Antenna For All Application By John D. Kraus, 3rd Edt .

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PrefaceThis Instructors’ Manual provides solutions to most of the problems in ANTENNAS:FOR ALL APPLICATIONS, THIRD EDITION. All problems are solved for whichanswers appear in Appendix F of the text, and in addition, solutions are given for a largefraction of the other problems. Including multiple parts, there are 600 problems in thetext and solutions are presented here for the majority of them.Many of the problem titles are supplemented by key words or phrases alluding to thesolution procedure. Answers are indicated. Many tips on solutions are included whichcan be passed on to students.Although an objective of problem solving is to obtain an answer, we have endeavoredto also provide insights as to how many of the problems are related to engineeringsituations in the real world.The Manual includes an index to assist in finding problems by topic or principle andto facilitate finding closely-related problems.This Manual was prepared with the assistance of Dr. Erich Pacht.Professor John D. KrausDept. of Electrical EngineeringOhio State University2015 Neil AveColumbus, Ohio 43210Dr. Ronald J. MarhefkaSenior Research Scientist/Adjunct ProfessorThe Ohio State UniversityElectroscience Laboratory1320 Kinnear RoadColumbus, Ohio 43212iii

Table of ContentsPrefaceiiiProblem Solutions:Chapter 2.Antenna Basics .1Chapter 3.The Antenna Family .17Chapter 4.Point Sources .19Chapter 5.Arrays of Point Sources, Part I .23Chapter 5.Arrays of Point Sources, Part II .29Chapter 6.The Electric Dipole and Thin Linear Antennas .35Chapter 7.The Loop Antenna .47Chapter 8.End-Fire Antennas: The Helical Beam Antenna and the Yagi-UdaArray, Part I .53Chapter 8.The Helical Antenna: Axial and Other Modes, Part II .55Chapter 9.Slot, Patch and Horn Antennas .57Chapter 10. Flat Sheet, Corner and Parabolic Reflector Antennas .65Chapter 11. Broadband and Frequency-Independent Antennas .75Chapter 12. Antenna Temperature, Remote Sensing and Radar Cross Section .81Chapter 13. Self and Mutual Impedances.103Chapter 14. The Cylindrical Antenna and the Moment Method (MM) .105Chapter 15. The Fourier Transform Relation Between Aperture Distributionand Far-Field Pattern .107Chapter 16. Arrays of Dipoles and of Aperture .109Chapter 17. Lens Antennas.121Chapter 18. Frequency-Selective Surfaces and Periodic StructuresBy Ben A. Munk .125Chapter 19. Practical Design Considerations of Large Aperture Antennas .127Chapter 21. Antennas for Special Applications .135Chapter 23. Baluns, etc. By Ben A. Munk .143Chapter 24. Antenna Measurements. By Arto Lehto andPertti Vainikainen .147Index153iv

1Chapter 2. Antenna Basics2-7-1. Directivity.Show that the directivity D of an antenna may be writtenE (θ , φ )max E (θ , φ )max 2rZD 1E (θ , φ )E (θ , φ ) 2r dΩ4π 4πZSolution: D,(Uθ,φUU (θ ,φ ) max S (θ ,φ ) max r 2 ,U av 14π)maxav π U (θ , φ )dΩ4E (θ ,φ )E (θ ,φ )U (θ ,φ ) S (θ ,φ )r , S (θ ,φ ) Z2ThereforeE (θ , φ )max E (θ , φ )max 2rZD 1E (θ , φ )E (θ , φ ) 2r dΩ4π 4πZq.e.d.Note that r 2 area/steradian, so U Sr 2 or (watts/steradian) (watts/meter2) meter22-7-2. Approximate directivities.Calculate the approximate directivity from the half-power beam widths of aunidirectional 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 for90 θ 180 . The patterns are independent of the azimuth angle φ.Solution:(a)θ HP 2 cos 1 (0.5) 2 60 o 120 o ,(b)θ HP 2 cos 1 ( 0.5 ) 2 45o 90o ,40,000 278 (ans.)(120) 240,000D 4.94 (ans.)(90) 2D

