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Unformatted text preview: Problem 9.5 A 2mlong centerfed dipole antenna operates in the AM broadcast
band at 1 MHZ. The dipole is made of copper wire with a radius of 1 mm.
(3) Determine the radiation efﬁciency of the antenna.
(b) What is the antenna gain in dB?
(c) What antenna current is required so that the antenna would radiate 80 W, and
how much power will the generator have to supply to the antenna? Solution: (a) Following Example 93, A : C/f : (3 X '108 m/s)/('106 Hz) : 300 m. As
UK : (2 m)/(300 m) : 6.7 X 10—3, this antenna is a short (Hertzian) dipole. Thus,
from respectively Eqs. (9.35), (9.32), and (9.31), z 2 ., a
R,ad : 80n2(i) : 80n~(6.7 >< '10—sz : 35 (m9), 6 i7
R1085 : f ' Mtfyc : 2m TE(10 Hz)(43‘t>< 10 H/m) : 83 (m9):
2R0 6C 2TE(10_3 m) 5.8 X 107 S/m Rm, _ 35 m9
Rmd +1210... _ 35 mg + 83 mo = 29.7%. a: (b) From Example 92, a Hertzian dipole has a directivity of 1.5. The gain, from
Eq. (9.29), is G: (ED : 0.297 X 1.5 : 0.44 : —3.5 dB.
(c) From Eq. (9.30a), 213ml 2(80 W)
Rmd 35 m9 and from Eq. (9.31), Problem 9.7 An antenna with a pattern solid angle of 1.5 (sr) radiates 60 W of
power. At a range of '1 km, what is the maximum power density radiated by the
antenna? Solution: From Eq. (9.23), D : 4TE/Qp, and from Eq. (9.24), D : 47ER25max/Pl3d.
Combining these two equations gives . _ Plad _ 60 _ 75 2
Smax—QPRZ —m—4X 10 (W/m Problem 9.13 For the short dipole with length i such that 1 << 1, instead of treating
the current Hz) as constant along the dipole, as was done in Section 91, a more realistic approximation that insures that the current goes to zero at the ends is to
describe [(2) by the triangular function 32), 10mm, forogzgz/z,
— 170(1l—22/flj for—i/zgzgoj as shown in Fig. 936 (P913). Use this current distribution to determine (a) the far
ﬁeld E(Rj 8, (b) the power density S(R,Bj¢), (c) the directivity D, and (d) the
radiation resistance Rmd. Solution:
(21) When the current along the dipole was assumed to be constant and equal to 10,
the vector potential was given by Eq. (9.3) as: N 7ij 3/2
A(R) :2ﬂ (L R 10d2. 4313 4/2 If the triangular current function is assumed instead, then 10 in the above expression
should be replaced with the given expression. Hence, N A 0 e—f'kR 1/2 22 0 22 A #010] e—ij
A: i 1 177 d 1 i d :
l4a< R / Z+/;/2 +/ Z 2 8a R ’ \\/I(Z)
\\
x
i I» 10
x
I; Figure P9.l3: Triangular current distribution on a short dipole (Problem 9.13). which is half that obtained for the constantcurrent case given by Eq. (9.3). Hence,
the expression given by (9.9a) need only be modiﬁed by the factor of l / 2: ~ M, . 'j {k rij
E 29139 : 9"7 Ugﬂno (6 R )sinB. (b) The corresponding power density is t IEBI2 1103721512  v
s R e : : — be.
( ’ l 2110 1287:3R2 3m (c) The power density is 4 times smaller than that for the constant current case, but
the reduction is true for all directions. Hence, D remains unchanged at 1.5.
((1) Since SIR, 8) is 4 times smaller, the total radiated power Prad is 4times smaller. Consequently, Rmd : ZEN/15 is 4 times smaller than the expression given
by Eq. (9.35); that is, Rmd : 2031:2(1 m2 (o). Problem 9.18 Determine the effective area of a halfwave dipole antenna at
"100 MHZ, and compare it to its physical cross section if the wire diameter is 2 cm. Solution: At f: 100 MHz, A = C/f : (3 x 108 m/s)/(100 X 106 Hz) 2 3 m. From
Eq. (9.47), a half wave dipole has a directivity of D = 1.64. From Eq. (9.64),
A. : 1813/47: : (3 m)2 X 1.64/41: : 1.17 m2. The physical cross section is: AP = id = ékd = m)(2 X 10—2 m) = 0.03 m2.
Hence, Ae/Ap : '39. ...
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 Spring '08
 HUFFAKER

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