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Unformatted text preview: has 32% x , 3 x
i/ f' g; Kr.» 2 w? “N ’ E k” f g/ 3 g , \Q
is"! ”‘ M’ \9/8 N. 3 huge 'E‘hih 333%th himga {h} ﬁgihg Zahs ghhciphhiiich pﬂﬂﬁiﬁgﬁﬁ {hc % ﬁehzi 3i aha @Eigih is given hy E E“ #7
#9 +2 #01 = [ﬁ‘mi‘xﬂ—i E 2 E} + $2 : "§27r(2) 2m) 47r Figs £3. Figure fea hmhiem ssh (a) I] ané I; in the same direction. (b) 11 and I2 in
opposite directions. (b) Thc B ﬁcid at point (1,130) can he wriiien as + x — ’ “ = 9
2mm «E "r yszzi «E
(c) Simiiaﬁy, the E ﬁcid at point (2,2,0) can he found as
..  _A M017 M01 _ “A M
E ‘ 3‘ +32 ‘ szz) + y27r(2) “[ X + ﬂ 4% (d) When the diE‘CCﬁOﬂ. of the current on ihc :3: axis is switched, ali the 31 hcids also switch
direction. Thus, the 3 held ai the origin is now given by ._ __ M017 M01 a Egg
ﬁ—El+§2—?2W(Z)+§2ﬁ(2) —§g+§q 4” The B ﬁeld at point (1,190) is Moi 1 M01 1 m+ ......._
27mm §2vr(x/§) 2
A #01 31 #03 1 $=B§+Bz=ﬁ Sig;th Ehhgh {if highchhh hi ihig and same cf ihc oihci‘ hmhiems h; Ehig Chapicr, h: i5 hgchii m mic thaw; 325a; {icgiccaﬁ its: Eighi‘c 5.5a and 65h, ihc h—hcﬁd 3i scmc dismth T hum ihc @éigﬁ hi“ h 47 , cc ax m Eéga 6.5a ﬁgure fee g 6% gggféiii} g ﬁeid a: a ﬁieﬁeeee r ﬁem
W “ 27?“ £342 + {2;}2 ﬁle Genie? 0f wire of Eeﬁgih 22) Ge this basis, the contributions of eeeh side of ﬁle square depicted in Figm‘e 6.56 can simpéy he
fomulaied as the supegpesiiion ef the B—ﬁeide ea: various dismmes from edges of wires of Eeegehe
L = (1/4 end L : 3a/4. We ﬁrst note the: {he conﬁrmations due E0 segments 3 and 4 me eqeeﬁe iheee Emma 2 mad Ejezéegg ﬁghiwham mﬁe, ii ﬁe dear mam 3E3 eegmewée e; ﬁﬁeéd eeieéi ? ﬁe W2: 3e» that 3:336: ﬂéﬁea’ew: eemeébméeee mm} ﬁeki. We Esme engage) 2 “(a/4&2 («z/4P + Gama r:a/4,L=a./4 r=a/4, L=3a/4
= 3) E9]. \/g+ 9E
2 27m 5 end geplsew: (E) %+ “ma/4)
2 “(a/4V 2VFW/492*x/C‘Ba/4>2n—“4753ET4>2
r=3a/4L=a/4 K z (.1. ) w. ( W at?)
