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Antenna__3rd_Edition__2002__-_Kraus_-_Solution_Manual

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Antenna__3rd_Edition__2002__-_Kraus_-_Solution_Manual

    Preface

     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 text and solutions are presented here for the majority of them.

     Many of the problem titles are supplemented by key words or phrases alluding to the solution procedure. Answers are indicated. Many tips on solutions are included which can be passed on to students.

     Although an objective of problem solving is to obtain an answer, we have endeavored to also provide insights as to how many of the problems are related to engineering situations in the real world.

     The Manual includes an index to assist in finding problems by topic or principle and to facilitate finding closely-related problems.

     This Manual was prepared with the assistance of Dr. Erich Pacht.

     Professor John D. Kraus

     Dept. of Electrical Engineering

     Ohio State University

     2015 Neil Ave

     Columbus, Ohio 43210

     Dr. Ronald J. Marhefka

     Senior Research Scientist/Adjunct Professor

     The Ohio State University

     Electroscience Laboratory

     1320 Kinnear Road

     Columbus, Ohio 43212

    iii

    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

    iv

    1

    Chapter 2. Antenna Basics

2-7-1. Directivity.

    Show that the directivity D of an antenna may be written

    ????E,E,;?;?2maxmaxrUZ ?D;?;?1E?,?E?,?2(rd??4?4Z?DSolution:

    ?avmaxU(?,?)12, , U(?,?)?S(?,?)rU?U(?,?)d(avmaxmax??4?4?

    EE;?;??,??,?2, S U(?,?)?S(?,?)r?(,)??Z

    Therefore

    ????E,E,;?;?2maxmaxrZ q.e.d. ?D;?;?1E?,?E?,?2(rd??4?4Z?

    222?Note that r?area/steradian, so or (watts/steradian) = (watts/meter) U?Sr2meter

    2-7-2. Approximate directivities. Calculate the approximate directivity from the half-power beam widths of a unidirectional 2antenna if the normalized power pattern is given by: (a) P = cos ?, (b) P = cos ?, (c) P nnn3n= cos ?, and (d) P = cos ?. In all cases these patterns are unidirectional (+z direction) n

    with P having a value only for zenith angles 0? ? ? ? 90? and P = 0 for 90? ? ? ? 180?. nn

    The patterns are independent of the azimuth angle ?.

Solution:

    40,0001oo(a) , (ans.) ??2cos(0.5)?2 ? 60?120D??278HP2(120)

2

    40,0001oo, (ans.) (b) ??2cos(0.5)?2 ? 45?90D??4.94HP2(90)

    40,0001oo3(c) , (ans.) ??2cos(0.5)?2 ? 37.47?74.93D??7.3HP2(75)

    2-7-2. continued

    10,0001n(d) , (ans.) ??2cos(0.5)D?HP12n(cos(0.5))

*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:

     2P(?,?) = P sin ? sin ? m

     3P(?,?) = P sin ? sin ? m

     23P(?,?) = P sin ? sin ? m

    P(?,?) has a value only for 0 ? ? ? ? and 0 ? ? ? ? and is zero elsewhere.

Solution:

    To find D using approximate relations,

    we first must find the half-power beamwidths.

    HPBWHPBW?90 or ?90??22

    HPBW1?? sin?sin90?For sin ? pattern, ?, ??22??

    HPBW1HPBW1????11o90sin, sin90, ? HPBW 120?????2222????

    HPBW1??222sinsin90For sin ? pattern, ???, ??22??

    HPBW1??osin90?, ? HPBW 90???22??

    3

    HPBW1??333sinsin90 ? pattern, , For sin?????22??

    HPBW1??osin90?, ? HPBW 74.9???322??

    *2-7-3. continued

    Thus,

    41,253 sq. deg.41,25340,000 (ans.) D? ??3.82??3.70??(120)(90)(120)(90)2HPHPfor P(?,?) = sin ? sin?

    41,25340,000??4.59??4.45 (ans.) (120)(74.9)(120)(74.9)3for P(?,?) = sin ? sin?

    41,25340,000??6.12??5.93 (ans.) 23(90)(74.9)(90)(74.9)for P(?,?) = sin ? sin?

    *2-7-4. Directivity and gain. (a) Estimate the directivity of an antenna with ? = 2?, ? = 1?, and (b) find the gain of HPHP

    this antenna if efficiency k = 0.5.

Solution:

    40,00040,0004(a) or 43.0 dB (ans.) D???2.0?10??(2)(1)HPHP

    44(b) or 40.0 dB (ans.) G?kD?0.5(2.0?10)?1.0?10

    2-9-1. Directivity and apertures. Show that the directivity of an antenna may be expressed as

    Ex,ydxdyEx,ydxdy;?;??????4ApAp D?2?;?;?Ex,yEx,ydxdy??Ap

    where E(x, y) is the aperture field distribution.

