Antennas & Wave Propagation 4. Luantity, symbots and Units of Waveguides 2 GATEACADEMY PUBLICATIONS ™ Parameter : ‘Symbol ahs |e ||] ele |= The OF Zep ‘GATE ACADEMY PUBLICATIONS 4-3 Waveguide: 2. TE (Transverse electic) Wave : If waveis propagating in z-direction then only tt clectric field is purely transverse to the direction of propagation and the magnet field isnot purely transverse.ie £,=0 and H,#0. 3, TM (Transverse magnetic) Wave : If wave is propagating in z-direction then'on the magnetic field is transverse to the direction of propagation and he elect field is not purely transverse.ie. £,#0 and H, 4. HE GHybrid) Wave: Here neither electric nor magnetic fields are pure the direction of propagation ic. £, #0 and 1,40. transverse to = “The field components inthe direction of wave propag: eee ‘components while those perpendicular tothe direction of propagation are defined TiO Za 5 Disease ‘Question 1 ~ A waveguide (WG) ‘an opera frequency and hercloreactsasa ws itis used for ultra high frequency (UHF) propagation te only above a certain frequen high-pass lier frequency called cutoff () Free space (ii) Parallel plane WG (b) Does nat in: a hollow WG. ¢ ‘wave is propagating in z-direction purely transverse to the direction of (ii) Transmission line tangular WG or Circular transverse component. ae eed ‘Question 3 ‘Compare waveguide with two-wire transmission tines. ‘Ans. Similarities: = = 1. Wave travelling in a waveguide has a phase velocity and will be attenuated asi transmission ine. 2. When the wave reaches the end of the waveguide it is reflected unless th lr ‘impedance is adjusted to absorb the wave. 3. Any irregularity in a waveguide produces reflection just like an irregularity i transmission line. 4. Reflected wave can be eliminated by proper impedance matching 2 i= transmission line. 5. When both incident and reflected waves are present in a waveguide, a stand ‘wave patter results as ina transmission line Dissimilaritics 1. There isa cut-off value for the frequency of transmission (/) depending up dimensions and shape of the waveguide. Only waves having frequencies © than cutoff frequency (f,) wil be propagated. Hence waveguide ses 252/ pass filter with J, a the cutof frequency. In a 2-wie lossless wansmission! all frequencies can pass through it. 2. Waveguide is a one conductor transmission system. waveguide acts as ground and the wave propagates throu from the walls of waveguide. 3, The velocity of propagation of the waves inside the wavegu from that through free space due to multiple reflections from U waveguide. 4. In waveguide, we define wave impedance analogous to the characteristic impedance 2, of “The whole body ugh mull fe ide is quite cil all (a function of frequen) een se vie transeission §Antennas & Wave Propagation 1-3 Characteristics 4-4 GATE ACADEMY PUBLICATIONS ™ 5. The system of propagation in waveguide isin accordance with “Field theory” while that in transmission line is in acconlance wide “Circuit theory”. Principle of cavity resonators. of Waveguide 4 | 7 becomes real and positive and equal to the attenuation constant “aie. the SENS Js completely attenuated and there is no phase change. Hence the wave (s) frequency at which 1 just becomes zero is defined asthe cut-off frequency (or threshold frequency) “J. Le. f, -»propagation takes place be At S= fu 1=0 oF w=25f =29f, 20, CY Bae 1G] “A However, at higher frequencies, ote (22). (GATE ACADEMY PUBLICATIONS ™ Waveguides 20b re ona? da wavegt Gone tn we ated and ose fe an 3, te inside the waveguide. [7E,, =TEy (m=1, 2=0)] ay prea | (2) Oe 5 i vel the wave in C5 Gade Wavclengh (3) Ws died we te dans vey order to undergo a phase shift of 2x radians. This is . shown in figure. ‘ Iti related to the phase constant by the relation,——————A, it is evident that 2, is imaginary which is nothing but no propagation in the waveguide. (© Group Velocity (»,) : @ ‘Ans. 1. Expression for v, :Phase velocity is givenby, ¥» = ‘cartier alone and of course maY (nx) slower than speed of light. The velocity of modulation envelope is called the wap rauen aoa -(22) G) Srosp velocity. This happens when a modulate signal waves ne marcas the modulation goes on slipping backward with respect to the canis aie <9P-(:: attenuation, «=0) = = Tis defined as thegate at which the wave propagates gh the waveguide and Ne oe a race : ees penta) e-em (4) Phase Velocity (v,) : We know that wave propagate in the waveguide when be —— «a Buide wavelength 2, is greater than the fre space wavehogtt No sce oe . oime=(22) «(58) ' ‘eloaty of propagation isthe producto. and fit fle atin wereeide re equation (HL we et Ye Auf where vi the phase velcy But the speed af light equ ore So oe OF Bnd. Tis, ger han te spa og nes Dae & A, 5 | Saye infinite extent in y-direction and 2-directions (as shown in figure. ‘To find the field configuration in the region between the planes we have to solve 9 4 ; Unfeespace iy and e=cjand 7 rss Fig. Propagation through parallel planes Sao ay ‘Maxwell's equation and apply the boundary conditions eel clan Fa=0, H.,=0 . ice. tangential component of elect field and normal component ofthe magnetic Bld Since %>%, 50 v, Vxit=(o+ joo aay eee ‘The wave equation are, VE=y'E, A=! a wher, 1= (e+ joe ) jon is called the propagation constant =a+ JB. Let E=E,+6,8, +6, and H12113,+H,3, +113, represent Band Hin rectangular co-ordinate system. . and v2 a oH, Iv =-0lpe for ¢=0) Its assumed that the wave is propagating in z-direction and the variation of all the field components in this direction is given by