2(c)θ HP 2 cos 1 (3 0.5 ) 2 37.47 o 74.93o ,D 40,000 7.3 (ans.)(75) 2D 10,000(cos 1 ( n 0.5 ))22-7-2. continued(d)θ HP 2 cos 1 ( n 0.5 ) ,(ans.)*2-7-3. Approximate directivities.Calculate the approximate directivities from the half-power beam widths of the threeunidirectional antennas having power patterns as follows:P(θ,φ) Pm sin θ sin2 φP(θ,φ) Pm sin θ sin3 φP(θ,φ) Pm sin2 θ sin3 φP(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.HPBWHPBW 90 θ or θ 90 22HPBW 1 For sin θ pattern, sin θ sin 90 ,2 2 HPBW 1 1 HPBW 1 1 90 sin , sin 90 , HPBW 120o222 2 HPBW 1 For sin2 θ pattern, sin 2 θ sin 2 90 ,2 2 HPBW 1 sin 90 , HPBW 90o 22 HPBW 1 For sin3 θ pattern, sin 3 θ sin 3 90 ,2 2

3HPBW 1 osin 90 3 , HPBW 74.92 2 *2-7-3. continuedThus,D 41,253 sq. deg.θ HPφHP41,25340,000 3.82 3.70 (ans.)(120)(90)(120)(90) for P(θ,φ) sin θ sin2φ 41,25340,000 4.59 4.45 (ans.)(120)(74.9)(120)(74.9)for P(θ,φ) sin θ sin3φ 41,25340,000 6.12 5.93 (ans.)(90)(74.9)(90)(74.9)for P(θ,φ) sin2 θ sin3φ*2-7-4. Directivity and gain.(a) Estimate the directivity of an antenna with θHP 2 , φHP 1 , and (b) find the gain ofthis antenna if efficiency k 0.5.Solution:40,000D (b)G kD 0.5(2.0 10 4 ) 1.0 10 4 or 40.0 dB (ans.)θ HPφ HP 40,000 2.0 10 4 or 43.0 dB (ans.)(2)(1)(a)2-9-1. Directivity and apertures.Show that the directivity of an antenna may be expressed asD 4πλ2 E (x, y )dxdy E (x, y )dxdy E (x, y )E (x, y )dxdy ApAp Apwhere 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 andthe power radiated is P. Equating effective powers

4*Eav EavApP′ 4πZD Dm 2 ApP λE ( x, y )E (x, y )dxdy ApZDm P ′ D P ,2-9-1. continuedE av whereD thereforewhere1Ap4πλ A 2ApE ( x, y )dxdypE ( x, y ) dxdy E ( x, y ) dxdyAp E ( x, y ) E ( x, y ) dxdy ApEav Eav Ap ApE ( x, y ) E ( x, y ) dxdy q.e.d.Eav Eav 1Ap E ( x, y ) E ( x, y ) dxdy Eav A ε ap e2( E )avAp2-9-2. 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:Ω A θ HPφHP 30o 35o ,Aem λ2ΩA 57.32λ 2 3.1λ 2 (ans.)oo30 35*2-9-3. Effective aperture and directivity.What is the maximum effective aperture of a microwave antenna with a directivity of900?Solution:D 4π Aem / λ 2 ,Aem Dλ 2 900 2 λ 71.6 λ2 (ans.)4π4π2-11-1. Received power and the Friis formula.What is the maximum power received at a distance of 0.5 km over a free-space 1 GHzcircuit consisting of a transmitting antenna with a 25 dB gain and a receiving antennawith a 20 dB gain? The gain is with respect to a lossless isotropic source. Thetransmitting antenna input is 150 W.