4

    ) and Solution: If the field over the aperture is uniform, the directivity is a maximum (= Dm

    ?the power radiated is . For an actual aperture distribution, the directivity is D and the P

    power radiated is P. Equating effective powers

    *EEavavAp??P4Z?, DDADP?DP??pmm2PE;?;?x,yEx,y?dxdy??ApZ

    2-9-1. continued

    1where E?E(x,y)dxdyav??ApAp

    ExydxdyExydxdy,,;?;??????4ApAptherefore q.e.d. ?D2?ExyExydxdy,,;?;???Ap

    ??EEAEEEAavavpavavavewhere ?????ap21()EA,,ExyExydxdy;?;?avp??,,ExyExydxdy;?;?Ap??Ap

2-9-2. Effective aperture and beam area.

    What is the maximum effective aperture (approximately) for a beam antenna having half-power widths of 30? and 35? in perpendicular planes intersecting in the beam axis? Minor lobes are small and may be neglected.

Solution:

    22?57.3oo22 (ans.) (?????3035,A????3.1?emAHPHPoo(30?35A

*2-9-3. Effective aperture and directivity.

    What is the maximum effective aperture of a microwave antenna with a directivity of 900?

    22?D90022Solution: (ans.) DA?4/,??A????71.6?emem44??

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

    5

    a 20 dB gain? The gain is with respect to a lossless isotropic source. The transmitting

    antenna input is 150 W.

Solution:

    22D?D?89tr ??????/310/100.3 m, , cfAA?eter44??

    2-11-1. continued

    222??AADD316?0.3?100etertr (ans.) PPP???150?0.0108 W?10.8 mWrtt2222222r??r??(4)(4)500

*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:

    2D?89???????/310/2.5100.12 m, cfAA?eter4?,,1210P(required)1001010 W???r

    2222222162?????rrr(4)(4)10(4)10PPPPans.??????1010966 W11 kW ()trrr2242262ADD100.12??et

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:

    2D?89 ???????/310/2100.15 m cfAA?eter4?,,1210P???1001010 Wr

    22222212????rr(4)(4) 910?10PPPans.????10158 W ()trr22242AAD4100.15????eter

6

    2-11-4. Mars and Jupiter links.

    (a) Design a two-way radio link to operate over earth-Mars distances for data and picture -19 W transmission with a Mars probe at 2.5 GHz with a 5 MHz bandwidth. A power of 10-1-17-1Hz is to be delivered to the earth receiver and 10 W Hz 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:

    89(a) ??????cf/310/2.5100.12 m

    ,,19613P(earth)10510510 W?????r

    ,,17611 P(Mars)10510510 W?????r

    22Take A(Mars)?(1/2)?1.5?3.5 m (??0.5)eap

    Take P(Mars)?1 kWt

    22Take A(earth)?(1/2)?15?350 m (??0.5) eap

    22r?P(earth)?P(Mars)trA(earth)A(Mars)etet

     822(360?3?10)0.1211P(earth)?5?10?6.9 MWt3.5?350

    To reduce the required earth station power, take the earth station antenna

    22 (ans.) A?(1/2) ? 50?3927 me

    so

    62 Pans.(earth)6.910(15/50)620 kW ()???t

    AA(Mars)(earth)3.5?3930314eterPP(earth)?(Mars)?10?8?10 Wrt22822r? (360?3?10)0.12

    13 13 which is about 16% of the required 5 x 10W. The required 5 x 10W could be

    obtained by increasing the Mars transmitter power by a factor of 6.3. Other alternatives

    7

     or would be (1) to reduce the bandwidth (and data rate) reducing the required value of Pr

    (2) to employ a more sensitive receiver.

    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 = 23 11.38 x 10JK.

     6For B = 5 x 10 Hz (as given in this problem) and T = 50 K (an attainable value),

    23615 P(noise)?1.38?10?50?5?10?3.5?10 W

2-11-4. continued

    14 The received power (8 x 10W) 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 2required transmitter powers 6.7 = 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 6 the Jupiter link by a factor of about 50, making B = (5/50) x 10= 100 kHz.

*2-11-5. Moon link.

    A radio link from the moon to the earth has a moon-based 5? long right-handed mono-

    filar 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 -14antenna in order to deliver 10 W to the receiver? Take the earth-moon distance as 1.27 light-seconds.

Solution:

    89 ??????cf/310/1.5100.2 m,

    From (8-3-7) the directivity of the moon helix is given by

    2?D(moon) and ?D?12?5?60Aet4?

    From Friis formula

8

    22221482PrPr(4)10(3?10?1.27)4????2rr or A????152 mRCPer22?60PDPA?ttet

     about 14 m diameter (ans.)

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:

    P101822t(a) (ans.) PV (at earth)???5.5?10 Wm?5.5 aWm262?r?44(380?10)

    21/2(b) or E?(SZ)PV?S?E/Z

    181/291 or (ans.) E?(5.5?10?377)?45?10?45 nVm

    68(c) (ans.) t?r/c?380?10/3?10?1.27 s

    3492434(d) Photon = hf , where ?6.63?10?2?10?1.3?10 Jh?6.63?10 Js

    1812This is the energy of a 2.5 MHz photon. From (a), PV ?5.5?10 Jsm

    185.5?10621Therefore, number of photons = (ans.) ?4.2?10 ms241.3?10

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:

    22EE12From (2-16-3) we have for rms fields that PVS??avZo

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