Question

Derive an expression for the electric field inside a rectangular cavity ( a ´ b ´ l ) made of a rectangular waveguide with I.D ( a ´ b ) in TE101 mode.

(b) Explain with neat sketches including field distributions the mode TE101 in a rectangular cavity made of a rectangular waveguide carrying dominant mode and closed at two ends by shorting plates.

(c) Discuss how you can determine the ‘Q’ factor of the cavity.

Answer 1

(6) field components for Temp Mode: TMmop modes of propagation On a rectangular waveguide is shown with components In fig.a. For Term wave to propagate Ezzo and Hz to. We know from maxwell's ean 7²/2 = - W2uEth se did Ey ↑ Mý 1. Z a - En fig = 1 im Z 2242 +22th 22 a22 +2 2 = (w? UE) ih -i Rg 4? = ßg² put & 222 22 22th +374 + (es2ue4=0 m² Applying, separation Cariable method 242 На ху |xy - Hz 7.02% +423 +(2k- PJ xy=0 – 2CA) dy? on dividing by xy of abowe egn d²x = - Kre? an ² d2y = - Kyh dyr pletting eq" (C) 90 22". 2CA)) kä ky + (W²ME- ß})=0 Let katk=ke then k² = W²ue- ßg 1-0 y 2

them solution for egn. (3) x = a sin(kon.ae) + C₂ cos(knin) Y = Ca sin (ky. y) + CA cos (ky. g) Acerroling to them H2 = Carsin kreast ce Cos (kr. 2) (Casin (kg. y) + Cq cos (keying) to the first brandoy condition for bottom walla Cie, Y=o) we know that the tangential component of electric field for Reactangular wave guide. En= – i will ath K² gy Ene - i will [co, sin (know + C2 (kr. a)]. (ky) [c3 cos (Kyig) _ cos ка ca sin (ky.y) at Yoo, Enzo them [03:0 How He=[ccisin (kriselt 62cos (Kr. ] [co cos(ky. y)] and boundry condition for left wall lienzo then Eys (Will 242 Rc2 Ey = WA Ckw [c, cos Ckm'de) _Ca Sin Ctross] [C3 8in (23,9)+G4C05 ():9) at n=o, Ey=0 then ci=0 Now solution become- ts CC, cos C:%)] [eu cos CK3.93] = C2C4 Cos Cka.m) COS Cky.y) on applying I ond boundary conclition at gab, Es=0 En=-(wall ath 엥 En=-1 [acacos (kron) Sin (ky. 9), ky at Enzo, ce gab lkynn b i wiele CO₂ Cy cos (Kein) Sin (ky, b)) k² sin ky.b=o=no Kyıb=niñ مان kynn b

Ey= M KR ra {tino-min - i P82 ē According to 4th boundary condition at n=a & Eys o iwe ath Ey = - wall x km [C2 CA Sin (kron) cos(ky. 4)] Ke ² Now feet Ey=0 then iwell kn (C2 CG Sin Cknia) cos (ky.y) K2 sin (kn.a)=0 Sin (kn.a)=0 = Kroa - mn knamn So, Hz = Hoz cos Cmtin) coscandy) if wave propagetting gn the Z-direction thermis- H2= Hötz cos como cos Conte) if wave as propagating an – ve z-disn then is H = Höz cos CMD cos (n) y e the amplitude along the tue Z-dern mit Hör and amplitele along then we z dira is Höz on Seperborition' of these two wave we Sbtain straging lesque such that. іEgz. Cos Crong) cos company) an abflying boundary condition at tao tangemdial component of En cual Ey=0 on Ey ill atz b et e Bgq iPg2) + Hоz the (Hot 2=0 Surface. -i8g2 ē 21. Mozeti Pyay an sin carne cos Cerra] Ey= i will [Hoz atazo Ey=o o pode Chose thaz] COTT Sincrons) cosCounty) then Hotzt Hözzo Mtz = - Höz if Motz = At and Moz=A