2 21m; 15 3 hath it tight: thith eieeelet tithe We eoestdei’ eeeh @éﬁlﬁﬁ ot‘ the leeh seemetehi eeti eeoiv sheegtesttioe The straight pottiotts tie hot etetioce may E heitiet poiiit sitter; along thﬁtt‘ ems. Assuming the z direction to be out oi” the gage the B ﬁehi at poiht P totiuceei
hy the cement in the shooter—etc pottioe with recites e is 7 p v e; = f“ hm _ _2#oi¢o
o 47? 032 47m, Similarly, the 3 ﬁeld produced by the current iii the circular etc with radius 5 is " 411‘ 52 " Z 47rb E _. ft ﬂtbdthxee “#01qu
0 Theieiore, the totai E produced by the wire loop is 3 =31 +32=2M0f¢0 .1. ~E
47r i) a see iielmheite eoils, Exempie 64:3, we hoew that the Eihehi. at at Eliﬁiﬂi along the axis of e
eitetiiet Qut‘t‘ﬁi‘lt etetggieg loop of tetiies e is hold2 A E 2 2(a2 t z2)3/2 E where z is the distance from the loop center, and the zaxis has chosen to be in the direction of
the thumb, when the tighthand ﬁngers move its the direction of the current ﬂow in the loop. (a)
Considei a point EKG, O, z) along the z axis at a distance of z from the center of the loop on the
left in Figure 6.58 on page 552 of the text. The distance between this point P and the center of
the loop on the right is then (d — z). The B—held at this point P is then given as the superpositioh of the contributions from the two lootvs, namely MoNlaz +ﬁ MONIGZ
2(a2 + 22)3/2 2[a2 + (d — z)2j3/2 poNlaz i + i
2 (a2 + z2)3/2 ta? + (d _— z)2]3/2 ep=e 2% where we have taken note of the fact that both of the loops have N turns. (b) We have
dB Nlaz 3  3 _
d z = “0 2 {jazz + zz) 5/2(22) + inf + (d — z)2l 5/2[2(d ~ m}
z zzd/Z zzd/Z
" hONIaZ (:33; + 303) W
"" 2h;2 + (d/2)2]5/2 2 2 ‘
(c) For d = a we have
A MONIaz 1 3
By = Z
2 {a2 + 2:2)3/2 [a2 + (a — z)2]3/2
and
3333 __ taltﬂieg “a :1 2 5X”,  i 2 5,52”;
dz .. 2 L 551a z E ate “ﬁe + to —— z} i 3 iiiei #GIVEQL {wga(a2 + zZ)‘5/2 +1522(a2 + ZZ)—7/‘2 * BEGZ + (a _ Z)2:§—5f2
dzz Z .r/ +155}; — ﬁgs? %~ ((1, m afﬁrm ' w...“ s: z = efZ we have 2 2 2 ~5/2 2 2 7/2 2 *5/2 a 2 2 “7/2
deziioNIa [_3<§i> +i5a 4(3):) +i5a 1 422 2 4 4 4 4 42.4 _ .uoNIaz [ 6 <5a2>_5/2 +3 <5a2>“5/Z+3 <5a2>‘5/2j “O
'3 2 _ 7353‘ T 7; ' Eoiiowii‘ig the same procedere as given shove, and with some patience, ii is a Esther straightforwsm
and usequ exercise for the student to differentiate the d2 B; /d22 expression given above and shew ihat d3Bz/dz3 : 0 for z = a/2. The natural question i0 ask at this point is whether the fourth derivative is aiso zero. It ean he
shows by expanding the expression 32(2) around z = (2/2 that given by 44402 — a/2)4 BM) 2 i25a4 so that d435,; /dz4 7% 0. The fact that ﬁne first three derivatives are zero provides for an extiemeiy
smooih arid unifomi magnetic ﬁelé, which is the basic motivation for using iiie Heimeitz eeii eonﬂguiation.
(d) We have Bz _ While? [ i + 1 j
‘ 2 (a2 + zap/2 [a2 + (d _ @253/2 370? d 2: a, the iieiei at the midpeiiit (z = (1/2) is B —ﬂONja2 Egg. —3/2+ E ~3/2
2‘” 2 4 d 2 3/2 ~3/2
= ,uoNIaz = so (5) M 2 0.8992 x 104571: d d a a {e} At the center of the eoii on the left (2 = O) we have 2 NI
BZ z MON“ [CE—3 .5. 24/2653] = 1%; i. 2Way a 0.8505 x 10—6—
2 2 a a which is qeiie eiose to the value m the midpoint, underscoring the purpose of the Heimeiis eeii
eeniigumiions nemeiy io esiehiisii e ﬁwiieiii wi’iieii is very semis? usiferm. Esme iiisi hf}! symmeiiyg Edieiei iii: iiie eeeiei“ ei iiie eoii es tiie iigiii {2: = a) mess iie iiie same. hetneen tern seieno‘idai eniist The in'nttnai indnctence hi2 hetnieen tern
hiecote r serihed in the ntohiern ean he written as ism N2 _
L~:——=——~ n «(is
l; I} {I 32 1 2 where A32 is the total this linkage through the area 5'2 of coii 2 (with 52 = 3 cm, 2a; = 2.5 cm,
and N2 = 3106) due to the B; ﬁeld produced by the current E in coil 1 (with 51 = 25 cm, 2n; = 2.5
cm and N1 = 1090). Note that the heid 31 can he assumed uniform over the iength of the shorter
coil (coii 2) approairnateiy given (from Section 6.2.2) by MOlel 11 since coii 2 is iocated at the center of the ionger coil with 51 > m. Thereforea substituting these
vaines into the mntnai inductance expression with dsz = hdsz yields N N
51171 52 ii I” 477 x iG‘7(iGQtl}{ihtl2 ‘. .J (nag) énitihiﬁﬁigi 2: eat? “is?
