 Next, from 20), we have β 10 ω µɛ si θ 10, whih, o substitutio, leads to P ab µ 4η E2 0 si θ 10 W with η ɛ The si θ 10 depedee demostrates the priiple of group veloity as eergy veloity or power). This was osidered i the disussio leadig to Eq. 23). 14.14. Show that the group dispersio parameter, d 2 β/dω 2, for give mode i a parallel-plate or retagular waveguide is give by d 2 β dω 2 ω ) [ 2 ω ) ] 2 3/2 ω ω ω where ω is the radia utoff frequey for the mode i questio ote that the first derivative form was already foud, resultig i Eq. 23)). First, takig the reiproal of 23), we fid dβ dω [ ω ) ] 2 1/2 ω Takig the derivative of this equatio with respet to ω leads to d 2 β dω 2 1 )[ ω ) ] 2 3/2 2ω 2 ) 2 ω ω 3 ω ) [ 2 ω ) ] 2 3/2 ω ω ω 14.15. Cosider a trasform-limited pulse of eter frequey f 10 GHz ad of full-width 2T 1.0 s. The pulse propagates i a lossless sigle mode retagular guide whih is air-filled ad i whih the 10 GHz operatig frequey is 1.1 times the utoff frequey of the TE 10 mode. Usig the result of Problem 14.14, determie the legth of the guide over whih the pulse broades to twie its iitial width: The broadeed pulse will have width give by T T 2 + τ, where τ β 2 L/T for a trasform limited pulse assumed gaussia). β 2 is the Problem 14.14 result evaluated at the operatig frequey, or ) [ β 2 d2 β dω 2 1 1 2 ) ] 1 2 3/2 ω10 GHz 2π 10 10 )3 10 8 ) 1.1 1.1 6.1 10 19 s 2 /m 0.61 s 2 /m Now τ 0.61L/0.5 1.2L s. For the pulse width to double, we have T 1 s, ad.05 + 1.2L 1 L 0.72 m 72 m 253
14.15. otiued) What simple step a be take to redue the amout of pulse broadeig i this guide, while maitaiig the same iitial pulse width? It a be see that β 2 a be redued by ireasig the operatig frequey relative to the utoff frequey; i.e., operate as far above utoff as possible, without allowig the ext higher-order modes to propagate. 14.16. A symmetri dieletri slab waveguide has a slab thikess d 10 µm, with 1 1.48 ad 2 1.45. If the operatig wavelegth is λ 1.3 µm, what modes will propagate? We use the oditio expressed through 77): k 0 d 2 2 2 m 1)π. Sie k 0 2π/λ, the oditio beomes 2d 2 1 λ 210) 2 2 m 1) 1.48) 1.3 2 1.45 4.56 m 1 Therefore, m max 5, ad we have TE ad TM modes for whih m 1, 2, 3, 4, 5 propagatig te total). 14.17. A symmetri slab waveguide is kow to support oly a sigle pair of TE ad TM modes at wavelegth λ 1.55 µm. If the slab thikess is 5 µm, what is the maximum value of 1 if 2 3.3 assume 3.30)? Usig 78) we have 2πd 2 1 λ 2 2 <π 1 < λ 2d + 2 2 1.55 25) + 3.30)2 3.32 14.18. 1 1.50, 2 1.45, ad d 10 µm i a symmetri slab waveguide ote that the idex values were reversed i the origial problem statemet). a) What is the phase veloity of the m 1 TE or TM mode at utoff? At utoff, the mode propagates i the slab at the ritial agle, whih meas that the phase veloity will be equal to that of a plae wave i the upper or lower media of idex 2. Phase veloity will therefore be v p utoff) / 2 3 10 8 /1.45 2.07 10 8 m/s. b) What is the phase veloity of the m 2 TE or TM modes at utoff? The reasoig of part a applies to all modes, so the aswer is the same, or 2.07 10 8 m/s. 14.19. A asymmetri slab waveguide is show i Fig. 14.24. I this ase, the regios above ad below the