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Unformatted text preview: 405 Chapter 9: Radiation and Antennas
Lesson #61
Chapter  Section: 91 Topics: Retarded potential, short dipole Highlights: Radiation by short dipole Farfield distance Special Illustrations: Exercise 9.1 406 Lesson #62
Chapter  Section: 92 Topics: Radiation characteristics Highlights: Antenna pattern Antenna directivity Antenna gain Special Illustrations: Example 92 Example 93 407 Lesson #63
Chapter  Section: 93 and 94 Topics: Halfwave dipole Highlights: Radiation pattern Directivity Radiation resistance Special Illustrations: CDROM Module 9.1 CDROM Demo 9.1 408 Lesson #64
Chapter  Section: 95, 96 Topics: Effective area, Friis formula Highlights: Receiving aperture of an antenna Relation of aperture to directivity Friis formula Special Illustrations: Example 95 Demo 9.2 409 Lessons #65 and 66
Chapter  Sections: 97 and 98 Topics: Aperture antennas Highlights: Aperture illumination Rectangular aperture Beamwidth and directivity Special Illustrations: CDROM Demo 9.3 410 Lessons #6769
Chapter  Sections: 99 to 911 Topics: Antenna arrays Highlights: Array factor Multiplication principle Electronic scanning Special Illustrations: CDROM Demo 9.4 The array pattern of an equallyspaced linear array can be steered in direction by applying linear phase across the array as shown. Note that = kd cos 0, with 0 measured from the +zaxis. Display the array pattern for the following values of the beam center angle: 0 = 90o (broadside) 0 = 60o (30o above xaxis) 0 = 30o (60o above xaxis) 0 = 120o (30o below xaxis) 0 = 150o (60o below xaxis) CHAPTER 9 411 Chapter 9
Sections 91 and 92: Short Dipole and Antenna Radiation Characteristics
Problem 9.1 A centerfed Hertzian dipole is excited by a current I0 20 A. If the dipole is 50 in length, determine the maximum radiated power density at a distance of 1 km. Solution: From Eq. (9.14), the maximum power density radiated by a Hertzian dipole is given by Problem 9.2 A 1mlong dipole is excited by a 1MHz current with an amplitude of 12 A. What is the average power density radiated by the dipole at a distance of 5 km in a direction that is 45 from the dipole axis? Problem 9.3 Determine the (a) direction of maximum radiation, (b) directivity, (c) beam solid angle, and (d) halfpower beamwidth in the xz plane for an antenna whose normalized radiation intensity is given by Suggestion: Sketch the pattern prior to calculating the desired quantities. Solution: The direction of maximum radiation is a circular cone 120 wide centered ^ around the zaxis. From Eq. (9.23), 2 2 60 0 0 sin d d cos 60 0 1 2 1 D 4 4 F d 4 4 2 F 1 0 for 0 60 and 0 elsewhere. 2 4 6 dB SR 2 0 k2 I0 l 2 sin2 322 R2 120 2 300 2 122 322 5 103 2 12 sin2 45 1 51 10 9 (W/m2 ) Solution: At 1 MHz, c f 3 108 106 300 m. Hence l therefore the antenna is a Hertzian dipole. From Eq. (9.12), 76 10 6 W/m2 S0 322 103 2 76 (W/m2 ) 2 0 k2 I0 l 2 322 R2 377 2 2 202 50 2 1 300, and 412 p 4 sr D 4 sr 4 (sr) CHAPTER 9 Problem 9.4 F Repeat Problem 9.3 for an antenna with 2 3 2 2 cos d 0 sin d 1 1 4 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. (a) 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: In the xz plane, 1 sin2 135 2. 0 and the half power beamwidth is 90 , since sin2 45 p 4 sr D 4 sr 6 2 (sr) 3 x 1 1 2 1 2 sin 2 2 2 x3 3 1 1 2 1 cos 2 d 2 1 22 D 4 4 F d 4 2 2 2 2 0 sin cos sin d d 4 4 1 x2 dx 4 4 3 6 7 8 dB Solution: The direction of maximum radiation is the 0). From Eq. (9.23), ^ xaxis (where sin2 cos2 0 The half power beamwidth is 120 . for 0 and elsewhere. 