«a K»
i. 231th Rectangninr teen and other, Since die h—neid produced by the wire is into the pater, its is
convenient to choose the area eiement (for the ioop) to also be into the paper, in which case the
associated (via the right hand rate) di element is outward at the terminal of the loop farther away
from the wire. Based on the discussion on pages 568 to 570 of the text in connection with Figure
7.4 of the text the generai role here is that? in order for {7.1} to he vaiids Wind must he defined to he positive on the terminal at which the di element, the polarity of which is associated with that
oi" ds via the right hand rote, points outward. With the loop moving to the right with a veiocity no, the distance between the left edge of the loop and the tong wire varies as ﬁt) = not + m, where m is the initial distance hetween the loop
and. the wire. The ﬂux iinked by the loop is given by i
i. 2'” Wm 7" 27? M0157 '”°t+“+a d?" MOI!) ( a > The induced voltage is then given hy d‘i’ noib ( vot+n ) [ —voa 1 ,uoIabvo
Vina “ ————— ————— = W if > 0
(it 2n vet + r; + a (mt + r02 2?!(?!{}t at whose e It a} .j nth. Extracting newer thorn a newer iinea The constraint doe to the length of the wire in hand can
he expressed as 2a+25=2€i0m ~> a=100veb Let the current ﬂowing in the power line he I = Io sin(27rft), where I0 = 4000 A. The magnetic
ﬂux linked hy the pickup loop is a? g b 20+“ at T : neonate:an <20+o\] 2 agroshsawfnmn (12:; mg
23 27W 27? 29 I; an“ I: irreiocesi mirage is their as? p . me w M
OVin = ’1}; =3 filo—IQ C05(27l‘]i)b iii < 2% j Power exirsere'i mssimised if ‘i/‘mé is memorized. Reece we heed is ﬁred she raise of h shes
maximizes {"éZG m were. This care he sierra»: h}; ﬁifferehrietieg with respecs re '5} error semi so zero er hf; simpie eiortihg or by rsiai she error. ’E’he resoir is i; 2 62 or. which reesrrs in
earn she a. :e 3% he. The choice of she opiimum Eoop dimensions maximizes the induced voitege. Since she wire has
nonzero resisrence Rwim, we can View she induced voisage and the wire as a. voltage source wish
voltage V; = “Vim and source resistance RS = ngre. To extract maximum amount of power from
such a source, we most use a Road marched” to the source, i.e., RL = ES. The wire resistance can
he caicuiated simpiy as l l 200 m
Rwire = —' = — = W 2 0,
0A O'copperﬁ'(d/Z)2 (5.8 x 107S_m—i)7r(4’1/10_3/2>2 261g; Thus we choose TL 2 0.2619. We then have 120* 62 (Vim = fﬂofo 008(271’ ft) [62131 < 62 j] 2 19.9 cos(27rft) The power extracred hy the iced under coeciiriorrs of maximum power transfer (re, RL = BS) is 1 (v.2 2 J (b) Try N = 2. Now our constraint is a + b = 50, or a = 50 — b. The ﬂux iinked is 20+a .
MOI ,uofo sm(27rft) 20 — 6
Nb 20 271‘?“ T 27r 2b In 29 Thus, she induced volrage Wind is maximizad if 2b in{{20 — b)/b] is maximized, which occurs for
b 2 29.16. {Isirrg this vaiue of b so caicuiate Wind as hefore, we ﬁnd a smaiier value for PL. holds one for either values of N Z 2. Therefore. we conciuoe that using more sores @0er her
show as so misses higher power. Messrssigh ershhsiorss. {a} We ieiiow so epprosch sihriisr so Exempie 743. Subsiihriihg the Fife};
phases: iriso she ehesor form of [7. i‘Sc]. we first ﬁnd the corresponding eiectric ﬁeid phasor 12(9) as i; 1 6H » E =—,—~—~V x 5% z I (3?; AW)
W60 jweo ﬁg
~'  4xi.83x10‘4 . _
mg. 345.83 x 10—46—34?! = ang—g—ﬂry mm 1 33255 meg Weir‘s. ” fresh 7. he; so ﬁne: SEE es i
i
i
x
i,
r
i
x 2 ’ 5%. “i A sxﬁmm 3 5 Ewen 3W9 ‘2 59s x 1 i .
