20102ee161_1_Homework07_solution

20102ee161_1_Homework07_solution - Problem 9.5 A 2-m-long...

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Unformatted text preview: Problem 9.5 A 2-m-long center-fed 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 9-3, 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 9-2, 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/Pl-3d. Combining these two equations gives . _ Pl-ad _ 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 9-1, 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(1-l—22/flj for—i/zgzgoj as shown in Fig. 9-36 (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 constant-current 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 4-times 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 half-wave 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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20102ee161_1_Homework07_solution - Problem 9.5 A 2-m-long...

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