2 2 2 and CHAPTER 9 413 (a) Following Example 93, c f 3 108 m/s 106 Hz 300 m. As 3 , this antenna is a short (Hertzian) dipole. Thus, l 2 m 300 m 6 7 10 from respectively Eqs. (9.35), (9.32), and (9.31), (b) From Example 92, a Hertzian dipole has a directivity of 1.5. The gain, from Eq. (9.29), is G D 0 297 1 5 0 44 3 5 dB. (c) From Eq. (9.30a), and from Eq. (9.31), Problem 9.6 Repeat Problem 9.5 for a 20cmlong antenna operating at 5 MHz. Solution: (a) At 5 MHz, c f 3 108 5 106 60 m. As l 0 2 60 3 , the antenna length satisfies the condition of a short dipole. From 3 33 10 Eqs. (9.35), (9.32), and (9.31), 3 I0 2Prad Rrad 2 80 8 76 10 135 2 A (b) For Hertzian dipole, D (c) From Eq. (9.30a), 1 5, and G D 0 32 15 0 48 3 2dB. Rloss l f uc 2a c Rrad Rrad Rloss 02 5 106 4 10 2 10 3 5 8 107 8 76 0 32 or 32% 8 76 18 57 7 Rrad 802 802 l 2 3 33 10 3 2 8 76 (m) 18 57 (m) Pt Prad 80 W 0 297 269 W I0 2Prad Rrad 2 80 W 35 m 67 6 A Rloss l f c 2a c Rrad Rrad Rloss 2m 106 Hz 4 10 7 H/m 2 10 3 m 5 8 107 S/m 35 m 29 7% 35 m 83 m Rrad 802 802 6 7 l 2 10 3 2 35 (m) 83 (m) 414 CHAPTER 9 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? 2 Problem 9.8 An antenna with a radiation efficiency of 90% has a directivity of 7.0 dB. What is its gain in dB? Alternatively, Problem 9.9 The radiation pattern of a circular parabolicreflector antenna consists of a circular major lobe with a halfpower beamwidth of 3 and a few minor lobes. Ignoring the minor lobes, obtain an estimate for the antenna directivity in dB. Problem 9.10 The normalized radiation intensity of a certain antenna is given by where is in radians. Determine: (a) the halfpower beamwidth, (b) the pattern solid angle, F exp 202 for 0 D dB 10 log D 10 log 4 58 In dB, 103 36 61 dB 4 58 103 Solution: A circular lobe means that xz yz we have 4 4 D xz yz 0 052 2 3 0 052 rad. Using Eq. (9.26), G dB dB D dB 10 log 0 9 70 0 46 G D 09 50 Solution: D 7 0 dB corresponds to D 5 0. 45 6 54 dB 70 Smax Prad p R 2 60 1 5 103 4 10 5 (W/m2 ) 6 54 dB Solution: From Eq. (9.23), D 4 p , and from Eq. (9.24), D Combining these two equations gives 4R2 Smax Prad . CHAPTER 9 415 F() 1 0.5  16 1 0 2 16 Figure P9.10: F versus . (c) the antenna directivity. ln exp 202 ln 0 5 20 2 0 693 Hence, 2 0 186 0 372 radians 21 31 . 0 693 20 1 2 F exp 202 05 0 186 radians Solution: (a) Since F is independent of , the beam is symmetrical about z 0 5, we have setting F 0. Upon 416 (b) By Eq. (9.21), CHAPTER 9 0 Numerical evaluation yields (c) Sections 93 and 94: Dipole Antennas
Problem 9.11 Repeat Problem 9.5 for a 1mlong halfwave dipole that operates in the FM/TV broadcast band at 150 MHz. Solution: (a) Following Example 93, (b) From Eq. (9.47), a halfwave dipole has a directivity of 1.64. The gain, from Eq. (9.29), is G D 0 993 1 64 1 63 2 1 dB. (c) From Eq. (9.30a), I0 2Prad Rrad 2 80 W 73 99 3% 1 48 A Rloss l f c 2a c Rrad Rrad Rloss 1m 2 10 3 m 73 73 0 5 150 106 Hz 4 10 5 8 107 S/m Rrad 73 As l 1m 2m (9.32), and (9.31), 1 2, this antenna is a halfwave dipole. Thus, from Eq. (9.48), c f 3 108 m/s 150 106 Hz D 4 p 4 0 156 p 0 156 sr 80 55 2 exp 202 sin d 0 0 exp 4 2 p F sin d d 202 sin d d 2m 7 H/m 05 CHAPTER 9
