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ANTENNA THEORY

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SOLUTIONS MANUAL TO ACCOMPANY

ANTENNA THEORY ANALYSIS AND DESIGN FOURTH EDITION

Constantine A. Balanis Arizona State University

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Copyright © 2016 by John Wiley & Sons, Inc. All rights reserved. Published by John Wiley & Sons, Inc., Hoboken, New Jersey. Published simultaneously in Canada. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording, scanning, or otherwise, except as permitted under Section 107 or 108 of the 1976 United States Copyright Act, without either the prior written permission of the Publisher, or authorization through payment of the appropriate per-copy fee to the Copyright Clearance Center, Inc., 222 Rosewood Drive, Danvers, MA 01923, (978) 750-8400, fax (978) 750-4470, or on the web at www.copyright.com. Requests to the Publisher for permission should be addressed to the Permissions Department, John Wiley & Sons, Inc., 111 River Street, Hoboken, NJ 07030, (201) 748-6011, fax (201) 748-6008, or online at http://www.wiley.com/go/permission. Limit of Liability/Disclaimer of Warranty: While the publisher and author have used their best efforts in preparing this book, they make no representations or warranties with respect to the accuracy or completeness of the contents of this book and specifically disclaim any implied warranties of merchantability or fitness for a particular purpose. No warranty may be created or extended by sales representatives or written sales materials. The advice and strategies contained herein may not be suitable for your situation. You should consult with a professional where appropriate. Neither the publisher nor author shall be liable for any loss of profit or any other commercial damages, including but not limited to special, incidental, consequential, or other damages. For general information on our other products and services or for technical support, please contact our Customer Care Department within the United States at (800) 762-2974, outside the United States at (317) 572-3993 or fax (317) 572-4002. Wiley also publishes its books in a variety of electronic formats. Some content that appears in print may not be available in electronic formats. For more information about Wiley products, visit our web site at www.wiley.com. Library of Congress Cataloging-in-Publication Data is available. 978-1-119-27374-5 Printed in the United States of America 10

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Contents

Preface

vii

2

Solution Manual

1

3

Solution Manual

57

4

Solution Manual

65

5

Solution Manual

125

6

Solution Manual

151

7

Solution Manual

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Solution Manual

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Solution Manual

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Solution Manual

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Solution Manual

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Solution Manual

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Solution Manual

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Solution Manual

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Solution Manual

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Preface

This Solutions Manual consists of solutions for all the problems found in Antenna Theory: Analysis and Design (4th edition, 2016) at the end of Chapters 2–16. There are 699 (103 new) problems, most of them with multiple parts. The degree of difficulty and length varies. While certain solutions need special functions, found in graphical form in the appendices, others require the use of the computer program. These computer programs are placed on a password protected website. All of the computer programs, especially those at the end of Chapters 6, 9, 11, 13 and 14 have been developed to design, respectively, uniform and nonuniform arrays, impedance transformers, log-periodic arrays, horns and microstrip patch antennas. In some cases, the computer programs also perform analysis on the designs. The programs at the end of Chapters 2, 4, 5, 7, 8, 10, 12, 15 and 16 are primarily developed for analysis. The problems have been designed to test the student’s grasp of this text’s material and to apply the concepts to the analysis and design of many practical radiators. In this fourth edition, more emphasis has been placed on design. To accomplish this, equations, procedures, examples, graphs, end-of-the-chapter problems, and computer programs have been developed. This manual has been prepared to assist the instructor in making homework and test assignments, and to provide one set of solutions for all of the problems. There maybe undoubtedly errors which have been overlooked. In addition, the solutions contained in this manual are not necessarily the simplest and/or the best. The author will, therefore, appreciate having errors brought to his attention and solicits alternate solutions to the problems. This Solutions Manual for the fourth edition has been prepared from the manuals of the first, second and third editions and many other new problems provided by the author.

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2

CHAPTER

Solution Manual

Exact 2.1. (a) dΩ = sin 𝜃 d𝜃 d𝜙 60◦

ΩA =

60◦

∫45◦ ∫30◦

dΩ =

Approximate )( ) 𝜋 𝜋 𝜋 𝜋 ΩA ≃ − − 3 4 3 6 ( )( ) 𝜋 𝜋2 𝜋 = ≃ 12 6 72 (

𝜋∕3

𝜋∕3

∫𝜋∕4 ∫𝜋∕6

sin 𝜃 d𝜃 d𝜙

ΩA ≃ 0.13708 sterads |𝜋∕3 |𝜋∕3 = (𝜙) |𝜋∕4 (− cos 𝜃)| |𝜋∕6 | ΩA ≃ (60 − 45)(60 − 30) ) ( 𝜋 𝜋 (−0.5 + 0.866) ≃ 450 (degrees)2 or error of = − 3 4 ( ) ( ) 450 − 314.5585 𝜋 (0.366) = 0.09582 sterads × 100 = 43.06% ΩA = 12 314.5585 { 0.09582 sterads )( ) ( ΩA = 180 180 = 314.5585 (degrees)2 0.09582 𝜋 𝜋 z

ΩA 30° 60°

y 60° 45° x

Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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SOLUTION MANUAL

(b) D0 =

4𝜋 4𝜋 = = 131.1456 (dimensionless) ΩA (sterads) 0.09582 = 10 log10 (131.1456) = 21.1775 dB

or )( ) 180 180 𝜋 𝜋 = 131.1456 (dimensionless) = 21.1775 dB ΩA (degrees)2 (

4𝜋 D0 = { D0 =

131.1456 (dimensionless) 21.1775 (dB)

] [ ] [ 2.2.  =  ×  = Re Eej𝜔t × Re Hej𝜔t ] [ ] [ Using the identity Re Aej𝜔t = 12 Eej𝜔t + E∗ e−j𝜔t The instant Poynting vector can be written as } { } 1 1 [Eej𝜔t + E∗ e−j𝜔t ] × [Hej𝜔t + H ∗ e−j𝜔t ] 2 2 { } 1 1 1 = [E × H ∗ + E∗ × H] + [E × Hej2𝜔t + E∗ × H ∗ e−j𝜔t ] 2 2 2 { } 1 1 1 = [E × H ∗ + (E × H ∗ )∗ ] + [E × Hej2𝜔t + (E × Hej𝜔t )∗ ] 2 2 2

=

{

Using the above identity again, but this time in reverse order, we can write that 1 [Re(E × H ∗ )] + 2 1 E2 52 = Re[E × H ∗ ] = â r = â 2 2𝜂 2(120𝜋) r =

2.3. (a) W rad

𝜋

2𝜋

(b) Prad = ∮ Wrad ds = ∫ s 0 𝜋

2𝜋

=

∫0

∫0

∫0

1 [Re(E × Hej2𝜔t )] 2 = 0.03315̂ar Watts∕m2

(0.03315)(r2 sin 𝜃 d𝜃 d𝜙)

(0.03315)(100)2 sin 𝜃 d𝜃 d𝜙

= 2𝜋(0.03315)(100)2

𝜋

∫0

sin 𝜃 d𝜃 = 2𝜋(0.03315)(100)2 ⋅ (2)

= 4165.75 Watts 2.4. (a) U(𝜃) = cos 𝜃 U(𝜃h ) = 0.5 = cos 𝜃h ⇒ 𝜃h = cos−1 (0.5) = 60◦ ⇒ Θh = 2(60◦ ) = 120◦ =

2𝜋 rads. 3

U(𝜃n ) = 0 = cos 𝜃n ⇒ 𝜃n = cos−1 (0) = 90◦ ⇒ Θn = 2(90◦ ) = 180◦ = 𝜋 rads.

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SOLUTION MANUAL

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(b) U(𝜃) = cos2 𝜃 U(𝜃h ) = 0.5 = cos2 𝜃h ⇒ 𝜃h = cos−1 (0.5)1∕2 = 45◦ ⇒ Θh = 2(45) = 90◦ = 𝜋∕2 rads U(𝜃n ) = 0 = cos2 𝜃n ⇒ 𝜃n = cos−1 (0) = 90◦ ⇒ Θn = 2(90◦ ) = 180◦ = 𝜋 rads (c) U(𝜃) = cos(2𝜃) 1 cos−1 (0.5) = 30◦ 2 ⇒ Θh = 2(30◦ ) = 60◦ = 𝜋∕3 rads

U(𝜃h ) = 0.5 = cos(2𝜃h ) ⇒ 𝜃h =

1 cos−1 (0) = 45◦ 2 ⇒ Θn = 2(45◦ ) = 90◦ = 𝜋∕2 rads

U(𝜃n ) = 0 = cos(2𝜃n ) ⇒ 𝜃n =

(d) U(𝜃) = cos2 (2𝜃) 1 cos−1 (0.5)1∕2 = 22.5◦ 2 𝜋 ⇒ Θh = 2(22.5◦ ) = 45◦ = rads 4 1 U(𝜃n ) = 0 = cos2 (2𝜃n ) ⇒ 𝜃n = cos−1 (0) = 45◦ 2 ⇒ Θn = 2(45◦ ) = 90◦ = 𝜋∕2 rads U(𝜃h ) = 0.5 = cos2 (2𝜃h ) ⇒ 𝜃h =

(e) U(𝜃) = cos(3𝜃) 1 cos−1 (0.5) = 20◦ 3 ⇒ Θh = 2(20◦ ) = 40◦ = 0.698 rads

U(𝜃h ) = cos(3𝜃h ) = 0.5 ⇒ 𝜃h =

1 cos−1 (0) = 30◦ 3 ⇒ Θn = 2(30◦ ) = 60◦ = 𝜋∕3 rads

U(𝜃n ) = cos(3𝜃n ) = 0 ⇒ 𝜃n =

(f) U(𝜃) = cos2 (3𝜃) 1 cos−1 (0.5)1∕2 = 15◦ 3 ⇒ Θh = 2(15◦ ) = 30◦ = 𝜋∕6 rads

U(𝜃h ) = 0.5 = cos2 (3𝜃h ) ⇒ 𝜃h =

1 cos−1 (0) = 30◦ 3 ⇒ Θn = 2(30◦ ) = 60◦ = 𝜋∕3 rads

U(𝜃n ) = 0 = cos2 (3𝜃n ) ⇒ 𝜃n =

2.5. Using the results of Problem 2.4 and a nonlinear solver to find the half power beamwidth of the radiation intensity represented by the transcentendal functions, we have that: { HPBW = 55.584◦ (a) U(𝜃) = cos 𝜃 cos(2𝜃) ⇒ FNBW = 90◦

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SOLUTION MANUAL

{ (b) U(𝜃) = cos2 𝜃 cos2 (2𝜃) ⇒

HPBW = 40.985◦ FNBW = 90◦

{

HPBW = 38.668◦ FNBW = 60◦ { HPBW = 28.745◦ (d) U = cos2 𝜃 cos2 (3𝜃) ⇒ FNBW = 60◦ { HPBW = 34.942◦ (e) U = cos(2𝜃) cos(3𝜃) ⇒ FNBW = 60◦ { HPBW = 25.583◦ (f) U = cos2 (2𝜃) cos2 (3𝜃) ⇒ FNBW = 60◦ (c) U = cos 𝜃 cos(3𝜃) ⇒

2.6. (a) D0 =

4𝜋Umax 4𝜋(200 × 10−3 ) = = 22.22 = 13.47 dB Prad 0.9(125.66 × 10−3 )

G0 = 𝜀cd D0 = 0.9(22.22) = 20 = 13.01 dB (b) D0 =

4𝜋Umax 4𝜋(200 × 10−3 ) = = 20 = 13.01 dB Prad (125.66 × 10−3 )

G0 = 𝜀cd D0 = 0.9 ⋅ (20) = 18 = 12.55 dB 2.7.

U = B0 cos2 𝜃 𝜋∕2

2𝜋

(a)

Prad =

∫0

∫0 𝜋∕2

= 2𝜋B0

∫0

U sin 𝜃 d𝜃 = 2𝜋B0

U= =

∫0

cos2 𝜃 sin 𝜃 d𝜃

cos2 𝜃 d (− cos 𝜃) [

𝜋∕2

Prad = −2𝜋B0

𝜋∕2

cos3 𝜃 || 3 ||0

= −2𝜋B0

] 2𝜋 −1 15 = B = 10 ⇒ B0 = 3 3 0 𝜋

| U| 15 cos2 𝜃 || 15 = 2 || = cos2 𝜃 ⇒ Wrad || 𝜋 𝜋 r |max r2 ||max |max 15 = 4.7746 × 10−6 Watts∕m2 @ 𝜃 = 0◦ 𝜋(103 )2

| Wrad || = 4.7746 × 10−6 Watts∕m2 @ 𝜃 = 0◦ |max 𝜋

2𝜋

(b)

ΩA (exact) = ΩA (exact) =

∫0

∫0

Un cos2 𝜃 sin 𝜃 d𝜃 d𝜙

2𝜋 steradians = 2.0944 sterads = 6, 875.51 (degrees)2 3

U = 0.5 = cos2 𝜃h ⇒ 𝜃h = cos−1 (0.5)1∕2 = 45◦ ( ΩA

Kraus’ approx

)

⇒ Θh = 2(45◦ ) = 90◦ = 𝜋∕2 rads = Θ2h = (𝜋∕2)2 =

𝜋2 = 2.4674 sterads = 8, 099.997 (degrees)2 4

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SOLUTION MANUAL

(c)

D0 (exact) = D0 (approx∕Kraus’) =

4𝜋 4𝜋 = = 6 = 7.782 dB ΩA (exact) 2𝜋∕3 4𝜋 4𝜋 16 = = 5.093 = 7.0697 = ΩA (approx) 𝜋 2 ∕4 𝜋

(d) G0 Assuming lossless antenna (Pin = Prad ) G0 (exact) = D0 (exact) = 6 = 7.782 dB G0 (approx) = D0 (approx) = 5.093 = 7.0697 dB U = B0 cos3 𝜃

(a)

(b)

( ) 𝜋 1 = B0 = 10 ⇒ B0 = 20∕𝜋 Prad = −2𝜋B0 − 4 2 | 20 1 20 Wrad || = = × 10−6 = 6.366 × 10−6 Watts∕m2 𝜋 𝜋2 𝜋 |max ΩA (exact) = (𝜋∕2) = 1.5708 sterads U = 0.5 = cos3 𝜃h ⇒ 𝜃h cos−1 (0.5)1∕3 = 37.467◦ ⇒ Θh = 2(37.467◦ ) = 74.934◦ = 1.30785 rads ΩA (approx) = (1.30785)2 = 1.71 sterads

(c)

D0 (exact) = 4𝜋∕(𝜋∕2) = 8 = 9.031 dB D0 (approx) =

4𝜋 = 7.347 = 8.66 dB 1.71

(d) Assuming lossless antenna ⇒ Gain = Directivity (see part c) −jkr

2.8. Ea = â 𝜃 Ea sin1.5 𝜃 e r ⇒ Un = (sin1.5 𝜃)2 = sin3 𝜃 Normalized Un | 4𝜋Umax (a) D0 = , Umax = Un | max = sin3 𝜃 || max = 1, 𝜃max = 90◦ Prad |𝜃=𝜃 max 𝜋

2𝜋

Prad =

∫0

∫0

Un sin 𝜃 d𝜃 d𝜙 =

𝜋

2𝜋

∫0

∫0

sin3 𝜃 sin 𝜃 d𝜃 d𝜙 = 2𝜋

𝜋

∫0

sin4 𝜃 d𝜃

] [ ( 𝜋 )𝜋 ] 𝜋 3 3 1 1 sin3 𝜃 cos 𝜃 || 2 = 2𝜋 − | + 4 ∫ sin 𝜃 d𝜃 = 2𝜋 4 2 − 4 sin(2𝜃) 0 4 |0 0 [ ( )] 2 3𝜋 3 𝜋 = = 2𝜋 4 2 4 4𝜋Umax 4𝜋(1) 16 = D0 = = 1.698 = 2.298 dB = 2 Prad 3𝜋 3𝜋 ∕4 (b)

[

Un = sin3 𝜃, Unmax = 1, 𝜃max = 90◦ ] [ | Un |𝜃=𝜃h = 0.5 = sin3 𝜃h ⇒ 𝜃h = sin−1 (0.54 )3 = sin−1 (0.794) = 52.533◦ | HPBW = Θh = 2(𝜃max − 𝜃h ) = 2(90 − 52.533) Θh = 2(37.467) = 74.934◦

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SOLUTION MANUAL

z

θh

Θh

θ max 90° y

(c) Because pattern is omnidirectional: D0 (McDonald) =

101 101 = 2 HPBW − 0.0027(HPBW) 74.934 − 0.0027(74.934)2

D0 (McDonald) =

101 101 = = 1.690 = 10 log10 (1.690) = 2.278 74.934 − 15.161 59.773

(d) Because pattern is omnidirectional: √ D0 (Pozar) = −172.4 + 191

√ 1 1 0.818 + = −172.4 + 191 0.818 + HPBW 74.934

= −172.4 + 191(0.912) = −172.4 + 174.150 = 1.750 P0 (Pozar) = 1.750 = 10 log10 (1.750) = 2.431 dB (e) Computer Program Directivity: D0 = 1.693 = 2.2864 dB Input parameters: ----------------The lower bound of The upper bound of The lower bound of The upper bound of

theta in degrees theta in degrees phi in degrees = phi in degrees =

= 0 = 180 0 360

Output parameters: ------------------Radiated power (watts) = 7.4228 Directivity (dimensionless) = 1.6930 Directivity (dB) = 2.2864

2.9. U(𝜃, 𝜙) = cosn (𝜃) 0 ≤ 𝜃 ≤ 𝜋∕2, 0 ≤ 𝜙 ≤ 2𝜋 (a) Un (𝜃n , 𝜙) = 0.5 = cosn (5◦ ) = [cos(5◦ )]n = (0.99619)n 0.5 = (0.99619)n log10 (0.5) = log[(0.99619)n ] = n log10 (0.99619) = n(−0.00166) −0.30103 = −0.00166n n = 181.34

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SOLUTION MANUAL

(b) U(𝜃, 𝜙) = cos181.34 (𝜃); Umax = 1, 𝜃 = 0◦ 𝜋∕2

2𝜋

Prad =

∫0

∫0

U(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙 = 2𝜋

𝜋∕2

∫0

cos181.34 (𝜃) sin 𝜃 d𝜃

]𝜋∕2 [ [ ] cos182.34 (𝜃) 2𝜋 1 2𝜋 = = −0 + = 0.03446 = 2𝜋 − 182.34 182.34 182.34 0 D0 =

4𝜋Umax 4𝜋(1) = (182.34) = 2(182.34) = 364.68 Prad 2𝜋 D0 = 364.68 = 25.62 dB

(c) Kraus’ Approximation (2.27): D0 ≃

41, 253 41, 253 = = 412.53 = 26.15 dB Θ1d Θ2d (10)(10) D0 ≃ 412.53 = 26.15 dB

(d) Tai & Pereira (2.30b): 72,815 72,815 72,815 = = = 364.075 = 25.61 dB 2 200 2(10)2 + Θ2d

D0 ≃

Θ21d

D0 ≃ 364.075 = 25.61 dB 2.10.

⎧1 ⎪ U(𝜃, 𝜙) = ⎨ 0.342 csc(𝜃) ⎪0 ⎩ 𝜋

2𝜋

Prad =

∫0

[

∫0

0◦ ≤ 𝜃 ≤ 20◦ ⎫ ⎪ 20◦ ≤ 𝜃 ≤ 60◦ ⎬ 0◦ ≤ 𝜙 ≤ 360◦ 60◦ ≤ 𝜃 ≤ 180◦ ⎪ ⎭

U(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙

20◦

60◦

sin 𝜃 d𝜃 + 0.342 csc(𝜃) × sin 𝜃 d𝜃 ∫0 ∫20◦ { } |𝜋∕9 |𝜋∕3 = 2𝜋 − cos 𝜃 | + 0.342 ⋅ 𝜃 | |0 |𝜋∕9 ] ( )} {[ ( ) 𝜋 𝜋 𝜋 + 1 + 0.342 = 2𝜋 − cos − 9 3 9 { ( )} 2 = 2𝜋 [−0.93969 + 1] + 0.342𝜋 9 = 2𝜋{0.06031 + 0.23876} = 1.87912

= 2𝜋

D0 =

4𝜋Umax 4𝜋(1) = = 6.68737 = 8.25255 dB Prad 1.87912

]

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SOLUTION MANUAL

41,253 41,253 2.11. (a) D0 ≃ Θ Θ = 30(35) = 39.29 = 15.94 dB 1d 2d Aem = (b) D0 ≃ Aem = 2.12. D0 = (a)

λ2 D 4𝜋 0 72,815 72,815 = = 34.27 = 15.35 dB (30)2 + (35)2 Θ21d + Θ22d λ2 D 4𝜋 0

4𝜋Umax Prad U = sin 𝜃 sin 𝜙

for

0 ≤ 𝜃 ≤ 𝜋, 0 ≤ 𝜙 ≤ 𝜋

U ||max = 1 and it occurs when 𝜃 = 𝜙 = 𝜋∕2. 𝜋

Prad =

∫0 ∫0

𝜋

U sin 𝜃 d𝜃 d𝜙 =

𝜋

sin 𝜙 d𝜙

∫0

𝜋

∫0

sin2 𝜃 d𝜃 = 2

( ) 𝜋 = 𝜋. 2

4𝜋(1) Thus D0 = = 4 = 6.02 dB 𝜋 The half-power beamwidths are equal to HPBW (az.) = 2[90◦ − sin−1 (1∕2)] = 2(90◦ − 30◦ ) = 120◦ HPBW (el.) = 2[90◦ − sin−1 (1∕2)] = 2(90◦ − 30◦ ) = 120◦ In a similar manner, it can be shown that for the following: (b) U = sin 𝜃 sin2 𝜙 ⇒ D0 = 5.09 = 7.07 dB HPBW (el.) = 120◦ , HPBW (az.) = 90◦ (c) U = sin 𝜃 sin3 𝜙 ⇒ D0 = 6 = 7.78 dB HPBW (el.) = 120◦ , HPBW (az.) = 74.93◦ (d) U = sin2 𝜃 sin 𝜙 ⇒ D0 = 12𝜋∕8 = 4.71 = 6.73 dB HPBW (el.) = 90◦ , HPBW (az.) = 120◦ (e) U = sin2 𝜃 sin2 𝜙 ⇒ D0 = 6 = 7.78 dB HPBW (az.) = HPBW (el.) = 90◦ (f) U = sin2 𝜃 sin3 𝜙 ⇒ D0 = 9𝜋∕4 = 7.07 = 8.49 dB HPBW (el.) = 90◦ , HPBW (az.) = 74.93◦ 2.13. U = sin 𝜃 cos2 𝜙, 0 ≤ 𝜃 ≤ 180◦ , 90◦ ≤ 𝜙 ≤ 270◦ | 4𝜋Umax , Umax = sin 𝜃 cos2 𝜙|| =1 (a) D0 = Prad | 𝜃=90◦◦ 𝜙=180

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SOLUTION MANUAL

𝜋

3𝜋∕2

Prad =

∫𝜋∕2 3𝜋∕2

=

∫𝜋∕2

∫0

U(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙 =

cos2 𝜙 d𝜙

𝜋

∫0

𝜋

3𝜋∕2

∫𝜋∕2

∫0

9

sin 𝜃 cos2 𝜙 sin 𝜃 d𝜃 d𝜙

sin2 𝜃 d𝜃

( )( ) 𝜋 𝜋2 𝜋 = 2 2 4 4𝜋(1) 16 D0 = 2 = 5.09296 = 10 log10 (5.09296) = 7.0697 dB = 𝜋 𝜋 ∕4

Prad =

D0 (exact) = 5.09296(dim) = 7.0697 dB (b) Azimuth (Horizontal) Principal Plane (𝜃 = 90◦ ): U(𝜃 = 90◦ ) = sin 𝜃 cos2 𝜙|𝜃=90◦ = cos2 𝜙

√ Uh = cos2 𝜙|𝜙=𝜙h = 0.5 ⇒ 𝜙h = cos−1 (± 0.5) = cos−1 (±0.707) = 135◦

Φh (az) = 2(180 − 135) = 2(45◦ ) = 90◦ Φh (az) = 90◦ (c) Elevation (vertical) Principal plane (𝜙 = 180◦ ): U(𝜙 = 180◦ ) = sin 𝜃 cos2 𝜙|𝜙=180◦ = sin 𝜃 Uh = sin 𝜃|𝜃=𝜃h = 0.5 ⇒ 𝜃h = sin−1 (0.5) = 30◦ Θh = 2(90◦ − 30◦ ) = 2(60) = 120◦ Θh (elev) = 120◦ (d) Either: D0 (Kraus) =

41,253 41,253 = ◦ = 3.8197 = 5.82 dB Φh Θh 90 (120◦ ) D0 (Kraus) = 3.8197 dim = 5.82 dB

or: D0 (Tai & Pereira) =

72,815 72,815 72,815 = = = 3.236 2 2 ◦ 2 ◦ 2 25,500 (90 ) + (120 ) Φh + Θh D0 (T&P) = 3.236 dim = 5.1 dB

2.14. Using the half-power beamwidths found in Problem 2.12, the directivity for each intensity using Kraus’ and Tai & Pereira’s formulas is given by U = sin 𝜃 sin 𝜙; 41253 41253 (a) D0 ≃ = = 2.86 = 4.57 dB Θ1d Θ2d 120(120)

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SOLUTION MANUAL

(b) D0 ≃

72,815 72,815 = = 2.53 = 4.03 dB (120)2 + (120)2 Θ21d + Θ22d

U = sin 𝜃 sin2 𝜙; (a) D0 ≃ 3.82 = 5.82 dB (b) D0 ≃ 3.24 = 5.10 dB U = sin 𝜃 sin3 𝜙; (a) D0 ≃ 4.59 = 6.62 dB (b) D0 ≃ 3.64 = 5.61 dB U = sin2 𝜃 sin 𝜙; (a) D0 ≃ 3.82 = 5.82 dB (b) D0 ≃ 3.24 = 5.10 dB U = sin2 𝜃 sin2 𝜙; (a) D0 ≃ 5.09 = 7.07 dB (b) D0 ≃ 4.49 = 6.53 dB U = sin2 𝜃 sin3 𝜙; (a) D0 ≃ 6.12 = 7.87 dB (b) D0 ≃ 5.31 = 7.25 dB 2.15. (a) D0 =

(b) D0 =

4𝜋 4𝜋 = = 5.5377 = 7.433 dB Θ1r Θ2r (1.5064)2 32 ln(2) Θ21r

2.16. (a) D0 =

+ Θ22r

32 ln(2) = 4.88725 = 6.8906 dB (1.5064)2 + (1.5064)2

4𝜋Umax U = max Prad U0 𝜋

2𝜋

Prad =

=

∫0

∫0 60◦

U sin 𝜃 d𝜃 d𝜙 = 2𝜋

𝜋

∫0

90◦

{ U sin 𝜃 d𝜃 = 2𝜋

∫0

}

(0.5) sin 𝜃 d𝜃 + (0.1) sin 𝜃 d𝜃 ∫30◦ ∫60◦ { ◦ ◦} ) ◦ ( cos 𝜃 |60 |30 |90 = 2𝜋 (− cos 𝜃)| + − | ◦ + (−0.1 cos 𝜃)| ◦ |30 |0 |60 2 ) ( )} { ( −0 + 0.5 −0.5 + 0.866 + = 2𝜋 (−0.866 + 1) + 2 10 = 2𝜋{−0.866 + 1 − 0.25 + 0.433 + 0.05} = 2𝜋(0.367) +

Prad

30◦

= 0.734𝜋 = 2.3059 1(4𝜋) D0 = = 5.4496 = 7.3636 dB 2.3059 (b) D0 (dipole) = 1.5 = 1.761 dB D0 (above dipole) = (7.3636 − 1.761) dB = 5.6026 dB D0 (above dipole) = 5.45 = 3.633 = 5.603 dB 1.5

sin 𝜃 d𝜃

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SOLUTION MANUAL

𝜋

2𝜋

2.17. (a) Prad =

Umax

∫0

U(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙 =

∫0 ( ) 𝜋 1 = = (𝜋) 5 5 = U(𝜃 = 0◦ , 𝜙 = 𝜋∕2) = 1

D0 =

2𝜋

∫0

sin2 𝜙 d𝜙

𝜋∕2

∫0

cos4 𝜃 sin 𝜃 d𝜃

4𝜋Umax 4𝜋 = = 20 = 13.0 dB Prad (𝜋∕5)

(b) Elevation Plane: 𝜃 varies, 𝜙 fixed ⇒ Choose 𝜙 = 𝜋∕2. U(𝜃, 𝜙 = 𝜋∕2) = cos4 𝜃, 0 ≤ 𝜃 ≤ 𝜋∕2. ] [ HPBW(el.) 1 = cos4 2 2 √ HPBW(el.) = 2 cos−1 { 0.5}1∕2 = 65.5◦ 2𝜋

2.18. (a) Prad =

𝜋

∫0 ∫ 0 { ◦ ⋅

30

∫0 {

U(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙 = 2𝜋 sin 𝜃 d𝜃 +

90◦

∫30◦

cos 𝜃 sin 𝜃 d𝜃 0.866

}

} 1 = 2𝜋 sin 𝜃 d𝜃 + cos 𝜃 sin 𝜃 d𝜃 ∫0 ∫𝜋∕6 0.866 { ( ) 𝜋∕2 } 1 cos2 𝜃 || 𝜋∕6 = 2𝜋 − cos 𝜃|0 + = 2𝜋[−0.866 + 1 + 0.433] − | 0.866 2 |𝜋∕6 𝜋∕6

𝜋∕2

Prad = 3.5626 D0 =

(b)

4𝜋Umax 4𝜋(1) = = 3.5273 = 5.4745 dB Prad 3.5626

cos(𝜃) = 0.5 ⇒ cos 𝜃 = 0.5(0.866) = 0.433, 𝜃 = cos−1 (0.433) = 64.34◦ 0.866 = 2(64.34) = 128.68◦ = 2.246 rad = Θ2r

U= Θ1r

D0 ≃

4𝜋 4𝜋 = = 2.4912 = 3.9641 dB Θ1r Θ2r (2.246)2

2.19. (a) 35 dB |E | | E | 35 (b) 20 log10 || max || = 35, log10 || max || = = 1.75 | Es | | Es | 20 | Emax | 1.75 | | = 56.234 | E | = 10 | s |

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SOLUTION MANUAL

𝜋

2𝜋

= D0 =

∫0

∫0

𝜋

2𝜋

2.20. (a) U = sin 𝜃, Umax = 1,

Prad =

∫0

∫0

U sin 𝜃 d𝜃 d𝜙

sin2 𝜃 d𝜃 d𝜙 = 𝜋 2

4𝜋Umax 4𝜋 4 = 2 = = 1.2732 Prad 𝜋 𝜋

(b) HPBW = 120◦ , 2𝜋∕3 The directivity based on (2-33a) is equal to, D0 =

101 = 1.2451 120◦ − 0.0027(120◦ )2

while that based on (2-33b) is equal to, √ D0 = −172.4 + 191

0.818 +

1 = 1.2245 120◦

(c) Computer Program: D0 = 1.2732 2.21. (a) U = sin3 𝜃, Umax = 1, Prad = D0 =

4𝜋 3 2 𝜋 4

=

𝜋

2𝜋

∫0

∫0

sin4 𝜃 d𝜃 d𝜙 =

3 2 𝜋 4

16 = 1.6976 3𝜋

(b) HPBW = 74.93◦ 101 = 1.68971 ◦ )2 (74.93◦ ) − 0.0027(74.93) √ 1 From (2-33b), D0 = −172.4 + 191 0.818 + = 1.75029 74.93◦ (c) Computer program: D0 = 1.693 The value of D0 = 1.693 is similar to that of (4-91) or 1.643 From (2-33a), D0 =

2.22. (a) U = J1 2 (ka sin 𝜃), 𝜋 a = λ∕10, ka sin 𝜃 = sin 𝜃. HPBW = 93.10◦ 5 2 From (2-33a): D0 = 101∕[(93.10) − √0.0027(93.10) ] = 1.449120 1 From (2-33b): D0 = −172.4 + 191 0.818 + = 1.477271 93.10 a = λ∕20, ka sin 𝜃 =

𝜋 sin 𝜃, HPBW = 91.10◦ 10

From (2-33a), D0 = 1.47033; From (2-33b), D0 = 1.502 (b)

a=

λ :P = 10 rad ∫0

𝜋

2𝜋

∫0

Umax = 0.0893, D0 =

J12 (ka sin 𝜃) ⋅ sin 𝜃 d𝜃 d𝜙 = 0.7638045

4𝜋(0.0893) = 1.469193 0.7638045

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SOLUTION MANUAL

a=

λ : Prad = ∫0 20

𝜋

2𝜋

∫0

Umax = 0.0240714, D0 =

13

J1 2 (𝜋∕10 sin 𝜃) sin 𝜃 d𝜃 d𝜙 = 0.202604 4𝜋(0.0240714) = 1.49257 0.202604

If the radius of loop is smaller than λ∕20, the directivity approaches 1.5. 2.23. Using the numerical techniques, the directivity for each intensity of Prob. 2.12, with 10◦ uniform divisions is equal to for U = sin 𝜃 sin 𝜙: 4𝜋Umax (a) Midpoint: D0 = Prad 18 ( ) 18 ∑ 𝜋 𝜋 ∑ Umax = 1: Prad = sin 𝜙j sin2 𝜃i 18 18 j=1 i=1 𝜋 𝜋 + (i − 1) , i = 1, 2, 3, ..., 18 36 18 𝜋 𝜋 + (j − 1) , j = 1, 2, 3, ..., 18 𝜙j = 36 18 ( )2 𝜋 Prad = (11.38656)(8.9924) = 3.119 18 4𝜋(1) D0 = = 4.03 = 6.05 dB 3.119 𝜃i =

(b) Trailing edge of each division: Trailing edge: 𝜃i = i(𝜋∕18), i = 1, 2, 3, … , 18 𝜙j = j(𝜋∕18), j = 1, 2, 3, … , 18 ( )2 𝜋 Prad = (11.25640)(8.96985) = 3.076 18 4𝜋(1) D0 = = 4.09 = 6.11 dB 3.119 In a similar manner: U = sin 𝜃 sin2 𝜙; (a) Prad = 2.463 ⇒ D0 = 5.10 = 7.07 dB (b) Prad = 2.451 ⇒ D0 = 5.13 = 7.10 dB U = sin 𝜃 sin3 𝜙; (a) Prad = 2.092 ⇒ D0 = 6.01 = 7.79 dB (b) Prad = 2.086 ⇒ D0 = 6.02 = 7.80 dB U = sin2 𝜃 sin 𝜙; (a) Prad = 2.469 ⇒ D0 = 4.74 = 6.76 dB (b) Prad = 2.618 ⇒ D0 = 4.80 = 6.81 dB U = sin2 𝜃 sin2 𝜙; (a) Prad = 2.092 ⇒ D0 = 6.01 = 7.79 dB (b) Prad = 2.086 ⇒ D0 = 6.02 = 7.80 dB U = sin2 𝜃 sin3 𝜙; (a) Prad = 1.777 ⇒ D0 = 7.07 = 8.49 dB (b) Prad = 1.775 ⇒ D0 = 7.08 = 8.50 dB

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SOLUTION MANUAL

2.24. Using the computer program Directivity of Chapter 2, the directivities for each radiation intensity of Problem 2.12 are equal to: (a) U = sin 𝜃 sin 𝜙; Prad = 3.1318 4𝜋 ⋅ Umax = 4.0125 ⇒ 6.034 dB Umax = 1; D0 = 3.1318 (b) U = sin 𝜃 sin2 𝜙; Prad = 2.4590 4𝜋 ⋅ 1 Umax = 1; D0 = = 5.110358 ⇒ 7.0845 dB 2.4590 (c) U = sin 𝜃 sin3 𝜙; Prad = 2.0870 4𝜋 ⋅ 1 Umax = 1; D0 = = 6.02124 ⇒ 7.80 dB 2.0870 2 (d) U = sin 𝜃 sin 𝜙; Prad = 2.6579 4𝜋 ⋅ 1 Umax = 1; D0 = = 4.72793 ⇒ 6.746 dB 2.6579 (e) U = sin2 𝜃 sin2 𝜙; Prad = 2.0870 4𝜋 ⋅ 1 Umax = 1; D0 = = 6.02126 ⇒ 7.7968 dB 2.0870 (f) U = sin2 𝜃 sin3 𝜙; Prad = 1.7714 4𝜋 ⋅ 1 Umax = 1; D0 = = 7.09403 ⇒ 8.5089 dB 1.7714 [ ] 𝜋 2.25. (a) E|max = cos (cos 𝜃 − 1) |max = 1 at 𝜃 = 0◦ . 4 [ ] 𝜋 0.707Emax = 0.707 ⋅ (1) = cos (cos 𝜃1 − 1) 4 { cos−1 (2) = does not exist 𝜋 𝜋 𝜋 (cos 𝜃1 − 1) = ± ⇒ 𝜃1 = cos−1 (0) = 90◦ = rad. 4 4 2 ( ) 𝜋 =𝜋 Θ1r = Θ2r = 2 2 4𝜋 4𝜋 4 D0 ≃ = 2 = = 1.273 = 1.049 dB Θ1r Θ2r 𝜋 𝜋 (b) Using the computer program Directivity of Chapter 2 D0 = 2.00789 = 3.027 dB Since the pattern is not very narrow, the answer obtained using Kraus’ approximate formula is not as accurate. [ ]| | 𝜋 2.26. (a) E|| = cos (cos 𝜃 + 1) || = 1 at 𝜃 = 𝜋. 4 |max |max ] [ 𝜋 0.707 = cos (cos 𝜃1 + 1) 4 { cos−1 (−2) → does not exist. 𝜋 𝜋 𝜋 (cos 𝜃1 + 1) = ± ⇒ 𝜃1 = cos−1 (0) → 90◦ → rad 4 4 2 ( ) 𝜋 =𝜋 Θ1r = Θ2r = 2 2 4𝜋 4 D0 ≃ 2 = = 1.273 = 1.049 dB 𝜋 𝜋

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SOLUTION MANUAL

(b) Computer Program Directivity: D0 = 2.00789 = 3.027 dB 𝜋∕2

2𝜋

2.27. (a) Prad = D0 =

∫0

∫0

𝜋 U0 sin(𝜋 sin 𝜃) sin 𝜃 d𝜃 d𝜙 = 2𝜋U0 J1 (𝜋) = U0 𝜋 2 J1 (𝜋) 2

4𝜋U0 4𝜋Umax 4 1 = = 4.4735 = 2 Prad U0 𝜋 J1 (𝜋) 𝜋 J1 (𝜋) 𝜋 J (𝜋) = 0.447 2 1

(b) Computer program Directivity: 𝜋∕2

2𝜋

Prad =

∫0

∫0

U0 sin(𝜋 sin 𝜃) sin 𝜃 d𝜃 d𝜙 = 2𝜋(0.447)

D0 = 4.4735 −jkr

2.28. E𝜙 = C0 sin1.5 𝜃 e r

(a) Un = |E𝜙 |2 = e20 sin3 𝜃, ⇒ Un| max = C02 𝜋

2𝜋

Prad = 𝜋

∫0

Prad

∫0

∫0

U sin 𝜃 d𝜃 d𝜙 = 2𝜋

𝜋

∫0

C02 sin3 𝜃 sin 𝜃 d𝜃 = C0 (2𝜋) 𝜋

𝜋

∫0

sin4 𝜃 d𝜃

𝜋

sin3 𝜃 cos 𝜃 |𝜋 4 − 3 3 sin2 𝜃 d𝜃 = sin2 𝜃 d𝜃 | + |0 4 4 ∫0 4 ∫0 [ ( ) ]𝜋 3𝜋 3 𝜋 3 𝜃 1 = − sin(2𝜃) = = 4 2 4 4 2 8 0 ( ) 2 3𝜋 2 3𝜋 = = 2𝜋C02 C 8 4 0

sin4 𝜃 d𝜃 = −

4𝜋C02 4𝜋Umax 16 D0 = = 2 = = 1.69765 = 2.298 dB 3𝜋 Prad 3𝜋 2 C0 4 D0 = 1.69765 = 2.298 dB (b) Un = C0 sin3 𝜃, Un ||max = C2 at 𝜃 = 90◦ , Un || = 0.5C02 = sin3 𝜃h C02 0 | |𝜃=𝜃n sin3 𝜃h = 0.5, 𝜃h = sin−1 (0.5)1∕3 = sin−1 (0.7937) = 52.5327◦ Θh = 2(90◦ − 52.5327◦ ) = 74.935◦ D0 (McDonald) =

101 101 = = 1.6897 = 2.278 dB 2 59.7738 74.935 − 0.0027(74.935)

D0 (McDonald) = 1.6897 dimensionless = 2.278 dB √ 1 D0 (Pozar) = −172.4 + 191 0.818 + = −172.4 + 191(0.91178) 74.935◦ D0 (Pozar) = −172.4 + 174.1502 = 1.7502 dimensionless = 2.431 dB

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SOLUTION MANUAL

2.29. (a) Using the computer program Directivity of Chapter 2. D0 = 14.0707 dimensionless = 11.48 dB ]2 sin(𝜋 sin 𝜃) (b) U ∣max = = 1 when 𝜃 = 0◦ . 𝜋 sin 𝜃 max [ ] sin(𝜋 sin 𝜃1 ) 2 1 1 U = Umax = (1) = 2 2 𝜋 sin 𝜃1 [

Iteratively we obtain 𝜃1 = 26.3◦ . Therefore Θ1d = Θ2d = 2(26.3◦ ) = 52.6◦ . 41, 253 = 14.91 dimensionless = 11.73 dB using the Kraus’ formula (52.6)2 (c) For Tai and Pereira’s formula and D0 ≃

D0 =

72,815 72,815 = = 13.16 dimensionless = 11.19 dB 2 2(52.6)2 2 ⋅ Θ1d

1 1 1 |E|2 = sin 𝜃 cos2 𝜙 ⇒ Umax = 2𝜂 2𝜂 2𝜂 ( ) ( ) 𝜋∕2 𝜋 𝜋 𝜋2 1 𝜋 1 (a) Prad = 2 ⋅ = sin2 𝜃 cos2 𝜙 d𝜃 d𝜙 = ∫0 ∫0 2𝜂 𝜂 4 2 8𝜂 ( ) 1 4𝜋 2𝜂 4𝜋Umax 16 = = = 5.09 = 7.07 dB D0 = prad 𝜋 𝜋2 8𝜂 1 at 𝜃 = 𝜋∕2, 𝜙 = 0 (b) Umax = 2𝜂 1 In the elevation plane through the maximum 𝜙 = 0 and U = sin 𝜃, the 3-dB point 2𝜂 occurs when ( ) 1 1 U = 0.5Umax = 0.5 = sin 𝜃1 ⇒ 𝜃1 = sin−1 (0.5) = 30◦ 2𝜂 2𝜂

2.30. U =

Therefore Θ1d = 2(90 − 30) = 120◦

1 In the azimuth plane through the maximum 𝜃 = 𝜋∕2 and U = cos2 𝜙, the 3-dB point 2𝜂 ( ) 1 1 occurs when U = 0.5Umax = 0.5 = cos2 𝜃1 ⇒ 𝜙1 = cos−1 (0.707) = 45◦ 2𝜂 2𝜂 Θ2d = 2(90◦ − 45◦ ) = 90◦ Therefore using Kraus’ formula: D0 ≃

41,253 = 3.82 dimensionless = 5.82 dB 120(90)

(c) Using Tai and Pereira’s formula: D0 ≃

72,815 72,815 = = 3.24 dimensionless = 5.10 dB 2 (120)2 + (90)2 + Θ2d

Θ21d

(d) Using the computer program Directivity of Chapter 2. D0 = 5.16425 = 7.13 dB

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SOLUTION MANUAL

17

[

] ] ] [ [ J1 (ka sin 𝜃) 2 J (ka sin 𝜃) 2 J (ka sin 𝜃) 2 = (ka)2 1 = U0 1 sin 𝜃 ka sin 𝜃 ka sin 𝜃 ( )2 U 1 (a) Umax = U0 = 0 and it occurs when ka sin 𝜃 = 0 ⇒ 𝜃 = 0◦ . The 3-dB point is 2 4 obtained using

2.31. U =

U=

] [ U J (ka sin 𝜃) J (ka sin 𝜃) 2 1 ⇒ 1 Umax = 0 = U0 1 = 0.3535 2 8 ka sin 𝜃 ka sin 𝜃

with the aid of the J1 (x)∕x of Appendix V. x = ka sin 𝜃1 = 1.61 ⇒ 𝜃1 = sin−1 (1.61∕2𝜋) = 14.847◦ ⇒ Θ1r = 29.694◦ (b) Since Θ1r = Θ2r = 29.694◦ , the directivity using Kraus’ formula is equal to D0 ≃

2.32.

41, 253 = 46.79 dimensionless = 16.70 dB (29.694)2

G0 = 16 dB ⇒ 16 = 10 log10 G0 (dimensionless) ⇒ G0 (dimensionless) = 101.6 = 39.81 r = 100 meters = 10, 000 cm = 104 cm Prad = ecd Pin = (1)Pin = 8 watts f = 1,900 MHz ⇒ λ = 30 × 109 ∕1.9 × 109 = 15.789 cm (a)

W0 = =

Prad 4𝜋r2

=

8 8 = 4𝜋(104 )2 4𝜋 × 108

2 × 10−8 = 0.6366 × 10−8 Watts∕cm2 𝜋

W0 = 0.6366 × 10−8 = 6.366 × 10−9 Watts∕cm2 Wmax = W0 G0 (dim) = 6.366 × 10−9 (39.81) = 253.438 × 10−9 . Wmax = 253.438 × 10−9 Watts∕cm2 (b) D0 (λ∕4 monopole) = 1.643 Aem =

1.643(15.789)2 λ2 λ2 D0 = (1.643) = = 32.5938 cm2 4𝜋 4𝜋 4𝜋

Aem = 32.5938 cm2 P(received) = Wmax Aem = (253.438 × 10−9 )(32.5938) P(received) = 8.2606 × 10−6 Watts 2.33. (a) Linear because Δ𝜙 = 0. (b) Linear because Δ𝜙 = 0.

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SOLUTION MANUAL

(c) Circular because 1. Ex = Ey 2. Δ𝜙 = 𝜋∕2. CCW because Ey leads Ex . AR = 1, 𝜏 = 90◦ (d) Circular because 1. Ex = Ey 2. Δ𝜙 = −𝜋∕2 CW because Ey lags Ex . AR = 1, 𝜏 = 90◦ (e) Elliptical because Δ𝜙 is not odd multiples of 𝜋∕2. CCW because Ey leads Ex . AR = OA/OB Letting Ex = Ey = E0 } √ OA = E0 [0.5(1 + 1 + √2)]1∕2 = 1.30656E0 1.30656 ⇒ AR = = 2.414 0.541196 OB = E0 [0.5(1 + 1 + 2)]1∕2 = 0.541196E0 ] [ ) ( ◦ 1.414 1 1 ◦ −1 2(1) cos(45 ) 𝜏 = 90 − tan = 90◦ − tan−1 2 1−1 2 0 1 = 90◦ − (90◦ ) = 45◦ 2 (f) Elliptical because Δ𝜙 is not odd multiples of 𝜋∕2. CW because Ey lags Ex . } From above OA = 1.30656E0 1.30656 = 2.414 ⇒ AR = OB = 0.541196E0 0.541196 From above 𝜏 = 90◦ − 12 (90◦ ) = 45◦ (g) Elliptical because 1. Ex ≠ Ey 2. Δ𝜙 is not zero or multiples of 𝜋. CCW because Ey leads Ex . OA = Ey { 12 [0.25 + 1 + 0.75]}1∕2 = Ey

}

1 =2 ⇒ AR = 0.5 OB = Ey { 12 [0.25 + 1 − 0.75]}1∕2 = 0.5Ey ) ( 0 1 1 = 90◦ − (180◦ ) = 0◦ 𝜏 = 90◦ − tan−1 2 −0.75 2 (h) Elliptical because 1. Ex ≠ Ey 2. Δ𝜙 is not zero or multiples of 𝜋. CCW because Ey lags Ex . From above OA = Ey OB = 0.5Ey

} ⇒ AR =

1 =2 0.5

1 𝜏 = 90◦ − (180◦ ) = 0◦ 2 2.34. x (z, t) = Re[Ex ej(𝜔t+kz+𝜙x ) ] = Ex cos(𝜔t + kz + 𝜙x ) y (z, t) = Re[Ey ej(𝜔t+kz+𝜙y ) ] = Ey cos(𝜔t + kz + 𝜙y ) where Ex and Ey are real positive constants.

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19

Choosing z = 0 and letting Δ𝜙 = 𝜙y − 𝜙x = 𝜙y − 0 = 𝜙 x (t) = Ex cos(𝜔t)

(1)

y (t) = Ey cos(𝜔t + 𝜙) and (t) =

√

x2 + y2 =

√ Ex2 cos2 (𝜔t) + Ey2 cos2 (𝜔t + 𝜙)

(2)

The maximum and minimum values of (2) are the major and minor axes of the polarization ellipse. Squaring (2) and using the half-angle identity, (2) can be written as  2 (t) =

1 2 {E + Ey2 + Ex2 cos(2𝜔t) + Ey2 cos2 [2(𝜔t + 𝜙)]} 2 x

(3)

Since Ex and Ey are constants, the maximum and minimum values of (3) occur when f (t) = Ex 2 cos(2𝜔t) + Ey 2 cos[2(𝜔t + 𝜙)] is maximum or minimum. These are found by differentiating (4) and setting it equal to zero. Thus df = −Ex 2 sin(2𝜔t) − Ey2 sin[2(𝜔t + 𝜙)] = 0 d(2𝜔t)

(4)

or Ex2 sin(2𝜔t) = −Ey2 sin[2(𝜔t + 𝜙)] = −Ey2 {sin 2𝜔t cos 2𝜙 + cos 2𝜔t sin 2𝜙}

(5)

Dividing (5) by cos(2𝜔t) yields Ex2 tan(2𝜔t) = −Ey2 [tan(2𝜔t) cos(2𝜙) + sin(2𝜙)] or tan(2𝜔t) =

−Ey2 sin(2𝜙) Ex2 + Ey2 cos(2𝜙)

from which we obtain that cos(2𝜔t) =

cos(2𝜔t + 2𝜙) =

Ex2 + Ey2 cos(2𝜙) ±𝜌 Ey2 + Ex2 cos(2𝜙) ±𝜌

(6)

(7)

where 𝜌=

√ Ex4 + Ey4 + 2Ex2 Ey2 cos(2𝜙)

(8)

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Substituting (6)–(8) into (3) yields ] [ 1 2 1 2 2  = E + Ey ± (𝜌 ) 2 x 𝜌 2

whose maximum value is }1∕2 1 2 [Ex + Ey2 + (Ex4 + Ey4 + 2Ex2 Ey2 cos 2𝜙)1∕2 ] 2 { }1∕2 1 2 = OB = [Ex + Ey2 − (Ex4 + Ey4 + 2Ex2 Ey2 cos 2𝜙)1∕2 ] 2

max = OA = max

{

The tilt angle 𝜏 can be obtained by expanding (1) and writing the two as x2 Ex2

−

2x y cos 𝜙 Ex Ey

+

y2 Ey2

= sin2 𝜙

(9)

which is the equation of a tilted ellipse. Choosing a coordinate system whose principal axes coincide with the major and minor axes of the tilted ellipse, we can write that x = x′ sin(z) − y′ cos(z)

(10)

y = x′ cos(z) + y′ sin(z)

where x′ and y′ are the new field values along the new principal axes x′ , y′ , z′ . Substituting (10) into (9) yields 2x′ y′ cos(z) sin(z) Ex2

−

2x′ y′ cos(z) sin(z) Ey2

−

2x′ y′ cos 𝜙 Ex Ey

(sin2 z − cos2 z) = 0

which when solved for the tilt angle 𝜏 reduces to [ ( )] 2Ex Ey cos 𝜙 𝜋 tan 2 −𝜏 = 2 Ex2 − Ey2 or 𝜋 1 𝜏 = − tan−1 2 2

(

2Ex Ey cos 𝜙

)

Ex2 − Ey2

For more details on the tilt angle derivation, see J.D. Kraus, Antennas, McGraw-Hill, 1950, pp. 464–476. 2.35. (a)

𝜌̂w = â x cos 𝜙1 + â y sin 𝜙1 𝜌̂a = â x cos 𝜙2 + â y sin 𝜙2 PLF = |𝜌̂w ⋅ 𝜌̂a |2 = |(̂ax cos 𝜙1 + â y sin 𝜙1 ) ⋅ (̂ax cos 𝜙2 + â y sin 𝜙2 )|2 = | cos 𝜙1 cos 𝜙2 + sin 𝜙1 sin 𝜙2 |2 = | cos(𝜙1 − 𝜙2 )|2

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(b)

21

𝜌̂w = â x sin 𝜃1 cos 𝜙1 + â y sin 𝜃1 sin 𝜙1 + â z cos 𝜃1 𝜌̂a = â x sin 𝜃2 cos 𝜙2 + â y sin 𝜃2 sin 𝜙2 + â z cos 𝜃2 PLF = |𝜌̂w ⋅ 𝜌̂a |2 = | sin 𝜃1 cos 𝜙1 sin 𝜃2 cos 𝜙2 + sin 𝜃1 sin 𝜙1 sin 𝜃2 ⋅ sin 𝜙2 + cos 𝜃1 ⋅ cos 𝜃2 |2 PLF = | sin 𝜃1 ⋅ sin 𝜃2 (cos 𝜙1 ⋅ cos 𝜙2 + sin 𝜙1 sin 𝜙2 ) + cos 𝜃1 cos 𝜃2 |2 PLF = | sin 𝜃1 sin 𝜃2 cos(𝜙1 − 𝜙2 ) + cos 𝜃1 cos 𝜃2 |2

2.36. Assuming electric field is x-polarized (a) Ew = â x E1 e−jkz ⇒ 𝜌̂w = â x

(

Ea = (̂a𝜃 − ĵa𝜙 )E0 f (r, 𝜃, 𝜙) ⇒ 𝜌̂a =

â 𝜃 − ĵa𝜙 √ 2

)

1 |̂a ⋅ â − ĵax ⋅ â 𝜙 |2 2 x 𝜃 since â 𝜃 = â x cos 𝜃 cos 𝜙 + â y cos 𝜃 sin 𝜙 − â z sin 𝜃 PLF = |𝜌̂w ⋅ 𝜌̂a |2 =

â 𝜙 = −̂ax sin 𝜙 + â y cos 𝜙 PLF = 12 (cos2 𝜃 cos2 𝜙 + sin2 𝜙) (b)

when Ea = (̂a𝜃 + ĵa𝜃 )E0 f (r, 𝜃, 𝜙), PLF is also PLF = 12 (cos2 𝜃 cos2 𝜙 + sin2 𝜙)

A more general, but also more complex, expression can be derived when the incident electric field is of the form Ew = (âax + b̂ay )e−jkz where a, b are real constants. It can be shown (using the same procedure) that PLF = √

1 2(a2 + b2 )

[(a cos 𝜃 cos 𝜙 + b sin 𝜃 sin 𝜙)2 + (a sin 𝜙 − b cos 𝜙)2 ]1∕2

z

Incident Wave

Antenna

x

E iw

y

2.37. (a) Ew = E0 (ĵay + 3̂az )e+jkx 1. Elliptical polarization; AR = 31 = 3; Left Hand (CCW) a. 2 components orthogonal to direction of propagation b. Not of same magnitude c. 90◦ phase difference between them d. y component is leading the z component or z component is lagging the y component

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SOLUTION MANUAL

(b) Ea = Ea (̂ay + 2̂az )e−jkx 1. Liner polarization; AR = ∞; No rotation a. 2 components orthogonal to direction of propagation. b. Not of same magnitude c. 0◦ phase difference between them, (c) PLF = |𝜌̂ ⋅ 𝜌̂ |2 w a ) ( ĵay + 3̂az √ +jkx +jkx Ew = E0 (ĵay + 3̂az )e = E0 10e √ 10 ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ ( 𝜌̂w =

ĵay + 3̂az √ 10

𝜌̂w

)

Ea = Ea (̂ay + 2̂az )e

( 𝜌̂a =

â y + 2̂az √ 5

) â y + 2̂az √ −jkx = E0 5e √ 5 ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ (

−jkx

𝜌̂a

)

| (ĵa + 3̂a ) (̂a + 2̂a ) |2 2 | y z y z | | = |j + 6| = 37 PLF = |𝜌̂w ⋅ 𝜌̂a | = || √ ⋅ √ | 50 50 | 10 5 || | 2

PLF =

37 = 0.740 = −1.31 dB 50

2.38. Eiw = (̂ax + ĵay )E0 e+jkz Ea = (̂ax + 2̂ay )E1

e−jkr || e−jkz = (̂ a + 2̂ a )E ◦ x y 1 | r |𝜃 =0 z z axis x

y

E iw z

( (a)

Eiw

=

â x +ĵay √ 2

)√ 2E0 e+jkz

Circular: 2 components, same amplitude, 90◦ phase difference (b) Clockwise (y component is leading the x component) ( ) â x +2̂ay √ e−jkz √ (c) Ea = 5E1 5 z Linear: 2 components, 0◦ phase difference (d) No rotation

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23

) â x + 2̂ay (e) 𝜌̂w = , 𝜌̂a = √ 5 )]2 [( ) ( â x + ĵay â x + 2̂ay |1 + j2|2 5 2 PLF = |𝜌̂w ⋅ 𝜌̂a | = = ⋅ = √ √ 10 10 5 2 (

PLF =

â x + ĵay √ 2

(

)

5 = 0.5 = 10 log10 (0.5) = −3 dB 10

e+jky 2.39. (a) Ew = (4̂az + j2̂ax )Ew = y

) 4̂az + j2̂ax √ e+jky 20Ew √ y 20 ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟

(

𝜌̂w

∙ Elliptical (2 components, not of same magnitude, 90◦ phase difference) (b) CW; x-components leads z-component by 90◦ ; rotate x into z while looking (observing) in the -y direction (from behind the wave). 4 (c) AR = = 2 2( ) 4̂az + j2̂ax (d) 𝜌̂w = ; 𝜌̂a = â z √ 20 ) |( |2 | 4̂az + j2̂ax | 16 2 ⋅ â z || = PLF = ||𝜌̂w ⋅ 𝜌̂a || = || = 0.8 = 10 log10 (0.8) √ 20 | | 20 | | PLF = 0.8 = −0.969 dB e−jkr 2.40. (a) Ea = E0 (ĵa𝜃 + 2̂a𝜙 )f0 (𝜃0 , 𝜙0 ) = E0 r ( 𝜌̂a =

ĵa𝜃 + 2̂a𝜙 √ 5

) ĵa𝜃 + 2̂a𝜙 √ e−jkr 5f0 (𝜃0 , 𝜙0 ) √ r 5 ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟

(

𝜌̂a

)

Elliptical, CW a^r a^ϕ a^θ y

x

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SOLUTION MANUAL

e+jkr (b) Ew = E1 (2̂a𝜃 + ĵa𝜙 )f1 (𝜃0 , 𝜙0 ) r ) ( 2̂a𝜃 + ĵa𝜙 √ e+jkr 5f1 (𝜃0 , 𝜙0 ) = E1 √ r 5 ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ ( 𝜌̂w =

𝜌̂w

2̂a𝜃 + ĵa𝜙 √ 5

)

Elliptical, CW (c)

|2 | |2 |( ĵa + 2̂a ) ( 2̂a + ĵa )|2 | | | 2j + j2 | | 4j | 𝜃 𝜙 𝜃 𝜙 || 2 | | | | | PLF = |𝜌̂a ⋅ 𝜌̂w | = | ⋅ √ √ | = | √ | = |√ | | | | 25 | | 25 | 5 5 | | | | | | PLF =

16 = 0.64 = −1.938 dB 25

1 −jkz ⇒ 𝜌̂w = √ (̂ax ± ĵay ) 2.41. (a) Ew = E0 (̂ax ± ĵay )e 2 1 Ea ≃ E1 (̂a𝜃 − ĵa𝜙 )f (r, 𝜃, 𝜙) ⇒ 𝜌̂a = √ (̂a𝜃 − ĵa𝜙 ) 2 PLF =

1| |2 1 | |2 |(̂a ± ĵay ) ⋅ (̂a𝜃 − ĵa𝜙 )| = |(̂ax ⋅ â 𝜃 ± â y ⋅ â 𝜙 ) − j(̂ax â 𝜙 ∓ â y â 𝜃 )| | | 2| x 2|