slab have uequal refrative idies, where 1 > 3 > 2 ote error i problem statemet). a) Write, i terms of the appropriate idies, a expressio for the miimum possible wave agle, θ 1, that a guided mode may have: The wave agle must be equal to or greater tha the ritial agle of total refletio at both iterfaes. The miimum wave agle is thus determied by the greater of the two ritial agles. Sie 3 > 2,wefidθ mi θ,13 si 1 3 / 1 ). b) Write a expressio for the maximum phase veloity a guided mode may have i this struture, usig give or kow parameters: We have v p,max ω/β mi, where β mi 1 k 0 si θ 1,mi 1 k 0 3 / 1 3 k 0. Thus v p,max ω/ 3 k 0 ) / 3. 14.20. A step idex optial fiber is kow to be sigle mode at wavelegths λ>1.2 µm. Aother fiber is to be fabriated from the same materials, but is to be sigle mode at wavelegths λ>0.63 µm. By what peretage must the ore radius of the ew fiber differ from the old oe, ad should it be larger or smaller? We use the utoff oditio, give by 80): λ> 2πa 2.405 2 2 2 254
14.20. otiued) With λ redued, the ore radius, a, must also be redued by the same fratio. Therefore, the peretage redutio required i the ore radius will be % 1.2.63 1.2 100 47.5% 14.21. A short dipole arryig urret I 0 os ωt i the a z diretio is loated at the origi i free spae. a) If β 1 rad/m, r 2m,θ 45, φ 0, ad t 0, give a uit vetor i retagular ompoets that shows the istataeous diretio of E: I spherial oordiates, the ompoets of E are give by 82) ad 83): E r I 0dη 2π E θ I 0dη 4π os θe j2πr/λ si θe j2πr/λ 1 r 2 + j 2π λr + 1 r 2 + λ ) j2πr 3 λ ) j2πr 3 Sie we wat a uit vetor at t 0, we eed oly the relative amplitudes of the two ompoets, but we eed the absolute phases. Sie θ 45, si θ os θ 1/ 2. Also, with β 1 2π/λ, it follows that λ 2π m. The above two equatios a be simplified by these substitutios, while droppig all amplitude terms that are ommo to both. Obtai 1 A r r 2 + 1 ) jr 3 e jr A θ 1 2 j 1 r + 1 r 2 + 1 ) jr 3 e jr Now with r 2 m, we obtai 1 A r 4 j 1 ) e j2 1 8 4 1.12)e j26.6 e j2 A θ j 1 4 + 1 8 j 1 ) e j2 1 16 4 0.90)ej56.3 e j2 The total vetor is ow A A r a r + A θ a θ. We a ormalize the vetor by first fidig the magitude: A A A 4 1 1.12 + 0.90 0.359 Dividig the field vetor by this magitude ad overtig 2 rad to 114.6, we write the ormalized vetor as A Ns 0.780e j141.2 a r + 0.627e 58.3 a θ I real istataeous form, this beomes A N t) Re A Ns e jωt) 0.780 osωt 141.2 )a r + 0.627 osωt 58.3 )a θ We evaluate this at t 0 to fid A N 0) 0.780 os141.2 )a r + 0.627 os58.3 )a θ 0.608a r + 0.330a θ 255
14.21a. otiued) Dividig by the magitude, 0.608 + 0.330 0.692, we obtai the uit vetor at t 0: a N 0) 0.879a r + 0.477a θ. We ext overt this to artesia ompoets: a Nx a N 0) a x 0.879 si θ os φ + 0.477 os θ os φ 1 2 0.879 + 0.477) 0.284 a Ny a N 0) a y 0.879 si θ si φ + 0.477 os θ si φ 0 sie φ 0 a Nz a N 0) a z 0.879 os θ 0.477 si θ 0.879 0.477) 0.959 2 The fial result is the a N 0) 0.284a x 0.959a z b) What fratio of the total average power is radiated i the belt, 80 <θ<100? We use the far-zoe phasor fields, 84) ad 85), ad first fid the average power desity: P avg 1 2 Re[E θsh φs ] I 2 0 d2 η 8λ 2 r 2 si2 θ W/m 2 We itegrate this over the give belt, a at radius r: P belt 2π 100 Evaluatig the itegral, we fid 0 P belt πi2 0 d2 η 4λ 2 I0 2d2 η 80 8λ 2 r 2 si2 θr 2 si θdθdφ πi2 0 d2 η 4λ 2 100 80 si 3 θdθ [ 13 ) ] os θ si 2 100 θ + 2 0.344) πi2 0 d2 η 80 4λ 2 The total power is foud by performig the same itegral over θ, where 0 <θ<180. Doig this, it is foud that P tot 1.333) πi2 0 d2 η 4λ 2 The fratio of the total power i the belt is the f 0.344/1.333 0.258. 