5Solution:λ c / f 3 108 /109 0.3 m,Aet Dt λ 2,4πAer Dr λ 24π2-11-1. continuedPr PtAet Aerr 2λ 2 PtDt λ 2 Dr λ 2(4π ) 2 r 2 λ 2 150316 0.3 2 100 0.0108 W 10.8 mW (ans.)(4π ) 2 500 2*2-11-2. Spacecraft link over 100 Mm.Two spacecraft are separated by 100 Mm. Each has an antenna with D 1000 operatingat 2.5 GHz. If craft A's receiver requires 20 dB over 1 pW, what transmitter power isrequired on craft B to achieve this signal level?Solution:λ c / f 3 108 / 2.5 109 0.12 m,Aet Aer Dλ24πPr (required) 100 10 12 10 10 WPt Pr162r 2λ 2(4π ) 2 r 2 λ 2r 2 (4π ) 2 10 10 (4π ) P P 10 10966 W 11 kW (ans.)rrAet 2D 2λ 4D 2λ 2106 0.1222-11-3. Spacecraft link over 3 Mm.Two spacecraft are separated by 3 Mm. Each has an antenna with D 200 operating at 2GHz. If craft A's receiver requires 20 dB over 1 pW, what transmitter power is requiredon craft B to achieve this signal level?Solution:λ c / f 3 108 / 2 109 0.15 mAet Aer Dλ24πPr 100 10 12 10 10 W212r 2λ 2(4π ) 2 r 2 λ 2 10 (4π ) 9 10Pt Pr Pr 10 158 W (ans.)Aet AerD 2λ 2λ 24 104 0.1522-11-4. Mars and Jupiter links.(a) Design a two-way radio link to operate over earth-Mars distances for data and picturetransmission with a Mars probe at 2.5 GHz with a 5 MHz bandwidth. A power of 10-19

6W Hz-1 is to be delivered to the earth receiver and 10-17 W Hz-1 to the Mars receiver. TheMars antenna must be no larger than 3 m in diameter. Specify effective aperture of Marsand earth antennas and transmitter power (total over entire bandwidth) at each end. Takeearth-Mars distance as 6 light-minutes. (b) Repeat (a) for an earth-Jupiter link. Take theearth-Jupiter distance as 40 light-minutes.2-11-4. continuedSolution:(a)λ c / f 3 108 / 2.5 109 0.12 mPr (earth) 10 19 5 106 5 10 13 WPr (Mars) 10 17 5 106 5 10 11 WTakeAe (Mars) (1/2) π 1.52 3.5 m 2 (ε ap 0.5)TakePt (Mars) 1 kWTakeAe (earth) (1/2) π 152 350 m 2 (ε ap 0.5)Pt (earth) Pr (Mars)Pt (earth) 5 10 11r 2λ 2Aet (earth) Aet (Mars)(360 3 10 8 ) 2 0.12 2 6.9 MW3.5 350To reduce the required earth station power, take the earth station antennaAe (1 / 2) π 502 3927 m 2 (ans.)soPt (earth) 6.9 106 (15 / 50) 2 620 kW (ans.)Pr (earth) Pt (Mars)Aet (Mars) Aer (earth)3.5 3930 103 8 10 14 W2 28 22r λ(360 3 10 ) 0.12which is about 16% of the required 5 x 10 13 W. The required 5 x 10 13 W could beobtained by increasing the Mars transmitter power by a factor of 6.3. Other alternativeswould be (1) to reduce the bandwidth (and data rate) reducing the required value of Pr or(2) to employ a more sensitive receiver.

7As discussed in Sec. 12-1, the noise power of a receiving system is a function of itssystem temperature T and bandwidth B as given by P kTB, where k Boltzmann’sconstant 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 W2-11-4. continuedThe received power (8 x 10 14 W) is about 20 times this noise power, which is probablysufficient for satisfactory communication. Accordingly, with a 50 K receiving systemtemperature 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 therequired transmitter powers 6.72 45 times as much or the required receiver powers 1/45as much.Neither appears feasible. But a practical solution would be to reduce the bandwidth forthe 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 1.5GHz. What should the polarization state and effective aperture be for the earth-basedantenna in order to deliver 10-14 W to the receiver? Take the earth-moon distance as 1.27light-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 byD 12 5 60andAet (moon ) D λ24πFrom Friis formulaAer Pr r 2λ2 Pr (4π ) r 2λ2 10 14 (3 108 1.27) 2 4π 152 m 2 RCP or22 60Pt DλPt Aetabout 14 m diameter (ans.)