Converting the spherical unit vectors to rectangular, as it was done in Problem 2.35, leads to PLF =

1 (cos 𝜃 ± 1)2 2

(b) When Ew = E0 (̂ax ± ĵay )e−jkz Ea ≃ E1 (̂a𝜃 + ĵa𝜙 ) f (r, 𝜃, 𝜙) the PLF is equal to PLF =

1 (cos 𝜃 ∓ 1)2 2

2.42. Ew = (̂a𝜃 cos 𝜙 − â 𝜙 sin 𝜙 cos 𝜃) f (r, 𝜃, 𝜙) or ⎤√ ⎡ ⎢ â 𝜃 cos 𝜙 − â 𝜙 sin 𝜙 cos 𝜃 ⎥ 2 Ew = ⎢ √ ⎥ cos2 𝜙 + sin 𝜙 cos2 𝜃 ⋅ f (r, 𝜃, 𝜙) ⎢ cos2 𝜙 + sin2 𝜙 cos2 𝜃 ⎥ ⎦ ⎣ â 𝜃 cos 𝜙 − â 𝜙 sin 𝜙 cos 𝜃 Thus 𝜌̂w = √ cos2 𝜙 + sin2 𝜙 cos2 𝜃

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SOLUTION MANUAL

25

and |⎛ |2 | â cos 𝜙 − â sin 𝜙 cos 𝜃 ⎞ | | | ⎜ ⎟ 𝜃 𝜙 PLF = |𝜌̂w ⋅ 𝜌̂a |2 = ||⎜ √ ⋅ a ̂ ⎟ x || |⎜ | |⎝ cos2 𝜙 + sin2 𝜙 cos2 𝜃 ⎟⎠ | | | Transforming the rectangular unit vector to spherical using â x = â r sin 𝜃 cos 𝜙 + â 𝜃 cos 𝜃 cos 𝜙 − â 𝜙 sin 𝜙 the PLF reduces to PLF =

cos2 𝜃 cos2 𝜙 + sin2 𝜙 cos2 𝜃

The same answer is obtained by transforming the spherical unit vectors to rectangular, as was done in Prob. 2.35. ) ( 2̂ax ± ĵay √ 5f (r, 𝜃, 𝜙) 2.43. Ea ≃ (2̂ax ± ĵay )f (r, 𝜃, 𝜙) = √ 5 x Antenna z Wave

( (a)

𝜌̂w = ( 𝜌̂a =

y

â x − ĵay √ 2

)

2̂ax ± ĵay √ 5

⇒ Wave is Right Hand (RH) )

PLF = |𝜌̂w ⋅ 𝜌̂a |2

(b)

⎧ 9 = −0.4576 dB using the + sign (Antenna is LH in receiving mode and RH in transmitting) ⎪ 10 =⎨ (Antenna is RH in receiving ⎪ 1 = −10 dB using the − sign ⎩ 10 mode and LH in transmitting) ( ) â x + ĵay 𝜌̂w = ⇒ Wave is Left Hand (LH) √ 2 ) ( 2̂ax ± ĵay 𝜌̂a = √ 5 PLF = |𝜌̂w ⋅ 𝜌̂a |2 ⎧ 1 = −10 dB using the + sign ⎪ 10 =⎨ ⎪ 9 = −0.4576 dB using the − sign ⎩ 10

(Antenna is LH in receiving mode and RH in transmitting) (Antenna is RH in receiving mode and LH in transmitting)

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SOLUTION MANUAL

2.44. For 𝜌̂w y

45° x

| â + â 4̂a + ĵa |2 â x + â y | x y x y| | 𝜌̂w = √ ; PLF = || √ ⋅ √ | | | 17 2 2 | | 1 1 PLF = |(̂a ⋅ 4̂ax ) + (̂ay ⋅ ĵay )|2 = |4 + j|2 = 0.5 dimensionless = −3 dB 34 x 34 2.45. (a) RHCP; 𝜌̂a =

â x − ĵay √ 2

PLF = |𝜌̂w ⋅ 𝜌̂a

(b) LHCP; 𝜌̂a =

|2

| 2̂a + ĵa â − ĵa |2 | x y x y| ⋅ √ || = 0.9 dimensionless = −0.46 dB = || √ | 5 2 || |

â x + ĵay √ 2

| 2̂a + ĵa â + ĵa |2 | x y x y| ⋅ √ || = 0.1 dimensionless = −10.0 dB PLF = |𝜌̂w ⋅ 𝜌̂a = || √ | 5 2 || | ( ) â x − ĵay √ 2.46. Ei = (̂ax − ĵay )E0 e−jkz = 2E0 e−jkz √ 2 |2

𝜌̂w =

â x − ĵay √ 2

CW

x

Ei

y

k^

z

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(a)

27

Ea = (̂ax + ĵay )E1 e+jkz ( ) â x + ĵay √ = 2E1 e+jkz √ 2 𝜌̂a =

â x + ĵay √ 2

CW

)2 |( â − ĵa ) ( â + ĵa )|2 ( | x 1 − j2 y x y | | | PLF = |𝜌̂w ⋅ 𝜌̂a | = | =1 ⋅ √ √ | = 2 | | 2 2 | | 2

PLF = 1 = 0 dB ( (b)

Ea = 𝜌̂a =

â x − ĵay √ 2

)

√ 2E1 e+jkz

â x − ĵay √ 2

|( â − ĵa ) ( â − ĵa )|2 | 2 |2 | x y x y | |1 + j | 2 | | PLF = |𝜌̂w ⋅ 𝜌̂a | = | ⋅ √ √ | = || 2 || = 0 | | 2 2 | | | | PLF = 0 = −∞ dB −jkr 2.47. Ea = (2̂a𝜃 + j4̂a𝜙 )Ea e = r

e+jkr = Ew = (j4̂a𝜃 + 2̂a𝜙 )Ew r

) 2̂a𝜃 + j4̂a𝜙 e−jkr 20Ea √ r 20 ⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟

(

𝜌̂a

) j4̂a𝜃 + 2̂a𝜙 e+jkr 20Ew √ r 20 ⏟⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏟

(

z a^r

𝜌̂w

Antenna a. Elliptical b. CCW c. AR = 42 = 2 Wave d. Elliptical e. CCW x f. AR = 42 = 2 ( ) ( )|2 | 2̂a + j4̂a g. j4̂ a + 2̂ a | | 𝜃 𝜙 𝜃 𝜙 | PLF = |𝜌̂a ⋅ 𝜌̂w |2 = || ⋅ √ √ | | | 20 20 | | 2 | j8 + j8 | | 16 | | = | | = (0.8)2 = 0.64 = || | | | | 20 | | 20 | PLF = 0.64 dimensionless = −1.9382 dB

a^ϕ a^θ y

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SOLUTION MANUAL

2.48. Ei = â x E0 e−jkz , 𝜌̂w = â x Ea = (̂ax + ĵay )E1 e+jkz = ( 𝜌̂a =

â x + ĵay √ 2

)

(

â x + ĵay √ 2

)

√ 2E1 e+jkz

λ2 λ2 eo D0 |𝜌̂a ⋅ 𝜌̂w |2 = G |𝜌̂ ⋅ 𝜌̂ |2 4𝜋 4𝜋 0 a w

(a) Aem =

(eo D0 = G0 ) At 10 GHz ⇒ λ =

3 × 108 c 3 × 108 = = 3 × 10−2 = 9 f 10 × 10 1010

G0 = 10 = 10 log10 G0 (dim) ⇒ G0 (dim) = 101 = 10 ( )|2 | −2 )2 + ĵ a a ̂ | | (3 × 10 x y λ2 | G |𝜌̂ ⋅ 𝜌̂ |2 = (10) ||â x ⋅ Aem = √ | 4𝜋 0 a w 4𝜋 | | 2 | | ( ) ( ) ( ) −4 −3 9 × 10 1 9 × 10 1 1 = = (0.7162 × 10−3 ) = (10) 4𝜋 2 4𝜋 2 2 Aem = 0.3581 × 10−3 m2 (b) PT = Aem W i = (0.3581 × 10−3 )(10 × 10−3 ) = 3.581 × 10−6 Watts PT = 3.581 × 10−6 Watts e+jky e−jky W , 𝜌̂w = â z , Ea = −̂az Ea , 𝜌̂a = −̂az , Winc = 100 × 10−3 2 y y cm (a) PLF = |𝜌̂w ⋅ 𝜌̂a |2 = |−̂az ⋅ â z |2 = 1 = 0 dB

2.49. Ew = â z Ew

(b) For the λ∕2 dipole (Za = 73 + j42.5) with a loss resistance RL of 5 ohms: Un = (E𝜃n )2 = (sin1.3 𝜃)2 = sin3 𝜃 ⇒ (Un )max = 1 D0 = 2𝜋 𝜋

Prad

4𝜋Umax Prad

𝜋 ⎞ ⎛ 𝜋 3 ⎟ ⎜ = U sin 𝜃d𝜃d𝜙 = sin 𝜃 sin 𝜃d𝜃 d𝜙 = 2𝜋 sin4 𝜃d𝜃 ∫ ∫ ∫ ⎜∫ ∫ ⎟ ⎠ 0 0 0 ⎝0 0 2𝜋

𝜋

𝜋

𝜋

𝜋 sin3 𝜃 cos 𝜃 || 4−1 3 sin 𝜃d𝜃 = − sin2 𝜃d𝜃 = sin2 𝜃d𝜃 | + | ∫ ∫ ∫ 4 4 4 |0 0 0 0 4

𝜋

𝜋

[ ] 3 3 𝜃 1 3𝜋 sin 𝜃d𝜃 = sin2 𝜃d𝜃 = − sin(2𝜃) = ∫ 4∫ 4 2 4 8 4

0

0

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SOLUTION MANUAL

𝜋

Prad = 2𝜋

∫

sin4 𝜃d𝜃 = 2𝜋

(

3𝜋 8

) =

3𝜋 2 4

0

∴ D0 =

4𝜋Umax 4𝜋(1) 16 = D0 = = 1.69765 (dimensionless) = 2.298 dB = 2 Prad 3𝜋 ∕4 3𝜋

Using the equivalent circuit of Figure 1.2 with Rr = 73 and RL = 5 ecd =

Rr 73 = = 0.9359 Rr + RL 73 + 5

∴ G0 = ecd D0 = 0.9359(1.69765) = 1.5888 = 2.011 dB ( ) | Zin − Zc | || Za + RL − Zc || | (73 + j42.5 + 5) − 50 | | | | = 50.8945 = 0.3774 |Γ| = | | = || ) | = |( | | Zin + Zc | || Za + RL + Zc || | (73 + j42.5 + 5) + 50 | 134.8712 ) ( |Γ|2 = (0.3774)2 = 0.1424 ⇒ 1 − |Γ|2 = (1 − 0.1424) = 0.8576 ) ( Gre0 = er G0 = 1 − |Γ|2 G0 = (0.8576) 1.5888 = 1.3626 (dim) = 1.344 dB ( Preceived = Aem (ecd )(1 − |Γ|2 )PLF = =

) λ2 D0 ecd (1 − |Γ|2 )PLF 4𝜋

[ ] 𝜋2 D0 ecd (1 − |Γ|2 ) (PLF)Winc 4𝜋 ⏟⏟⏟ G0

⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏟ Gre0

λ2 30 × 109 λ= = 3 cm Gre (PLF), 4𝜋 10 × 109 ( 2) (3) 10−1 (9)(1.3562) −3 = (100 × 10 ) (1.3562) (1) = 4𝜋 ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ 4𝜋 ⏟⏞⏟⏞⏟ ⏟⏟⏟ ⏟⏞⏟⏞⏟ Gre0 Winc PLF

Preceived = (Winc )

32 ∕4𝜋

Preceived = 0.0981 Watts = 98.1 mWatts = 98.1 × 10−3 Watts ] [ ) (3)2 ( −3 Preceived = 100 × 10 (1.3626) (1) = 97.59 mW = 97.59 × 10−3 W ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ 4𝜋 ⏟⏞⏟⏞⏟ ⏟⏟⏟ ⏟⏟⏟ Gre0 PLF W inc

32 ∕4𝜋

2.50. Ea = (2̂ax ± ĵay )Ee−jkz 𝜌̂a =

2̂ax ± ĵay √ 5

(a) Ew = â x Ew ⇒ 𝜌̂w = â x | |2 | | 2 4 PLF = |𝜌̂w ⋅ 𝜌̂a |2 = || √ || = = 0.8 dimensionless = −0.9691 dB 5 | 5| | |

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SOLUTION MANUAL

(b) Ew = â y Ew ⇒ 𝜌̂w = â y PLF = |𝜌̂w ⋅ 𝜌̂a

|2

|2 | | 1 | 1 | = | √ || = = 0.2 dimensionless = −6.9897 dB 5 | 5| | |

2.51. (a) Ey = E′ + E′′ = 3 cos 𝜔t + 2 cos 𝜔t = 5 cos 𝜔t y y ) ( ) ( 𝜋 𝜋 + 3 cos 𝜔t − Ex = Ex′ + Ex′′ = 7 cos 𝜔t + 2 2 = −7 sin 𝜔t + 3 sin 𝜔t = −4 sin 𝜔t 5 AR = = 1.25 4 (b) At 𝜔t = 0, E = 5̂ay At 𝜔t = 𝜋∕2 ⇒ E = −4̂ax ⇒ Rotation in CCW 1 independent of 𝜓 → must have CP 2 ∴ AR = 1. (b) Polarization will be elliptical with major axis aligned with x-axis. Guess: AR = 2 √ Verify: 𝜌̂w = (2̂ax + jay )∕ 5 |2 | 2 2 | 2 cos 𝜓 + j sin 𝜓 | 2 | = 4 cos 𝜓√+ sin 𝜓 | PLF = |𝜌̂w ⋅ 𝜌̂a | = | √ | | | 5 5 | | 𝜓 = 0 : PLF = 0.8 𝜓 = 90◦ : PLF = 0.2 (c) PLF = 1 at 𝜓 = 45◦ and 225◦ PLF = 0 at 𝜓 = 135◦ and 315◦ Polarization must be linear at an angle of 45◦ ∴ AR = ∞

2.52. (a) PLF =

2.53.

Ig =

2 2 = (50 + 1 + 73) + j(25 + 42.5) 124 + j67.5

= (12.442 − j6.7724) × 10−3 = 14.166 × 10−3 ∠ − 28.56◦

Rg = 50 Vg = 2

(a) Ps = (b) Pr =

Xg = 25

RL = 1

Ig

1 Re(Vg ⋅ Ig∗ ) = Re(12.442 + j6.7724) × 10−3 2 1 |I |2 Rr = 7.325 × 10−3 W 2 g 1 |I |2 RL = 0.1003 × 10−3 W 2 g

Rr = 73 X = 42.5

= 12.442 × 10−3 W

(c) PL = The remaining supplied power is dissipated as heat in the internal resistor of the generator, or Pg =

1 |I |2 R = 5.0169 × 10−3 W 2 g g

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SOLUTION MANUAL

31

Thus Pr + PL + Pg = (7.325 + 0.1003 + 5.0169) × 10−3 = 12.4422 × 10−3 = Ps 2.54. The impedance transfer equation of [ Zin = Zc

ZL + jZc tan(kl) Zc + jZL tan(kl)

]

reduces for l = λ∕2 to Zin = ZL . Therefore the equivalent load impedance at the terminals of the generator is the same as that for Problem 2.53. Thus the supplied, radiated, and dissipated powers are the same as those of Problem 2.53. 2.55. (a) Zin = Ig =

(100)2 10000 = (50 − j50)2 = 100 − j100 Ω 50 + j50 5000 10 10 = 0.05546∠33.7◦ A = 150 − j100 180.3∠ − 33.7◦ 50 Ω

+ 10 V –

Z0 = 100 Ω

ZA = 50 + j 50 Ω

λ/4

1 1 Re{Vg Ig∗ } = × 10 × 0.05546 × cos(33.7◦ ) = 0.231 W 2 2 1 1 (c) PA = |Ig |2 Re{Zin } = × (0.05546)2 × 100 = 0.1538 W 2 2 Prad = ecd PA = 0.96 × 0.1538 = 0.148 W

(b) Ps =

2.56.

Gain =

Prad (Directivity) Paccepted

Realized Gain =

Prad (Directivity) Pavailable

P Gain = available Realized Gain Paccepted ( )2 Pavailable =

1 2

V √s 2

Z0

=

Vs2 4Z0

V(x) = A[e−jkx + Γ(0)ejkx ] I(x) =

A −jkx [e − Γ(0)ejkx ] Z0

V(0) = A[1 + Γ(0)] I(0) =

A [1 − Γ(0)] Z0

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SOLUTION MANUAL

Z0 = R0 Z0

Vs

Z0* ≡

Z0

VT

I0 + VT

Z0

–

+ V0 –

Fig. 1

From Fig. 1: −Vs + I(0)Z0 + V(0) = 0 −Vs +

A [1 − Γ(0)]Z0 + A[1 + Γ(0)] = 0 Z0

−Vs + A − AΓ(0) + A + AΓ(0) = 0 Vs 2 = Re[V(0)I ∗ (0)]

2A = Vs → A = Paccepted

Vs [1 + Γ(0)] 2 V I(0) = s [1 − Γ(0)] 2Z0

V(0) =

Zin − Z0 Zin + Z0 ( ) V Z − Z0 ⇒ V(0) = s 1 + in 2 Zin + Z0 ( ) V R + jXin − Z0 = s 1 + in 2 Rin + jXin + Z0 ( ) Vs Rin + jXin + Z0 + Rin + jXin − Z0 = 2 Rin + jXin + Z0 Γ(0) =

Vs (Rin + jXin ) Rin + jXin + Z0 ( ) ( ) V V Z − Z0 Zin + Z0 − Zin + Z0 I(0) = s 1 − in = s 2Z0 Zin + Z0 2Z0 Zin + Z0

V(0) =

Vs Vs = Zin + Z0 Rin + jXin + Z0 [ ] Vs Rin + jVs Xin Vs ∗ Re[V(0)I(0) ] = Re × Rin + Z0 + jXin Rin + Z0 − jXin I(0) =

Zin

Z0*

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SOLUTION MANUAL

( Paccepted = Re

Vs2 (Rin + jXin ) 2 (Rin + Z0 )2 + Xin Vs2 4Z0

Gain = Realized Gain

=

Vs2 Rin

) =

2.57. (a) RL = Rhf =

√

2 (Rin + Z0 )2 + Xin

2 (Rin + Z0 )2 + Xin

2 (Rin +Z0 )2 +Xin

l = C

Vs2 Rin

4Z0 Rin

𝜔𝜇0 2𝜎 √

λ∕60 ⋅ 2𝜋(λ∕200)

2𝜋 × 109 (4𝜋 × 10−7 ) 2(5.7 × 107 )

RL = 0.4415 × 10−2 = 0.004415 ohms ( )2 ( )2 l 1 = 80𝜋 2 = 0.21932 λ 60 ⇒ Rin = Rr = 0.21932 ohms (because of assumed constant current) Rr 0.21932 = = 0.98027 (c) ecd = RL + Rr 0.21932 + 0.004415 ecd = 98.027% ZL = (RL + Rin ) + jXin = (0.21932 + 0.004415) + jXin (d) (b) Rr = 80𝜋 2

= 0.2237 + jXin

) ] [ ( λ∕60 −1 ln λ∕100 ln(l∕2a) − 1 Xin ≃ −120 = −120 ) ( kl 2𝜋 λ tan( ) tan 2 2λ 60 ] [ 0.51003 − 1 = +1, 120.03 = −120 0.05241 Z − Zc (0.2237 + j1, 120.03) − 50 |Γ| = L = = 0.9999 ZL + Zc (0.2237 + j1, 120.03) + 50 VSWR =

1 + |Γ| 1 + 0.9999 = = 9, 999 1 − |Γ| 1 − 0.9999

2.58. Radiation Efficiency of a dipole ] [ 𝜋 Iz (z) = I0 cos z′ , −l∕2 ≤ z′ ≤ l∕2 l [ ] I 𝜋 H𝜙 (r = a)|at the surface = 0 cos z 2𝜋a l ds = a d𝜙 dz ⇒ differential patch of area. dW ⇒ power loss into this patch. 1 |H |2 R a d𝜙 dz 2 𝜙 s (time ave) (Rs = skin resistance) ) ( [ ] I0 2 Rs 𝜋 ⋅ cos2 z a d𝜙 dz dW = 2𝜋a 2 l

dW =

ds

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SOLUTION MANUAL

l∕2

W(total loss) =

W=

2𝜋

I0 2 Rs

∫−l∕2 ∫𝜙=0 8𝜋 2 ⋅ a2

cos2

[

] 𝜋 z a d𝜙 dz l

[ ] l∕2 I0 2 l ⋅ Rs ( 1 ) 2 𝜋 (2𝜋a)R cos z dz = s ∫−l∕2 l 4𝜋 a 2 8𝜋 2 a2 I0 2

1 2 I RL 2 0 ( ) 1 lRs RL = 2 2𝜋a W=

⎧1 ⎪ 2.59. E = ⎨ 0 ⎪1 ⎩2 (a)

0 < 𝜃 ≤ 45◦ 45◦ < 𝜃 ≤ 90◦ 90◦ < 𝜃 ≤ 180◦ r2 |E|2 r2 r2 E2 1 = , Umax = = 2𝜂 𝜂 𝜂 120𝜋 [ ] 2𝜋 45◦ 180◦ r2 1 = d𝜙 sin 𝜃 d𝜃 + sin 𝜃 d𝜃 ∫0 ∫90◦ 4 𝜂 ∫0

U= Prad

] [ ◦ 1 r2 180◦ + [2𝜋] − cos 𝜃|45 (− cos 𝜃)| ◦ 0 90 𝜂 4 [ ] 2r2 𝜋 1 1 − cos 45◦ + cos 0◦ − cos 180◦ + cos 90◦ = 𝜂 4 4 =

Prad = 0.54289

2𝜋r2 𝜂

( 4𝜋

D=

r2 𝜂

)

4𝜋Umax = = 3.684 Prad 0.54289(2𝜋)r2 ∕𝜂

(b) When the electric field is equal to 10 V/m, for 𝜃 = 0◦ . ⎧10 V∕m 0 < 𝜃 ≤ 45◦ ⎪ 45◦ < 𝜃 ≤ 90◦ ⇒ E = ⎨0 1 ⎪ × 10 V∕m 90◦ < 𝜃 ≤ 180◦ ⎩2 } ] { [ 2𝜋 45◦ 180◦ r2 2 2 Prad = |E| sin 𝜃 d𝜃 + |E| sin 𝜃 d𝜃 d𝜙 ∫90◦ ∫0 𝜂 ∫0 ( ) 2𝜋 Prad = r2 (0.54289) |10|2 = 36,193 𝜂 1 2 |I| Rr = |Irms |2 ⋅ Rr 2 36,193 36,193 ⇒ Rr = = = 1,447.72 25 |Irms |2 Prad =

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SOLUTION MANUAL

35

e−jkr ⇒ Un = (cos3 𝜃)2 = cos6 𝜃 r = cos6 𝜃|max = 1, 𝜃max = 0◦

2.60. Ea = â 𝜃 Ea cos3 𝜃 (a) Un |max

Un |𝜃=𝜃h = cos6 𝜃h = 0.5 ⇒ 𝜃h = cos−1 [(0.5)1∕6 ] = cos−1 (0.891) = 27.01◦ Θh = HPBW = 2𝜃h = 2(27.01) = 54.02◦ (b) Exact Directivity: 𝜋∕2

2𝜋

Prad =

∫0 ∫0

Un sin 𝜃 d𝜃 d𝜙 =

𝜋∕2

= −2𝜋

∫0

2𝜋

𝜋∕2

cos6 𝜃 sin 𝜃 d𝜃 d𝜙 = 2𝜋

∫0 ∫0

(

(cos 𝜃) d(cos 𝜃) = −2𝜋 6

cos7 𝜃 7

)𝜋∕2 0

𝜋∕2

∫0

cos6 𝜃 sin 𝜃 d𝜃

) ( 1 = 2𝜋∕7 = −2𝜋 0 − 7

4𝜋Umax 4𝜋(1) D0 = = = 14 = 10 log10 (14) = 11.46 dB Prad 2𝜋∕7 (c) Since the HPBW = 54.02◦ > 39.77◦ (n = 6 < 11.48), then Kraus’ approximate formula is the more accurate for the maximum directivity. Thus D0 (Kraus) =

41,253 || 41,253 41,253 = = = 14.137 | 2 Θ1h Θ2h |Θ1h =Θ2h (54.02)2 Θh

D0 (Kraus) = 14.137 (dimensionless) = 10 log10 (14.137) = 11.50 dB D0 (Tai & Pereira) =

72,815 || | Θ21h + Θ22h ||Θ

=

1h =Θ2h

72,815 72,815 = = 12.476 2 2(54.02)2 2Θh

D0 (T & P) = 12.476 (dimensionless) = 10 log10 (12.476) = 10.961 dB By comparsion, the Kraus’ approximate formula D0 is more accurate, compared to the exact D0 , for this problem. (d) Using the computer program Directivity, the maximum directivity is D0 = 13.9637 (dimensionless) = 11.45 dB Basically identical to the exact value. λ2 λ2 D0 |exact = (14) = 1.141λ2 (e) Aem = 4𝜋 4𝜋 Input parameters: ----------------The lower bound of The upper bound of The lower bound of The upper bound of

theta in degrees theta in degrees phi in degrees = phi in degrees =

= 0 = 90 0 360

Output parameters: ------------------Radiated power (watts) = 0.8999 Directivity (dimensionless) = 13.9637 Directivity (dB) = 11.4500

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SOLUTION MANUAL

2.61. In general, D𝜃o =

4𝜋(U𝜃 )max ; (Prad )𝜃 + (Prad )𝜙

D𝜙o =

U𝜃 = |E𝜃 |2 ;

4𝜋(U𝜙 )max (Prad )𝜃 + (Prad )𝜙

U𝜙 = |E𝜙 |2

U = U𝜃 + U𝜙 = |E|2 = |E𝜃 |2 + |E𝜙 |2 However for this problem Umax (𝜃 = 0◦ ; 𝜙 = 0◦ or 90◦ or any value 0 ≤ 𝜙 ≤ 2𝜋) = |E|2max = |E𝜃 |2max = |E𝜙 |2max 𝜋∕2

2𝜋

(Prad )𝜃 =

∫0 ∫0 𝜋∕2

2𝜋

(Prad )𝜙 =

∫0 ∫0

U𝜃 sin 𝜃 d𝜃 d𝜙 = U𝜙 sin 𝜃 d𝜃 d𝜙 =

2𝜋

Prad = (Prad )𝜃 + (Prad )𝜙 =

𝜋∕2

∫0 ∫0

∫0 ∫0 𝜋∕2

2𝜋

∫0 ∫0

U sin 𝜃 d𝜃 d𝜙 = 2𝜋

Prad = (Prad )𝜃 + (Prad )𝜙 =

𝜋∕2

2𝜋

|E𝜃 |2 sin 𝜃 d𝜃 d𝜙 |E𝜙 |2 sin 𝜃 d𝜃 d𝜙

2𝜋

𝜋∕2 [

∫ 0 ∫0

] U𝜃 + U𝜙 sin 𝜃 d𝜃 d𝜙

𝜋∕2 [

∫0 ∫0

] |E𝜃 |2 + |E𝜙 |2 sin 𝜃 d𝜃 d𝜙

However, since for this problem Umax (𝜃 = 0◦ ; 𝜙 = 0◦ or 90◦ or any value 0 ≤ 𝜙 ≤ 2𝜋) = |E|2max = |E𝜃 |2max = |E𝜙 |2max D0 = D𝜃0 = D𝜙0 ; NOT D0 = D𝜃0 + D𝜙0 However, in general, for any problem, other than special cases like Problem 2.61 D0 = D𝜃0 + D𝜙0 if Umax = |E|2max = |E𝜃 |2max + |E𝜙 |2max ≠ |E𝜃 |2max ≠ |E𝜙 |2max Input parameters: ----------------The lower bound of The upper bound of The lower bound of The upper bound of

theta in degrees theta in degrees phi in degrees = phi in degrees =

= 0 = 90 0 360

Output parameters: ------------------Radiated power (watts) = 0.1566 Partial Directivity (theta) (dimensionless) = 80.2511 Partial Directivity (theta) (dB) = 19.0445 Partial Directivity (phi) (dimensionless) = 80.2511 Partial Directivity (phi) (dB) = 19.0445 Directivity (dimensionless) = 80.2511 Directivity (dB) = 19.0445

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SOLUTION MANUAL

Using Table 12.1 a = 3λ, b = 2λ ( ) ab D0 = 4𝜋 2 = 4𝜋(6) = 24𝜋 λ D0 = 75.398 = 18.774 dB Since the maximum |E𝜃 | = |E𝜙 | = |E| then the maximum directivity D0 = D𝜃 = D𝜙 2.62. Input parameters: ----------------The lower bound of The upper bound of The lower bound of The upper bound of

theta in degrees theta in degrees phi in degrees = phi in degrees =

= 0 = 90 0 360

Output parameters: -----------------Radiated power (watts) = 0.0330 Partial Directivity (theta) (dimensionless) = 62.4635 Partial Directivity (theta) (dB) = 17.9563 Partial Directivity (phi) (dimensionless) = 62.4635 Partial Directivity (phi) (dB) = 17.9563 Directivity (dimensionless) = 62.4635 Directivity (dB) = 17.9563

Using Table 12.1 a = 3λ, b = 2λ ) ( ab D0 = 0.81 4𝜋 2 = 0.81(24𝜋) λ = 61.072 = 17.858 dB Since the maximum |E𝜃 | = |E𝜙 | = |E|, then the maximum directivity D0 = D𝜃 = D𝜙 2.63. Input parameters: ----------------The lower bound of The upper bound of The lower bound of The upper bound of

theta in degrees theta in degrees phi in degrees = phi in degrees =

= 0 = 90 0 360

Output parameters: -----------------Radiated power (watts) = 0.4863 Partial Directivity (theta) (dimensionless) = 4.2443 Partial Directivity (theta) (dB) = 6.2780

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SOLUTION MANUAL

Partial Directivity (phi) (dimensionless) = 4.2443 Partial Directivity (phi) (dB) = 6.2780 Directivity (dimensionless) = 4.2443 Directivity (dB) = 6.2780

Using Table 12.1 2.286 λ = 0.762λ 3 1.016 λ = 0.3387λ b= 3 ) ( ab D0 = 0.81 4𝜋 2 = 0.81(4𝜋)(0.762)(0.3387) λ = 2.627 = 4.194 dB

f = 10 GHz ⇒ λ = 3 cm ⇒ a =

Since the maximum |E𝜃 | = |E𝜙 | = |E|, then the maximum directivity D0 = D𝜃 = D𝜙 2.64. Input parameters: ----------------The lower bound of The upper bound of The lower bound of The upper bound of

theta in degrees theta in degrees phi in degrees = phi in degrees =

= 0 = 90 0 360

Output parameters: -----------------Radiated power (watts) = 0.0338 Partial Directivity (theta) (dimensionless) = 92.9470 Partial Directivity (theta) (dB) = 19.6824 Partial Directivity (phi) (dimensionless) = 92.9470 Partial Directivity (phi) (dB) = 19.6824 Directivity (dimensionless) = 92.9470 Directivity (dB) = 19.6824

Using Table 12.2 a = 1.5λ

) ( 4𝜋 2𝜋a 2 2 (𝜋a ) = = 9𝜋 2 λ λ2 D0 = 88.826 = 19.485 dB

D0 =

Since the maximum |E𝜃 | = |E𝜙 | = |E|, then the maximum directivity D0 = D𝜃 = D𝜙 2.65. Input parameters: ----------------The lower bound of The upper bound of The lower bound of The upper bound of

theta in degrees theta in degrees phi in degrees = phi in degrees =

= 0 = 90 0 360

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SOLUTION MANUAL

Output parameters: -----------------Radiated power (watts) = 0.0418 Partial Directivity (theta) (dimensionless) = 75.1735 Partial Directivity (theta) (dB) = 18.7606 Partial Directivity (phi) (dimensionless) = 75.1735 Partial Directivity (phi) (dB) = 18.7606 Directivity (dimensionless) = 75.1735 Directivity (dB) = 18.7606

Using Table 12.2 a = 1.5λ

) 2𝜋a 2 = 0.836 (9𝜋 2 ) λ D0 = 74.2589 = 18.71 dB (

D0 = 0.836

Since the maximum |E𝜃 | = |E𝜙 | = |E|, then the maximum directivity D0 = D𝜃 = D𝜙 2.66. Input parameters: ----------------The lower bound of The upper bound of The lower bound of The upper bound of

theta in degrees theta in degrees phi in degrees = phi in degrees =

= 0 = 90 0 360

Output parameters: -----------------Radiated power (watts) = 0.4952 Partial Directivity (theta) (dimensionless) = 6.3439 Partial Directivity (theta) (dB) = 8.0236 Partial Directivity (phi) (dimensionless) = 6.3439 Partial Directivity (phi) (dB) = 8.0236 Directivity (dimensionless) = 6.3439 Directivity (dB) = 8.0236

Using Table 12.2 f = 10 GHZ ⇒ λ = 3 cm ⇒ a =

1.143 λ = 0.381λ 3

) 2𝜋a 2 = 0.836[2𝜋(0.381)]2 λ D0 = 4.791 = 6.804 dB (

D0 = 0.836

Since the maximum |E𝜃 | = |E𝜙 | = |E|, then the maximum directivity D0 = D𝜃 = D𝜙

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SOLUTION MANUAL

2.67. Ea = (̂a𝜃 + j2̂a𝜙 ) sin 𝜃 E0

e−jkr ◦ (0 ≤ 𝜃 ≤ 180◦ , 0◦ ≤ 𝜙 ≤ 360◦ ) r z

Outgoing wave Eϕ

y Eθ

x

(a) Elliptical because 1. 2 components transverse to the direction of wave propagation 2. Both components not of the same magnitude 3. 90◦ phase difference between the 2 components 4. AR = 2/1 = 2 because ellipse aligned with principal axes 5. CCW (E𝜙 leads E𝜃 ) due to the 90◦ phase difference between the two. 4𝜋(U𝜙 )max 4𝜋(U𝜃 )max (b) (D0 )𝜃 = ; (D0 )𝜙 = (Prad )𝜃 + (Prad )𝜙 (Prad )𝜃 + (Prad )𝜙 (Ut )n = (U𝜃 )n + (U𝜙 )n = |E𝜃 |2n + |E𝜙 |2n = (1 + 4) sin2 𝜃|E0 |2 = 5 sin2 𝜃|E0 |2 (Ut )n = 5 sin2 𝜃|E0 |2 (U𝜃 )n = |E𝜃 |2n = sin2 𝜃|E0 |2 ; (U𝜃 )nmax = |E0 |2 ;

(U𝜙 )n = |E𝜙 |2n = 4 sin2 𝜃|E0 |2

(U𝜙 )nmax = 4|E0 |2 2𝜋

(Prad )t = (Prad )𝜃 + (Prad )𝜙 = 2𝜋

∫0 ∫0

𝜋

(Ut )n sin 𝜃 d𝜃 d𝜙

𝜋

2𝜋

𝜋

5 sin2 𝜃|E0 |2 sin 𝜃 d𝜃 d𝜙 = 5|E0 |2 d𝜙 sin3 𝜃 d𝜃 ∫0 ∫0 ∫0 ∫0 ( ) 𝜋 40𝜋 4 = sin3 𝜃 d𝜃 = 10𝜋|E0 |2 |E0 |2 = 5(2𝜋)|E0 |2 ∫0 3 3

=

(Prad )t = (D0 )𝜃 =

(D0 )𝜙 =

40𝜋 |E0 |2 3 4𝜋|E |2 4𝜋(U𝜃 )max 3 = 40𝜋 0 = = 0.3 = −5.2288 dB 2 (Prad )t 10 |E0 | 3 4𝜋(U𝜙 )max (Prad )t

=

4𝜋(4)|E0 |2 40𝜋 |E0 |2 3

=

12 = 1.2 = 0.79181 dB 10

(c) (D0 )t = (D0 )𝜃 + (D0 )𝜙 = 0.3 + 1.2 = 1.5 = 1.761 dB

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SOLUTION MANUAL

2.68. f = 150 MHz, λ = 2 m ⇒ 1 m dipole is 2λ in electrical length ⇒ Rr = 73 ohms, Zin = 73 + j42.5 ohms [see (4-93), Chapter 4]

0.625 RLoss Rs

73 Ω

Rr

42.5

XA

50 ohms Vs = 100 V

Vs = 0.765∠ − 18.97◦ A 50 + 73 + 0.625 + j42.5 1 (b) Pdissip = PLoss = |Iant |2 ⋅ RLoss = 189 mW 2 1 (c) Prad = |Iant |2 ⋅ Rr = 21.36 W 2 Rr 73 (d) Ecd = = = 99% Rr + RLoss 73 + 0.625 (a) Iant =

2.69. E = â 𝜃 E𝜃 ≃ â 𝜃 j𝜂

kI0 l −jkr kI e−jkr sin 𝜃 = −j𝜂 0 e [− â 𝜃 l sin 𝜃 ] 4𝜋r 4𝜋r ⏟⏟⏟ le

(a) le = −̂a𝜃 l sin 𝜃 (b) |le |max = | − â 𝜃 l sin 𝜃|max = l @ 𝜃 = 90◦ (c) |le |max ∕l = 1 ( ) ⎡ ⎤ 𝜋 cos cos 𝜃 I0 ⎢ ⎥ 2 E = â 𝜃 E𝜃 = â 𝜃 j𝜂 ⎥ 2𝜋r ⎢⎢ sin 𝜃 ⎥ ⎣ ⎦ ( ) ⎡ ⎤ 2 cos 𝜋2 cos 𝜃 ⎥ kI0 e−jkr ⎢ −̂a = j𝜂 ⎥ 4𝜋r ⎢⎢ 𝜃 k sin 𝜃 ⎥ ⎣ ⎦ e−jkr

2.70.

⎡ )⎤ ( 𝜋 ⎢ cos 2 cos 𝜃 ⎥⎥ kI0 e−jkr ⎢ λ = j𝜂 − â ⎥ 4𝜋r ⎢⎢ 𝜃 𝜋 sin 𝜃 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⎥ ⎥ ⎢ le ⎦ ⎣ ) ) ( ( cos 𝜋2 cos 𝜃 cos 𝜋2 cos 𝜃 λ = −̂a𝜃 0.3183λ le = −̂a𝜃 𝜋 sin 𝜃 sin 𝜃

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SOLUTION MANUAL

|le |max

) ( | cos 𝜋2 cos 𝜃 || | | | = |−̂a𝜃 0.3183λ = 0.3183λ @ 𝜃 = 90◦ | | | sin 𝜃 | | | |max

|le |max 0.3183λ = = 0.6366 = 63.66% @ 𝜃 = 90◦ l λ∕2 2.71.

le = −̂a𝜃 l sin 𝜃, l = λ∕50, f = 10 GHz ⇒ λ = 3 cm √ 1 W= |E|2 = 10−3 W∕cm ⇒ |E| = 2𝜂W 2𝜂 √ = 2(377)(10−3 ) = 0.8683 V∕cm ( ) λ = 52.1 × 10−3 Volts Voc |max = |Ei ||le |max = (0.8683) 50

2.72. Since |le |max = l∕2 ⇒ |Voc |max = Voc |max =

1 (52.1 × 10−3 ) 2

1 2

(Voc of dipole with uniform current)

= 26.05 × 10−3 Volts (see Problem 2.71)

2.73. |le |max = 0.3183λ ⇒ |Voc | = |le |max |Ei |. From Problem 2.71 solution |Voc | = 0.8683(0.3183λ) = 0.27638λ = 0.27638(3) = 0.82914 Volts 2.74. Using (2.94), the effective aperture of an atenna can be written as Ae =

|VT |2 ⋅ RT , where Wi = |E|2 ∕2𝜂 2Wi |Zt |2

Defining the effective length le as VT = E ⋅ le reduces Ae to

Ae =

𝜂RT le2 |Zt |2

√ ⇒ le =

Ae |Zt |2 𝜂RT

For maximum power transfer and lossess antenna (RL = 0) √ Thus le = 2.75.

XA = −XT , Rr = RT ⇒ |Zt | = 2Rr = 2RT √ √ 4Aem ⋅ R2T Aem RT Aem Rr =2 =2 𝜂RT 𝜂 𝜂 (

Aem = 2.147 =

λ2 4𝜋

) ⋅ ecd ⋅ (1 − |Γ|2 ) ⋅ |𝜌̂w ⋅ 𝜌̂a |2 ⋅ D0

75 − 50 3 × 108 =3m = 0.2; λ = 75 + 50 100 × 106 2.147 ∴ D0 = 2 = 3.125 3 2] [(1 − (0.2) 4𝜋 Γ=

3 × 108 = 0.1 m 2.76. d = 1 m, f = 3 GHz, 𝜀ap = 68% ⇒ λ = 3 × 109 ( )2 𝜋(1)2 d 𝜋 d2 (a) Ap = 𝜋r2 = 𝜋 = = = 𝜋4 = 0.785 m2 2 4 4

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SOLUTION MANUAL

(b) 𝜀ap =

Aem ⇒ Aem = 𝜀ap Ap Ap

Aem = 𝜀ap Ap = 0.68(0.785) = 0.534 m2 λ2 4𝜋 D ⇒ D0 = 2 Aem 4𝜋 0 λ 4𝜋 4𝜋 4𝜋 D0 = 2 Aem = (0.534) = (0.534) = 671.044 0.01 λ (0.1)2

(c) Aem =

D0 = 671.044 = 28.268 dB (d) PL = Aem Wi = 0.534(10 × 10−6 ) PL = 5.34 × 10−6 Watts 2.77.

Wi = 10 × 10−3 W∕cm2 ,

l = λ∕2,

D0 = 2.148 dB, |Γ| = 0.2

f = 1 GHz ⇒ λ = 𝜐∕f = 30 × 10 ∕109 = 30 cm 9

(a)

D0 = 2.148 dB = 10 log10 D0 (dim) ⇒ D0 (dim) = 102.148∕10 = 1.6398 Aem =

2 λ2 : 1 = λ D (dim) [1 − |Γ|2 ]   *  (ecd ) 1 (er )  D0  PLF 4𝜋 4𝜋 0

Aem =

λ2 λ2 λ2 (1.6398)(1 − |0.2|2 ) = (1.6398)(1 − 0.04) = (1.6396)(0.96) 4𝜋 4𝜋 4𝜋

Aem = 0.12527λ2 ( ) λ2 λ λ = = 0.00167λ2 2 300 600 A 0.12527λ2 (c) 𝜀ap = em = = 75.162 = 7, 516.2% Ap 0.00167λ2 (d) P = W A = 10 × 10−3 W (0.12527λ2 ) = 10−2 [0.12527(30)2 ] L i em cm2 (b) Ap = ld =

PL = 112.743 × 10−2 = 0.112743 × 10+1 Watts = 1.12743 Watts PL = 1.12743 Watts 2.78.

Wi = 10−3 W∕m2 Aem = λ=

λ2 D , D = 20 dB = 10 log10 D0 (dim) ⇒ D0 (dim) = 100 4𝜋 0 0 c 3 × 108 = 0.03 m = 3 × 10−2 m = f 10 × 109

(3 × 10−2 )2 9 × 10−4 ⋅ 100 = ⋅ (100) = 0.716 × 10−2 = 7.16 × 10−3 4𝜋 4𝜋 ( ) 9 × 10−2 9 × 10−5 −3 = 10 ⋅ = = 0.716 × 10−5 = 7.16 × 10−6 Watts 4𝜋 4𝜋

Aem = Prec

Prec = 7.16 × 10−6 Watts.

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SOLUTION MANUAL

1 = 0.5 = −3 dB 2 D0 = 12 dB = 15.849 (dimensionless); ZA = 100; Zc = 50; ecd = 75% = 0.75

f = 10 GHz, W i = 10 × 10−3 Watts∕cm2 ; PLF =

(a)

Aem =

λ2 D , 4𝜋 0

Aem =

(3)2 (15.849) = 11.351 4𝜋

λ=

30 × 109 = 3 cm 10 × 109

Aem = 11.351 cm2 (b)

Γ=

ZA − Zc 100 − 50 50 1 = = = = 0.3333 ZA + Zc 100 + 50 150 3

er = (1 − |Γ|2 ) = (1 − |0.3333|2 ) = 0.88889 Aem (lossy) = Aem (lossless)(er )(ecd )(PLF) = 11.351(0.88889)(0.75)(0.5) = 3.78367 Aem (lossy) = 3.78367 cm2 (c)

Preceiver = W i Aem = 10 × 10−3 (3.78367) = 37.8367 × 10−3 Preceiver = 37.8367 × 10−3 Watts

2.80. Ap = 10 cm2 , f = 10 GHz ⇒ λ = 30 × 109 ∕10 × 109 = 3 cm, W i = 10 × 10−3 W∕cm2 λ2 λ2 (a) Aem = D0 = G = Ap = 10 4𝜋 4𝜋 0 4𝜋(10) 4𝜋(10) ⇒ G0 = = = 13.96 = 11.45 dB λ2 (3)2 (b)

Pr = Aem W i (PLF) =

1 (10)(10 × 10−3 ) = 100 × 10−3 ∕2 = 0.05 Watts 2

Pr = 0.05 Watts ( )2 | â x + ĵay || | | =1 PLF = ||â x ⋅ √ | 2 | | 2 | | 2.81. Ew = (ĵax + 2̂ay)e+jkz Ea = ĵay e−jkz

x

+jkz

(a) Ew = (ĵax + 2̂ay )e

Ea

(ĵax + 2̂ay ) √ +jkz = 5e √ 5 ⏟⏟⏟

z

𝜌̂w

Elliptical polarization, AR = 2, CCW because: 1. 2 components not of equal magnitude 2. 90◦ phase difference between the two 3. x-component is leading the y-component

Ew

y

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SOLUTION MANUAL

(b) Ea = j â y e−jkz ⏟⏞⏟⏞⏟ 𝜌̂a

Linear polarization, AR = ∞, no rotation because one component. λ2 (c) Aem = D , D = 2.15 dB = 10 log10 D0 (dimensionless) 4𝜋 0 0 D0 (dimensionless) = 102.15∕10 = 100.215 = 1.641 D0 (dimensionless) = 1.641 Aem =

(d) Aem

Aem

λ2 λ2 D0 = (1.641) = 0.131λ2 = 0.410𝜋λ2 4𝜋 4𝜋

| (ĵa + 2̂a ) |2 | x | y λ2 λ2 2 2 | ⋅ (̂ay )|| = D0 = (1 − |Γ| )PLF = (1.641)(1 − |0.5| ) | √ 4𝜋 4𝜋 | | 5 | | )2 ( ( ) 2 4 = 0.079λ2 = 0.131λ2 (1 − 0.25) √ = 0.131λ2 (0.75) 5 5

/ (e) PL = Aem Wi = 0.079λ2 (10 × 10−3 λ2 ) = 0.79 × 10−3 Watts PL = 0.79 × 10−3 Watts = 0.79 mWatts 2.82. W rad = W ave ≃ C0

1 cos4 (𝜃)̂ar r2 𝜋∕2

2𝜋

(a) Prad =

∫0

∫0

𝜋∕2

2𝜋

= C0

Prad

∫0

W rad ⋅ ds =

∫0

(0 ≤ 𝜃 ≤ 𝜋∕2, 0 ≤ 𝜙 ≤ 2𝜋) 𝜋∕2

2𝜋

∫0

∫0

â r Wrad ⋅ â r r2 sin 𝜃 d𝜃 d𝜙

cos4 𝜃 sin 𝜃 d𝜃 d𝜙 = 2𝜋C0

)𝜋∕2 cos5 𝜃 = 2𝜋C0 − 5 0 ) ( 2𝜋 1 = = 2𝜋C0 0 + C = 1.2566C0 5 5 0

(b) D0 = D0 =

(

𝜋∕2

∫0

cos4 𝜃 sin 𝜃 d𝜃

4𝜋Umax ⇒ Umax = r2 Wrad |max = C0 cos4 𝜃|max = C0 Prad 4𝜋C0 = 10 = 10 log10 (10) = 10 dB 2𝜋C0 ∕5

(c) D0 = 10 toward 𝜃 = 0◦ 2 (d) Aem = λ D0 4𝜋

Aem =

λ=

c 3 × 108 m∕sec = 0.3 m = f 1 × 109

(0.3)2 0.09 0.225 (10) = (10) = = 0.0716 m2 4𝜋 4𝜋 𝜋

(e) PL = Aem W i = 0.0716 × (10 × 10−3 ) = 0.716 × 10−3 Watts

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SOLUTION MANUAL

2.83. D0 (λ∕2) = 2.286 dB = 100.2286 = 1.69278 (dim) D0 (λ∕4) = 5.286 dB = 100.5286 = 3.37754 (dim) 30 × 109 = 15.78947 cm 1.9 × 109 Prad 10 10 = = = 0.07958 × 10−9 (a) Wrad (isotropic) = 4𝜋r2 4𝜋(1, 000 × 100)2 4𝜋 × 1010 Prad = 10 watts, f = 1, 900 MHz ⇒ λ =

= 79.58 × 10−12 W∕cm2 Wrad (λ∕2) = Wrad (isotropic)D0 = 79.58 × 10−12 (1.69278) = 134.711 × 10−12 Wrad = (λ∕2) = 134.711 × 10−12 W∕cm2 (b) D0 (λ∕4) = 5.286 dB = 3.37754 dim. Aem =

(15.78947)2 λ2 D0 = (3.37754) = 67 cm2 4𝜋 4𝜋

(c) Prec = Wrad (λ∕2)Aem (λ∕4) = 134.711 × 10−12 (67) = 9,025.637 × 10−12 Prec = 9.0256 × 10−9 Watts 2.84. Aem = (a)

λ2 eD 4𝜋 t 0

=

λ2 G 4𝜋 0

G0 = 14.8 dB ⇒ G0 (power ratio) = 101.48 = 30.2 f = 8.2 GHZ ⇒ λ = 3.6585 cm Aem =

(3.6585)2 (30.2) = 32.167 cm2 4𝜋

The physical aperture is equal to Ap = 5.5(7.4) = 40.7 cm2 (b) G0 = 16.5 dB ⇒ G0 (power ratio) = 101.65 = 44.668 f = 10.3 GHz ⇒ λ = 2.912 cm Aem =

(2.912)2 (44.668) = 30.142 cm2 4𝜋

(c) G0 = 18.0 dB ⇒ G0 (power ratio) = 101.8 = 63.096 f = 12.4 GHz ⇒ λ = 2.419 cm Aem =

(2.419)2 (63.096) = 29.389 cm2 4𝜋

2.85. Pin = 100 Watts; Zc = 75 ohms; Zin = ZA = 100; ecd = 50% U(𝜃, 𝜙) = B0 sin 𝜃; (a)

Γ=

0 ≤ 𝜃 ≤ 180◦ , 0 ≤ 𝜙 ≤ 360◦

ZA − Zc 100 − 75 25 1 = = = = 0.14286 ZA + Zc 100 + 75 175 7

er = (1 − |Γ|2 ) = (1 − |0.1428|2 ) = (1 − 0.0204) = 0.9796 = 97.96% er = 0.9796 = 97.96%

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SOLUTION MANUAL

(b) e0 = ecd er = 0.5(0.9796) = 0.4898 = 48.98% 𝜋

2𝜋

(c)

Prad = Prad

∫0

∫0

U sin 𝜃 d𝜃 d𝜙 = B0

𝜋

2𝜋

∫0

sin2 𝜃 d𝜃 d𝜙 = 2𝜋B0

∫0 ∫0 ] [ ] 𝜋[ 𝜋 1 − cos(2𝜃) 1 = 2𝜋B0 d𝜃 = 𝜋B0 𝜃 − sin(2𝜃) = 𝜋 2 B0 ∫0 2 2 0

𝜋

sin2 𝜃 d𝜃

Prad = Pin (er ecd ) = Pin (e0 ) = 100(0.4898) = 48.98 watts ) ( 48.98 = 4.9627 48.98 = 𝜋 2 B0 ⇒ B0 = 𝜋2 (d)

U = B0 sin 𝜃 = 4.9627 sin 𝜃 ⇒ Umax = 4.9627 D0 =

4𝜋Umax 4𝜋(4.9627) = = 1.2732 Prad 48.98

D0 = 1.2732 = 1.049 dB (e) Wrad|max =

Umax r2

=

4.9627 4.9627 = = 4.9627 × 10−10 [1,000(100)]2 (105 )2 Wrad| max = 0.49627 × 10−9 Watts∕cm2

λ2 D 4𝜋 0 From Problem 2.61: Computer Program Directivity: D0 = 80.2511 ⇒ Aem =

2.86. Aem =

Table 12.1: D0 = 75.398 ⇒ Aem =

λ2 4𝜋

λ2 (80.2511) 4𝜋

= 6.386λ2

λ2 (62.4635) 4𝜋

= 4.971λ2

(75.398) = 6λ2

λ2 D 4𝜋 0 From Problem 2.62: Computer Program Directivity: D0 = 62.4635 ⇒ Aem =

2.87. Aem =

Table 12.1: D0 = 61.072 ⇒ Aem =

λ2 4𝜋

(61.072) = 4.86λ2

λ2 D 4𝜋 0 From Problem 2.63: Computer Program Directivity: D0 = 4.2443 ⇒ Aem =

2.88. Aem =

Table 12.1: D0 = 2.627 ⇒ Aem =

λ2 4𝜋

(2.627) =

λ2 (4.2443) 4𝜋 2 0.20905λ

λ2 D 4𝜋 0 From Problem 2.64: Computer Program Directivity: D0 = 92.947 ⇒ Aem =

= 0.3378λ2

2.89. Aem =

Table 12.2: D0 = 88.826 ⇒ Aem = 2.90. Aem =

λ2 D 4𝜋 0

λ2 4𝜋

(88.826) =

λ2 (92.947) 4𝜋 2 7.068λ

= 7.396λ2

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SOLUTION MANUAL

From Problem 2.65: Computer Program Directivity: D0 = 75.1735 ⇒ Aem = Table 12.2: D0 = 74.2589 ⇒ Aem =

λ2 4𝜋

(74.2589) =

λ2 (75.1735) 4𝜋 2 5.909λ

λ2 D 4𝜋 0 From Problem 2.66: Computer Program Directivity: D0 = 8.0236 ⇒ Aem =

= 5.982λ2

2.91. Aem =

Table 12.2: D0 = 4.791 ⇒ Aem =

λ2 (4.791) 4𝜋

λ2 (8.0236) 4𝜋

= 0.638λ2

= 0.3813λ2

2.92. Gain = 30 dB, f = 2 GHZ, Prad = 5 W Receiving antenna VSWR = 2, efficiency = 95% ER = (2̂ax + ĵay )FR (𝜃, 𝜙), Use Friis transmission formula (2.118) ) ( λ 2 Dt (𝜃t , 𝜙t )Dr (𝜃r , 𝜙r ) ⋅ PLF Pr = Pt ecdt ecdr (1 − |Γt |2 )(1 − |Γr |2 ) 4𝜋R Pr = 10−14 W, ecdt = 1 (we assume that), ecdr = 0.95, 1 − |Γr |2 = 1 | VSWR − 1 | 2 − 1 1 |= = , (1 − |Γr |2 ) = 8∕9 Since VSWR = 2 ⇒ |Γr | = || | | VSWR + 1 | 2 + 1 3 3 × 108 = 0.15 m, R = 4000 × 103 m, 2 × 109 ( )2 ) ( λ 2 0.15 Hence = = 8.9 × 10−18 4𝜋R 4𝜋4000 × 103 λ=

⎧ 1 2 ⎪ 𝜌̂t = √2 (̂ax + ĵay ) ⇒ |𝜌̂t ⋅ 𝜌̂r | = 0.1 Dt = 30 dB = 10 , PLF ⇒ ⎨ 1 ⎪ 𝜌̂r = √5 (2̂ax + ĵay ) ⎩ ( ) 8 (8.9 × 10−18 )(103 )Dr (0.1) ⇒ 10−14 = 5(1)(0.95)(1) 9 Dr = 2.661 3

λ2 2.661 = 0.00476 m2 4𝜋 { 4 } cos 𝜃, 0◦ ≤ 𝜃 ≤ 90◦ 2.93. U(𝜃, 𝜙) = 0◦ ≤ 𝜙 ≤ 360◦ 0, 90◦ ≤ 𝜃 ≤ 180◦ Hence Aem =

λ2 D 4𝜋 0 4𝜋Umax D0 = Prad

Aem =

𝜋

2𝜋

Prad =

∫0

(

∫0

4𝜋Umax Prad

)

2𝜋 1 = 5 5 4𝜋(1) = = 10 2𝜋∕5

Prad = 2𝜋 −0 + D0 =

U(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙 = 2𝜋

𝜋∕2

∫0

[

cos5 𝜃 cos (𝜃) sin 𝜃 d𝜃 = 2𝜋 − 5

]𝜋∕2

4

0

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SOLUTION MANUAL

Aem =

λ2 λ2 10λ2 3 × 108 = 3 × 10−2 = 0.03 m D0 = (10) = , λ= 4𝜋 4𝜋 4𝜋 1010

Aem =

10(0.03)2 10(3 × 10−2 )2 10(9 × 10−4 ) = = = 7.16197 × 10−4 4𝜋 4𝜋 4𝜋

Aem = 7.16197 × 10−4 2.94. 1 status mile = 1609.3 meters, 22,300(status miles) = 3.588739 × 107 m P 8 × 10−14 (a) Pi = rad2 = = 4.943 × 10−16 Watts∕m2 . 4𝜋 × (3.58874) 4𝜋R λ2 (b) Aem = D , (D0 = 60 dB = 106 ), (λ = 0.15 m) 4𝜋 0 (0.15)2 (106 ) = 1790.493 m2 Aem = 4𝜋 Preceived = Aem ⋅ Pi = (1790.493)(4.943 × 10−16 ) = 8.85 × 10−13 Watts. 2.95. A = 0.7162 m2 em ( )2 λ Aem = ecd (1 − |Γ|2 )|𝜌̂w ⋅ 𝜌̂a |2 D0 4𝜋 A 75 − 50 3 × 108 , Γ= D0 = ( )2 em = 3m = 0.2, λ = 75 + 50 100 × 106 λ 2 (1 − 1Γ| ) 4𝜋 0.7162 D0 = 2 3 (1 − |0.2|2 ) 4𝜋 D0 = 1.0417 ( 2.96. P = W A = W e (1 − |Γ|2 ) r i em i cd

λ2 4𝜋

)2

Wi = 5 W∕m2 , ecd = 1(lossless), Γ =

D0 |𝜌̂w ⋅ 𝜌̂a |2 Zin − Z0 73 − 50 = = 0.187 Zin + Z0 73 + 50

3 × 108 = 30 m, D0 = 2.156 dB = 1.643, PLF = 1 10 × 106 ( 2) 30 Pr = (5)(1)[1 − (0.187)2 ] (1.643)(1) = 567.78 Watts 4𝜋 λ=

Pr = 567.78 Watts. 2.97.

Pr ( λ )2 = G0r G0t , G0r = G0t = 16.3 ⇒ G0 (power ratio) = 42.66 Pt 4𝜋R f = 10 GHz ⇒ λ = 0.03 meters. VSWR − 1 1.1 − 1 0.1 = = = 0.0476 VSWR + 1 1.1 + 1 2.1 Pt = 200 m watts = 0.2 Watts

VSWR = 1.1 ⇒ |Γ| =

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[ (a) R = 5 m: Pr =

0.03 4𝜋(5)

]2 (42.66)2 (0.2)[1 − |Γ|2 ]2

= 82.9[1 − (0.0476)2 ]2 = 82.9(0.9977)2 = 82.5 (b) R = 50 m : Pr = 0.825 𝜇Watts (c) R = 500 m : Pr = 8.25 nWatts ) ( λ 2 2.98. pr = |𝜌̂ ⋅ 𝜌̂ |2 G0t G0r t r pt 4𝜋R G0t = 20 dB ⇒ G0t (power ratio) = 102 = 100 G0r = 15 dB ⇒ G0r (power ratio) = 101.5 = 31.623 f = 1 GHz ⇒ λ = 0.3 meters R = 1 × 103 meters (a) For |𝜌̂t ⋅ 𝜌̂r |2 = 1 ( Pr =

0.3 4𝜋 × 103

)2 (100)(31.623)(150 × 10−3 ) = 270.344 𝜇Watts

(b) When transmitting antennas is circularly polarized and receiving antenna is linearly polarized, the PLF is equal to |2 |( â ± ĵa ) | | x y 1 2 | ⋅ â x || = |𝜌̂t ⋅ 𝜌̂r | = | √ 2 | | 2 | | Thus Pr =

1 (270.344 × 10−6 ) = 135.172 × 10−6 = 135.172 𝜇Watts 2

2.99. Lossless: ecd = 1, polarization matched: |𝜌̂w ⋅ 𝜌̂a |2 = 1, line matched: (1 − |Γ|2 ) = 1 D0 = 20 dB = 102 = 100 = D0r = D0t ) )2 ( ( λ 2 λ Pr = Pt D0t D0r = 10 (100)(100) = 0.253 Watts 4𝜋R 4𝜋 ⋅ 50λ Pr = 0.253 Watts 2.100. Lossless: ecd = 1, PLF = 1. Line matched: (1 − |Γ|2 ) = 1. D0 = 30 dB = 103 = 1000 = D0r = D0t )2 ( ( )2 λ 1 Pr = Pt (1000)2 = 20 100 = 12.665 Watts 4𝜋 ⋅ 100λ 4𝜋 8 2.101. G0r = 20 dB = 100, G0t = 25 dB = 316.23, λ = 3 × 10 = 0.1 m 3 × 109 )2 ( λ G0r G0t Pr = Pt |𝜌̂t ⋅ 𝜌̂r |2 4𝜋R )2 ( 0.1 = 100(1) (100)(316.23) 4𝜋 × 500 Pr = 8 × 10−4 Watts

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SOLUTION MANUAL

f = 10 GHz → λ =

2.102.