e"™wheve y=<4jB For the time varying field, te field variations are given by, e™ €™ | " ss gle 0 (attenuation constant) then, the field variations are given bye"? which | represents the propagation without any attenvation.4 GATE ACADEMY PUBLICATIONS ™ 4-44 Waveguid} +] ‘The wave equation will become | P oi, 2 & Zeanbe replaced by ‘-y' and J ebin equation Gv) to (bx). Those equations v become . ¥6,=- sou, 08) a) ~~ j (i) ‘ = bo) ro) 2, 2, Se pout, { = | where, Rap tole 7 From equation (xi), we get] Astennas & Wave Propagation 4-12 GATE ACADEMY PupLicaTIONS ™ Equation (xi) will bocome = %, WE, Ste jonxZOE g, 2, & ye 8, -F =o, = -, (reste) 5-2, Ra where Pep sohe From equation (xiv) we get, HL VE, 1 Pe H, ==les 2 eee ae ns oe plneveareguese 5GaS Lea But, for TE wave, we know that £, =O and H, #0, ‘Hence from above equations, we get (xix) NTE ACADEMY PUBLICATIONS ™ 4-13 Wavegui As E,=E ge? & WE, whichis a standard differential equation whose solution i given by, E,=(Gsinhx+C, cosh xje™ ‘Where the constants C, end are to be found from the boundary conditions. fe, Ey, =0 at the surface of the perfect conductor for all values of 2’. E,=0at x20 anda At 320, 20> E,=Csiahe™. At From the above equation, we get sinha=0 = ha=me 1, Wave equation for TE Wave are E,=Gyin( Es} 4, -Taa(BEJer, Hasta, jouer 2. Sioa, Wave oqsion for THE Wave i: ee oe ce” =r tee-(: Atlower frequencies, oye (2 rece a and ps ive and equal tothe attenuation constant ‘a’ ie. the wave i pletely attenuated is no : ot sd there {no phate change. Hence the wave anne However, at higher frequencies, ue > y engnmes imaginary, tere wil be phase change Band hence the wave propagates, hoes aiton becomes zee andthe propagation ist stars, Thetegueny a Which 7 just becomes zero is defined asthe cu = lefined asthe cut-off frequency (or threshold frequency) ie At f= ‘Question 9 What do you mean by attenuation of wave in parallel plane waveguide? Ans. The field expression is obtained for propagating wave for the frequency 7> [by replacing ¥= JP by assuming the media lossless and perfect media. This holds good for all types of waves (TE, TM and TEM) by assuning (0=0. But in actual guide, the conductor and the dielectric will have finite conductivity and the energy losses will ot be negligible. Then in such a case a¥0and it is equal to a=a,+a, The Propagation constant y=a+ jpand hence e* xe"*v*%h—-t! wherea,—>attenuation constant for a lossy dielectrc,a,>this arises due to the power loss in the conducting plates of the parallel plate guide. From the transmission line theory, voltage and current phasor along the line correspond to the sinusoidally time varying voltage and current can be expressed as GATE ACADEMY PUBLICATIONS ™* 4245 Wavegui Fayette?” = inte ‘Time average power transmited is given by, Wy one a re St ene 3% ten. Rate of decrease of transmitted power along the lines given by, ‘The decrease of transmitted power/unit length ofthe ine is ~AM%, =20'7, From concervation of energy, itis dear that ‘The decrease of transmitted power/unit length of line = power lostfunit lengths of = -20'm, & line ; Time avergepowerlosnitenghot tele 20H a1, os Time average power wansmited W 11 Time verge powerlstnitleathoftetine a “Time average power transmitted i ‘conducting plates situated ¢ ‘Ans. Consider a parallel plate waveguide with perfectly el planes x=Oand za, that is having a fixed spacing between them, as shor figure a). @ (o) © @ © © : Fig, For illustrating phenomenon of cut-off frequency in a parallel plate waves? “Then for TE,, wave guided by the plates, we have m. 208Antennas & Ws i itennas & Wave Propagation 1-16 GATE ACADEMY PUBLICATIONS ™ ‘Thus, wave of different wavelengths Plates at different values ofthe angle @ (or frequencies) bounce obliquely between the * cos(ar Jp, lz) Refer] Guide wavelength is given by, 4 228, 2e GATE ACADEMY PUBLICATIONS ™ 4-47 Waveguides ence the field equations are Es sisi or-,2) ands! sin(or-,z)0, +=2* sin( ™* )sin(we-B,2) (=) mi 1 [Lisccireatar Waveguide 1. TEM mode does not exist inside cylindrical waveguide. 2. TM mode does not exist inside the waveguide but TM,, for m= 1, 2,3 inside cylindrical waveguide. 3. TE,, mode does not exist inside the waveguide but TE. for n = 1, 2,3 .. inside cylindrical waveguide. 4. TE,mode is dominant mode inside cylindrical Waveguide. 341 5. ForTEmote a2 And = or 2,228 For TM mode, = 6. Itis easy to manufacture. 7. They are used in rotational coupling. 8. TM, mode is preferred to TE,as it requires a smaller diameter for the same cut- off wavelength. 9. TE, does not have practical application. 10. The main disadvantage is that its cross-section is larger than that of a rectangular waveguide for carrying the same signal. 11. The space occupied by circular waveguides is more than that of a rectangular waveguides. ‘Question 12 Explain method of excitations of modes in circular waveguide induding slotted ‘waveguide. Give neat sketches for illustrations. ‘Ans. TE modes have no z component of an electric field, and TM modes have no 2 ‘component of magnetic intensity. If a device is inserted in a circular waveguide in such a way that it excites only a z component of electric intensity, the wave propagating through the guide will be the TM mode, on the other hand, if a device is placed in a circular guide in such a manner that only the z component of magnetic intensity exists, the traveling wave will be the TE mode, The methods of excitation for various modes in circular waveguides are shown i. igure (I) aeGATE ACADEMY PUBLICATIONS 7% ‘Cex ine direction