2g _ («374927—483 x 1046?"?
jwﬂﬂ 9W€0 is .
2 3.83 x 30453411 Am“
W #060 A =3 Note that for this expression to he the same as the Egg) expression given as
my) 2 21.83 x io4eJ'4y A—m‘l the foiiowihg eondition must hoid: 16 4 3 9 ~1
2 :1 —> w: :4x3x10=1.2x10md—s
w #060 x/Moﬁé) With this value of w, Raj) and its counterpart Ety) satisfy both [7.18% and [7.18c]. In addition
E(y) satisﬁes [7.18M (since E(y) = ﬁEﬁy) and therefore V  B = 50V ' E = 0) and EE(y) satisﬁes
[7.18d] (since Eﬂy) = stew); and so V B = NOV  H = 0). (b) Substituting w = 1.2 x 109 tad—3‘1 into the My) expression found in part (21) yields E(y) 2 400696141! V—m“ (c) The electric ﬂux density B and the displacement—current density Ed in phases form can he
found as o = 60E 2 41:06} 1.3“1'49 pC—m‘z 2:237me 2 «see x tomsss x 16"”)(Ooii >< to"12s*34ys:em2>
:3 w seesMW”) teem—2 7345532. EEecteomegssesée ease its free some. We start with
so, 2, t) = 334.9 cos(i.8 x 10% ~ on: — 250,2) Vm‘l
Using Maxwell’s equation {7.1831 and noting that “ﬁx and “(ﬁg are both zero, we have _ es as
Vx‘é=ﬁ<~—5f+ﬁfé—f>
= —s4.9(—2.5a)] — sin(i.8 >< 109m —— cm:  2.5%);
+ ﬁ4.9(—a)[— sin(i.8 x 10%: — a1: —. 2.5m]
379.3%, z, t) = (4.5M: + a§)(4.9) sin(1.8 x 10%: ~ m: —— 25412:) =  at from which @(m, z, t) can he found by integration over time. We ﬁnd
393%, z, t) = ~(25ait + e§}(4.9)(i.8 X @9704 cosﬂﬁ >< iﬁga‘rt  om ~ 2.5oz)
We can now use Maxweh’s equatioo 7.18c] to ﬁnd Fees, 2, t) from @3703, 2, t}. We have §£0%($, z,t) _ iv X a):
at  #0 1 2 2 g A , r, 9 m
x 4“ 2‘13. E W §§ﬁ(3.8 K 735 “' {Lﬁ “w was:ng + e2} W €03€L3 X 3,3937% — as: W KIT—£34
€0I££G\ '7 i I e ﬂeetﬁs" m m a”; as”; 5 , . ,3 Y.“ “‘1' n * °
ﬁwfe E30 Ems magma expsessmn E03 @333 z? 3:} gives} as“; gas‘emem séeﬁemes’g‘é: 9 me}: messimsmg mm ﬁzhe ewes/semi fee gaze, if; we 2233": B772! \ E u ~59; (masses: s L81§ﬁ®sﬂﬁ x 3.9% ~ ems w 274%} seeﬂ E! Eissiﬁses’e gﬁegse wees.) {3;} Fee free space; we have _ _wm21rf_2fr
EWWVMGE “EM 5: m3; 3 $inee x Q SHE and e '2 3 x m3 m—s‘ , we eeieuéeie ihe gshase censmm as N 27mm x 109 where 27 is the intrinsic impedance of free space which is n = ‘/ yo/eo 1: 120% 2 377g. As a
zesuit, the magnesia ﬁeld phases Eﬁy) can be feunci as _ ~1
E s. 3 X 308 — 3207 rad—m
and the waveﬁength as
27? 271‘
= —~ 2 —————— 2 0.61 = .
A ,8 1207? 67 m 1 67 cm (b) Using {use} my) = — .——v x my) = ~ . (wag A”) We quo 8y zgw = g Eisejﬁy = @éejﬂy WHO #0 77 my) : sawrieﬂzm Am" gang‘s Seﬁﬁ‘eées“ gsésessesa (a) Fires, We wsﬁ’ae the pﬁeser expressien fer t) z 2100 cos(wt + 4.87m: — 3.6714; + 49) m‘Vm_l Egg, 3;) 2 2543, y) :: ﬁiﬁﬂeﬂ'gmewgf’weﬂ REV—m”); N€K§9 we substimte Em, y) into the Helmhohz wave equation {88¢} 82E; WE;
+
8:62 ayz +,62Ez :0 Where ,6 = cue/Moe = w/e = wa/c = ZW/A where c z 3 x 108 m—s'l. Performing the paras} derivatives yields {34,st + {—j3.é7rj§2 a» 53 r. 0 ma» /3 = erg: sad—m"i iter wisieis tine 2: eﬁéiﬁn‘} m 98% Mtge end; A :2 it fit tn t’esneetisety. exegessien $53}:9 gag? t} with t m {3 anti 3; 2 2n n @722 m; see new: eke, on, on = stoo (308(0487r ~ exm + a) 2 280.9 asymmi see
——>’ «432° e 6 z "i — 2 36°
ms 100 wnieh resuits in 6 2 792° = 0.4477 rad.