3 × 108 = 0.03 m 1010

G0t = G0r = 15 dB = 101.5 = 31.62 (dimensionless) R = 10 km = 104 m Pr ≥ 10 nW = 10−8 W |𝜌̂t ⋅ 𝜌̂r |2 = −3 dB =

1 2

Friis Transmission Equation: ) ( Pr λ 2 = G0t G0r |𝜌̂t ⋅ 𝜌̂r |2 Pt 4𝜋R ( )2 ( ) 1 0.03 1.5 2 = 2.85 × 10−11 = (10 ) 4 2 4𝜋 × 10 Pr 2.85 × 10−11

Pt =

Pr ≥ 10−8 W → (Pt )min = 351 W ) ( Pr λ 2 = (PLF)et er D0t D0r Pt 4𝜋R ) ( λ 2 = (PLF)(ert ecdt )(err ecdr ) D0t D0r 4𝜋R ) ( Pr λ 2 = (1)[ert (1)][err (1)] D0t D0r Pt 4𝜋R

2.103.

c 3 × 108 = 3 m, R = 10 × 103 = 104 = f 108 ( )2 ( ) ( )2 λ 2 3 3 −4 = = × 10 4𝜋R 4𝜋 4𝜋 × 104 λ=

= (0.2387 × 10−4 )2 = 5.699 × 10−2 × 10−8 (

λ 4𝜋R

)2

= 5.699 × 10−10 (

ert = err = (1 − |Γ|2 ) =

| 73.3 − 50 |2 | 1 − || | | 73.3 + 50 |

)

( =

= [1 − (0.18897)2 ] = (1 − 0.0357) = 0.9643 ecdt = ecdr = 1 D0t = D0r = 1.643 Pr = (0.9643)2 (1.643)2 (5.699 × 10−10 ) Pt = (0.92987)(2.699)(5.699 × 10−10 ) = 2.51(5.699 × 10−10 ) = 14.305 × 10−10

| 23.3 |2 | 1 − || | | 12.3 |

)

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SOLUTION MANUAL

Pt =

Pr = 6.99 × 10−2 × 1010 (1 × 10−6 ) 14.305 × 10−10

= 6.99 × 102 Pt = 699 Watts 2.104.

Pr ( λ )2 3 × 108 3 × 108 1 = G0t G0r , λ = = = 9 Pt 4𝜋R 30 9 × 10 90 × 108 R = 10,000 meter =

10,000 λ = 3 × 105 λ 1∕30

[ ]2 Pr λ 10 × 10−6 2 = G = = 10−6 0 Pt 10 4𝜋(3 × 105 λ) G0 2 = 10−6 (4𝜋 × 3 × 105 )2 G0 = 10−3 (4𝜋 × 3 × 105 ) = 12𝜋 × 102 = 1200𝜋 G0 = 1200𝜋 = 3,769.91 = 10 log10 (3,769.91) dB G0 = 3,769.91 = 35.76 dB 2.105.

R = 16 × 103 m, f = 2 GHz, G0t = 20 dB, Pt = 100 watts, Pr = 5 × 10−9 Watts ⇒ G0r =? G0t = 20 dB = 10 log10 [G0t (dim)] ⇒ G0t (dimensionless) = 102 = 100 G0t (dimensionless) = 100 f = 2 GHz ⇒ λ =

3 × 108 = 0.15 meters 2 × 109

Friis Transmission Equation

(2-119):

( )( ) ) ( ) ( Pr Pr 4𝜋R 2 λ 2 1 1 = G0t G0r PLF ⇒ G0r = Pt 4𝜋R Pt G0t λ PLF ( ) [ 4𝜋(16 × 103 ) ]2 ( ) 1 2 5 × 10−9 G0r = 100 100 0.15 1 ] [ 2 10 × 10−9 × 106 4𝜋(16) = = 10−6 (1, 340.413)2 0.15 104 G0r = 1, 796, 706.65 × 10−6 = 1.7967 = 2.545 dB G0r = 1.7967 = 2.545 dB 2.106.

𝜎 = 𝜋a2 = 25𝜋λ2 G0t = G0r = 16.3 dB ⇒ G0t (power ratio) = 101.63 = 42.66 f = 10 GHz ⇒ λ = 0.03 m ( )2 G0t G0r Pr λ =𝜎 Pt 4𝜋 4𝜋R1 R2

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SOLUTION MANUAL

(a) R1 = R2 = 200λ = 6 meters: [ ]2 2 λ 2 (42.66) Pr = 25(𝜋λ ) (0.2) = 9.00 nWatts 4𝜋 4𝜋(200λ)2 (b) R1 = R2 = 500λ = 15 meters: Pr = 0.23 nWatts [ ]2 G G λ 2.107. Pr = Pt 𝜎 0t 0r , 4𝜋 4𝜋R1 R2 [ ]2 1502 0.06 Pr = 105 (3) 4𝜋 4𝜋(106 )

λ=

3 × 108 = 0.06 m 5 × 109

Pr = 1.22 × 10−8 Watts [ ] ]2 [ G G P (4𝜋) 4𝜋R1 R2 2 Pr λ 2.108. = 𝜎 0r 0t ⇒𝜎= r Pt 4𝜋 4𝜋R1 R2 Pt G0r G0t λ 3 × 108 = 0.03 m 10 × 109 [ ]2 4𝜋(104 )(104 ) 10−16 (4𝜋) ∴𝜎 = = 3.445 m2 1000(80)(80) 0.03 ] [ Pr 4𝜋 4𝜋R1 R2 2 2.109. 𝜎 = Pt G0r G0t λ λ=

3 × 108 = 0.1 m 3 × 109 [ ]2 10−16 (4𝜋) 4𝜋(104 )(104 ) 𝜎= = 0.31 m2 100(80)(80) 0.1 λ=

2.110.

𝜎 = 0.85λ2 G G Pr = 𝜎 0t 0r Pt 4𝜋

(

λ 4𝜋R1 R2

)2 |𝜌̂w ⋅ 𝜌̂r |2

𝜎 = 0.85λ2 , G0t = G0r = 15 dB ⇒ G0t = G0r = 31.6228 (dimensionless) R1 = R2 = 100 meter ⇒ R1 = R2 = 1, 000λ f = 3 GHz ⇒ λ =

3 × 108 = 0.1 meters 3 × 109

|𝜌̂w ⋅ 𝜌̂r |2 = 1 dB ⇒ |𝜌̂w ⋅ 𝜌̂r |2 = 0.7943 ( )2 2 Pr λ 2 (31.6228) = 0.85λ (0.7943) Pt 4𝜋 4𝜋 × 106 λ2 =

0.85(31.6228)2 (0.7943) = 0.3402 × 10−12 (4𝜋)3 (1012 )

Pr = 0.3402 × 10−12 (102 ) = 0.3402 × 10−10 = 34.02 × 10−12 Watts Pr = 34.02 pWatts

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2.111.

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SOLUTION MANUAL

Ta = TA e−2𝛼l + T0 (1 − e−2𝛼l ) TA = 5 K 5 (72 − 32) + 273 = 295.2 K 9 −4 dB = 20 log10 e−𝛼 = −𝛼(20) log10 e = −𝛼(20)(0.434) T0 = 72◦ F =

𝛼=

4 = 0.460 Nepers∕100 ft = 0.0046 Nepers∕ft. 8.68

(a) l = 2 feet: 2

2

Ta = 5e−2(0.0046) + 295.2[1 − e−2(0.0046) ] = 4.91 + 5.38 = 10.29 K (b) l = 100 feet; Ta = 5e−2(0.0046)100 + 295.2[1 − e−2(0.0046)100 ] = 179.72 K d

2.112. Ta = TA e− ∫0

d 2𝛼(z) dz

+

∫0 If 𝛼(z) = 𝛼0 = Constant d

Ta = TA e−2𝛼0 d +

∫0

d

𝜀(z)Tm (z)e− ∫z

2𝛼(z′ ) dz′

dz

𝜀(z)Tm (z)e−2𝛼0 (d−z) dz d

Ta = TA e−2𝛼0 d + e−2𝛼0 d +

∫0

𝜀(z)Tm (z)e+2𝛼0 z dz

If Tm (z) = T0 = Constant and 𝜀(z) = 𝜀0 = constant

−2𝛼0 d

Ta = TA e

∫0

e2𝛼0 z dz

𝜀 + 0 T0 e−2𝛼0 d (e2𝛼0 d − 1) 2𝛼0

For 𝜀0 = 2𝛼0 : Ta = TA e−2𝛼0 d + T0 e−2𝛼0 d (e2𝛼0 d − 1) = TA e−2𝛼0 d + T0 (1 − e−2𝛼0 d )

dz

Ta α (z), ε (z),Tm(z)

Ta = TA e−2𝛼0 d + 𝜀0 T0 e−2𝛼0 d

d

d

z

TA

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3

Solution Manual

If H e = j𝜔𝜀∇ × Πe Maxwell’s curl equation ▽ × Ee = −j𝜔𝜇H e can be written as

3.1.

(1)

∇ × Ee = −j𝜔𝜇H e = −j𝜔𝜇(j𝜔𝜀∇ × Πe ) = 𝜔2 𝜇𝜀∇ × Πe or ∇ × (Ee − 𝜔2 𝜇𝜀Πe ) = ∇ × (Ee − k2 Πe ) = 0 where k2 = 𝜔2 𝜇𝜀 Letting Ee − k2 Πe = −∇𝜙e ⇒ Ee = −∇𝜙e + k2 Πe

(2)

Taking the curl of (1) and using the vector identity of (3-8) leads to ∇ × H e = j𝜔𝜀∇ × ∇ × Πe = j𝜔𝜀[∇(∇ ⋅ Πe ) − ∇2 Πe ]

(3)

Using Maxwell’s equation ∇ × H e = J + j𝜔𝜀Ee reduces (3) to J + j𝜔𝜀Ee = j𝜔𝜀[∇(∇ ⋅ Πe ) − ∇2 Πe ]

(4)

Substituting (2) into (4) reduces to ∇2 Πe + k2 Πe = j

J 𝜔𝜀

+ [∇(∇ ⋅ Πe ) + ∇𝜙e ]

(5)

Letting 𝜙e = −∇ ⋅ Πe simplifies (5) to ∇2 Πe + k2 Πe = j

J 𝜔𝜀

(6)

Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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SOLUTION MANUAL

and (2) to Ee = ∇(∇ ⋅ Πe ) + k2 Πe

(7)

Comparing (6) with (3-14) leads to the relation Πe = −j

1 A 𝜔𝜇𝜀

(8)

If Em = −j𝜔𝜇∇ × Πm

3.2.

(1)

Maxwell’s curl equation ∇ × H m = j𝜔𝜀Em can be written as ∇ × H m = j𝜔𝜀(−j𝜔𝜇∇ × Πm ) = 𝜔2 𝜇𝜀∇ × Πm or ∇ × (H m − 𝜔2 𝜇𝜀Πm ) = ∇ × (H m − k2 Πm ) = 0

where k2 = 𝜔2 𝜇𝜀

Letting H m − k2 Πm = −∇𝜙m ⇒ H m = −∇𝜙m + k2 Πm

(2)

Taking the curl of (1) and using the vector identity of (3-8) leads to ∇ × Em = −j𝜔𝜇∇ × ∇ × Πm ) = −j𝜔𝜇[∇(∇ ⋅ Πm ) − ∇2 Πm ]

(3)

Using maxwell’s equation ∇ × Em = −M − j𝜔𝜇H m

(4)

reduces (3) to −M − j𝜔𝜇H m = −j𝜔𝜇[∇(∇ ⋅ Πm ) − ∇2 Πm ] Substituting (2) into (4) reduces to ∇2 Πm + k2 Πm = j

M 𝜔𝜇

+ [∇(∇ ⋅ Πm ) + ∇𝜙m ]

(5)

Letting 𝜙m = −∇ ⋅ Πm simplifies (5) to ∇2 Πm + k2 Πm = j

M 𝜔𝜇

(6)

and (2) to H m = ∇(∇ ⋅ Πm ) + k2 Πm Comparing (6) with (3-25) leads to the relation Πm = −j

1 F 𝜔𝜇𝜀

(7)

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SOLUTION MANUAL

3.3.

57

e−jkr A = â z Az1 = â z C1 r Substituting the above into (3-34) leads to the following terms: [ ( )] [ ] d d e−jkr d e−jkr e−jkr C = C −jk − 1 1 dr dr r dr r dr2 r2 [ ] e−jkr e−jkr e−jkr e−jkr + jk 2 − jk 2 + 2 3 = C1 (−jk)2 r r r r ( ) 2 2 dAz1 e−jkr e−jkr = C1 −jk − 2 r dr r r r d2 Az1

=

k2 Az1 = k2 C1

e−jkr r

The sum of the above three terms is equal to zero, and it then satisfies (3-34). The same conclusion is derived using A = â z Az2 = â z C2

e−jkr r

as a solution. 3.4. The solution of ∇2 Az = −𝜇Jz can be inferred from the solution of Poisson’s equation ∇2 𝜙 = −

𝜌 𝜀

(1)

for the potential 𝜙. 𝜌(x′ , y′ , z′ ) represents the charge density. We begin with Green’s theorem

∫v

(𝜓∇2 𝜙 − 𝜙∇2 𝜓)d𝜐′ =

∮Σ

(𝜓∇𝜙 − 𝜙∇𝜓) ⋅ n̂ da

(2)

where 𝜓 and 𝜙 are well behaved functions (nonsingular, continuous, and twice differentiable). For 𝜓 we select a solution of the form 𝜓=

1 R

(3)

where R=

√ (x − x′ )2 + (y − y′ )2 + (z − z′ )2

(3a)

By considering the charge at the origin of the coordinate system, it can be shown that (provided r ≠ 0) ∇2 𝜓 =

1 𝜕 r2 𝜕r

( r2

𝜕𝜓 𝜕r

) +

1 𝜕 r2 sin 𝜃 𝜕𝜃

( sin 𝜃

𝜕𝜓 𝜕𝜃

) +

( ) 1 1 𝜕2𝜓 2 1 =0 = ∇ r r2 sin2 𝜃 𝜕𝜙2

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SOLUTION MANUAL

Thus (2) reduces to

∫V

𝜓∇2 𝜙 dv′ = −

𝜌(x′ , y′ , z′ ) ′ 1 dv 𝜀 ∫V r

(4)

To exclude the r = 0 singularity of 𝜓, the observation point x′ , y′ , z′ is surrounded by a sphere of radius r′ and surface Σ′ . Therefore the volume V is bounded by the surfaces Σ and Σ′ , and (2) is broken into two integrals; one over Σ0 and the other of Σ′ . Using (4) reduces (2) to −

𝜌 ′ 1 (𝜓∇𝜙 − 𝜙∇𝜓) ⋅ n̂ da + (𝜓∇𝜙 − 𝜙∇𝜓) ⋅ n̂ da dv = ∮Σ0 ∮Σ0 𝜀 ∫V r

(5)

and ∮Σ′

(𝜓∇𝜙 − 𝜙∇𝜓) ⋅ n̂ da =

[ ∮Σ′

=−

] 1 ̂ da ∇𝜙 − 𝜙(∇𝜓) ′ ⋅n r=r r′

(5a)

𝜕𝜙 1 1 𝜙 da da − ′2 ′ r ∮Σ′ 𝜕r r ∮Σ′

Since r′ is arbitrary, it can be chosen small enough so that 𝜙 and 𝜕𝜙 are essentially constant at 𝜕r ′ ′ every point on Σ . If we make r progressively smaller, 𝜙 and its normal derivative approach their limiting values at the center (by hypothesis, both exit and are continuous functions of position). Therefore, in the limit as r′ → 0, both can be taken outside the integral and we can write that ∮Σ′

(𝜓∇𝜙 − 𝜙∇𝜓) ⋅ n̂ da = −4𝜋𝜙(x, y, z)

(6)

since 𝜕𝜙 1 1 da = lim lim ′ ′ ′ ∮ r →0 r r →0 r ′ Σ′ 𝜕r

(

𝜕𝜙 𝜕r

)

1 da = lim ′ ∮ r →0 r ′ r=r′ Σ′

(

𝜕𝜙 𝜕r

) (4𝜋r′2 ) = 0 r=r′

Substituting (6) into (5) reduces it to 𝜙(x, y, z) =

[ ( )] 𝜌 ′ 1 1 1 1 ⋅ n̂ da dv + ∇𝜙 − 𝜙∇ 4𝜋𝜀 ∫v r 4𝜋 ∮Σ r r

(7)

The first term on the right side of (7) accounts for the contribution from the charges within Σ while the second term for those outside Σ. Expansion of Σ to include all charges makes the second term to vanish and to reduce (7) to 𝜙(x, y, z) =

𝜌(x′ , y′ , z′ ) ′ 1 dv 4𝜋𝜀 ∫v r

(8)

By comparing ∇2 Az = −𝜇Jz with (1), we can write that Az (x, y, z) =

Jz (x′ , y′ , z′ ) ′ 𝜇 dv 4𝜋 ∫v r

(9)

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SOLUTION MANUAL

59

For more details see D.T. Paris and F.K. Hard, Basic Electromagnetic Theory, McGraw-Hill, 1969, pp. 128-131. For the details of the solution of (3-31) see R.E. Collin, Field Theory of Guided Waves, McGraw-Hill, 1960, pp. 35-39. It can be shown that Az =

𝜇 e−jkr Jz (x′ , y′ , z′ ) dv 4𝜋 ∫v r

Because of the length of the derivation, it will not be repeated here. A ≃ [̂ar A′r (𝜃, 𝜙) + â 𝜃 A′𝜃 (𝜃, 𝜙) + â 𝜃 A′𝜙 (𝜃, 𝜙)]

3.5.

E = −j𝜔A − j 𝜓 =∇⋅A =

e−jkr r

1 ∇(∇ ⋅ A) 𝜔𝜇𝜀

1 𝜕 2 1 𝜕 1 𝜕A𝜙 A ) − sin 𝜃) + (r (A r 𝜃 r sin 𝜃 𝜕𝜃 r sin 𝜃 𝜕𝜙 r2 𝜕r

e−jkr e−jkr e−jkr + 2 [⋯ ⋯] + 3 [⋯ ⋅ ⋅] + ⋯ ⋅ r r r 𝜕𝜓 1 𝜕𝜓 1 𝜕𝜓 ∇(∇ ⋅ A) = ∇𝜓 = â r + â 𝜃 + â 𝜙 𝜕r r 𝜕𝜃 r sin 𝜃 𝜕𝜙 { } 1 1 1 = â r [−𝜔2 𝜇𝜀e−jkr A′r (𝜃, 𝜙)] + 2 [⋯ ⋅] + 3 [⋯ ⋅] + ⋅⋅ r r r { } 1 1 1 (0) + 2 [⋯ ⋅] + 3 [⋯ ⋅] + ⋯ ⋅ +̂a𝜃 r r r { } 1 1 1 (0) + 2 [⋯ ⋅] + 3 [⋯ ⋅] + ⋯ +̂a𝜙 r r r

𝜓 = ∇ ⋅ A = −jk

Therefore E = −j𝜔A − j

1 ∇(∇ ⋅ A) 𝜔𝜇𝜀

e−jkr E ≃ −j𝜔[̂ar A′r + â 𝜃 A′𝜃 + â 𝜙 A′𝜙 ] r ] { [ −jkr e 1 1 1 â r 𝜔2 𝜇𝜀 + 2 (⋯) + 3 (⋯ ⋅) + ⋯ ⋅ −j 𝜔𝜇𝜀 r r r [ ] 1 1 1 +̂a𝜃 (0) + 2 (⋯) + 3 (⋯) + ⋯ ⋅ r r r [ ]} 1 1 1 + â 𝜙 (0) + 2 (⋯) + 3 (⋯) + ⋯ r r r or E≃

1 1 1 {−j𝜔e−jkr [̂ar (0) + â 𝜃 A′𝜃 + â 𝜙 A′𝜙 ]} + 2 [⋯] + 3 [⋯] ⋅ ⋅ r r r

In a Similiar manner, it can be shown that } { 1 1 1 1 𝜔 j e−jkr [̂ar (0) + â 𝜃 A′𝜙 − â 𝜙 A′𝜃 ] + 2 [⋯] + 3 [⋯] + ⋅⋅ H = ∇×A= 𝜇 r 𝜂 r r

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SOLUTION MANUAL

3.6. Let us assume that within a linear and isotropic medium, but not necessarily homogeneous, there exist two sets of sources J 1 , M 1 and J 2 , M 2 which are allowed to radiate simultaneously or individually inside the same medium at the same frequency and produce E1 , H 1 and E2 , H 2 , respectively. For the fields to be valid, they must satisfy Maxwell’s equations ̇ 1 − M1 ∇ × E1 = −zH

(1)

̇ 1+ j ∇ × H 1 = yE

(2)

1

̇ 2 − M2 ∇ × E2 = −zH

(3)

̇ 2+ j ∇ × H 2 = yE

(4)

2

where ż = j𝜔(𝜇 ′ − j𝜇 ′′ )

(5)

ẏ = 𝜎 + j𝜔(𝜀′ − j𝜀′′ )

(6)

If we dot multiply (1) by H 2 and (4) by E1 , we can write ̇ 2 ⋅ H1 − H2 ⋅ M1 H 2 ⋅ ∇ × E1 = −zH

(7)

̇ 1 ⋅ E2 + E1 ⋅ J 2 E1 ⋅ ∇ × H 2 = yE

(8)

Subtracting (7) from (8) reduces to ̇ 1 ⋅ E2 + zH ̇ 2 ⋅ H 1 + E1 ⋅ J 2 + H 2 ⋅ M 1 E1 ⋅ ∇ × H 2 − H 2 ⋅ ∇ × E1 = yE

(9)

which by using the vector identity ∇ ⋅ (A × B) = B ⋅ (∇ × A) − A ⋅ (∇ × B)

(10)

̇ 1 ⋅ E2 + zH ̇ 2 ⋅ H 1 + E1 ⋅ J 2 + H 2 ⋅ M 1 ∇ ⋅ (H 2 × E1 ) = −∇ ⋅ (E1 × H 2 ) = yE

(11)

can be writen as

In a similar manner, if we dot multiply (2) by E2 and (3) by H 1 , we can write ̇ 2 ⋅ E1 + E2 ⋅ J 1 E2 ⋅ ∇ × H 1 = yE

(12)

̇ 1 ⋅ H2 − M1 ⋅ M2 H 1 ⋅ ∇ × E2 = −zH

(13)

Subtracting (13) from (12) leads to ̇ 2 ⋅ E1 + zH ̇ 1 ⋅ H 2 + E2 ⋅ J 1 + H 1 ⋅ M 2 E2 ⋅ ∇ × H 1 − H 1 ⋅ ∇ × E2 = yE

(14)

which by using (10) can be written as ̇ 2 ⋅ E1 + zH ̇ 1 ⋅ H 2 + E2 ⋅ J 1 + H 1 ⋅ M 2 ∇ ⋅ (H 1 × E2 ) = −∇ ⋅ (E2 × H 1 ) = yE

(15)

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SOLUTION MANUAL

61

Substracting (15) from (11) leads to −∇ ⋅ (E1 × H 2 − E2 × H 1 ) = E1 ⋅ J 2 + H 2 ⋅ M 1 − E2 ⋅ J 1 − H 1 ⋅ M 2

(16)

which is known as the Lorentz Reciprocity Theorem in differential form. Taking the volume integral of both sides of (16) and using the divergence theorem on the left side, we can write (16) as −

∯ S

(E1 × H 2 − E2 × H 1 ) ⋅ ds =

∭

(E1 ⋅ J 2 + H 2 ⋅ M 1 − E2 ⋅ J 1 − H1 ⋅ M2 ) dv

V

which is known as the Lorentz Reciprocity Theorem in integral form.

(17)

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CHAPTER

4

Solution Manual √ √ 1 − cos2 𝜓 = 1 − |̂az ⋅ â r |2 √ = 1 − (sin 𝜃 cos 𝜙)2

4.1. (a) sin 𝜓 =

In far-zone fields kI0 le−jkr kI le−jkr √ 1 − (sin 𝜃 cos 𝜙)2 sin 𝜓 = j𝜂 0 4𝜋r 4𝜋r E𝜓 kI le−jkr H𝜒 = j 0 sin 𝜓 = 4𝜋r 𝜂 E𝜓 = j𝜂

z χ

ψ

0

y

x

(b)

U = U0 (1 − sin2 𝜃 cos2 𝜙) 𝜋

2𝜋

∴ Prad = U0

∫0

∫0

(1 − sin2 𝜃 cos2 𝜙) sin 𝜃d𝜃 d 𝜙 = U0

8𝜋 3

4𝜋U0 3 = = 1.5 8𝜋 2 U0 3 √ √ 4.2. (a) sin 𝜓 = 1 − cos2 𝜓 = 1 − |̂ay ⋅ â r |2 √ = 1 − sin2 𝜃 sin2 𝜙 D0 =

Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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SOLUTION MANUAL

In far-zone, the fields are √ kI0 le−jkr kI le−jkr 1 − sin2 𝜃 sin2 𝜙 sin 𝜓 = j𝜂 0 4𝜋r 4𝜋r √ E𝜓 kI e−jkr l 1 − sin2 𝜃 sin2 𝜙 H𝜒 ≃ ≃j 0 𝜂 4𝜋r E𝜓 = j𝜂

z χ

ψ y x

(b)

U = U0 (1 − sin2 𝜃 sin2 𝜙) 𝜋

2𝜋

Prad = U0 = U0 = U0 D0 =

4.3. (a) A =

∫0

∫0 2𝜋 [

∫0 [ ∫0

∫0 2𝜋

(1 − sin2 𝜃 sin2 𝜙) sin 𝜃 d𝜃 d𝜙 𝜋

] sin 𝜃 − sin3 𝜃 sin2 𝜙 d𝜃 d𝜙

4 2 d𝜙 − 3 ∫0

2𝜋

]

[ ] 4 8 sin 𝜙 d𝜙 = U0 4𝜋 − 𝜋 = 𝜋U0 3 3 2

4𝜋 ⋅ U0 3 = = 1.5 8𝜋 2 U0 ⋅ 3

+l∕2 I 𝜇 e−jkr +l∕2 ′ 𝜇 𝜇 e−jKR ′ e−jkr ′ â x I0 dx Ie dl = dx = â x 0 4𝜋 ∫ R 4𝜋 ∫−l∕2 r 4𝜋 4𝜋r ∫−l∕2

A = â x

𝜇I l l𝜇I0 e−jkr = â x Ax ⇒ Ax = 0 e−jkr 4𝜋r 4𝜋r ⎛ Ar ⎞ ⎛ l sin 𝜃 cos 𝜙 ⎜ A𝜃 ⎟ = ⎜ cos 𝜃 cos 𝜙 ⎜ ⎟ ⎜ ⎝ A𝜙 ⎠ ⎝ − sin 𝜙

sin 𝜃 sin 𝜙 cos 𝜃 ⎞ ⎛ Ax ⎞ cos 𝜃 sin 𝜙 − sin 𝜃 ⎟ ⎜ 0 ⎟ ⎟⎜ ⎟ cos 𝜙 0 ⎠⎝ 0 ⎠

Ar = Ax sin 𝜃 cos 𝜙 =

𝜇I0 le−jkr sin 𝜃 cos 𝜙 4𝜋r

A𝜃 = Ax cos 𝜃 cos 𝜙 =

𝜇I0 le−jkr cos 𝜃 cos 𝜙 4𝜋r

A𝜙 = −Ax sin 𝜙 = −

𝜇I0 le−jkr sin 𝜙 4𝜋r

(4-5)

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SOLUTION MANUAL

In far-field: ⎫ Hr ≃ 0 Er ≃ 0 ⎫ ⎪ −jkr E𝜃 Er ≃ 0 ⎪ (3-58a) ⇒ E = −j 𝜔𝜇I0 le ⎪ H𝜙 = cos 𝜃 cos 𝜙 𝜃 𝜂 E𝜃 ≃ −j𝜔A𝜃 ⎬ ⎬ 4𝜋r ⎪ E𝜙 ≃ −j𝜔A𝜙 ⎪ 𝜔𝜇I0 le−jkr E ⎭ (3-58b) E𝜙 = −j sin 𝜙 ⎪ H𝜃 = − 𝜙 4𝜋r ⎭ 𝜂 (b)

r2 [|E |2 + |E𝜙 |2 ] (2-12a) 2𝜂 𝜃 ( ) 𝜔𝜇I0 l 2 1 U= [cos2 𝜃 cos2 𝜙 + sin2 𝜙] 4𝜋 2𝜂 ( ) see 3-D 2 2 2 = B0 [cos 𝜃 cos 𝜙 + sin 𝜙] plot U=

]2 [ ) ( )2 𝜔𝜇I0 l 2 𝜂𝜔𝜇I0 l 1 𝜂𝜔𝜇I0 l 1 = = √ 4𝜋 2𝜂 𝜂4𝜋 2𝜂 4𝜋 𝜇∕𝜀 ] [ √ [ ] ( )2 ( )2 𝜂 kI0 l 𝜂 2 kI0 l 1 𝜂𝜔 𝜇𝜀 1 𝜂kI0 l = = I0 l = = 2𝜂 4𝜋 2𝜂 4𝜋 2𝜂 4𝜋 2 4𝜋

1 B0 = 2𝜂

B0 =

𝜂 2

(

(

kI0 l 4𝜋

)2

U = B0 (cos2 𝜃 cos2 𝜙 + sin2 𝜙) ⇒ Umax = B0 when 𝜙 = 90◦ , 270◦ ; 0 ≤ 𝜃 ≤ 180◦ 𝜋

2𝜋

Prad =

∫0

∫0

U sin 𝜃 d𝜃 d𝜙

⎧ ⎫ 2𝜋 𝜋 ⎪ 2𝜋 𝜋 ⎪ 2 2 2 cos 𝜃 cos 𝜙 sin 𝜃 d𝜃 d𝜙 + cos 𝜙 sin 𝜃 d𝜃 d𝜙⎬ = B0 ⎨ ∫ ∫ ∫ ∫ 0 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ 0 0 ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⎪ ⎪ 0 ⎩ ⎭ I1 I2 2𝜋

I1 =

∫0

cos2 𝜙 d𝜙 2𝜋

(

𝜋

∫0

cos2 𝜃 sin 𝜃 d𝜃 =

1 + cos(2𝜙) 2

)

2𝜋

∫0

cos2 𝜙 d𝜙

𝜋

∫0

cos2 𝜃 d(− cos 𝜃)

𝜋

(cos 𝜃)2 d(cos 𝜃) ∫0 ∫0 [ ] [ 3 ]𝜋 1 1 cos 𝜃 2𝜋 = − (𝜙 + sin 2𝜙)0 2 2 3 0 ) ( ( ) 1 2𝜋 1 1 1 2 = (2𝜋) = I1 = − [(2𝜋)] − − 2 3 3 2 3 3 =−

2𝜋

I2 =

∫0

𝜋

d𝜙

cos2 𝜙 sin 𝜃 d𝜃 d𝜙 =

2𝜋

cos2 𝜙 d𝜙

∫0 ∫0 ) ( 2𝜋 𝜋 1 + cos(2𝜙) = sin 𝜃 d𝜃 d𝜙 ∫0 ∫0 2

𝜋

∫0

sin 𝜃 d𝜃

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SOLUTION MANUAL

I2 =

|𝜋 1 1 1 | = (2𝜋)[−(−1) + 1] = 2𝜋 (− cos 𝜃) [𝜋 + sin 2𝜙]2𝜋 | 0 2 2 |0 2

2𝜋 8𝜋 + 2𝜋 = 3 3 ( ) 8𝜋 Prad = B0 (I1 + I2 ) = B0 3 4𝜋(B0 ) 4𝜋Umax 3 D0 = = = 1.761 dB = 8𝜋 Prad 2 (B ) 3 0 I1 + I2 =

D0 = 1.5 = +1.761 dB (c) Computer Program Directivity: D0 = 1.4980 = 1.7551 dB

z

Dipole

x

y

4.4. From Example 4.5 Er ≃ 0 E𝜃 ≃ −j𝜔A𝜃 = −j

𝜔𝜇I0 le−jkr cos 𝜃 sin 𝜙 4𝜋r

E𝜙 ≃ −j𝜔A𝜙 = −j

𝜔𝜇I0 le−jkr cos 𝜙 4𝜋r

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SOLUTION MANUAL

D0 =

(a)

67

4𝜋Umax Prad

( )2 r2 1 𝜔𝜇I0 l [cos2 𝜃 sin2 𝜙 + cos2 𝜙] [|E𝜃 |2 + |E𝜙 |2 ] = 2𝜂 2𝜂 4𝜋 )2 ( 𝜂𝜔𝜇I0 l 1 U(𝜃, 𝜙) = [cos2 𝜃 sin2 𝜙 + cos2 𝜙] = B0 (cos2 𝜃 sin2 𝜙 + cos2 𝜙) √ 2𝜂 4𝜋 𝜇∕𝜀 U(𝜃, 𝜙) =

( 1 B0 = 2𝜂

)2 √ 𝜂𝜔 𝜇𝜀I0 l 4𝜋

=

𝜂 2

(

kI0 l 4𝜋

)2

U(𝜃, 𝜙) = B0 (cos2 𝜃 sin2 𝜙 + cos2 𝜙) ⇒ Umax = B0 @𝜙 = 0◦ , 180◦ 𝜋

2𝜋

(b) Prad =

∫0

∫0

U(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙 = B0

𝜋

2𝜋

∫0

∫0

(cos2 𝜃 sin2 𝜙 + cos2 𝜙) sin 𝜃 d𝜃 d𝜙

⎧ ⎫ ⎪ ⎪ 2𝜋 𝜋 2𝜋 𝜋 ⎪ ⎪ 2 2 2 cos 𝜃 sin 𝜙 sin 𝜃 d𝜃 d𝜙 + cos 𝜙 sin 𝜃 d𝜃 d𝜙⎬ = B0 ⎨ ∫ ∫ ∫ ∫ 0 0 0 0 ⎪⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⎪ ⎪ ⎪ I1 I2 ⎩ ⎭ 2𝜋

I1 =

∫0

sin2 𝜙 d𝜙

𝜋

∫0

cos2 𝜃 sin 𝜃 d𝜃

) 𝜋 1 − cos 2𝜙 cos2 𝜃 d(− cos 𝜃) d𝜙 ∫0 ∫0 2 [ ) ]2𝜋 ( 3 )𝜋 ( 1 2𝜋 1 −1 − 1 cos 𝜃 1 = = − (2𝜋) = − 𝜙 − sin 2𝜙 2 2 3 2 3 3 0 0 ) ( 2𝜋 𝜋 2𝜋 𝜋 1 + cos 2𝜙 cos2 𝜙 sin 𝜃 d𝜃 d𝜙 = sin 𝜃 d𝜃 I2 = d𝜙 ∫0 ∫ 0 ∫0 ∫0 2 [ ]2𝜋 1 1 1 𝜙 + sin 2𝜙 = (− cos 𝜃)𝜋0 = (2𝜋)(2) = 2𝜋 2 2 2 0 ( ( ) ) 2𝜋 8𝜋 Prad = B0 (I1 + I2 ) = B0 + 2𝜋 = B0 3 3 2𝜋

(

=

(c) D0 =

4𝜋B0 4𝜋Umax 3 = = (same as in Problem 4.2 or any other infinitesimal dipole) 8𝜋 Prad 2 B 3 0

(d) Input parameters: -------The lower bound of theta in degrees = 1 The upper bound of theta in degrees = 180 The lower bound of phi in degrees = 0 The upper bound of phi in degrees = 360 Output parameters: -------Radiated power (watts) = 8.4122

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SOLUTION MANUAL

Directivity (dimensionless) = 1.4938 Directivity (dB) = 1.7430

4.5. (a)

𝜙 = 0◦ (x − z plane) kI le−jkr √ 1 − sin2 𝜃 E𝜓 = j𝜂 0 4𝜋r kI0 le−jkr cos 𝜃 4𝜋r At 𝜙 = 0◦ , E𝜓 has only â 𝜃 direction. ≃ j𝜂

E𝜓 ⇝ E𝜃 polarization z

y

0

ψ

z

ϕ = 0°

x

EA(θ )

kI0 le−jkr 1 4𝜋r At 𝜙 = 90◦ , (y − z plane), E𝜓 has only â 𝜙 direction. E𝜓 ≃ j𝜂

E𝜓 ⇝ E𝜙 polarization

1

𝜙 = 90◦ (y − z plane)

0.8

(b)

0.6

(a)

0.4

0.2

x

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SOLUTION MANUAL

z

ϕ = 90° Eϕ (θ )

x y

0

(b)

(c)

𝜃 = 90◦ (x − y plane) kI0 le−jkr √ kI le−jkr 1 − cos2 𝜙 = j𝜂 0 sin 𝜙 4𝜋r 4𝜋r At 𝜃 = 90◦ (x − y plane) , E𝜓 has only â 𝜙 direction. E𝜓 = j𝜂

E𝜓 ⇝ E𝜙 polarization y

θ = 90°

90 1 120

0.8

60

Eϕ (θ )

0.6 150

30

0.4 0.2

180

0

210

330

240

300 270 (c)

4.6. (a)

𝜙 = 0◦ (x − z plane) kI0 e−jkr (1) 4𝜋r At 𝜙 = 0◦ , E𝜓 direction has only â 𝜙 component E𝜓 = j𝜂

E𝜓 ⇝ E𝜙 polarization

x

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SOLUTION MANUAL

z

ψ y

0 z

ϕ = 0° x

Eϕ (θ ) x

(b)

𝜙 = 90◦ (y − z plane) k I le−jkr √ 1 − sin2 𝜃 E𝜓 = j𝜂 0 0 4𝜋r k0 I0 le−jkr cos 𝜃 4𝜋r At 𝜙 = 90◦ , E𝜓 direction has only â 𝜃 component = j𝜂

E𝜓 ⇝ E𝜃 polarization z 1 0.8 0.6

ϕ = 90° Eϕ (θ )

0.4 0.2

y

(c)

𝜃 = 90◦ (x − y) plane. −jkr

k0 I0 le cos 𝜙 4𝜋r At 𝜃 = 90◦ , E𝜓 direction has only â 𝜙 component E𝜓 = j𝜂

E𝜓 ⇝ E𝜙 polarization

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y

θ = 90°

1

Eϕ (θ )

0.8 0.6 0.4 0.2

x

−jkr

4.7. E𝜃 = −j

𝜔𝜇I0 le 4𝜋r

−jkr

cos 𝜃 cos 𝜙, E𝜙 = −j

𝜔𝜇I0 le 4𝜋r

sin 𝜙

−jkr

(a) 𝜙 = 0 :

E𝜃 = −j

𝜔𝜇I0 le 4𝜋r

cos 𝜃, E𝜙 = 0 (same as in Problem 4.5) −jkr

𝜔𝜇I0 le (same as in Problem 4.5) 4𝜋r −jkr 𝜔𝜇I0 le (c) 𝜃 = 90◦ : E𝜃 = 0, E𝜙 = −j sin 𝜙 (same as in Problem 4.5) 4𝜋r 4.8. From Example 4.5 (b) 𝜙 = 90◦ :

E𝜃 = 0, E𝜙 = −j

−jkr

E𝜃 = −j

𝜔𝜇I0 le 4𝜋r

E𝜙 = −j

𝜔𝜇I0 le 4𝜋r

cos 𝜃 sin 𝜙

−jkr

cos 𝜙

−jkr

(a) 𝜙 = 0 : (b) 𝜙 = 90◦ : (c) 𝜙 = 90◦ :

𝜔𝜇I0 le (same as in Problem 4.6) 4𝜋r −jkr 𝜔𝜇I0 le E𝜃 = −j cos 𝜃 (same as in Problem 4.6) 4𝜋r −jkr 𝜔𝜇I0 le E𝜃 = 0j E𝜙 = −j cos 𝜙 4𝜋r

E𝜃 = 0j E𝜙 = −j

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SOLUTION MANUAL

4.9. (a) Using (4-26a)–(4-26c) and the duality of Table 3.2, the fields of an infinitesimal magnetic dipole of length l and magnetic current lm are given by Er = E𝜃 = H𝜙 = 0 [ ] kI l 1 E𝜙 = −j m sin 𝜃 1 + e−jkr 4𝜋r jkr ] [ I l cos 𝜃 1 Hr = m e−jkr 1 + jkr 2𝜋𝜂r2 [ ] kI l 1 1 H𝜃 = j m sin 𝜃 1 + − e−jkr 4𝜋𝜂r jkr (kr)2 (b) Since the pattern of the magnetic dipole is the same as that of the electric, the directivities are also identical and equal to

D0 =

3 (dimensionless) = 1.761 dB 2

4.10. (a) When the element is placed along the x-axis √ √ 1 − cos2 𝜓 = 1 − |̂ax ⋅ â r |2 √ = 1 − |̂ax ⋅ (̂ax sin 𝜃 cos 𝜙 + â y sin 𝜃 sin 𝜙 + â z cos 𝜃)|2

sin 𝜓 =

and the fields can be written as kI le−jkr E𝜒 = −j m 4𝜋r E𝜒 H𝜓 = − 𝜂

√ kI le−jkr 1 − sin2 𝜃 cos2 𝜙 = −j m sin 𝜓 4𝜋r

(b) In a similar manner, when the element is placed along the y-axis

sin 𝜓 =

√

1 − cos2 𝜓

√ √ 2 = 1 − |̂ay ⋅ â r | = 1 − sin2 𝜃 sin2 𝜙

and the fields can be written as kIm le−jkr kI le−jkr sin 𝜓 = −j m 4𝜋r 4𝜋r E𝜒

E𝜒 = −j H𝜓 = −

𝜂

√ 1 − sin2 𝜃 sin2 𝜙

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4.11.

kI0 l −jkr sin 𝜓 e 4𝜋r kI l Hx = j 0 e−jkr sin 𝜓 4𝜋r

E𝜓 = j𝜂

z

y

ϕ = 45° x

Convert 𝜓 to spherical coordinates √ √ ( )2 √ â x + â y √ 2 ⋅ â r sin 𝜓 = 1 − cos 𝜓 = √1 − √ 2 ( ) â x + â y â y â x ⋅ â r = √ + √ ⋅ (̂ax sin 𝜃 cos 𝜙 + â y sin 𝜃 sin 𝜙 + â z cos 𝜃) √ 2 2 2 =

sin 𝜃 cos 𝜙 sin 𝜃 sin 𝜙 1 + = √ sin 𝜃(cos 𝜙 + sin 𝜙) √ √ 2 2 2

Thus √ kI0 l −jkr 1 1 − [sin2 𝜃(cos 𝜙 + sin 𝜙)2 ] e E𝜓 = j𝜂 4𝜋r 2 √ kI0 l −jkr 1 1 − [sin2 𝜃(cos 𝜙 + sin 𝜙)2 ] H𝜒 = j e 4𝜋r 2 4.12. H𝜓 = j

kIm l −jkr sin 𝜓 e 4𝜋𝜂r

E𝜒 = −j

kIm l −jkr sin 𝜓 e 4𝜋r

Convert 𝜓 to spherical coordinates 1 sin 𝜓 = √ sin 𝜃(sin 𝜙 + cos 𝜙) 2

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z

y

ϕ = 45° x

Thus kI l H𝜓 = j m e−jkr 4𝜋𝜂r

√

1 1 − [sin2 𝜃(cos 𝜙 + sin 𝜙)2 ] 2 √ kIm l −jkr 1 E𝜒 = −j 1 − [sin2 𝜃(cos 𝜙 + sin 𝜙)2 ] e 4𝜋r 2

4.13. Since the dipole is tilted 45◦ on the yz-plane, it can be decomposed into two dipoles r one along the z-direction r the other along the y-direction each with an effective current of

I √0 2

= 0.707. Now we can use superposition to find the

total field. z

I0 45°

I0 sin 45° =

I0 cos 45° =

I0 √2

I0 √2

y

I0

(a) For the z-directed part of the dipole, the electric and magnetic fields are obtained using (4-26a)–(4-26c) or Er ≃ E𝜙 = Hr = H𝜃 = 0 kI le−jkr E𝜃 ≃ j𝜂 √0 sin 𝜃, 2 4𝜋r

kI le−jkr H𝜙 ≃ j √0 sin 𝜃 2 4𝜋r

For the y-directed part of the dipole, the electric and magnetic fields are obtained from Example 4.5 or Er ≃ Hr ≃ 0

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75

𝜔𝜇 I le−jkr E𝜃 ≃ −j √ 0 cos 𝜃 sin 𝜙 2 4𝜋r 𝜔𝜇 I le−jkr E𝜙 ≃ −j √ 0 cos 𝜙 2 4𝜋r 𝜔𝜇 I le−jkr H𝜃 ≃ j √ 0 cos 𝜙 2 4𝜋r 𝜔𝜇 I le−jkr H𝜙 = +j √ 0 cos 𝜃 sin 𝜙 2 4𝜋r √ √ 𝜇 𝜔𝜇 = k𝜂 = 𝜔 𝜇𝜀 𝜀 The total electric and magnetic fields are: Er ≃ Hr ≃ 0 kI le−jkr E E𝜃 ≃ j𝜂 √ 0 [sin 𝜃 − cos 𝜃 sin 𝜙], H𝜙 = 𝜃 𝜂 2 (4𝜋r) E𝜙 I le−jkr E𝜙 ≃ −j𝜂 √0 [cos 𝜙], H𝜃 = − 𝜂 2 (4𝜋r) (b) Since it is an infinitesimal dipole, its directivity is D0 = 1.5 = 1.761 dB. (c) Since it is an infinitesimal dipole, its polarization is linear but tilted at an angle of 45◦ . [ ] kI l sin 𝜃 1 1 ∇ × H where Hr = H𝜃 = 0, H𝜃 = j 0 1+ e−jkr 4.14. E = j𝜔𝜀 4𝜋r jkr Since H is not a function of 𝜙 E=

1 1 ∇×H = j𝜔𝜀 j𝜔𝜀

{

â r

}

1 𝜕 1 𝜕 + (H𝜙 sin 𝜃) − â 𝜃 (rH𝜙 ) + â 𝜙 (0) r sin 𝜃 𝜕𝜃 r 𝜕r

which reduces using the H𝜙 from above to Er = 𝜂

I0 l cos 𝜃

E𝜃 = j𝜂

2𝜋r2

[ 1+

] 1 e−jkr jkr

[ ] kI0 l sin 𝜃 1 1 1+ − e−jkr 4𝜋r jkr (kr)2

E𝜙 = 0 1 1 Re[E × H ⋆ ] = Re[̂a𝜃 E𝜃 × â 𝜙 H𝜙⋆ ] 2 2 [ ] E𝜃⋆ |E |2 1 1 W ave = â r Wr = Re â 𝜃 E𝜃 × â 𝜙 = â r Re(|E𝜃 |2 ) = â r 𝜃 2 𝜂 2𝜂 2𝜂 [ ] 𝜂 || kI0 l ||2 sin2 𝜃 𝜂 || kI0 l ||2 sin2 𝜃 Wr = = W , where W = 0 0 2 || 4𝜋 || 2 || 4𝜋 || r2 r2

4.15. W ave =

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𝜋

2𝜋

Prad = Prad =

∫0

∫0

W ave â r r2 sin 𝜃 d𝜃 d𝜙 = 2𝜋W0

𝜋

∫0

sin3 𝜃 d𝜃 = 2𝜋W0

( ) 4 3

( ) |I l 8𝜋 𝜋 | 0 | W0 = 𝜂 3 3 || λ ||

|2

4.16. A = â z Az = â z

𝜇 I l 𝜇I0 l −jkr ⇒ Az = 0 0 e−jkr e 4𝜋r 4𝜋r

Using (4-6a)–(4-6c) Ar = Az cos 𝜃 =

𝜇I0 l e−jkr 𝜇I l e−jkr cos 𝜃 = A′r (𝜃) ⇒ A′r = 0 cos 𝜃 4𝜋r r 4𝜋

A𝜃 = −Az sin 𝜃 =

−𝜇I0 l e−jkr −𝜇I0 l e−jkr sin 𝜃 = A′0 (𝜃) ⇒ A′0 = sin 𝜃 4𝜋r r 4𝜋

A𝜙 = 0 ⇒ A′𝜙 = 0 Substituting these into (3–57) and (3–57a) reduces to Er = 0 E𝜃 = −j𝜔

𝜔𝜇I0 le−jkr kI le−jkr e−jkr ′ A𝜃 = j sin 𝜃 = j𝜂 0 sin 𝜃 r 4𝜋r 4𝜋r

E𝜙 = −j𝜔

e−jkr ′ A =0 r 𝜙

Hr = 0 H𝜃 = j

𝜔 e−jkr ′ A =0 𝜂 r 𝜙

H𝜙 = −j

𝜔𝜇I0 le−jkr kI le−jkr 𝜔 e−jkr ′ A𝜃 = j sin 𝜃 = j 0 sin 𝜃 𝜂 r 4𝜋𝜂r 4𝜋r

which are identical to (4-26a)–(4-26c) [ ( )]1∕2 −2𝜋z′ cos 𝜃 + z′2 4.17. R = [r2 + (−2rz′ cos 𝜃 + z′2 )]1∕2 = r 1 + r2 Using the binomial expansion of (a + b)n =

an b0 nan−1 b1 an−2 ⋅ b2 an−3 b3 + + (n)(n − 1) + (n)(n − 1)(n − 2) + .. 0! 1! 2! 3!

it can be shown by letting a = r2 b = (−2rz′ cos 𝜃 + z′2 ) n=

1 2

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that R = r − z′ cos 𝜃 + +

1 r

(

z′2 sin2 𝜃 2

) +

1 r2

(

z′3 cos 𝜃 sin2 𝜃 2

] [ 1 z′4 2 4 𝜃 − 5 cos 𝜃) +⋯ (−1 + 6 cos r3 8

Therefore the fifth term of (4-41) is ] [ 1 z′4 2 4 𝜃 − 5 cos 𝜃) (−1 + 6 cos r3 8 4.18. For maximum phase error of 𝜋∕8 radians (22.5◦ ) √ 0.62 D3 ∕λ ⩽ r ⩽ 2D2 ∕λ (a) For a maximum phase error of 𝜋∕16 radians (11.25◦ ) √

√ 2(0.385) D3 ∕λ ⩽ r ⩽ 4D2 ∕λ √ 0.8775 D3 ∕λ ⩽ r ⩽ 4D2 ∕λ

(b) For a maximum phase error of 𝜋∕4 radians (45◦ ) √

0.385 2

√

D3 ∕λ ⩽ r ⩽ D2 ∕λ

√ 0.43875 D3 ∕λ ⩽ r ⩽ D2 ∕λ (c) For a maximum phase error of 18◦ radians 18◦ ⇝ √

𝜋 radians 10

√ D3 ∕λ ⩽ r ⩽ (1.25) ⋅ 2D2 ∕λ √ 0.6937 D3 ∕λ ⩽ r ⩽ 2.5D2 ∕λ

1.25(0.385)

(d) For a maximum phase error of 15◦ radians 15◦ ⇝ √

√

𝜋 radians 12

1.5(0.385) D3 ∕λ ⩽ r ⩽ (1.5)(2)D2 ∕λ √ 0.7599 D3 ∕λ ⩽ r ⩽ 3D2 ∕λ

)

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SOLUTION MANUAL

4.19. l = 5λ0 ⇒ z′ |max = 2.5λ (a) Far-Field (Fraunhofer)

r=

2(5λ)2 2(25λ2 ) 2l2 = = = 50λ λ λ λ

k Δ𝜙e = r

(

2

z′ sin2 𝜃 2

)

2𝜋 [ ] | (2.5)2 λ2 1 λ | = = 0.0982 rads = 5.6250◦ | 2 4 |𝜃=30◦ , z′ =2.5λ, r=50λ 50λ

(b) Fresnel √ √ √ r = 0.62 l3 ∕λ = 0.62 (5λ)3 ∕λ = 0.62λ 125 = 6.9318λ ) ( | k z′3 2𝜋 (2.5λ)3 Δ𝜙e = 2 = cos 𝜃 sin2 𝜃 || 𝜃 = 30◦ (cos 30◦ )(sin 30◦ )2 2 λ (6.9318λ)2 2 r | z′ = 2.5λ r = 6.9318λ

Δ𝜙e =

4.20.