and from Eq. (9.31), 417 Problem 9.12 Assuming the loss resistance of a halfwave dipole antenna to be negligibly small and ignoring the reactance component of its antenna impedance, calculate the standing wave ratio on a 50 transmission line connected to the dipole antenna. Solution: According to Eq. (9.48), a half wave dipole has a radiation resistance of 73 . To the transmission line, this behaves as a load, so the reflection coefficient is and the standing wave ratio is Problem 9.13 For the short dipole with length l such that l , instead of treating the current I z 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 I z by the triangular function as shown in Fig. 936 (P9.13). Use this current distribution to determine (a) the farfield E R , (b) the power density S R , (c) the directivity D, and (d) the radiation resistance Rrad . Solution: (a) When the current along the dipole was assumed to be constant and equal to I 0 , the vector potential was given by Eq. (9.3) as: l 2 If the triangular current function is assumed instead, then I0 in the above expression should be replaced with the given expression. Hence, 0 l 2 R R A ^ z I0 1 1 ^ z 0 4 e jkR l 2 2z dz l 0 2z dz l 0 I0 l 8 R AR ^ z 0 4 e jkR l 2 I0 dz I z I0 1 I0 1 2z l 2z l for 0 z l 2 for l 2 z 0 S 1 1 1 1 0 187 0 187 1 46 Rrad Rrad Z0 Z0 73 73 50 50 0 187 e
jkR Pt Prad 80 W 0 993 80 4 W 418
I(z)
~ CHAPTER 9 l I0 Figure P9.13: 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 1 2: R (b) The corresponding power density is (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. (d) Since S R is 4 times smaller, the total radiated power Prad is 4times 2 smaller. Consequently, Rrad 2Prad I0 is 4 times smaller than the expression given by Eq. (9.35); that is, Problem 9.14 For a dipole antenna of length l 3 2, (a) determine the directions of maximum radiation, (b) obtain an expression for S max , and (c) generate a plot of the normalized radiation pattern F . Compare your pattern with that shown in Fig. 9.17(c). Solution: (a) From Eq. (9.56), S for an arbitrary length dipole is given by S 2 15I0 cos R2 l cos cos sin Rrad 202 l 2 () l SR E 2 20 2 0 k2 I0 l 2 sin2 1282 R2 2 E ^ E ^ jI0 lk0 8 e jkR sin CHAPTER 9 419 Solving for the directions of maximum radiation numerically yields two maximum directions of radiation given by (c) The normalized radiation pattern is given by Eq. (9.13) as Using the expression for S from part (a) with the value of S max found in part (b), The normalized radiation pattern is shown in Fig. P9.14, which is identical to that shown in Fig. 9.17(c). F cos 3 cos 1 2 1 96 sin F S Smax Smax 2 15I0 1 96 R2 2 (b) From the numerical results, it was found that S Thus, 2 15I0 R2 1 96 at max . max1 42 6 max2 137 4 S 2 15I0 cos 3 cos 2 R2 sin For l 3 2, S becomes
2 420 CHAPTER 9 x z Figure P9.14: Radiation pattern of dipole of length 3 2. Solving for the directions of maximum radiation numerically yields (b) From the numerical results, it was found that S 2 15I0 R2 2 91 at max . max1 90 max2 270 sin 2 15I0 R2 cos 3 4 cos 1 2 2 S 2 15I0 R2 Solution: (a) For l 3 4, Eq. (9.56) becomes cos
3 4 cos cos sin 3 4 2 Problem 9.15 Repeat parts (a)(c) of Problem 9.14 for a dipole of length l 3 4. CHAPTER 9
Thus, 421 x (c) The normalized radiation pattern is given by Eq. (9.13) as Using the expression for S from part (a) with the value of S max found in part (b), The normalized radiation pattern is shown in Fig. P9.15. Problem 9.16 Repeat parts (a)(c) of Problem 9.14 for a dipole of length l sin F 1 2 91 cos 3 4 cos F S Smax 1 2 2 Figure P9.15: Radiation pattern of dipole of length l Smax 2 15I0 2 91 R2 Z 3 4. . 422 CHAPTER 9
, Eq. (9.56) becomes Solving for the directions of maximum radiation numerically yields x Z 2 (b) From the numerical results, it was found that S 15I0 R2 4 at max . Thus, 2 60I0 Smax R2 (c) The normalized radiation pattern is given by Eq. (9.13), as Using the expression for S from part (a) with the value of S max found in part (b), F 1 cos cos 4 sin F S Smax 1 2 max1 90 max2 270 Figure P9.16: Radiation pattern of dipole of length l . S 2 15I0 R2 cos cos cos sin 2 2 15I0 R2 cos cos sin Solution: For l 1 2 CHAPTER 9
