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Thus the incident field from the magnetic surface current sOurce M s onto cylinder q for 115 - pql < 115' - pql is, from (12.3.32) and (12.3.34),
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Thus the incident field coefficients onto cylinder q are
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After the exciting field coefficients are determined, we can find the far field scattered field. Asymptotically
pA H TE(k p, k z, - z ) - ry { - cos (k zz )k P + ~ sm (k z Z )k Z } n p, - 1 kZA k p ein"v ik i(k P p-2C)-in 2 '!' -e 4 1rp
H n (kp,kz,p,z)
1 --cos(kzz) e m '!'
'(k pP _2C)_.tn 2C 2 4
The total scattered field is
_ ~~ (N) TM(p)-TM k -_~) - LJLJTm W Zm H m (kpz, zZ,p Pp,z 2
p=l m,l
(M) _ _ d) + LJLJTm W TE(p)-TE( kpz ,kzz ,p-Pp,z 2 Zm H m
p=l m,Z
The far field in direction ks = cos sx + sin sy is
3.4 Excitation of Magnetic Ring Currents
3.4 Excitation of Magnetic Ring Currents
We consider the case when there is a ring source of magnetic surface current centered at cylinder j (Fig. 12.3.3). Specifically at Zl = d/2,
M su P
=-1- - Vju n-;Lb Ppj -11 I P Pj
fora::; p -Pj ::;b
and at
= -d/2 ,
M sb (_') - P -
I-I - Vjb 11n-;L ) , I !lor a::; IpI - b'+'ppj ' P Pj
Pj I ::; b
(1 2.3.61 )
We first consider the excitation terms of a'fnM(j) and a'fnE(j) for cylinder j.
dp' m-n(k pz , kzz, pi - Pj)e-mrPplpj . M s (pi)
~:; 6no [H6 1) (kpzb) a
H6 ) (kpza)]
where Vj = Vju for M su , and Vj = Vjb for M sb ' In (12.3.62), the Kronecker delta onO arises because there is no azimuthal dependence of the source. Because no(k pz1 k zz , pi - Pj) = O. Thus
dp'n-n(k pZ1 k zl , pi - Pj)e -inc/>-;7;; . M s (pi) = 0
We then have
TM(j)=_7]iwE(-1)1/27fVj" [H(l)(k b)-H(l)(k )] a1n 2d k 2 Jl lnQ UnO 0 pi 0 pia pZ a a Tz~E. (j) = 0 "
(12364) .. (12 .3. 6) 5
Next we calculate the excitation terms of aTnM(q) and afE(q) for cylinder q =1= j. We use translation addition theorem, since Ip' - Pql < Ipj - pq 1
a (q) __ 7]ZWE (l)n+1 + In -
k2 pI
dp'm_n(kpl, k zl , P'
- Pq)e -in pf pq . M s (p')
m+n (k pI Pq
= _ 7]ZWE (l)n+1 -
k2 pI
dp'Rgm m k, P'
~ m=-oo
"" H(l)
1- _-J.1) e -i(rn+n) pqej P
- Pj
eim~ . - s (15') pf eq M
= _ 7]ZWE
(l)n+1 00 + "" H(l) (k k2 JI ~ m+n pI Pq pI m=-oo
1- _-.1) e -i(m+n) pqej P
21fV --tfJmo [ In a
[Jo (kplb)
- J o (kpla)]
Thus for q =1= j
a ln
= _ ik
(_l)n+1 + H(l) (k k2 JI n pI Pq pi
1- _- .1) e-in pqej P
21fV --t [ In a
[Jo (kplb)
- J o (kpla)]
(12.3.67) (12.3.68)
We use these incident field coefficients of aTnM(q) in (12.3.48). After the exciting field coefficients wTnM(q) are determined, the surface current on cylinder pis
-(p) Js
= " " wTM(p)-TM( kpl,kzl,z d j 2) L Js lm
_ " " TM(p) ( -z k pI A) -~Wlm
~ (1)
H m (kpla)
( cos [k zlz dj2)] eim
3.4 Excitation of Magnetic Ring Currents
The far field scattered field is
Hs =
-~ L LT~N)w~M(P) cos [kzl(Z d/2)] s
p=l n,1
A, etn'f'se- tOk pI k- soPP
I ") _ P eto(k p1P-4 -m 2
The primary field from source j is
Hp=-2d: ~~fIHo (kpl ,kzl ,p-Pj,z d/2)
TJiwc"" (_1)1
--t [Jo (kplb) - J o (kpla)] In 21rV
In the far field the primary field from source j is
_ Hp
ik "" (_1)1 21rVj 2 d ~ ~ fzQ ln a TJ I pi
[Jo (kplb) - Jo (kpla)]
cos [kzl(z d/2)] cPse- tOk pI ksOPj
I ") _ P eto(k p1P-4
The total field is
H = Hp
In the far field, the power per unit length in the of cPs is
z direction per unit angle
O's (cPs)
TJ H ="2 1-1
and the total power per unit length in the
z direction is
r io
dcPsO's( cPs)
3.4.1 First Order Solution
We calculate the first order solution. Since the cylinder j is close to the source at j, we combine the field from source j and the scattered field from cylinder j as the incident field on other cylinders. Let superscript (1) denote first order solution. Thus
TM(j)(l) _ TM(j) win - a ln
~ 2d (_1)1 J-FI 21rVJ UnO ki k2 ln Q
[H(l) 0
(k b) _
H(l) 0
(12 3 76)
For q
i- j,
= a1n
+ '"'" ~
I-p. - -Pq l)e-i(n-m)<p-;;-;pqT(N)a (j) J m 1m
Thus, putting (12.3.64) in (12.3.77), and noting that cP PJpq = cP Pq l2J have 1 TM(q)(l) _ ik (-I t + jH (1) (k -in -o-::o-::-2JrVj win - - 2d k21 I n Pq - Pj e Pqe In Q x
+ Jr,
1- - I)
{[.fo (kplb) - Jo (kpla)] + TJN)
(kplb) -
(kp1a)]} (12.3.78)
= -Jo (kp1a) j H6 1 ) (kp1a) , putting it in (12.3.78) gives
ik (_1)n+1 fH(l) (k k2 Jl n Pq-Pj pi
1- -I) e-in<!>o=2JrVj X Pqe ln
Jo (kp1b)
(kp1a) - Jo (kp1a)
3.4.2 Numerical Results
We illustrate numerical results of the radiation as given by (J(cPs) of (12.3.74). The fractional volume of cylinders is at f = 0.0005, and the size of the cylinders is such that ka = 0.05, or a = 0.00796)., where .\ is the wavelength. The size ofthe aperture is b = 2a = 0.0159.\ (see Fig. 12.3.3). The separation between the parallel plates is such that kd = 0.3. In Figs. 12.3.4a-c, we show results of (J( cPs) for N = 16, 64 and 256 cylinders, respectively. The positions of the cylinders are chosen stochastically using the shuffling algorithm in Section 2.1 of 9. For each case, the size of the square area in which the cylinders are placed are (2.52>.)2, (5.04.\)2, and (10.08>' respectively for N = 16, 64, and 256 cylinders. The source cylinder j is chosen that it is the cylinder that is closest to the center of the square. The voltage is Vj = 1 volt. The magnetic ring source is placed at z = dj2. We plot results for (J(cPs) with the observation point at z = -dj2. The total power per unit 27r length in direction, P = Jo dcPs(Js(cPs) are 0.10,0.11, and 0.11 respectively for 16, 64, and 256 cylinders. We note that when there are more cylinders, the radiation pattern has more angular fluctuations.