A = â z

𝜋(2.5)3 (6.9318)2

(0.866)(0.5)2 = 0.2212 rads = 12.6724◦

l 𝜇I0 l −jkz′ e−jkR ′ 𝜇I ′ e e−jk(1−cos 𝜃)z dz′ dz ≅ â z 0 e−jkr ∫0 4𝜋 ∫0 R 4𝜋r

𝜇I0 e−jkr l e−jk(1−cos 𝜃)z d[−jk(1 − cos 𝜃)z′ ] 4𝜋r ∫0 −jk(1 − cos 𝜃) [ ]l 𝜇I0 e−jkr e−jk(1−cos 𝜃)z′ 𝜇I le−jkr −jz sin(z) = 0 Az ≃ e 4𝜋r −jk(1 − cos 𝜃) 0 4𝜋r z ′

Az ≃

kl (1 − cos 𝜃) 2 ⎧ E ≃ −j𝜔A Ar = Az cos 𝜃 ⎫ 𝜃 ⎪ ⎪ 𝜃 (a) A𝜃 = −Az sin 𝜃 ⎬ ⇒ For far-field ⇒ ⎨ E𝜙 ≃ −j𝜔A𝜙 ⎪ ⎪ Er ≃ 0 A𝜙 = 0 ⎭ ⎩

where z =

Therefore 𝜔𝜇I0 le−jkr −jz sin(z) e sin 𝜃 4𝜋r z E E𝜙 = 0 = H𝜃 , H𝜙 ≃ 𝜃 𝜂 Er ≃ 0 ≃ Hr , E𝜃 ≃ j

(b) W ave = W rad = 12 Re[E × H ∗ ] = =

1 2𝜂

| 𝜔𝜇I0 l sin (z) |2 | | sin 𝜃 | 4𝜋r | z | |

1 |E |2 2𝜂 𝜃

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SOLUTION MANUAL

4.21. (a) A =

79

∞ ∞ −jkR 𝜇I 𝜇 e−jkr ′ e I(z′ ) dz = â z 0 dz′ 4𝜋 ∫−∞ R 4𝜋 ∫−∞ R

where

√ R=

(x − x′ )2 + (y − y′ )2 + (z − z′ )2 |x′ =y′ =0 =

√ x2 + y2 + (z − z′ )2

Making a change of variable of the form, z − z′ = −p, dz′ = dp reduces the potential to √ 𝜇I0 ∞ e−jk 𝜌2 + p2 dp Az = √ 4𝜋 ∫−∞ 𝜌2 + p2

where 𝜌2 = x2 + y2

Using ∞

∫−∞

√ e−j𝛽 b2 + t2 dt = −j𝜋H0(2) (b𝛽) √ b2 + t2

We can write the potential as Az = −j (b) H =

( √ ) 𝜇I0 (2) 𝜇I H0 (k𝜌) = −j 0 H0(2) k x2 + y2 4 4

1 1 ∇ × A and E = ∇×H 𝜇 j𝜔𝜀

Since A𝜌 = A𝜙 = 0, in cylindrical coordinates 1 1 H = ∇×A= 𝜇 𝜇

(

𝜕Az −̂a𝜙 𝜕𝜌

) = â 𝜙 j

I0 𝜕 (2) H (k𝜌) 4 𝜕𝜌 0

Using Equation (V-19), we can write the H-field as H = â 𝜙 H𝜙 = −̂a𝜙 j

kI0 (2) H (k𝜌) 4 1

where H1(2) (k𝜌) is the Hankel function of the second kind of order one and argument k𝜌. The electric field can be obtained using ] [ ( ) 𝜕H𝜙 H𝜙 1 1 1 𝜕 1 ∇ × H = â z (𝜌H𝜙 ) = â z + j𝜔𝜀 j𝜔𝜀 𝜌 𝜕𝜌 j𝜔𝜀 𝜕𝜌 𝜌 ] [ kI kI 𝜕 1 = â z −j 0 H1(2) (k𝜌) − j 0 H1(2) (k𝜌) j𝜔𝜀 4 𝜕𝜌 4𝜌

E=

Since

𝜕 (2) 1 H1 (k𝜌) = kH0(2) (k𝜌) − H1(2) (k𝜌), by using (V-18) then 𝜕𝜌 𝜌 ] I k kI0 (2) kH0 (k𝜌) = −̂az 𝜂 0 H0(2) (k𝜌) − 4𝜔𝜀 4

[ E = â z

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4.22. Prad = 𝜂

|I0 |2 I 4𝜋 int

where [ 𝜋

Iint =

( cos

kl 2

) ( )]2 cos 𝜃 − cos kl2

∫0

sin 𝜃

d𝜃

which can also be written as 𝜋∕2

Iint = 2 { Letting

∫0

[cos2 ( kl2 cos 𝜃) + cos2 ( kl2 ) − 2 cos( kl2 cos 𝜃) cos( kl2 )] sin 𝜃

l cos 𝜃 = u − sin 𝜃 d𝜃 = du ⇒ d𝜃 = −

d𝜃

du sin 𝜃

reduces Iint to 1

Iint = 2

∫0

1 − u2 [cos2 ( kl2 u) + cos2 ( kl2 ) − 2 cos( kl2 u) cos( kl2 )]

1

=

∫−1

Iint =

[cos2 ( kl2 u) + cos2 ( kl2 ) − 2 cos ( kl2 u) cos ( kl2 )]

(1 + u)

du

du

1 1 cos[ kl (1 + u)] + cos[ kl (1 − u)] [1 + cos(klu) + 1 + cos(kl)] 1 2 2 du − du ∫−1 2 ∫−1 (1 + u) 1+u

Making another change of variable of the form (1 + u)kl = v ⇒ du =

dv kl

we can write that Iint

1 = 2 ∫0 kl

=

2kl

kl 1 cos[ kl (1 − u)] 2 + cos(kl) + cos(klv) cos v 2 dv − dv − du ∫0 ∫−1 v v 1+u

2kl 1 1 + cos(kl) − cos(v) − cos(kl) + cos(v − kl) cos[kl(1 − v)] 1 dv + dv − dv ∫ ∫ v 2 0 v v 0

∫0

provided v =

1+u 2

If z = klv, kl

Iint =

∫0

2kl 1 + cos(kl) − cos(v) − cos(kl) + cos(v) cos(kl) + sin(v) sin(kl) 1 dv + dv v 2 ∫0 v kl

−

∫0

cos(kl) cos(z) + sin(kl) sin(z) dz z

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SOLUTION MANUAL

kl

Iint = [1 + cos(kl)]

∫0

2kl

+ sin(kl)

∫0

1 − cos v dv − 2 ∫0 v

sin v dv − cos(kl) ∫0 v

kl

2kl

sin(v) sin(kl) dv v

1 − cos v dv v

which reduces to Iint =

{ 1 C + ln(kl) − Ci (kl) + sin(kl)[Si (2kl) − 2Si (kl)] 2 ]} [ ( ) 1 kl + cos(kl) C + ln + Ci (2kl) − 2Ci (kl) 2 2

where C = 0.5772 and Prad = 𝜂

|I0 |2 I is identical to (4 − 68) 4𝜋 int

From (4-88)

Prad

( ) 2 𝜋 |I0 |2 𝜋 cos 2 cos 𝜃 =𝜂 d𝜃 4𝜋 ∫0 sin 𝜃

Letting u = cos 𝜃 du = − sin 𝜃 d𝜃

} ⇒ sin2 𝜃 = 1 − cos2 𝜃 = 1 − u2

We can write Prad

2 𝜋 2 𝜋 |I0 |2 0 cos ( 2 u) |I0 |2 1 cos ( 2 u) = −𝜂 du = 𝜂 du 2𝜋 ∫1 1 − u2 2𝜋 ∫0 1 − u2

which can also be written as Prad

[ ] 1 cos2 ( 𝜋 u) 1 cos2 ( 𝜋 u) |I0 |2 2 2 =𝜂 du + du ∫0 4𝜋 ∫0 1−u 1+u

Making another change of variable of the form } } v=1−u v=1+u for the first integral, for the second integral dv = −du dv = du We can write Prad as { } 2 𝜋 1 sin2 ( 𝜋 v) 2 sin2 ( 𝜋 v) |I0 |2 |I |2 2 sin ( 2 v) 2 2 Prad = 𝜂 dv + dv = 𝜂 0 dv ∫0 ∫1 4𝜋 v v 4𝜋 ∫0 v ( ) [ Using the half-angle identity sin2 𝜋2 v = Prad = 𝜂

1−cos(𝜋v) 2

] reduces Prad to

|I0 |2 2 [1 − cos(𝜋v)] dv 8𝜋 ∫0 v

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SOLUTION MANUAL

By letting y = 𝜋v, dy = 𝜋 dv we can write Prad as

Prad

4.23. (a)

( ⎧ I0 1 + ⎪ Iz (z′ ) = ⎨ ( ⎪ I0 1 − ⎩

] [ |I0 |2 2𝜋 1 − cos(y) |I |2 =𝜂 dy = 𝜂 0 Cin (2𝜋) 8𝜋 ∫0 y 8𝜋 ) 2 ′ −l z , < z′ < 0 l ) 2 2 ′ z , 0 < z′ < l∕2 l

𝜇 e−jkr l∕2 ′ Iz (z′ )ejkâ r ⋅r dz′ ∫ 4𝜋 r −l∕2 ) ( |z′ | jkz′ cos 𝜃 ′ 𝜇 e−jkr l∕2 = â z dz e 1−2 4𝜋 r ∫−l∕2 l ( ) ⎫ ⎧ sin kl cos 𝜃 l∕2 ′ −jkr ⎪ |z | jkz′ cos 𝜃 ′ ⎪ 𝜇 e 2 l ( = â z dz ⎬ e ) − 2∫ kl 4𝜋 r ⎨ −l∕2 l ⎪ ⎪ cos 𝜃 ⎭ ⎩ 2

A(r) ≅ â z

l∕2

∫−l∕2

l∕2 ′ 0 |z′ | jkz′ cos 𝜃 ′ z jkz′ cos 𝜃 ′ z′ jkz′ cos 𝜃 ′ dz = dz − dz e e e ∫0 ∫−l∕2 l l l

=

l∕2 ′ z

∫0

l

′ cos 𝜃

ejkz

dz′ +

l∕2 ′ z

∫0

l

e−jkz

′ cos 𝜃

}

dz′

[ ] 1 l kl 𝜉 cos 𝜉 cos 𝜃 d𝜉 ∫0 l 2 ∫0 2 ) ) ( ( kl ⎧ sin kl cos 𝜃 cos 𝜃 − 1 ⎫ cos ⎪ l⎪ 2 2 = ⎨ + ⎬ ( ) 2 kl 2⎪ kl ⎪ cos 𝜃 cos 𝜃 ⎭ ⎩ 2 2 ( ) ⎧ 1 − cos kl cos 𝜃 ⎫ ⎪ 𝜇l e−jkr ⎪ 2 ∴ A(⃗r) = â z ⎬ ( )2 4𝜋 r ⎨ kl ⎪ ⎪ cos 𝜃 ⎭ ⎩ 2 ( ) ⎧ 1 − cos kl cos 𝜃 ⎫ ⎪ ⎪ 𝜇l e−jkr 2 A𝜃 = â 𝜃 ⋅ A = − sin 𝜃 ⎨ ( ⎬ ) 2 4𝜋 r kl ⎪ ⎪ cos 𝜃 ⎭ ⎩ 2 A𝜙 = â 𝜙 ⋅ A = 0 =2

In the far-zone Er ≃ 0

l∕2 ′ z

cos[kz′ cos 𝜃]dz′ =

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SOLUTION MANUAL

E𝜃 ≃ j𝜔𝜇

l 4𝜋

e−jkr r

) ( ⎫ ⎧ kl cos 𝜃 1 − cos ⎪ ⎪ 2 sin 𝜃 ⎨ ( ⎬ )2 kl ⎪ ⎪ cos 𝜃 ⎭ ⎩ 2

E𝜙 ≃ 0 Hr ≃ 0 H𝜃 ≃ 0 H𝜙 ≃ E𝜃 ∕𝜂 (b) From (4-58a) ] [ l∕2 ke−jkr ′ jkz′ cos 𝜃 ′ I(z ) e dz sin 𝜃 E𝜃 = j𝜂 ∫−l∕2 4𝜋r ( ′) l∕2 ′ ke−jkr 𝜋z cos sin 𝜃 I0 ejkz cos 𝜃 dz′ E𝜃 = j𝜂 ∫ 4𝜋r l −l∕2 Let a = jk cos 𝜃 and b = 𝜋l , use following integral formula

∫ then

cos bzeaz dz =

eaz (a cos bz + b sin bz) a2 + b2

{

[ ]}l∕2 ′ 𝜋z′ 𝜋 ejkz ⋅cos 𝜃 𝜋z′ k cos 𝜃 cos j + sin l l l ( 𝜋l )2 − k2 cos2 𝜃 −l∕2 [ ] ke−jkr ejkl∕2 cos 𝜃 𝜋 e−jkl∕2 cos 𝜃 𝜋 = j𝜂 sin 𝜃 I0 𝜋 + 𝜋 2 2 2 4𝜋r ( l ) − k cos 𝜃 l ( l )2 − k2 cos2 𝜃 l ) ) ( ( kl 𝜋 cos 𝜃 cos 𝜃 2 cos cos −jkr −jkr I ke I e 2 2 𝜋 E𝜃 = j𝜂 0 = j𝜂 0 sin 𝜃 𝜋 2 2 2 4𝜋r l ( ) − k cos 𝜃 2𝜋r sin 𝜃 l ) ) ( ( kl 𝜋 cos 𝜃 cos 𝜃 2 cos cos −jkr −jkr I ke I e 2 2 𝜋 sin 𝜃 𝜋 =j 0 H𝜙 = j 0 4𝜋r l ( )2 − k2 cos2 𝜃 2𝜋r sin 𝜃 l ke−jkr E𝜃 = j𝜂 sin 𝜃 I0 4𝜋r

[ ] ( ) l∕2 ′ 𝜋 ke−jkr (c) E𝜃 = j𝜂 I cos2 sin 𝜃 z′ ejkz cos 𝜃 dz′ ∫−l∕2 0 4𝜋r l 𝜋 Let a = jk cos 𝜃 and b = , use the following integral formula l

∫

cos2 bz ⋅ eaz dz =

eaz eaz + 2 2a a + 4b2

(

a cos 2bz + b sin 2bz 2

)

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SOLUTION MANUAL

then ⎧ ′ ( ′ ⎪ ejkz cos 𝜃 jk cos 𝜃 ke−jkr ejkz cos 𝜃 2𝜋 ′ E𝜃 = j𝜂 sin 𝜃 I0 ⎨ k cos 𝜃 + ( )2 ⋅ cos z 4𝜋r 2 l 2j 2𝜋 2 cos2 𝜃 ⎪ − k ⎩ l l∕2

)⎫ 𝜋 2𝜋 ′ ⎪ + sin z ⎬ l l ⎪ ⎭−l∕2 ) ) ( ( ⎧ ⎫ kl kl cos 𝜃 cos 𝜃 sin sin ⎪ ⎪ 2 2 = j𝜂 sin 𝜃 I0 ⎨ + k cos 𝜃 ( )2 ⎬ 4𝜋r k cos 𝜃 2𝜋 2 cos2 𝜃 ⎪ ⎪ − k ⎩ ⎭ l ( ) ) ( ⎧ ⎫ kl sin kl2 cos 𝜃 ⎪ sin 2 cos 𝜃 ⎪ ke−jkr sin 𝜃 I0 ⎨ + k cos 𝜃 ( )2 H𝜙 = j ⎬ 4𝜋r k cos 𝜃 2𝜋 2 cos2 𝜃 ⎪ ⎪ − k ⎩ ⎭ l ke−jkr

R − Z0 1 + |Γ| , R = , Γ = in 1 − |Γ| Rin + Z0 in

Rr ( ) , Z0 = 50 kl sin 2 (a) l = λ∕4, kl∕2 = 𝜋∕4, kl = 𝜋∕2, 2kl = 𝜋 ( )]} { ( )[ 𝜋 1 𝜋 Si (𝜋) − 2Si Rr = 60 C + ln(𝜋∕2) − Ci (𝜋∕2) + sin 2 2 2 { } 1 Rr = 60 0.5772 + 0.45158 − 0.470 + [1.85 − 2(1.3698)] = 6.8388 2 Rr 6.8388 Rin = ( )= ( ) = 13.6776 kl sin2 2 sin2 𝜋4

4.24. VSWR =

Γ=

2

13.6776 − 50 1 + 0.5704 = −0.5704 ⇒ VSWR = = 3.6555 13.6776 + 50 1 − 0.5704

(b) l = λ∕2 : kl∕2 = 𝜋∕2, kl = 𝜋, 2kl = 2𝜋 { ]} [ ( ) 1 𝜋 + Ci (2𝜋) − 2Ci (𝜋) Rr = 60 C + ln(𝜋) − Ci (𝜋) + cos(𝜋) C + ln 2 2 } { 1 = 60 0.5772 + 1.14473 − 0.059 − [0.5772 + 0.45158 − 0.0227 − 2(0.059)] 2 Rr 73.13 Rr = 73.13 ⇒ Rin = ( ) = 73.13 ( )= 𝜋 2 kl sin2 sin 2 2 73.13 − 50 1 + 0.18785 Γ= = 0.18785 ⇒ VSWR = = 1.4626 73.13 + 50 1 − 0.18785 (c) l = 3λ∕4; kl∕2 = 3𝜋∕4, kl = 3𝜋∕2, 2kl = 3𝜋 ( ) ( )]} { ( ) ( )[ 1 3𝜋 3𝜋 3𝜋 3𝜋 − Ci + sin Si (3𝜋) − 2Si Rr = 60 0.5772 + ln 2 2 2 2 2 { } 1 = 60 0.5772 + 1.5502 − (−0.19839) − [1.67473 − 2(1.611)] 2

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SOLUTION MANUAL

Rr = 185.965 ⇒ Rin = Γ=

85

185.965 = 371.93 sin2 (3𝜋∕4)

371.93 − 50 1 + 0.7630 = 0.7630 ⇒ VSWR = = 7.4386 371.93 + 50 1 − 0.7630

(d) l = λ; kl∕2 = 𝜋, kl = 2𝜋, 2kl = 4𝜋 1 cos(2𝜋)[0.5772 + ln(𝜋) + Ci (4𝜋) − 2Ci (2𝜋)]} 2 = 60 {0.5772 + 1.8378 − (−0.0227)

Rr = 60 {0.5772 + ln(2𝜋) − Ci (2𝜋) +

1 + (1)[0.5772 + 1.14473 − 0.006 − 2(−0.0227)]} 2 199.099 =∞ Rr = 199.099 ⇒ Rin = sin2 (𝜋) Γ=

4.25.

∞ − 50 1 − 50∕∞ = = 1 ⇒ VSWR = ∞ ∞ + 50 1 + 50∕∞ ( )2 l , a = 10−4 λ, f = 10 MHz, b = 5.7 × 107 S∕m λ √ √ √ √ 𝜔𝜇0 𝜔𝜇0 2𝜋 × 107 (4𝜋 × 10−7 ) l 𝜔𝜇 l l l = = = = p 2𝜎 C 2𝜎 2𝜋a 2𝜎 2𝜋 × 10−4 λ 2(5.7 × 107 ) ( ) Rr l , ecd = = 1.3245 λ RL + Rr

Rr = 80𝜋 2 RL = Rhf RL = Rhf

(a)

(b)

) λ 2 = 0.316 ohms 50λ ( ) 1 = 0.02649 RL = Rhf = 1.3245 50 Rr 0.316 × 100 ecd = × 100 = = 92.26% RL + Rr 0.02649 + 0.316 l = λ∕50; Rr = 80𝜋 2

(

l = λ∕4; From Prob. 4.24 Rr = 6.8388 1.3245 = 0.3311 4 6.8388 × 100 = = 95.38% 6.8388 + 0.3311

RL = Rhf = ecd

(c)

l = λ∕2; From Prob. 4.24, Rr = 73.13 1.3245 = 0.66225 2 73.13 × 100 = = 99.10% 73.13 + 0.66225

RL = Rhf = ecd

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SOLUTION MANUAL

(d)

l = λ; From Prob 4.24, Rr = 199.099 RL = Rhf = 1.3245

199.099 × 100 = 99.34% 199.099 + 1.3245 ] [ ) ( l∕2 ′ ke−jkr 𝜋 ′ 4.26. H𝜃 = j I cos sin 𝜃 z ejkz cos 𝜃 dz′ ∫−l∕2 m 4𝜋r𝜂 l ecd =

H𝜃 = j

) ( l∕2 kIm e−jkr ′ 𝜋 ′ cos sin 𝜃 z ejkz cos 𝜃 dz′ ∫−l∕2 4𝜋r𝜂 l

Using the same formula in Problem 4.23 (b). ( ) kl 2 cos cos 𝜃 ) ( I 𝜋 2 H𝜃 = j m sin 𝜃 𝜂4𝜋r l ( 𝜋 )2 − k2 cos2 𝜃 l ( ) 𝜋 2 cos cos 𝜃 −jkr I ke l 2 =j m 𝜂4𝜋r k sin 𝜃 ( ) 𝜋 cos cos 𝜃 −jkr I e 2 H𝜃 = j m 𝜂2𝜋r sin 𝜃 ( ) 𝜋 Im e−jkr cos 2 cos 𝜃 E𝜙 = −𝜂H𝜃 = −j 2𝜋r sin 𝜃 1 + |Γ| VSWR − 1 || 2 − 1 || || 1 || 4.27. (a) VSWR = ⇒ |Γ| = = = 1 − |Γ| VSWR + 1 || 2 + 1 || || 3 || ke−jkr

Zin ⎧ || 2 − 1 || ⎪ || 2 + 1 || ⇒ Z = 2 c | 1 | | Z − Zc | | Zin ∕Zc − 1 | ⎪ | = ⎨| 1 |=| |Γ| = || || = || in | | | | Z | 3 | | Zin + Zc | | Zin ∕Zc + 1 | ⎪ || 2 − 1 || 1 ⇒ in = ⎪ || 1 + 1 || Zc 2 ⎩| 2 | Largest Zin = 2 ⇒ Zin = 2Zc = 100 Zc (b)

Rin = 11.14G4.17

λ∕2 < l < 2λ∕𝜋

4.17

𝜋∕2 < kl∕2 < 2

100 = 11.14G

100 = G4.17 , 8.9767 = G4.17 11.14 log10 (8.9767) = 4.17 log10 (G), 0.953 = 4.17 log10 G 0.2286 = log10 G,

G = 100.2286 = 1.6928 =

kl = 96.99◦ 2

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SOLUTION MANUAL

kl = 2(1.6928),

2(1.6928)λ 1.6928 = λ = 0.5388λ 2𝜋 𝜋

l=

l = 0.5388λ (c) Rin =

( ) Rr 2 kl = 100 sin2 (96.99◦ ) ( ) ⇒ Rr = Rin sin 2 2 kl sin 2

Rr = 100(0.9926)2 = 100(0.9852) = 98.52 ohms Rr = 98.52 ohms [ ] ) 1 1 𝜔2 𝜇2 sin2 𝜃 2 ( 2 2 2 k I A + 4A (|E𝜃 |2 + |E𝜙 |2 ) = 0 1 2 2𝜂 2𝜂 16𝜋 2 r2 ] 2𝜋 𝜋 2 2 2 [ [ 2 2 ] 1 𝜔 𝜇 I0 3 2 k = sin 𝜃 d𝜃 d𝜙 A + 4A 1 2 2𝜂 16𝜋 2 ∫0 ∫0 ( 2𝜋 𝜋 ) 𝜔2 𝜇2 I0 2 (k2 A1 2 + 4A2 2 ) 8𝜋 3 = sin 𝜃 d𝜃 d𝜙 = ∫ 0 ∫0 12𝜋𝜂 3

4.28. Wav = Prad Prad

⇒ Rrad =

2Prad I0 2

=

𝜔2 𝜇2 (k2 A1 2 + 4A2 2 ) 6𝜋𝜂

Elliptical polarization since ⃗ = −𝜔𝜇k sin 𝜃 I0 A1 sin(𝜔t − kr)̂a𝜃 + 𝜔𝜇k sin 𝜃 I0 A2 cos(𝜔t − kr)̂a𝜙 E(t) 4𝜋r 2𝜋r e−jkr ⇒ U ∼ |E𝜃 |2 = C0 sin3 𝜃 r 4𝜋Umax , Umax = C0 @ 𝜃 = 90◦ D0 = Prad

4.29. E𝜃 ≃ C0 sin1.5 𝜃 (a)

2𝜋 𝜋

Prad =

∫ ∫ 0

0

𝜋

U sin 𝜃 d𝜃 d𝜙 = 2𝜋 [

= 2𝜋C0 Prad = D0 =

∫

C0 sin4 𝜃 d𝜃

0

(

1 sin 𝜃 cos 𝜃 3 1 + 𝜃 − sin 2𝜃 − 4 4 2 4 3

)]𝜋 = 2𝜋C0 0

[ ( )] 3 𝜋 4 2

3𝜋 2

C 4 0 4𝜋(C0 )

3𝜋 2 ∕4C0

=

16 = 1.69765 = 2.298 dB 3𝜋

(b) U = C0 sin3 𝜃 U = 0.5 C0 = C0 sin3 𝜃h ⇒ sin3 𝜃h = 0.5 ⇒ sin 𝜃h = (0.5)1∕3 = 0.7937 𝜃h = sin−1 (0.7937) = 52.533◦ ⇒ Θh = 2(90 − 52.533) = 2(37.467) Θh = 74.934

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SOLUTION MANUAL

McDonald: D0 =

101 101 = = 1.6897 = 2.27 2 HPBW − 0.0027(HPBW) 74.934 − 0.0027(74.934)2

Pozar: √ D0 = −172.4 + 191

√ 1 1 0.818 + = −172.4 + 191 0.818 + HPBW HPBW

= 1.75 = 2.43 dB (c) Dipole length is l = λ∕2. See Section 4.6, Equation (4-87) (d) Zin = 73 + j42.5 4.30. Dipole with l = λ∕2 ) ( ⎧ cos 𝜋 cos 𝜃 ⎫ I ⎪ ⎪ 2 Ea ≃ â 𝜃 j𝜂 0 ⎨ ⎬ 2𝜋r ⎪ sin 𝜃 ⎪ ⎩ ⎭ ) ( 𝜋 ⎧ ⎫ I0 ke−jkr ⎪ − cos 2 cos 𝜃 ⎪ ≃ −̂a𝜃 j𝜂 ⎬ 2𝜋r ⎨ k sin 𝜃 ⎪ ⎪ ⎩ ⎭ ( ) 𝜋 ⎧ ⎫ cos cos 𝜃 −jkr I0 ke ⎪ λ ⎪ 2 − ≃ −̂a𝜃 j𝜂 ⎬ 2𝜋r ⎨ 2𝜋 sin 𝜃 ⎪ ⎪ ⎩ ⎭ ( ) 𝜋 ⎧ cos cos 𝜃 ⎫ −jkr I ke ⎪ ⎪ λ 0 2 a E ≃ −j𝜂 −̂a𝜃 ⎨ ⎬ 4𝜋r ⎪ 𝜋 sin 𝜃 ⎪ ⎩ ⎭ e−jkr

(

) 𝜋 cos 𝜃 λ 2 (a) le (𝜃) = −̂a𝜃 𝜋 sin 𝜃 ( ) 𝜃=90◦ 𝜋 | cos cos 𝜃 || | λ λ | | | | 2 (b) |le (𝜃)| = |−̂a = = 0.3183λ ⋅ | | |max | 𝜃 𝜋 | sin 𝜃 𝜋 | | | |max |l (𝜃)| λ∕𝜋 2 | e |max (c) = = = 0.6366 l = λ∕2 λ∕2 𝜋 cos

which is 63.66% of l = λ∕2. ( ) 𝜋 | | cos cos 𝜃 | 10−3 || λ | | 2 i| = |−̂a (−̂a𝜃 ) V| (d) Voc = |le ⋅ E | | | |𝜃=90◦ | 𝜃 𝜋 sin 𝜃 λ | | | |𝜃=90◦ ( −3 ) λ 10 10−3 = = = 3.183 × 10−4 Volts 𝜋 λ 𝜋

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SOLUTION MANUAL

4.31. λ∕2 dipole ⇒ (Prad = Pin = 1 Watt, D0 = 1.643 = 2.1564 dB) Zin = 73 + j42.5, f = 1,900 MHz ⇒ λ = 3 × 108 ∕1.9 × 109 = 0.15789 meters Prad 1 = = 0.07958 Watts/sterad 4𝜋 4𝜋 Udipole = U0 D0 = 0.07958(1.643) = 0.130745 Watts/unit solid angle (sterad)

(a) U0 =

(b) Wdipole =

4.32.

Udipole r2

=

λ 2

0.130745 = 5.229 × 10−9 Watts/m2 (5 × 103 )2 λ 2

200 m

𝜃 = 90◦ , 𝜙 = 40◦ At f = 300 MHz, λ =

c = 1m f

( )2 λ 2D2 2 = = 0.5 m ⇒ λ λ r = 200 m ≫ 0.5 m ) ) ( ( λ 2 λ 2 Pr = G0t G0r = D0t D0r 4𝜋r 4𝜋r 2

for lossless antenna. λ Now, since D0t = D0r = 1.643 for dipole 2 )2 ( 1 (1.643)(1.643) 600 = 0.2564 mW Pr = 4𝜋 ⋅ 200 ( ) 1 |E|2 4.33. The time average power density Wav = 2 𝜂 ( ( ) ) ⎡ cos2 𝜋 cos 𝜃 ⎤ 2 𝜋 cos 𝜃 cos 𝜋 2 2 |I | ⎢ |I0 | ⎥ 2 2 Wav = 𝜂 02 2 ⎢ d𝜃 ⎥ , Prad = 𝜂 4𝜋 ∫ 2 8𝜋 r ⎢ sin 𝜃 sin2 𝜃 0 ⎥ ⎦ ⎣ [ ] 𝜂 1 C + ln(2𝜋) − Ci (2𝜋) = 30[0.5772 + 1.838 + 0.02] Prad = Rrad |I0 |2 , Rrad = 2 4𝜋 Rrad = 73.0523. Prad = (0.5 ⋅ 100) = 50 Watts. ⇒ 50 =

1 (73.0523)|I0 |2 ⇒ |I0 |2 = 1.36888 2

At r = 500 m, 𝜃 = 60◦ , 𝜙 = 0◦ Wav

( ) ⎡ cos2 𝜋 cos 60◦ ⎤ ⎥ 1.36888 ⎢ 2 = 120𝜋 2 ⎢ ⎥ 2 2 8𝜋 (500) ⎢ sin 60◦ ⎥ ⎣ ⎦

= 15

1.36888 (0.6667) 𝜋(25)104

Wav = 1.743 × 10−5 Watts∕m2

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SOLUTION MANUAL

4.34. l = λ∕20 ⇒ triangular current distribution; a = λ∕400, f = 30 MHz ⇒ λ = 0.1 meters ( ) ( )2 l 1 = 0.4935 ohms = 20𝜋 2 (a) Rr = Rin = 20𝜋 2 λ 20 [ ( ) ] ] [ ( ) λ 1 400 l ln −1 −1 ln [ln(20) − 1] 20 2 λ 2a Xin = j120 = −j120 ( ) = −j120 ) ( 𝜋l 𝜋 λ tan(𝜋∕20) tan tan λ λ 20 Xin = −j986.935 Zin = 0.4935 − j986.935 (capacitive) (b)

Rr Rr + RL Since element is PEC ⇒ 𝜎 = ∞ ⇒ RL = 0 ecd =

ecd =

Rr = 1 = 100% Rr

(c) Must use an inductor in series to resonate the element with a reactance of XL = 𝜔L = 2𝜋fL = 2𝜋(30 × 106 )L = 986.35 L=

986.35 = 5.236 × 10−6 Henries 2𝜋(30 × 106 ) L = 5.236 × 10−6 Henries

4.35. Za = 73 + j42.5, Zc = 75, f = 100 MHz 42.547|92.694 Z − Zc 73 + j42.5 − 75 −2 + j42.5 (a) Γ = a = = = Za + Zc 73 + j42.5 + 75 148 + j42.5 153.981|16.02 Γ = 0.2763|76.674 ⇒ |Γ| = 0.2763, 𝜙 = 76.674◦ = 1.338(rads) (b) VSWR =

1 + |Γ| 1 + 0.2763 1.2763 = = = 1.76358 1 − |Γ| 1 − 0.2763 0.2763

(c) Za = 73 + j42.5 Need a capacitor in series to resonate. Xc = 42.5 (d) Xc = C=

1 1 1 = = 42.5 ⇒ C = 𝜔C 2𝜋fC 2𝜋f (42.5) 1 = 0.00374 × 10−8 = 37.4 × 10−12 farads 2𝜋(42.5)(108 )

(e) Zin = Za − jXc = 73 + j42.5 − j42.5 = 73 Zin = 73 Γ=

Zin − Zc 73 − 75 −2 = = = −0.0135 Zin + Zc 73 + 75 148

Γ = −0.0135 ⇒ |Γ| = 0.0135

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SOLUTION MANUAL

VSWR =

1 + |Γ| 1 + 0.0135 = = 1.027 1 − |Γ| 1 − 0.0135 VSWR = 1.027

4.36. See below:

1 + |Γ|max (

1 + 0.5

1

4.37. λ∕2 dipole ⇒ Zin = 73 + j42.5, f = 1.9 × 109 Hz | Z − Zc | | 73 + j42.5 − 50 | | 23 + j42.5 | 48.324 |=| | | | |Γ| = || in (a) | | 73 + j42.5 + 50 | = | 123 + j42.5 | = 130.1355 = 0.371 Z + Z | in | | | | c| 1 + |Γ| 1 + 0.371 VSWR = = = 2.17965 1 − |Γ| 1 − 0.371

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1 𝜔CT

= 42.5

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SOLUTION MANUAL

(b) Capacitance ⇒ XT = ⇒ CT =

1 1 1 = = 42.5𝜔 42.5(2𝜋f ) 42.5(2𝜋 × 1.9 × 109 )

CT = 1.971 × 10−12 f (c) C0 = 2CT = 2(1.971 × 10−12 ) = 3.942 × 10−12 f C0

C0

CT

1 1 Z 1 = + = CT C0 C0 C0 C0 = 2CT (d)

| Z − Zc | 73 − 50 23 |= = = 0.18699 |Γ| = || in | | Zin + Zc | 73 + 50 123 1 + |Γ| 1 + 0.18699 VSWR = = = 1.46 1 − |Γ| 1 − 0.18699

[ ( )] 4.38. (a) I = I sin k l + |z| in 0 2 l = λ∕4, z = λ∕8 )] ] [ ( [ ] [ ( ) λ λ λ 2𝜋 λ 𝜋 = I0 sin k = I0 sin = I0 sin = 0.707I0 ± Iin = I0 sin k 4 8 8 λ 8 4 ( )2 ( )2 I0 I0 Rin = Rr = Rr = 2Rr = 2(73) = 146 Iin 0.707I0 ( )2 ( )2 I0 I0 Xm = Xm = 2Xm = 2(42.5) = 85 Xin = Iin 0.707I0 Zin = Rin + jXin = 146 + j85 (b) Yin =

146 − j85 146 − j85 1 = = (5.115 − j2.978) × 10−3 146 + j85 146 − j85 168.941

Ym = +j2.978 × 10−3 ⇒ Xm =

1 = −j335.776 Ym

(Capacitive)

(c) Y ′ = 5.115 × 10−3 ⇒ Z = 5.115 × 10−3 = 195.503 ohms in in | 195.503 − 300 | 104.4966 |= |Γ| = || = 0.21 | 495.503 | 195.503 + 300 | VSWR = (1 + 0.211)∕(1 − 0.211) = 1.5346

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SOLUTION MANUAL

Y'in

4.39.

l = λ∕2,

Ym

Yin

146 + j 85

Zc = 50 ohms

Zin = 73 + j42.5,

Yin =

73 − j42.5 1 1 = Zin 73 + j42.5 73 − j42.5

Yin = 0.01023 − j0.0059563 = (10.23 − j5.9563) × 10−3 = Gin − jBin Bin = 𝜔Cin = 2𝜋fCin ⇒ Cin =

Bin 5.9563 × 10−3 = = 0.94797 × 10−12 2𝜋f 2𝜋(10 × 108 )

∴ Cin = 0.94797 pF Gin = 10.23 × 10−3 Rin =

1 = 97.75, Gin

VSWR =

Γin =

Rin − Zc 97.75 − 50 = = 0.3232 Rin + Zc 97.75 + 50

1 + |Γin | 1 + 0.3232 = = 1.955 1 − |Γin | 1 − 0.3232

4.40. (a) Zin = 4Zin (l = λ∕2) = 4(73 + j42.5) = 292 + j170 292 − j170 292 − j170 1 (b) Y = 1 = = in Zin 292 + j170 292 − j170 114,164 Yin = (2.5577 − j1.4891) × 10−3 Need a capacitor. (c)

Bc = 𝜔C = 2𝜋fC = 1.4891 × 10−3 C=

1.4891 × 10−3 = 0.237 × 10−11 = 2.37 × 10−12 2𝜋(100 × 108 ) C = 2.37 × 10−12

′ = 2.5577 × 10−3 (d) Yin

r Z′ = 1 = ′ in Yin

1 = 390.98 2.5577 × 10−3

r Γ = 390.98 − 300 = 90.98 = 0.1317 in 390.98 + 300

VSWR =

690.98

1 + |Γin | 1 + 0.1317 1.1317 = = = 1.3033 1 − |Γin | 1 − 0.1317 0.8683

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SOLUTION MANUAL

( 4.41.

E = −̂a𝜃 j

𝜔𝜇bI0 4𝜋r

e−jkr

sin

kb cos 𝜃 2 kb cos 𝜃 2

) | | | |𝜃=90◦

= −̂a𝜃 j

𝜔𝜇bI0 e−jkr 4𝜋r

𝜔𝜇I0 e−jkr 𝜔𝜇I0 e−jkr b = −j le (𝜃) 4𝜋r 4𝜋r le (𝜃) = â 𝜃 b

E|𝜃=90◦ = −̂a𝜃 j

Einc |𝜃=90◦ = â 𝜃 j𝜂

| kI0 le−jkr sin 𝜃 || 4𝜋r |𝜃=90◦

= â 𝜃 j𝜂

kI0 le−jkr 4𝜋r

| bkI0 l |2 |𝜂 | | 4𝜋r | |le | | pe = = =1 | bkI0 l |2 |le (𝜃)|2 |Einc |2 | |b|2 ||𝜂 | | 4𝜋r | (𝜃) ⋅ Einc |2

pe (dB) = 10 log10 (1) = 0 dB [ ] + ĵ a a ̂ ◦ x y j20 = C[̂ay ] ⋅ at z = 0 √ 4.42. V1 = 4e 2 √ √ ◦ ◦ ◦ 1 V1 = 4ej20 = jC √ ⇒ C = −j4 2ej20 ⇒ C = 4 2e−j70 2 ( [ ) ] j30◦ √ √ â x + ĵay −j70◦ j30◦ −j70◦ 2 + je V2 = (4 2e )[10(2̂ax + â y e )] ⋅ = 40 2e √ √ 2 2 ◦

= 40e−j70 [2 + j(cos 30◦ + j sin 30◦ )] ◦

◦

◦

= 40e−j70 [1.5 + j0.866] = 40e−j70 [1.73ej30 ] ◦

V2 = 70e−j40 = 53.6 − j45◦ 4.43.

l = 3 cm, λ = 5 cm, I = 10ej60

◦

2D2 2 × 32 18 = = = 3.6 cm ⇒ 10 cm is in the far field. λ 5 5 3 kl l = = 0.6 ⇒ length of dipole is finite, = 𝜋 = 0.6𝜋 5 2 λ ) ( ) ( ⎡ cos kl cos 𝜃 − cos kl ⎤ ] −jkr −jkr [ I e ⎢ cos(0.6𝜋 cos 𝜃) + 0.309 2 2 ⎥ = j𝜂 I0 e ≃ j𝜂 0 ⎥ 2𝜋r ⎢⎢ sin 𝜃 2𝜋r sin 𝜃 ⎥ ⎣ ⎦ ) ( E𝜃 cos(0.6𝜋 cos 45◦ ) + 0.309 || ≃ = 0.7703 , | 𝜂 sin 45◦ |𝜃=45◦

r> l λ E𝜃

H𝜙

e−jkr ⇒ kr =

2𝜋 2𝜋 r= 10 = 4𝜋 = 12.5663 rad λ 5

⇒ E𝜃 = j120𝜋

I0 ej60 e−j4𝜋 (0.7703) = 4,620ej11.52 2𝜋(0.1m)

|E𝜃 | = 4620 V∕m, |H𝜙 | =

4,620 = 12.25 amperes∕meter 120𝜋

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SOLUTION MANUAL

4.44. Using equation (4-79) Rin =

4.45.

Rr 120 ohms = 132.668 ohms ( )= 2 2 kl sin (0.6𝜋) sin 2

kl 3𝜋 3𝜋 = , kl = , 2kl = 3𝜋, a = 0.03λ 2 4 2 (a) Using (8-60a), (8-60b) Rr = 185.808, Xr = 192.7967 (b) Using (8-61a), (8-61b) Rin =

(c) Γ =

185.808 192.7967 = 371.617, Xin = = 385.5936 2 sin (3𝜋∕4) sin2 (3𝜋∕4)

371.617 − 300 = 0.10663 371.617 + 300 VSWR =

1 + 0.10663 = 1.2387 1 − 0.10663

Computer Program: Dipole Rr = 185.8086

Rin = 371.6172

Xm = 192.7968

Xin = 385.5937

4.46. l = 0.625λ (a) Using (8-57a), (8-57b) Rr = 131.9415,

Xr = 146.131638

(b) Using (8-58a), (8-58b) Rin = 154.579,

Xin = 171.203

1 + |0.3199| 154.579 − 300 = −0.3199 ⇒ VSWR = = 1.9407 154.579 + 300 1 − 1 − 0.31991 4.47. (a) l = 200 m, a = 1m, f = 150 KHz → λ = 2000 meters. Using (11-37), (8-29) or (11-37) (c) Γ =

] [ ( ) l −1 ln ( )2 ( )2 [ln(100) − 1] l 1 2a 2 Zin ≃ 20𝜋 2 − j120 − j120 ( ) ≃ 20𝜋 l λ 10 tan(𝜋∕10) tan 𝜋 λ Zinput = 2 + Zin = 2 + 1.9739 + j1,377.07 Zinput = 3.9739 + j1,377.07 (b) Radiation efficiency = 100

Rr 1.9739 = 100 = 49.67% RL + Rr 3.9739

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SOLUTION MANUAL

Rr 1.9739 = 1.4335 × 10−3 = |Im(Zinput )| 1.377 × 10+3 (d) X = −Im(Zinput ) = −1,377.07 √ √ Rr + RL 3.9739 = n= = 0.282 Z0 50 (c) RPF =

(e) The answer to this part was found by manually entering values of X until |Γ| = 0.333 was obtained. The values obtained are X1 = 0.99803 X2 = 1.00198 The corresponding percent bandwidth is BW = (X2 − X1 ) × 100% = 0.395% 0.9 0.8 0.7 0.6 0.5 0.4 0.3 0.2 0.1 0 0.99

0.995

1

) 2̂ax − ĵay √ = 5E0 e+jkz √ 5 ⏟⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏟ 𝜌̂w (

4.48. Ew = (2̂ax − ĵay )E0

e+jkz

) 2̂ax − ĵay (a) 𝜌̂w = √ 5 ( ( ) ) â x + ĵay −ĵax + â y (b) 𝜌̂a = or 𝜌̂a = √ √ 2 2 (c) 1. Elliptical, AR = 2 2. CCW (

1.005

1.01

1.015

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97

(d) 1. Circular, AR = 1 2. CCW |( 2̂a − ĵa ) ( â + ĵa )|2 | |2 | |2 | | 2 − j2 1 | |2 + 1| x y x y | | | | | | (e) PLF = |𝜌̂w ⋅ 𝜌̂a |2 = | ⋅ √ √ | | =| √ | =|√ | | | | | | | 5 10 | 2 | | | 10 | | 9 = = −0.4576 dB 10 or |2 | |2 |( 2̂a − ĵa ) ( −ĵa + â )|2 | | | −j2 − j | | −j3 | x y x y | | | | | | | ⋅ PLF = |𝜌̂w ⋅ 𝜌̂a | = | √ √ | =| √ | = |√ | | | | | | | 5 10 | 2 | | | | 10 | 9 = = −0.4576 dB 10 2

4.49. Wi = 2 𝜇W∕m2 = 2 × 10−6 W∕m2 (a) Eiw = (3̂az + ĵay )Eo e+jkx ) ( 3̂az + ĵay i 10Eo e+jkx Ew = √ 10 ) ( 3̂az + ĵay 𝜌̂w = √ 10

z

z

Elliptical CCW AR = 3/1 = 3

y

E iw y

𝜋 I0 e−jky cos( 2 cos 𝜃) (b) Ea = â 𝜃 j𝜂 2𝜋r sin 𝜃

x

cos( 𝜋2 cos 𝜃) | | = â 𝜃 E0 | sin 𝜃 |𝜃=𝜋∕2

Ea =

â 𝜃 E0 ⏟⏟⏟ 𝜌̂a

𝜌̂a = â 𝜃 ⇒ Linear 𝜌̂a = [̂ax cos 𝜃 cos 𝜙 + â y cos 𝜃 sin 𝜙 − â z sin 𝜃]𝜃=90◦ 𝜌̂a = −̂az

|( 3̂a + ĵa ) |2 | | z y 9 | (c) PLF = |𝜌̂w ⋅ 𝜌̂a | = | ⋅ (−̂az )|| = = 0.9 = −0.4576 dB √ 10 | | 10 | | 2

PLF = −0.4576 dB = 0.9

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SOLUTION MANUAL

(d)

λ= Aem =

3 × 108 = 1m 300 × 106 λ2 1 1.643 D0 = D0 = = 0.1307 m2 4𝜋 4𝜋 4𝜋

PL = Aem Wi (PLF) = 0.1307(2 × 10−6 )(0.9) = (0.2614)(0.9) × 10−6 PL = 0.2353 × 10−6 Watts E𝜃 = j𝜂

4.50.

W ave =

kI0 le−jkr sin 𝜃 cos(kh cos 𝜃); 0 ⩽ 𝜃 ⩽ 𝜋∕2, 0 ⩽ 𝜙 ⩽ 2𝜋 2𝜋r

â 𝜂 | kI l |2 1 Re[E × H ∗ ] = r |E𝜃 |2 = â r || 0 || sin2 𝜃 cos2 (kh cos 𝜃) 2 2𝜂 2 | 2𝜋r | 𝜋∕2

2𝜋

Prad = =

∫0

∫0

W ave ⋅ â r r2 sin 𝜃 d𝜃 d𝜙

𝜂 || kI0 l ||2 2𝜋 𝜋∕2 3 sin 𝜃 cos2 (kh cos 𝜃) d𝜃 d𝜙 2 || 2𝜋 || ∫0 ∫0

𝜋∕2 | kI0 l |2 | | ⋅ sin3 𝜃 cos2 (kh cos 𝜃) d𝜃 | 2 | ∫ | | 0 [ ] 𝜂 || kI0 l ||2 𝜋∕2 3 1 + cos (2kh cos 𝜃) sin 𝜃 d𝜃 = | 𝜋 | 2 || ∫0 2 } { 𝜋∕2 𝜋∕2 𝜂 || kI0 l ||2 = sin3 𝜃d𝜃 + sin3 𝜃 ⋅ cos(2kh cos 𝜃)d𝜃 ∫0 ∫0 2𝜋 || 2 ||

=

Prad =

𝜂 𝜋

𝜂 2𝜋

| kI0 l |2 | | | 2 | {I1 + I2 } | |

𝜋∕2

where I1 =

∫0 𝜋∕2

I2 =

∫0

|𝜋∕2 2 1 sin3 𝜃 d𝜃 = − cos 𝜃(sin2 𝜃 + 2)|| = 3 3 |0 sin3 𝜃 cos (kh cos 𝜃) d𝜃 = Let

u = sin2 𝜃

du = 2 sin 𝜃 cos 𝜃 d𝜃

𝜋∕2

∫0 v=−

dv = −

sin2 𝜃 cos (kh cos 𝜃) sin 𝜃 d𝜃 1 sin (2kh cos 𝜃) 2kh

cos(2kh cos 𝜃) d(2kh cos 𝜃) 2kh

Thus I2 = −

𝜋∕2 |𝜋∕2 sin2 𝜃 2 cos 𝜃 sin(2kh cos 𝜃) sin 𝜃 d𝜃 sin(2kh cos 𝜃)|| + 2kh 2kh ∫0 |0

Let u = cos 𝜃

1 sin(2kh cos 𝜃)d(2kh cos 𝜃) 2kh 1 v= cos(2kh cos 𝜃) 2kh

dv = −

du = − sin 𝜃 d𝜃

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SOLUTION MANUAL

99

} { 𝜋∕2 |𝜋∕2 2 1 cos 𝜃 I2 = 0 + + cos(2kh cos 𝜃) sin 𝜃 d𝜃 cos(2kh cos 𝜃)|| 2kh 2kh 2kh ∫0 |0 { } { } |𝜋∕2 cos(2kh) sin(2kh) 2 1 1 | = sin(2kh cos 𝜃)| + − cos(2kh) − =2 − 2kh 2kh (2kh)2 (2kh)2 (2kh)3 |0 Therefore Prad

𝜂 = 2𝜋

[ ] | kI0 l |2 | |2 | | {I + I } = 𝜋𝜂 | I0 l | 1 − cos(2kh) + sin(2kh) 2 | 2 | 1 |λ| 3 (2kh)2 (2kh)3 | | | |

kI0 le−jkr 2𝜋r = C1 sin 𝜃 cos(kh cos 𝜃)|𝜃=30◦ = 0 ⇒ cos(kh cos 𝜃)|𝜃=30◦ = 0

4.51. E𝜃 = C1 sin 𝜃 cos(kh cos 𝜃), where C1 = j𝜂 (a) E𝜃 |𝜃=30◦

2𝜋 𝜋 1 h(0.867) = cos−1 (0) = ⇒ h = λ = 0.288λ λ 2 4(0.867) ( ) 2𝜋 2 (0.288λ) = 3.632 (b) D0 = [ ] , 2kh = 2 λ 1 cos(2kh) sin(2kh) + − 3 (2kh)2 (2kh)3 kh cos(30◦ ) =

D0 = [

2 2 ]=[ ] 1 1 cos(3.632) sin(3.632) + 0.06689 − 0.00983 + − 3 3 (3.632)2 (3.632)3

D0 = 5.12 = 7.1 dB ] ( )2 [ l 1 cos(3.632) sin(3.632) (c) Rr = 2𝜋𝜂 + − λ 3 (3.632)2 (3.632)3 ( )2 1 [0.39] = 0.37 ohms Rr = 2𝜋(377) 50 kI le−jkr E𝜃 = C1 sin 𝜃 cos(kh cos 𝜃), where C1 = j𝜂 0 4.52. 2𝜋r E𝜃 |h=2λ = C1 sin 𝜃n cos(kh cos 𝜃n )|h=2λ = 0 ⇒ sin 𝜃n = 0, cos(kh cos 𝜃n )|h=2λ = 0 sin 𝜃n = 0 ⇒ 𝜃n = 0◦ cos(kh cos 𝜃n )|h=2λ = cos(4𝜋 cos 𝜃n ) = 0 ) ( 2n + 1 𝜋, n = 0, 1, 2, ... ⇒ 4𝜋 cos 𝜃n = cos−1 (0) = ± 2 𝜃n = cos−1 [±(2n + 1)∕8], n = 0, 1, 2, 3, 4, .... 𝜃0 = cos−1 (± 18 ) = 82.82◦ ⎫ ⎪ n = 1; 𝜃1 = cos−1 (± 38 ) = 67.98◦ ⎪ for 0◦ ≤ 𝜃 ≤ 90◦ ⎬ ◦ ◦ n = 2; 𝜃2 = cos−1 (± 58 ) = 51.32◦ ⎪ (for 90 ≤ 𝜃 ≤ 180 , the field is zero) n = 3; 𝜃3 = cos−1 (± 78 ) = 28.96◦ ⎪ ⎭ ( ) 9 = does not exist. The same holds for n ≥ 5. n = 4; 𝜃4 = cos−1 ± 8 n = 0;

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SOLUTION MANUAL

Therefore where the field vanishes for 0◦ ≤ 𝜃 ≤ 90◦ , are 𝜃 = 0◦ , 28.96◦ , 51.32◦ , 67.98◦ , and 82.82◦ 4.53.

kI0 e−jkr l 2𝜋r ◦ ◦ E𝜃 |𝜃=60◦ = C1 sin(60 ) cos(khn cos(60 )) = 0 ⇒ cos(khn cos(60◦ )) = 0 ) ( ) ( 𝜋 1 2n + 1 = hn = cos−1 (0) = ± 𝜋, n = 0, 1, 2, 3, .... khn cos(60◦ ) = khn 2 λ 2

E𝜃 = C1 sin 𝜃 cos(kh cos 𝜃), where C1 = j𝜂

Choosing the positive values ) 2n + 1 λ, n = 0, 1, 2, 3, .... 2 hn = 0.5λ, 1.5λ, 2.5λ, 3.5λ, 4.5λ (

hn =

4.54.

E𝜃 (4 − 99) ≃ C sin 𝜃[2 cos(kh cos 𝜃)] ⇒ AF = 2[cos(kh cos 𝜃)]max = ±2 cos(kh cos 𝜃m ) = ±1 (a)

kh cos 𝜃m = cos−1 (±1) ⇒ cos 𝜃m = cos

[

] ] [ 1 m𝜋 cos−1 (±1) = cos ± kh kh

⎤ ⎡ ] [ ⎢ ±m𝜋 ⎥ m −1 , m = 0, 1, 2, ... cos 𝜃m = cos ⎢ ± ( ) ⎥ ⇒ 𝜃m = cos 3 ⎢ 2𝜋 3λ ⎥ ⎣ λ 2 ⎦ m = 0; 𝜃0 = cos−1 (±0) = 90◦ { cos−1 (1∕3) = 70.5288◦ −1 m = 1; 𝜃1 = cos (±1∕3) = cos−1 (−1∕3) = 2◦ (⇒ below ground plane) { cos−1 (2∕3) = 48.1897◦ −1 m = 2; 𝜃2 = cos (±2∕3) = cos−1 (−2∕3) = 3◦ (⇒ below ground plane) { cos−1 (1) = 0◦ m = 3; 𝜃3 = cos−1 (±1) = cos−1 (−1) = 0◦ (⇒ below ground plane) m = 4; 𝜃4 = cos−1 (±4∕3) ⇒ does not exist (b) E𝜃m = C sin 𝜃[2 cos(kh cos 𝜃)]max = ±2C, where 𝜃 = 90◦ (c)

E𝜃 = sin 𝜃 cos(kh cos 𝜃) E𝜃m 𝜃 = 0◦ :

E𝜃 E = 0 ⇒ 𝜃 = 20 log10 (10) = −∞ dB E𝜃m E𝜃m

𝜃 = 48.1897◦ :

E𝜃 E = sin 𝜃 cos(kh cos 𝜃)|h= 3λ = 0.7454 ⇒ 𝜃 E𝜃m E𝜃m 2

= 20 log10 (0.7454) E𝜃 = −2.55 dB E𝜃m

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101

𝜃 = 70.5288◦ : E𝜃 ∕E𝜃m = sin 𝜃 cos(kh cos 𝜃)|h= 3λ = 0.9428 2

E ⇒ 𝜃 = 20 log10 (0.9428) = −0.5115 dB E𝜃m | E | = 1 ⇒ 𝜃 = 20 log10 (1) = 0 dB 𝜃 = 90◦ : E𝜃 ∕E𝜃m = sin 𝜃 cos(kh cos 𝜃)| | 3λ E 𝜃m |h= 2

4.55.

120°

60°

60°

4.56. (a) Since the equivalent problem, based on Figure 4.16(a), is that of 2 sources of the same magnitude but 180◦ phase difference, the normalized array factor is the same as that of (4-115) or AFn = sin(kh cos 𝜃) (b)

h = 0.5λ AFn = sin(kh cos 𝜃n ) = sin

[

2𝜋 λ

( ) ] λ cos 𝜃n = 0 2

sin(𝜋 cos 𝜃n ) = 0 ⇒ 𝜋 cos 𝜃n = sin−1 (0) = n𝜋, n = 0, ±1, ±2, .... cos 𝜃n = n ⇒ 𝜃n = cos−1 (n), n = 0, ±1, ±2, ±....

n = 0: n = +1 : n = −1 : (c)

𝜃n = cos−1 (0) = 90◦ 𝜃+1 = cos−1 (+1) = 0◦ 𝜃−1 = cos−1 (−1) = 180◦

𝜃 = 𝜃n = 60◦ [

]

AFn = sin(khn cos 𝜃)|𝜃=60◦ = sin khn (0.5) = sin khn = sin−1 (0) = n𝜋, n = +1, +2, .... 2 2𝜋hn 𝜋hn = = n𝜋 2λ λ hn = nλ, n = 1, 2

(

khn 2

) =0

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z

Im θ

PEC ( σ = ∞) (a) Physical problem

Actual source θ

h

h Image

(b) Equivalent problem

h1 = λ h2 = 2λ

4.57.

kI0 le−jkr sin 𝜃[2 cos(kh cos 𝜃)] 4𝜋r |AF|max = | cos(kh cos 𝜃)|max = 1 when kh cos 𝜃max = 𝜋 E𝜃 ≃ j𝜂

kh cos(60◦ ) = 𝜋 ( ) 2𝜋 1 = 𝜋, h = λ h λ 2

kh cos 𝜃max = 𝜋,

No matter what the height is when 𝜃 = 90◦ , it is a maximum. So you always have a maximum at 𝜃 = 90◦ . If you want a maximum at 𝜃 = 60◦ , then kh cos 𝜃 = n𝜋, (n = 1, 2, 3, ...) leads to maximum at 𝜃 = 60◦ . n = 1 : kh cos 𝜃|max = 𝜋, h = λ leads to maxima at 𝜃 = 90◦ , 60◦ If you check closely, it also leads to a maximum at 𝜃 = 0◦ . So you cannot only have one maximum at 𝜃 = 60◦ 4.58.

| =0 E𝜃 ∼ C1 sin 𝜃 cos(kh cos 𝜃)| |𝜃=80◦ 𝜋 2𝜋 𝜋 | | | cos(kh cos 𝜃)| = 0, kh cos 𝜃 | = , = h cos 𝜃 | |𝜃=80◦ |𝜃=80◦ |𝜃=80◦ 2 λ 2

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h=

λ | λ λ = = = 1.4397λ | 4 cos 𝜃 |𝜃=80◦ 4(0.1736) (0.6946)

30 × 107 3 × 108 = = 6 meters 50 × 106 5 × 107 h = 1.4397λ = 1.4397(6) = 8.6382 meters h = 1.4397λ, λ =

h = 8.6382 meters 1 1 Z (l = λ)|free space ≃ (Rim + jXim )|l=λ 2 im 2 Form Problem 4.24 ⇒ Rim = R = 199.099 r) ( λ ≃ 62.5 | From Figure 4.23 ⇒ Xim l = 2 above ground plane Therefore

4.59. (a) Zim (l = λ∕2)|above ground plane =

Zim (l = λ∕2)|above ground plane =

199.099 + j62.5 = 99.5495 + j62.5 2

Feed Referred to feed at center of λ /2.

(b) Zin =

(c)

Γ=

Zim 99.5495 + j62.5 =∞ ( )= sin2 (𝜋) 2 kl sin 2 Zin − Zc ∞ − 50 1 − 50∕∞ = = =1 Zin + Zc ∞ + 50 1 + 50∕∞ VSWR =

1 + |Γ| 1 + 1 = =∞ 1 − |Γ| 1 − 1

Xim can also be obtained using (8-57b). For l = λ ⇒ kl = 2𝜋, 2kl = 4𝜋. Thus Xim (l = λ∕2)|above ground plane =

1 X (l = λ)|free space 2 im

𝜂 {2Si (kl) + cos(kl)[2Si (kl) − Si (2kl)]} 8𝜋 120𝜋 {2Si (2𝜋) + cos(2𝜋)[2Si (2𝜋) − Si (4𝜋)]} = 80𝜋 = 15{2(1.418) + [2(1.418) − 1.492]} = 62.7

=

4.60. (a) Array factor, h = 1.5λ; two sources; separated by 2h; image 180◦ . (AF)n = sin(kh cos 𝜃)

(4-115)

103

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θ h PEC (0 ≤ θ ≤ 90°)

h

Image

±n𝜋 (b) | sin(kh cos 𝜃m )| = 1 ⇒ kh cos 𝜃m = sin−1 (±1) = , n = 1, 3, 5... [ ] n𝜋 2 ] [ ) ( ± ±n𝜋∕2 λ , n = 1, 3, 5.. 𝜃m = cos−1 = cos−1 2𝜋2 = cos−1 ±n kh 4h h 2 ) ( ( ) { 1 λ = 80.406◦ below GP n = 1 : 𝜃1 = cos−1 ± cos−1 ± 4(2.5λ) 6 ) ( ( ) { 3λ 1 −1 −1 = cos = 60.00◦ below GP ± ± n = 3 : 𝜃3 = cos 6λ 2 ) ( ( ) { 5λ 5 −1 −1 = cos = 33.537◦ below GP ± ± n = 5 : 𝜃5 = cos 6λ 6 ) ( ( ) 7λ 7 = cos−1 ± = does not exist n = 7 : 𝜃7 = cos−1 ± 6λ 6 (c) | sin(kh cos 𝜃n )| = 0 ⇒ kh cos 𝜃n = sin−1 (0) = ±n𝜋,

𝜃n = cos

−1

n = 0, 1, 2, 3..

⎛ ⎞ ] ) ( ( ) n𝜋 nλ || n −1 ⎜ ±n𝜋 ⎟ = cos ⎜ ± = cos−1 ± = cos−1 ± ⎟ | 2𝜋 kh 2h 3 |h=1.5λ ⎜ h⎟ ⎝ λ ⎠

[

n = 0 : 𝜃0 = cos−1 (0) = 90◦ ( ) { 1 −1 = 70.53◦ below GP ± n = 1 : 𝜃1 = cos 3 ( ) { 2 = 48.190◦ below GP n = 2 : 𝜃2 = cos−1 ± 3 { n = 3 : 𝜃3 = cos−1 (±1) = 0◦ below GP ( ) 4 = does not exist n = 4 : 𝜃4 = cos−1 ± 3 4.61. |Γ|max = 0.2 (a) E𝜃 of a λ∕8 monopole is the same as that of a λ∕4 dipole. E𝜃 (l = λ∕8 monopole) = E𝜃 (l = λ∕4 dipole)

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) ( ) ⎤ cos 𝜃 − cos kl2 ⎥ ⎥ sin 𝜃 ⎥ ⎦l=λ∕4 ) ) ( ( ⎡ 2𝜋 λ 2𝜋 λ ⎤ I0 e−jkr ⎢ cos 2λ 4 cos 𝜃 − cos 2λ 4 ⎥ = j𝜂 ⎥ 2𝜋r ⎢⎢ sin 𝜃 ⎥ ⎣ ⎦ ( ) ⎡ 𝜋 𝜋 ⎤ I0 e−jkr ⎢ cos 4 cos 𝜃 − cos( 4 ) ⎥ = j𝜂 ⎥ 2𝜋r ⎢⎢ sin 𝜃 ⎥ ⎣ ⎦ ( ) 𝜋 2𝜋 λ = = 12.35 G2.5 , G = kl = (4-109b) λ 8 4 ( )2.5 𝜋 = 12.35 = 12.35(0.54667) = 6.7514 ohms 4

⎡ I0 ⎢ cos E𝜃 (l = λ∕8 monopole) = j𝜂 2𝜋r ⎢⎢ ⎣

(

e−jkr

(b) Rin (monopole)l≥λ∕8

Computer Program: Rin (λ∕8 monopole) =

(c) |Γ|max

#1 :

kl 2

[ ] ] [ Zin − Zc 1 − |Γ|max 1 − 0.2 = |Γ|max ⇒ Zc = Zin = 6.7514 Zin + Zc 1 + |Γ|max 1 + 0.2 ( ) ( ) 2 0.8 Zc = 6.7514 = 6.7514 = 4.5 1.2 3 Zc = 4.5

#2 :

] [ ) ( |Γ|max + 1 Zc − Zin 0.2 + 1 = |Γ|max ⇒ Zc = Zin = 6.7514 Zc + Zin −|Γ|max + 1 −0.2 + 1 ( Zc = 6.7514

1.2 0.8

) = 6.7514

( ) 3 = 10.125 2

Zc = 10.125 4.62. AF = cos(kh cos 𝜃), f = 1 GHz ⇒ λ = (a)

3 × 108 = 0.3 meters 1 × 109

|(AF)|𝜃n =30◦ = | cos(kh cos 30◦ )| = | cos(0.866kh)| = 0 ⇒ 0.866kh = cos−1 (0) =

(4-26a)

( ) λ 1 Rin l = dipole = 6.72025 ohms 2 4

| | | | | | | | | | | | |Z − Z | |Z − Z | | c | in c| in | =| |=| | = 0.2. Either of the two answers is okay. | Zin + Zc | | Zc + Zin | | | | | |⏟⏞⏟⏞⏟| |⏟⏞⏟⏞⏟| | #1 | | #2 | | | | |

n𝜋 , n = 1, 2, 3, ... 2

105

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h1 =

(b) 1.

2.

n𝜋∕2 n𝜋∕2 n(3) || nλ = 0.0866 meters = = = 0.866k 0.8662𝜋∕λ 0.866(4𝜋) 4(0.866𝜋) ||n=1

h1 = 0.0866 meters )| ( | 2𝜋 | cos(kh cos 𝜃)|h=0.3m=λ = ||cos λ cos 𝜃n || = | cos(2𝜋 cos 𝜃n )| = 0 λ | | n𝜋 −1 ⇒ n = 1, 3, 5, ... 2𝜋 cos 𝜃n = cos (0) = 2 ) ( ( ) n −1 n𝜋∕2 , n = 1, 3, 5, ... = cos−1 𝜃n = cos 2𝜋 4 ( ) 1 = 75.52◦ n = 1 : 𝜃1 = cos−1 4 ( ) 3 = 41.41◦ n = 3 : 𝜃3 = cos−1 4 ( ) 5 = does not exist n = 5 : 𝜃5 = cos−1 4 )| ( | 2𝜋 | cos(kh cos 𝜃m )|h=0.3m=λ = ||cos λ cos 𝜃m || = | cos(2𝜋 cos 𝜃m )| = 1 λ | | 2𝜋 cos 𝜃m = cos−1 (1) = m𝜋, m = 0, 1, 2, 3, ... 𝜃m = cos−1

(

m𝜋 2𝜋

)

= cos−1

( ) m 2

m = 0 : 𝜃0 = cos−1 (0) = 90◦ ( ) 1 = 60◦ m = 1 : 𝜃1 = cos−1 2 m = 2 : 𝜃2 = cos−1 (1) = 0◦ ( ) 3 = does not exist m = 3 : 𝜃3 = cos−1 2 3 × 108 f = 200 MHz ⇒ λ = = 1.5 meters 4.63. 2 × 108 ( ) 𝜋 cos cos 𝜃 2 cos(kh cos 𝜃) E𝜃 (normalized) = sin 𝜃 ( ) 𝜋 cos cos 𝜃 2 has a null only toward 𝜃 = 0◦ , the only way to place a null toward Since sin 𝜃 𝜃 = 60◦ will be through cos(kh cos 𝜃). | cos(kh cos 𝜃)|𝜃=𝜃h =60◦ = | cos(kh cos 𝜃n )| = | cos(kh cos 60◦ )| = 0 )| | ( )| ( | |cos 2𝜋 h 1 | = |cos 𝜋h | = 0 | λ 2 || || λ || | 𝜋h n𝜋 = cos−1 (0) = , n = 1, 3, 5, ... λ 2 ( ) nλ n𝜋 λ = h= , n = 1, 3, 5, ... 2 𝜋 2

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107

( ) 3 λ 3 1 = = 0.75 meters = 2 2 2 4 3λ (b) h|n=3 = h3 = = 2.25 meters 2 5λ (c) h|n=5 = h5 = = 3.75 meters 2 4.64. (a) Since the dipole is vertical and placed a height h above a PEC ground plane, its array factor is that of (4-99), or (a)

h|n=1 = h1 =

(AF)n = cos(kh cos 𝜃)

2 sources, same amplitude, same phase.