The normalized radiation pattern is shown in Fig. P9.16. 423 Problem 9.17 A car antenna is a vertical monopole over a conducting surface. Repeat Problem 9.5 for a 1mlong car antenna operating at 1 MHz. The antenna wire is made of aluminum with c 0 and c 3 5 107 S/m, and its diameter is 1 cm. Solution: 3 108 m/s 106 Hz 300 m. As (a) Following Example 93, c f l 2 1 m 300 m 0 0067, this antenna is a short (Hertzian) monopole. From Section 93.3, the radiation resistance of a monopole is half that for a corresponding dipole. Thus, (b) From Example 92, a Hertzian dipole has a directivity of 1.5. The gain, from 0 3 dB. Eq. (9.29), is G D 0 62 1 5 0 93 (c) From Eq. (9.30a), and from Eq. (9.31), Sections 95 and 96: Effective Area and Friis Formula
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, c f 3 108 m/s 100 106 Hz 3 m. From Eq. (9.47), a half wave dipole has a directivity of D 1 64. From Eq. (9.64), Ae 2 D 4 3 m 2 1 64 4 1 17 m2 . Pt Prad 80 W 0 62 129 2 W I0 2Prad Rrad 2 80 W 17 7 m 95 A Rloss l f c 2a c Rrad Rrad Rloss 1m 106 Hz 4 10 7 H/m 2 m 10 3 5 107 S/m 17 7 m 62% 17 7 m 10 7 m Rrad 2 1 2 80 402 0 0067 l 2 2 17 7 (m) 10 7 m 424 CHAPTER 9 Problem 9.19 A 3GHz lineofsight microwave communication link consists of two lossless parabolic dish antennas, each 1 m in diameter. If the receive antenna requires 10 nW of receive power for good reception and the distance between the antennas is 40 km, how much power should be transmitted? Solution: At f 3 GHz, c f 3 108 m/s 3 109 Hz 0 10 m. Solving the Friis transmission formula (Eq. (9.75)) for the transmitted power: Problem 9.20 A halfwave dipole TV broadcast antenna transmits 1 kW at 50 MHz. What is the power received by a home television antenna with 3dB gain if located at a distance of 30 km? Solution: At f 50 MHz, c f 3 108 m/s 50 106 Hz 6 m, for which a half wave dipole, or larger antenna, is very reasonable to construct. Assuming the TV transmitter to have a vertical half wave dipole, its gain in the direction of the home would be Gt 1 64. The home antenna has a gain of G r 3 dB 2. From the Friis transmission formula (Eq. (9.75)):
2 2 R 2 2 Problem 9.21 A 150MHz communication link consists of two vertical halfwave dipole antennas separated by 2 km. The antennas are lossless, the signal occupies a bandwidth of 3 MHz, the system noise temperature of the receiver is 600 K, and the desired signaltonoise ratio is 17 dB. What transmitter power is required? Solution: From Eq. (9.77), the receiver noise power is Prec Sn Pn 50 2 48 For a signal to noise ratio Sn 17 dB 50, the received power must be at least 10
14 W 1 24 10 12 W Pn KTsys B 1 38 10 23 600 3 106 4 4 30 103 m 2 48 10 14 W Prec Pt 103 2 G r G t 6m 2 1 64 2 83 10 7 W 1 1 4 1m 2 4 1m 2 10 8 0 100 m 2 Pt Prec 2 R 2 t r A t A r 40 103 m
2 25 9 10 2 W 259 mW The physical cross section is: A p Hence, Ae Ap 39. ld 1 2 d 1 2 3m 2 10 2 m 0 03 m2 . CHAPTER 9 425 Since the two antennas are halfwave dipoles, Eq. (9.47) states D t Dr 1 64, and since the antennas are both lossless, G t Dt and Gr Dr . Since the operating 3 108 m/s 150 106 Hz 2 m. Solving frequency is f 150 MHz, c f the Friis transmission formula (Eq. (9.75)) for the transmitted power: Problem 9.22 Consider the communication system shown in Fig. 937 (P9.22), with all components properly matched. If Pt 10 W and f 6 GHz: (a) what is the power density at the receiving antenna (assuming proper alignment of antennas)? (b) What is the received power? (c) If Tsys 1 000 K and the receiver bandwidth is 20 MHz, what is the signal to noise ratio in dB?