(e) Using [7.183.], the phasoa‘ magnetic ﬁeld can be found as i i 6E2 6E2
Hwy) — jWDV >< Remy) — m (it ﬂy  t 533) E’erfonning the patio! derivatives yields ﬁ(~j3.67r) — 3011.870
jwilo E83? = 1006j4.81rxe~j3.6ny€j6 mA_m—1 Substituting jwpzo :: j7106,vthe magnetic field phasot is given by
not, y) 2 [—20.159 «a $70.212}ej4‘8me—j3'67ryejg yamm“i
Therefore, the corresponding tealtime expression 53w, y, t) can be written as saw, y, t) 2 {40.159  @0212] cos(wt + 4.87m“ —— 3.67734 + 0.4417?) mA—ml where w =_187r x 103 rad—m4. (s) The time—average power carried by this wave can be caleuiated as 3.23 N Wt s _. 21.33 10*5w— *2
2 ’9’; 2 377 X m Aismnet‘i’eet eentﬂttetee in'tet‘t‘neet We start with em = 6Oe’j4°"z{y — 3's) y—mf1 (a) By competing the given expression with [8.10], it is clear that ,8 = 40%. We thus have 2 2% 2n
ﬂ”? “i A=g~4m07T—0.05m c smog 9 f—X— 0°05 —6><10—6GHZ (b) Once again comparing with {8.10}, it is clear that the titeetien of propagation is 3%. Thus, we
have _ 1 1 A  ‘
s >< n12: “easement = gamma?) em” E' .’ m
1‘31!) 3!? get: wait i
i
3
i
1
z
E {e} E‘Eie teiieeteti stave iieitts ate ere) 3 seems — it) = éeett‘tmet + it)
69 em = giee x e = ﬁeﬂ‘mme we
a.“ (d) The total eieettie tieid its air is E1 = e; + E, = 6Ge”j4°"z(§ — 3e) + 606+j40”(—§ + 32)
= y 003—32 sin(4~07rx)i + e j60[2j siti(407rm)1 = (—ynzoj sin(407r:c) + (@120 steam”) "he magnitude of each of the two components is equal and varies with distance as 3.20! sie(4t)7rie}1.
the sketch 0:? tttis satietieii is itientieai t0 the [Elna/)1 in Figure 8.1% 0f the teat? exeeet tee fact that 3: z i {€25} : W35 m. gwéis ﬁeetesee eeeigtio {a} fat tite eetttet t’teqeeiiey 0f the Xeaiid weather tadat which is 9.4 GEEi2, the
wavelength in the foam material with 627. = 2 is 3
A1: 6 ~———iX——i§——~O.0226m=2.26cm t/e‘zi 72/63 " (9.4 x109)\/§ _ So, the minimum thickness of the tadome (which makes it comeletely invisibie to the radar sigaei
at 9.4 GHZ) is d :2 Az/Z 21.133 cm. (b) At 8.5 $st A1 = 3 x 1010/35 x 109 2 3.53 em, A2 = ABA/“52¢ 2 sea/Vi 2 2.50 cm, tee
tanwgd) = tanﬁZw/Aﬁd} 2 tanKZa/ZSGXLBH 2 41319. Also noting m = m 2 377373 anti the inttinsic impedance of the foam material is 712 = m A /62 2 377/ 1/2 2 2679, we have item
[8.52] A2: 773 + 7772 mtﬂzd)
772 27773 tantﬂzdi 77 + j26'7(—~0.3i0)
2(267567 + j377(~0.339)
38661124"
2918«j23.7° 2353eju‘3052 Using {8.53}, the effective reﬂection coefﬁcient is Z2('~d) =772 2(267) (347 — 377) — 369.3
gm};
75.7.371‘114"
2 7276454?