At f = 300 MHz ⇒ λ = 3 × 108 ∕3 × 108 = 1 meter (b) According to the geometry of the figure 𝜓=

𝜋 𝜋 𝜋 − 𝜃 = ⇒ 𝜃 = = 45◦ 2 4 4

Since the element pattern cos( 𝜋2 cos 𝜃) | E𝜃 | = = 0 only when 𝜃 = 0◦ |λ∕2 dipole sin 𝜃 the only way to place nulls at 𝜃 = 45◦ is to adjust the height h so that the array factor goes to zero. AF|𝜃=45◦ = cos(khn cos 𝜃)|𝜃=45◦ = cos[khn (0.707)] = 0 n𝜋 0.707khn = cos−1 (0) = , n = 1, 3, 5, ... 2 n𝜋 n𝜋 ( ) n𝜋 λ nλ 2 2 hn = = = ( )= 0.702k 0.707 2𝜋 2 2𝜋(0.707) 0.707(4) λ hn = h1 (n = 1) : h3 (n = 3) :

n(1) nλ || = 2.828 ||λ(300 MHz)=1 meter 2.828

1 = 0.3536 meters 2.828 3 h3 = = 1.0608 meters 2.828 h1 =

4.65. G0 (dB) = 10 log10 G0 (dimensionless) ⇒ 16 = 10 log10 G0 ⇒ G0 (dimensionless) = 101.6 = 39.81 Prad = e0 Pin = (1)(8) = 8 Watts W0 =

Prad 4𝜋r2

=

8 8 2 = = × 10−8 = 0.6366 × 10−8 𝜋 4𝜋(100 × 100)2 4𝜋 × 108

= 6.366 × 10−9 Watts∕cm2 Wmax = W0 G0 (dimensionless) = 39.81(6.366 × 10−9 ) = 2.534 × 10−7 = 0.2534 × 10−6 Watts∕cm2

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4.66. l = λ∕4, f = 1.9 GHz, Wi = 10−6 W∕m2 ⇒ λ =

3 × 106 = 0.15789 m 1.9 × 109

(a) The power pattern of a λ∕4 monopole above a PEC is equivalent to that of a λ∕2 dipole in free space. Since the same power radiated by the monopole above the PEC is concentrated only in the upper hemisphere, instead over the entire free space, its radiation intensity will be twice as strong/intense as that of the λ∕2 dipole radiating in free space. Since the directivity is given by D0 =

4𝜋Umax Prad

The Umax of the monopole will be twice that of the dipole, or D0 (l = λ∕4) = 2(1.643) = 3.286 = 5.17 dB Using the computer program Directivity it gives D0 (l = λ∕4) = 3.3365 = 5.2329 dB (b) Aem =

(0.15789)2 λ2 D0 = (3.286) = 6.52 × 10−3 m2 4𝜋 4𝜋

PL = Aem Wi = 6.52 × 10−3 (10−6 ) = 6.52 × 10−9 PL = 6.52 × 10−9 Watts 4.67. (a) D0 = 2(1.5) = 3 = 4.7712 dB ( ) 3λ2 λ2 λ2 1 (b) Aem = = D0 (PLF) = (3) 4𝜋 4𝜋 2 8𝜋 3λ2 W = 10 × 10−6 8𝜋 i 8𝜋 80𝜋 Wi = 2 (10 × 10−6 ) = 2 × 10−6 3λ 3λ

Prec = Aem Wi =

30 × 109 = 3 × 103 cm 10 × 106 80𝜋 80𝜋 80𝜋 Wi = × 10−6 = × 10−6 = × 10−12 3 2 6 27 3(3 × 10 ) 3(9) × 10 λ=

Wi = 9.3084 × 10−12 Watts/cm2 4.68. f = 900 MHz, Prad = 1,000 Watts (a) Isotropic Wr0 ≤

Prad

4𝜋r2 Prad 1,000 100 r2 ≥ = = = 7.9558 4𝜋Wr0 4𝜋(10) 4𝜋 r ≥ 2.821 meters

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109

(b) λ∕4 monopole D0 (monopole) = 2(1.643) = 3.286 Wrad ≤ D0 Wr0 = D0 r2 ≥ D0

Prad 4𝜋Wrad

Prad

4𝜋r2 ( ) 1,000 = 3.286 = 26.1492 4𝜋(10)

r ≥ 5.114 meters

4.69. Using the coordinate system of Figure 4.27 the total field is given by (4-115) or −jkr √

kI le E𝜓 = j𝜂 0 4𝜋r

1 − sin2 𝜃 sin2 𝜙[2j sin(kh cos 𝜃)], 0 ⩽ 𝜃 ⩽ 𝜋, 0 ⩽ 𝜙 ⩽ 2𝜋

However if we rotate the axes so that the z axis is parallel to the axis of the element and y is vertical to the ground, the total E-field can be written as

E𝜃 = j𝜂

kI0 e−jkr sin 𝜃[2j sin(kh sin 𝜃 sin 𝜙)], and 4𝜋r 𝜋

Prad = Prad = I= where I1 = = =

∫0 ∫0

𝜋

W ave ⋅ â r r2 sin2 𝜃 d𝜃 d𝜙 =

𝜋

𝜋

1 |E𝜃 |2 r2 sin 𝜃 d𝜃 d𝜙 2𝜂 ∫0 ∫0

𝜂 || kI0 l ||2 𝜋 𝜋 3 𝜂 || kI0 l ||2 2 sin 𝜃 sin (kh sin 𝜃 sin 𝜙) d𝜃 d𝜙 = I 2 || 2𝜋 || ∫0 ∫0 2 || 2𝜋 || { 𝜋 } 𝜋 𝜋 3 2 sin 𝜃 sin (kh sin 𝜃 sin 𝜙) d𝜙 d𝜃 = sin3 𝜃[I1 ] d𝜃 ∫0 ∫0 ∫0 } { 𝜋 𝜋 𝜋 1 2 sin (kh sin 𝜃 sin 𝜙) d𝜙 = d𝜙 − cos(2kh sin 𝜃 ⋅ sin 𝜙 d𝜙 ∫0 ∫0 2 ∫0 { ) } 𝜋( y2 y4 y6 1 𝜋− + − + ⋅ ⋅ ⋅ d𝜙 , where y = 2kh sin 𝜃 sin 𝜙 1− ∫0 2 2! 4! 6! } { 𝜋 𝜋 𝜋 1 1 𝜋 1 1 y2 d𝜙 + y4 d𝜙 − y6 d𝜙 + ⋅ ⋅ ⋅⋅ − 𝜋− 2 2 2 ∫0 (2 × 2)! ∫0 (2 × 3)! ∫0

∞ ∞ 𝜋 (y)2n (2𝛼)2n 𝜋 2n 1∑ 1∑ (−1)n+1 (−1)n+1 sin 𝜙 d𝜙 d𝜙 = ∫0 2n! 2 n=1 2 n=1 2n! ∫0 ] [ ∞ 𝜋∕2 2n 1∑ 2n n+1 (2𝛼) I1 = (−1) sin 𝜙 d𝜙 2 ∫0 2 n=1 2n!

I1 =

=

∞ ∑ (2𝛼)2n 𝜋∕2 2n (−1)n+1 sin 𝜙 d𝜙 2n! ∫0 n=1

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SOLUTION MANUAL

( From Mathematical Handbook of Formulas and Tables

a = kh sin 𝜃 y = 2𝛼 sin 𝜙

) Schaum’s Outline

Series, pg. 96 Equation 15-30. 𝜋∕2

∫0 Thus I1 =

sin2n x dx =

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅(2n − 3)(2n − 1) 𝜋 , n = 1, 2, 3, 4, ⋅ ⋅ ⋅ 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ ⋅(2n − 2)(2n) 2

∞ ∑ (2kh sin 𝜃)2n 𝜋 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅(2n − 3)(2n − 1) (−1)n+1 ⋅ ⋅ A2n , where A2n = 2n! 2 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ ⋅(2n − 2)(2n) n=1

and 𝜋

I=

∫0

∞ 𝜋 ∑ (2kh)2n ( 𝜋 ) A2n sin 𝜃[I1 ] d𝜃 = (−1)n+1 (sin 𝜃)2n+3 d𝜃 ∫0 (2n)! 2 n=1 3

∞ 𝜋∕2 ∑ (2kh)2n (−1)n+1 ⋅ (sin 𝜃)2n+3 d𝜃 ⋅ A2n =𝜋 ∫0 (2n)! n=1

Using Series equation of the previous reference, or 𝜋∕2

∫0

(sin x)2n+3 dx =

2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ ⋅ ⋅ (2n − 2)(2n)(2n + 2) , n = 1, 2, 3, ... 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅(2n − 1)(2n + 1)(2n + 3)

we can write that I=𝜋 A2n+3 =

∞ ∑ (2kh)2n (−1)n+1 (A2n )(A2n+3 ), (2n)! n=1

2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ ⋅(2n − 2)(2n)(2n + 2) 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅(2n − 1)(2n + 1)(2n + 3)

However A2n ⋅ A2n+3 =

1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ (2n − 3)(2n − 1)2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ ⋅(2n − 2)(2n)(2n + 2) 2 ⋅ 4 ⋅ 6 ⋅ ⋅ ⋅ (2n − 2)(2n) 1 ⋅ 3 ⋅ 5 ⋅ ⋅ ⋅ ⋅(2n − 1)(2n + 1)(2n + 3)

= (2n + 2)∕[(2n + 1)(2n + 3)] Therefore I=𝜋

∞ ∑

(−1)n+1

n=1

[ =𝜋

(2kh)2n (2n + 2) (2n)! (2n + 1)(2n + 3)

(2kh)2 4 (2n + 2) (2kh)2n (2kh)4 6 (2kh)6 8 − + − ⋅ ⋅ ⋅ ⋅ +(−1)n+1 2! 3 ⋅ 5 4! 5 ⋅ 7 6! 7 ⋅ 9 (2n)! (2n + 1)(2n + 3)

which when expanded can be written as ) ) ( ( [ 1 1 2 2 1 1 1 1 + (2kh)4 + ⋅⋅ − + (2kh)2 − + − − + 3 3 3! 4! 5! 5! 6! 7! )]} ( 1 1 1 ±(2kh)2n − + (2n + 1)! (2n + 2)! (2n + 3)!

I=𝜋

{

]

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SOLUTION MANUAL

111

Recombining appropriate terms, we have that { [ ] ] [ ∞ ∞ 2n+1 2n ∑ (2kh)2 ∑ 2 1 1 n (2kh) n (2kh) I=𝜋 (−1) − (2kh) + (−1) − + 1− 3 (2kh) (2n + 1)! 2! (2n)! (2kh)2 n=1 n=2 [ ]} ∞ 2n+3 (2kh)3 ∑ 1 n+1 (2kh) + (−1) (2kh) − + 3! (2n + 3)! (2kh)3 n=1 which reduces when expanded to [ I=𝜋

2 sin(2kh) cos(2kh) sin(2kh) + − − 3 (2kh) (2kh)2 (2kh)3

]

Therefore the radiated power can be written as

Prad =

4.70. E𝜓 = C2

𝜂 || kI0 l ||2 𝜋 I=𝜂 2 || 2𝜋 || 2

[ ] | I0 l |2 2 sin(2kh) cos(2kh) sin(2kh) | | + − − |λ| 3 (2kh) (2kh)2 (2kh)3 | |

√ kI le−jkr 1 − sin2 𝜃 sin2 𝜙[sin(kh cos 𝜃)], C2 = −𝜂 0 2𝜋r

◦ (a) E𝜓 (𝜙 = 90◦ ) ||𝜃=45◦ = C2 cos 𝜃 sin(kh cos 𝜃)|| 𝜃=45◦ = C2 cos(45 ) sin

(

) kh √ 2

=0

kh −1 √ = sin (0) = ±n𝜋, n = 0, 1, 2, 3, … 2 Choosing the positive values and excluding the n = 0 value, we have the smallest height of (n = 1) √ 2𝜋 2𝜋 λ = λ = √ = 0.707λ k 2𝜋 2

√ h=

( ) √ λ 2𝜋 λ (b) h = √ ⇒ 2kh = 2 √ = 2 2𝜋 = 8.88576 λ 2 2 [ ] √ √ √ ( )2 1. 1 2 sin(2 2𝜋) cos(2 2𝜋) sin(2 2𝜋) 2 Rr = 120𝜋 + − − √ √ √ 50 3 2 2𝜋 (2 2𝜋)2 (2 2𝜋)3 ( )2 [ ] 1 2 Rr = 120𝜋 2 − 0.057765 + 0.0108694 + 0.0007316 = 0.294 50 3 2. kh =

√ 2𝜋

D0 = [

4(−0.9639)2

2 3

4(0.9291) = 5.9893 ] = 0.6205 − 0.57765 + 0.0108694 + 0.0007316

D0 = 5.9893 = 7.774 dB

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4.71.

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SOLUTION MANUAL

kI0 le−jkr 2𝜋r = C2 cos 𝜃n sin(0.707λk cos 𝜃n ) = 0

E𝜓 (𝜙 = 90◦ ) = C2 cos 𝜃 sin(kh cos 𝜃), C2 = −𝜂 E𝜓 (𝜙 = 90◦ )|h=0.707λ

Null due to element factor: cos 𝜃n = 0 ⇒ 𝜃n = cos−1 (0) = 90◦ Nulls due to array factor: sin(0.707λk cos 𝜃n ) = sin(1.414𝜋 cos 𝜃n ) = 0 ⇒ 1.414𝜋 cos 𝜃n = sin−1 (0) 1.414𝜋 cos 𝜃n = sin−1 (0) = ±n𝜋, n = 0, 1, 2, 3, … ) ( n , n = 0, 1, 2, 3, … 𝜃n = cos−1 ± 1.414 ⎫ n = 0 : 𝜃n = cos−1 (0) = 90◦ ⎪ ◦ ◦ ) ( 1 ⎬ for 0 ≤ 𝜃 ≤ 90 = 45◦ ⎪ n = ±1 : 𝜃n = cos−1 ± 1.414 ⎭ ) ( 2 = does not exist. n = ±2 : 𝜃n = cos−1 ± 1.414 The same holds for n ≥ 3. The null at 𝜃 = 90◦ is due to both the element factor and array factor. 4.72. Since the horizontal dipole is placed a distance of 2λ above the PEC, then its image must also be a distance of 2λ below the PEC. This makes the separation between the actual source and its image to be 4λ. Since the minimum far-field distance is equal to r = 2D2 ∕λ where D is the large distance, which in this case is the hypotenuse, or √ D = (4λ)2 + (λ∕50)2 = 4.00005λ ≃ 4λ Then r = 2(4λ)2 ∕λ = 32λ Since λ at 300 MHz the wavelength is 1 meter, then r = 32λ|λ=1 = 32 meters Source

4.73.

H𝜃

d

kI le−jkr1 =j m sin 𝜃1 𝜂4𝜋r

H𝜃 r = −j

kIm le−jkr2 sin 𝜃2 𝜂4𝜋r

h PEC h Image

σ=∞

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SOLUTION MANUAL

r1 = r − h cos 𝜃 r2 = r + h cos 𝜃 H𝜃 = j

4.74.

E𝜃

d

} phase

r = r1 = r2 , for amplitude 𝜃1 = 𝜃2 = 𝜃

113

} ⇒ far field

kIm le−jkr sin 𝜃[2j sin(kh cos 𝜃)] 𝜂4𝜋r Source

kI le−jkr1 = j𝜂 0 sin 𝜃1 4𝜋r1

h

σm = ∞

kI0 le−jkr2 sin 𝜃2 4𝜋r2 } r1 = r − h cos 𝜃 phase Far field : r2 = r + h cos 𝜃 E𝜃 r = −j𝜂

PMC h Image

(r = r1 = r2 , 𝜃 = 𝜃1 = 𝜃2 ) amplitude E𝜃 = j𝜂

kI0 le−jkr sin 𝜃[ejkh⋅cos 𝜃 − e−jkh cos 𝜃 ] 4𝜋r

E𝜃 = j𝜂

kI0 le−jkr sin 𝜃[2j sin(kh cos 𝜃)] 4𝜋r

30 × 109 = 60 cm 5 × 108 (a) AF = 2 cos(kh cos 𝜃) (4-99) Same as that of a vertical dipole above a PEC. (b) AF = 2 cos(kh cos 𝜃)|𝜃=60◦ = 0 [ ] 2𝜋 cos h(0.5) = 0 λ ( ) 𝜋h =0 cos λ 𝜋h n𝜋 ⇒ = cos−1 (0) = , n = 1, 3, 5, … λ 2 nλ hn = , n = 1, 3, 5, … 2

4.75. f = 500 MHz ⇒ λ =

θ

h

σm = ∞

PMC

Source h h Image

Smallest h (n = 1): h1 =

4.76.

λ 60 = = 30 cm 2 2

H𝜃 total = H𝜃d + H𝜃r H𝜃 d = j H𝜃

r

le−jkr1

kIm 𝜂4𝜋r1

Source h

sin 𝜃1

kI le−jkr2 =j m sin 𝜃2 𝜂4𝜋r2

PMC

h Image

σm = ∞

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SOLUTION MANUAL

r1 ≃ r − h cos 𝜃 r2 ≃ r + h cos 𝜃

For far field.

} phase, (r1 = r2 = r) → amplitude

𝜃 = 𝜃 1 = 𝜃2 H𝜃total = j

4.77. (a)

E𝜃 ≃ j𝜂

kIm le−jkr sin 𝜃[2 cos(kh cos 𝜃)] 𝜂 ⋅ 4𝜋r

kl e−jkr I sin 𝜃[j2 sin(kh cos 𝜃)] 4𝜋 0 r

Source h

sin(kh cos 60◦ ) = 0 → khn cos 60◦ = n𝜋, n = 1, 2, 3, … hn =

PMC

n𝜋 nλ = = nλ k cos 60◦ 2 ⋅ cos 60◦

h

Smallest h ⇒ n = 1 ⇒ h = λ |E |2 2 𝜂(kl)2 0 2 (kl) Wav ≃ | | ≃ |I | sin2 𝜃[4 sin2 (kh cos 𝜃)] (b) 0 2𝜂 32𝜋 2 r2 𝜂 ( l )2 |I0 |2 sin2 𝜃 sin2 (kh cos 𝜃) U(𝜃, 𝜙) = lim r2 Wav = r→∞ 2 λ 𝜋∕2

2𝜋

Prad =

∫0

∫0

σm = ∞ Image

U(𝜃, 𝜙) sin 𝜃 d𝜃 d𝜙

( )2 𝜋∕2 l |I0 |2 sin3 𝜃 sin2 (kh cos 𝜃) d𝜃 ∫0 λ { } ( )2 1 cos(2kh) sin(2kh) l 2 |I0 | − + = 𝜋𝜂 λ 3 (2kh)2 (2kh)3

= 𝜋𝜂

Prad = 𝜋𝜂 ↑ kh=2𝜋

1. D0 (𝜃 = 45◦ , 𝜙) =

{ } ( )2 ( )2 l 1 1 1 |I0 |2 |I0 |2 {0.3397} + = 𝜋𝜂 λ 3 (4𝜋)2 λ

4𝜋U(𝜃 = 45◦ , 𝜙) 2 sin2 (45◦ ) sin2 (2𝜋 cos 45◦ ) = Prad 0.3397

= 2.74 = 4.37 dB ( )2 2Prad l = 2𝜋𝜂 {0.3397} 2. Rr = λ |I0 |2 Rr = 2𝜋 × 10−4 × 0.3397 = 2.13 × 10−4 𝜂 4.78. Since d ≪ a tan 𝜓 ≃ d1 (+h′1

h′1 d1

≃

+ h′2 )

h′2 d2

=

=

h′1 d

h′2 d − d1 ⇒ d1 =

⇒ h′1 (d − d1 ) = d1 h′2 h′1 d h′1 + h′2

=

5(20 × 103 ) = 99.5 meters 5 + 1,000

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SOLUTION MANUAL

( 𝜓 = tan−1

h′1

)

d

= tan−1

(

5 99.5

)

115

= 2.87669◦

𝜎 10−2 18 = × 10−2 = 3.6 × 10−2 ≪ 1 = 9 𝜔𝜀 2𝜋 × 10 (5 × 10−9 ∕(36𝜋)) 5 √

𝜇1 √ , k1 = 𝜔 𝜇1 𝜀1 𝜀1 The divergence factor is equal to (a = 5280 miles = 8.497368 × 106 m) Therefore the earth is a good dielectric ⇒ 𝜂1 ≃

[ D≃ 1+2

]−1∕2

h′1 h′2 ad tan3 𝜓

[

2(5)(1,000) = 1+ 8.497368 × 106 × 2 × 104 (0.05)3

]−1∕2

D = (1 + 0.000463)−1∕2 = 0.99977 and the reflection coefficient is equal to √

√ j𝜔𝜇0 𝜇0 ≃ 𝜎1 + j𝜔𝜀1 𝜀1 √ 𝛽 𝜀0 𝛾0 sin 𝜃i = 𝛾1 sin 𝜃t ⇒ 𝛽0 sin 𝜃i = 𝛽1 sin 𝜃t ⇒ sin 𝜃t = 0 sin 𝜃i = sin 𝜃i 𝛽1 𝜀1 √ √ √ 𝜀 sin2 𝜃i cos 𝜃t = 1 − sin2 𝜃t = 1 − 0 sin2 𝜃i = 1 − 𝜀1 𝜀r 𝜂 cos 𝜃i − 𝜂1 cos 𝜃t , where 𝜂0 = Rv = 0 𝜂0 cos 𝜃i + 𝜂1 cos 𝜃t

√

𝜇0 , 𝜂 = 𝜀0 1

Therefore 𝜂 cos 𝜃i − cos 𝜃i − 1 cos 𝜃t 𝜂0 = Rv = 𝜂 cos 𝜃i + 1 cos 𝜃t cos 𝜃i + 𝜂0

√ 1 − sin2 𝜃i ∕𝜀r

√ 𝜀r − sin2 𝜃i = √ √ 1 2 𝜀 cos 𝜃 + 𝜀r − sin2 𝜃i 1 − sin 𝜃 ∕𝜀 √ r i i r 𝜀r 1 √ 𝜀r

𝜀r cos 𝜃i −

𝜃i = 90 − 𝜓 = 90 − 2.87669◦ = 87.12331◦ ⇒ sin 𝜃i = 09987, cos 𝜃i = 0.0502

Thus Rv =

5(0.0502) −

√ √

5 − (0.9987)2

=

−1.749649 = −0.777 2.251649

5(0.0502) + 5 − (0.9987)2 ( ) 𝜋 cos cos 𝜃 [ ′ ] 2 I e−jkr ′ ejkh1 cos 𝜃 + DRv e−jkh1 cos 𝜃 E𝜃 ≃ j𝜂 0 2𝜋r sin 𝜃 𝜃=𝜃i ≃87.12331◦ √ √ r ≃ d2 + (h′2 − h′1 )2 = (20,000)2 + (1,000 − 99.5)2 = 20,020.26 m = 66, 734.207λ h′1 = 5 m = 16.667λ, h′2 = 1,000 m = 3, 333.3333λ

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SOLUTION MANUAL

(

) 𝜋 ◦ 2𝜋 cos cos(87.12 ) [ jkh′ cos 𝜃 ] 2 I e−j λ (66,734.207λ) ′ i + DR e−jkh1 cos 𝜃i 1 e E𝜃 = j120𝜋 0 v 2𝜋(20, 020.26) sin(87.12◦ )

e−j [ cos

2𝜋 (66,734.207λ) λ

= e−j2𝜋(0.207) = e−j1.3 = cos(74.52◦ ) − j sin(74.52◦ )

]

= 0.2669 − j0.9637 = 1∠ − 74.52◦

𝜋 cos(87.12◦ ) = 0.996887, sin(87.12◦ ) = 0.99874 2

ejkh1 cos 𝜃i = ej ′

2𝜋 (16.667λ)(0.0502) λ

= ej2𝜋(16.667)(0.0502) = ej5.257

= cos(301.2◦ ) + j sin(301.2◦ ) = 1∠301.2◦ = 0.5181 − j0.8553 e−jkh1 cos 𝜃i = 1∠ − 301.2◦ = 0.5181 + j0.8553 ′

DRv e−jkh1 cos 𝜃i = 0.99977(−0.777)[0.5181 + j0.8553] = −(0.4025 + j0.6644) ′

Thus ejkh1 cos 𝜃i + Rv De−jkh1 cos 𝜃i = (0.5181 − j0.8553) − (0.4025 + j0.6644) ′

′

= 0.1156 − j1.5197 = 1.5241∠ − 85.65◦ Therefore E𝜃 ≃ (1∠90◦ )(120𝜋)

I0 (1∠ − 74.52◦ )(0.996887) (1.5241∠ − 85.65◦ ) 2𝜋(0.99874)(20, 020.26)

E𝜃 ≃ 4.5592 × 10−3 I0 ∠ − 70.17 or |E𝜃 | = 4.5592 × 10−3 |I0 |

Volts∕m

4.79. Use Friis Transmission Equation of (2-118) with: r e = e = 1 because of lossless. cdt cdr

r Z = 73 because of resonant. a [ ]2 cos( 𝜋2 cos 𝜃) | r D =D| t r |𝜃=45◦ = D0 sin 𝜃

𝜃=45◦

[ = 1.643

cos( 𝜋2 cos 𝜃) sin 𝜃

] 𝜃=45◦

| 0.44417 |2 | = 1.643(0.62824)2 = 1.643(0.3947) = 1.643 || | | 0.707 | Dt (𝜃 = 45◦ ) = Dr (𝜃 = 45◦ ) = 1.643(0.3947) = 0.648

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SOLUTION MANUAL

P. S. You could also use: | = 1.643(0.3536) = 0.581. Dt (𝜃 = 45◦ ) = Dr (𝜃 = 45◦ ) ≃ 1.643 sin3 𝜃 | |𝜃=45 √ R = 2(1,000) = 1, 414 meters λ(1 GHz) =

𝜐 3 × 108 = 0.1 meters = f 3 × 109

Pt = 100 × 10−3 watts 2 2 PLF = ||𝜌̂t ⋅ 𝜌̂r || , ||â 𝜃 ⋅ â 𝜃 || = 1 ] [ Pr λ 2 = ecdt ecdr (1 − |Γt |2 )(1 − |Γr |2 ) Dt Dr (PLF) Pt 4𝜋R

| Z − Zc | | 73 − 50 | 23 | |=| |Γt | = |Γr | = || a | | 73 + 50 | = 123 = 0.187 Z + Z | | | a c| |Γt |2 = |Γr |2 = |0.187|2 = 0.035 (1 − |Γt |2 ) = (1 − |Γr |2 ) = (1 − 0.035) = 0.965 [ ]2 Pr 0.1 = (1)(1)(0.965)(0.965) (0.648)(0.648)(1) Pt 4𝜋(1, 414) Pr = 0.931228(5.6278 × 10−6 )(0.4199) Pt Pr = 0.931228(31.67438 × 10−12 )(0.4199)(100 × 10−3 ) = 12.3854 × 10−13 Pr = 1.23854 × 10−12 Watts 4.80. From calibration; C P Pr 10 × 10−6 = 21 → C1 = r R2 = × (10 × 103 )2 = 200 m2 Pt Pt 5 R on asteroid C Pr = 21 |1 + DRv e−j2kht cos 𝜃 |2 Pt R Approximate geometry; s' ht = 1.5 m

s

θi ψ

ψ

hr = 1.5 m

2000 m

𝜓 = tan−1 𝜃i =

(

1.5 1000

)

≃ 1.5 × 10−3 = 0.086◦

𝜋 − 𝜓; cos 𝜃 i = sin 𝜓 ≃ 1.5 × 10−3 2

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SOLUTION MANUAL

𝜂0 𝜂 Rv = 𝜂1 0 𝜂1

√ i 2 2 sin 𝜃 1 sin 𝜃 t = ≃ ; cos 𝜃 t ≃ 3 3 3 √ 2 2 −3 cos 𝜃 i − cos 𝜃 t 3(1.5 × 10 ) − 3 = √ = −0.9905 i t 2 2 cos 𝜃 + cos 𝜃 3(1.5 × 10−3 ) + 3

s′ ≃ s ≃ 1000 m; a = 106 m [ ]−1∕2 [ ]−1∕2 (1000)(1000) ss′ D≃ 1+2 ≃ 1+2 6 = 0.7746 ad tan 𝜓 10 (2000)1.5 × 10−3 λ=

3 × 108 c =1m = f 300 × 106

ht |2 | |1 + DRv e−j2kht cos 𝜃 |2 ≃ ||1 − (0.7746)(0.9905)e−j4𝜋 λ || | | 2 | | = |1 − (0.7746)(0.9905)e−j4𝜋 | | |

|1 + DRv e−j2kht cos 𝜃 |2 = 0.0541772 Pr =

200 (0.0541772)(5) = 1.3544 × 10−5 W = 13.5 𝜇W (2 × 103 )2

4.81. Prad = 10 Watts, r = 3.7 × 107 m, (a)

D0 = 50 dB ⇒ 105 (

4𝜋Umax 4𝜋r2 |E|2 = D0 = = 105 , Prad 2𝜂10 ⇒ E2 =

since Umax =

2 r2 Emax

2𝜂

) ; 𝜂 = 120𝜋

105 × 2 × 120𝜋 × 10 = 4.4 × 10−8 4𝜋(3.7 × 107 )2 E = 2 × 10−4 V∕m

(b) Use Friis Transmission equation ) ( Pr ( λ )2 λ 2 = G0t G0r = D0t G0r Pt 4𝜋R 4𝜋R (since we assume 100% efficiency) At 10 GHz, λ = 0.03 m [ ]2 Pr 0.03 (10,000)(1.643) = 10 4𝜋(3.7 × 107 ) Pr = 6.84 × 10−15 Preceived =

V2 . Since Rr = 73 = Rin for λ∕2 dipole then 8Rin √ V = 8(Preceived )(Rin ) = 2𝜇V

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SOLUTION MANUAL

4.82.

S0 dA0 = SdA,

dA0 S = S0 dA

Far zone ⇒ S =

1 |E|2 2𝜂

|E|

For spherical wave :

|E| |E0 |

|E0 |

√ =

S02 (S + S0 )2

√ =

dA0 dA

=

S0 S + S0

For plane wave : |E|∕|E0 | = 1

At S0

Wavefront (eikonal surface)

Point source S

In general, it can be shown that for a wave front eikonal surface we have |E| |E0 |

√ =

𝜌1 𝜌2 (𝜌1 + s)(𝜌2 + s)

𝜌1 and 𝜌2 are radii of curvature of wavefront; e.g. spherical wave ⇒ 𝜌1 = 𝜌2 = S0 |E| |E0 |

√ =

S0 2 (S + S0

)2

=

S0 S + S0

Plane wave: ⇒ 𝜌1 = 𝜌2 = ∞ |E| |E0 | ρ1

=1

ρ2

Wavefront Wavefront at S + S0 at S0

When the wave front is reflected from a surface we have √ √ √ √ √ 𝜌r1 𝜌r2 |E| √ 1 = √( r ( )( ) )( r ) =√ √ |E0 | 𝜌1 + s 𝜌 2 + s s √ 1+ s 1 + 𝜌r1 𝜌r2

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SOLUTION MANUAL

E is field at observation point. E0 is field at reflection point. Flat surface Observation

Source s s' s'

Radius of curvature of wavefront not changed by reflection. 𝜌r1 = 𝜌r2 = 𝜌1 = 𝜌2 = s′

|E| |E0 |

=

s′

s′ 1 = + s 1 + s∕s′

Spherical surface Observation s ψ : grazing angle

ψ

Source s'

θ

i

ψ

a

1 1 1 1 1 1 = ′+ ; r = ′+ 𝜌r1 s f1 𝜌2 s f2 In physics, we always used f1 = f2 = a∕2. This is not valid here because that f was valid for near normal incidence; we have near grazing incidence here. a cos 𝜃 i (perpendicular to the plane of incidence = elevation plane) 2 a : (parallel to the plane of incidence = azimuthal plane) f2 = 2 cos 𝜃i f1 =

Thus

1 2 1 1 2 cos 𝜃 i 1 = ′+ ; r = ′+ r 𝜌1 s a cos 𝜃i 𝜌2 s a

√ √ 1 =√ )} ( )} { ( √{ |E0 | √ 2 1 1 2 cos 𝜃 i 1+s ′ + 1+s ′ + s a cos 𝜃1 s a |E|

|E| |E0 |

1 = √ √ s 2s s 2s cos 𝜃i 1+ ′ + 1+ i + i s a a cos 𝜃 s

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|E| |E0 |

|E| |E0 |

= (

= (

|E| |E0 |

1 )√ √ √ 2ss′ √ √ √ 2ss′ cos 𝜃i √ √ √ a cos 𝜃i √ a 1+ 1+ s + s′ s + s′ ( ) 𝜋 cos 𝜃i = cos − 𝜓 = sin 𝜓 2 1

s 1+ ′ s

1 s 1+ ′ s

)√

[ ≃ 1+

1+

1

√

2ss′ a(s + s′ ) sin 𝜓

2ss′ a(s + s′ ) sin 𝜓

]−1∕2

1+

2ss′ sin 𝜓 a(s + s′ )

1 (1 + s∕s′ )

Near grazing neglect divergence in azimuthal plane.

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5

CHAPTER

Solution Manual

5.1. From (5-17) ⇒ A = â 𝜙 A𝜙 (r, 𝜃) = â 𝜙 j

k𝜇a2 I0 sin 𝜃 4r

[ 1+

1 jkr

]

e−jkr

(a) Using (3-2a) and (VII-26) ⎧ 1 1⎪ 1 H = ∇ × A = ⎨â r 𝜇 𝜇 ⎪ r sin 𝜃 ⎩

0 ⎡ 7 ⎤  𝜕A 𝜕 𝜃 ⎢ (A sin 𝜃) −  ⎥ ⎢ 𝜕𝜃 𝜙 𝜕𝜙 ⎥⎦ ⎣

0 0 ⎫ ⎤ ⎡ ⎡ 7 ⎤⎪  7 𝜕A 0 𝜕 1 ⎢ 1 𝜕Ar 1 𝜕 r *  + â 𝜃 (rA −  ⎥⎬  − (rA𝜙 )⎥ + â 𝜙 ⎢  𝜃) ⎥ r ⎢ sin 𝜃 𝜕𝜙 𝜕r r ⎢ 𝜕r 𝜕𝜙 ⎥⎦⎪ ⎦ ⎣ ⎣ ⎭

which reduces to H=

1 𝜇

{

â r

}

1 𝜕 1 𝜕 (A𝜙 sin 𝜃) − â 𝜃 (rA ) r sin 𝜃 𝜕𝜃 r 𝜕r 𝜙

Using the A𝜙 from above H=

[ ( ) ] 2 1 𝜕 k𝜇a I0 sin 𝜃 1 j 1+ e−jkr r sin 𝜃 𝜕𝜃 4r jkr [ ( ) ]} 2 1 𝜕 k𝜇a I0 sin 𝜃 1 − â 𝜃 j 1+ e−jkr r 𝜕r 4 jkr 1 𝜇

{ â r

which can be written as ) 1 Hr = j e−jkr 1+ jkr 2r2 [ ] (ka)2 I0 sin 𝜃 1 1 H𝜃 = − 1+ − e−jkr 4r jkr (kr)2 ka2 I0 cos 𝜃

(

H𝜙 = 0 Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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(b) Using Equation (3-10) with J = 0 along with the H-field components from above 1 1 E= ∇×H = j𝜔𝜀 j𝜔𝜀

{ [ ]} 𝜕Hr 1 𝜕 â r (0) + â 𝜃 (0) + â 𝜙 (rH𝜃 ) − r 𝜕r 𝜕𝜃

which reduces to Er = 0 E𝜃 = 0

[ ] (ka)2 I0 sin 𝜃 1 1+ e−jkr E𝜙 = 𝜂 4r jkr The same expressions can be obtained using (3-15) with the A𝜙 from part a. 5.2. According to the duality theorem and the dual quantities as outlined in Table 3.2 Electric Dipole

E H Ie 𝜀 𝜇 𝜅 𝜂 1∕𝜂

Magnetic Dipole

⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔ ⇔

H −E Im 𝜇 𝜀 𝜅 1∕𝜂 𝜂

Thus applying the above to the fields of an electric dipole, as given by (4-8a)–(4-10c), we obtain the fields of a magnetic dipole given by Er = 0 E𝜃 = 0

( ) kIm l sin 𝜃 1 1+ e−jkr 4𝜋r jkr ) ( 1 Im l cos 𝜃 1 Hr = e−jkr 1+ 𝜂 2𝜋r2 jkr [ ] 1 kIm l sin 𝜃 1 1 H𝜃 = j 1+ − e−jkr 𝜂 4𝜋r jkr (kr)2 E𝜙 = −j

H𝜙 = 0 which are identical to (5-20a)–(5-20d) 5.3. a = λ∕30, b = λ∕1,000 = 10−3 λ, f = 10 MHz ⇒ λ = 30 meters, 𝜎 = 5.7 × 107 s∕m ) ( )4 ( ( )4 C 2𝜋a 4 2𝜋 = 20𝜋 2 = 20𝜋 2 (a) Rr = 20𝜋 2 λ λ 30 = 20𝜋 2 (0.2094)4 = 0.3798 ohms

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√ √ C 𝜔𝜇0 2𝜋a 2𝜋f 𝜇0 (b) RL = Rhf = P 2𝜎 2𝜋b 2𝜎 √ √ λ∕30 𝜋(107 )4𝜋 × 10−7 a 𝜋f 𝜇0 = 0.02774 = = b 𝜎 λ∕1, 000 5.7 × 107 RL = Rhf = 0.02774

]} { [ ( ) a −2 (c) XA = 𝜔LA = 2𝜋fLA = 2𝜋f 𝜇0 a ln 8 b ]} { ( ) [ ( 1, 000 ) λ = 2𝜋 × 107 4𝜋 × 10−7 ln 8 −2 30 30 ( ) 30 [ln(266.667) − 2] = 8𝜋 2 (5.58599 − 2) = 283.139 XA = 8𝜋 2 30 [ √ ] √ √ 𝜔𝜇0 a a 𝜋f 𝜇0 a 2𝜋f 𝜇0 = = Xi = 𝜔Li = 𝜔 𝜔b 2𝜎 b 2𝜎 b 𝜎 √ λ∕30 𝜋(107 )4𝜋 × 10−7 Xi = λ∕1, 000 5.7 × 107 Xi =

1, 000 1 (2𝜋) × 104 × 10−7 √ = 0.02774 30 57

XT = XA + Xi = 283.139 + 0.02774 = 283.1667 (d) Zin = (Rr + RL ) + j(XA + Xi ) = (0.3798 + 0.02774) + j(283.1667) Zin = 0.40754 + j283.1667 (e) ecd =

Rr 0.3798 = = 0.9319 = 93.19% Rr + RL 0.3798 + 0.02774

5.4. The pattern of a small circular loop of uniform current is given by E𝜙n ∼ sin 𝜃 ⇒ U ∼ sin2 𝜃 which is omnidirectional. (a) 4𝜋Umax D0 = Prad 𝜋

2𝜋

Prad =

∫0

∫0 𝜋

U sin 𝜃 d𝜃 d𝜙 =

∫0

sin3 𝜃 d𝜃 ∫0 ( ) 8𝜋 4 = Prad = 2𝜋 3 3 4𝜋(1) 3 D0 (exact) = = = 1.5 = 1.761 dB 8𝜋∕3 2 = 2𝜋

𝜋

2𝜋

∫0

sin2 𝜃 sin 𝜃 d𝜃 d𝜙

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(b) Half-power beamwidth of sin2 𝜃 is 1 ⇒ sin 𝜃h = 0.707 ⇒ 𝜃h = 45◦ 2 ΘH = 2𝜃h = 90◦ = HPBW

sin2 𝜃h =

D0 (McDonald) = =

101 HPBW(degrees) − 0.0027 [HPBW (degrees)]2 101 101 = = 1.48246 90 − 21.87 90 − 0.0027(90◦ )2

D0 (McDonald) = 1.48246 = 1.7098 dB √ (c) D0 (Pozar) = −172.4 + 191 0.818 + √ = −172.4 + 191

0.818 +

1 HPBW (degrees) 1 90

= −172.4 + 191(0.91055) = −172.4 + 173.916 = 1.5116 D0 (Pozar) = 1.516 = 1.807 dB 5.5. C = λ∕4 = 2𝜋a ⇒ a = λ∕8𝜋 < λ∕6𝜋 ⇒ small loop ( )4 ( )4 C 1 20𝜋 2 2 (a) Rr = 20𝜋 2 N 2 = 20𝜋 2 N2 = N = 300 λ 4 256 ( )1∕2 300(256) ⇒N= = 19.72 ≃ 20 20𝜋 2 20𝜋 2 (b) Rin = Rr = (20)2 = 308.425 ohms 256 R − Zc 308.425 − 300 (c) Γ = in = = 0.01385 Rin + Zc 308.425 + 300 1 + |Γ| 1 + 0.01385 (d) VSWR = = = 1.0281 1 − |Γ| 1 − 0.01385 ) â y + 2̂az √ = (̂ay + 2̂az = 5e−jkx 5.6. √ 5 (a) Linear: Two components in phase. (b) AR = ∞ (

Eiw

)e−jkx

(c) E = â 𝜙 E𝜙 = â 𝜙 C sin 𝜃,

| â 𝜙 = (−̂ax sin 𝜙 + â y cos 𝜙)|| 𝜃 = 𝜋∕2 = â y | 𝜙 = 0

| E|| 𝜃 = 90◦ = â y C ⇒ Polarization: Linear in y direction | 𝜙 = 0 |( â + 2̂a ) |2 | y | z 1 ⋅ â y || = = −6.99 dB (d) PLF = || √ 5 | | 5 | | 9 30 × 10 (e) = 30 cm f = 1 GHz ⇒ λ = 1 × 109 ( ) (30)2 ( ) 3 λ2 λ2 3 = = 107.4296 cm2 Aem = D0 = 4𝜋 4𝜋 2 4𝜋 2

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SOLUTION MANUAL

Pr = Aem W i (PLF) = 107.4296(5 × 10−3 )

( ) 1 = 107.4296 × 10−3 5

Pr = 107.4296 × 10−3 Watts 5.7. (a) Et = Co (ĵa𝜃 + â 𝜙 ) sin 𝜃

e−jkr r

(b) Circular(

) ĵa𝜃 + â 𝜙 𝜌̂a = √ 2 ( ) â 𝜃 + â 𝜙 𝜌W = √ 2

(c)

|( ĵa + â ) ( â + â )|2 | √ |2 | | j + 1 |2 | 2 | 𝜃 𝜙 𝜃 𝜙 || 1 2 | | | | | | | PLF = |𝜌̂a ⋅ 𝜌̂w | = | ⋅ √ √ | =| 2 | =| 2 | = 2 | | | | | | 2 2 | | | | PLF =

5.8. Rr (1 turn) = 20𝜋 2

1 = −3 dB 2

( )4 ( )4 C 1 = 20𝜋 2 = 0.31583 ohms λ 5

Rr (4 turn) = N 2 Rr (1 turn) = 42 (0.31583) = 5.0532 ohms √ √ 2𝜋 × 107 (4𝜋 × 10−7 ) a 𝜔𝜇0 1 RL (1 turn) = Rhf (1 turn) = = b 26 10𝜋 × 10−3 2(5.7 × 107 ) RL = Rhf = 0.0265

) ( Rp Na +1 Rs b R0 √ √ 𝜔𝜇0 2𝜋 × 107 × (4𝜋 × 10−7 ) Rs = = = 8.3223 × 10−4 2𝜎 2(5.7 × 107 )

RL (4 turn) = Romic =

R0 = Rp R0

NRs 4(8.3223 × 10−4 ) = = 0.5298 2𝜋b 2𝜋(10−3 )

≃ 0.5 from Fig. 5.3

4(8.3223 × 10−4 ) (0.5 + 1) = 0.15724 4𝜋 × 10−3 0.3158 × 100 and ecd (1 turn) = 100Rr ∕(Rr + RL ) = = 92.26 = 92.26% 0.3158 + 0.0265 5.0532(10) = 96.98% ecd (4 turn) = 100Rr ∕(Rr + RL ) = 5.0532 + 0.15724 𝜋SI0 e−jkr sin 𝜃 where S = 𝜋a2 5.9. H𝜃 = − λ2 r Thus

RL = Rohmic =

E𝜙 = −𝜂H𝜃 = 𝜂

𝜋SI0 e−jkr λ2 r

sin 𝜃

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SOLUTION MANUAL

1 1 1 Re(E × H ∗ ) = Re(̂a𝜙 E𝜙 × â 𝜃 H𝜃∗ ) = Re(−̂a𝜙 𝜂H𝜃 × â 𝜃 H𝜃∗ ) 2 2 2 𝜂 𝜂 | 𝜋SI | sin2 𝜃 1 = â r Re(𝜂|H𝜃 |2 ) = â r |H𝜃 |2 = â r || 2 0 || 2 = â r Wr 2 2 2| λ | r

W ave = W ave

Prad = ○ W ave ds = ∫∫ ∫0

∫0

S0

𝜋

= 2𝜋

Prad

∫0

𝜋

2𝜋

â r Wr ⋅ â r r2 sin 𝜃 d𝜃 d𝜙

Wr r2 sin 𝜃 d𝜃

2 | 𝜋SI0 |2 𝜋 3 4𝜋𝜂 || (𝜋a)2 I0 || | | sin 𝜃 d𝜃 = = 𝜋𝜂 | 2 | | | 3 || λ2 || | λ | ∫0 𝜋 = 𝜂 (ka)4 |I0 |2 12

( ) k𝜇a2 I0 sin 𝜃 k𝜇a2 I0 e−jkr 1 5.10. A = â 𝜙 j 1+ e−jkr ≃ â 𝜙 j sin 𝜃 4r jkr 4r from equation (5-17) and r → large. Using (3-58a) Er ≃ E𝜃 ≃ 0

(

k𝜇a2 I0 e−jkr sin 𝜃 E𝜙 ≃ −j𝜔A𝜙 = −j𝜔 j 4r where S = 𝜋a2 , 𝜂 =

) =𝜂

𝜋SI0 e−jkr λ2 r

sin 𝜃

√ 𝜇∕𝜀. Also using (3-58b)

Hr ≃ H𝜙 ≃ 0 𝜔 𝜔 H𝜃 ≃ j A𝜙 = j 𝜂 𝜂

( j

𝜇ka2 I0 e−jkr sin 𝜃 4r

) =−

𝜋SI0 e−jkr λ2 r

sin 𝜃

5.11. a = λ∕8𝜋, b = 10−4 λ∕2𝜋, 𝜎 = 5.7 × 107 S∕m Assuming uniform current ( )4 ( ) λ C λ = , C = 2𝜋a = 2𝜋 (5-24) λ 8𝜋 4 ) ( )4 ( 197.392 λ 1 = = 20𝜋 2 = 0.771 Rr = 20𝜋 2 4λ 256 256 λ √ √ 2𝜋(108 )4𝜋 × 10−7 a 𝜔𝜇0 RL = Rhf = = 8𝜋 −4 b 2𝜎 2(5.7 × 107 ) 10 λ 2𝜋

(a) Rr = 20𝜋 2

RL =

20𝜋 5𝜋 15.708 104 (2𝜋) × 10−3 = √ =√ = √ = 6.5794 √ 4 5.7 4 5.7 5.7 5.7

ecd =

Rr 0.771 = = 0.10489 = 10.489% Rr + RL 0.771 + 6.5794

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(b) D0 = 3∕2 = 1.5 = 1.761 dB

Uniform current

G0 = ecd D0 = 0.10489(1.5) = 0.15734 G0 = 0.15734 = −8.03 dB 5.12. C = 𝜋λ = 2𝜋a ⇒ a = λ∕2 b = 10−4 λ, 𝜎 = 107 S∕m, f = 100 MHz, uniform current ( ) C = 0.667(𝜋) = 2.127 = 3.278 dB (5-63b) (a) D0 = 0.667 λ √ √ √ C 𝜔𝜇0 2𝜋a 𝜔𝜇0 a 𝜔𝜇0 (b) RL = Rhf = = = (2-90b) P 2𝜎 2𝜋b 2𝜎 b 2𝜎 √ √ 2𝜋f 𝜇0 λ∕2 104 2𝜋(103 )4𝜋 × 10−7 = = 10𝜋 = 31.416 RL = −4 2𝜎 2 10 λ 2(107 ) ( ) C = 60𝜋 2 (𝜋) = 60𝜋 3 = 1, 860 (c) Rr = 60𝜋 2 (5-63a) λ Rr 1, 860 = = 0.98339 = 98.339% ecd = Rr + RL 1, 860 + 31.416 (d) G0 = ecd D0 = 0.98339(2.127) = 2.092 = 3.205 dB 5.13. (a)

) ( )4 ( C 2𝜋a 4 3 × 108 = 20𝜋 2 , λ= = 30 m λ λ 107 ) ( 2𝜋a 4 0.73 = 20𝜋 2 ⇒ a = 0.03924λ = 1.177 meters λ Rr = 20𝜋 2

(b) 0.73N 2 = 300 ⇒ N = 20.272 ≃ 20 Rr (20 turns) = 0.73(20)2 = 292

( ) λ2 λ2 3 (1 − |Γ|2 )10−6 D0 e0 (10−6 ) = 4𝜋 4𝜋 2 ( ) | 292 − 300 |2 (30)2 ( 3 ) | 10−6 = 0.1074 × 10−3 Watts = 1 − || | 4𝜋 2 | 292 + 300 |

(c) PL = Aem Wi eo =

5.14. a = λ∕30, b = λ∕300,

2C = λ∕100 ⇒ C = λ∕200,

N = 6,

f = 5 × 107 Hz

(a) Since a = λ∕30 ≪ λ D0 = 1.5 = 1.761 dB ( )4 ( )4 (b) R = 20𝜋 2 c = 20𝜋 2 𝜋 = 20𝜋 2 (1.924 × 10−3 ) = 0.3798 ohms r λ 15 ( ) 𝜋 λ = λ C = 2𝜋a = 2𝜋 30 15 Rr (single turn) = 0.3798 ohms Rr (6 turn) = 13.673 ohms ) ( Rp Na RL = +1 R b s R0

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SOLUTION MANUAL

Rp λ∕200 3 = 0.65 = = 1.5 ⇒ λ∕300 2 R0 √ √ √ 𝜔𝜇0 2𝜋f (4𝜋 × 10−7 ) 4𝜋 2 f Rs = = × 10−7 = 7 2𝜎 5.7 2(5.7 × 10 ) √ √ 4𝜋 2 (50 × 106 ) 50 −7 = × 10 = 2𝜋 × 10−4 = 18.609 × 10−4 5.7 5.7 ( ) λ∕30 RL = 6 18.609 × 10−4 (0.65 + 1) λ∕300 c∕b =

RL = 6(10)(18.609)(1.65) × 10−4 = 1, 842.31 × 10−4 RL (6 turns) = 0.184231 √ λ∕30 (Single) RL = λ∕300 √ = 2𝜋(10)

√ 2𝜋f (4𝜋 × 10−7 ) 4𝜋 2 f × 10−7 = 10 5.7 2(5.7 × 107 ) 50 × 10−4 = 186.0919 × 10−4 5.7

(6 turns) RL = 186.0919(6)(1.65) × 10−4 = 1, 842.31 × 10−4 ecd =

Rr 13.673 = × 100 = 98.67% Rr + RL 13.673 + 0.184231

| (R + RL ) − 50 | | 13.857 − 50 | | −36.14277 | |=| |=| | = 0.566 (c) |Γ| = || r | | | | | | (Rr + RL ) + 50 | | 13.857 + 50 | | 63.857 | er = (1 − |Γ|2 ) × 100 = (1 − |0.566|2 ) × 100 = (1 − 0.32) × 100 = 68% (d) G0 = ecd D0 = (0.9867)D0 = (0.9867)(1.5) G0 = 1.48005

(total maximum gain does not include the reflection loss)

e−jkr , r (a) D = 4𝜋Umax 0 Prad

5.15. E𝜙 ≃ C0 cos2 𝜃

C = λ, uniform current

U ≃ C0 cos4 𝜃, ⇒ Umax = C0 𝜋∕2

2𝜋

Prad =

∫0

∫0

U sin 𝜃 d𝜃 d𝜙 = C0

𝜋∕2

2𝜋

∫0

𝜋∕2

∫0

cos4 𝜃 sin 𝜃 d𝜃 d𝜙

cos4 𝜃 sin 𝜃 d𝜃 ∫0 ]𝜋∕2 [ ] [ 2𝜋 1 cos5 𝜃 z = = 2𝜋C0 0 + C = 2𝜋C0 − 5 5 5 0 0 = 2𝜋C0z

D0 =

4𝜋C0 4𝜋Umax = = 10 = 10 dB 2𝜋 Prad C0 5

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131

(b) From Fig. 5.15 Zin ≃ 100 − j90

(c) Γ =

Zin − Zc 100 − j90 − 300 −200 − j90 219.317⌊204.228 = = = Zin + Zc 100 − j90 + 300 400 − j90 410.00⌊347.3196

Γ = 0.53492⌊−143.09◦ = 0.53492(−0.79960 − j0.60053) Γ = 0.53492⌊−143.09◦ = −0.42722 − j0.32123 (d) G0 =  ecd > 1 D0 = D0 = 10 = 10 dB (lossless loop)  (e) Gre0 = (1 − |Γ|2 )G0 = (1 − |0.53492|2 )10 = (1 − 0.28614)10 = 0.71386(10) = 7.13861 = 8.536 dB Gre0 = 7.1386 = 8.536 dB 5.16. f = 30 MHz → λ = 10 m,

ka =

2𝜋 (0.15) = 0.03𝜋 = 0.09425 (rad) 10

𝜋𝜂0 𝜋 ⋅ 120 (ka)4 = 64 × × (0.03𝜋)4 = 0.9968 ohms 6 6 1 1 𝛿= √ =√ 𝜋f 𝜇0 𝜎 𝜋 × 30 × 106 × 4𝜋 × 10−7 × 5.7 × 107

Rr = N 2

𝛿 = 1.217 × 10−5 m ≪ b 1-turn: 8-turn:

a 0.15 = 0.2162 ohms = 𝜎b𝛿 5.7 × 107 × 0.001 × 1.217 × 10−5 ( ) Rp Rp RL = 8 × RL (1-turn) × + 1 , c∕b = 1.8 ⇒ = 0.5 R0 R0

RL =

∴ RL = 8 × (0.2162) × 1.5 = 2.594 Ω ecd =

Rr 0.9968 = = 0.278 = 27.8% Rr + RL 0.9968 + 2.594

5.17. Since the small circular loop area is parallel to the y-z plane, its electrical equivalent is an infinitestimal magnetic dipole directed along the x-axis. (a) Thus, using the procedure of Example 4.5, we can write the electric and magnetic fields for the infinitesimal electric dipole of length l directed along the x-axis as Er ≃ 0

Er ≃ 0

E𝜃 ≃ −j𝜔A𝜃

E𝜃 ≃ −j

E𝜙 ≃ −j𝜔A𝜙

E𝜙 ≃ +j

Hr ≃ 0 𝜔𝜇Io le−jkr cos 𝜃 cos 𝜙 4𝜋r

𝜔𝜇 Io le−jkr sin 𝜙 4𝜋r

H𝜙 ≃

E𝜃 𝜂

H𝜃 ≃ −

E𝜙 𝜂

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Using duality and Table 3.2, the fields of an x-directed infinitesimal magnetic dipole of constant current Im can be written as Hr ≃ 0

𝜔𝜀Im le−jkr cos 𝜃 cos 𝜙 H𝜃 ≃ −j 4𝜋r

H𝜙 ≃ +j

𝜔𝜀 Im le−jkr sin 𝜙 4𝜋r

Er ≃ 0

E𝜙 ≃ −𝜂H𝜃 = +j𝜂

E𝜃 ≃ +𝜂H𝜙 = +j𝜂

𝜔𝜀Im le−jkr cos 𝜃 cos 𝜙 4𝜋r

−jkr 𝜔𝜀 Im le sin 𝜙 4𝜋r

Since the infinitesimal magnetic dipole directed along the x−aixs is equivalent to a small circular loop, with its area parallel to the y-z plane, we can write the fields of the circular loop by making in the above equations the substitution lIm = jS𝜔𝜇I0 = j(𝜋a2 )𝜔𝜇I0

Thus the far-zone electric fields can be written as

Er ≃ 0

E𝜃 ≃ +j𝜂

𝜔𝜀I0(jS𝜔𝜇)e−jkr 𝜔𝜀I0(j𝜋a2𝜔𝜇)e−jkr sin 𝜙 sin 𝜙 = +j𝜂 4𝜋r 4𝜋r

E𝜙 ≃ +j𝜂

𝜔𝜀I0 (jS𝜔𝜇)e−jkr 𝜔𝜀I0 (j𝜋a2 𝜔𝜇)e−jkr cos 𝜃 cos 𝜙 = +j𝜂 cos 𝜃 cos 𝜙 4𝜋r 4𝜋r

≃𝜂

𝜔2 𝜇𝜀a2 I0 e−jkr (ka)2 I0 e−jkr sin 𝜙 = 𝜂 sin 𝜙 4r 4r

≃ −𝜂

𝜔2 𝜇𝜀a2 I0 e−jkr (ka)2 I0 e−jkr cos 𝜃 cos 𝜙 = −𝜂 cos 𝜃 cos 𝜙 4r 4r

while the far-zone magnetic fields can be expressed as Hr ≃ 0; H𝜃 ≃ −

E𝜙 𝜂

; H𝜙 ≃

E𝜃 𝜂

(b) Since the far-field pattern of the antenna is the same as that of a loop with an area parallel to the x-y plane, or in infinitesimal magnetic dipole oriented along the z-axis, their directivities are the same. Thus D0 = 3∕2 = 1.5. 5.18. Using the results of Problem 5.17 √ a2 𝜔𝜇kI0 e−jkr √ a2 𝜔𝜇kI0 e−jkr (a) E𝜒 ≅ 1 − |̂ay ⋅ â r |2 = 1 − sin2 𝜃 sin2 𝜙 4r 4r E𝜒 H𝜓 ≅ 𝜂 (b) Directivity = D0 =

3 2 5.19. Using the computer program Loop Uniform of Chapter 5. (a) a = λ∕50 = 0.02λ D0 = 1.4988 = 1.7575 dB, Rr = 0.049 ohms

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(b)

a = λ∕10 = 0.1λ D0 = 1.4699 = 1.6731 dB,

(c)

a = λ∕4 = 0.25λ D0 = 1.2969 = 1.1291 dB,

(d)

Rr = 28.41 ohms

Rr = 723.938 ohms

a = λ∕2 = 0.5λ D0 = 1.7968 = 2.5449 dB,

Rr = 2, 202.528 ohms

5.20. According to (5-54b) E𝜙 ≃

ak𝜂I0 e−jkr J1 (ka sin 𝜃) ∼ J1 (ka sin 𝜃) 2r

Therefore the nulls of the pattern occur when J1 (ka sin 𝜃n ) = 0 ⇒ ka sin 𝜃n = 0, 3.84, 7.01, 10.19, ... Excluding 𝜃 = 0 ] [ ) ( ⎧ 3.84 ⎪ sin−1 3.84 = sin−1 = sin−1 (0.4889) = 29.27◦ ka 2𝜋(1.25) ⎪ 𝜃n = ⎨ ] [ ) ( 7.01 ⎪ sin−1 7.01 = sin−1 = sin−1 (0.8925) = 63.19◦ ⎪ ka 2𝜋(1.25) ⎩ 5.21. Since E𝜙 ∼ J1 (ka sin 𝜃) (a) E | = J (ka sin 𝜃)| 𝜙 𝜃=0

1

𝜃=0

= J1 (0) = 0

E𝜙 |𝜃=𝜋∕2 = J1 (ka sin 𝜃)|𝜃=90◦ = J1 (ka) = 0 ⇒ ka = 3.84 Thus a =

3.84 3.84λ = = 0.61115λ k 2𝜋

(b) Since a = 0.61115λ > 0.5λ, use large loop approximation. According to (5-63a) Rr = 60𝜋 2 (C∕λ) = 60𝜋 2

(

2𝜋a λ

)

= 60𝜋 2 [2𝜋(0.61115)] = 2,273.94

(c) The directivity is given by (5-63b), or D0 = 0.677

( ) ( ) 2𝜋a C = 0.677 = 0.677(2𝜋)(0.61115) = 2.6 λ λ

5.22. E𝜙 ∼ J1 (ka sin 𝜃) (a) E𝜙 |𝜃=30◦ = J1 (ka sin 𝜃)|𝜃=30◦ = J1

(

ka 2

)

=0⇒

ka = 3.84 2

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From the Table for J1 (x) in Appendix V. Thus a= (b)

2(3.84) 2(3.84) = λ = 1.222λ k 2𝜋

E𝜙 |max = E𝜙 |ka sin 𝜃=1.84 = J1 (1.84) = 0.58152 = −4.709 dB [ ] 2𝜋 E𝜙 |𝜃=90◦ = J1 (ka) = J1 (1.222λ) = J1 (7.678) = 0.175 = −15.139 dB λ Thus ΔE = E𝜙 |𝜃=90◦ − E𝜙 |max = −15.139 − (−4.709) = −10.43 dB

5.23. E𝜙 ∼ J1 (ka sin 𝜃) (a) According to J1 (x) in the Appendix V J1 (x) = 0

when x = 0, 3.84, 7.01, 10.19, ....