Gt = 20 dB Gr = 23 dB Pt Tx 20 km Figure P9.22: Communication system of Problem 9.22. 2 (b) (c) Pn KTsys B 1 38 10 23 103 2 107 2 76 10 13 W Prec Pt Gt Gr 10 100 200 4R 2 5 10 2 4 2 104 2 7 92 Sr Gt Pt 4R2 102 10 4 2 104 2 10 7 Solution: (a) Gt 20 dB 100, Gr 23 dB 200, and c f 2m 2 1 64 1 64 Prec Rx 5 cm. From Eq. (9.72), (W/m2 ) 10 Pt Prec 1 24 10 12 4 2 R2 2 G r G t 4 2 2 103 m 2 75 (W) 9 W 426 Sn CHAPTER 9 13 Sections 97 and 98: Radiation by Apertures Solution: The radiation intensity of a uniformly illuminated antenna is given by Eq. (9.90): F sinc2 lx sin sinc2 with and Problem 9.24 The 10dB beamwidth is the beam size between the angles at which F is 10 dB below its peak value. Determine the 10dB beamwidth in the xz plane for a uniformly illuminated aperture with length l x 10. The peak value of F is 1, and the 10dB level below the peak corresponds to when F 0 1 (because 10 log 0 1 10 dB). Hence, we set F 0 1 and solve for : 01 sinc2 10 sin F sinc2 lx sin sinc2 10 sin Solution: For a uniformly illuminated antenna of length l x null 2 5 73 sin 1 or 1 20 1 20 sin 2 87 The first zero of the sinc function occurs when For lx 20, 20 sin lx sin 1, as shown in Fig. 923. Hence, 10 Eq. (9.90) gives Problem 9.23 A uniformly illuminated aperture is of length l x the beamwidth between first nulls in the xz plane. 20. Determine Prec Pn 7 92 10 2 76 10 9 2 87 104 44 6 dB CHAPTER 9 427 From tabulated values of the sinc function, it follows that the solution of this equation is 10 sin 2 319 or Hence, the 10dB beamwidth is Problem 9.25 A uniformly illuminated rectangular aperture situated in the xy plane is 2 m high (along x) and 1 m wide (along y). If f 10 GHz, determine (a) the beamwidths of the radiation pattern in the elevation plane (xz plane) and the azimuth plane (yz plane), and (b) the antenna directivity D in dB. Solution: From Eqs. (9.94a), (9.94b), and (9.96), Problem 9.26 An antenna with a circular aperture has a circular beam with a beamwidth of 3 at 20 GHz. (a) What is the antenna directivity in dB? (b) If the antenna area is doubled, what would be the new directivity and new beamwidth? (c) If the aperture is kept the same as in (a), but the frequency is doubled to 40 GHz, what would the directivity and beamwidth become then? Solution: (a) From Eq. (9.96), 180
2 3 D 4 2 4 4 59 103 36 6 dB D 104 yz 1 51 xz 0 88 lx 0 88 ly 4 xz yz 3 10 2 1 32 10 2 rad 2 0 88 3 10 2 2 64 10 2 rad 1 4 3 61 1 32 10 2 2 64 10 2 0 88 2 8 46 0 75 45 6 dB 4 23 428 (b) If area is doubled, it means the diameter is increased by beamwidth decreases by 2 to CHAPTER 9 22 2 The directivity increases by a factor of 2, or 3 dB, to D 36 6 3 39 6 dB. (c) If f is doubled, becomes half as long, which means that the diameter to wavelength ratio is twice as large. Consequently, the beamwidth is half as wide: 3 15 2 and D is four times as large, or 6 dB greater, D 36 6 Problem 9.27 A 94GHz automobile collisionavoidance radar uses a rectangularaperture antenna placed above the car's bumper. If the antenna is 1 m in length and 10 cm in height, (a) what are its elevation and azimuth beamwidths? (b) what is the horizontal extent of the beam at a distance of 300 m? Solution: 3 108 94 109 3 2 mm. The elevation (a) At 94 GHz, 2 rad beamwidth is e 0 1 m 3 2 10 1 8 . The azimuth beamwidth is a 1 m 3 2 10 3 rad 0 18 . (b) At a distance of 300 m, the horizontal extent of the beam is Problem 9.28 A microwave telescope consisting of a very sensitive receiver connected to a 100m parabolicdish antenna is used to measure the energy radiated by astronomical objects at 20 GHz. If the antenna beam is directed toward the moon and the moon extends over a planar angle of 0 5 from Earth, what fraction of the moon's cross section will be occupied by the beam? Solution: 3 ant moon 2 1 5 10 8 73 10 4 2 03 For the moon, moon 0 5 180 cross section occupied by the beam is 8 73 10 3 rad. Fraction of the moon's 10 3 or 0 03% ant d 15 10 100 2 15 10 4 rad y a R 32 10 3 300 0 96 m 6 42 6 dB. 3 2, and therefore the CHAPTER 9 429 0.5 Figure P9.28: Antenna beam viewing the moon. Sections 99 to 911: Antenna Arrays