of wave propagation. The magnetic ‘TM modes in the guide. However, when coaxial line, a discontinuity problem at ratio on the line and eventually ic field from the coaxial line will excite the the guide is connected to the source by a the junction will increase the standing-wave lcerease the power transmission. Itis often necessary to place a turing device around the junction in order to suppress the reflection Signal sub Single su Weveeie = fz me Coxsatioe JT] ll Wavenuide @ Fig. 2 Methods of exciting TM modes in a circular wave 4 slotted waveguide, (b) Coaxial line in series with © -guide, (a) Coaxial line with a circular waveguide. Ans. Fora Maxwell's equation, Vi, =-o'neHl, Expanding V"H, in cylindrical co-ordinates. OM, 12H, 12H, an, ap ap oT age” ar OH, F obiva, Let S=07 +0) (anoperator 2H, 10H, 1 2n, 2H 18 AH a tye, 20 one Bp ae a a veh “ But Proteel Therefore, equation (i) becomes 2H, oH, 1 2H, 2H, 19H, 1 PH ay 2 i 3° “pop “pf ag ‘ ‘This is a partial differential equation, whose solution by separation of variables method is assumed to be H,=?Q iv) GATE ACADEMY PUBLICATIONS 1-19 where, P isa function of p only and @ isa function of ¢ only. Substituting for H, in equation (i), it reduces to FCO) 1 +L ZU, 2p9=0 a pap Der ag TH Pe 222, FPO, argo Ca pap wag ere = Multiplying throughout by >, we get a BaP, of Pa ‘The terms (i), i) and (iv) of equation (v) are functions of p only and the term (i) function of ¢ only. 1299, ie! ooF where risa constant. Equation (9) then becomes FOP OT ott nao OP OP ty nyp a Mulipying twoughoutby P, we get 855+ p+ ('H —n) This is similar to the Besse’ equation ofthe form Pe, Baia aiych x. oF a y: whose solution is y=C.J(2), where J,(s) represents the n® order Bess’ x of the first kind and C, is a constant. To bring equation (v) into standard Bese POP gas wecanwrite, (ph aha +(ph —nP=0 te, OM Sart Phage tommy Therefore, the solution of this equation is given by, P= 4 Also, ‘The general solution of this equation is given by, Q=A,cosng+ B,sinng Equation (vi) is the Bessel's solution for P. ‘The n® order Bessel function J,(ph) of the first kind are plotted in figure for some integer values of ‘Therefore the complete solution becomes as per equation (iv),Waveguides i) 21 Antennas & Wave Propagation 4-20 GATE ACADEMY PUBLICATIONS ™ | GATEACADEMY PUBLICATIONS™ 4-2 aad ‘The permissible values of his given from equation (i). laa. z Hy =Car ccnp ane SG r + aks * s ‘The various field components F,,,, E,, H,, H,and H, can be obtained by using the =CJ.H-C.coent (voeat-nbow' deat] H, =GJioN corm (wtereC, le we considera sinusoidal variation along 2" 4H, Soe ‘The various field components are given by, g (= ponenger, : 1-824 leplanter, 20 Muauirteeecae cae Ans. The cut-off wavelength is that mode, for which the mode propagation constant, (TE, mode (TM, mode OTEm Fig. Field patterns circular wave guide for different TE and TM mo ieAntennas & Wave Propagation 1-24 GATE ACADEMY PUBLICATIONS 1% Phase Velocity, Group Velocity, Guide Wavelength and Wave Impedance The relations for phase velocity, group velocity and Buide wavelength remain the sa MiP ease ola rectangular waveguide both for TEand TM modan ie, Phase velocity: v, = ®=—__"> any Groep velocity: vg = ay TA Guide wavelength: 4, « OF Wave impedance: Z,, 28. + Zang =LenfimGgTaF B aay (hy 12, ideo GATE ACADEMY PUBLICATIONS ™ 1-26 Wavogui 4 Substituting this operator in equation (i), we get Chae nas, FE FE satus, Bet GENT +0) Let, 77 +" pe = i”, be a constant, then equation (i) can be rewritten as 25, 26, 5h. 7B Pe, 20 for Mwave ey Sila fH, 2M, oe) ‘ 2th PH pi, «0 for TE wave Ath gt oy _ ing the above partial diferential equations, we get solutions for £, and H,. By solving the diferent equa at Using Maxwell's equation i is possble to find the various components along x and 4 directions [£,.11, 1) From Maxwell's 1* equation, wehave,Vx f= jock Expanding Vii, = 9 system with its. breadth along x-axis, width along y. aus andthe wave assumed to propagate slong the zdiecton, Wavepuide is led with ai as dicecic a shown in Fig. The wave equation for TE and TM waves are given by, Expanding Since the wave is propagating in the 2 of ee “2 direction we have the operator. = joe[3,£, 43,6, +0,6,) Equating oeient of 2,2, and (ter expanding) we get sat a, aH, i, = jock, pets Avi) 2, 8H | oes, (vill) a 3 Similarly from Maxwell's >¥cquaton, wehave ¥xE=-jon/? a ~r{an operat) Expanding VE, we get {with 2 >GATE ACADEMY PUBLICATIONS -1 |=-sova.H, 4 g ‘Ans. ForTMwave, H, ss eal eae a ‘The wave equation of a TM waveis given by, i 6 PE ue 20 te seit) eee aE, ‘This isa portal differential equation (p.d.e) which can be solved to get the differ: GENE, = sou, Je components E,, ,, H, and H, by separation of variables method. 