Since we want a null in the plane of the loop (𝜃 = 0◦ ) and two additional ones for 0◦ ≤ 𝜃 ≤ 90◦ , then ka sin 𝜃|max = ka sin 𝜃|𝜃=90◦ = ka = 7.01 Thus a=

7.01 7.01 = λ = 1.1157λ k 2𝜋

(b) The nulls will occur at 𝜃 = 0◦ and 180◦ 𝜃 = 90◦ and ka sin 𝜃|a=1.1157λ = 3.84 ] [ 3.84 ⇒ 𝜃 = sin−1 = 33.21◦ 2𝜋(1.1157) and 𝜃 = 180◦ − 33.21◦ = 146.79◦ 5.24. E = â 𝜙 C1 J1 (ka sin 𝜃) where C1 is a constant ⇒ 𝜌̂w = â 𝜙 and PLF = |𝜌̂w ⋅ 𝜌̂a |2 = |̂a𝜙 ⋅ 𝜌̂a |2 By inspection, the PLF is maximized if the probe antenna is also linearly polarized in the 𝜙 direction. This can be accomplished by using as a probe antenna another loop antenna so that 𝜌̂a = â 𝜙

and

PLF = |̂a𝜙 ⋅ â 𝜙 |2 = 1

It can also be accomplished by using a linear dipole as a probe antenna with its length parallel to the plane of the loop and tangent to its curvature. Some specific examples would be [using

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135

the transformation of VII-7b] 𝜌̂a = â x |𝜙=90◦ ⇒ PLF = |̂a𝜙 ⋅ â x |𝜙=90◦ = |̂a𝜙 ⋅ (̂a𝜌 cos 𝜙 − â 𝜙 sin 𝜙)|2𝜙=90◦ = |̂a𝜙 ⋅ (−̂a𝜙 )|2 = 1 𝜌̂a = â y |𝜙=0◦ ⇒ PLF = |̂a𝜙 ⋅ â y |𝜙=0◦ = |̂a𝜙 ⋅ (̂a𝜌 sin 𝜙 + â 𝜙 sin 𝜙)|2𝜙=0◦ = |̂a𝜙 ⋅ â 𝜙 |2 = 1 and many others. 5.25. A very small loop of constant current is equivalent to a magnetic dipole. Since the loop is placed for both parts (a and b) perpendicular to the xy-plane (the plane of the loop is perpendicular to the xy-plane), the axis of the linear magnetic dipole will also be parallel to the xy-plane. Therefore according to Figure 4.15a, the image of the horizontal magnetic dipole will be as shown in this figure. In turn the array factor for both parts (a and b) of this problem will be the same as that of the vertical electric dipole of Figure 4.16 or AF = 2 cos(kh cos 𝜃) Since the actual source and the image are oriented in the same direction. Therefore according to (5-27a)–(5-27c) Actual Source h xy – plane

σ=∞

h

Image

(a) Plane of the loop is parallel to the xz-plane √ √ (ka)2 I0 e−jkr sin 𝜓(AF), sin 𝜓 = 1 − cos2 𝜓 = 1 − |̂ay ⋅ â r |2 4r √ = 1 − sin2 𝜃 sin2 𝜙

E𝜒 = 𝜂

(ka)2 I0 e−jkr sin 𝜓[2 cos(kh cos 𝜃)] 4r √ (ka)2 I0 e−jkr cos(kh cos 𝜃) 1 − sin2 𝜃 sin2 𝜙 E𝜒 = 𝜂 2r E𝜒 H𝜓 = − 𝜂 =𝜂

(b) Plane of the loop is parallel to the yz-plane. The fields for this problem are the smae as those in part (a) above except that sin 𝜓 =

√

1 − cos2 𝜓 =

√

√ 1 − |̂ax ⋅ â r |2 =

(For Alternate Solution see the end of the solution manual)

1 − sin2 𝜃 cos2 𝜙

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5.26. (a)

(ka)2 I0 e−jkr sin 𝜃 4r |AF| = |2j sin(kh cos 𝜃)| E𝜙 = 𝜂

E𝜙 = 𝜂

𝜋SI0 e−jkr λ2 r

sin 𝜃, S = 𝜋a2

(E𝜙 )t = E𝜙 (AF) = 𝜂

𝜋SI0 e−jkr

sin 𝜃[2j sin(kh cos 𝜃)] λ2 r above ground plane total field.

h

σ=∞

⇒

σ=∞ h h σ=∞

(b) h = λ, kh = 2𝜋 sin 𝜃[2j sin(2𝜋 cos 𝜃)] = 0, sin(2𝜋 cos 𝜃) = 0 2𝜋 cos 𝜃 = n𝜋, n = 0, 1, 2 n cos 𝜃n = , n = 0, 1, 2. 2 𝜃0 = cos−1 (0) = 90◦ ( ) 1 = 60◦ 𝜃1 = cos−1 2 𝜃2 = cos−1 (1) = 0◦ (c) (E𝜙 )t = C sin 𝜃 sin(kh cos 𝜃)|𝜃=60◦

(√ ) ) ( 3 2𝜋 1 =0=C sin h 2 λ 2

√ C ( sin

𝜋h λ

)

=0⇒

( ) 3 𝜋h =0 sin 2 λ

𝜋h = sin−1 (0) = n𝜋, n = 0, 1, 2, 3, ... λ

h = ±n ⇒ physical nonzero height ⇒ h = nλ, n = 1, 2, 3, ... λ 5.27. (a) Since the equivalent to a circular loop is a vertical magnetic dipole which is placed vertically to a PMC ground plane, the normalized array factor is that of (4-99), or (AF)n = cos(kh cos 𝜃)

Two sources, equal magnitude, same phase

According to Figure 4.16(b).

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(b) (AF)max = | cos(kh cos 𝜃m )| = 1 ⇒ cos(kh cos 𝜃m ) = ±1 kh cos 𝜃m = cos−1 (±1) = m𝜋, m = 0, 1, 2, ... cos 𝜃m =

( ) m𝜋 mλ 2 m𝜋 mλ || = = m = 2𝜋 = ) ( | 3λ 3λ kh 2h |h= 3 (h) 4 2 λ 4 ( ) 2 , m = 0, 1, 2, ... 𝜃m = cos−1 m 3 𝜃0 = cos−1 (0) = 90◦ ( ) 2 = 48.2◦ 𝜃1 = cos−1 3 ( ) 4 = does not exist 𝜃2 = cos−1 3

m = 0: m = 1: m = 2:

5.28. (a) AF for 2 sources of the same magnitude and of the same phase (AF)n = 2 cos(kh cos 𝜃) Total Field: E|total = E|single element ⋅ (AF) (ka)2 I0 e−jkr E𝜙 |total = 𝜂 sin 𝜃 [2 cos(kh cos 𝜃)] 4r ⏟⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏟ ⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟⏞⏞⏞⏞⏞⏞⏞⏞⏞⏞⏟ AF

single element

E𝜙 |total = 𝜂

(ka)2 I0 e−jkr sin 𝜃 cos(kh cos 𝜃) 2r

θ electric loop

h

equivalent vertical magnetic dipole

h h

PMC (a) Actual problem

(b) (E𝜙 )|total norm = 𝜂

(b) Equivalent problem

(ka)2 I0 e−jkr sin 𝜃 cos(kh cos 𝜃) 2r {

(E𝜙 )norm = sin 𝜃 cos(kh cos 𝜃)|𝜃=𝜃n =0◦ ,30◦ = 0 ⇒ ( cos(kh cos 𝜃n )|𝜃n =30 = 0 = cos

sin 𝜃n = 0 ⇒ 𝜃 = 0

cos(kh cos 𝜃n ) = 0 ( √ ) √ ) 3 𝜋 3 2𝜋 h = cos h =0 λ 2 λ

√ 𝜋 3 n𝜋 h = cos−1 (0) = , n = 1, 3, 5, ... λ 2

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)

( n𝜋 h= 2

λ √ 3𝜋

(

) λ √

=n

, n = 1, 3, 5, ...

2 3

n=1: h=

λ = 0.2886λ 3.464

5.29. (a) Array Factor = 2 cos(kh cos 𝜃) h

h

σ=∞

h

(b) AF = 2 cos(kh cos 𝜃n ) = 0 ⇒ kh cos 𝜃n = cos−1 (0) = n𝜋∕2, n = ±1, ±3, ±5, .... ( n𝜋 ) ] [ ( )| ( ) n𝜋∕2 nλ | n = cos−1 = cos−1 2𝜋2 𝜃n = cos−1 = cos−1 | kh 4h 2 |h=λ∕2 h (

𝜃1 = cos−1 ±

1 2

)

λ

( ) 3 = 60◦ , 𝜃3 = cos−1 ± = does not exist 2

5.30. h = 0, f = 100 MHz, C = λ∕10, b ≪ λ λ = 30 × 109 ∕108 = 300 cm z

PMC Im PMC

(a) Since the equivalent of a small electric circular loop is a magnetic dipole and the image of a vertical magnetic dipole above a PMC is at the same magnitude and phase, then the loop will not be shorted, and it will radiate. Since the loop is radiating only in half of a sphere (hemisphere), its directivity is twice that when radiating into an infinite medium. Thus D0 = 2(1.5) = 3 = 4.77 dB (b) Using a PLF = 1∕2 Aem =

(3 × 102 )2 λ2 27 D0 (PLF) = (3) = × 104 = 10.743 × 103 cm2 4𝜋 4𝜋(2) (2)4𝜋

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(c)

139

Wi = 10−4 W∕cm2 Prad = Aem Wi = 10.743 × 103 (10−4 ) = 1.0743 Watts

5.31. Since the small circular loop area is parallel to the x-z plane, its electrical equivalent is an infinitesimal magnetic dipole directed along the y-axis placed a height h above the PEC. Also its image is at a depth h below the PEC interface. The image is in the same direction as the actual source (the same magnitude and phase). (a) Therefore its normalized array factor is (AF)n = cos(kh cos 𝜃) whose maximum value is unity. (b) To find the two smallest heights, other than h = 0, where the maximum will be directed along 𝜃 = 0◦ , we set the normalized array factor to unity, or [AFn (𝜃 = 0◦ ]max = [cos(kh cos 𝜃)|𝜃=0◦ ]max = cos(kh)|max = 1 kh = cos−1 (1) = m𝜋 h=

m𝜋λ λ m𝜋 = = m, m = 0, 1, 2, 3... k 2𝜋 2 m = 0: h = 0 λ 2 m = 2: h = λ

m = 1: h =

5.32. (a) AF = 2 cos(kh cos 𝜃)|𝜃=90◦ = 2 (which is max). So the minimum height is h = 0 because the power on the upper half plane is doubled and the maximum is still maintained along the horizon (𝜃 = 90◦ ). Thus the directivity along 𝜃 = 90◦ is doubled. (b) Using the image theory and duality, the directivity of the cicular loop is twice that of free space, or using (4-104) and Figure 4.20: kh = 2.881(h = 0.4585λ) D0 = 2(1.5) = 3 = 4.7712 dB

(c) Aem =

Aem = Prec = Wi = λ=

D0 = 6.566 = 8.173 dB

( ) λ2 λ2 1 D0 (PLF) = (3) 4𝜋 4𝜋 2 3(30 × 102 )2 = = 107.424 × 104 cm2 8𝜋 8𝜋 3λ2 Aem Wi = (W ) = 10 × 10−6 8𝜋 i 8𝜋 80𝜋 (10 × 10−6 ) = 2 × 10−6 2 3λ 3λ 30 × 109 3 = 3 × 10 cm 10 × 106 3λ2

PMC

λ2 D (PLF) 4𝜋 0 (30 × 102 )2 (6.566)(1) = 4𝜋 Aem = 235.127 × 104 cm2 P 10 × 10−6 Wi = r = Aem 235.127 × 104 Aem =

Wi = 4.253 × 10−12 W∕cm2

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Wi =

80𝜋 80𝜋 80𝜋 × 10−6 = × 10−6 = × 10−6 3(3 × 103 )2 3(9) × 106 27 × 106

Wi =

80𝜋 10 × 10−6 = 9.3084 × 10−12 × 10−12 = 9.3084 × 10−12 = 27 107.429 × 104 Wi = 9.3084 × 10−12 Watts∕cm2

5.33. From Problem 5.19(a) √ E𝜒 | 𝜙=90◦ = C1 cos(kh cos 𝜃) 𝜃=45◦

E𝜒 | 𝜙=90◦ 𝜃=45◦

1 − sin2 𝜃 sin2 𝜙| 𝜃=45◦

𝜙=90◦

= C1 cos(kh cos 𝜃) cos(𝜃)|𝜃=45◦ ( ) kh 𝜋 kh = 0.707C1 cos √ = 0 ⇒ √ = cos−1 (0) = n, n = 1, 3, 5, ... 2 2 2

For the smallest height √ √ 2𝜋 2 kh 𝜋 = λ = 0.3535 λ √ = ⇒h= 2 2 k 4 2 ) λ λ = = 0.1λ ⇒ Assume uniform current. 20𝜋 10 (a) The small loop, with a uniform current, can be represented by a horizontal magnetic dipole perpendicular to the area of the loop. Then, according to Fig. 4.16(a), the image to accounts for the reflections has the same magnitude and the same phase as the actual source. Thus the normalized array factor of two sources with the same magnitude and phase separated by 2h is (same as vertical electric dipole above PEC), or

5.34. a = λ∕20𝜋 ⇒ C = 2𝜋a = 2𝜋

(

(AF)n = cos(kh cos 𝜃) (b) cos(kh cos 𝜃)|𝜃=𝜃 = 0 = cos(kh cos 𝜃n ) n kh cos 𝜃n = cos−1 (0) =

n𝜋 , n = 1, 3, 5, ... 2

h PEC

For the smallest h: kh cos 𝜃n = 𝜋∕2 h=

h

Horizontal Magnetic Dipole (HMD)

𝜋∕2 λ 𝜋 𝜋 𝜋 = | = = ◦ = 2𝜋 k cos 𝜃n 𝜃=60 2k(0.5) k 2 λ h=

λ 2

PEC HMD h

HMD

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141

5.35. The far-zone electric field radiated by a small circular loop of uniform current is given by (5-27b), or Eloop = â 𝜙 E𝜙 = â 𝜙 𝜂

(ka)2 I0 e−jkr sin 𝜃 ⇒ 𝜌̂loop = â 𝜙 4r

(a) For an infinitesimal dipole oriented along the z-axis, its far-zone electric field is given by (4-26a), or z-axis: Edipole = â 𝜃 j𝜂

klI0 e−jkr sin 𝜃 ⇒ 𝜌̂dipole = â 𝜃 4𝜋r

PLF = |𝜌̂loop ⋅ 𝜌̂dipole |2 = |̂a𝜙 ⋅ â 𝜃 |2 = 0 = −∞ dB (b) For an infinitesimal dipole oriented along the y-axis, its far-zone electric field is given, according to Example 4.5, by y-axis: Edipole = â 𝜃 E𝜃 + â 𝜙 E𝜙 = j

𝜔𝜇I0 le−jkr [̂a𝜃 cos 𝜃 sin 𝜙 + â 𝜙 cos 𝜙] 4𝜋r

Toward the loop (𝜃 = 90◦ , 𝜙 = 180◦ ), the field of the y-directed dipole reduces to Edipole |

𝜃=90◦ 𝜙=180◦

= â 𝜃 E𝜃 + â 𝜙 E𝜙 = −j = +̂a𝜙 j

𝜔𝜇I0 le−jkr [̂a𝜃 (0) + â 𝜙 (−1)] 4𝜋r

𝜔𝜇I0 le−jkr ⇒ 𝜌̂dipole = â 𝜙 4𝜋r

PLF = |𝜌̂loop ⋅ 𝜌̂dipole |2 = |̂a𝜙 ⋅ â 𝜙 |2 = 1 = 0 dB √ √ √ 2𝜋(3 × 108 )(4𝜋 × 10−7 ) a 𝜔𝜇0 𝜋 12 1 = = × 10 5.36. (a) RL = −4 7 b 2𝜎 20 5 ⋅ 7 20(10 ) 2 ⋅ 5 ⋅ 7 × 10 ( (b) Rr = 120𝜋

2 𝜋 3

)(

kS λ

)2

RL = 2.27915 ohms )2 ( )( 2 2𝜋 2 = 120𝜋 𝜋 3 (20)2

4𝜋 6 = 1.92278 ohms (400)2 ( ( )2 ) 1 S=𝜋 20

RL = 80

(c) inductive reactance XA = 𝜔LA ] ( )[ ( ) ] [ ( ) 1 λ 8a 1 − 2 = 4𝜋 × 10−7 ln LA = 𝜇0 a ln ⋅ −4 − 2 b 20 20 10 ( ) λ = 2.648 × 10−7 a = 0 , b = 10−4 λ0 ← λ = 1 m, f = 3 × 108 20 XA = 2𝜋fLA = 2𝜋(3 × 108 )(2.648 × 10−7 ) = 499.158 ∴ XA ≫ (RL or Rr )

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5.37. From equation (5-24) ( ) ( ) ( ) 𝜋 2𝜋 kS 2 2𝜋 2𝜋S 2 (k2 a2 )2 = 𝜂 = 120𝜋 6 3 λ 3 λ2 ( ) ( 2) ( ) ( )2 2 S S2 S 4(𝜋 4 ) 2 = 31,170.909 = 120 ≃ 31,171 3 λ λ4 λ4

Rr = 𝜂

(a) Area ( S = ab,

Rr = 31,170.909

a2 b2 λ4

)

( ≃ 31,171

a2 b2 λ4

)

b a

(b) Area ( )( ) b a , S=𝜋 2 2

(

𝜋 2 a2 b2 Rr = 31,170.909 16λ4 ) ( 𝜋ab 2 Rr = 31,171 4λ2

)

( ≃ 31,171

𝜋 2 a2 b2 16λ4

b

a

5.38. f = 100 MHz ⇒ λ = c∕f = 3 × 108 ∕108 = 3 meters, = 𝜎 = 5.7 × 107 S∕m λ∕20 λ 3 C = = = = 0.0239 m = 0.00796λ 2𝜋 2𝜋 40𝜋 40𝜋 ( )4 ( )4 C 1 20𝜋 2 = 20𝜋 2 = × 10−4 = 1.2337 × 10−3 ohms (a) Rr = 20𝜋 2 λ 20 16 √ √ λ∕40𝜋 2𝜋 × 108 (4𝜋 × 10−7 ) a 𝜔𝜇0 = RL = = 0.00838 b 2𝜎 λ∕400𝜋 2(5.7 × 107 )

C = 2𝜋a ⇒ a =

Rin = Rr + RL = 0.0012337 + 0.00838 = 0.0096137 ⎡ ⎛ 8 λ ⎞ ⎤ ] [ ( ) 8a (b) La = 𝜇0 a ln − 2 = 4𝜋 × 10−7 (0.0239) ⎢ln ⎜ 40𝜋 ⎟ − 2⎥ ⎢ ⎜ λ∕400 ⎟ ⎥ b ⎣ ⎝ ⎦ ⎠ ] [ ( ) 80 − 2 = 0.3 × 10−7 [3.2373 − 2] = 0.3 × 10−7 ln 𝜋 La = 37.12 × 10−9 Henries Xa = 𝜔La = 2𝜋fLa = 2𝜋(108 )(37.12 × 10−9 ) = 23.323 ohms

)

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Li =

a 𝜔b

√

λ

𝜔𝜇0 40𝜋 = ( ) λ 2𝜎 2𝜋(108 ) 400

√ 2𝜋(108 )(4𝜋 × 10−7 ) = 0.1333 × 10−10 2(5.7 × 107 )

Xi = 2𝜋fLi = 2𝜋(108 )(0.1333 × 10−10 ) = 0.83771 × 10−2 = 0.0084 ohms Xt = Xa + Xi = 23.323 + 0.0084 = 23.3314 ohms (inductive) (c) Capacitance Xc = C=

1 = 23.3314 2𝜋fC 1 = 6.82 × 10−11 = 68.2 × 10−12 Farads 23.3313(2𝜋 × 108 )

5.39. From the solution of Problem 5.37, the radiation resistance of a loop is Rr = 31,171

(Area)2 (S)2 = 31,171 λ4 λ4

Thus for rectangular and elliptical loops: (a) Area S = ab,

Rr ≃ 31,171

a2 b2 λ4

b a

(b) Area S=𝜋

( )( ) b a , 2 2

Rr ≃ 31,171

𝜋 2 a2 b2 16λ4

b

a

5.40. In Far-Field (kr ≫ 1) region Ea = E𝜙 â 𝜙 = −j𝜂

kIin −jkr le 4𝜋r e

(le : vector effective length)

k2 a2 I0 e−jkr 𝜋SI0 e−jkr sin 𝜃 = 𝜂 sin 𝜃 4r λ2 r kI (jkS sin 𝜃) −jkr E𝜙 = −j𝜂 0 e 4𝜋r ∴ le = jkS sin 𝜃 â 𝜙 E𝜙 ≃ 𝜂

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5.41. C = 2𝜋a = 1.4λ ⇒ a = 1.4λ = 0.2228λ 2𝜋 ) ( ) ( 0.2228 a = 2 ln 2𝜋 = 9.0 Ω = 2 ln 2𝜋 b 0.01555 (a) From Figure 5.15 Zin = Rin + jXin = 320 − j40 (b)

| Z − Zc | | 320 − j40 − 300 | | = 0.0718 |=| |Γ| = || in | | | | Zin + Zc | | 320 − j40 + 300 | 1 + |Γ| 1 + 0.0718 VSWR = = = 1.155 1 − |Γ| 1 − 0.0718 1 1 = = (3.0769 + j0.3846) × 10−3 = Gc + jBc Zin 320 − j40 To resonate the circuit, the unknown element must have an inductive admittance of

(c) Yin =

Yunknown = −j0.3846 × 10−3 = −j

1 1 ⇒L= 𝜔L 0.3846 × 10−3 (2𝜋f ) = L=

G = 3.0769 × 10–3

1 0.3846 × 10−3 (2𝜋 × 108 ) 10−5 = 4.138 × 10−6 H 0.769𝜋

Bc = 0.3846 × 10–3

?

Therefore the unknown element across the terminals of the loop must be an inductor of L = 4.138 × 10−6 Henries. 5.42. (a) From Figure 5.15 Zin = 90 − j110 (b) Inductor; XL = +110 = 𝜔L = 2𝜋fL L= (c)

110 110 110 = × 10−9 Henries = 2𝜋f 2𝜋 2𝜋(109 )

Zin = 90 | Z − Zc | | 90 − 78 | 12 |=| | |Γ| = || in | | 90 + 78 | = 168 = 0.0714 Z + Z | | in | c| 1 + |Γ| 1 + 0.0714 1.0714 VSWR = = = = 1.1538 1 − |Γ| 1 − 0.0714 0.9285

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5.43.

145

C = λ = 2𝜋a ⇒ a = C∕2𝜋 = λ∕2𝜋; b = 2.47875 × 10−3 λ; Zc = 75; f = 500 MHz [ ( )] [ ( )] [ ( )] λ∕2𝜋 a a = 2 ln 2𝜋 = 2 ln 2𝜋 Ω = 2 ln 2𝜋 b b 2.47875 × 10−3 λ ) ( 1 = 2 ln 2.47875 × 10−3 Ω = 2 ln(403.429) = 2(6) = 12 (a) From Figure 5.15 Rin ≃ 100, Xin = −100, Zin ≃ 100 − j100 (b) Capacitive (c) Zin ≃ 100 − j100. Since the lumped element must be placed in parallel, it is better to work with admittances. Therefore Yin =

100 + j100 100 + j100 100 + j100 1 + j 1 1 1 = = = = = Zin 100 − j100 100 − j100 100 + j100 200 104 + 104 2 × 104 Yin = 5 × 10−3 (1 + j)

Need an inductor, in parallel, with a susceptance BL . (d) 1 1 1 1 BL = 5 × 10−3 = = = = 𝜔L 2𝜋fL 2𝜋(5 × 108 )L 𝜋 × 109 L L=

1 1 = 63.66 × 10−9 H = 5 × 10−3 (𝜋 × 109 ) 5𝜋 × 106 L = 63.66 × 10−9 H

(e)

′ ′ Yin = 5 × 10−3 ⇒ Zin =

1 1 = = 200 ′ Yin 5 × 10−3

′ = 200 Zin

(f)

Γ= VSWR =

′ −Z Zin c ′ +Z Zin c

=

200 − 75 125 = = 0.45455 200 + 75 275

1 + |Γ| 1 + 0.45455 = = 2.667 1 − |Γ| 1 − 0.45455 VSWR = 2.667

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SOLUTION MANUAL

5.44. (a) From Figure 5.15(b) Zin = Rin + jXin = Rin ⇒ Xin = 0 when a ⎧ 6 ⎪ 12 ⇒ 2𝜋(a∕b) = e = 403.429 ⇒ b = 64.21 ⎪ ) ⎪ 11 ⇒ 2𝜋(a∕b) = e5.5 = 244.692 ⇒ a = 38.94 ( a b =⎨ Ω = 2 ln 2𝜋 b ⎪ 10 ⇒ 2𝜋(a∕b) = e5 = 148.413 ⇒ a = 23.62 b ⎪ ⎪ 9 ⇒ 2𝜋(a∕b) = e4.5 = 90.017 ⇒ a = 14.23 ⎩ b (b) These occur when the smallest circumference of the loop is from Figure 5.15(b) Ω = 12 ⇒ C = 2𝜋a ≃ 1.08λ ⇒ a = 0.1719λ ⇒ b = 0.1719λ∕64.21 = 2.68 × 10−3 λ Ω = 11 ⇒ C = 2𝜋a ≃ 1.10λ ⇒ a = 0.175λ ⇒ b = 0.175λ∕38.94 = 4.496 × 10−3 λ Ω = 10 ⇒ C = 2𝜋a ≃ 1.14λ ⇒ a = 0.1814λ ⇒ b = 0.1814λ∕23.62 = 7.68 × 10−3 λ Ω = 9 ⇒ C = 2𝜋a ≃ 1.28λ ⇒ a = 0.2037λ ⇒ b = 0.2037λ∕14.33 = 14.216 × 10−3 λ 5.45. E𝜙 =

ak𝜂I0 e−jkr J1 (ka sin 𝜃) 2r

For small radius a (small argument ka sin 𝜃), the Bessel function can be approximated by (5-66b), or a≪λ

J1 (ka sin 𝜃) ≃

ka sin 𝜃 2

Thus E𝜙 = 𝜂

(ka)2 I0 e−jkr sin 𝜃 4r

The radiated power can then be written as 𝜋

2𝜋

Prad = = = Prad = Rr =

2𝜋

𝜋

1 â |E |2 ⋅ â r r2 sin 𝜃 d𝜃 d𝜙 ∫0 ∫0 r 2𝜂 𝜙 [ ]2 2𝜋 𝜋 2𝜋 𝜋 (ka)2 1 1 2 |E𝜙 | sin 𝜃 d𝜃 d𝜙 = |I0 |2 sin3 𝜃 d𝜃 d𝜙 𝜂 ∫0 ∫0 2𝜂 ∫0 ∫0 2𝜂 4 ( ) 𝜋 |I0 |2 2 (ka)4 𝜋 4 𝜋 = 𝜂 (ka)4 |I0 |2 sin3 𝜃d𝜃 = 𝜂 (ka)4 |J0 |2 𝜂 (2𝜋) ∫0 2𝜂 16 16 3 12 ( ) 𝜋 1 (ka)4 |I0 |2 = 10𝜋 2 (ka)4 |I0 |2 ≡ |I0 |2 Rr (120𝜋) 12 2 ( ( )4 )4 2𝜋 C 20𝜋 2 (ka)4 = 20𝜋 2 a = 20𝜋 2 λ λ ∫0

∫0

W ave ⋅ â r r2 sin 𝜃 d𝜃 d𝜙 =

5.46. I(𝜙) = I0 cos 𝜙 2𝜋 2𝜋 ′ 𝜇I0 e−jkr ′ 𝜇I0 e−jkr â 𝜙 cos 𝜙′ â 𝜙 cos 𝜙′ ejkrâ ⋅r d𝜙′ a d𝜙 ≃ a (a) A(r) = 4𝜋 ∫0 R 4𝜋 r ∫0 { 2𝜋 𝜇I0 e−jkr ′ = cos 𝜙′ sin 𝜙′ ejka sin 𝜃 cos(𝜙−𝜙 ) d𝜙′ a −̂ax ∫ 4𝜋 r 0 } 2𝜋 2 ′ jka sin 𝜃 cos(𝜙−𝜙′ ) ′ + â y cos 𝜙 e d𝜙 ∫0

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𝜇I0 a e−jkr 8𝜋 r

A𝜙 =

+ â y

2𝜋

∫0

{

2𝜋

−̂ax

∫0

sin(2𝜙′ )ejka sin 𝜃 cos(𝜙−𝜙 ) d𝜙′ ′

cos(2𝜙′ ) + 1)ejka sin 𝜃 cos(𝜙−𝜙 ) d𝜙′ ′

}

𝜇I0 a e−jkr {̂ax J2 (ka sin 𝜃) sin 2𝜙 − â y J2 (ka sin 𝜃) cos 2𝜙 4 r + â y J0 (ka sin 𝜃)}

A𝜙 =

𝜇I0 a e−jkr 1 {−̂a𝜙 J2 (ka sin 𝜃) cos 𝜙 + â y [J2 (ka sin 𝜃) + J0 (ka sin 𝜃)]} 2 r 2 { } 𝜇I0 a e−jkr J1 (ka sin 𝜃) A𝜙 = −̂a𝜙 J2 (ka sin 𝜃) cos 𝜙 + â y 2 r ka sin 𝜃 { } −jkr −𝜇I0 a e J (ka sin 𝜃) A𝜙 ≃ J2 (ka sin 𝜃) − 1 cos 𝜙 2 r ka sin 𝜃 =

𝜇I0 a e−jkr ′ J (ka sin 𝜃) cos 𝜙 2 r 1 𝜇I a e−jkr J1 (ka sin 𝜃) A𝜃 ≃ 0 cos 𝜃 sin 𝜙 2 r ka sin 𝜃 j𝜂ka e−jkr ′ E𝜙 ≃ I J (ka sin 𝜃) cos 𝜙 2 0 r 1 j𝜂ka e−jkr J1 (ka sin 𝜃) E𝜃 ≃ I cos 𝜃 sin 𝜙 2 0 r ka sin 𝜃 =

(b) 𝜃 = 0, 𝜙 = 𝜋∕2 E𝜙 = 0 E𝜃 =

j𝜂ka e−jkr I 4 0 r

|E|2 𝜂 (ka)2 = I 2𝜂 32 0 r2 ) ( 𝜂 𝜋 = U 𝜃 = 0, 𝜙 = I (ka)2 2 32 0 Wav ≃

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6

Solution Manual −jkr

−jkr1

6.1. (a) Et = E1 + E2 + E3 = 2E0 e r + E0 e r

1

−jkr2

+ E0 e r

2

where the center element is placed at the origin. For far-field observations r1 ≃ r − d cos 𝜃 r2 ≃ r + d cos 𝜃 r1 ≃ r2 ≃ r

} for phase variations for amplitude variations

and Et = E0

e−jkr {2 + ejkd cos 𝜃 + e−jkd cos 𝜃 } r { [ ]} 1 2 1 + (ejkd cos 𝜃 + e−jkd cos 𝜃 ) 2

≃ E0

e−jkr r

= E0

e−jkr {2[1 + cos(kd cos 𝜃)]} r

Thus the array factor is equal to AF(𝜃) = 2[1 + cos(kd cos 𝜃)] = 4 cos2

(

kd cos 𝜃 2

)

which in normalized form can also be written as AF(𝜃)n = 1 + cos(kd cos 𝜃) = 2 cos2

(

kd cos 𝜃 2

)

(b) The nulls of the pattern can be found using either of the above forms for the array factor. For example:

Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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SOLUTION MANUAL

One form

kd cos 𝜃n = cos−1 (−1) = n𝜋, n = ±1, ±3, …

The other form ( ) kd 2 cos2 cos 𝜃n = 0 2 kd n𝜋 cos 𝜃n = cos−1 (0) = 2 2 n = ±1, ±3, …

𝜃n = cos−1 (nλ∕2d), n = ±1, ±3, ±5, …

𝜃n = cos−1 [nλ(2d)], n = ±1, ±3, …

AF(𝜃) = 1 + cos(kd cos 𝜃n ) = 0 cos(kd cos 𝜃n ) = −1

which are of identical form. Therefore both forms yield the same results. Thus for d = λ∕4

𝜃n = cos−1

(

nλ 2d

) d=λ∕4

= cos−1 (2n), n = ±1, ±3, … ⇒ No nulls exist.

(c) Similarly the maxima of the pattern can be found using either of the two forms for the array factor. For example Other form

One form AF(𝜃) = 1 + cos(kd cos 𝜃m ) = 2 cos(kd cos 𝜃m ) = 1 kd cos 𝜃m = cos−1 (1) = 2m𝜋 m = 0, ±1, … , ( ) mλ , m = 0, ±1, ±2, … , 𝜃m = cos−1 d

( ) kd AF(𝜃) = 2 cos2 cos 𝜃m = 2 ) 2 ( kd cos cos 𝜃m = ±1 2 kd cos 𝜃m = cos−1 (±1) = m𝜋 2 m = 0, ±1, … ( ) mλ 𝜃m = cos−1 , m = 0, ±1, ±2, … d

which are of identical form. Therefore both yield the same results. Thus for d = λ∕4. { 𝜃m = cos (4m),

m = 0, ±1, ±2, →

−1

m = 0:

𝜃0 = cos−1 (0) = 90◦

m = ±1:

𝜃1 = cos−1 (4) ⇒ Does not exist

The same is true for other values of m (i.e, m = ±2, ±3, …). Therefore the only maximum occurs at 𝜽 = 90◦ . (d) Computer Program Directivity: When d = λ∕4 ( ) ) kd 𝜋 cos 𝜃 = 4 cos2 cos 𝜃 2 4 ( ) 𝜋 cos 𝜃 Un = cos4 4

AF(𝜃) = 4 cos2

(

D0 = 1.4384 = 1.5787 dB

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SOLUTION MANUAL

6.2. One dipole: E𝜃 = j𝜂

kI0 le−jkr sin𝜃 4𝜋r z

Array Factor: [ 𝜋 λ 2𝜋 ] λ 2𝜋 (AF)2 = E0 ej 2 e−j 8 ⋅ λ cos 𝜓 + e+j 8 λ cos 𝜓 ] 𝜋 [ 𝜋 𝜋 = E0 ej 4 e−j 4 (cos 𝜓−1) + ej 4 (cos 𝜓−1) –d/2 ( ) 𝜋 j 𝜋4 (cos 𝜓 − 1) = E0 e 2cos 4 ( ) 𝜋 x 𝜋 (sin 𝜃 sin 𝜙 − 1) (AF)2 = E0 ej 4 2 cos 4 (̂ay ⋅ â r = sin 𝜃 sin 𝜙 = cos 𝜓) = sin 𝜃

ψ 0

y

d/2

ϕ

At y-z plane (𝜙 = 90◦ ):

( )| | 𝜋 | , (x-z plane) (1) |E𝜃 (𝜃)|𝜙=0◦ ∝ ||sin 𝜃 cos 4 || | 0 < 𝜃 < 𝜋 − 𝜋 THMAX = 90.0 DE GREES EXCITATION COEFFICIENTS FOR THE ARRAY DESIGN 1 9.964 2 15.311 3 6.347 NORMALIZED EXCITATION COEFFICIENTS 1 1.570 2 2.412 3 1.000 ***NOTE: THE NORMALIZED ARRAY FACTOR (in dB) IS STORED IN AN OUTPUT FILE CALLED ......... ArrFac.dat ==============================================

6.76. R0 = 40 dB = 20 log10 Rovr ⇒ Rovr = 102 = 100; N = 20, P = 19 (a) z0 = 1 2

[(

)1∕19 ( )1∕19 ] √ √ 2 2 100 + (100) − 1 + 100 − (100) − 1 z0 =

dmax ≤

1 (1.32161899 + 0.7566477) = 1.03913 2

( ) ) ) ( ( λ 1 λ 𝜋 1 λ = (164.226◦ ) cos−1 − = cos−1 − 𝜋 z0 𝜋 1.03913 𝜋 180 dmax = 0.91236λ [

(b) Θh (uniform) = 2

)] [ )] ( ( 𝜋 1.391λ 1.391 × 2 𝜋 =2 − cos−1 − cos−1 2 𝜋Nd 2 20𝜋

Θh (uniform) = 2(90◦ − 87.462◦ ) = 5.075◦

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SOLUTION MANUAL

{

}2 √ 2 cosh (cosh−1 R0 )2 − 𝜋 2 R0 [ ] √ cosh−1 (R0 ) = ln 100 ± (100)2 − 1 = 5.29829

(c) f = 1 + 0.636

}2 √ 2 cosh (5.29829)2 − 𝜋 2 100 { ( }2 )2 2 2 = 1 + 0.636 cosh(4.2644) = 1 + 0.636 35.639 100 100 {

f = 1 + 0.636

f = 1 + 0.323 = 1.323 Θh (Tschebyscheff) = Θh (uniform) f = 5.075(1.323) = 6.714◦ (d) D0 =

2R20 1 + (1002 − 1)(1.323) (

1 ) 19 1 + 2 2

=

20,000 = 15.11 = 11.79 dB 1,323.868

( ) ( ) 1 d = 2(20) = 20 = 13 dB λ 2 6.77. N = 4, linear, broadside, Dolph-Tschebyscheff, d = 3λ∕8 D0 = 27.959 dB = 25(dimensionless) = R0 (a) D0 = 27.959 dB = 20 log10 D0 (vr) ⇒ D0 (vr) = 1027.959∕20 = 101.398 = 25 (e) D0 = 2N

1 z0 = 2 1 = 2 =

[( [(

)1∕3 ( )1∕3 √ √ + R0 − R20 − 1 R0 + R20 − 1

]

)1∕3 ( )1∕3 ] √ √ 2 2 25 + 25 − 1 + 25 − 25 − 1

] 1 1[ (49.98)1∕3 + (0.02)1∕3 = (3.6835 + 0.27148) = 1.9775 2 2 z0 = 1.9775 [

Alternate: z0 = cosh

1 cosh−1 (25) 3

]

] [ cosh−1 (25) = ln 25 ± (252 − 1)1∕2 = ln[25 ± 24.98] = 3.9116 [ ] 1 3.6835 + 0.2715 z0 = cosh (3.9116) = cosh(1.30387) = 3 2 z0 = 1.9775 dmax

( ) ) ( λ 1 λ 1 λ −1 (120.3771) = 0.6688λ = ≤ cos − = cos−1 − 𝜋 z0 𝜋 1.9775 180 dmax = 0.6688λ

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{ (b)

f = 1 + 0.636

2 cosh R0

207

[√ ]}2 (cosh−1 R0 )2 − 𝜋 2

{

[√ ]}2 2 −1 2 2 = 1 + 0.636 (cosh (25)) − 𝜋 cosh 25 ]}2 { [ ]2 [√ √ 2 2 = 1 + 0.636 (3.9116)2 − 𝜋 2 = 1 + 0.636 cosh cosh 5.4310 25 25 [ [ ]2 ]2 2 2 f = 1 + 0.636 cosh(2.3305) = 1 + 0.636 (5.1899) = 1 + 0.636(0.4152)2 25 25 f = 1 + 0.636(0.1724) = 1 + 0.1096 = 1.1096 Θh = 34.3366(1.1096) = 38.1◦ (c) D0 =

2R20 ( ) 1 + R20 − 1 f

λ L+d

2(25)2

=

1 + (252 − 1)(1.1096)

2(625)

λ (3 + 1)

3λ 8

( ) 2 1 + (624)(1.1096) 3 2(625) D0 = = 2.7022 = 4.3171 dB 1 + 461.5936 =

D0 = 2.7022 = 4.3171 dB 6.78. (a) Order of polynomial Tm (z) = T5 (z) (b) N = 6 ⇒ P = N − 1 = 6 − 1 = 5 (c) 40 = 20 log10 Rvr ⇒ Rvr = 102 = 100 z0 =

1 2

[( 100 +

)1∕5 ( )1∕5 ] √ √ (100)2 − 1 + 100 − (100)2 − 1

1 (2.8854 + 0.3466) = 1.616 2 ( ) ) ( λ 1 λ 1 λ −1 = (128.23◦ = 2.238 rads) ≤ cos cosh−1 − − = 𝜋 z0 𝜋 1.616 𝜋 z0 =

dmax

dmax = 0.7124λ 6.79. Dolph-Tschebyscheff, N = 10, d = λ∕2, R0 = −26 dB, z0 = 1.0851 (a) 𝛽 = 0◦ (broadside) (b) Tn (x) = TN−1 (x) = T10−1 (x) = T9 (x) Number of minor lobes (complete) = 4 ( ) ) ) ( ( λ 1 λ 𝜋 1 λ −1 = (157.1573◦ ) = 0.8731λ − = cos−1 − (c) dmax ≤ cos 𝜋 z0 𝜋 1.0851 𝜋 180 dmax ≤ 0.8731λ

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SOLUTION MANUAL

6.80. N = 6, d = λ∕4, R0 = 25 = 20 log10 Rovr ⇒ Rovr = 1025∕20 = 17.7828 [√ ]}1∕2 { 2 −1 2 2 (cosh R0 ) − 𝜋 cosh−1 (R0 ) (a) f = 1 + 0.636 R cosh 0 [ ] √ = cosh−1 (17.7828) = ln R0 ± R20 − 1 [ ] √ = ln 17.7828 ± (17.7828)2 − 1 = 3.57059 ]}2 [√ 2 (3.57059)2 − 𝜋 2 cosh 17.7828 { { }2 }2 2 2 = 1 + 0.636 cosh(1.69733) = 1 + 0.636 (2.8217) 17.7828 17.7828 {

f = 1 + 0.636

f = 1 + 0.636(0.31732)2 = 1 + 0.064 = 1.064 )] ( [ 1.391λ = 2[90◦ − cos−1 (0.29518)] Θh (uniform) = 2 90◦ − cos−1 𝜋Nd Θh (uniform) = 2(90◦ − 72.8317◦ ) = 34.337◦ Θh (Tschebyscheff) = Θh (Uniform) f = 34.337◦ (1.064) = 36.535◦ (b) D0 =

2R20 λ 1 + (R20 − 1)f (L + D)

2(17.7828)

=

1 + [(17.7828)2 − 1]1.064 (

λ 5 4

+

1 4

) λ

2(17.7828)2 = 2.8168 = 4.498 dB 1 + 223.531

D0 =

6.81. N = 6, R0 = 50 dB ⇒ Rovr = 1050∕20 = 316.228 { f = 1 + 0.636

2 cosh R0

= ln[316.228 ± { f = 1 + 0.636 { = 1 + 0.636

[√ }2 (cosh−1 R0 )2 − 𝜋 2 ⇒ cosh−1 (y) = cosh−1 (316.228)

√ (316.228)2 − 1] = ±6.4496

}2 √ 2 cosh[ (6.4496)2 − 𝜋 2 ] 316.228 {

}2

2 cosh(5.63274) 316.228

= 1 + 0.636

}2

2 (139.713) 316.228

f = 1 + 0.636(0.7807) = 1.4966 ≃ 1.5 )] [ )] ( ( [ 1.391λ 1.391 × 2 (a) Θh (uniform) ≃ 2 90◦ − cos−1 = 2 90◦ − cos−1 𝜋Nd 6𝜋 Θh (uniform) = 2(90◦ − 81.5128◦ ) = 16.975◦

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SOLUTION MANUAL

Θh (Tschebyscheff) = Θh (uniform)f = 16.975(1.5) = 25.4625◦ Δ = Θh (Tschebyscheff) − Θh (uniform) = (25.4625◦ − 16.975) Δ = 8.4875◦ (greater because of lower sidelobe − 50 dB vs. − 13.5 dB)

(b)

D0 =

2R20 1 + (R20 − 1)f

λ L+d

=

2(316.228)2 1 + [(316.228)2 − 1]1.5(1∕3)

D0 = 3.99996 ≃ 4 = 6.02 dB

D0 (uniform) = 2N

( ) ( ) λ d = 2(6) =6 λ 2λ

D0 (uniform) = 7.78 dB (greater because smaller HPBW) 6.82. Binomial Design (a) r 0 or −∞ dB because d = λ∕z r HPBW = 21.4725◦ r 5.0922 = 7.069 dB (b) Dolph-Tschebyscheff design (30 dB = 20 log10 R0 ⇒ R0 = 101.5 = 31.623) r For a uniform array, Θ = 5.3438◦ . The beam broadening factor is: h { f = 1 + 0.636

2 cosh R0

[√ ]}2 (cosh−1 R0 )2 − 𝜋 2

√ √ cosh−1 R0 = ln[R0 + R20 − 1] = ln[31.623 + (31.623)2 − 1] = 4.1468 √ √ (cosh−1 R0 )2 − 𝜋 2 = (4.1468)2 − 𝜋 2 = 2.7067 cosh 2.7067 =

e2.7067 + e−2.7067 14.9797 + 0.0668 = = 7.5232 2 2 {

f = 1 + 0.636

2 (7.5232) 31.623

}2 = 1 + 0.144 = 1.144

f = 1.144

r Therefore the HPBW of the Dolph-Tschebyscheff design is HPBW(Tschebyscheff) = HPBW (uniform) f = (11.2992)1.144 HPBW(Tschebyscheff) = 12.9263◦

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SOLUTION MANUAL

r D (DT) = 0 D0 (DT) =

2R20 ( ) 1 + R20 − 1 f 254.9716

2(31.623)2

) ( λ λ 1 + [(31.623)2 − 1]1.144 (L + d) 4.5λ = 7.8441 = 8.9454 dB =

(c) The binomial HPBW (21.4725◦ ) is larger than that of the Dolph-Tschebyscheff because of the more severe amplitude tapering of the binomial which leads to a wider HPBW; yes, this is what we expected. (d) The binomial D0 (max)(7.069 dB) is smaller than that of the Dolph-Tschebysheff (8.9454 dB) because the binomial has wider HPBW, thus smaller D0 (max), due to the severe amplitude tapering in the binomial array.

6.83. (a) Uniform Dolph-Tschebyscheff Taylor Cosine-Squared Binomial

(b) Binomial (−∞dB) Cosine-Squared (−31.5 dB) Taylor (−25 dB) Dolph-Tschebyscheff (−25 dB) Uniform (−13.5 dB)

6.84. (a) HPBW = 50.6 = 50.6λ ⇒ d = 50.6 = 50.6 = 50.6 d∕λ λ λ HPBW 1 ⇒ d = 50.6λ

◦ (b) HPBW(𝜃 = 𝜃0 ) = HPBW(𝜃 = 0) sec 𝜃0 = HPBW(𝜃 = 0 ) cos𝜃0

𝜃0 = 60◦ : HPBW(𝜃0 = 60◦ )cos(60◦ ) = HPBW(𝜃 = 0◦ ) HPBW(𝜃 = 0◦ ) = cos(60◦ )HPBW(𝜃0 = 60◦ ) = 0.5(1) = 0.5◦ Thus d 50.6 50.6 = = = 101.2 ◦ λ HPBW(𝜃 = 0 ) 0.5

d = 101.2λ 6.85. (a) Tschebyscheff (b) 30 = 20 log (Rovr ) ⇒ Rovr = 103∕2 = 31.623 10 z0 =

1 [(R + 2 0

√ √ R20 − 1)1∕P + (R0 − R20 − 1)1∕P ]

=

√ √ 1 [(31.623 + (31.623)2 − 1)1∕2 + (31.623 − (31.623)2 − 1)1∕2 ] 2

=

1 [(31.623 + 31.607)1∕2 + (31.623 − 31.607)1∕2 ] 2

z0 =

1 [7.952 + 0.1265] = 4.039 2

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SOLUTION MANUAL

211

2M + 1 = 3 ⇒ M = 1, M + 1 = 2 a1 = (−1)1−1+1 (4.039)2(1−1)

(1 + 2 − 2)!2 2(1 − 1)!(1 + 1 − 2)!(1 − 1 + 1)!

+ (−1)1−2+1 (4.039)2(2−1) = (−1)(1)

(2 + 1 − 2)!2 2(2 − 1)!(2 + 1 − 2)!(1 − 2 + 1)!

1⋅2 1⋅2 + (1) ⋅ (4.039)2 2(1)(1) 2(1)(1)(1)

a1 = −1 + 16.314 = 15.314 a2 = (−1)1−2+1 (4.039)2(2−1) a2 = (1)(4.039)2

(2 + 1 − 2)!2! 2(2 − 2)!(2 + 2 − 2)!(1 − 2 + 1)!

(1)(2) = 16.314∕2 = 8.157 2(1)(2)(1)

}2 2 cosh[(cosh−1 31.623)2 − 𝜋 2 ] 31.623 ]}2 { [√ 1 = 1 + 0.636 (4.1468)2 − 𝜋 2 cosh 31.623 { { }2 }2 2 2 f = 1 + 0.636 cosh(2.707) = 1 + 0.636 7.525 31.623 31.623 {

(c) f = 1 + 0.636

f = 1.144

] ] [ [ λ λ − cos−1 cos𝜃0 + 0.443 HPBW = Θh = cos−1 cos𝜃0 − 0.443 L + d 𝜃0 =90◦ L + d 𝜃0 =90◦ ] ] [ [ λ λ = cos−1 −0.443 − cos−1 0.443 = 126.20◦ − 53.8◦ 3λ∕4 3λ∕4 Θh = 72.4◦

Θh (Tschebyscheff) = 72.4(1.144) = 82.8256 2(1,000)

( ) = 1.312 = 1.1793 dB 4 1 + (999)1.144 3 ( 𝜋x ) n 6.86. an = 1 + cos L (d) D0 =

( 𝜋x )

1 = 1 + cos(0) = 2 ⇒ a1 = 1 L ( ) ( 𝜋x ) ( ) 𝜋 𝜋L 2 |x2 =L∕4 = 1 + cos = 1 + cos = 1.707 a2 = 1 + cos L 4L 4 ( 𝜋x ) ( ) 𝜋 3 a3 = 1 + cos =1 = 1 + cos L x3 =L∕2 2

2a1 = 1 + cos

∴ a1 = 1 a2 = 1.707 a3 = 1

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SOLUTION MANUAL

6.87. The length of the array is obtained using an iterative procedure of (6-22) or its graphical solution of Figure 6.12. ) ) ( ( 2.782 2.782 − cos−1 cos𝜃0 + 𝜃h = cos−1 cos𝜃0 − Nkd Nkd (a) 𝜃0 = 30◦ , using iterative procedure: ) ) ( ( 𝜋 2.782 𝜋 2.782 2◦ −1 −1 − cos cos cos × 𝜋 = cos − + 180◦ 6 x 6 x Nkd = x = 318.95, Nkd = 318.95, N = (L + d)∕d 2𝜋 = 318.95, L + d = 50.76λ λ ∴ L = (50.76λ − d)

∴ (L + d)

From Fig. 6.12, L + d = 50λ

∴ L = (50λ − d)

(b) 𝜃0 = 45◦ , using iterative procedure: Nkd = x = 225.46, L + d = 35.88λ From Fig. 6.12, L + d = 35λ

∴ L = (35.88λ − d)

∴ L = (35λ − d)

(c) 𝜃0 = 60◦ , using iterative procedure: Nkd = x = 184.07, L + d = 29.2956λ, From Fig. 6.12, L + d = 30λ,

∴ L = 29.2956λ − d

∴ L = (30λ − d)

6.88. 𝜃0 = 60◦ , 𝜙0 = 90◦ , dx = dy = λ, 0 ≤ 𝜃 ≤ 180◦ , 0◦ ≤ 𝜙 ≤ 360◦ We need to find 𝜙 and 𝜃 which angle satisfy both (6-49a) and (6-49b) i.e. sin 𝜃 cos 𝜙 − sin 𝜃0 cos 𝜙0 = ± sin 𝜃 sin 𝜙 − sin 𝜃0 sin 𝜙0 = ± sin 𝜃 cos 𝜙 = ±m;

mλ = ±m (← dx = λ) dx

nλ = ±n (← dy = λ) dy

m = −1, 0, 1

√ sin 𝜃 ⋅ sin 𝜙 −

3 = ±n; 2

n = 0, 1

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SOLUTION MANUAL

In case of

In case of

Again, we found the angle of grating lobe ⟨ 𝜃 = 90◦ → Major lobe ⎧ ◦ At, 𝜙 = 90 ⎪ 𝜃 = 120◦ → Grating lobe ⎪ n = 0, m = 0 ⎨ ⟨ } ◦ 𝜃 = 240 ⎪ ⇒ Not physical. ⎪ 𝜙 = 270◦ 𝜃 = 300◦ ⎩ { n = −1, m = 0

{ ◦

𝜙 = 90

𝜃 = −7.699◦ 𝜃 = −172.301◦

at 𝜙 = 270◦ , 𝜃 = 7.699◦ , ↑ Grating lobe

}} ⇒ Not physical

𝜃 = 172.301◦ . ↑ Grating lobe

Therefore, three grating lobes exist. n = 0, m = 0, n = −1, m = 0,

A (𝜃 = 120◦ , 𝜙 = 90◦ ) ○

(𝜃 = 7.699◦ , 𝜙 = 270 ◦ ) (𝜃 = 172.301◦ , 𝜙 = 270 ◦ ) B C ○ ○

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SOLUTION MANUAL

1

(7.699°, 270°) B

The maximum value is not 1 at (θ = 90°, ϕ = 352°) θ = 90° (0.98026) ϕ = 180°

C

(172.301, ϕ = 270°)

A (θ = 120°, ϕ = 90°)

0.8 Major lobe

0.6 0.4 0.2 0 200 × 2 150 × 2 =300° 100 × 2 =200° 50 × 2 ϕ =100°

150° 100° 50° 0

0°

)

eta θ (Th

Figure P6-88

6.89. According to (6-4a), the normalized array factor of a two-element array with the elements along the z-axis and of the same amplitude, 𝛽-phase difference, and d separation between them ] [ 1 AF = cos (kd cos 𝜃 + 𝛽) 2 (a) For a two-element array with the elements along the x-axis [ (AF)n = cos

]

1 (kd sin 𝜃 cos 𝜙 + 𝛽x ) 2 x

(b) For a two-element array with the elements along the y-axis [ (AF)n = cos

]

1 (kd sin 𝜃 sin 𝜙 + 𝛽y ) 2 y

(c) For a four-element array with the elements along the x- and y-directions [ (AF)n = cos

] [ ] 1 1 (kdx sin 𝜃 cos 𝜙 + 𝛽x ) + cos (kdy sin 𝜃 sin 𝜙 + 𝛽y ) 2 2

(d) Maximum along the +z axis: 𝛽 x = 𝛽y = 0 Minimum along the +z axis: 𝛽x = 𝛽y = 180◦

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SOLUTION MANUAL

215

6.90. dx = dy = λ∕8, M = 10, N = 8, 𝜃0 = 10◦ , 𝜙0 , 90◦ 2𝜋 λ (a) 𝛽x = −kdx sin 𝜃0 cos 𝜙0 = − sin (10◦ ) cos (90◦ ) = 0 λ 8 2𝜋 λ sin (10◦ ) cos (90◦ ) = −0.1364 rad = −7.81◦ 𝛽y = −kdy sin 𝜃0 sin 𝜙0 = − λ 8 (b) D0 = 𝜋 cos 𝜃0 Dx Dy ( ) ( ) dx 1 = 2.5 = 3.98 dB Dx = 2N = 2(10) λ 8 ( ) ( ) dy 1 = 2.0 = 3.01 dB Dy = 2N = 2(8) λ 8 D0 = 𝜋 cos (10◦ )(2.5)(2) = 15.47 = 11.89 dB [ −1 cos 𝜃 − 0.443 Θ = cos (c) x0 0 [

λ L + dx

] 𝜃0 =90◦

] λ − cos 𝜃0 + 0.443 L + dx 𝜃 =90◦ 0 ) ) ( ( 1 1 −1 −1 − cos = 110.76◦ − 69.24◦ −0.443 0.443 = cos 1.25 1.25 cos−1

Θx0 = 41.52◦ = 0.7245 rads Also from Table 6.2 [ ( )] )] ( [ 1.391 × 8 1.391λ = 41.49◦ = 2 90◦ − cos−1 Θx0 = 2 90◦ − cos−1 𝜋Mdx 10𝜋 Θx0 = 0.724 rads ( −1 Θy0 = cos cos 𝜃0 − 0.443

Θy0

) ( ) λ λ −1 − cos cos 𝜃0 + 0.443 L + dy 𝜃 =90◦ L + dy 𝜃 =90◦ 0 0 ) ) ( ( 1 1 − cos−1 0.443 = 116.3◦ − 63.70◦ = 52.59◦ = cos−1 −0.443 1 1 = 0.918 rads

Also from Table 6.2 [ ( )] )] ( [ 1.391 × 8 1.391λ = 2 90◦ − cos−1 Θy0 = 2 90◦ − cos−1 𝜋Ndy 8𝜋 Θy0 = 52.56◦ = 0.917 rads Therefore √ Θh =

cos2 𝜃0 [Θ−2 x 0

Θy 0 | 52.59◦ 1 | = = ◦ | 2 cos 𝜃0 cos(10◦ ) | 𝜙0 =90◦ cos2 𝜙0 + Θ−2 y sin 𝜙0 ] 𝜃0 =10

0

◦

Θh = 53.40 = 0.932 rads

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SOLUTION MANUAL

√ Ψh =

Θ−2 x 0

| 1 ◦ | ◦ = Θx0 = 41.52 = 0.7245 rads | sin 𝜙0 + Θ2y cos2 𝜙0 | 𝜙0 =90◦ 2

𝜃0 =10

0

ΩA = Θh Ψh = 53.40(41.52) = 2,217.17 (degrees)2 D0 ≃

32,400

=

ΩA (degrees)2

32,400 = 14.61 = 11.65 dB 2,217.17

and it agrees with the more accurate values above. D0 = 15.7755 = 11.9798 dB

Computer program Arrays: 6.91. N = 4 = M; Binomial planar array 𝜃0 = 30◦ , 𝜙0 = 45◦ (a) Θh = Θx0 sec 𝜃0 = Θy0 sec 𝜃0

From Problem 6.57, Θx0 = Θy0 = 35.06◦ Θh = 35.06◦ sec(30◦ ) = 35.06◦ (1.1547) = 40.484◦ Θh = 40.484◦ Ψh = Θx0 = Θy0 = 35.06◦ (b) D0 = 𝜋 cos 𝜃0 Dx Dy Dx = Dy = 3.2 from Problem 6.57 D0 = 𝜋 cos(30◦ )(3.2)(3.2) = 27.86 = 14.45 dB D0 = 14.45 dB = 27.86 (c)

ΩA = Θh Ψh = (35.06)(40.484) = 1,419.369 (degrees)2 D0 ≃

32,400 32,400 = = 22.827 = 13.585 dB 2 1,419.369 ΩA (degrees) D0 ≃ 22.827 = 13.585 dB

D0 (Kraus) ≃

41,253 41,253 = = 29.064 = 14.634 dB 1,419.369 ΩA (degrees)2 D0 = 29.064 = 14.634 dB

6.92. dx = dy = λ∕4, N = M = 10, 𝜃0 = 10◦ , 𝜙0 = 45◦ , R0 = 26 dB = 20 { f = 1 + 0.636 { = 1 + 0.636

2 cosh R0 2 cosh 20

[√ ]}2 (cosh−1 R0 )2 − 𝜋 2 [√

]}2 (cosh

−1

(20))2

− 𝜋2

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SOLUTION MANUAL

]}2 [√ 2 (3.69)2 − 𝜋 2 cosh 20 [ [ ]2 ]2 2 2 f = 1 + 0.636 cosh(1.93) = 1 + 0.636 (3.52) = 1 + 0.079 = 1.079 20 20 {

f = 1 + 0.636

(a) D0 = 𝜋 cos 𝜃0 Dx Dy , Lx = Ly = 9

Dx =

2R20 ( ) 1 + R20 − 1 f

Dx = Dy =

λ Lx + dx 2(20)2

( ) 9 λ = λ = 2.25λ 4 4 2R20

= Dy =

1 1 + (202 − 1)1.079 2.5

λ Ly + dy

1 + (R20 − 1)f

= 4.62 = 6.65 dB

D0 = 𝜋cos(10◦ )(4.62)(4.62) = 66.04 = 18.20 dB (b) For a square array ⇒ Θh = Θx0 sec 𝜃0 = Θy0 sec 𝜃0 , Ψh = Θx0 = Θy0 { Θx0 = Θy0 =

[ cos−1 cos 𝜃0 − 0.443

[ − cos

−1

λ Lx + dx

λ cos 𝜃0 + 0.443 Lx + dx

{ =

[ cos−1 cos 𝜃0 − 0.443 [

− cos

=

−1

λ Ly + dy

λ cos 𝜃0 + 0.443 Ly + dy

] 𝜃0 =90◦

}

]

⋅f

𝜃0 =90◦

] 𝜃0 =90◦

}

]

⋅f 𝜃0 =90◦

) )} ( ( { 1 1 − cos−1 0 + 0.443 ⋅ (1.079) cos−1 0 − 0.443 2.5 2.5

Θx0 = Θy0 = [100.21 − 79.79]1.079 = 20.41(1.079) = 22.03◦ = 0.3844 rads Also from Table 6.2 [ ( Θx0 = Θy0 = 2 90◦ − cos−1

1.391λ 𝜋 × 10 × λ∕4

)] (1.079) = 2(10.20)(1.079)

= 22.01◦ = 0.3842 rads Thus Θh = Θx0 sec 𝜃0 = Θy0 sec 𝜃0 = 22.03 sec (10◦ ) = 22.03(1.015) = 22.37◦ = 0.390 rads Ψh = Θx0 = Θy0 = 22.03◦ = 0.3844 rads (c) ΩA = Θh Ψh = 22.37(22.03) = 492.81 (degrees)2 = 0.1501 (rads)2

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SOLUTION MANUAL

6.93. dx = dy = λ∕8, M = 10, N = 8, 𝜃0 = 10◦ , 𝜙0 = 90◦ , R0 = 30 dB = 31.623 ( ) 2𝜋 λ sin(10◦ ) cos(90◦ ) = 0 (a) 𝛽x ≃ −kdx sin 𝜃0 cos 𝜙0 = − λ 8 ( ) 2𝜋 λ sin(10◦ ) sin(90◦ ) = −0.1364 rad = −7.81◦ 𝛽y ≃ −kdy sin 𝜃0 cos 𝜙0 = − λ 8 (b) D0 = 𝜋 cos 𝜃0 Dx Dy { f = 1 + 0.636 { = 1 + 0.636

2 cosh 31.623

[√ ]}2 (cosh−1 (31.623))2 − 𝜋 2

]}2 [ [√ ]2 2 2 (4.147)2 − 𝜋 2 = 1 + 0.636 cosh (7.523) 31.623 31.623

f = 1.144 Dx =

2R20 1 + (R20 − 1)f

λ Lx + dx

2(31.623)2

=

1 + (31.623)2 − 1)(1.144)

1 1.25

Dx = 2.185 = 3.395 dB Dy =

2R20 1 + (R20 − 1)f

λ Ly + dy

=

2(31.623)2 1 + (31.6232 − 1) (1.144)

1 1

Dy = 1.748 = 2.426 dB D0 = 𝜋cos(10◦ )(2.185)(1.748) = 11.817 = 10.725 dB ′ ′ (c) Θx0 = Θ′x |uniform f = 41.52(1.144) = 47.50◦ ⎫ Θx0 |uniform and Θy0 |uniform ⎪ 0 ⎬ were obtained from the ′ Θy0 = Θy |uniform f = 52.56(1.144) = 60.13◦ ⎪ 0 ⎭ solution of Problem 6.90 From the solution of Problem 6-90

Θy0

60.13◦ = 1.0154(60.13◦ ) = 61.05◦ = 1.066 rads cos(10)◦

Θh | 𝜙

=

Ψh | 𝜙

= Θx0 = 47.50◦ = 0.829 rads

◦ 0 =90 𝜃0 =10◦ ◦ 0 =90 𝜃0 =10◦

cos 𝜃0

=

ΩA = Θh Ψh = 61.09(47.50) = 2,901.775 (degrees)2 = 0.884 (rads)2

D0 ≃

32,400 32,400 = = 11.17 = 10.48 dB and it agrees closely with the 2 2,901.775 ΩA (degrees) more accurate value of 11.817 or 10.725 dB

6.94. In the design of an array, the maximum accurs at 𝜃 = 𝜃0 at the design frequency f = f0 which has been used to determine the progressive phase between the elements. As the shifts from f0 , the maximum also shifts to some other angle 𝜃 array from 𝜃0 .