20104575703" Heme, the percentage of the incident power that reﬂects back at 8.5 GHZ can be eaietiiateti as i (Sax/)1» !
itgaviil At tat eitz, A; = 3 x two/(10.3 x 309) g 2.01 cm. i2 = AIM—‘52? 2 2.91/02 2 2.00 em.
and tamﬁzd) 2 tati[(27r/2.06)(i.i3)] 2 0.310. Using these vaittes with {852}a we have ‘2 377 e j26?(®.3itijt
’2s7 t $700310} x me = 1176102 x 300 2 (0.3047)2 x 100 2 3.085% thwd} 2 test? i
2
i {iii ere ehesre is the eerrtpiers eeejegete
'riiz. errci ee remit. ei’iieetiee reiieetieii eeeiiiciettt wen.
1 Size
W
m Vi???
mg
if?)
C!)
m
M .
S3
W
m
be at i03 GHZ wiii he the compiex conjugate oi the effective reﬂection ceefﬁcient at 85 GHz, i.e.. we can write I‘m at 10.3 GEiz directly from Pee at 8.5 Gi—iz es Fee 2 0. were? 108°. Therefere. the
percentage oi the incident power that reﬂects back at 10.3 Giiz is aiso ~1.i)85%. Air Air
n 1=377;Q 7114779
. kt
w”? ——B* TiM Fig. 8.5. Figure for i‘rehiem 841. The comhihation of mediums 3 and 4 (Radome and
air) is equivalent to air, since the impedance n34 = 3779. (c) As shown iii Figure 8.5, the radome—eir comhinatiori on the right behaves just like air, since the
radome was designed to eliminate reﬂections. in ether words, we have 7734 2 3779.. Hence, we
can repiece the radomeair combination with just air. resulting in the three layer problem shown
in the righthand panei of Figure 8.5. Using the given properties of the layer medium, namely
er 2 4.1 and ten 66 = 0.04, we can calcuiete its intrinsic impedance from {8.22}. We ﬁrid = 311/...— mejiian‘VOO‘i) ~ 186 1ej1~15°
‘72 [1+(o.04)2]1/4 ’“ ' We now have a simple three layer probiem which can be dealt with simply by using {8.50} for
the case when the second medium is lossy, nameiy r f. z W
e (712 + twins + 772) + (772 ~ "Dim * ’72)e‘2‘120l
where
. .w .w . ,27r 9.4 x 109} _
72=9w¢ﬁaé=yg¢e=2g 64—Jtanécet=J—~———~———(3 X m. 4.1 J<0%4)(4oi> Using this value at 72., the eta; found above, and m = 7:13 = 377g, we ﬁnd ‘3’ a. r
. 3» ~ ree e ti.§?% timiii. A riieiectrie eeeteri rei’iecter eeterrrre. This is e three—Eagles" prehierri iii which medium 3 is e perfect eerieecter. with intririsic impeeertce 5:73 2 t}. it cart he solved by rising [8.56] for the case
when the second medium is lossy. nameiy e. (it: E mite “r i" it}; e rhiteg ~ age—272d i sit ” ,
he: i tithes. e mi e mg m 9.373 c. riﬂe—ETM fire intrinsie eta: can ne toned from {8.22}, transeiy _ no _ 47r X 10—7
m "’ e; — jag! “ 4.5 ~ 30.009 Using this value and m = 377%. we ﬁnd reg c: seesaw31° e % Power Reﬂected = 1m? x 100 = 99.82% in in principle we now have a four layer probiern, two of the layers of which are iossy. “tire
general solution is quite difﬁcuit, involving impedance transformation on iossy lines. However;
we can observe from part (a) that the coating has very iittie effect, since its ioss tangent is qnire
srnail and it is thin (compared to wavelength). Thus, we may ignore the coating and treat the
tour layer protoiern as if it is a three layer airwater—metal problem. In that case, we have none
again in = O, and m = 377E), whereas in can be found using 6T = 55.4 — 335.45. For a thickness
oi" the water iayer of 0.}. inn}, we find almost complete reﬂection (99.92% of incident power is
reﬂected). However, it the thickness of the water iayer was 1 mm (instead of 0.1 mm) only 46.6%
of the incident power is reﬂected. Thus? condensation and rain may indeed seriously degrade tire
eeri’ormance of the reﬂector antenna. 1.75mm Fig. 3.9, Figure tor Praiaiern 349., ...
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This note was uploaded on 01/17/2011 for the course ECE 305 taught by Professor Staff during the Summer '08 term at Michigan State University.
 Summer '08
 STAFF

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