Problem 9.29 A twoelement array consisting of two isotropic antennas separated by a distance d along the zaxis is placed in a coordinate system whose zaxis points eastward and whose xaxis points toward the zenith. If a 0 and a1 are the amplitudes of the excitations of the antennas at z 0 and at z d respectively, and if is the phase of the excitation of the antenna at z d relative to that of the other antenna, find the array factor and plot the pattern in the xz plane for (a) a0 a1 1, 4, and d 2, (b) a0 1, a1 2, 0, and d , (c) a0 a1 1, 2, and d 2, (d) a0 a1 , a1 2, 4, and d 2, and (e) a0 a1 , a1 2, 2, and d 4. Solution: (a) Employing Eq. (9.110), Fa 1 2 ji jikd cos A plot of this array factor pattern is shown in Fig. P9.29(a). 1 ej cos 4 2 4 cos2 1 ej 2 2 cos 4 2 i 0 ai e e 4 cos 8 1 430 CHAPTER 9 x z Figure P9.29: (a) Array factor in the elevation plane for Problem 9.29(a). (b) Employing Eq. (9.110), Fa A plot of this array factor pattern is shown in Fig. P9.29(b). 1 1 2e j 2 cos 0 2 2e j2cos 2 5 4 cos 2 cos i 0 ai e 1 2 ji jikd cos e CHAPTER 9 431 x z Figure P9.29: (b) Array factor in the elevation plane for Problem 9.29(b). A plot of the array factor is shown in Fig. P9.29(c). 4 cos2 cos 2 4 1 ej cos 2 2 1 e j 2 j 2 2 cos e i 0 Fa d ai e 1 2 ji jikd cos e 2 (c) Employing Eq. (9.110), and setting a0 2, we have a1 1, 0, 1 2 and 432 CHAPTER 9 x Z Figure P9.29: (c) Array factor in the elevation plane for Problem 9.29(c). A plot of the array factor is shown in Fig. P9.29(d). 5 4 cos cos 4 1 2e j cos 4 2 1 2e j 4 e j 2 2 cos i 0 Fa ai e 1 2 ji jikd cos e 2 (d) Employing Eq. (9.110), and setting a0 and d 2, we have 1, a1 2, 0 0, 1 4, CHAPTER 9 433 x Z Figure P9.29: (d) Array factor in the elevation plane for Problem 9.29(d). A plot of the array factor is shown in Fig. P9.29(e). 5 4 cos cos 2 2 1 2e j cos 2 2 1 2e j 2 e j 2 4 cos i 0 Fa ai e 1 2 ji jikd cos e 2 5 4 sin cos 2 (e) Employing Eq. (9.110), and setting a0 and d 4, we have 1, a1 2, 0 0, 1 2, 434 CHAPTER 9 x Z Figure P9.29: (e) Array factor in the elevation plane for Problem 9.29(e). Problem 9.30 If the antennas in part (a) of Problem 9.29 are parallel vertical Hertzian dipoles with axes along the xdirection, determine the normalized radiation intensity in the xz plane and plot it.
x ' z d Figure P9.30: (a) Two vertical dipoles of Problem 9.30. CHAPTER 9 435 x z Figure P9.30: (b) Pattern factor in the elevation plane of the array in Problem 9.30(a). Solution: The power density radiated by a Hertzian dipole is given from Eq. (9.12) by Se S0 sin2 , where is the angle measured from the dipole axis, which in the present case is the xaxis (Fig. P9.30). Hence, 2 and Se S0 sin2 1 S0 cos2 . Then, from 2 Eq. (9.108), the total power density is the product of the element pattern and the array factor. From part (a) of the previous problem: This function has a maximum value of 3 52S0 and it occurs at max 135 5 . The maximum must be found by trial and error. A plot of the normalized array antenna pattern is shown in Fig. P9.30. S Se Fa 4S0 cos2 cos2 4 cos 8 1 436 CHAPTER 9 Problem 9.31 Consider the twoelement dipole array of Fig. 9.29(a). If the two dipoles are excited with identical feeding coefficients (a 0 a1 1 and 0 1 0), choose d such that the array factor has a maximum at 45 . Fa is a maximum when the argument of the cosine function is zero or a multiple of . Hence, for a maximum at 45 , The first value of n, namely n 0, does not provide a useful solution because it requires d to be zero, which means that the two elements are at the same location. While this gives a maximum at 45 , it also gives the same maximum at all angles in the yz plane because the twoelement array will have become a single element with an azimuthally symmetric pattern. The value n 1 leads to Problem 9.32 Choose d so that the array pattern of the array of Problem 9.31 has a null, rather than a maximum, at 45 . For n 0, d 1 2 cos 45 0 707 d cos 45 2 n n 012 Fa is equal to zero when the argument of the cosine function is Hence, for a null at 45 , Fa 1 ej 2d cos 2 4 cos2 Solution: With a0 a1 1 and 0 1 0, d cos d 1 cos 