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219

At a frequency fh , the maximum of the array factor will be 0.707 of the normalized maximum value of unity. The frequency fh is referred to as the half-power frequency, and it is used to determine the frequency bandwidth over which the pattern varies over an amplitude of 3-dB from the maximum at f0 . To determine the frequency fh and the 3-dB frequency bandwidth, the normalized array factor of (6-10c) is written using (6-21) as [

] N𝜋d (cos 𝜃 − cos 𝜃0 ) 1 λ AF = [ ] 𝜋d N sin (cos 𝜃 − cos 𝜃0 ) λ sin

The frequency is obscured in the array factor. To be valid over a band of frequencies, the wavelengths λ and λ0 and the frequencies f and f0 should be shown explicitly. Using the relation 𝜐 = λf , the array factor can be written as [

] 𝜋Nd (f cos 𝜃 − f0 cos 𝜃0 ) 1 𝜐 AF = [ ] 𝜋d N sin (f cos 𝜃 − f0 cos 𝜃0 ) 𝜐 sin

which peaks at 𝜃 = 𝜃0 when f = f0 . At any other frequency, the array factor peaks when f cos 𝜃 − f0 cos 𝜃0 = 0 ⇒ cos 𝜃 =

f0 cos 𝜃0 f

The half-power of the array factor is obtained by letting 𝜃 = 𝜃0 and occurs N𝜋d N𝜋d (f cos 𝜃0 − f0 cos 𝜃0 ) = cos 𝜃0 (fh − f0 ) = 1.391 𝜐 h 𝜐 or (fh − f0 ) =

1.391𝜐 0.886𝜐 0.886𝜐 = = N𝜋 d cos 𝜃 Nd cos 𝜃0 (L + d) cos 𝜃

Therefore the 3-dB frequency bandwidth is BW(3-dB) =

0.886𝜐 0.886𝜐 = Nd cos 𝜃0 (L + d) cos 𝜃0

Therefore the bandwidth of an array depends not on the frequency operation but rather on the array length and scan angle. This is a fundamental constraint on wide-instantaneous bandwidth of arrays.

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7

Solution Manual

7.1. Using (7-4), the array factor can be written as AF = a1 + a2 z + a3 z2 = 1 + 2z + z2 = (1 + z)2 which has two roots and both occur at z = −1. (a) The nulls of the array factor can be found by setting z equal to −1. Thus z = ej(kd cos 𝜃+𝛽) = −1 ⇒ kd cos 𝜃 + 𝛽 = ±(2n + 1)𝜋, n = 0, 1, 2, … For d = λ∕4 ( ) λ 𝜋 cos 𝜃 + 𝛽 = cos 𝜃 + 𝛽 = ±(2n + 1)𝜋, n = 0, 1, 2, … 4 2 [ ] 2 (−𝛽 + (2n + 1)𝜋) , n = 0, 1, 2, … 𝜃 = cos−1 𝜋

kd cos 𝜃 + 𝛽 =

2𝜋 λ

For 𝛽 = 0; 𝜃 = cos−1 [±2(2n + 1)] = does not exist; no zeros. For 𝛽 =

𝜋 ; 2 𝜃 = cos−1 [2(−0.5 ± (2n + 1))] For n = 0 ⇒ 𝜃 = cos−1 [2(−0.5 ± 1)]

Using the positive sign between the two terms 𝜃 = cos−1 (+1) = 0◦ For 𝛽 = 𝜋; 𝜃 = cos−1 [2(−1 ± (2n + 1))] For n = 0 ⇒ 𝜃 = cos−1 [2(−1 ± 1)] = cos−1 (0) = 90◦ Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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SOLUTION MANUAL

For 𝛽 = 3𝜋∕2; 𝜃 = cos−1 [2(−1.5 ± (2n + 1))] For n = 0 ⇒ 𝜃 = cos−1 [2(−1.5 ± 1)] = cos−1 (−1) = 180◦ λ The computed patterns for d = and 𝛽 = 0, 𝜋∕2, 𝜋, 3𝜋∕2 are shown plotted in Fig. P7.1. 4 The nulls do occur at the computed angles. 0 30

30

beta = 0 beta = 90 beta = 180 beta = 270

60

60

90

90

–30 dB

–20 dB 120

120 –10 dB

150

0 dB

150

180

Figure P7.1

7.2. For 𝛽 = 0 and d = λ∕4 𝜓 = kd cos 𝜃 + 𝛽 =

2𝜋 λ

( ) λ 𝜋 cos 𝜃 = cos 𝜃 4 2

which reduces to 𝜃 = 0◦ : 𝜃 = 60◦ : 𝜃 = 120◦ :

𝜋 2 𝜋 𝜓 = 𝜓2 = 2 𝜋 𝜓 = 𝜓3 = 2

𝜓 = 𝜓1 =

⇒ z1 = j ( ) √ 1 𝜋 = ⇒ z2 = (1 + j)∕ 2 2 4 ( ) √ 1 𝜋 − = − ⇒ z3 = (1 − j)∕ 2 2 4

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SOLUTION MANUAL

Therefore the array factor of (7-5) can be written as (

)(

1 AF = (z − z1 )(z − z2 )(z − z3 ) = (z − j) z − √ (1 + j) 2 √ √ = −j + (1 + j 2)z − ( 2 + j)z2 + z3

)

1 z − √ (1 − j) 2

AF = (1)ej3𝜋∕2 + (1.732)ej0.9553 z + (1.732)ej(3.757) z2 + (1)z3 Another Form is AF = (1)e−j𝜋∕2 + (1.732)ej0.9553 z + (1.732)e−j2.526 z2 + (1)z3 AF = a1 + a2 z + a3 z2 + a4 z3 (a) Four (N = 4) elements are required. (b) The excitation coefficients are equal to a1 = (1)ej3𝜋∕2 = e−j𝜋∕2 = 1 ∠3𝜋∕2 = 1 ∠ − 𝜋∕2 a2 = (1.732)ej0.9553 = 1.732 ∠0.9553 a3 = (1.732)ej3.757 = (1.732)e−j2.526 = 1.732 ∠3.757 = 1.732 ∠ − 2.526 a4 = 1 = 1 ∠0 (c) The array factor is given by any of the two above forms. (d) The array factor is plotted and it is shown below. 0 30

30

60

60

90

90

–30 dB

–20 dB 120

120 –10 dB

150

0 dB 180

Figure P7.2

150

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SOLUTION MANUAL

7.3. (a) Schelkunoff Method 𝜓=

𝜋 cos 𝜃 2

Fig. 7.2(b)

𝜋 𝜋 cos 𝜃|𝜃=0 = ⇒ z1 = +j 2 2 ( ) 𝜋 𝜋 𝜋 1 1 = ⇒ z2 = √ (1 + j) = cos 𝜃|𝜃=60◦ = 2 2 2 4 2 ( ) 𝜋 𝜋 𝜋 𝜋 − 12 = − ⇒ z3 = √ (1 − j) = cos 𝜃|𝜃=120◦ = 2 2 4 2

𝜓|𝜃=0 = 𝜓|𝜃=60◦ 𝜓|𝜃=120◦

𝜓|𝜃=180◦ =

AF =

N ∑

𝜋 𝜋 𝜋 cos 𝜃|𝜃=180◦ = (−1) = − ⇒ z4 = −j 2 2 2

an zn−1 = (z − z1 )(z − z2 )(z − z3 )(z − z4 )

n=1

][ ] 1 1 = (z − j) z − √ (1 + j) z − √ (1 − j) (z + j) 2 2 [( ) ] [( ) ] 1 1 1 1 2 2 z− √ − j√ z− √ + j√ = (z − j ) 2 2 2 2 ( )2 ⎡ ⎤ ] [ √ 1 = (z2 + 1) ⎢ z − √ + 12 ⎥ = (z2 + 1) z2 − 2z + 12 + 12 ⎢ ⎥ 2 ⎣ ⎦ √ √ √ = (z2 + 1)(z2 − 2z + 1) = (z4 − 2z3 + z2 + z2 − 2z + 1) 4

AF = (z −

[

√

N √ ∑ 2z + 2z − 2z + 1) = an zn−1 3

2

n=1

N=5 √ √ 2 3 4 2 3 4 (b) AF = a1 + a2 z + a3 z + a4 z + a5 z = 1 − 2z + 2z − 2z + z √ √ a1 = 1, a2 = − 2 = −1.414, a3 = 2, a4 = − 2 = −1.414, a5 = 1 2𝜋 7.4. For d = λ∕2, 𝛽 = 0 ⇒ 𝜓 = kd cos 𝜃 + 𝛽 = λ 𝜓 = 𝜋 cos 𝜃 𝜃 = 60◦ : 𝜃 = 90◦ : 𝜃 = 120◦ :

( ) λ cos 𝜃 = 𝜋 cos 𝜃 2

𝜓1 = 𝜋 cos(60◦ ) = 𝜋(1∕2) = 𝜋∕2 𝜓2 = 𝜋 cos(90◦ ) = 0 𝜓3 = 𝜋 cos(120◦ ) = 𝜋(−1∕2) = −𝜋∕2 ⇒ s1 = +j ⎫ AF = (z − s1 )(z − s2 )(z − z3 ) ⎪ = (z − j)(z − 1)(z + j) = (z − 1)(z2 + 1) ⇒ s2 = 1 ⎬ ⇒ s3 = −j ⎪ AF = z3 − z2 + z − 1 ⎭

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SOLUTION MANUAL

(a) 4 elements (b) AF = a1 + a2 z + a3 z2 + a4 z3 = −1 + z − z2 + z3 a1 = −1, a2 = +1, a3 = −1, a4 = +1 (c) Nulls @: Element Factor: 𝜃 = 0◦ , 180◦ Array Factor: 𝜃 = 60◦ , 90◦ , 120◦ 7.5. d = 3λ∕8, 𝛽 = 0 𝜓 = kd cos 𝜃 + 𝛽 = 𝜓=

2𝜋 λ

(

) 3λ cos 𝜃 8

3𝜋 cos 𝜃 4

[ ] 3𝜋 1 𝜓= = 135◦ z1 = √ (−1 + j) 4 2

𝜃 = 0◦ : 𝜃 = 90◦ :

𝜓 = 0 [z2 = 1] ] [ 3𝜋 1 ◦ 𝜃 = 180 : 𝜓 = − z3 = √ (−1 − j) 4 2

y

z - plane z = x + jy

Visible Region

θ = 0° z1 = – 1 + j 1 √2 √z

135°

ψ

z3 = – 1 + j 1 √2 √2

θ = 180°

–135°

θ = 90° z2 (1, 0)

x

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SOLUTION MANUAL

(a) AF = (z − z1 )(z − z2 )(z − z3 ) [ ] [ ] 1 1 = z − √ (−1 + j) [z − 1] z − √ (−1 − j) 2 2 √ = (z − 1)(z2 − 2z + 1) AF = z3 + 0.414z2 − 0.414z − 1 AF = a4 z3 + a3 z2 + a2 z + a1 Number of elements = N = 4 (b) a1 = −1, a2 = −0.414, a3 = 0.414, a4 = 1 ) ( 3𝜋 5𝜋 (c) 135◦ < 𝛽 < 225◦ cos 𝜃 cos 𝜙 + My cos 𝜃 sin 𝜙 −   > sin 𝜃]ejk(y′ sin 𝜃 sin 𝜙+z′ cos 𝜃) dy′ dz′  [ M M x z

Sa

= 2E0 cos 𝜃 sin 𝜙

a∕2

∫−a∕2

ejky

′ sin 𝜃 sin 𝜙

b∕2

dy′

∫−b∕2

′ cos 𝜃

ejkz

dz′

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SOLUTION MANUAL

] [ ka sin Y sin Z kb ,Y = L𝜃 = 2E0 ab cos 𝜃 sin 𝜙 sin 𝜃 sin 𝜙, Z = cos 𝜃 Y Z 2 2 0

L𝜙 =

∬

> sin 𝜙 + My cos 𝜙]ejk(y′ sin 𝜃⋅sin 𝜙+z′ cos 𝜃) dy′ dz′  [− M x

Sa

] [ sin Y sin Z = 2E0 ab cos 𝜙 Y Z Using (12-10a)-(12-10f)

Er ≃ 0 ] 0 abkE0 e−jkr [ sin Y sin Z ke−jkr cos 𝜙 N 𝜃 ] = −j [L𝜙 + 𝜂 4𝜋r 2𝜋r Y Z [ ] −jkr −jkr 0 abkE0 e sin Y sin Z ke − cos 𝜃 sin 𝜙 E𝜙 ≃ j N 𝜙 ] = −j [L𝜃 − 𝜂 4𝜋r 2𝜋r Y Z E𝜙 E Hr ≃ 0, H𝜃 ≃ − , H𝜙 ≃ + 𝜃 𝜂 𝜂 E𝜃 ≃ −j

12.9. Ea = â x E0 , M s = −2̂n × Ea = −2̂ay × â x E0 = â z 2E0 Thus Mz = 2E0 , Mx = My = Jx = Jy = Jz = 0 N𝜃 = N𝜙 = 0 0

L𝜃 =

∬

0

> cos 𝜃 cos 𝜙 +   > cos 𝜃 sin 𝜙 − Mz sin 𝜃]ejk(x′ sin 𝜃 cos 𝜙+z′ cos 𝜃) dx′ dz′  [ M M x y

Sa

= −2E0 sin 𝜃

a∕2

∫−a∕2

′ cos 𝜃

ejkz

b∕2

dz′

∫−b∕2

ejkx

′ sin 𝜃 cos 𝜙

dx′

] [ kb sin x sin Z ka , X= L𝜃 = −2E0 ab sin 𝜃 sin 𝜃 cos 𝜙, Z = cos 𝜃 x Z 2 2 0

L𝜙 =

∬

0

> sin 𝜙 +   > cos 𝜙]ejk(x′ sin 𝜃 cos 𝜙+z′ cos 𝜃) dx′ dz′ = 0  [− M M x y

Sa

Using (12-10a)-(12-10f)

Er ≃ 0,

E𝜃 ≃ −

0 jke−jkr 0 N𝜃 ] = 0 [L

𝜙 + 𝜂 4𝜋r

] 0 abkE0 e−jkr [ sin X sin Z ke−jkr sin 𝜃 N 𝜙 ] = −j [L𝜃 − 𝜂 4𝜋r 2𝜋r X Z E𝜙 E𝜃 Hr ≃ 0, H𝜃 ≃ − , H𝜙 ≃ =0 𝜂 𝜂

E𝜙 ≃ j

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SOLUTION MANUAL

12.10. The only difference between this problem and Problem 12.8 is that for the y variations the integral reduces to ( ) ka cos sin 𝜃 sin 𝜙 ) ′ ( ( ) a∕2 𝜋 ′ jky sin 𝜃 sin 𝜙 ′ 𝜋a 2 cos dy = − y e )2 ( )2 ∫−a∕2 a 2 ( ka 𝜋 sin 𝜃 sin 𝜙 − 2 2 Thus Er ≃ 0

[ ] abkE0 e−jkr ke−jkr sin Z cos Y ka L𝜙 = j cos 𝜙 2 , Y= sin 𝜃 sin 𝜙 4𝜋r 4r 2 (Y) − (𝜋∕2)2 Z [ ] abkE0 e−jkr ke−jkr sin Z cos Y E𝜙 ≃ j L𝜃 = +j − cos 𝜃 sin 𝜙 2 4𝜋r 4r (Y) − (𝜋∕2)2 Z E𝜃 ≃ −j

kb cos 𝜃 2 Hr ≃ 0, H𝜃 ≃ −E𝜙 ∕𝜂, Z=

H𝜙 ≃ E𝜃 ∕𝜂

12.11. The only difference between this problem and Problem 12.9 is that for the z variations the integral reduces to ( ) ka cos cos 𝜃 ) ′ ( ( ) a∕2 𝜋 ′ jkz cos 𝜃 ′ 𝜋a 2 cos dz = − z e )2 ( )2 ∫−a∕2 a 2 ( ka 𝜋 cos 𝜃 − 2 2 Thus Er ≃ 0 E𝜃 ≃ 0 abkE0 e−jkr ke−jkr E𝜙 ≃ j L𝜃 = j 4𝜋r 4r

⎤ ⎡ ⎥ ⎢ sin X cos Z ⎢sin 𝜃 X ( )2 ⎥ 𝜋 ⎥ ⎢ (Z)2 − ⎣ 2 ⎦

kb ka sin 𝜃 cos 𝜙, Z = cos 𝜃 2 2 Hr ≃ 0, H𝜃 ≃ −E𝜙 ∕𝜂, H𝜙 ≃ E𝜃 ∕𝜂 = 0 X=

12.12. (a) Ea = â z E0 , M s = −̂n × Ea = â y E0 ⇒ Mx = Mz = 0, My = E0 ( ) E E E H a = −̂ay 0 , J s = n̂ × H a = â x × −̂ay 0 = −̂az 0 𝜂 𝜂 𝜂 ⇒ Jx = Jy = 0, Jz = −E0 ∕𝜂 From Prob. 12.8. ] [ ka sin Y sin Z kb , Y= L𝜃 = E0 ab cos 𝜃 sin 𝜙 sin 𝜃 sin 𝜙, Z = cos 𝜃 Y Z 2 2

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SOLUTION MANUAL

353

] [ sin Y sin Z L𝜙 = E0 ab cos 𝜙 Y Z 0

N𝜃 =

0

 cos 𝜃 cos 𝜙 +
J
y cos 𝜃 sin 𝜙 − Jz sin 𝜃]ejk(y [J
∬
x

′ sin 𝜃 sin 𝜙+z′ cos 𝜃)

dy′ dz′

Sa

=+

] E0 ab [ sin Y sin Z sin 𝜃 𝜂 Y Z 0

N𝜙 =

0

x sin 𝜙 +
J
y cos 𝜙]ejk(y [−
J
∬

′ sin 𝜃 sin 𝜙+z′ cos 𝜃)

dy′ dz′ = 0

Sa

Er ≃ 0 ] kabE0 e−jkr [ sin Y sin Z ke−jkr (cos 𝜙 + sin 𝜃) [L𝜙 + 𝜂N𝜃 ] = −j 4𝜋r 4𝜋r Y Z ] −jkr [ −jkr 0 kabE e sin Y sin Z ke 0 − cos 𝜃 sin 𝜙 E𝜙 ≃ j N 𝜙 ] = −j [L𝜃 − 𝜂 4𝜋r 4𝜋r Y Z Hr ≃ 0 E𝜃 ≃ −j

H𝜃 ≃ −E𝜙 ∕𝜂 H𝜙 ≃ E𝜃 ∕𝜂 (b) E = â x E0 , a

M s = −̂n × Ea = â z E0 ⇒ Mx = My = 0, Mz = E0 ( ) E E0 E0 E H a = −̂az , J s = n̂ × H a = â y × −̂az = −̂ax 0 ⇒ Jy = Jz = 0, Jx = − 0 𝜂 𝜂 𝜂 𝜂 From Problem 12.9 ] [ sin X sin Z L𝜃 = −E0 ab sin 𝜃 X Z L𝜙 = 0 X=

kb ka sin 𝜃 cos 𝜙, Z = cos 𝜃 2 2 0

N𝜃 =

∬

0

[Jx cos 𝜃 cos 𝜙 +
J
y cos 𝜃 sin 𝜙 −
J
z sin 𝜃]ejk(x

′ sin 𝜃 cos 𝜙+z′ cos 𝜃)

Sa

=− N𝜙 =

] E0 [ sin X sin Z ab cos 𝜃 cos 𝜙 𝜂 X Z

∬

[−Jx sin 𝜙 + Jy cos 𝜙]ejk(x

′ sin 𝜃 cos 𝜙+z′ cos 𝜃)

dx′ dz′

Sa

=

] E0 [ sin X sin Z ab sin 𝜙 𝜂 X Z

Er ≃ 0 −jkr

E𝜃 ≃ −j

kabE0 e ke−jkr 
0 [L
+ 𝜂N𝜃 ] = j 4𝜋r 𝜙 4𝜋r

[

cos 𝜃 cos 𝜙

sin X sin Z X Z

]

dx′ dz′

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SOLUTION MANUAL

] kabE0 e−jkr [ sin X sin Z ke−jkr −(sin 𝜃 + sin 𝜙) [L𝜃 − 𝜂N𝜙 ] = j 4𝜋r 4𝜋r X Z Hr ≃ 0, H𝜃 ≃ −E𝜙 ∕𝜂, H𝜙 ≃ E𝜃 ∕𝜂

E𝜙 ≃ j

(c) Using a comparison between the fields of Problems 12.8 and 12.12(a). We can write by referring to the fields of Problem 12.10 that

Er ≃ 0

[ ] abkE0 e−jkr sin Z cos Y E𝜃 ≃ j (cos 𝜙 + sin 𝜃) 2 8r (Y) − (𝜋∕2)2 Z [ ] abkE0 e−jkr cos Y − cos 𝜃 sin 𝜙 E𝜙 ≃ j 8r (Y)2 − (𝜋∕2)2 Y=

kb ka sin 𝜃 sin 𝜙, Z = cos 𝜃, Hr ≃ 0, H𝜃 ≃ −E𝜙 ∕𝜂, H𝜙 ≃ E𝜃 ∕𝜂 2 2

(d) Using a comparison between the fields of Problems 12.9 and 12.12(b) we can write by referring to the fields of Problem 12.11 that

Er ≃ 0

[ ] kabE0 e−jkr sin X cos Z E𝜃 ≃ j cos 𝜃 cos 𝜙 8r X (Z)2 − (𝜋∕2)2 [ ] kabE0 e−jkr sin X cos Z −(sin 𝜃 + sin 𝜙) E𝜙 ≃ j 8r X (Z)2 − (𝜋∕2)2 ka ka sin 𝜃 cos 𝜙, Z = cos 𝜃 2 2 Hr ≃ 0, X=

H𝜃 ≃ −E𝜙 ∕𝜂 H𝜙 ≃ E𝜃 ∕𝜂 ) ) ( 𝜋 ′ 𝜋 ′ x ⇒ M s = −̂n × Ea = â x E0 cos x a( a ) 𝜋 ′ Thus Mx = E0 cos x , My = Mz = 0 a

12.13. Ea = â y E0 cos

(

b∕2

L𝜃 =

a∕2

∫−b∕2 ∫−a∕2

Mx cos 𝜃 cos 𝜙ejk(x

= E0 cos 𝜃 cos 𝜙

(

a∕2

∫−a∕2

cos

′ sin 𝜃 cos 𝜙+y′ sin 𝜃 sin 𝜙)

dx′ dy′

) b∕2 ′ 𝜋 ′ jkx′ sin 𝜃 cos 𝜙 ′ dx ejky sin 𝜃 sin 𝜙 dy′ x e ∫ a −b∕2

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SOLUTION MANUAL

Since

∫−a∕2

cos

b∕2

∫−b∕2

) ( ) cos(X) 𝜋 ′ jkx′ sin 𝜃 cos 𝜙 ′ 𝜋a dx = − x e a 2 (X)2 − (𝜋∕2)2

(

a∕2

ejky

′ sin 𝜃 sin 𝜙

dy′ = b

sin Y , Y

X=

ka kb sin 𝜃 cos 𝜙, Y = sin 𝜃 sin 𝜙 2 2

then

L𝜃 = −

] [ 𝜋ab sin Y cos X E0 cos 𝜃 cos 𝜙 2 2 (X) − (𝜋∕2)2 Y

Similarly b∕2

L𝜙 =

a∕2

∫−b∕2 ∫−a∕2

−Mx sin 𝜙ejk(x

′ sin 𝜃 cos 𝜙+y′ sin 𝜃 sin 𝜙)

dx′ dy′

) b∕2 ′ 𝜋 ′ jkx′ sin 𝜃 cos 𝜙 ′ dx ejky sin 𝜃 sin 𝜙 dy′ x e ∫−a∕2 ∫ a −b∕2 ] [ 𝜋ab sin Y cos X L𝜙 = + E sin 𝜙 2 2 2 0 (X) − (𝜋∕2) Y ( ) ) ( E E E E 𝜋 ′ x H a ≃ −̂ax a ⇒ J s = n̂ × H a = â z × −̂ax a = −̂ay a = −̂ay 0 cos 𝜂 𝜂 𝜂 𝜂 a = −E0 sin 𝜙

cos

Jx = Jz = 0, Jy = − b∕2

N𝜃 =

a∕2

∫−b∕2 ∫−a∕2

(

a∕2

) ( E0 𝜋 ′ cos x 𝜂 a

Jy cos 𝜃 sin 𝜙ejk(x

′ sin 𝜃 cos 𝜙+y′ sin 𝜃 sin 𝜙)

dx′ dy′

) ( a∕2 b∕2 E0 ′ 𝜋 ′ jkx′ sin 𝜃 cos 𝜙 ′ cos dx ejky sin 𝜃 sin 𝜙 dy′ cos 𝜃 sin 𝜙 x e ∫−a∕2 ∫−b∕2 𝜂 a [ ] 𝜋abE0 sin Y cos X =+ cos 𝜃 sin 𝜙 2 2𝜂 (X) − (𝜋∕2)2 Y =−

b∕2

N𝜙 =

a∕2

∫−b∕2 ∫−a∕2

=+

Jy cos 𝜙ejk(x

′ cos 𝜙 sin 𝜃+y′ sin 𝜃 sin 𝜙)

[ ] 𝜋abE0 sin Y cos X cos 𝜙 2 2𝜂 (X) − (𝜋∕2)2 Y

dx′ dy′

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SOLUTION MANUAL

Using (12.10a)-(12.10f), we can write that Er ≃ 0

[ ] abkE0 e−jkr ke−jkr sin Y cos X E𝜃 ≃ −j [L + 𝜂N𝜃 ] = −j sin 𝜙(1 + cos 𝜃) 2 4𝜋r 𝜙 8r (X) − (𝜋∕2)2 Y [ ] abkE0 e−jkr ke−jkr sin Y cos X E𝜙 ≃ j [L − 𝜂N𝜙 ] = −j cos 𝜙(1 + cos 𝜃) 2 4𝜋r 𝜃 8r (X) − (𝜋∕2)2 Y Hr ≃ 0, H𝜃 = −E𝜙 ∕𝜂, H𝜙 = E𝜃 ∕𝜂 12.14. a = 4λ, b = 3λ (a) From Appendix I ] ] [ [ 2(7.7) 2(7.7) kb sin 𝜃s ≃ 7.7 ⇒ 𝜃s = sin−1 = sin−1 = 54.785◦ 2 kb 2𝜋(3) Θs = 2𝜃s = 2(54.785) = 109.57◦ kb sin 𝜃s = 7.7 ⇒ E𝜃 = 0.12833 or E𝜃 = −17.83 dB 2 ) ( (c) 0.443 41,253 = 16.98◦ ⎫ D ≃ Θh (E-plane) = 114.6 sin−1 ⎪ 0 16.98(12.72) = 191 3 ) ( ⎬ 0.443 = 12.72◦ ⎪ Θh (H-plane) = 114.6 sin−1 = 22.81 dB ⎭ ( )4 ab From Table 12.1, D0 = 10.2 2 = 10.2(3)(4) = 122.4 = 20.88 dB λ

(b) From Appendix I, at

12.15.

a = 0.9′′ = 2.286 cm, b = 0.4′′ = 1.016 cm, f = 10 GHz ⇒ λ = 30 × 109 ∕10 × 109 = 3 cm (a) D0 =

4𝜋 4𝜋 4𝜋 Aem = 2 [0.81(ab)] = [0.81(2.286)(1.016)] = 2.6268 λ2 λ (3)2 = 4.194 dB (Table 12.1)

1. From Table 12.1: D0 = 2.6268 = 4.194 dB 2. From Table 12.1: E(HPBW) =

50 50 = = 147.638◦ b∕λ 1.016∕3

H(HPBW) =

68.8 68.8 = = 90.2887◦ a∕λ 2.286∕3

D0 (Kraus) =

41,253 = 3.0947 = 4.9062 dB (147.638)(90.2887)

(b) Aem = 0.81(ab) = 0.81(2.286)(1.016) = 1.881286 cm2 PL = Wi Aem = 10 × 10−3 (1.881286) = 18.81286 × 10−3 Watts

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SOLUTION MANUAL

12.16. G0 = 11 dB, Ap = 2λ2 ⇒ G0 (dimensionless) = 101.1 = 12.5893 4𝜋 4𝜋 4𝜋 (a) G0 = ecd D0 = D0 = 2 Aem = 2 (𝜀ap AP ) = 2 (𝜀ap 2λ2 ) = 𝜀ap 8𝜋 = 12.5893 λ λ λ 12.5893 𝜀ap = = 0.5 = 50% 8𝜋 (b) PL = Aem Wi = 𝜀ap Ap Wi = 0.5(2λ2 )(10 × 10−3 ) = 10 × 10−3 λ2 f = 10 GHz ⇒ λ = 30 × 109 ∕10 × 109 = 3 cm PL = 10 × 10−3 (3)2 = 90 × 10−3 = 90 m Watts 12.17. a = b = 3λ 4𝜋 Using (12-37), D0 = 2 ab = 4𝜋(3)2 = 113.1 = 20.53 dB λ Using the computer program Directivity of Chapter 2. D0 = 119.46 = 20.77 dB 12.18. a = b = 3λ 4𝜋 Using (12-37), D0 = 2 ab = 4𝜋(3)2 = 113.1 = 20.53 dB λ Using the computer program Directivity of Chapter 2. D0 = 119.38 = 20.77 dB 12.19. Using the computer program Aperture of Chapter 12. (a) a = 3λ, b = 2λ; D0 = 62.437 = 17.95 dB (b) a = b = 3λ; D0 = 93.174 = 19.69 dB Using Table 12.1 (a) a = 3λ, b = 2λ; D0 = 0.81(4𝜋)(3)(2) = 61.07 = 17.85 dB (b) a = b = 3λ; D0 = 0.81(4𝜋)(3)(3) = 91.61 = 19.62 dB 12.20. Using the computer program Aperture of Chapter 12. (a) a = 3λ, b = 2λ; D0 = 63.961 = 18.06 dB (b) a = b = 3λ; D0 = 94.306 = 19.75 dB 12.21. a = 3λ, b = 2λ (a) Θh (E-plane) = 50.6∕2 = 25.30◦ (b) Θh (H-plane) = 68.8∕3 = 22.93◦ (c) Θn (E-plane) = 114.6∕2 = 57.30◦ (d) Θn (H-plane) = 171.9∕3 = 57.30◦ (e) E𝜃 (E-plane) = −13.26 dB (f) E𝜙 (H-plane) = −23 dB Using the data from Figures 12.13 and 12.14 (a) (b) (c) (d)

Θh (E-plane) = 25.6◦ Θh (H-plane) = 21◦ Θn (E-plane) = 60◦ Θn (H-plane) = 60◦

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SOLUTION MANUAL

(e) E𝜃 (E-plane) = −13.26 dB (f) E𝜙 (H-plane) = −26 dB 12.22. From Figure 12.15, for a 90% efficiency u=

ka kb sin 𝜃1 ≃ 3.18 = sin 𝜃1 2 2

For 𝜃1 = 37∕2 = 18.5◦ a=b= 12.23.

) ( E0 𝜋 ′ cos x 𝜂 a ( ) 2 |E | 𝜋 ′ 1 = Re(E × H ∗ ) = â z 0 cos2 x 2 2𝜂 a

Ea = â y E0 cos W ave

(

) 𝜋 ′ x , a

Prad

H a ≃ −̂ax

) ( 2𝜋 ′ ⎤ ⎡ x ( ) a∕2 1 + cos 2 2b |E |E | | ⎢ ⎥ ′ 𝜋 ′ 1 0 a = cos2 x dx′ dy′ = 0 ⎥ dx 2 ∫−b∕2 ∫−a∕2 𝜂 a 2𝜂 ∫−a∕2 ⎢⎢ 2 ⎥ ⎣ ⎦ [ )] ( 2 2 |E | b ′ ab|E0 | a 2𝜋 ′ a∕2 x + = 0 = sin x 4𝜂 2𝜋 a 4𝜂 −a∕2 b∕2

Prad

2(3.18) = 3.19λ k sin(18.5◦ )

a∕2

From Table 12.1, at 𝜃 = 0◦ −jkr −jkr 2 abkE0 e 2 abkE0 e 2abe−jkr sin 𝜙, E𝜙 |max = j cos 𝜙 = j cos 𝜙 𝜋 2𝜋r 𝜋 2𝜋r 𝜋λr √ ( ) 2(ab)2 |E0 |2 4 ab 2 r2 |E|max = |E𝜃 |2max + |E𝜙 |2max = |E0 | 2 ⇒ Umax = |E|2max = 2𝜂 r 𝜋λ (𝜋λ)2 𝜂 [ ( )] 4𝜋[2(ab)2 |E0 |2 ]∕[(𝜋λ)2 𝜂] 4𝜋Umax 4𝜋 8 ab 2 = D0 = = Prad ab|E0 |2 ∕4𝜂 𝜋2 λ

E𝜃 |max = j

Aem = 12.24.

λ2 8 8 (ab) = 2 Ap = 𝜀ap Ap D = 4𝜋 0 𝜋 2 𝜋

4𝜋 4𝜋 4𝜋 Aem = 2 𝜀ap Ap = 𝜀ap 2 (ab) λ2 λ λ ( 2) 4𝜋(ab) 4𝜋 λ 𝜋 D0 = 𝜀ap = 𝜀ap 2 = 𝜀ap 2 8 2 λ λ

D0 =

(a) Triangular: 𝜀ap = 75% = 3∕4 ( ) 3 𝜋 = 1.1781 = 0.7118 dB 4 2 D0 = 1.1781 = 0.7118 dB

D0 =

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SOLUTION MANUAL

(b) Cosine Square: 𝜀ap = 66.67% = 2∕3 ( ) 𝜋 𝜋 2 𝜋 = = 1.0472 = 0.2 dB = 2 3 2 3 D0 = 1.0472 = 0.2 dB

D0 = 𝜀ap

12.25.

D0 = 23 dB = 102.3 = 199.526 e0 = ecd er = 0.9(1) = 0.9 G0 = e0 D0 = 0.9(199.526) = 179.5736 = 22.54 dB Aem =

λ2 λ2 e0 D0 = G 4𝜋 4𝜋 0

Aem =

λ2 32 (179.5736) = (179.5736) = 128.61 cm2 4𝜋 4𝜋

3 × 108 = 0.03 m = 3 cm 10 × 109 A 128.61 = em = = 0.6431 = 64.31% Ap 200

λ= 𝜀ap

12.26. (a) λ =

3 × 1010 = 3 cm 10 × 109

a = 0.9′′ = 2.286 cm = 0.762λ b = 0.4′′ = 1.016 cm = 0.339λ Prad =1W

f =10 GHz

#1

#2

Load

10 km

Power density for isotropic source; Prad

W0 =

4𝜋R2

=

1 Watt = 7.96 × 10−10 W∕m2 4𝜋(10 × 103 )2

Directivity from Table 12.1, 12.2. D0 =

[ ] 8 4𝜋 32 ab = (0.762)(0.339) = 2.63 2 2 𝜋 𝜋 λ

Incident power density Wi = W0 D0 = (7.96 × 10−10 W∕m2 )(2.63) ⇒ Wi = 2.09 × 10−9 W∕m2 (b) The maximum power that can be delivered to a matched load. Aem = 𝜀ap Ap = 0.81ab = 1.88 × 10−4 m2 Pmax = Wi Aem = (2.09 × 10−9 W∕m2 )(1.88 × 10−4 m2 ) = 3.94 × 10−13 W Pmax = 3.94 × 10−13 W

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SOLUTION MANUAL

(

kb cos 𝜃 𝜔𝜇bI0 2 12.27. E = â 𝜃 j kb 4𝜋r cos 𝜃 2 𝜔𝜇I0 e−jkr (a) E = −j le (𝜃) 4𝜋r ( ) kb sin cos 𝜃 2 le (𝜃) = −̂a𝜃 b kb cos 𝜃 2 e−jkr

)

sin

= −j

𝜔𝜇I0 4𝜋r

e−jkr

( ) kb ⎫ ⎧ sin cos 𝜃 ⎪ ⎪ 2 ⎬ ⎨−̂a𝜃 b kb ⎪ cos 𝜃 ⎪ ⎭ ⎩ 2

x b

a

z

y

(

kb cos 𝜃 2 (b) le (𝜃)|max = −̂a𝜃 b kb cos 𝜃 2 kb when cos 𝜃 = 0 ⇒ 𝜃 = 90◦ 2 |l (𝜃) ⋅ Einc |2 (c) pe = e |le (𝜃)|2 |Einc |2 kI le−jkr Einc = â 𝜃 j𝜂 0 sin 𝜃 8𝜋r sin

)

| | = −̂a𝜃 b(1) | | |max

| |2 [ ] −jkr | | kI le | | sin 𝜃 |le (𝜃) ⋅ Einc |2 = |(−̂a𝜃 b) ⋅ â 𝜃 j𝜂 0 | 𝜋 | 8𝜋r 𝜃 = || | | 2| 2 | | bkI0 l |2 kI le−jkr || | | = |−j𝜂b 0 | = ||𝜂 | | 8𝜋r || | 8𝜋r | | |le (𝜃)|2 = |b|2 |2 | | kI le−jkr | | kI0 l |2 | | | 0 | |Einc |2 = |j𝜂 sin 𝜃 | | = ||𝜂 | | | 𝜋| 8𝜋r | 8𝜋r | | |𝜃 = | 2| | kbI0 l |2 |𝜂 | | 8𝜋r | | | = 1 = 0 dB pe = | kI0 l |2 | |b|2 ||𝜂 | | 8𝜋r |

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SOLUTION MANUAL

12.28. Aem =

[ ( )] 4𝜋 λ2 λ2 0.81ab 2 = 0.81ab D0 = 4𝜋 4𝜋 λ

Aem = 0.81(0.02286)(0.0106) = 0.81(2.32257 × 10−4 ) = 1.88 × 10−4 m2 The maximum power that can be delivered to matched load is Pmax = Wi Aem = (10−4 Watts∕m2 )(1.88 × 10−4 m2 ) = 1.88 × 10−8 Watts = 0.0188 𝜇Watts. ( ) 2 ′ a ⎧ a ̂ 2E + 1 , − ⩽ x′ ⩽ 0 x ⎪ y 0 a 2 12.29. (a) Ea = ⎨ ( ) 2 a ⎪ â y 2E0 − x′ + 1 , 0 ⩽ x′ ⩽ ⎩ a 2 M s = −2̂n × Ea = −2̂az × â y Ey = â x 2Ey My = Mz = Jx = Jy = Jz = 0 ( ) 2 ′ a ⎧ + 1 , − ⩽ x′ ⩽ 0 2E x ⎪ 0 a 2 Mx = ⎨ ( ) 2 a ⎪ 2E0 − x′ + 1 , 0 ⩽ x′ ⩽ ⎩ a 2 y

Sa

b

x a/2

a/2 z

Figure P12.29(a)

Using (12-12a)–(12-12d) N𝜃 = N𝜙 = 0 0

L𝜃 =

∬ s

L𝜃 = 2E0 cos 𝜃 cos 𝜙 a∕2 (

+

0

> cos 𝜃 sin 𝜙 −   > sin 𝜃]ejkr  [Mx cos 𝜃 cos 𝜙 +  M M y z

∫0

b{

∫0

0

∫−a∕2

(

′ cos 𝜓

) ′ 2 ′ x + 1 ejkx sin 𝜃 cos 𝜙 dx′ a }

) ′ 2 − x′ + 1 ejkx sin 𝜃 cos 𝜙 dx′ a

ejky

′ sin 𝜃 sin 𝜙

dy′

ds′

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SOLUTION MANUAL

( b

∫0

ejky

′ sin 𝜃 sin 𝜙

dy′ =

′ ejky sin 𝜃 sin 𝜙

jk sin 𝜃 sin 𝜙 (

0

∫−a∕2

=e

−j kb 2

sin 𝜃 sin 𝜙

sin b

)

kb sin 𝜃 sin 𝜙 2 kb sin 𝜃 sin 𝜙 2

) ′ 2 ′ x + 1 ejkx sin 𝜃 cos 𝜙 dx′ a

⎡( ⎤ ′ ) jkx′ sin 𝜃 cos 𝜙 |0 0 2 ′ e 2 ejkx sin 𝜃 cos 𝜙 ′ ⎥ | ⎢ = − x +1 dx | ⎢ a jk sin 𝜃 cos 𝜙 ||−a∕2 a ∫−a∕2 jk sin 𝜃 cos 𝜙 ⎥ ⎣ ⎦ u= du = 0

(

∫−a∕2

2 ′ x +1 a

dv = ejkx

′ sin 𝜃 cos 𝜙

dx′

ejkx sin 𝜃 cos 𝜙 jk sin 𝜃 cos 𝜙 ′

2 ′ dx a

v=

) ′ 2 ′ x + 1 ejkx sin 𝜃 cos 𝜙 dx′ = a

{

ka

}

ka

}

−j sin 𝜃 cos 𝜙 ] 1 2 [1 − e 2 + jk sin 𝜃 cos 𝜙 a (k sin 𝜃 cos 𝜙)2

Similarly a∕2 (

) ′ 2 − x′ + 1 ejkx sin 𝜃 cos 𝜙 dx′ = a

∫0

{

j sin 𝜃 cos 𝜙 ] −1 2 [1 − e 2 + 2 jk sin 𝜃 cos 𝜙 a (k sin 𝜃 cos 𝜙)

Combining terms, we can write [ ( )] ka 2 1 − cos sin 𝜃 cos 𝜙 sin Y 2 2 L𝜃 = 2E0 cos 𝜃 cos 𝜙 b e−jY Y a (k sin 𝜃 cos 𝜙)2 ( ) ka 2 sin2 sin 𝜃 cos 𝜙 ( ) b sin Y 4 cos 𝜃 cos 𝜙e−jY = 8E0 a Y (k sin 𝜃 cos 𝜙)2 ( ) ka sin2 sin 𝜃 cos 𝜙 kb sin Y 4 L𝜃 = abE0 cos 𝜃 cos 𝜙e−jY ( )2 , Y = 2 sin 𝜃 sin 𝜙 Y ka sin 𝜃 cos 𝜙 4 {

}

Similarly 0

L𝜙 =

∬

> cos 𝜙]ejkr′ cos 𝜓 ds′  [−Mx sin 𝜙 +  M y

Sa

(

)

ka sin 𝜃 cos 𝜙 sin Y 4 L𝜙 − abE0 sin 𝜙e−jY ( )2 Y ka sin 𝜃 cos 𝜙 4 sin2

The electric field components are obtained using (12-10a)–(12-10c).

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SOLUTION MANUAL

363

Thus Er = 0 (

) ka sin 𝜃 cos 𝜙 0 abE0 ke−jkr sin Y 4 N 𝜃 ] = +j [L𝜙 + 𝜂 sin 𝜙e−jY E𝜃 = −j ( )2 4𝜋r 2λr Y ka sin 𝜃 cos 𝜙 4 ( ) 2 ka sin sin 𝜃 cos 𝜙 −jkr 0 abE0 e ke−jkr sin Y 4 E𝜙 = j N𝜙 ] = j [L𝜃 − 𝜂 cos 𝜃 cos 𝜙e−jY ( )2 4𝜋r 2λr Y ka sin 𝜃 cos 𝜙 4 sin2

e−jkr

According (12-13a) ( ) ⎧ ⎤⎫ ⎡ 2 ka sin sin 𝜃 cos 𝜙 |E0 |2 ( ab )2 ⎪ ⎥⎪ ⎢ sin Y 4 2 2 2 U(𝜃, 𝜙) = (sin 𝜙 + cos 𝜃 cos 𝜙) ⎢ ( )2 ⎥⎬ 2𝜂 2λ ⎨ Y ka ⎪ ⎥⎪ ⎢ sin 𝜃 cos 𝜙 ⎦⎭ ⎣ ⎩ 4 |E |2 (ab)2 | U(𝜃, 𝜙)| = Umax (𝜃 = 0) = 0 2 |max 8𝜂λ Using (12-39a) Prad

|E |2 = W ave ⋅ ds = 0 ∯ 2𝜂 s b

+

Prad

D0 =

b

{ 2b

∫0 ∫−a∕2

4𝜋Umax = Prad

) 2 − x′ + 1 dx′ dy′ a

0

(

(

0

a∕2 (

∫0 ∫ 0

|E |2 = 0 2𝜂

{

dx

∫−a∕2 [ ] |E0 |2 ( ab )2 4𝜋 8𝜂 λ |E0 |2 ab 6𝜂

dx′ dy′

}

}

)2

2 ′ x +1 a

)2

2 ′ x +1 a

′

=

=

|E0 |2 ab 6𝜂

3𝜋ab λ2

Using (12-40) 𝜀ap

λ2 D0 Aem λ2 3𝜋ab 3 λ2 = = 4𝜋 = = 0.75 = 75% = D0 = Ap ab 4𝜋ab 4𝜋ab λ2 4

as compared to 81% for the cosine distribution. ( ) (b) E = â y E0 cos2 𝜋 x′ = â y Ey a a ( ) 𝜋 Ey = E0 cos2 x′ a M s = −2̂n × Ea = −2̂az × â y Ey = â x 2Ey Jx = Jy = Jz = My = Mz = 0

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SOLUTION MANUAL

Mx = 2Ey = 2E0 cos2

(

𝜋 ′ x a

)

N𝜃 = N𝜙 = 0 L𝜃 =

∬

Mx cos 𝜃 cos 𝜙ejkr

′ cos 𝜓

ds′

Sa

) b∕2 ′ 𝜋 ′ jkx′ sin 𝜃 cos 𝜙 ′ dx ejky sin 𝜃 sin 𝜙 dy′ x e ∫−a∕2 ∫−b∕2 a ( ) kb sin sin 𝜃 sin 𝜙 b∕2 ′ kb sin Y 2 ejky sin 𝜃 sin 𝜙 dy′ = b , Y= sin 𝜃 sin 𝜙 =b ∫−b∕2 kb Y 2 sin 𝜃 sin 𝜙 2 a∕2

L𝜃 = 2E0 cos 𝜃 cos 𝜙

cos2

(

y a Sa

b

x

z

Figure P12.29(b)

Using the identity ∫

2𝛽 2 e𝛼x 𝛼[𝛼 2 + (2𝛽)2 ]

edx cos2 (𝛽x) dx =

We can write that a∕2

∫−a∕2

ejkx

′ sin 𝜃 cos 𝜙

cos2

(

(

)

𝜋 ′ a x dx′ = a 2

(

𝜋2

sin

)

ka sin 𝜃 cos 𝜙 2 ka sin 𝜃 cos 𝜙 2

)2 ka sin 𝜃 cos 𝜙 2 2 sin(X) 𝜋 ka a , X= sin 𝜃 cos 𝜙 = 2 2 2 𝜋 − (X) X 2 𝜋2 −

Thus L𝜃 = abE0 cos 𝜃 cos 𝜙

𝜋2

𝜋2 sin X sin Y ka kb , X= sin 𝜃 cos 𝜙, Y = sin 𝜃 cos 𝜙 Y 2 2 − (X)2 X

In a similar manner L𝜙 =

∬

−Mx sin 𝜙ejkr

′ cos 𝜓

ds′

−sa

L𝜙 = −abE0 sin 𝜙

𝜋2

𝜋2 sin X sin Y Y − (X)2 X

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SOLUTION MANUAL

365

Thus Er = 0 E𝜃 = −j E𝜙 = j U(𝜃, 𝜙) =

0 abE0 e−jkr ke−jkr sin X sin Y 𝜋2 N𝜃 ] = j [L𝜙 + 𝜂 sin 𝜙 2 2 4𝜋r 2λr Y 𝜋 − (X) X

0 abE0 e−jkr ke−jkr sin X sin Y 𝜋2 N𝜙 ] = j [L𝜙 − 𝜂 cos 𝜃 cos 𝜙 2 2 4𝜋r 2λr Y 𝜋 − (X) X

(ab)2 |E0 |2 8𝜂λ2 X=

Umax (𝜃 = 0) =

[sin2 𝜙 + cos2 𝜃 cos2 𝜙]

𝜋2

𝜋2 sin X sin Y 2 Y − (X) X

ka kb sin 𝜃 cos 𝜙, Y = sin 𝜃 sin 𝜙 2 2

(ab)2 |E0 |2 8𝜂λ2

Using (12-39a) Prad =

∯

W ave ⋅ ds =

s

= a∕2

Since

∫−a∕2

cos4

(

( ) |E0 |2 b∕2 a∕2 𝜋 ′ cos4 x dx′ dy′ 2𝜂 ∫−b∕2 ∫−a∕2 a ( ) b|E0 |2 a∕2 𝜋 ′ cos4 x dx′ 2𝜂 ∫−a∕2 a

) 𝜋 ′ 3a x dx′ = a 8 3 ab|E0 | 16 𝜂

2

Prad =

D0 =

4𝜋Umax = Prad

[ 4𝜋

(ab)2 |E0 |2 8𝜂λ2

2 3 ab|E0 | 16 𝜂

] =

8𝜋 ab 3 λ2

Thus 𝜀ap =

Aem Ap

λ2 D0 λ2 8𝜋ab 2 λ2 = 4𝜋 = = 66.67% = D0 = ab 4𝜋ab 4𝜋ab 3λ2 3

as compared to(81% )for the ( cosine ) distribution. ) ( ) ( 𝜋 𝜋 ′ 𝜋 𝜋 ′ ′ ′ (c) E = â y E0 cos x cos y = â y Ey , Ey = E0 cos x cos y a a b a b N𝜃 = N𝜙 = 0 ( ′) a∕2 ′ 𝜋x cos L𝜃 = 2E0 cos 𝜃 cos 𝜙 ejkx sin 𝜃 cos 𝜙 dx′ ∫−a∕2 a ( ′) b∕2 𝜋y ′ cos ejky sin 𝜃 sin 𝜙 dy′ ∫−b∕2 b

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SOLUTION MANUAL

[ L𝜙 = 2E0 cos 𝜃 cos 𝜙

(

b∕2

∫−b∕2

cos

𝜋y′ b

]

)

jky′ sin 𝜃 sin 𝜙

e

dy′

( ) ⎤ ⎡ ka cos sin 𝜃 cos 𝜙 ) ( ⎥ ⎢ 2 𝜋a × ⎢− ⎥ ) ( ) ( 2 2 2 𝜋 ⎥ ka ⎢ sin 𝜃 cos 𝜙 − ⎣ 2 2 ⎦ 𝜋2 4

cos Y cos X ( )2 ( )2 . 𝜋 𝜋 2 2 X − Y − 2 2 ka kb X= sin 𝜃 cos 𝜙, Y = sin 𝜃 sin 𝜙 2 2

L𝜙 = 2E0 cos 𝜃 cos 𝜙ab

In similar procedure L𝜙 = −2E0 sin 𝜙ab

𝜋2 4

cos Y cos X ( )2 ( )2 𝜋 𝜋 X2 − Y2 − 2 2

The fields are kabE0 ke−jkr E𝜃 = −j L =j sin 𝜙 4𝜋r 𝜙 2𝜋r

(

𝜋2 4

)

cos Y cos X ( )2 ( )2 𝜋 𝜋 X2 − Y2 − 2 2 ( 2) −jkr kabE ke cos Y 𝜋 cos X 0 E𝜙 = j L =j cos 𝜃 cos 𝜙 ( )2 ( )2 4𝜋r 𝜃 2𝜋r 4 𝜋 𝜋 X2 − Y2 − 2 2 Umax occurs at 𝜃 = 0◦ Umax = r

21

|E|2

2 𝜂

11 = 2𝜂

(

k2 a2 b2 2 E 4𝜋 2 0

)(

𝜋2 4

)2 ( )4 ( )4 |E |2 8 2 2 = 0 2 4 a2 b2 𝜋 𝜋 𝜂λ 𝜋

Using (12-39a) Prad =

Prad =

𝜀ap

∯

Wave ⋅ ds =

|E0 |2 ab. Thus 8𝜂

|E0 |2 b∕2 a∕2 |E |2 b a 𝜋 𝜋 cos2 x′ cos2 y′ dx′ dy′ = 0 2𝜂 ∫−b∕2 ∫−a∕2 a b 2𝜂 2 2

D0 =

4𝜋Umax = Prad

4𝜋

|E0 |2 8 2 2 a b 𝜂λ2 𝜋 4 |2

|E0 ab 8𝜂

=

64 ab 4𝜋 𝜋 4 λ2

λ2 D0 Aem λ2 64 ab 64 λ2 = = 4𝜋 4𝜋 = 4 = D0 = Ap ab 4𝜋ab 4𝜋ab 𝜋 4 X 2 𝜋

64 = 0.657 = 65.7% 𝜋4 ( ) ( ) ( ) ( ) 𝜋 ′ 𝜋 ′ 𝜋 ′ 𝜋 ′ (d) Ea = â y E0 cos2 x cos2 y = â y Ey , Ey = E0 cos2 x cos2 y a b a b 𝜀ap =

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SOLUTION MANUAL

Using the same procedure as in Part(b), we have that, N𝜃 = N𝜙 = 0.

