L15_Two_Dimensional_Greens_Functions

L15_Two_Dimensional_Greens_Functions - Two-Dimensional...

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Unformatted text preview: Two-Dimensional Green’s Function Consider the parallel plate waveguide excited by a line source at (x’,y’): We are interested in finding the Green’s function of this configuration subject to the boundary conditions G = 0 at y : 0, b and the radiation conditions at x —} it». We have 826 +82 G i 819+ ayz H306" ‘5lx x ) 50’ — 39’): [301 k0:wo«./1u€ and we propose the solution (by invoking separation of variables) G = GxGy : lexgc’) Gy(y,y') As done for the one-dimensional case we may choose (this sum is most appropriate for Gy since run will be sinusoidal and not decaying exponentials) on Gy : Z anWJIly) with “#02 )+B,11Pn(y )20 1 WRZOatyZQb mt 2 mt are A — — ' — w' :_ 111,1 (y )2 sin 5) :fl sm E: 1th [3,; b Note that 111,102) are the eigenfunctions and [3,1 are the associated eigenvalues of the problem. We have chosen A so that It follows that [bv(y)vm<y)dywaa:{ 3) 2;: i.e., urn are orthonormal eigenfunctions. Then G:\/; 2 anG )sin iii—Ey— 2 an Willy 1x) Substituting this into the DE. satisfied by G gives Bani: {Grlwit‘l'Gx ‘izll’ +B0Gx1l’nl— 8(xwxf) SDI—y!) Next we multiply both sides by trim and integrate with respect to y to get (this step eliminates 6(3) — y') and the y dependence) arniGiri+ii3t2lmB§1)Gxiz _8(x_ xi) mef) This leads to N N (3+(%—BQG¢:7&xmfi) where E? : G an X I wn (3}!) HP“ constant Clearly, we have now reduced the 2D problem down to a corresponding ID problem. The solution of this latter D.E. was already given earlier and we have (since only the radiation condition need be satisfied) e’jfigflixe‘i'jflgnx‘ ~—--— x > x’ 5 szgn €_‘iBg”lx_x!| x — e‘ngnx'6+jfignx a ZngII ——_—~— A: < x’ 2] Bgn with l3 _ if I38 — [5% [30 > Bu: Bo : ll50 : (Dex/#080 an — "j fig—[3% B0<BR Returning back to G, we now have G 3 GxGy : Zaanb’) Gr 2 Xaanb’) [mi/IOU] air an e—jflgnix—xii Bngn 2 . mt . rm ,e‘stt'lx'qfl : G : EWIIO’) Wnb”) G=— — —— —-— b ,1 Sin [9 ysm b Zjfign when [30 = 0 (static case), it follows that 1 3173 ME 1 G: u 'n— ' _ ’ “"3an716! ém‘cSl bysrnbye and as expected G decays in the direction of in since there is no propagation at DC. Each of the summands in the expression for G corresponds to a mode. The specific modal functions or eigenfunctions are given by (RWY) = C5111 "Eye , C is a constant These modes propagate in the ix direction, and since (1),. "—- 0 at y : 0,19 (on the PEC surfaces), they can represent the electric field component EZ (TE wave) or Ex (TM). The mode phase velocity is _ (I) 7 ‘00}J Co an — B— u gm in which CO is the speed of light. Note that “omD is mode dependent and frequency dependent (i.e., the waveguide is dispersive). Also, ”D I _ (D _ 0.) _ 1 M C0 0P”:O_i3g0ml30_ #S—N Ifwe define ml: 2 Co ”WM 1“ m : ”Wm. and we observe that Neq S N ”unp = mode velocity velocity e «S enveiepe or K group velocity 90° This is the basis for designing microwave lenses. cylindrical phase fronts —-—>~\ o n n n .9, ‘l—--——. __> planar phase fronts I! I 5—m_.._" I x r' r‘—~\ —’: I I ’———._\ 1 I I \ I f." beam Because the phase velocity in each guide formed by a pair of plates is greater than pop for a propagating mode (say TEI), a cylindrical wave (or abeam) may exit as a plane wave on the right side. Whether a mode propagates or attenuates as it travels along the x direction, this depends on the value of the propagation constant B _ V [3% g :21 [30 > B11 an — —j\/ 53*53 B0< Br: eiiv 5‘ 333‘ (no attenuation) For [30 > I3” the propagation factor is and for [3,; > [30 we have 6— V 531433 N (attenuating) The transition from a propagating to an attenuating mode occurs when mag—ta :0 ways: (”ff :> (Hot- = L (E) : Zflfoie => 1lye b n n cutoff frequency for the nth mode yaw—e That is, if the excitation frequency is below foc (f < fgc) the nth mode will not propagate. The associated cutoff wavelength is fOc: l _ 2112 i 2% _ 7&0 g BE” 2— 7&0 2 B0 1-(“3‘”) 1-0) (’30 Dr XOZEEWZTCUOP Group velocity Each mode in the guide can be rewritten as C (grow/buy + e—jm/bjy) eijflw 2i This constitutes the superposition of two plane waves (13+ and q: where n. 7 e‘ ,1 . mi: 2 $3: 3 6' J'[Bgnx=|:( TE/b)}’} Bgn : I30— (_ b _) Recalling the typical plane wave expression (3) N e—jfio(xcos¢:f:ysin¢) we introduce the substitutions ? 2 BO Slfltx 2}. [39.13 [3% — [3% 31112 (x 2 BO c0305 Thus (133': g —-j[30(xcosoc:i:y since) The energy or group velocity for this mode/wave is given by 2 CU Bgn C0 RTE n :1) , on: — — — 1— m— <1) 6’ 01°05 ‘(N. Bo NV (Bob) — “P \V/ 130p cosoc which is also a function of the excitation frequency. Note that in general ‘1) deDo _ 6 deg d:S—g dCDQ where [3g or Big” is the propagation constant of the guided wave. Since %:%(%(?jafi(we$fl (none 03%;!8- (%)2 This expression agrees with the above, which was derived geometrically. IfN : ‘ hirer is also a function of 0) (material dispersion), then (note (0(2)st : [5(2) and ya = szcfi) ng_ d 2 an _ d 2 mm2 MWEUBO—b "doe We“) m figfizflimfius) [if] 4 — 1 (“WOW”) ,__ N2 Zdo) 2 2 0 wéue— (”r“) d 2 203% dN m— e 2 2a) a ———m dwO (mall ) 0‘11 (3% dwo 2 28 dN as? dN : 2 G 0— : 2 __0___ wGHe+ N203 (1030 (”we N2 51030 Thus, ng N (00 (UV 1 N [3060 dN 1 + Z r“ 1+ —m dCOU C0 C0 deo m: 2 CG N2 dCDO m1; 1— —— 1— m— (5017) (Bob) and ...
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