14.22. Prepare a urve, r vs. θ i polar oordiates, showig the lous i the φ 0 plae where: a) the radiatio field E θs is oe-half of its value at r 10 4 m, θ π/2: Assumig the far field approximatio, we use 84) to set up the equatio: E θs I 0dη 2λr si θ 1 2 I 0dη 2 10 4 λ r 2 104 si θ b) the average radiated power desity, P r,av, is oe-half of its value at r 10 4 m, θ π/2. To fid the average power, we use 84) ad 85) i P r,av 1 2 Re{E θsh φs }1 2 I0 2d2 η 4λ 2 r 2 si2 θ 1 2 1 I0 2d2 η 2 4λ 2 10 8 ) r 2 10 4 si θ 256
14.22. otiued) The polar plots for field r 2 10 4 si θ) ad power r 2 10 4 si θ) are show below. Both are irles. 14.23. Two short ateas at the origi i free spae arry idetial urrets of 5 os ωt A, oe i the a z diretio, oe i the a y diretio. Let λ 2π m ad d 0.1 m. Fid E s at the distat poit: a) x 0,y 1000,z 0): This poit lies alog the axial diretio of the a y atea, so its otributio to the field will be zero. This leaves the a z atea, ad sie θ 90, oly the E θs ompoet will be preset as 82) ad 83) show). Sie we are i the far zoe, 84) applies. We use θ 90, d 0.1, λ 2π, η η 0 120π, ad r 1000 to write: E s E θs a θ j I 0dη 2λr si θe j2πr/λ a θ j 50.1)120π) e j1000 a θ 4π1000) j1.5 10 2 )e j1000 a θ j1.5 10 2 )e j1000 a z V/m b) 0, 0, 1000): Alog the z axis, oly the a y atea will otribute to the field. Sie the distae is the same, we a apply the part a result, modified suh the the field diretio is i a y : E s j1.5 10 2 )e j1000 a y V/m ) 1000, 0, 0): Here, both ateas will otribute. Applyig the results of parts a ad b, we fid E s j1.5 10 2 )a y + a z ). d) Fid E at 1000, 0, 0) at t 0: This is foud through Et) Re E s e jωt) 1.5 10 2 ) siωt 1000)a y + a z ) Evaluatig at t 0, we fid E0) 1.5 10 2 )[ si1000)]a y + a z ) 1.24 10 2 )a y + a z ) V/m. e) Fid E at 1000, 0, 0) at t 0: Takig the magitude of the part d result, we fid E 1.75 10 2 V/m. 257
14.24. A short urret elemet has d 0.03λ. Calulate the radiatio resistae for eah of the followig urret distributios: a) uiform: I this ase, 86) applies diretly ad we fid R rad 80π 2 d λ 80π 2.03 0.711 b) liear, Iz) I 0 0.5d z )/0.5d: Here, the average urret is 0.5I 0, ad so the average power drops by a fator of 0.25. The radiatio resistae therefore is dow to oe-fourth the value foud i part a,orr rad 0.25)0.711) 0.178. ) step, I 0 for 0 < z < 0.25d ad 0.5I 0 for 0.25d < z < 0.5d: I this ase the average urret o the wire is 0.75I 0. The radiated power ad radiatio resistae) are dow to a fator of 0.75 times their values for a uiform urret, ad so R rad 0.75 0.711) 0.400. 