82-16-1. Spaceship near moon.A spaceship at lunar distance from the earth transmits 2 GHz waves. If a power of 10 Wis radiated isotropically, find (a) the average Poynting vector at the earth, (b) the rmselectric field E at the earth and (c) the time it takes for the radio waves to travel from thespaceship to the earth. (Take the earth-moon distance as 380 Mm.) (d) How manyphotons per unit area per second fall on the earth from the spaceship transmitter?2-16-1. continuedSolution:Pt10 5.5 10 18 Wm 2 5.5 aWm 2 (ans.)26 24π r4π (380 10 )(a)PV (at earth) (b)PV S E 2 / Zor E ( SZ )1 / 2or E (5.5 10 18 377)1 / 2 45 10 9 45 nVm 1 (ans.)(c)t r / c 380 10 6 / 3 10 8 1.27 s (ans.)(d) Photon hf 6.63 10 34 2 10 9 1.3 10 24 J , where h 6.63 10 34 JsThis is the energy of a 2.5 MHz photon. From (a), PV 5.5 10 18 Js 1m 25.5 10 18 4.2 10 6 m 2 s 1 (ans.)Therefore, number of photons 241.3 102-16-2. More power with CP.Show that the average Poynting vector of a circularly polarized wave is twice that of alinearly polarized wave if the maximum electric field E is the same for both waves. Thismeans that a medium can handle twice as much power before breakdown with circularpolarization (CP) than with linear polarization (LP).Solution:From (2-16-3) we have for rms fields that PV S av For LP,E2 (or E1 ) 0,so Sav E12ZoFor CP,E1 E2 ,so Sav 2 E12ZoThereforeSCP 2 SLP (ans.)E12 E22Zo

92-16-3. PV constant for CP.Show that the instantaneous Poynting vector (PV) of a plane circularly polarizedtraveling wave is a constant.Solution:ECP Ex cos ωt E y sin ωtwhere Ex E y Eo2-16-3. continuedECP ( Eo2 cos 2 ω t Eo2 sin 2 ω t )1/ 2 Eo (cos 2 ω t sin 2 ω t )1/ 2 Eo (a constant)Eo2Therefore PV or S (instantaneous) Z(a constant) (ans.)*2-16-4. EP wave powerAn elliptically polarized wave in a medium with constants σ 0, μr 2, εr 5 has Hfield components (normal to the direction of propagation and normal to each other) ofamplitudes 3 and 4 A m-1. Find the average power conveyed through an area of 5 m2normal to the direction of propagation.Solution:S av 111Z ( H12 H 22 ) 377( μ r / ε r )1 / 2 ( H12 H 22 ) 377(2 / 5)1 / 2 (32 42 ) 2980 Wm 2222P AS av 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 isthe polarization of the radiation perpendicular to the plane of the dipoles if the currentsare (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.)(c)From (2-17-3)sin 2ε sin 2γ sin δ

10whereγ tan 1 ( E2 / E1 ) 45δ 45ε 22 1 2AR cot ε 1/ tan ε 2.41 (EP).(ans.)*2-17-2. Polarization of two LP waves.A wave traveling normally out of the page (toward the reader) has two linearly polarizedcomponentsE x 2 cos ωtE y 3 cos(ωt 90)(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:, AR 3 / 2 1.5 (ans.)(a)From (2-15-8)(b)τ 90o(c)At t 0, E Ex ; at t T / 4, E E y , therefore rotation is CW (ans.)(ans.)2-17-3. Superposition of two EP waves.A wave traveling normally outward from the page (toward the reader) is the resultant oftwo elliptically polarized waves, one with components of E given byE ′y 2 cos ωtandE x′ 6 cos(ωt π2 )and the other with components given byE ′y′ 1 cos ωt andE x′′ 3 cos(ωt π2 )(a) What is the axial ratio of the resultant wave?(b) Does E rotate clockwise or counterclockwise?Solution:E y E ′y E ′′y 2 cos ω t cos ω t 3cos ω tEx Ex′ Ex′′ 6 cos(ω t π / 2) 3cos(ω t π / 2) 6sin ω t 3sin ω t 3sin ω t(a)Ex and Ey are in phase quadrature and AR 3 / 3 1 (CP) (ans.)