28, _ 96, Let us assume a solution | ey lent, enol) E=aY \ i ti Xisa function of‘ x" only. Combining equation (¥) and equation () to eliminate, we get an expresion foe om 7 tea pare function of" only. Seas Since X and Yare independent variables, \ Hy» 1p SE, PON, 2X 4 intaWiek oe. = ar = Substituting Te value of H, in cquation (vi), we gat Pe, van, yor Mh, 8B eo sce ey OF & "Jon & "Joy “188 Using these two in equation (i), we get # [roe Hoot he xBwareo jou |” Jon Multiplying by jou, we get Dvn uopouty ep E[-etve rare jan Yar tex, a isa pure function of x’ only. E[-0 vate) a1 jon Se Xap #3Pu ¥" where, Peatyenst pH pure tunction oy’ only. Dividing by -h’ throughout, we get ‘The sum of these are a constant. Hence each term must be equal 0+ £, = 2126, _ jon aH, ‘separately since X and Y are independent variables, RoR Oy Similarly, Weuse: of variables method to solve the differential equaton(n} separation Lex op baad meee and where, =A? and -B* are constants. Substituting equation (v) and equation (vi) in equation (v), we B* WB aA eH 0 or aaa4 Antennas & Wave Propagation 4-28 GATE ACADEMY PUBLICATIONS ™ Equations (v) and equation (vi) a eae soltinwal whshantere a) Ae OMA 2nd ede ctl equation, de X= C005 Br+C, sin Bx T=Gyoasy+C,sin dy where, G, G and G, are boundary conditions ili) Sabstiuting the values of Xand fom equation (vii) and equation (a), we get z ~Gyo0e + €, 0 Bs] [Cc08 Ay +Csin Ay] 4 Fig. Rectangualar waveguide for TM wave stands for “forall” and x-+0 toa means x varying between 0 toa. 2 boundary condition [Left side plane or left side wall En0nz-0vy-+005 ‘> boundary condition: [Top plane or top wall} £,=0a yebuxs0we ‘# boundary condition : (Right side plane or right side wall] E,-0uz-0~y+005 1. Substituting 1« boundary condition in equation (x) is given by, E, =[C,cos Br + C, sin Bx] [C,c0s Ay+C, sin Ay] Webave, £, at yn Ome 0000 On (6, cos Be +, sin 24] C, [+ cos =1, sino=0} GATE ACADEMY PUBLICATIONS ™ 4-29 4 ) Waveguides Thisis true for all (v)x-¥0 to a Gens e+ C,sinBe 20 Using this in equation (x), the solution reduces to E, =[CjcosBr+ C,sin Be] [C,sin Ay] faa} 2. Substituting 2+ boundary condition in equation (xi), we get E,=0=C,Cysin dy y+ 0 0b Since, sin Ay #0 and C,#0,C,=0 ‘Now using this in equation (xi), the solution further reduces to |, CC, sin Brsin Ay [> c0s0=1 and sind=0] 3. Substituting 3+ boundary condition in equation (xii) above, we get E,=0=C,CsiaBrsin Ab [a y=b,02-+00} Since, Be #0, C,#0,C, +0, otherwise there would be no solution sin Ab =0 . Ab=nn where, nis a constant n =0,1,2,.. (Gv) Substituting 4% boundary condition in equation (xi), we get E,=0=C,CsinBasindy [at r=0,4y-+0t0b] Since, Ay #0, C,#0,C,#0, sinBa=0, Ba=mx ‘where, mis another constant, m = 0, 1,2, 4 Now te complete satons ey, equation) E, =C,C, sin Brsin Ay were Aan asin eution a an gut ves BaCcaie Maer where © = propagation alg inion, 6 = sl aston? Use €=6.Gysome cer ena z, Cams) nEy)em* =v) oo) ‘Question 19 Drv the expression propagation of TE wan ectngilr waves se PER ot ICSVTU Dec 2012] Tn other The TE,, modes in a rectangular waveguide are characterised by words, the ‘2° component of the magnetic field, /H,, must exists in order to have energy transmission inthe guile. [24, #0)Antennas & Wave Propagation 4-30 GATE ACADEMY PUBLICATIONS ‘The wave equation (Helmholtz equation) for TE waves given by, WH, =-0wcH, ie, 2H, ZH, FH, Fe Shan solu, 0 [ 2H, @ Beer sotuo,-0 2H, ano ‘Thisis a partial differential equation whose solution can be assumed Assume solution H, = XY. i Where, Xisa pure function of Y’ only, Yis.a pure function of ‘y’ only. Substituting for H, in equation (), we get £1] i ex er re xGetarao Dividing throughout by XY, we get lex 1er xe rath? 12x Here Prey function fx and 12% is purely afuncon of y 2 zi 1x ge ana LOY Equating each ofthese items to a constant, we get 2+ at where, -B and -4* are constants. Substituting these in equation (i) above, we get -B-A' +P =0 = Rake Solving for X and Y by separation of variable method, we get X =C,cosBx+C,sinBr, ¥=C,cosAy+C,sindy ‘Therefore, the complete solution is, H, = XY ile. H, =(Ccos Br+C, sin Br)(C, cos Ay+ C,sin Ay) sneodiv) where G,, CG, G and C, are constants which can be evaluated by applying boundary conditions. Boundary Conditions: As in case of TM waves, we have four boundaries for TE waves also, as shown in figure. GATE ACADEMY PUBLICATIONS ™ 4-31 Wavery 4 i} Fig, Rectangular waveguide for TE wave Here since we are considering aTE wave, ,=0 but webave component along xand y direction. . E,=0 all along bottom and top walls ofthe waveguide. £20 allalong left and right walls ofthe waveguide. Oat y=Owx->0%04 (bottom wall) Oat yebrx->0 102 (top wall ‘Boundary Condition:E, =Oat x=O%r-y-»0 10 (eftside wal) “Boundary Condition: E, =O at x= 0% y->0 0 (cightside wall) () Substtting 1 Boundary Condition in equation (v) ‘Since, 1* Boundary Cofition is given by, Z,0at y-O~2-0104,letus waite E, in termsof H,. 28, _ jon dH, a Since £,=0, the Isttern =0. o gz =H 2c cone +Csin B (C28 47+ C849} ,==AE (Geos Be+ C, sin Bs) (AC 80 Ay + ACOA) Wehave, E, =F {For TE wave] ie. ‘Substituting 1* Boundary condition in the above equation, we get Js (¢ cos Bx +C,sin Br) (0+ AC.) F Since (C,cos Br+C, sin Bx) #0, A#0,C,=0 Substituting the value of C, in equation (jv), the solution reduces to H, = (C,c0s Br+C, sin Bs)(C, 008 Ay) (i) 3* Boundary condition: E, =O at x=0-+y> 0108 From general equation of RWG we have, E, Since E,=0 and substituting the value of H, from equation (¥), wet jon a 12H 2 (6,005 Be+C, sin Bx)C, 008.49) EP gi lCrees 2 Be Ic0G, sin Bx + BC, 08 Bx)C, 054]Antennas & Wave ion snnas & Wave Propagati 4-82 GATE ACADEMY PUBLICATIONS ™ Substituting the 3* boundary condition, we get ¥=0,9-40 06 in the above equation. LEH BC) Cyoosdy Since, c0s Ay #0, B#0,C, #0, C,=0 ‘Substituting the value of C, in equation (v) the solution now reduces to i H, =CCcos Breas dy (vi) (i> Boundary Condition, =0x y=b 1-20 a Wetave £, «2726, Jeu2l, ~jou 9 EP lee cceumcondy] fe £0) LPC docs Bis dy Substituting ™ Boundary condition inthe above equation, we get 0 S88. tes isn a cos Br #0,G,C, 0, [sin Ab =0, oF b= nn} ST where, n=0,1,2, 00.01" ri (viiy (iv) 4% Boundary condition : E,=Oatz=avy 40105, £, =, Joust, "Re RP & jou 2 £, = TP ZIG C cos Brcos dy] [> E,=and H, =CC,cosBx00s 4] ie g Le cosa ecoly Substituting the boundary condition, = 2H, Bsin hacen Ay > = y +010 05 Ay #0, Gy, C#0, sin Ba =0 2 Ba= mn where m=0,1,2,. emt Seg fran 8 on etn endo o 1 Go(2s}or( Jere = acer mst Ht, Com ME )on( yom sendin) ) we get, Let (GATE ACADEMY PUBLICATIONS ™ 4-33 feat ae It can be seen that for a TM wave E, has sine-sine components and for a TE wave H, has cosine-cosine component. "Ans. No, TEM wave does not exist ina rectangular wave guide. ‘The wave equations for RWG are: For TEM wave, Ei=0 and Hi=0. Substituting E:=0 and H:~0 in above equations we get, =_ BH 0, y= 0, Hen0,Hy=Oand EO and Him 0 Soa Hence all the field components vanish and therefore, TEM wave does not exist in rectangular waveguide. Question 24 : Why do rectangular waveguide behave as high passfilter?