Exam (elaborations) TEST BANK FOR Antennas for all Applications By John D. Kraus (Solution Manual)

Exam (elaborations) TEST BANK FOR Antennas for all Applications By John D. Kraus (Solution Manual) Table of Contents Preface iii Problem Solutions: Chapter 2. Antenna Basics.............................................................................................1 Chapter 3. The Antenna Family...................................................................................17 Chapter 4. Point Sources .............................................................................................19 Chapter 5. Arrays of Point Sources, Part I...................................................................23 Chapter 5. Arrays of Point Sources, Part II .................................................................29 Chapter 6. The Electric Dipole and Thin Linear Antennas .........................................35 Chapter 7. The Loop Antenna .....................................................................................47 Chapter 8. End-Fire Antennas: The Helical Beam Antenna and the Yagi-Uda Array, Part I................................................................................................53 Chapter 8. The Helical Antenna: Axial and Other Modes, Part II..............................55 Chapter 9. Slot, Patch and Horn Antennas ..................................................................57 Chapter 10. Flat Sheet, Corner and Parabolic Reflector Antennas................................65 Chapter 11. Broadband and Frequency-Independent Antennas.....................................75 Chapter 12. Antenna Temperature, Remote Sensing and Radar Cross Section ............81 Chapter 13. Self and Mutual Impedances....................................................................103 Chapter 14. The Cylindrical Antenna and the Moment Method (MM).......................105 Chapter 15. The Fourier Transform Relation Between Aperture Distribution and Far-Field Pattern................................................................................107 Chapter 16. Arrays of Dipoles and of Aperture...........................................................109 Chapter 17. Lens Antennas..........................................................................................121 Chapter 18. Frequency-Selective Surfaces and Periodic Structures By Ben A. Munk ......................................................................................125 Chapter 19. Practical Design Considerations of Large Aperture Antennas.................127 Chapter 21. Antennas for Special Applications...........................................................135 Chapter 23. Baluns, etc. By Ben A. Munk ..................................................................143 Chapter 24. Antenna Measurements. By Arto Lehto and Pertti Vainikainen ....................................................................................147 Index 153 1 Chapter 2. Antenna Basics 2-7-1. Directivity. Show that the directivity D of an antenna may be written ( ) ( ) ( ) ( ) ∫∫ Ω = ∗ ∗ π θ φ θ φ π θ φ θ φ 4 2 max max 2 , , 4 1 , , r d Z E E r Z E E D Solution: av U D U max (θ ,φ ) = , 2 max max U(θ ,φ ) = S(θ ,φ ) r , = ∫∫ Ω π θ φ π 4 ( , ) 4 U 1 U d av U(θ ,φ ) = S(θ ,φ )r2 , ( ) ( ) Z S E E θ φ θ φ (θ ,φ ) , , ∗ = Therefore ( ) ( ) ( ) ( ) ∫∫ Ω = ∗ ∗ π θ φ θ φ π θ φ θ φ 4 2 max max 2 , , 4 1 , , r d Z E E r Z E E D q.e.d. Note that r2 = area/steradian, so U = Sr2 or (watts/steradian) = (watts/meter2) × meter2 2-7-2. Approximate directivities. Calculate the approximate directivity from the half-power beam widths of a unidirectional antenna if the normalized power pattern is given by: (a) Pn = cos θ, (b) Pn = cos2 θ, (c) Pn = cos3 θ, and (d) Pn = cosn θ. In all cases these patterns are unidirectional (+z direction) with Pn having a value only for zenith angles 0° ≤ θ ≤ 90° and Pn = 0 for 90° ≤ θ ≤ 180°. The patterns are independent of the azimuth angle φ. Solution: (a) 1 o o HP θ = 2 cos − (0.5) = 2 × 60 = 120 , 278 (120) 40,000 2 D = = (ans.) (b) 1 o o HP θ = 2cos− ( 0.5) = 2× 45 = 90 , 4.94 (90) 40,000 2 D = = (ans.) 2 (c) 1 3 o o HP θ = 2cos− ( 0.5) = 2× 37.47 = 74.93 , 7.3 (75) 40,000 2 D = = (ans.) 2-7-2. continued (d) 2cos 1( 0.5) HP θ = − n , (cos 1( 0.5))2 10,000 n D − = (ans.) *2-7-3. Approximate directivities. Calculate the approximate directivities from the half-power beam widths of the three unidirectional antennas having power patterns as follows: P(θ,φ) = Pm sin θ sin2 φ P(θ,φ) = Pm sin θ sin3 φ P(θ,φ) = Pm sin2 θ sin3 φ P(θ,φ) has a value only for 0 ≤ θ ≤ π and 0 ≤ φ ≤ π and is zero elsewhere. Solution: To find D using approximate relations, we first must find the half-power beamwidths. = 90 −θ 2 HPBW or 2 θ = 90 − HPBW For sin θ pattern, 2 1 2 sin sin 90 HPBW =     θ =  − ,    − −  2 sin 1 2 90 HPBW 1 , 90 2 sin 1 2 HPBW 1 −     − −  , ∴HPBW =120o For sin2 θ pattern, 2 1 2 sin2 sin2 90 HPBW =     θ =  − , 2 1 2 sin 90 HPBW =      − , ∴HPBW = 90o For sin3 θ pattern, 2 1 2 sin3 sin3 90 HPBW =     θ =  − , 3 3 2 1 2 sin 90 HPBW =      − , ∴HPBW = 74.9o *2-7-3. continued Thus, 3.70 (120)(90) 3.82 40,000 (120)(90) 41,253sq. deg. 41,253 HP HP = = = ≅ = θ φ D (ans.) 4.45 (120)(74.9) 4.59 40,000 (120)(74.9) = 41,253 = ≅ = (ans.) 5.93 (90)(74.9) 6.12 40,000 (90)(74.9) = 41,253 = ≅ = (ans.) *2-7-4. Directivity and gain. (a) Estimate the directivity of an antenna with θHP = 2°, φHP = 1°, and (b) find the gain of this antenna if efficiency k = 0.5. Solution: (a) 4 HP HP 2.0 10 (2)(1) = 40,000 = 40,000 = × θ φ D or 43.0 dB (ans.) (b) G = kD = 0.5(2.0×104 ) = 1.0×104 or 40.0 dB (ans.) 2-9-1. Directivity and apertures. Show that the directivity of an antenna may be expressed as ( ) ( ) ∫∫ ( ) ( ) ∫∫ ∫∫ ∗ ∗ = Ap Ap Ap E x y E x y dxdy E x y dxdy E x y dxdy D , , 4 , , λ2 π where E(x, y) is the aperture field distribution. Solution: If the field over the aperture is uniform, the directivity is a maximum (= Dm) and the power radiated is P′ . For an actual aperture distribution, the directivity is D and the power radiated is P. Equating effective powers for P(θ,φ) = sin θ sin2φ for P(θ,φ) = sin θ sin3φ for P(θ,φ) = sin2 θ sin3φ 4 D P′ = D P m , ( ) ( ) ∫∫ ∗ = ′ = Ap p p dxdy Z E x y E x y A Z E E A P D D P , , 4 * av av m λ2 π 2-9-1. continued where = ∫∫ Ap p E x y dxdy A E 1 ( , ) av therefore ( ) ( ) 2 ( ) ( ) 4 , , , , Ap Ap Ap E x y dxdy E x y dxdy D E x y E x y dxdy π λ ∗ ∗ = ∫∫ ∫∫ ∫∫ q.e.d. where ( ) ( ) ( ) ( ) av av av av av 2 av , , 1 , , ( ) p e ap Ap p p E E A E E E A E x y E x y dxdy E x y