14.25. A dipole atea i free spae has a liear urret distributio. If the legth is 0.02λ, what value of I 0 is required to: a) provide a radiatio-field amplitude of 100 mv/m at a distae of oe mile, at θ 90 : With a liear urret distributio, the peak urret, I 0, ours at the eter of the dipole; urret dereases liearly to zero at the two eds. The average urret is thus I 0 /2, ad we use Eq. 84) to write: E θ I 0dη 0 4λr si90 ) b) radiate a total power of 1 watt? We use I 0 0.02)120π) 4)5280)12)0.0254) 0.1 I 0 85.4 A P avg ) ) 1 1 4 2 I 0 2 R rad where the radiatio resistae is give by Eq. 86), ad where the fator of 1/4 arises from the average urret of I 0 /2: We obtai P avg 10π 2 I 2 0 0.02)2 1 I 0 5.03 A. 14.26. A moopole atea i free spae, extedig vertially over a perfetly odutig plae, has a liear urret distributio. If the legth of the atea is 0.01λ, what value of I 0 is required to a) provide a radiatio field amplitude of 100 mv/m at a distae of 1 mi, at θ 90 : The image atea below the plae provides a radiatio patter that is idetial to a dipole atea of legth 0.02λ. The radiatio field is thus give by 84) i free spae, where θ 90, ad with a additioal fator of 1/2 iluded to aout for the liear urret distributio: E θ 1 2 I 0 dη 0 2λr I 0 4r E θ 45289)12.0254)100 10 3 ) 85.4A d/λ)η 0.02)377) b) radiate a total power of 1W: For the moopole over the odutig plae, power is radiated oly over the upper half-spae. This redues the radiatio resistae of the equivalet dipole atea by a fator of oe-half. Additioally, the liear urret distributio redues the radiatio resistae of a dipole havig uiform urret by a fator of oe-fourth. Therefore, R rad is oe-eighth the value obtaied from 86), or R rad 10π 2 d/λ. The urret magitude is ow I 0 [ 2Pav R rad ] 1/2 [ 21) ] 1/2 10π 2 d/λ 2 10 π.02) 7.1A 258
14.27. The radiatio field of a ertai short vertial urret elemet is E θs 20/r)si θe j10πr V/m if it is loated at the origi i free spae. a) Fid E θs at Pr 100,θ 90,φ 30 ): Substitutig these values ito the give formula, fid E θs 20 100 si90 )e j10π100) 0.2e j1000π V/m b) Fid E θs at P if the vertial elemet is loated at A0.1, 90, 90 ): This plaes the elemet o the y axis at y 0.1. As a result of movig the atea from the origi to y 0.1, the hage i distae to poit P is egligible whe osiderig the hage i field amplitude, but is ot whe osiderig the hage i phase. Cosider lies draw from the origi to P ad from y 0.1 to P. These lies a be osidered essetially parallel, ad so the differee i their legths is. l 0.1 si30 ), with the lie from y 0.1 beig shorter by this amout. The ostrutio ad argumets are similar to those used i the disussio of the eletri dipole i Se. 4.7. The eletri field is ow the result of part a, modified by iludig a shorter distae, r, i the phase term oly. We show this as a additioal phase fator: E θs 0.2e j1000π e j10π0.1 si 30 0.2e j1000π e j0.5π V/m ) Fid E θs at P if idetial elemets are loated at A0.1, 90, 90 ) ad B0.1, 90, 270 ): The origial elemet of part b is still i plae, but a ew oe has bee added at y 0.1. Agai, ostrutig a lie betwee B ad P, we fid, usig the same argumets as i part b, that the legth of this lie is approximately 0.1 si30 ) loger tha the distae from the origi to P. The part b result is thus modified to ilude the otributio from the seod elemet, whose field will add to that of the first: E θs 0.2e j1000π e j0.5π + e j0.5π ) 0.2e j1000π 2 os0.5π) 0 The two fields are out of phase at P uder the approximatios we have used. 259