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Unformatted text preview: ﬁnial. La) For M¢X1mum pave)’ 0’6 (ft/f IK/ CHAPTER 2 55 a. Pm = gmdﬁfﬂ = émqmace—ﬂ‘ﬁﬁ“ x 3.249—1'0491 _ 143.6 x 3.24 2 cos[21.62°} =216 ( (d)
r =
[70+ = V (v):
(V).
__ Vg+ .54“
_ _,._.. = _J
IL _ 20 50 2) 2 49 (A),
PL = émemjé] = %m¢[1803"' 4° x . ' °] = 216 (W)
PL 2 Pm, which is as expected bec 655; power inpm t0 the line
ends up in the load.
(It) 1 A. P3 = quﬁga] = .4 (W). mqﬁﬁzg] = émqﬁffzg] = gm? = g (3.24)2 x 50 = 262.4 (W). l: P =Pz +Pm =478.4 W.
E s Problem 2.32 If the twoanlenna conﬁguration shown in Fig. 241 (P232) is
connected to a generator with VS 2 250 V and Z3 : 50 Q, how much average power
is delivered to each antenna? Solution: Since line 2 is M2 in length, the input impedance is the same as
ZL. = 75 Q. The same is true for line 3. At junction C—D, we now have two 759
impedances in parallel, whose combination is ?5/2 = 37.5 9. Line I is M2 long.
Hence at A—C, input impedance of line 1 is 37.5 $1, and ~ Vg 250 Ii = 28 +2... = 50+37.5 = 2'86 (AL 56 CHAPTER 2 2L, = 75 o
' {Antennal} M“Eh : :51 Hr Figure P232: Antenna conﬁguration for Problem 2.32. (2.86}2 x 37.5 1 ””1 1 ~N‘ Mt
pm : 5914M ] = 59%[1’i1i 2“,] = 2 = 153.37 (W). This is divided equally between the two antennas. Hence, each antenna receives
L323? = 76.68 (W). : nu ission line as equivalent to a load
‘ . e rest ofthe circuit), ZL = Z] = 100 Q. _zL—zo 100—50 F1 ‘2L+zo=1oo+so ‘ 3‘ F 58 CHAPTER 2
Solution: From Eqs. (2.66) and (2.61), V+ = ﬂzin l
0 2g + 2i" all“ + reJW V320 [(1 + Fri“) /{1 _ renown
W When EL 2 (75 j25) (I, from Eq. (2.493), _ zL—zo _ (75+125m—5oo
‘ zL+za ' (75+j25) n+500 so 9., :20w (1 — m1) =2ow (1 41.2772) = 18.46 w. Section 2—9: Smith Chart Problem 2.35 Use the Smith chart to ﬁnd the reﬂection coefﬁcient corresponding
to a load hnpcdancc: (a) ZL = 320, (b) 2L = (2 — 2320. (‘3 Z]. = —2}'Zo. (d) ZL = 0 (short circuit). Solution: Refer to Fig. P235.
(a) PointA is 2L = 3 + jO. r = 0590"
(b) Point B is zL z 2 — 12. r = 0.62.9493"
(c) Point C is 2L = 0 ~12. r = Log—534°
('1) Point D is 2L = 0 + jO. r = 1.091801?" CHAPTER 2 Figure P235: Solution of Problem 2.35. 59 Pro 2.36 Use the Smith chart to ﬁnd the normalized load impedance corresponding ﬂectlon coeﬁic1ent:
(a) F = 0.5,
(b) l" = 0.5.ﬂ,
(c) F = —1.
(cl) F = 0.3%,
(e) F = 0, (ﬂ F= '
S ulion: Refer to Fig. P236. 60 CHAPTER 2 l H.“ uv—cv 1—H” m
 —6—.
"* a! J "I ' u. m4: igure P236: Solution of Problem 2.36. (a) PointA' i =0.5 atzL =3+j0. (b)Poiut .~ isr=o.sef6°° atzL=1+jl.15.
(c)Poin ’isl"=~latzL=0+j0. (d) 1a.: ”3' is 1" = nae£30" at 2,, = 1.50— 113.53.