E x y dxdy E A A ε ∗ ∗ ∗ ∗ ∗ = = = = ∫∫ ∫∫ 2-9-2. Effective aperture and beam area. What is the maximum effective aperture (approximately) for a beam antenna having halfpower widths of 30° and 35° in perpendicular planes intersecting in the beam axis? Minor lobes are small and may be neglected. Solution: o o HP HP 30 35 , A Ω ≅θ φ = × 2 2 o o 2 2 3.1 30 35 57.3 λ λ λ = × ≅ Ω = A Aem (ans.) *2-9-3. Effective aperture and directivity. What is the maximum effective aperture of a microwave antenna with a directivity of 900? Solution: 4 / 2 , em D = π A λ 2 2 71.6 4 900 4 2 λ λ π π λ Aem = D = = (ans.) 2-11-1. Received power and the Friis formula. What is the maximum power received at a distance of 0.5 km over a free-space 1 GHz circuit consisting of a transmitting antenna with a 25 dB gain and a receiving antenna with a 20 dB gain? The gain is with respect to a lossless isotropic source. The transmitting antenna input is 150 W. 5 Solution: 2 2 / 3 108 /109 0.3 m, , 4 4 t r et er c f A D A D λ λ λ π π = = × = = = 2-11-1. continued 0.0108 W 10.8mW (4 ) 500 150 316 0.3 100 (4 ) 2 2 2 2 2 2 2 2 2 2 = = × × = = = π λ π λ λ λ r D D P r A A P P t r t et er r t (ans.) *2-11-2. Spacecraft link over 100 Mm. Two spacecraft are separated by 100 Mm. Each has an antenna with D = 1000 operating at 2.5 GHz. If craft A's receiver requires 20 dB over 1 pW, what transmitter power is required on craft B to achieve this signal level? Solution: 2 / 3 108 / 2.5 109 0.12 m, et er 4 c f A A D λ λ π = = × × = = = 12 10 2 2 2 2 2 2 2 1 6 2 10 2 2 4 2 2 6 2 (required) 100 10 10 W (4 ) (4 ) 10 10 (4 ) 10966 W 11 kW ( ) 10 0.12 r t r r r et P P P r P r P r ans. A D D λ π λ π π λ λ − − − = × = = = = = = ≅ 2-11-3. Spacecraft link over 3 Mm. Two spacecraft are separated by 3 Mm. Each has an antenna with D = 200 operating at 2 GHz. If craft A's receiver requires 20 dB over 1 pW, what transmitter power is required on craft B to achieve this signal level? Solution: 2 / 3 108 / 2 109 0.15 m et er 4 c f A A D λ λ π = = × × = = = 12 10 2 2 2 2 2 2 12 10 2 2 2 4 2 100 10 10 W (4 ) 10 (4 ) 9 10 158 W ( ) 4 10 0.15 r t r r et er P P P r P r ans. A A D λ π λ π λ λ − − − = × = × = = = = × × 2-11-4. Mars and Jupiter links. (a) Design a two-way radio link to operate over earth-Mars distances for data and picture transmission with a Mars probe at 2.5 GHz with a 5 MHz bandwidth. A power of 10-19 6 W Hz-1 is to be delivered to the earth receiver and 10-17 W Hz-1 to the Mars receiver. The Mars antenna must be no larger than 3 m in diameter. Specify effective aperture of Mars and earth antennas and transmitter power (total over entire bandwidth) at each end. Take earth-Mars distance as 6 light-minutes. (b) Repeat (a) for an earth-Jupiter link. Take the earth-Jupiter distance as 40 light-minutes. 2-11-4. continued Solution: (a) λ = c / f = 3×108 / 2.5×109 = 0.12 m 19 6 13 17 6 11 (earth) W (Mars) W r r P P − − − − = × × = × = × × = × Take (Mars) (1/2) 1.5 3.5m ( ap 0.5) = π 2 = 2 ε = Ae Take Pt (Mars) =1kW Take (earth) (1/2) 15 350m ( ap 0.5) = π 2 = 2 ε = Ae 6.9MW 3.5 350 (earth) 5 10 (360 3 10 ) 0.12 (earth) (Mars) (earth) (Mars) 8 2 2 11 2 2 = × × × = × = − t et et t r P A A P P r λ To reduce the required earth station power, take the earth station antenna = (1/ 2)π 502 = 3927 m2 Ae (ans.) so (earth) 6.9 106 (15 / 50)2 620 kW ( ) t P = × = ans. 8 10 W (360 3 10 ) 0.12 (earth) (Mars) (Mars) (earth) 10 3.5 3930 14 8 2 2 3 2 2 = × − × × × = = r λ P P Aet Aer r t which is about 16% of the required 5 x 10−13 W. The required 5 x 10−13 W could be obtained by increasing the Mars transmitter power by a factor of 6.3. Other alternatives would be (1) to reduce the bandwidth (and data rate) reducing the required value of Pr or (2) to employ a more sensitive receiver. 