___ [CVU Dec 2008] ‘Ans. There is a cut-off value for the frequency of transmission () of the electromagnetic ‘waves, depending upon the dimension and shape of the waveguide. Only waves having frequencies greater than cutoff frequency (J,) will be propagated. Hegce ‘waveguide acts as a high pass filter with f, as the cut-off frequency. Since propagation constant y=<+ /B and ForRWG) When ©<0,:7=real #0 and B=0-+ No propagation possible. ©>0.:7 imaginary ;@ =O and B «0+ Propagation possible, Question 22 Draw thea, B vs o curvein recangular waveguides showing the cutoff frequency. [CSVTU May 2008) ‘Ans. In Maxwell's equations for the region within rectangular or drcular waveguides, the assumptions were made that the dielectric was lossless and that the walls ofthe guide were perfectly conducting. The propagation constant yeYFooe = hap tow ll The quantity #? is » real number, the value of which depends upon the guide diversions and the oxder ofthe mode being consideced. For example, for rectangular uides,Antennas & Wave Propagation 4-34 __ GATEACADEMY PUBLICATIONS $ Ati) For frequencies below cut-off frequencies ofc is les than Hand y is areal number, ‘which is then put equal toa, ie. below cutoff frequency ean fP—ahe Butat cut-off wavelength 1? =alus.7=0 ‘ rea eicae alm (2) 1 > =I Nepecimeter renee aa] ae ‘ 2e 228 itto>a, tien an2 AA 2% Nepereter ae Now if the waveguide has perfectly conducting walls and the medium filling the ‘guide is lossless, there is no attenuation at frequencies greater than cut-off. Hence ‘a=0for a rectangular waveguidgse y= 8 pa wre wet (2) mt ‘This phase constant Bis not only true for the rectangular waveguide but for the hollow single-conductor guides of any cross-sectional shape. ‘A waveguide behaves as a high pass filter. Therefore, there is heavy attenuation for frequencies below cut-off although the waveguide itself is virtually lossless. A wave- ‘guide below cut-off is often used as an adjustable, calibrated attenuator for UHF and ‘microwave applications. Waveguides operated at wavelength greater than cut-off are termed as waveguide attenuators and are normally used as attenuators in signal generators. () Vasa 54569 & 1015 3040080 Frequency (Cigaherts) —> Fig. 1 Attenuation vs frequency curves for various modes in 2 typical rectangular brass guide GATE ACADEMY PUBLICATIONS ™ 4-35 emi, In general for frequencies well below cutoff, Gare a lage number andy decrease exponentially ata rapid rate, At the cut-off frequency.y become 82 ‘zero for all frequencies above cut-off. However, whereas the dielecic cet guide may be very nearly loses (i, for ample) the walls ofan actual peo ‘Rave some loss. Therefore a finite value of attentuation would be expected inte, of frequencies above cut-off. | ‘Attenuation vs frequencies curves are sketched for typical rectangular by guides in figure (1) and that for circular guides in figure (2). The atenation ig ye high near the cutoff frequency, but decreases to a quite low value at somewhat above cut-off. For TM waves in cylindrical guides of any shape, thee frequency of minimum attenuation. In general for still higher frequencies ¢ attenuation again increases. An exception is waves in circular guides. For these way in perfectly circular guides the wall currents decreases as frequency increases, nd attenuation theoretically decreases indefinitely with increasing frequen ‘Unfortunately slight deformation of the guide produces additional wal currents t nullify this desirable characteristic. TE, Ty ) Astensaton (48m) —> ; Frequency (GH) Fig. 2 Attenuation vs frequency in a circular guide ferries otrney Ina ctrcalee SolS ‘Question 23. lain wave impedance in TE and TM waves... (CSVTU Dec 2009, 2012, May 28 "Ans. Wave impedance is defined as the ratio of the strength of elecie Bald i transverse direction tothe sttength ofthe magnetic field along the other Wane direction as shown in igure y —= =e x Fig, Magnetic field Vs electric field curve‘Antennas & Wave Propagation 4-38 GATE ACADEMY PUBLICATIONS le at (ea 2m 1. Wave impedance for a TM waves in rectangular waveguide : a Zn = ERT F =O Forait, f= [bot . | Axl esemxi —[Forar . © Yee, ¥1/36xx10" =2x6nx10=1202=3770=9 where, is the intrinsic impedance of fee space. 2 Zu =ni-G Since, 4 is always less than 2, for wave propagation Zny <1- This shows that wave impedance for a TM wave is always less than free space impedance. 2. Waveimpedance of TE waves in rectangular waveguide. 