Quality factor of a cavity Resonator- Quality factor e is defined as Energy Max stored Power dissipated per cycle 277 -X 21T T р * * کی (20) р where 'w' and 'p' is is maximum Energy stored Average power loss Pewt Ped t Pie (32 Power loss due to cavity wall/ wave Guide where PLW Pld Power loss due to dielectric medium, PH power loss due to coupling

ab - AL sin²gx sin²2 dady ofz 2 o 0 b d = 2 O 아 Siv? ( 1x) sin sin(772) dx dy dz wa Jasin ( 4 4x 1944 **** (13) dz 1 e ( % ) (6) (d) 2 E abd 8 Now 10 Resonator calculate power loss due to walls of p = Pio P4 PLE PLZ - + + Y TOP Back FRONT side walls 6 BOTTOM Power loss due to front and Back sides walls at z=0 and 2=d a P4 12 daty 0 0 PG RS je do 14h+t9 + + 2? I dudy at ZED z=d and at Hy = 0 2=0 & a=d H₂=0 D and sin Br Cos D2 then b Rs O iwell sin2 Bx w2D2 202) dx dy sin? ( MIx) dx dy w? M² от 2 D а ob Py RS $$ at z=0,d of dr = D = PIT 0

sin? (FT) Cody Per pr2 Rs dazu? Rs p²72 of we u² Rs PIT 2 ab (32. PG 2d2 Power loss due والوں to and Top Bottom at Rs fa e Hep dada y = 0 Y = b Ha to I H2 to Hy at y=0, y=bl O and ao at yo, y=b PLZ O f{1Hx² + y ²2 dxdz Pis - foto siren.como2} + fer un cosa siro2 } ] deck of custoz dz + 80 [°C32pxdx a sin Dadzi Plz Rs W242 D² sin²B x dx a d + PLZ Rs wh 4² 2 2 2 2 RS lz wa uz at { 8² +8²} ad Rs whe? { m22 92 p22 d2 } PLE 4 2 m² TT + PL2 = RG ad 4 92 side ماه power loss due to rb rd Rr O JO at The 12 dydz x=a कि 火Q at x=0, 0 and Hoe =0 Ну rbd sin B dy dz PLZ Rs wru² 2 ta RsB² dy sin B2 de Puz wizuz

Piz .RsB2 درجن Piz Rs Wall (54) ) bd m²2 Rs of u² (s) a2 Pe Py + Pas + PLZ then using from equation cus, (3) & (5) Rg π (m² m² w 42 ) ²) Pab + p2 d2 ad 4 (bd) 2d2 a2 92 Now from equation co Quality factor W PL Wy sabd 8 Rs ( 2 ) E m² GCY ائرل 92 Q2 p² ab a02 Wy € abd with u² ī 2. m² p2 mi + + ad 4 + per 92 2. Wy u abd [ 278 8 Rs T2 + plab 242 ad 4 a2 d2 a2 . wiz 2 HE ( 27 ) 2 1 ME | 27 27 do Wy 25 dar Jut atu abd 472 ad Lue 8 Rs ² (Ar) 3 ( peab (m² a p3 da + m² (bd 2 22 ( bd) a2 Tu abd Rssue (dx)3 [ p²ab m2 ad + 1 + p2 d2 ( bd)] 2d2 a² 4 az

กh ab Rs (dr)3 p²ab 202 TE mode in R-cavity This is the expression for & -factor for Resonator. for Dominant mode mal no and Quality factor for Dominant p=1 Then mode -18) Tim abd Rs (r.)3 242 ab + ad ( 1 + 1) + bd for Cubic in Equation ) a Resonator a=b=d. Tha3 Rs (13 P² + + Els m2 4 793 utina (9) 3 Rs der (p²+ m2) This expression for TE mode in cubic Resonator. for dominant mode Iza. then uth a 3 Rs 212 a3 (2) Tn 312 Rs. Hence proved dimension of cut So a - factor does not depend on the Resonator in Telo mode