7 As discussed in Sec. 12-1, the noise power of a receiving system is a function of its system temperature T and bandwidth B as given by P = kTB, where k = Boltzmann’s constant = 1.38 x 10−23 JK−1. For B = 5 x 106 Hz (as given in this problem) and T = 50 K (an attainable value), P(noise) =1.38×10−23 ×50× 5×106 = 3.5×10−15 W 2-11-4. continued The received power (8 x 10−14 W) is about 20 times this noise power, which is probably sufficient for satisfactory communication. Accordingly, with a 50 K receiving system temperature at the earth station, a Mars transmitter power of 1 kW is adequate. (b) The given Jupiter distance is 40/6 = 6.7 times that to Mars, which makes the required transmitter powers 6.72 = 45 times as much or the required receiver powers 1/45 as much. Neither appears feasible. But a practical solution would be to reduce the bandwidth for the Jupiter link by a factor of about 50, making B = (5/50) x 106 = 100 kHz. *2-11-5. Moon link. A radio link from the moon to the earth has a moon-based 5λ long right-handed monofilar axial-mode helical antenna (see Eq. (8-3-7)) and a 2 W transmitter operating at 1.5 GHz. What should the polarization state and effective aperture be for the earth-based antenna in order to deliver 10-14 W to the receiver? Take the earth-moon distance as 1.27 light-seconds. Solution: λ = c / f = 3×108 /1.5×109 = 0.2 m, From (8-3-7) the directivity of the moon helix is given by D =12× 5 = 60 and π λ 4 (moon) D 2 Aet = From Friis formula 152m RCP 2 60 (4 ) 10 (3 10 1.27) 4 2 14 8 2 2 2 2 2 2 = × × × = = = − π λ λ π λ P D P r P A A P r t r t et r er or about 14 m diameter (ans.) 8 2-16-1. Spaceship near moon. A spaceship at lunar distance from the earth transmits 2 GHz waves. If a power of 10 W is radiated isotropically, find (a) the average Poynting vector at the earth, (b) the rms electric field E at the earth and (c) the time it takes for the radio waves to travel from the spaceship to the earth. (Take the earth-moon distance as 380 Mm.) (d) How many photons per unit area per second fall on the earth from the spaceship transmitter? 2-16-1. continued Solution: (a) 18 2 2 2 6 2 5.5 10 Wm 5.5 aWm 4 (380 10 ) 10 4 PV (at earth) = × − − = − × = = π r π Pt (ans.) (b) PV = S = E2 / Z or E = (SZ)1/ 2 or E = (5.5×10−18 × 377)1/ 2 = 45×10−9 = 45 nVm−1 (ans.) (c) t = r / c = 380×106 / 3×108 = 1.27 s (ans.) (d) Photon = hf = 6.63×10−34 × 2×109 = 1.3×10−24 J , where h = 6.63×10−34 Js This is the energy of a 2.5 MHz photon. From (a), PV = 5.5×10−18 Js−1m−2 Therefore, number of photons = 6 2 1 24 18 4.2 10 m s 1.3 10 5.5 10 − − − − = × × × (ans.) 2-16-2. More power with CP. Show that the average Poynting vector of a circularly polarized wave is twice that of a linearly polarized wave if the maximum electric field E is the same for both