12, _ jou dt, GATE ACADEMY PUBLICATIONS 7 For TE waves E,=0 and y= jp ~souati, Em Jou _ Jou ou ay BB Wa Fo Viela-oF ilar muy at oan Therefore Z,,>n as ,<2, for wave propagation. This shows that wave {impedance for a TE wave is always greater than free space impedance. =ni-Ou/ 2 2, a 2m Oy! a OT 3. For TEM waves between parallel planes oan ordinary parallel wire or co-axial transnssion lines the cut-off frequency is"zero and wave impedance for TEM ‘waveis the fre space impedance itself. ies 2,(TEM)=1 eel fGuestion 24 Seo “Draw the impedance vs frequency curves for TE and TM modes of rectangular waveguide and write the corresponding equations. Why TE and TM mode impedance ‘show opposite characteritics? ‘Ans. ‘There are two types of impedance terms generally used in transmission line theory > 1. In conventional transmission line theory, the impedance is defined as the rato of voltage to the current for an infinitely long transmission line known as characteristics impedance denoted by Z, 2. The impedance associated with a waveguide and defined as the ratio of transverse components of electric to magnetic field strength is known as wave impedance. ‘The formula for wave impedance is written below for various waves or modes S 2g Ai) (@) For TEM wave: tau = =n, Ss © TT FT Since, f, > tye >My (9 ForTMwave: ny=nft ( 7 (i)Waves 4:30 we ‘Astannas & Ware Propasation es oars acapmurrumuicarions™ Te ACADEMY PUBLICATIONS ™ SLPS San % ai ¥) a ‘The wave impedance vs frequency curve is drawn in below figure. = yes © 7 rey i atl 4 {£) fif{> e Al) I (4) (i) a 9 n +(e) Jt Squaring both sides equation (), we get [ ‘Sol. The expression forthe propagation constant ina rectangular waveguide is BR where, Bis given by the following expression efemtEl 45) [ye eeme-(2)3]] peietie «in (] [7 Guitenaveghingventy, «28 c "iene oe (ef ar (4) 1 vp=k phase velocit oT ty given by, y={ien(o+ jos) and # =y* +a pe where (=) (PE) 7M,, modes are defined as dominant mode. ‘Question 28 ‘What do you understand by degenerated mode in waveguide? ‘Ans. Whenever two or more modes have the same cut-off frequency, they are said to be degenerated modes. In a rectangular waveguide the corresponding TE, and 7M, mode are always degenerated. ‘Question 29 What are the dominant modes for rectangular waveguide when TE and TM at propagate through waveguide? : ‘Ans. Dominant modes for RWG TE mode >TE, (ifa > 8) TE, (i> @) TM mode 97M, Generally, TEy is taken as overall dominant mode. Fa ates ——, wi es ee a “cern 1L1.9 Numericals ‘Question 31 ea rates ait A parallel plate waveguide consists of two. P ing infinite Steers eat ree esteem ‘Bol. (i) Cutoff frequency for TE, mode: aaa ree Cut-off trequéhicy is given by, cf ee) NT le) Ca oe Sone of) (2 os i ‘Question 32 When the dominant mide is propagated in an air fled rectangular waveguide, the ‘guide wavelength fot a frequency of 9000 MHz is 4 cm. Calculate dimensions of the guide? ‘Sol. Given: f= 9000 MHz= 9x10'Hz, 2, For a rectangular waveguide the dominant mode is the TE;, mode. TE, mode can propagate ata lower frequency. For TE, mode 2, =2a. “ 4GATE ACADEMY PUBLICATIONS ™ ‘were, cathe velocity oflightinonface =3x10" ance | Soon 4B 330360 Substnatng the valuesof 1 andy in equation (we get i 3333 . | a oc (ios) =3383 Foamy «re Squaring both sides ofthe above equation we Bt 6 [1-0333/4,9 J=0335" na at Be +e tat Bo Ssaa3 + 2 ody the wave propagates and 2, =2afor TE mode = 260290 4 4 anke=3.015en 2 Breadth =3.0145 cm *3.cn_ ‘Sol, Given:a=10an, f=25 GHz Ina rectangular waveguide the dominant mode is the TEjgmode- 2x10=20 em ase velody FE From the calculation of 7,, we know that, fi=Qy/2,.)" = yl=(12/20)' =08 Saisie oO 15x10! emvsce ‘Ans. GATE ACADEMY PUBLICATIONS ™ 4-43 ‘Bol. Given: a=25am b= 1am, f= 8.6GHz =8.6x10 He ‘Case 1: Consider TE modes propagation. ‘The condition forthe wave to propagate is that 2ab Behe. oper Since, 2, <2y TE does not propagate. For TE\, mode 4, Since 4, > hy TE is a possible mode. Cut-off frequency is given by, 2, for TE, and TM, tr Sean. [E, and TA, is depeerasts Ay for TEs and TM, = aa 856.em as, for TE, and 73 2, Ay For TE waves: 2, = Ret 2 Bansereae6em Therefore, for TE =6en ‘Ans. ‘Since %.=Gam for TE, is not greater than 2,, so wave will not propagate for TE, mode. ForTéy —n2a > Tiy=2x4=8on ‘Ans. Since 2,(1E,) >A, S5'TE,, isa possible mode. “= Se eae ne aa a= = =, Vow le 5 ‘As A.(TE,) <2» 90 TE, wave doesnot propagate. ‘The TE,, mode is propagated in a rectangular waveguide of dimensions a™6 cm and b= 4 om By means of a travelling detector, the distance between a maximum and ‘minimum is found to be 4.55 cm. Find the of the wave. 6ans; b=4an Distance between maximum and minimum = 4.45 em = Fora TE, mode 2, =2a=2x6=12en | Guide wavelength ea ae, ia 1G 331.24 ATE ACADEMY PUBLICATIONS 1-45 Waveguides 1=23{6944x10? +3.01895%107] = 2, 210em esx Sip eI He “ 2S ant fe: ‘ We aod f y= 45 em For TE, mode which is the dominant mode a PAS AS eae, pa Vay Woasnay 0455 _3x10° Ty 36H" ani ‘Ans. [ae] Vee. Pre gp 2 One? “Oy pe NG ORTAy = TRIE 2x BS NO OF “5 00-7035 =1.25 radians. Ans. Altemative method: p=24 Question 38.77; 3 E Consider a guide of -8emx4 em. Given critical wavelength of TE, -16cm; TMy=7.16em and Tif, =$6 em. What modes are propagated at a free space _Wwavelength of () 10 emt and (i) Se? Given za=8em,b=tanAntennas & Wave Propagation 4:48 GATE ACADEMY PUBLICATIONS For any wave to be propagated, the condition to be met is A, >2y- ab 0 =100n d= oo ee Sik oe T16em=A, TM %, and all other modes thatis TM,, and TM,, do not propagate. Ans. Gi) 4, =5 an Wem=h, TE>2, be TM, > TB, mode propagates = Thi, mode propagates TH, mode propagates TM, mode propagates 4-47 A A ANS es “The impressed frequency of 8 GHz is less than the cutoff frequen Signal will not pass tough the guide. Alteatively the wavelength signlisgivendy, = 221 00375 ma3.75 em hich song than the eatotf wavelength ( on) and Hence no