(e) ‘oimE’ isF=0atzL= l+j0. ( PointF" is F=jatzL=0+jL Problem 2.37 On a lossless transmission line terminated in a load 2;, 2 100 $2, the standingwave ratio was measured to be 2.5. Use the Smith chart to ﬁnd the two
possible values of Zg. CHAPTER 2 61 Solution: Refer to Fig. P237. S = 2.5 is at point L1 and the constant SWR
circle is shown. 2L is real at only two places on the SWR circle, at Ll, where
2L = S = 2.5. and L2, whore 2L = 1/8 = 0.4. so 201 = ZL/zu :100 Qj2.5 = 40 ﬂ
and Zn; =ZL/3L2 =100 9/034 = 250 Q. a in a load with CHAPTER 2 69 at 0. 0.3571. — 0.500}. = 0.1071. on the WT : 0.32 — 10.39. Therefore 2 o = (0.82 — 10.39) X so 9 = . —j19.5] n. Problem 2.44 At an operating frequency of 5 GHz, :3. 509 iossless coaxial line
with insulating material having a relative permittivity er = 2.25 is terminated in an
antenna with an impedance ZL = 150 52. Use the Smith chart to ﬁnd Zin. The line
length is 30 cm. Solution: To use the Smith chart the line length must be converted into wavelengths.
Since [3 = ZTE/l and up = {Jo/B, x—ZE—Z’WP—c— 3x103mfs —004m
B (0 J8? \/2.ZS><(5><109H2) ' '
0.30m Hence, I = MA = 7.57m Since this is an integral number of half wavelengths, in=ZL= 150 Q. Section 210: Impedance Matching Problem 2.45 A 509 lossless line 0.63. long is terminated in a load with
ZL = (50 +j25) (1. At 0.31 from the load, a resistor with resistance R z 30 (2 is
connected as shown in Fig. 2—43 (P2.45(a)). Use the Smith chart to ﬁnd Zin. 2... *‘e 20 = so 12 30 9 2.1 a 50 n H— o.3t——»l~—o.3?t—l
2]. = {50 +125) :2 Figure P145: (3) Circuit for Problem 2.45. 70 CHAPTER2
... Aw”: " . "*2“ a .
”r“ u ‘  H ‘51
only" . ~ '
\ .. x
f?“ a Y
n! _t J fr. . _ a \
Kg/ar r “x?
{'3 3 / {f
f r
a z “k;
n‘ . A
‘ \t
w 5 K
3?’ *1" LOAD it
if 5 ‘
.
if 
il’é _" '"l‘; a . : "ﬁr;
l 4 l
is ' is
”la 5 . _ E
l * , T—LUAD ’  " '
'4"  IN \ €
is: a 3‘1 ’ 'I g
\‘3 ‘3?  . . f
‘3‘ g! a:
.\ . . .
it ~ ‘3‘ .x x »’
\‘ Q} z )‘1 . {32°
by  / .r '  § 6
a! ah. ah + a ,ﬂ‘
“at? In ,. ‘  . r “a";
“n.“h if . q u o‘ "JR.
“V1,, n. m. 3 r g: w‘ﬁ' U
h. ”I '3 O
"' J“ ”new?” m
 fruit? K" .3003. Figure P245: (b) Solution of Problem 2.45. Solution: Refer to Fig. P2.4S(b). Since the 309 resistor is in parallel with the input
impedance at that point, it is advantageous to convert all quantities to admittances.
‘ é _ {50+j25) Q
21 ' zo ‘ 50 o and is located at point Z—LOAD. The corresponding normalized load admittance is
at point Y—LOAD, which is at 0.39% on the WTG scale. The input admittance of
the load only at the shunt conductor is at 0.394l+ 0.3001  0.5001 = 0.1942. and is
denoted by point A. It has a value of =1+j0.5 ym = 1.37 +1045. CHAPTER 2 71 The shunt conductance has a normalized conductance 509
g— m — 1.67. The normalized admittance of the shunt conductance in parallel with the input
admittance of the load is the sum of their admittances: yinB :g‘l’yjm =1.6'7+l.37+j0.45 = 3.04+j’0.45 and is located at point B. On the WTG scale, point B is at 0.2421. The input
admittance of the entire circuit is at 0.2421+ 0.3001— 0.500).. = 0.0421. and is
denoted by point Y—IN. The corresponding normalized input impedance is at ZIN
and has a value of zin = 1.97114.
Thus,
in =zinZo = (1.9—jl.4) x 50 Q : (95—j70) Q.