waves. This means that a medium can handle twice as much power before breakdown with circular polarization (CP) than with linear polarization (LP). Solution: From (2-16-3) we have for rms fields that o 2 2 2 PV 1 Z S E E av + = = For LP, 2 1 2 1 o (or ) 0, so av E E S E Z = = For CP, 2 1 1 2 o , so 2 av E E S E Z = = Therefore SCP = 2SLP (ans.) 9 2-16-3. PV constant for CP. Show that the instantaneous Poynting vector (PV) of a plane circularly polarized traveling wave is a constant. Solution: ECP = Ex cosωt + Ey sinωt where Ex = Ey = Eo 2-16-3. continued 2 2 2 2 1/ 2 2 2 1/ 2 CP o o o o E = (E cos ω t + E sin ω t) = E (cos ω t + sin ω t) = E (a constant) Therefore Z S E 2 PV or (instantaneous) = o (a constant) (ans.) *2-16-4. EP wave power An elliptically polarized wave in a medium with constants σ = 0, μ r = 2, ε r = 5 has Hfield components (normal to the direction of propagation and normal to each other) of amplitudes 3 and 4 A m-1. Find the average power conveyed through an area of 5 m2 normal to the direction of propagation. Solution: 2 1/ 2 2 2 2 2 2 1 2 1/ 2 2 2 1 377(2/5) (3 4 ) 2980 Wm 2 377( / ) ( ) 1 2 ( ) 1 2 S = 1 Z H + H = H + H = + = − av μr ε r P = ASav = 5× 2980 =14902 W =14.9 kW (ans.) 2-17-1. Crossed dipoles for CP and other states. Two λ/2 dipoles are crossed at 90°. If the two dipoles are fed with equal currents, what is the polarization of the radiation perpendicular to the plane of the dipoles if the currents are (a) in phase, (b) phase quadrature (90° difference in phase) and (c) phase octature (45° difference in phase)? Solution: (a) LP (ans.) (b) CP (ans.) 10 (c) From (2-17-3) sin 2ε = sin 2γ sinδ 1 2 1 1 2 where tan ( / ) 45 45 22 AR cot 1/ tan 2.41 (EP)...( ) E E ans. γ δ ε ε ε = − = = = = = =    *2-17-2. Polarization of two LP waves. A wave traveling normally out of the page (toward the reader) has two linearly polarized components E t x = 2 cosω E = 3cos( t + 90 ) y ω (a) What is the axial ratio of the resultant wave? (b) What is the tilt angle τ of the major axis of the polarization ellipse? (c) Does E rotate clockwise or counterclockwise? Solution: (a) From (2-15-8) , AR = 3/ 2 =1.5 (ans.) (b) τ = 90o (ans.) (c) At 0, ; x t = E = E at / 4, y t = T E = −E , therefore rotation is CW (ans.) 2-17-3. Superposition of two EP waves. A wave traveling normally outward from the page (toward the reader) is the resultant of two elliptically polarized waves, one with components of E given by E t y ′ = 2 cosω and ( ) 2 E′ = 6cosωt + π x and the other with components given by E t y ′′ = 1cosω and ( ) 2 E′′ = 3cosωt − π x (a) What is the axial ratio of the resultant wave? (b) Does E rotate clockwise or counterclockwise? Solution: 2cos cos 3cos 6cos( / 2) 3cos( / 2) 6sin 3sin 3sin y y y x x x E E E t t t E E E t t t t t ω ω ω ω π ω π ω ω ω = ′ + ′′ = + = = ′ + ′′ = + + − = − + = − 11 (a) Ex and Ey are in phase quadrature and AR = 3/ 3 =1 (CP) (ans.) (b) At t = 0, E = yˆ3 , at t = T / 4, E = −xˆ3, therefore rotation is CCW (ans.) *2-17-4. Two LP components. An elliptically polarized plane wave traveling normally out of the page (toward the reader) has linearly polarized components Ex and Ey. Given that Ex = Ey = 1 V m-1 and that Ey leads Ex by 72°, (a) Calculate and sketch the polarization ellipse. (b) What is the axial ra