Pony, Ii the waveguide is loaded with dielectric of ¢, =4, then the wavelengths do 5m 52.5 6m ae whichis less than 2 and less than cut-off wavelength. Now the signal yg. frequency will passthrough the dielectric loaded waveguide. where, 1=the intrinsic impedance =120% deat id eee 2x3x2__ 12 TE Ee Ba en 0x10 “The characteristic wave impedance 2p, 712021 Og/A,F = 1203/3 3287 Zpq =1202%0,4329=163.2.0 =0.03 m=3em Ans. ‘Question 40 Show that for a TE, mode, a frequency of 8 GHz will pass through the waveguide of dimensions = 15 an, b = 1 cm if a dielectric with ¢,=4 is inserted into the ICSVTU May 2008] waveguide. Gol. Given:a~15am,b=1om,f=8GHz, ¢,=4 For TE 2, =2a=2%15=30m, 10 Giz ‘Thus the 6 GHz signal will not pass through the waveguide but will get atena, amount, the attenuation is given by, : “tla $+0-nx6x 10") xdnx10" x8.854%10" 3a) 0.015, = 43864 908 =(12n) x10" x4nx107 «8.85410 = 3864.90 (12m) «4n%0.8854 = E3864 90815812908 (a= 167.5 Nplm = 1454.79 dBm ipim =8.68648/m) ‘Question 42 Given a rectangular waveguide 3emx] cm operating at a frequency of 9 Git mode. Calailate the maximum power handling capacity of the waveguitig ‘maximum potential gradient of the signal is 3 kV/om. =3kViem Sol. Given:a=30m,b=1em,f+9 GHz, Eq. = 2A 1 20.0333 m=3.33 em iWaveguides ‘Antennas & Ware Propagation 4243 GATE ACADEMY PUDLICATIONS GATE ACADEMY PUBLICATIONS ™ 4-49 For TE ee ae Sol. Given: a=2cm,b=1em, f=30 GHz P=0.5 hp=373 W dy 333 The field components of the dominant ‘Tay Veoae Power handling capacity ofthe rectangular waveguide i given by, Pecesmsioetat{ ~Gasnironvyeo(22) P=1423kW Question 43 ‘An airfilled rectangular waveguide of inside dimensions 7 emx3.5 cm operates in the dominant TE,, mode as shown in fig. ee 2. Find the cutoff frequency. bb. Desermine the phase veloity ofthe wavein the guide ata frequency of 35 GHz. Determine the guided wavelength atthe same frequency ol. Given:4~7 am, b= 35am f~35 GHz (0) fake ka eruate — (2,¢06,)=20) Ans, < 3x0! : 2 =3.79x10 mvs : © on Ieeuny oo 10 /G5x10) hyo = PILE SHI0) «1055 om ‘am. OG Jems auestion 44 ‘An air filled waveguide with a eross section 2 emxl em transports energy in the TE rode at the rate of 05 hp. The impressed frequency is 30 GHz. What is the peak value of the electric field occurring in the guide? (Refer to Fig.) rode TE, can be obtained by substituting = mal and n=l. Then B,-0 = esol em Boti(Z Jem 1,20 bl | BO Hetyox(S)e% ptm where, 2, =n /By ecg ee ‘The phase constant f, can be found from > Se #1935 =608:36 radi ‘The power delivered in the z-direction by the guide is given by, Be eff [ext] aa, ‘ a1 pg 1935n00°) x10") * ae x10 (41) 5, = 53.87 kVin The peak value ofthe electric intensity is $3.87 KV /m. Ans. Question 45 [= Caleulate critical (cutoff) frequency and. guide wavelength with waveguide dimensions 7.62 emx2.54 cm for normal i, (otTE,,) mode at a frequency of 3 GH. [cSVTU Dec 2008) [: thp=746 Ww] Sol. Given: a= 7.62em,b=254em and f=3 lla Also the mode is Mf, m=1and n=0 Tofind OL wiAntennas & Wave Propagation 4-50 GATE ACADEMY PUBLICATIONS ™ @ Lf: Cutoff frequency f, is given by, e[(mnY (my r ss ac) Sy \ Now substituting the values of m, n, and b in the above equation, we get sei0" [193.14] . Le eee A91GH2 = f,-197GH2 Ans. Guide wavelength 4 is givenby, 2 ‘g70up velocity for the dominant modes in a rectangular waveguide (23cmx 1.0cm) at 10 GHz. ICSVTU Dee 2008] Given :a=23am b= Lam f= 10GHz. Tofind:() y, and (i) ¥, © % (phase velocity): Phase velocity »,is given by But /, is unknown so it has to be calculated and itis calculated using the formula SoS 2) +(Z) rasan ote m= 120030] £310 6s GHz “46x10 3x10 65200) 1x10" vy, =3.9572%10" mis Ans. (GATE ACADEMY PUBLICATIONS ™ 4-51 Group velocity is given by, y_-22047x10' mis 3 Tox10" “700 =F" >a5%or3>a>2 or 3>a> avaFho>a>} or 30515 Let, a=2am then b-<% or b=lom(say) 22 36-22 4em for m=1, n=0 Guide wavelength is, fre by, lary er 3 ae ‘Yoas7s o6e1as 799" ‘ Now the velocity of propagation in aren gen by FT Fa Fa jeoe 10" =4.5356«10'm/s f-05605 “Yeas 1Atcritial frequency @, ove-(=) Hence Ba forne— asus = arue js = Bala Hey- Jef ee are ee 0757 14_41536 s B=138.46 radian/metres For waveguide propagation P=/ft (Here is denoted byP) a P=f3es6ndin Alternative method : Ans. Question 48 ‘A hollow rectangular waveguide has dimensions a~ 4 cm, b= 26 cm. The frequency of impressed signal is7 GHz, compute the following for possible modes : : @% wy i) Zs Zry —__ (CSVTU May 2009] Sol Given: a=4an,b-26an f=7GHz To find :() 4, (i) A (ii) Zee GY) Zr @ 2, (Cut-off wavelength) = 2 ———— 3) (3) = 2,=008m Alo Xf 00028, Ans. doh TE, is one of the possible mode. GATE ACADEMY PUBLICATIONS 53 dye 208 wo.0s06m h-| (2) h-{ sect) y (3) H (ior) wo eee FB F 2x71 % (@) a om) 7123362 nin (@) The phase velocity is given by, : xT "BB 12386 (©) Thenext possible mode is TE, for which m=0, and n=1 Sothat 2, = 240026 3.55xl0' ms =0.052m, = Also 2h, s0 TE, can beg-possible mod¢ = (aan) ay [Vay £28700 a B=83.221 radi Dy Bg arstm (eae) 0.052, 2axTxi0 ¥, =0.5285%10" misee. (©) For TE, mode which is also equal to TM,, mode 4 is given by, “Toor omg PO" dk Hence, this can also be a possible mode. satu f soon f (2) < yl sa \' looney, =27942 rad!q SATE ACADEMY PUBLICATIONS GATE ACADEMY PUBLICATIONS ™ Antennas & Wave Propagation 1-52 Phase constant B= fo7us-| (==) [n=0) ea oe ‘Attica frequency @. oin-() 2 J -(exaay : Hence Teel a, 2 2 4 Fey yt Hee is Bayer | aC} moshaf 2123862 radi () The phase velocity is given by, 22 IE sua = y (_2ex10x10" “a0 - 7 BB 123.862 = a) ine 2 a (0) Thenext possible mode is TE, or which m=0, and 1 = “4, 3 ‘So that 2 _2x0.026 22 240.026 9.052 = 32008 Foes 2008 JaagTs = 2OAOGSINII.IA_ 4153 : T = 3 Also A (Cototf wavelength) : le = pet > i,="formodeTE, a hyn AO N06 _0.08360m Z - (0.04) +(0.026¢ : ah 20, a Hence, this can also bea possible mode. 7 hae 7 J sant (ey Ocaea +) x -( See) f=27.942radim doh TE, is one ofthe possible mode.