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Unformatted text preview: TwoDimensional 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 onedimensional 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‘”) 10)
(’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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 Summer '10
 Ramprasad

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