45 1 414 d cos 45 n n 012 Fa 1 ej 2d cos 2 4 cos2 Solution: With a0 a1 1 and 0 1 0, d cos 2 n . CHAPTER 9 437 Problem 9.33 Find and plot the normalized array factor and determine the halfpower beamwidth for a fiveelement linear array excited with equal phase and a uniform amplitude distribution. The interelement spacing is 3 4. Solution: Using Eq. (9.121), and this pattern is shown in Fig. P9.33. The peak values of the pattern occur at 90 . From numerical values of the pattern, the angles at which Fan 05 are approximately 6.75 on either side of the peaks. Hence, 13 5 . x z Figure P9.33: Normalized array pattern of a 5element array with uniform amplitude distribution in Problem 9.33. Problem 9.34 A threeelement linear array of isotropic sources aligned along the zaxis has an interelement spacing of 4 Fig. 938 (P9.34). The amplitude excitation of the center element is twice that of the bottom and top elements and the phases Fan sin2 Nd cos N 2 sin2 d cos sin2 15 4 cos 25 sin2 3 4 cos 438 CHAPTER 9 are 2 for the bottom element and 2 for the top element, relative to that of the center element. Determine the array factor and plot it in the elevation plane. 1 /4 2 /4 1 /2 0 /2 Figure P9.34: (a) Threeelement array of Problem 9.34. Solution: From Eq. (9.110), Fa This normalized array factor is shown in Fig. P9.34. Fan 1 4 1 cos 1 cos 2 41 cos 1 2 1 2 1 cos 2 e e 2 e j1 e j 2 cos 2 j 2 j 2 cos e j 2 e j 2 cos 2 ej 2e j1 e j ej 1 2 2 4 cos a0 e j0 a1 e j1 e jkd cos i 0 ai e 2 2 ji jikd cos e z a2 e j2 e j2kd cos 2 1 2 e j2 2 4 cos 2 CHAPTER 9 439 x z Figure P9.34: (b) Normalized array pattern of the 3element array of Problem 9.34. Problem 9.35 An eightelement linear array with 2 spacing is excited with equal amplitudes. To steer the main beam to a direction 60 below the broadside direction, what should be the incremental phase delay between adjacent elements? Also, give the expression for the array factor and plot the pattern. Combining Eq. (9.126) with (9.127) gives The pattern is shown in Fig. P9.35. cos 0 64 sin2 1 2 cos 1 2 3 Fan 1 2 Nkd cos 2 sin2 1 kd cos N 2 sin2 cos 0 sin2 4 cos kd cos 0 2 72 rad 2 cos 150 2 Solution: Since broadside corresponds to 0 150 . From Eq. (9.125), 90 , 60 below broadside is 155 9 1 2 3 440 CHAPTER 9 x z Figure P9.35: Pattern of the array of Problem 9.35. Problem 9.36 A linear array arranged along the zaxis consists of 12 equally spaced elements with d 2. Choose an appropriate incremental phase delay so as to steer the main beam to a direction 30 above the broadside direction. Provide an expression for the array factor of the steered antenna and plot the pattern. From the pattern, estimate the beamwidth. Combining Eq. (9.126) with (9.127) gives cos 0 Fan 1 2 12kd cos 144 sin2 1 kd cos 2 sin2 cos 0 sin2 6 cos 0 5 144 sin2 cos 0 5 2 kd cos 0 1 57 rad 2 cos 60 2 90 Solution: Since broadside corresponds to From Eq. (9.125), 90 , 30 above broadside is 0 60 . CHAPTER 9 441 x z Figure P9.36: Array pattern of Problem 9.36. Problem 9.37 A 50cm long dipole is excited by a sinusoidally varying current with an amplitude I0 5 A. Determine the time average power radiated by the dipole if the oscillating frequency is: (a) 1 MHz, (b) 300 MHz. Solution: (a) At 1 MHz, 108 300 m 106 Hence, the dipole length satisfies the "short" dipole criterion (l 3 The pattern is shown in Fig. P9.36. The beamwidth is 10 . 50). 442 Using (9.34), CHAPTER 9 (b) At 300 MHz, 3 108 1m 3 108 Hence, the dipole is 2 in length, in which case we can use (9.46) to calculate Prad : Problem 9.38 The configuration shown in the figure depicts two vertically oriented halfwave dipole antennas pointed towards each other, with both positioned on 100mtall towers separated by a distance of 5 km. If the transit antenna is driven by a 50MHz current with amplitude I0 2 A, determine: (a) The power received by the receive antenna in the absence of the surface. (Assume both antennas to be lossless.) (b) The power received by the receive antenna after incorporating reflection by the ground surface, assuming the surface to be flat and to have r 9 and conductivity 10 3 (S/m).