& Wave Propagation 4-54 GATE ACADEMY PUBLICATIONS ype Be2S _S28x7HI0 BB” 79n vpa1ST33H10 wis Yah 1S 733810x ATED M8028 om 2 (@) For TE tod TM,, made 1, is givenby, 2x0.08%0.025 157310 “Fee Fou coeny pe Here, A,2 ie. 0.12>0.10 Hence itis possible mode. (b) The guide wavelength is given by, a 0.10) Ay p(ey My ye(ai) aA = = 01 a.18mes Vim oas3y Ji-0.093885 Joserit 0.5533 (0 The phase velocity is given by, vy Pesaro St =S4x10' m/s Calculate the cut-off frequency for Ty, TE, and TM,, modes in a rectangular waveguide of dimension 2cm x1 em. ICSVTU May 2010) (@) The wave impedance for 7, mode is given by, h-( >’ 1-230) Sy Hey For ae eit =15Gi nxtstennas & Ware Propaation 1-58 GATE ACADEMY PUBLICATIONS Sol Given: e-Somb-25am A=45em © Phase constant Bis given by, ey js unknown so it can be calculated as or (gs the dominant maeis 76, becuse) 2e5s10" =5x107 = Sem =0.05 m 2st | (SH) con ae ‘asxt07 (S107 o 2ne (Gy Phase velocity is given by, ya =A 2x3.14x3x10" yp a PENS 6 8e10'mis Ans. 55x10 60.83 ‘Question 54 ‘Anair filled rectangular waveguide has cross-sectional dimensions of a= 100 mm and ‘b=50 mm. Calaulate the cut-off frequencies for the modes TEy, TE TE TE TE smd TE,and also the ratio of group velocity y, to the velocity in free space for each of the above modesif f=2f,. Sol, Given: a= 100mm, b=50mm, f =2f, Gf, for TEy,TE,y,TE,q TE, TE, andTE,, modes {CSVTU May 20111 Treratioot % foreach modesiff=24, #28, 2).() ‘Substituting the v sof a, bandc OATEACADEATYPUBLICATIONS™ 4-69 We get L ar {(orsy+(raye - ae (mx 50x10)" + (nx100x10”) ea ToT OAH + onto) Now for mode TE.y;m=1 and n=0 a TOY fe ‘PatOoxto? xsoxt07 "SO iT £0) Lond" Eat Sut0" He andin=0 Set _ fans} v0 ‘Deora aoa NOHO} +0 310d e310 Sadik acto" He For TE,, mode; m=O and n= = 210 _ femme z Fm so a{tetcoxso*} mk acigr So=d0% ch aant0" For TE, mode; m=0 andi 1 F=0 ori(Qx100-00°) -3510° x} = 6:00 He For TE, mode jm=1 and n=1 J.=dx10"[(ixs0s10°) +(Det0007F = 3.10 & 20 SF.=3354%10" He 10" (2xs0«107)' +(ix100x10°) =3.10° x22 L-A2Ox10 He For TE mode; m=1 and n=2 Sari (010) +Qxioo) =ac1e So=6.185.10° Itz on summarizing we have For TE.mode, f,=1S00\MIlzor1 $GHte For Teymode, f,=30tlz For Tey moe, f,=3G1z For TE mode, f, =6GH2 faveguides 4Antannas & Wave Propagation (GATE ACADEMY PUBLICATIONS ™ For TE, mode, f.=3354Gi, For TE, mode, f,=618SMHz=6.185Giz For TE, mode, f,=4243GHz Out of the seven modes shown, itis clear that TE, mode's the dominant mode having the smallest cut-off frequency i.e. S=15GHe Gi) Phase velocity » is given by: 4-00 Mee nas tected hen, eee be0Bem, 020, hia nand code, (2 3 H, =2sia (2}=() 22) (0s —fe) vn Determine: @ The mode of operation (i) The cut-off frequency (ii) The phase constant B (iv) The propagation constant [eSVTUDec 207, 20121 (¥) The intrinsic wave impedance ‘Sol. Given: 2=15am, 6=0.8cm, o=0,n=14,and €=4e, (@ The mode of operation : From the given expression for H1,, from the coefficient of angles of sine and cosine term, m=1, n=3, that i, the guide is operating at TM, or TE, and ¢=4¢y- In i) The cat-off frequency: e ee ie. where wee, a Vee Tete 2 Hence, for Tif, ot TE;, - = 28.57GHz may TRF (aot pasecontant: poo -(4) lire f-(E Ans. See GATE ACADEMY PUBLICATIONS ™ 4-61 a x2 (4) mast Pere: (4) 100x107 onayensto! or =O =50GHe 10") f 3x10 (Gv) The propagation constant : (0) The intrinsic wave: TF. Nas = =n 2 (22) y a1s470 2857 (2857) aes = For TEymode, 2, = 2x1=2em Wor __ 12081208 agg ne 7 oe VrosF 0866 Given, P= 1 hp=746 watt => wo Eg (ab) ‘Question 65 ‘If the dimension of the waveguide are 3.5 emx2.0cm and the frequency of operation possible TE and TM modes that ean be propagated in this Gis, determing all the. es “ACSVTU Dec 20141 ‘Sol. Given: a=35em, b=2em, f=10GH2 Case I: Consider TE modes of propagation ¢_ 3x10" _ “7 Toe ~ ‘The condition for the wave to propagate is that de> Py 2ab Vinten GATE ACADEMY PUBLICATIONS 4-67 Waveguides For TE, AFB rmereason Since, 4, >yy TE propagates. For TE, mode 2, =20=2%35=76m Since 2, >Aq, TE, is posible mode. Cutoff frequency is given by, Biot as =42GHz Ans, pce a 213552 arom = er eos is ASA, for TE, and TM, is greater han both TE, and TA, wil propagate as 2,>4, is these modes i “Fe AT From the above analysis we conclude that TE,,,7M,, TE, TE modes are possible. Case Il: For TM,, mode: 2, for TM, is equal to 2, for TE, mode =3472 em which means that 7M, wil propagates because 2, >), £: Thy TEg,TMyTE, are possible modes that propagate. =3320em a 4 i 2004-68 _ GATEACADEMY PUBLICATIONS ™ Last Minute Revision (LMR) 1 Cato fguency: : se) coll lao ‘where, €= Vela fight (free pace eocty) Te sar Sis male propesion b= dimension of waveguide 2 Cat off ave length: ae J 2 nm) a2, iT liane 2 sPre sec waength = $= 2 |g Phasecetocity: | yetefaR peace: v=none Cin mit) {2 F (4) ‘Sm opeaing funy (in He) sng pereability in fre space | a ‘6, permittivity in fre space and v,», =< (GATE ACADEMY PUBLICATIONS ™ 4-69 Wave, o Be fees 7. Wave impedance: o “Ty HA (Zp =" anere ny =Fee space impetinse = ft vf 8. Cut off wave length : . az (for circular waveguide) Where = Radius of ireularozoequide RyxiSsl or TE, mode) Note: Standard danant mode: For Grevar waveguide» TE, ie.m=1,n=1 For Recangula wnoeguide=TE,,it.m=1,n=0