11(b)At t 0, E yˆ 3 , at t T / 4, E xˆ 3 , therefore rotation is CCW (ans.)*2-17-4. Two LP components.An elliptically polarized plane wave traveling normally out of the page (toward thereader) has linearly polarized components Ex and Ey. Given that Ex Ey 1 V m-1 andthat Ey leads Ex by 72 ,(a) Calculate and sketch the polarization ellipse.(b) What is the axial ratio?(c) What is the angle τ between the major axis and the x-axis?Solution:(b) γ tan 1 ( E2 / E1 ) 45o , δ 72o(c)From (2-17-3),ε 36o , therefore AR 1 / tan ε 1.38 (ans.)From (2-17-3),sin 2 τ tan 2ε / tan δor τ 45 o (ans.)2-17-5. Two LP components and Poincaré sphere.Answer the same questions as in Prob. 2-17-4 for the case where Ey leads Ex by 72 asbefore but Ex 2 V m-1 and Ey 1 V m-1.Solution:(b)γ tan 1 2 63.4oδ 72oε 24.8o and AR 2.17 (ans.)(c)τ 11.2o (ans.)*2-17-6. Two CP waves.Two circularly polarized waves intersect at the origin. One (y-wave) is traveling in thepositive y direction with E rotating clockwise as observed from a point on the positive yaxis. The other (x-wave) is traveling in the positive x direction with E rotating clockwiseas observed from a point on the positive x-axis. At the origin, E for the y-wave is in thepositive z direction at the same instant that E for the x-wave is in the negative z direction.What is the locus of the resultant E vector at the origin?Solution:

12Resolve 2 waves into components or make sketch as shown. It is assumed that the waveshave equal magnitude.*2-17-6. continuedLocus of E is a straight line in xy plane at an angle of 45o with respect to x (or y) axis.*2-17-7. CP waves.A wave traveling normally out of the page is the resultant of two circularly polarizedcomponents E right 5e jωt and Eleft 2e j (ωt 90 ) (V m-1). Find (a) the axial ratio AR, (b)the tilt angle τ and (c) the hand of rotation (left or right).Solution:(a)(b)2 5 7 / 3 2.33 (ans.)2 5From diagram, τ 45o (ans.)AR [Note minus sign for RH (right-handedpolarization)]

13(c)Since E rotates counterclockwise as a function of time, RH. (ans.)2-17-8. EP wave.A wave traveling normally out of the page (toward the reader) is the resultant of twolinearly polarized components E x 3 cos ωt and E y 2 cos(ωt 90 ) . For the resultantwave find (a) the axial ratio AR, (b) the tilt angle τ and (c) the hand of rotation (left orright).Solution:(a)AR 3/2 1.5 (ans.)(b)τ 0o (ans.)(c)CW, LEP (ans.)*2-17-9. CP waves.Two circularly polarized waves traveling normally out of the page have fields given byEleft 2e jωt and E right 3e jωt (V m-1) (rms). For the resultant wave find (a) AR, (b) thehand of rotation and (c) the Poynting vector.Solution:2 3 5 (ans.)2-3(a)AR (b)REP (ans.)(c)PV EL2 ER2 4 9 0.034 Wm 2 34 mWm 2 (ans.)Z3772-17-10. EP waves.A wave traveling normally out of the page is the resultant of two elliptically polarized(EP) waves, one with components E x 5 cos ωt and E y 3 sin ωt and another withcomponents E r 3e jωt and El 4e jωt . For the resultant wave, find (a) AR, (b) τ and(c) the hand of rotation.Solution:(a)

14E x 5 cos ω t 3 cos ω t 4 cos ω t 12 cos ω tE y 3 sin ω t 3 sin ω t 4 sin ω t 2 sin ω t2-17-10. continuedAR 12 / 2 6 (ans.)(b)Since Ex and Ey are in time-phase quadrature with Ex(max) Ey(max), τ 0o.Or from (2-17-3), sin 2 τ tan 2ε / tan δ , ε tan 1 (1 / AR ) 9.46obut δ 90o so tan δ Therefore τ 0o (ans.)(c)E at t T/4At t 0, Ex 12, E y 0CCWAt t T / 4 (ω t 90o )

This Instructors’ Manual provides solutions to most of the problems in ANTENNAS: FOR ALL APPLICATIONS, THIRD EDITION. All problems are solved for which answers appear in Appendix F of the text, and in addition, solutions are given for a large fraction of the other problems. Including multiple parts, there are 600 problems in the

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