Direct h = 100 m i Reflected 100 m 5 Km Solution: (a) Since both antennas are lossless, Prec Pint Si Aer Thus, at the higher frequency, the antenna radiates 915 27 3 times as much power as it does at the lower frequency! Prad 2 36 6I0 36 6 52 915 W 402 52 Prad 2 402 I0 l 2 05 300 2 27 4 mW 10 3 33 516 5 CHAPTER 9 443 where Si is the incident power density and A er is the effective area of the receive dipole. From Section 93, 2 15I0 Si S0 R2 and from (9.64) and (9.47), Hence, 2 15I0 1 642 3 6 10 6 W R2 4 (b) The electric field of the signal intercepted by the receive antenna now consists of a direct component, Ed , due to the directly transmitted signal, and a reflected component, Er , due to the ground reflection. Since the power density S and the electric field E are related by E2 S 20 it follows that where the phase of the signal is measured with respect to the location of the transmit antenna, and k 2 . Hence, The electric field of the reflected signal is similar in form except for the fact that R should be replaced with R , where R is the path length traveled by the reflected signal, and the electric field is modified by the reflection coefficient . Thus, From the problem geometry R 2 25 103 2 100 2 5004 0 m Er 300 I0 e R jkR Ed 0 024e j120 300 I0 e R jkR (V/m) 20 2 15I0 e R2 Ed 20 Si e jkR jkR Prec Aer 1 64 2 D 4 2 4 1 642 4 444 CHAPTER 9 Since the dipole is vertically oriented, the electric field is parallel polarized. To calculate , we first determine 2 50 10
12 9 From Table 71, From (8.66a), From the geometry, R 2 87 71 The reflected electric field is The total electric field is 0 02e j73 3 (V/m) 0 024e E Ed Er
j120 0 018e j0 6 0 018e j0 6 (V/m) Er 300 I0 e R jkR Hence, 0 3 0 3 0 94 0 94 2 1 0 (air) 0 3 0 0 0 04 0 04 0 77 t sin 1 i sin i r 19 46 cos i h 2 cos t 2 cos t 1 cos i 1 cos i 100 2502 0 04 c 0 r 0 9 0 3 0 r 10 3 106 8 85 0 04 CHAPTER 9
The received power is 445 Problem 9.39
d The figure depicts a halfwave dipole connected to a generator through a matched transmission line. The directivity of the dipole can be modified by placing a reflecting rod a distance d behind the dipole. What would its reflectivity in the forward direction be if: (a) d 4, (b) d 2. Solution: Without the reflecting rod, the directivity of a halfwave dipole is 1.64 (see 9.47). When the rod is present, the wave moving in the direction of the arrow Prec Si Aer E 2 1 642 20 4 2 5 10 6 W 446 consists of two electric field components: CHAPTER 9 E1 where E1 is the field of the radiated wave moving to the right and E 2 is the field that initally moved to the left and then got reflected by the rod. The two are essentially equal in magnitude, but E2 lags in phase by 2kd relative to E 1 , and also by because the reflection coefficient of the metal rod is 1. Hence, we can write E at any point to the right of the antenna as The directivity is proportional to power, or E 2 . Hence, D will increase by a factor of 4 to D 1 64 4 6 56 Thus, the antenna radiation pattern will have a null in the forward direction. E E1 1 (b) For d 2, 2kd 2. 1 0 E E1 1 e (a) For d 2 4, 2kd 2 4 . j E1 1 e j 2kd E E1 E1 e j e j2kd E E1 E2 E2 2E1 CHAPTER 9 447 Problem 9.40 A fiveelement equally spaced linear array with d 2 is excited with uniform phase and an amplitude distribution given by the binomial distribution where N is the number of elements. Develop an expression for the array factor. Solution: Using the given formula, Application of (9.113) leads to: Fa N 1 i 0 2 Fa 6 8 cos cos 2 cos 2 cos With d 2, 2 2 cos cos ,
2 6 8 cos 2 cos 2 2 e j2 e j2 4e j 6 1 4e j 6e j2 4e j3 ai e a4 a3 a2 a1 4 6 4 1 ji 2d cos e j4 4e j
2 2 e j2 a0 1 5 1 0!4! 4! 1!3! 4! 2!2! 4! 3!1! 4! 0!4! ! note that 0! 1 ai N 1! i! N i 1 ! i 01 N 1 ...
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This note was uploaded on 02/09/2011 for the course EE 172 taught by Professor . during the Spring '10 term at UCLA.
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