P : cm 2.46 A 500 lossless line is to be matched to an antenna with 2L = (75 — 1'20} 9 using a shorted stu . ' ' . ' d the distance
between the antenna  : Solution: Refer to Fig. P2.
solutions. admittance coordinates, yL is . tted as point Y—LO .
0.041%. on the WTG scaI For the ﬁrst soluti in Fig. P2.46(a). point Y LOAD ~
at which g = l u the SWR circle of the load. Y—LOAD—IN—l
WTG scale, to the stub should be located at 0.1451 — 0.04M = i 041 from the
load (or 2' e multiple of a half wavelength further). At Y LOAD—I , b : 0.52,
so a ' b with an input admittance of ysmb : 0 ~— jl).52 is required. This point is
YSTUBIN~I and is at 0.423}. on the WTG scale. The short circuit admittance in both ﬁgures. Y 4.0111.) is at represents the point
 at 0.1451 on the CHAPTER 2 '73 Figure P246: (b) Second soluiion to Problem 2.46. l  gcr). Problem 2.47 Repeat Problem 2.46 for a load with ZL = (100+j50} (2. Solution: Refer to Fig. P2.47(a) and Fig. P2.47(b), which represent two different
solutions. a_ﬁm+ﬁoo “:f‘ SM} and is located at point Z—LOAD in both ﬁgures. Since it is advantageous to Work in admittance coordimtes, yL is plotted as point YLOAD in both ﬁgures. Y«LOAD is at
0.463% on the WTG scale. =2+j1 74 CHAPTER 2 1302i
. vow ' 25$:
4..._ I Figure P247: (:1) First solution to Problem 2.47. For the ﬁrst solution in Fig. P2.47(a), point Y—LOAD—ﬂV—l represents the point
at which g = l on the SWR circle of the load. YLOAD—DVl is at 0.162% on the
WTG scale, so the stub should be located at 0.1627» — 0.4637t+ 0.500). = 0.1991
from the load (or some multiple of a half wavelength further). At YLOADINl,
b = I, so a stub with an input admittance of yam], = 0 — fl is required. This point
is YSTUB—W—l and is at 0.375}. on the WTG scale. The short circuit admittance
is denoted by point Y—SHT, located at 0.2501. Therefore, the short stub must be
0.3751 — 0.2501 = 0.1251 long (or some multiple of a half wavelength longer). For the second solution in Fig. P2.47(b), point YLOADINZ represents the point
at which g = 1 on the SWR circle of the load. Y LOADIN2 is at 0.338% on the CHAPTER 2 7S , “m:  win" or H U' 
1— “ IN ‘1‘ 44
‘ ‘ «a...» + ‘
IR: rﬁw—H : Figure P147: (1)) Second solution to Problem 2.47. WTG scale, so the stub should be located at 0.33% — 0.463IL+ 0.500% = 0.3751
from the load (or some multiple of a half wavelength further). At YLOADINZ,
b = +1, so a stub with an input admittance ofym]J = 0 +j1 is required. This point
is Y—SYUB—INZ and is at 0.1251 on the WTG scale. The short circuit admittance
is denoted by point Y—SHT, located at 0.2501. Therefore, the short stub must be
0.125%. — 0.2507L+ 0.500%. = 0.375}. long (or some multiple of a half wavelength
longer). 0 ﬁnd Zia of the feed line shown in Fig. 2—44 364 CHAPTER 8 Problem 8.9 The three regions shown in Fig. 832 (P83) contain perfect
dielectrics. For a wave in medium I incident normally upon the boundary at z = —a’,
what combination of 3,2 and (I produce no reﬂection? Express your answers in terms
of an , er3 and the osciliation frequency of the wave, f. H—d—d
Medium 1 _—.___. Figure P89: Three dielectric regions. Solution: By analogy with the transmissionline case, there will be 110 reﬂection at
z : —d if medium 2 acts as a quarterwave transformer, which requires that 12
d r r
and
712 = vnms.
The second condition may be rewritten as
1/2
Tia [ 110 110
= — — 1 01' 8r : v8: 8 ,
«8—1; ﬂ «6T3 2 I If]
M c c
FL:1 Z EVE: renew“ and
c d:————.
4f(er1€f3)]/4 ...
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 Winter '07
 Williams

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