L𝜃 = 2E0 cos 𝜃 cos 𝜙 b∕2

∫−b∕2

cos2

(

a∕2

∫−a∕2

cos2

(

) 𝜋 ′ jkx′ sin 𝜃 cos 𝜙 ′ dx x e a

) 𝜋 ′ jky′ sin 𝜃 sin 𝜙 ′ dy y e b

L𝜃 =

ab 𝜋2 𝜋2 sin X sin Y E0 cos 𝜃 cos 𝜙 2 2 Y 𝜋 − (X)2 𝜋 2 − (Y)2 X

L𝜙 =

ab 𝜋2 𝜋2 sin X sin Y E0 sin 𝜙 2 2 2 2 Y 𝜋 − (X) 𝜋 − (Y)2 X

ka sin 𝜃 cos 𝜙 2 kb sin 𝜃 sin 𝜙 Y= 2

X=

Thus

E𝜃 = j

abE0 e−jkr 𝜋2 sin X sin Y 𝜋2 sin 𝜙 2 2 2 4λr Y 𝜋 − (X) 𝜋 − (Y)2 X

E𝜙 = j

abE0 e−jkr 𝜋2 sin X sin Y 𝜋2 cos 𝜃 cos 𝜙 2 4λr Y 𝜋 − (X)2 𝜋 2 − (Y)2 X

U(𝜃, 𝜙) =

(ab)2 |E0 |2 32𝜂 λ2

[sin2 𝜙 + cos2 𝜃 cos2 𝜙]

𝜋2 𝜋2 sin X sin Y Y 𝜋 2 − (X)2 𝜋 2 − (Y)2 X

Umax occurs at 𝜃 = 0◦ ⇒ Umax (𝜃 = 0◦ ) = (ab)2 |E0 |2 ∕(32𝜂λ2 ) ( ) ( ) b∕2 |E0 |2 a∕2 𝜋 ′ 𝜋 ′ cos4 cos4 x dx′ y dy′ ∫−b∕2 2𝜂 ∫−a∕2 a b ) ( ) ( ( ) ( ) a∕2 b∕2 3a 3b 4 𝜋 ′ ′ 4 𝜋 ′ ′ cos cos x dx = ; x dx = ∫−b∕2 ∫−a∕2 a 8 b 8

Prad =

9 ab|E0 | 128 𝜂

2

Prad =

Thus D0 =

𝜀ap

[ ]/[ 2] (ab)2 |E0 |2 4𝜋Umax 16𝜋ab 9 ab|E0 | = 4𝜋 = Prad 128 𝜂 32𝜂 λ2 9λ2

λ2 D0 Aem λ2 16𝜋ab 4 λ2 = = 4𝜋 = = 44.44% = D0 = Ap ab 4𝜋ab 4𝜋ab 9λ2 9

12.30. a = 0.9′′ = 2.286 cm = 0.763λ b = 0.4′′ = 1.02 cm = 0.340λ

367

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SOLUTION MANUAL

(a) Using the computer program Aperture of Chapter 12. D0 = 4.264 = 6.298 dB (b) From Table 12.1 D0 = 0.81[4𝜋(0.763)(0.340)] = 2.64 = 4.217 dB 12.31. X-band rectangular waveguide (TE10 mode) a = 2.286 cm, b = 1.016 cm, Pr (satellite) = 10 Watts, R = 100 kilometers, f = 10 GHz λ = 30 × 109 ∕10 × 109 = 3 cm W0 (isotropic) =

Pr 10 10 10−9 = = = 2 3 2 10 4𝜋 4𝜋R 4𝜋(100 × 10 ) 4𝜋(10 )

W0 = 0.0796 × 10−9 = 79.6 × 10−12 W∕m2 = 79.6 × 10−12 (×10−4 ) W∕cm2 W0 = 79.6 × 10−12 W∕m2 = 79.6 × 10−16 W∕cm2 ( ) [ ( )] TE10 mode ab From Table 12.1 D0 = 0.81 4𝜋 2 waveguide λ [ ( )] 2.286 × 1.016 = 0.81 4𝜋 = 2.6268 = 4.1942 dB (3)2 Aem =

λ2 D (PLF) 4𝜋 0

(a) Linearly polarized: Aem =

λ2 λ2 D0 (1) = (2.6268) = 0.2090λ2 4𝜋 4𝜋

PL = Aem W0 = 0.2090 λ2 (79.6 × 10−16 W∕cm2 ) = 0.2090(3)2 (79.6 × 10−16 ) PL = 149.7276 × 10−16 Watts (b) Circularly polarized: Aem =

( ) 1 λ2 = 0.1045 λ2 D0 4𝜋 2

PL = Aem W0 = 149.7276∕2 × 10−16 = 74.864 × 10−16 Watts PL = 74.864 × 10−16 Watts Alternate Procedure X-band rectangular waveguide (TE10 mode) a = 2.286 cm, b = 1.016 cm, Pr (satellite) = 10 Watts, R = 100 kilometers f = 10 GHz ⇒ λ = 30 × 109 ∕10 × 109 = 3 cm = 3 × 10−2 m ( D0

TE10 mode waveguide

)

[ ( )] ab = 0.81 4𝜋 2 λ

From Table 12.1

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SOLUTION MANUAL

[

(

)] 2.286 × 1.016 D0 = 0.81 4𝜋 = 2.6268 = 4.1942 dB (3)2 ) ( Pr λ 2 = er et Dt Dr |̂ 𝜌t ⋅ 𝜌̂r |2 Pt 4𝜋R (a) Linearly polarized |̂ 𝜌t ⋅ 𝜌̂r | = 1, ( Pr = (1)(1)

3 × 10−2 4𝜋 × 105

(2-118)

Dt = 1, Dr = 2.6268 Pt = 10

)2 (1)(2.6268)(10) =

9(2.6268) × 10−3 = 0.1497 × 10−13 16𝜋 2 × 1010

Pr = 149.71 × 10−16 Watts (b) Circularly polarized |̂ 𝜌t ⋅ 𝜌̂r | = 1∕2 Pr =

1 (149.71 × 10−16 ) = 74.854 × 10−16 Watts 2

12.32. a = 0.42′′ = 1.067 cm = 0.711λ b = 0.17′′ = 0.432 cm = 0.288λ (a) Using the computer program Aperture of Chapter 12. D0 = 3.981 = 5.999 dB (b) From Table 12.1 D0 = 0.81[4𝜋(0.711)(0.288)] = 2.084 = 3.189 dB 12.33. E = â y E0 , H = −̂ax

369

E0 𝜂 Js = 0 Ms Js Ms Js = 0 Ms

(a) Outside the aperture J s = n̂ × H = 0 because tangential H vanishes next to PMC. M s is unknown. On the aperture: J s = known = n̂ × H a = â z × (−̂ax Hx ) = −̂ay Hx = −̂ay

E0 𝜂

M s = known = −̂n × Ea = −̂az × (̂ay E0 ) = +ax E0 Utilizing a PMC conductor shorts out M s both over and outside the aperture.

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SOLUTION MANUAL

An electric current source over the aperture next to a PMC results, due to image, to a J s = 2̂n × H a = −̂ay

2E0 𝜂

So the equivalent is: J s = 2̂n × H a = −̂ay Js = 0 M s = 0 everywhere (b)

2E0 𝜂

over the aperture elsewhere

M s = 0 ⇒ L𝜃 = L𝜙 = 0 J s = −̂ay

2E 2E0 ⇒ Jy = − 0 , Jx = Jz = 0 𝜂 𝜂

N𝜃 =

Jy cos 𝜃 sin 𝜙ejkr

∬

′ cos 𝜙

ds′ , N𝜙 =

∬

Jy cos 𝜙ejkr

′ cos 𝜙

ds′

] 2E0 [ sin X sin Y ab cos 𝜃 sin 𝜙 𝜂 X Y ] [ 2E sin X sin Y = − 0 ab cos 𝜙 𝜂 X Y ( ) [ ] −jkr −jkr 2E0 jke jk𝜂e sin X sin Y =− 𝜂N𝜃 = − − ab cos 𝜃 sin 𝜙 4𝜋r 4𝜋r 𝜂 X Y ] abkE0 e−jkr [ sin X sin Y =j cos 𝜃 sin 𝜙 2𝜋r X Y [ ] −jkr abkE0 e sin X sin Y ke−jkr cos 𝜙 = +j (−𝜂N𝜙 ) = j 4𝜋r 2𝜋r X Y ka = sin 𝜃 cos 𝜙 2 kb sin 𝜃 sin 𝜙 = 2

N𝜃 = − N𝜙 E𝜃

E𝜙 X Y 12.34.

a = 0.9′′ = 2.286 cm = 0.7620λ b = 0.4′′ = 1.016 cm = 0.3387λ d = 0.85λ R0 = −30 dB = 31.6228 f = 10 GHz ⇒ λ = 30 × 109 ∕(10 × 109 ) = 3 cm Dt = De Da ] [ (0.7620λ)(0.3387λ) ab De = 0.81 4𝜋 2 = 0.81(4𝜋) = 2.6270 = 4.1947 dB λ λ2 Da =

2R0 2 1 + (R0 2 − 1) f {

f = 1 + 0.636

λ L+d

}2 √ 2 −1 2 2 cosh[ (cosh R0 ) − 𝜋 ] R0

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SOLUTION MANUAL

371

{

}2 √ 2 cosh[ (4.1468)2 − 𝜋 2 ] = 1.1440 31.6228 √ −1 −1 cosh R0 = cosh (31.6228) = ln[31.6228 ± (31.6228)2 − 1] = ln(63.2297) = 4.1468 f = 1 + 0.636

L = 3(0.85λ) = 2.55λ 2(31.6228)2

Da =

1 + [(31.6228)2 − 1](1.144)

λ (2.55 + 0.85)λ

=

2(1, 000) 1 + 336.1346

Da = 5.9323 = 7.7322 dB Dt = De Da = 2.6270(5.9323) = 15.5843 = 11.9269 dB 12.35. Dt = Dw [waveguide] Da [array] From Table 12.1 Dw =

λ=

[ ] [ ] ab ab 8 4𝜋 = 0.81 4𝜋 𝜋2 λ2 λ2

0.9 30 × 109 0.4 = 3 cm = 1.1811′′ ⇒ a = λ = 0.7620λ, b = λ = 0.3387λ 9 1.1811 1.1811 10 × 10 [ ] 0.7620λ(0.3387λ) Dw = 0.81 4𝜋 = 0.81(3.2429) = 2.6268 = 4.1942 dB λ2 Da = 𝜋 cos 𝜃0 Dx Dy |𝜃0 =00 = 𝜋Dx Dy [From (6-103)] ( ) ( ) 0.85λ d = 2(8) = 13.6 (From Table 6-8) Dx = Dy ≃ 2N λ λ Da = 𝜋Dx Dy = 𝜋(13.6) = 581.069 = 581.07 = 27.642 dB Dt ≃ Dw Da = 2.6268(581.069) = 1,526.35 = 31.84 dB = 4.1942 (dB) + 27.642 (dB) = 31.836 dB

12.36. The results can be obtained by using a comparative analogy between the fields of a rectangular aperture when not mounted and mounted on an infinite ground plane. These are listed in Table 12.1. Therefore we can write that the fields of the circular aperture of Section 12.6.1, when it is not mounted on an infinite ground plane, are [by using (12-53a)–(12-53c)] Er ≃ 0 ka2 E0 e−jkr E𝜃 ≃ j 2r ka2 E0 e−jkr E𝜙 ≃ j 2r Hr ≃ 0 H𝜃 = −E𝜙 ∕𝜂 H𝜙 = E𝜃 ∕𝜂

{ {

J (ka sin 𝜃) (1 + cos 𝜃) sin 𝜙 1 ka sin 𝜃

}

J (ka sin 𝜃) (1 + cos 𝜃) cos 𝜙 1 ka sin 𝜃

}

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SOLUTION MANUAL

) ( ′ E0 ′ 𝜌 12.37. E𝜌 = ′ J1 𝜒11 sin 𝜙′ 𝜌 a ) ( ′ ′ 𝜌 E𝜙 = E0 J1′ 𝜒11 cos 𝜙′ , a

J1′ =

𝜕′ J 𝜕𝜌 1

Ez = 0 (a) Using (VII-7b) ) ) ( ( } ′ ′ 1 ′ 𝜌 ′ ′ ′ 𝜌 ′ J 𝜒11 sin 2𝜙 − J1 𝜒11 sin 2𝜙 𝜌′ 1 a a ) ) { ( ( } ′ ′ 1 2 ′ ′ 𝜌 ′ ′ 𝜌 2 ′ J 𝜙 + J 𝜙 sin cos Ey = E𝜌 sin 𝜙′ + E𝜙 cos 𝜙′ = E0 𝜒 𝜒 11 a 1 11 a 𝜌′ 1

E Ex = E𝜌 cos 𝜙 − E𝜙 sin 𝜙 = 0 2 ′

Ly =

∬

My ejkr

′ cos 𝜓

ds′ =

Sa

Lx =

∬

{

′

∬

′ cos 𝜓

𝜌′ d𝜌′ d𝜙′

′ cos 𝜓

𝜌′ d𝜌′ d𝜙′

My ejk𝜌

Sa

Mx ejkr

′ cos 𝜓

ds′ =

Sa

∬

Mx ejk𝜌

Sa

M = −2̂n × Ea = −2̂az × (̂ax Ex + â y Ey ) = −̂ay 2Ex + â x 2Ey Mx = 2Ey , My = −2Ex J = 0 ⇒ Jx = Jy = Jz = 0 Therefore ) ) ( ( } ′ ′ ′ 1 ′ 𝜌 ′ ′ ′ 𝜌 ′ J − J sin 2𝜙 sin 2𝜙 𝜒 𝜒 ejk𝜌 cos 𝜓 𝜌′ d𝜌′ d𝜙′ 11 a 1 11 a ∫0 ∫0 𝜌′ 1 ) { 2𝜋 ( } a ′ ′ ′ 𝜌 = −E0 J1 𝜒11 sin 2𝜙′ ejk𝜌 cos 𝜓 d𝜙′ ∫0 ∫0 a ) ( a 2𝜋 ′ ′ ′ 𝜌 J1′ 𝜒11 sin 2𝜙′ ejk𝜌 cos 𝜓 𝜌′ d𝜙′ d𝜌′ + E0 ∫0 a ∫0 a{

2𝜋

Ly = −E0

However 2𝜋

I0 =

∫0

′ cos 𝜓

sin 2𝜙ejk𝜌

d𝜙′ =

1 2j

{

2𝜋

ej(k𝜌

∫0 2𝜋

−

ej(k𝜌

∫0

′ sin 𝜃 cos(𝜙′ −𝜙)+2𝜙′ )

′ sin 𝜃 cos(𝜙′ −𝜙)−2𝜙′ )

d𝜙′ }

d𝜙′

Letting 𝜙′ − 𝜙 = 𝛽 ⇒ d𝜙′ = d𝛽, we can write 1 I0 = 2j

{

2𝜋−𝜙 j2𝜙

e

∫−𝜙

j(k𝜌′ sin 𝜃 cos 𝛽+2𝛽)

e

2𝜋−𝜙 −j2𝜙

d𝛽 − e

∫−𝜙

j(k𝜌′ sin 𝜃 cos 𝛽−2𝛽)

e

} d𝛽

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SOLUTION MANUAL

373

with the aid of (V-35), it reduces to I0 =

} 1 { j2𝜙 −e 2𝜋J2 (k𝜌′ sin 𝜃) + e−j2𝜙 2𝜋J−2 (k𝜌′ sin 𝜃) 2j

Using (V-10) we can write it as 2𝜋

I0 =

′ cos 𝜓

d𝜙′ = −2𝜋 sin 2𝜙J2 (k𝜌′ sin 𝜃). Thus

sin 2𝜙ejk𝜌

∫0

a

Ly = E0 (2𝜋) sin 2𝜙

[

) )] ( ( ′ ′ 1 ′ 𝜌 ′ ′ 𝜌 𝜌 J2 (k𝜌 sin 𝜃) ′ J1 𝜒11 − J1 𝜒11 d𝜌′ 𝜌 a a ′

∫0

′

Using (V-19), it reduces to Ly =

′ sin 2𝜙 2𝜋𝜒11

a

a

E0

∫0

) ( ′ ′ 𝜌 𝜌′ J2 (k𝜌′ sin 𝜃)J2 𝜒11 d𝜌′ a

Since ∫

𝛾xJp (𝛼x)Jp−1 (𝛾x) − 𝛼xJp−1 (𝛼x)Jp (𝛾x)

xJp (𝛼x)Jp (𝛾x)dx =

𝛼2 − 𝛾 2

then Ly = E0

′ ) sin 2𝜙 2𝜋(𝜒11

{

} ′ J (ka sin 𝜃)J (𝜒 ′ ) − ka sin 𝜃J (ka sin 𝜃)J (𝜒 ′ ) 𝜒11 2 1 11 1 2 11 ′ ∕a)2 (k sin 𝜃)2 − (𝜒11

a

′ ) = −J (𝜒 ′ ) − Because J1′ (𝜒11 2 11

Ly = −2𝜋aE0 sin 2𝜙 ⋅

1 ′ ′ ′ J (𝜒 ′ ) = 0 ⇒ J1 (𝜒11 ) = 𝜒11 J2 (𝜒11 ) ′ 1 11 𝜒11

′ ) J1 (𝜒11

{

′ )2 J (ka sin 𝜃) − ka sin 𝜃J (ka sin 𝜃) (𝜒11 2 1

}

′ )2 1 − (ka sin 𝜃∕𝜒11

′

𝜒112

Using a similar procedure, it can be shown that Lx =

∬

′ cos 𝜓

Mx ejk𝜌

Sa

𝜌′ d𝜌′ d𝜙′

) ) ( ( } ′ ′ ′ 1 2 ′ ′ 𝜌 ′ ′ 𝜌 2 ′ = 2E0 J 𝜒11 sin 𝜙 + J1 𝜒11 cos 𝜙 ejk𝜌 cos 𝜓 𝜌′ d𝜌′ d𝜙′ ∫0 ∫0 𝜌′ 1 a a [ ] ′ )2 (𝜒11 ⎫ ⎧ 2 ′ )2 J ′ (ka sin 𝜃) − cos2 𝜙(𝜒11 ( ′ ) ⎪sin 𝜙J1 (ka sin 𝜃) ka sin 𝜃 − ⎪ 1 ka sin 𝜃 −4𝜋aJ1 𝜒11 E0 ⎪ ⎪ Lx = ⎬ ⎨ ′ )2 ′ )2 ] (𝜒11 [1 − (ka sin 𝜃∕𝜒 ⎪ ⎪ 11 ⎪ ⎪ ⎭ ⎩ 2𝜋

a{

L𝜃 = Lx cos 𝜃 cos 𝜙 + Ly cos 𝜃 sin 𝜙,

L𝜙 = −Lx sin 𝜙 + Ly cos 𝜙

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SOLUTION MANUAL

′ L𝜃 = 4𝜋aJ1 (𝜒11 )E0 cos 𝜃 cos 𝜙 ′ )E0 sin 𝜙 L𝜙 = −4𝜋aJ1 (𝜒11

J1′ (ka sin 𝜃) ′ )2 ] [1 − (ka sin 𝜃∕𝜒11

J1 (ka sin 𝜃) ka sin 𝜃

Thus using (12-10a)–(12-10f) Er ≃ 0 E𝜃 ≃ j E𝜙 ≃ j

Hr ≃ 0

′ )e−jkr kaE0 J1 (𝜒11

r ′ )e−jkr kaE0 J1 (𝜒11 r

J1 (ka sin 𝜃) ka sin 𝜃 J1′ (ka sin 𝜃) cos 𝜃 cos 𝜙 ′ )2 ] [1 − (ka sin 𝜃∕𝜒11 sin 𝜙

H𝜃 ≃ E𝜙 ∕𝜂 H𝜙 ≃ E𝜃 ∕𝜂

(b) The results can be obtained by using a comparative analogy between the fields radiated by a rectangular aperture when not mounted and mounted on an infinite ground plane. These are listed in Table 12.1. Also compare with Problem 12.36. Therefore we can write that the fields of the circular aperture of Section 12.6.2, when it is not mounted on an infinite ground plane, are (by using the results of part a) Er ≃ 0 ′ )e−jkr kaE0 J1 (𝜒11

Hr ≃ 0

J1 (ka sin 𝜃) 2r ka sin 𝜃 ′ )e−jkr kaE0 J1 (𝜒11 J1′ (ka sin 𝜃) E𝜙 ≃ j (1 + cos 𝜃) cos 𝜙 ′ )2 ] 2r [1 − (ka sin 𝜃∕𝜒11 E𝜃 ≃ j

12.38.

(1 + cos 𝜃) sin 𝜙

H𝜃 ≃ −E𝜙 ∕𝜂 H𝜙 ≃ E𝜃 ∕𝜂

C2 J (Z) sin 𝜙(1 + cos 𝜃) 1 2 Z J1′ (Z) C E𝜙 = 2 cos 𝜙(1 + cos 𝜃) ′ )2 2 1 − (Z∕𝜒11 E𝜃 =

′ Z = ka sin 𝜃, 𝜒11 = 1.841, J1′ (Z) = J0 (Z) − J1 (Z)∕Z [ ] sin 𝜙 cos 𝜙 Et = â 𝜃 (1 + cos 𝜃) + â 𝜙 (1 + cos 𝜃) 2 2

1 + cos 𝜃 [̂a𝜃 sin 𝜙 + â 𝜙 cos 𝜙] 2 Er = (̂a𝜃 sin 𝜙 + â 𝜙 cos 𝜙 sin 𝜙) =

(̂a𝜃 sin 𝜙 + â 𝜙 cos 𝜙) â t = √ = (̂a𝜃 sin 𝜙 + â 𝜙 cos 𝜙) 2 2 sin 𝜙 + cos 𝜙 â 𝜃 sin 𝜙 + â 𝜙 cos 𝜃 cos 𝜙 â r = √ sin2 𝜙 + cos2 𝜃 cos2 𝜙 |2 | | | 2 2 2 2 2 | sin 𝜙 + cos 𝜃 cos 𝜙 | 2 | = (sin 𝜙 + cos 𝜃 cos 𝜙) | PLF = |̂at ⋅ â r | = | √ | | | sin2 𝜙 + cos2 𝜃 cos2 𝜙 | sin2 𝜙 + cos2 𝜃 cos2 𝜙 | | |

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SOLUTION MANUAL

375

12.39. (a) Using Table 12.1 and comparing the fields of a rectangular aperture mounted on a PEC and not a ground plane, we can write the fields of a circular aperture not mounted on a ground plane (based on the fields of the same aperture mounted on a PEC and Table 12.2) as

Er ≃ Hr ≃ 0 C J (Z) E𝜃 ≃ 2 sin 𝜙(1 + cos 𝜃) 1 2 Z J1′ (Z) C2 E𝜙 ≃ cos 𝜙(1 + cos 𝜃) ′ )2 2 1 − (Z∕𝜒11

H𝜃 ≃ −E𝜙 ∕𝜂 ⎫ ⎪ ⎪ H𝜙 ≃ +E𝜃 ∕𝜂 ⎪ 0◦ ≤ 𝜃 ≤ 180◦ ⎬ ◦ 0 ≤ 𝜙 ≤ 360◦ Z = ka sin 𝜃 ⎪ ⎪ ′ = 1.841 ⎪ 𝜒11 ⎭

(b) From Table 12.2 Aem = 0.836Ap = 0.836(𝜋a2 ), a = 1.125 cm, f = 10 GHz λ=

30 × 109 1.125 = 3 cm, a = λ = 0.375λ 3 10 × 109

Aem = 0.836[𝜋(0.375λ)2 ] = 0.369λ2 = 0.369(3)2 = 3.324 cm2 (PL )max = Aem Wi (1 − |Γ|2 ), Wi = 100 W∕m2 = 100(10−4 ) W∕cm2 = 10 × 10−3 W∕cm2 | Z − Zc |2 | ZL = 350 + j400, Zc = 300 ⇒ |Γ|2 = || L | | ZL + Zc | | 350 + j400 − 300 |2 ( 403.1129 )2 | = = || = 0.27897 | 763.2169 | 350 + j400 + 300 | PL = 3.324(10 × 10−3 )(1 − 0.27897) = 23.966 × 10−3 Watts 12.40. Circular Waveguide: a = 2 cm Rectangular Waveguide: a = 2.286 cm, b = 1.016 cm Frequency = 10 GHz ⇒ λ = c∕f = 30 × 109 ∕10 × 109 = 3 cm (a) Transmitting Antenna: Rectangular waveguide [ ] [ ] 2.286(1.016) ab Dt = 𝜀ap D0 = 𝜀ap 4𝜋 2 = 0.81 4𝜋 = 0.81(3.243) λ (3)2 Dt = 2.627 = 4.194 dB (b) W0 (isotropic source) = W0 (isotropic) =

Pt 100 100 = = 4𝜋R2 4𝜋(100 × 103 × 100)2 4𝜋(107 )2 10−12 = 7.958 × 10−14 W∕cm2 4𝜋

W(waveguide)max = Dt W0 = 7.958(2.627) × 10−14 = 20.905 × 10−14 W(waveguide)max = 20.905 × 10−14 W∕cm2

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SOLUTION MANUAL

(c) Receiving Antenna: Circular Waveguide (Aem )r = 𝜀ap Ap = 0.836(𝜋a2 ) = 0.836[𝜋(2)2 ] (Aem )r = 0.836[4𝜋] = 0.836(12.566) = 10.505 (Aem )r = 10.505 cm2 (d) ∙

( Pr = Wmax (Aem )r = 20.905 × 10

) W 10.505(cm2 ) cm2

Pr = 2.196 × 10−12 Watts ∙

Pr = (PLF)Wmax (Aem )r =

1 (20.905 × 10−14 )(26.465) 2

Pr = 1.098 × 10−12 Watts 12.41. G0 = 15 dB ⇒ G0 (dimensionless) = 101.5 = 31.623; Ap = 𝜋a2 = 25 cm2 f = 10 GHz ⇒ λ = 3 cm (a) G0 = ecd D0 =

4𝜋 4𝜋 4𝜋 4𝜋 Aem = 2 𝜀ap Ap = 2 (𝜀ap )(25 cm2 ) = (𝜀ap )(25) = 31.623 λ2 λ λ (3)2 𝜀ap =

9(31.623) = 0.906 = 90.6% 4𝜋(25)

(b) PL = Wi Aem = Wi 𝜀ap Ap = 30 × 10−3 (0.906)(25) = 679.5 × 10−3 Watts PL = 0.6795 Watts 12.42. G0 = 9 dB ⇒ G0 (dimensionless) = 100.9 = 7.94328 ( ) 4𝜋 (a) From Table 12.2 D0 = G0 = 0.836 2 Ap = 7.94328 λ ⇒ Ap = 𝜋a2 =

7.94328(λ2 ) = 0.7561λ2 0.836(𝜋)(4)

(b) Aem = 𝜀ap Ap = 0.836(0.7561)λ2 = 0.6321λ2 (c) 𝜀ap = 0.836 = 83.6% (d) The TE11 -mode is less efficient (83.6% vs. 100%) than the uniform distribution. 12.43. Using the computer program Aperture of Chapter 12. (a) a = 0.5λ ⇒ D0 = 11.24 = 10.51 dB (b) a = 1.5λ ⇒ D0 = 92.704 = 19.67 dB (c) a = 3.0λ ⇒ D0 = 357.278 = 25.53 dB Using Table 12.2 (a) a = 0.5λ ⇒ D0 = (2𝜋a∕λ)2 = [2𝜋(0.5)]2 = 9.8696 = 9.943 dB (b) a = 1.5λ ⇒ D0 = (2𝜋a∕λ)2 = [2𝜋(1.5)]2 = 88.826 = 19.485 dB (c) a = 3.0λ ⇒ D0 = (2𝜋a∕λ)2 = [2𝜋(3)]2 = 355.306 = 25.506 dB They compare quite well except for a = 0.5λ

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SOLUTION MANUAL

12.44. Using the computer program Aperture of Chapter 12. (a) a = 0.5λ ⇒ D0 = 11.264 = 10.517 dB (b) a = 1.5λ ⇒ D0 = 93.197 = 19.694 dB (c) a = 3.0λ ⇒ D0 = 356.815 = 25.524 dB 12.45. With ground plane Using the computer program Aperture of Chapter 12. (a) a = 0.5λ ⇒ D0 = 9.922 = 9.966 dB (b) a = 1.5λ ⇒ D0 = 75.161 = 18.759 dB (c) a = 3.0λ ⇒ D0 = 295.096 = 24.699 dB Using Table 12.2 (a) a = 0.5λ ⇒ D0 = 0.836(2𝜋a∕λ) = 0.836(9.8696) = 8.251 = 9.165 dB (b) a = 1.5λ ⇒ D0 = 0.836(2𝜋a∕λ) = 0.836(88.826) = 74.259 = 18.707 dB (c) a = 3.0λ ⇒ D0 = 0.836(2𝜋a∕λ) = 0.836(355.306) = 297.036 = 24.728 dB Without ground plane Using the computer program Aperture of Chapter 12. (a) a = 0.5λ ⇒ D0 = 8.8244 = 9.457 dB (b) a = 1.5λ ⇒ D0 = 75.9458 = 18.805 dB (c) a = 3.0λ ⇒ D0 = 296.959 = 24.727 dB 12.46. From Table 12.2 (a) HPBW (E-plane) = 29.2∕1.5 = 19.47◦ (b) HPBW (H-plane) = 37.0∕1.5 = 24.67◦ (c) FNBW (E-plane) = 69.9∕1.5 = 46.6◦ (d) FNBW (H-plane) = 98.0∕1.5 = 65.33◦ (e) FSLMM (E-plane) = −17.6 dB (f) FSLMM (H-plane) = −26.2 dB

From Figs. 12.19 & 12.20 HPBW ≃ 20◦ HPBW ≃ 23.8◦ FNBW ≃ 49◦ FNBW ≃ 68◦ FSLMM ≃ −17 dB FSLMM ≃ −28.5 dB

12.47. From Table 12.2 (a) Aperture with infinite PEC ground plane J1 (Z) Z J (Z) E𝜙 = jC1 cos 𝜃 cos 𝜙 1 Z H𝜃 = −E𝜙 ∕𝜂 H𝜙 = E𝜃 ∕𝜂 E𝜃 = jC1 sin 𝜙

Z = ka sin 𝜃 C1 = Constant

Aperture without PEC ground plane ) J (Z) 1 + cos 𝜃 sin 𝜙 1 2 Z ) ( J (Z) 1 + cos 𝜃 cos 𝜙 1 E𝜙 = C1 2 Z (

E𝜃 = C1

(b)

∙ HPBW(E-plane) =

29.2 29.2 = = 9.7333◦ a∕λ 3

∙ HPBW(H-plane) =

29.2 29.2 = = 9.733◦ a∕λ 3

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[

]2

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SOLUTION MANUAL

(

2𝜋a (Table 12.2): D0 = λ (Kraus): D0 = (T&P): D0 =

)2

2𝜋(3)λ = λ

= (6𝜋)2 = (18.8496)2 = 355.306 = 25.52 dB

41,253 41,253 = = 435.44 = 26.39 dB ΘE ΘH (9.7333)2

72,815 72,815 72,815 = = = 384.30 = 25.85 dB 2 2 2 2(9.733) 2(Θ) ΘE + ΘH

T&P is more accurate because HPBW is less than 39.77◦ . 12.48. (a) Ea = â y E0 [1 − (𝜌′ ∕a)2 ] ⇒ M s = −2̂n × Ea = â x 2E0 [1 − (𝜌′ ∕a)2 ] L𝜃 =

( ∬

0 0 ) > cos 𝜃 sin 𝜙 −   > sin 𝜃 ejkr′ cos 𝜓 ds′  Mx cos 𝜃 cos 𝜙 +  M M y z

s

= 2E0 cos 𝜃 cos 𝜙

a

∫0

𝜌′ [1 − (𝜌′ ∕a)2 ]

2𝜋

∫0

′ sin 𝜃 cos 𝜙′

ejk𝜌

d𝜙′ d𝜌′

Using (V-35) it reduces, after separating into two integrals, to [ L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙

a

∫0

𝜌′ J0 (k𝜌′ sin 𝜃) d𝜌′ −

a

1 𝜌′ 3J0 (k𝜌′ sin 𝜃) d𝜌′ a2 ∫0

]

Making a change of variable of the form x = k𝜌′ sin 𝜃 ⇒ dx = k sin 𝜃 d𝜌′ we can write it as [ L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙

a

∫0

1 1 𝜌 J0 (k𝜌 sin 𝜃) d𝜌 − 2 a (k sin 𝜃)4 ∫0 ′

′

′

ka sin 𝜃

] 3

x J0 (x) dx

Using (V-22) and (V-24) it reduces to [ 2

L𝜃 = 8𝜋a E0

J (ka sin 𝜃) cos 𝜙 cos 𝜃 2 (ka sin 𝜃)2

]

In a similar manner, it can be shown that L𝜙 =

∬

[−Mx sin 𝜙 + My cos 𝜙]ejkr

s

′ cos 𝜓

[ ] J (ka sin 𝜃) ds′ = −8𝜋a2 E0 sin 𝜙 2 (ka sin 𝜃)2

Since N𝜃 = N𝜙 = 0, we can write the radiated fields, using (12-10a)–(12-10f), as Er ≃ 0

[ ] J (ka sin 𝜃) e−jkr sin 𝜙 2 r (ka sin 𝜃)2 [ ] J2 (ka sin 𝜃) e−jkr 2 cos 𝜃 cos 𝜙 E𝜙 ≃ j2ka E0 r (ka sin 𝜃)2 E𝜃 ≃ j2ka2 E0

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SOLUTION MANUAL

Hr ≃ 0 H𝜃 ≃ −E𝜙 ∕𝜂 H𝜙 ≃ E𝜃 ∕𝜂 (b) Following a similar procedure, it can be shown that when Ea = â y E0 [1 − (𝜌′ ∕a)2 ]2 the radiated field are given by Er ≃ 0

[ ] J3 (ka sin 𝜃) e−jkr E𝜃 ≃ j16 ka E0 sin 𝜙 r (ka sin 𝜃)3 [ ] J (ka sin 𝜃) e−jkr cos 𝜃 cos 𝜙 3 E𝜙 ≃ j16 ka2 E0 r (ka sin 𝜃)3 2

Hr ≃ 0, H𝜃 ≃ −E𝜙 ∕𝜂, H𝜙 ≃ E𝜃 ∕𝜂 12.49. Ea = â y E0 (1 − 𝜌′ ∕a) ⇒ M s = â x 2E0 (1 − 𝜌′ ∕a), J = 0 ⇒ N𝜃 = N𝜙 = 0 L𝜃 =

∬

Mx cos 𝜃 cos 𝜙ejkr

′ cos 𝜓

ds′ = 2E0 cos 𝜃 cos 𝜙

s 2𝜋

×

∫0 2𝜋

Using (12-48)

∫0

′ sin 𝜃 cos(𝜙−𝜙′ )

ejk𝜌

′ sin 𝜃 cos(𝜙−𝜙′ )

ejk𝜌

[

a

∫0

𝜌′ (1 − 𝜌′ ∕a)

d𝜙′ d𝜌′

d𝜙′ = 2𝜋J0 (k𝜌′ sin 𝜃)

] a 1 ′2 ′ ′ 𝜌 J0 (k𝜌 sin 𝜃) d𝜌 − 𝜌 J0 (k𝜌 sin 𝜃) d𝜌 L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙 ∫0 a ∫0 [ ] |a 1 a ′2 𝜌′ ′ ′ ′ | 𝜌 J0 (k𝜌 sin 𝜃) d𝜌 J (k𝜌 sin 𝜃)| − L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙 k sin 𝜃 1 |0 a ∫0 a

′

′

′

x 1 ⇒ d𝜌′ = dx let x = k𝜌′ sin 𝜃 ⇒ 𝜌′ = k sin 𝜃 k sin 𝜃 { } ka sin 𝜃 a 1 2 x J0 (x) dx J (ka sin 𝜃) − L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙 k sin 𝜃 1 ak3 sin3 𝜃 ∫0 Using V-30, V-27 { L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙

J1 (x) ||ka sin 𝜃 a J1 (ka sin 𝜃) − x2 | k sin 𝜃 ak3 sin3 𝜃 |0 } ka sin 𝜃 1 + xJ1 (x) dx ak3 sin3 𝜃 ∫0

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SOLUTION MANUAL

{

a a J1 (ka sin 𝜃) − J (ka sin 𝜃) k sin 𝜃 k sin 𝜃 1 { [ }]} ka sin 𝜃 |ka sin 𝜃 1 | −xJ0 (x) | + + J0 (x) dx | ∫0 k3 a sin3 𝜃 |0 { } ka sin 𝜃 1 L𝜃 = 4𝜋E0 cos 𝜃 ⋅ cos 𝜙 J0 (ka sin 𝜃) + J0 (x) dx ∫0 k2 sin2 𝜃

L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙

ka sin 𝜃

J0 (x) dx cannot be evaluated in closed form.

∫0

L𝜙 also has same except for front term. { } ka sin 𝜃 1 L𝜙 = −4𝜋E0 sin 𝜙 J0 (ka sin 𝜃) + J0 (x) dx ∫0 k2 sin2 𝜃 1 V V 1 = −̂a𝜌 C ′ , where C = ′ 𝜀 ln(b∕a) 𝜌 𝜌 𝜀 ln(b∕a) ( ) C 2C C M s = −2̂n × Ea = −2̂az × −̂a𝜌 ′ = â 𝜙 (2) ′ ⇒ M𝜙 = ′ 𝜌 𝜌 𝜌

12.50.

Ea = −̂a𝜌

Thus using (12-42c) 0

L𝜃 =

∬

0

> cos 𝜃 cos(𝜙 − 𝜙′ ) + M𝜙 cos 𝜃 sin(𝜙 − 𝜙′ ) −   > sin 𝜃]ejkr′ cos 𝜓 ds′  [ M M 𝜌 z

Sa

b[

= 2C cos 𝜃

∫a

2𝜋

∫0

′ sin 𝜃 cos(𝜙−𝜙′ )

sin(𝜙 − 𝜙′ )ejk𝜌

] d𝜙′ d𝜌′

Because of azimuthal symmetry, the field is not a function of 𝜙. Choosing 𝜙 = 0: L𝜃 = 2C cos 𝜃

b[

2𝜋

−

∫a

∫0

′ jk𝜌′ sin 𝜃 cos 𝜙′

sin 𝜙 e

] d𝜙 d𝜌′ = 2C cos 𝜃 ′

b

∫a

[0]d𝜌′ = 0

0

L𝜙 =

∬

> sin(𝜙 − 𝜙′ ) + M𝜙 cos(𝜙 − 𝜙′ )]ejkr′ cos 𝜓 ds′  [− M 𝜌

s

= 2C L𝜙 = 2C

b[

∫a

2𝜋

∫0

b[

∫a

𝜋

∫0

′ sin 𝜃 cos 𝜙′

cos 𝜙′ ejk𝜌

′ jk𝜌′ sin 𝜃 cos 𝜙′

cos 𝜙 e

] d𝜙′ d𝜌′ 2𝜋 ′

d𝜙 +

∫𝜋

′ jk𝜌′ sin 𝜃 cos 𝜙′

cos 𝜙 e

] ′

d𝜙 d𝜌′

Let 𝜃 ′′ = 𝜙′ − 𝜋 ⇒ d𝜙′′ = d𝜙′ ] b[ 𝜋 𝜋 ′ jk𝜌′ sin 𝜃 cos 𝜙′ ′ ′′ jk𝜌′ sin 𝜃 cos 𝜙′′ ′′ L𝜙 = 2C cos 𝜙 e d𝜙 − cos 𝜙 e d𝜙 d𝜌′ ∫a ∫0 ∫0 b

= 2C

∫a

[J1 (k𝜌′ sin 𝜃) − J1 (−k𝜌′ sin 𝜃)]d𝜌′ = 4C

b

∫a

J1 (k𝜌′ sin 𝜃) d𝜌′

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SOLUTION MANUAL

by using (V-36) and (V-10). With the aid of (V-23) [ L𝜙 = −4C

1 (J (kb sin 𝜃) − J0 (ka sin 𝜃)) k sin 𝜃 0

]

If however the slot is very thin, L𝜙 can be approximated by b

L𝜙 = 4C

∫a

J1 (k𝜌′ sin 𝜃) d𝜌′ = 4C J1 (ka′ sin 𝜃)

∫a ) ( a+b where a′ = 2

b

d𝜌′ = 4C(b − a)J1 (ka′ sin 𝜃)

Using (12-10a)–(12-10f), we can write the radiated fields as Er = 0

[ ] 0 ke−jkr Ce−jkr J0 (kb sin 𝜃) − J0 (ka sin 𝜃) E𝜃 = −j N 𝜃 ] = +j [L + 𝜂 4𝜋r 𝜙 𝜋r sin 𝜃 ≃ +j

(a − b)Cke−jkr J1 (ka′ sin 𝜃) 𝜋r

ke−jkr [L − 𝜂N𝜙 ] = 0 4𝜋r 𝜃 Hr = 0, H𝜃 = −E𝜙 ∕𝜂, H𝜙 = E𝜃 ∕𝜂

E𝜙 = +j

12.51. Using Figure 12.21 u = ka sin 𝜃1 =

2𝜋 3.3λ a sin 𝜃1 = 3.3 ⇒ a = = 1.05λ λ 2𝜋 sin(30◦ )

2𝜋 (2λ) sin(20◦ ) = 4.298 ≃ 4.3 λ Using Figure 12.21 ⇒ Beam efficiency ≃ 97 ∼ 98%

12.52. u = ka sin 𝜃1 =

12.53. (a) For a square aperture 𝜃ce = 𝜃ch . Therefore the optimum dimension a=b=

λ λ λ 2 = √ = √ = 0.577λ 2 sin(60◦ ) 2 3 3

(b) The maximum directivity, according to (12-59), is D0 =

4 𝜋 𝜋 = 𝜋 = 4.189 = 6.221 dB = 2 2 ◦ 3 sin 𝜃c sin (60 )

(c) The directivity at 𝜃 = 60◦ is −3.922 dB from the maximum at 𝜃 = 0◦ . λ λ = = 0.338λ 3.413 sin 𝜃c 3.413 sin(60◦ ) 1.086𝜋 1.086𝜋 (b) D0 = = 4.55 = 6.58 dB = 2 sin 𝜃c sin2 (60◦ )

12.54. (a) a =

381

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SOLUTION MANUAL

(c) P(𝜃 = 0)|max = (2𝜋a)2 } } { { 2J1 (ka sin 𝜃) 2 2J1 (1.8392) 2 2 ◦ 2 P(𝜃)|𝜃=60◦ = (2𝜋a) , P(𝜃 = 60 ) = (2𝜋a) ka sin 𝜃 1.8392 }2 { 2(0.5819) P(𝜃 = 60◦ ) = (2𝜋a)2 = (2𝜋a)2 (0.4004) 1.8392 P(𝜃 = 𝜃c ) || = 0.4004 = −3.975 dB P(𝜃 = 0) ||𝜃c =60◦ | λ λ | = = 0.42268λ | 2.7318 sin(𝜃c ) |𝜃c =60◦ 2.3658 1.263𝜋 1.263𝜋 = = 5.2904 = 7.235 dB (b) D0 = 2 0.75 sin (𝜃c ) { } 8J2 (ka sin 𝜃) 2 2 (c) P(𝜃) = 0.75(2𝜋a) [ka sin 𝜃]2 2 P(𝜃 = 0) = 0.75(2𝜋a) {8J2 (x)∕x2 }2 |x=0 = 0.75(2𝜋a)2 (1)

12.55. (a) a =

because

8J2 (x) || =1 x ||x=0 ( ) 2 ⎧ 8J 2𝜋 (0.42268λ) sin 60◦ ⎫ ⎪ 2 ⎪ P(𝜃 = 60◦ ) = 0.75(2𝜋a)2 ⎨ [ λ ]2 ⎬ ⎪ 2𝜋 (0.42268λ) sin 60◦ ⎪ ⎩ ⎭ λ { } 2 = 0.75(2𝜋a)2 8J2 (2.3)∕(2.3)2 = 0.75(2𝜋a)2 (0.39182) P(𝜃 = 60◦ ) || = 0.39182 = −4.069134 dB P(𝜃 = 0◦ ) ||𝜃c =60◦

λ λ = = 0.37696λ 3.06317 sin 𝜃c 3.06317(0.8660) (b) To find the maximum directivity, we need to drive the far-field for a circular aperture with a parabolic taper on 10 dB pedestal. [ )2 ] ( { } 𝜌′ ⇒ M s = −2̂n × Ea = â x 2E0 1 − [𝜌′ ∕(1.2097a)]2 E0 = â y 1 − 1.2097a [ )2 ] ( 2𝜋 a 𝜌′ L𝜃 = 2E0 cos 𝜃 cos 𝜙 1− 𝜌′ d𝜌 d𝜃 ∫0 ∫0 1.2097a [ a 𝜌′ J0 (k𝜌′ sin 𝜃) d𝜌′ L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙 ∫0 ] a 1 ′3 ′ ′ − 𝜌 J0 (k𝜌 sin 𝜃) d𝜌 (1.2097a)2 ∫0 Making a change of variable of the form ⇒ X = k𝜌′ sin 𝜃 ⇒ dX = k sin 𝜃 d𝜌′ , we can write it as { ka sin 𝜃 1 X J0 (X) dx L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙 (k sin 𝜃)2 ∫0 ]} [ ka sin 𝜃 1 1 3 − X J0 (X) dx (1.2097a)2 (k sin 𝜃)4 ∫0

12.56. (a) a =

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SOLUTION MANUAL

[

a2 J(X) J(X) a2 2a2 J2 (X) L𝜃 = 4𝜋E0 cos 𝜃 cos 𝜙 − + 2 X (1.2097) X (1.2097)2 X 2

(

XJ0 (X) dX = X J1 (X) + C,

∫

∫

3

3

] )

2

X J0 (X) dX = X J1 (X) − 2X J2 (X) + C

X = ka sin 𝜃 Using a similar procedure: [( L𝜙 = 4𝜋E0 sin 𝜙a 1−

] J2 (X) J1 (X) 2 + X (1.2097)2 X 2 [ ] kE0 a2 e−jkr J1 (X) J2 (X) ke−jkr ∴ E𝜃 = −j L =j sin 𝜙 0.316647 + 1.3667 2 4𝜋r 𝜙 r X X [ ] 2 −jkr kE a e J (X) J (X) E𝜙 = j 0 cos 𝜃 cos 𝜙 0.316647 1 + 1.3667 2 2 r X X 2

)

1 (1.2097)2

X = ka sin 𝜃 Radiated power reduces to Prad

[ ( )2 ]2 |E0 |2 2𝜋 a 𝜌′ = W av ⋅ ds = 𝜌′ d𝜌′ d𝜙 1− ∯ 2𝜂 ∫0 ∫0 1.2097a s

Umax

[ 2 ] |E0 |2 |E0 |2 1 a 1 a2 a2 + 2𝜋 − (2𝜋)0.23615(a2 ) = = 2𝜂 2 2 (1.2097)2 6 (1.2097)4 2𝜂 ] ( )2 [ |E |2 J (X) 2 || J (X) 2𝜋 = 0 (a4 ) + 1.3667 2 2 0.316647 1 | | 2𝜂 λ X X |X=0 ⎛ ← J1 (X) || = 0.5 ⎜ X ||X=0 ⎜ J (X) | 1 ⎜ ← 2 || = ⎝ X 2 |X=0 8

( )2 |E |2 2𝜋 = 0 (a4 ) (0.32916)2 2𝜂 λ

D0 =

4𝜋Umax = Prad

4𝜋

|E0 |2 ( 2𝜋 )2 (0.1083469)a4 2𝜂 λ 2𝜋

|E0 |2 (0.23615)a2 2𝜂

4𝜋 (𝜋)(0.458805) λ2 4𝜋 D0 = 0.91761 2 (𝜋a2 ) λ

D0 = 2(a2 )

For a circular aperture with parabolic taper with −10 dB pedestral, the normalized power pattern multiplied by the maximum directivity can be written as { 2

P(𝜃) = 0.91761(2𝜋a)

[ (3.038026)

2

J (X) J (X) + 1.3667 2 2 0.316647 1 X X

X = ka sin 𝜃

]2 }

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SOLUTION MANUAL

For any other angle 𝜃 = 𝜃c , the maximum of the pattern occurs when [

ka sin 𝜃c = 2.0512

J (X) ⇒ 0.316647 J1 (X) + 1.3667 2 X

∴ Optimum Radius: a =

] has maximum

λ λ = 2𝜋 3.06317 sin 𝜃c sin 𝜃c 2.0512

Directivity is 4𝜋 D0 = 0.91761 2 (𝜋) λ

(

λ 3.06317 sin 𝜃c

)

( The maximum directivity with aperture radius a a = D0 =

=

1.228247𝜋 (sin 𝜃c )2

λ 3.06317 sin 60◦

) is

1.228247𝜋 = 5.144869 = 7.1137 dB. (sin 60◦ )2

(c) The value of the directivity at the edge of the desired coverage (𝜃 = 𝜃c = 60◦ ), relative to its maximum value 𝜃 = 0, is P(𝜃 = 𝜃c ) P(𝜃 = 0) P(𝜃 = 0) = 0.91761(2𝜋a)2 ( ) 2𝜋 λ ka sin 𝜃c |𝜃c =60◦ = sin 𝜃c = 2.0512 λ 3.06317 sin 𝜃c P(𝜃c = 60◦ )

{

[ ]2} J (2.0512) (2.0512) J 1 2 + 1.3667 = 0.91761(2𝜋a2 ) (3.038026)2 0.316647 (2.0512) (2.0512)2 } { = 0.91761(2𝜋a2 ) (3.038026)2 [0.316647(0.27930) + 1.3667(0.086548)]2

= 0.91761(2𝜋a2 )[(3.038026)(0.20672522)]2 P(𝜃c = 60◦ ) = 0.91761(2𝜋a2 )[0.39443] P(𝜃 = 𝜃c ) = 0.39443 = −4.0402979 dB P(𝜃 = 0) 12.57. For circular aperture with parabolic-taper illumination, the power pattern (relative power pattern times the antenna directivity), is given by { 2

P(𝜃) = 0.75(𝜋2a)

8J2 (ka sin 𝜃) [ka sin 𝜃]2

}2

where a is the radius of the aperture in wavelengths, and J2 is the second-order Bessel function.

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SOLUTION MANUAL

385

The maximum value of P(𝜃), for a given 𝜃, occurs when the function [J2 (X)∕X] is a maximum; i.e., ka sin 𝜃c = 2.3 λ λ 2𝜋 = (a) sin 𝜃c = 2.3 ⇒ a = 2𝜋 λ 2.7318 sin 𝜃c sin 𝜃c 2.3 ( ) 4𝜋 D0 = 0.75 2 (𝜋a2 ) λ ( 2)[ ]2 (4𝜋) 4𝜋 𝜋 λ D0 = 0.75 = 0.75 2.7318 sin(𝜃c ) λ2 (2.7318)2 [sin(𝜃c )]2 𝜋 D0 = 0.75(1.68388) [sin(𝜃c )]2 D0 = 1.2629𝜋∕ sin2 𝜃c E.O.C. (Relative power at 𝜃c to maximum power), using P(𝜃 = 0) from Problem 12.55(c): [

] J2 (2.3) 2 8 P(𝜃 = 𝜃c ) (2.3)2 = P(𝜃 = 0) 0.75[𝜋(2)a]2 [ ] 8J2 (2.3) 2 = = [8(0.078244)]2 (2.3) 0.75[𝜋(2)a]2

= (0.625952)2 = 0.3918159 P(𝜃 = 𝜃c ) = −4.069 dB P(𝜃 = 0) 12.58. (a) b = λ∕2 sin 𝜃ce = λ∕2 sin(30◦ ) = λ

λ = 0.707λ 2(0.707) 4𝜋 4𝜋 4𝜋 4𝜋 (b) D0 = 2 Aem = 2 Ap = 2 (a)(b) = 2 (λ)(0.707λ) = 8.886 λ λ λ λ D0 = 8.886 = 9.487 dB @ 𝜃 = 0◦ ) 2 ( kLx ⎡ ⎤ ⎢ sin 2 sin 𝜃 ⎥ ⎥ (c) De = D0 ⎢ . kLx ⎢ ⎥ sin 𝜃 ⎢ ⎥ 2 ⎣ ⎦ 𝜃=𝜃ce Lx =b ( ) | kLx 𝜋 kb 2𝜋 λ = = 90◦ = sin 𝜃 || sin 𝜃ce = 𝜃=𝜃ce 2 2 2λ 2 2 | L =b x ] [ ( )2 ( ) sin(𝜋∕2) 2 2 4 = 0.4052(8.886) De = D0 = D0 = D0 𝜋∕2 𝜋 𝜋2 De = 3.6 = 5.564 dB a = λ∕2 sin 𝜃ch = λ∕2 sin 45◦ =

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SOLUTION MANUAL

) 2 ( kLy ⎡ ⎤ ⎢ sin 2 sin 𝜃 ⎥ ⎥ (d) Dh = D0 ⎢ . kLy ⎢ ⎥ sin 𝜃 ⎢ ⎥ 2 ⎣ ⎦ 𝜃=𝜃ch Lx =a | kLy ka 2𝜋 𝜋 | = sin 𝜃 | sin 𝜃ch = (0.707λ)(0.707) = = 90◦ | 𝜃=𝜃ch 2 2 2λ 2 | L =a x ] [ ( )2 ( ) sin(𝜋∕2) 2 2 4 = 0.4052(8.886) Dh = D0 = D0 = D0 𝜋∕2 𝜋 𝜋2 Dh = 3.6 = 5.564 dB 12.59. Circular aperture; uniform distribution; D0 (𝜃 = 35◦ ) maximized with respect to 𝜃 = 0◦ . | λ λ | = = 0.511λ (a) a = 3.413 sin 𝜃c ||𝜃c =35◦ 3.413(0.57358) a = 0.511λ 1.086(𝜋) 1.086𝜋 || = = 10.37 = 10.158 dB (b) D0 = | 2 | sin 𝜃c |𝜃 =35◦ (0.57358)2 c

(c) The directivity @ 𝜃 = 35◦ is −3.985 dB from the maximum @ 𝜃 = 0◦ . Therefore D0 (𝜃c = 35◦ ) = D0 (𝜃 = 0◦ ) − 3.985 = 10.158 − 3.985 = 6.173 dB D0 (𝜃c = 35◦ ) = 6.173 dB

12.60. (a) One method that can be used combines the results of a vertical dipole in the presence of a thin, plane, infinite, perfectly conducting electric screen with a horizontal opening and those of a vertical dipole in the presence of a flat, thin, perfectly conducting electrical vertical strip as shown in part b of the figure. The strip has been rotated to represent the magnetic equivalent of the screen’s opening. (b) Another method combines the results of a vertical dipole in the presence of an electric conducting screen with a horizontal dipole in the presence of a horizontal electric conducting strip, as shown in part c of the figure. The dipole has been rotated to interchange the E and H-fields and obtain the magnetic equivalent of the actual source. ( ) ( ) −b∕2 < y < b∕2 a = 0.1λ 12.61. E = −E0 â x −a∕2 < x < a∕2 b = 0.5λ E = −̂ax E0 M = −̂n × 2E = −̂n × (−2̂ax E0 ) = â y 2E0 sin X sin Y ka , X= sin 𝜃 cos 𝜙 X Y 2 sin X sin Y kb E𝜙 = C cos 𝜃 sin 𝜙 , Y= sin 𝜃 sin 𝜙 X Y 2 E𝜃 = −C cos 𝜙

C=j

abkE0 e−jkr 2𝜋r

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SOLUTION MANUAL

Vertical Electric Dipole

ε0

(a)

Vertical Electric Dipole

P

μ0

E0, H0

Vertical Electric Dipole

ε0 μ0 ’

Sa

P

Electric Conductor

Electric Conductor

ε0 μ0 ’

P

Sa

+

E1’ H1

E2, H2

(b) E0 = E1 + E2’ H0 = H1 + H2

(c) E0 = E3 + E4, H0 = H3 + H4

ε0 μ0

Horizontal Electric Dipole

Vertical Electric Dipole

P

Sa Electric Conductor

Electric Conductor

Sa

+

ε 0’ μ 0

P E4, H4

E3, H3

Vertical electric dipole in an unbounded free-space and Babinet’s principle equivalents.

Figure P12.60

(x−z plane)

(a) |E𝜃 (𝜃)|at 𝜃=0◦

| sin X | |≈1 ∝ || | | X |

X = 0.1(𝜋) sin 𝜃 cos 0◦ = 0.1(𝜋) sin 𝜃 ⇒

sin(0.1𝜋 sin 𝜃) ≈1 (0.1𝜋 sin 𝜃)

Y=0

⇒

(b) At 𝜙 = 90◦ (y-z plane) |E𝜃 (𝜃)|at 𝜙=90◦ = 0 X = 0 ⇒ sin X∕X = 1 𝜋 Y = sin 𝜃 ⇒ sin Y∕Y 2 z

–a/2

–b/2

x

a/2

b/2

y

sin Y =1 Y

387

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SOLUTION MANUAL

z

z

│Eθ (θ )│ϕ = 0° = 1

│Eθ (θ )│ϕ = 90°

0

0 x

y

0

│Eθ (ϕ )│

│Eϕ (θ )│≈ 0 ϕ = 0°

θ = 90°

z

0

0.8

0.6

0.4

0.2

x

y

x

│Eϕ (θ )│ϕ = 90°

z

θ = 90° │Eϕ (ϕ )│≈ 0

0

0.8

0.6

0.4

0.2

y

y

x

Figure P12.61

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SOLUTION MANUAL

(c) At 𝜃 = 90◦ (x-y plane) | sin X sin || |E𝜃 (𝜙)|at 𝜃=90◦ ∝ ||cos 𝜙 X Y || | X = 0.1(𝜋) cos 𝜙 𝜋 Y = sin 𝜙 2 (d) At 𝜙 = 0◦ (x-y plane) |E𝜙 (𝜃)|at 𝜙=0◦ ∝ |0| → zero (e) At 𝜙 = 90◦ (y-z plane) | sin Y || |E𝜙 (𝜃)|at 𝜙=90◦ ∝ ||cos 𝜃 Y || | Y=

𝜋 sin 𝜃 2

(f) At 𝜃 = 90◦ (x-z plane) |E𝜙 (𝜙)|at 𝜃=90◦ ∝ |0| → zero 12.62. Referring to Fig. 5-15 Zloop = 100 − j100 According to Babinet’s Principle 𝜂02

Zc Zs = Zslot Zloop = Zslot =

𝜂02

(

1

4 Zloop

=

(377) 4

) 2

4

(377)2 1 1 = 100 − j100 4 141.42∠ − 45◦

= 251.27∠45◦ Zslot = 177.65 + j177.65 (

) 𝜋 x a From (12-113a) and (12-113b) fx = 0

12.63. Ea = â y E0 cos

b∕2

fy = E0

∫−b∕2 ∫−a∕2 b∕2

= E0

∫−b∕2

(

a∕2

ejky

cos

) 𝜋 ′ jk(x′ sin 𝜃 cos 𝜙+y′ sin 𝜃 sin 𝜙) ′ ′ dx dy x e a ∫−a∕2

( ∫

(

a∕2

′ sin 𝜃 sin 𝜙

Bx

cos Axe

cos

) 𝜋 ′ jk(x′ sin 𝜃 cos 𝜙) ′ ′ dx dy x e a

eBx (B cos Ax + A sin Ax) dx = A2 + B2

)

⎞ ⎛ a∕2 ⎜ ka sin 𝜃 cos 𝜙 ⎟ 𝜋 ′ jkx′ sin 𝜃 cos 𝜙 ′ 𝜋 cos X cos x e dx = − a , X= ⎟ ⎜∴ ∫ a 2 2 ( 𝜋 )2 2 −a∕2 ⎟ ⎜ X − ⎠ ⎝ 2

389

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SOLUTION MANUAL

( ) 𝜋 fy = E0 − ab 2

ka sin 𝜃 cos 𝜙 kb sin 𝜃 sin 𝜙 sin Y cos X , Y= ( )2 Y , X = 2 2 𝜋 X2 − 2

The â 𝜃 and â 𝜃 component of E-field can be written 𝜋 (ab)E0 ke−jkr sin 𝜙 E𝜃 = −j 2 2𝜋r

= −j

abE0 −jkr sin 𝜙 ke 4r

cos X sin Y ( )2 Y 𝜋 X2 − 2

cos X sin Y ( )2 Y 𝜋 X2 − 2

𝜋 abE0 ke−jkr cos 𝜃 cos 𝜙 E𝜙 = −j 2 2𝜋r

= −j

a ⋅ bE0 −jkr cos 𝜃 cos 𝜙 ke 4r

cos X sin Y ( )2 Y 𝜋 X2 − 2 cos X sin Y ( )2 Y 𝜋 X2 − 2

12.64. b = λ∕20 < λ∕10, W = 10 cm; f = 10 GHz ⇒ λ = 30 × 109 ∕3 × 109 = 3 cm ( )] [ 2𝜋 λ 2 ⎤ ⎡ [ ] 2 ⎥ (kb) W𝜋 10𝜋 ⎢ 1− = 1 − λ 20 (a) Ga = ⎢ ⎥ 𝜂λ 24 120𝜋(3) ⎢ 24 ⎥ ⎣ ⎦ [ ] 2 (𝜋∕10) 1 1 1− = [1 − 0.00411] = 36 24 36 Ga = 27.892 × 10−3 ( )] 10(𝜋) [ 2𝜋λ W𝜋 1 − 0.636 ln (b) Ba = [1 − 0.636 ln(kb)] = 𝜂λ[ 20λ ( )] 120𝜋(3) 𝜋 1 1 − 0.636 ln = 36 10 1 1 Ba = [1 − 0.636(−1.15786)] = [1 + 0.7364] = 48.233 × 10−3 36 36 Ya = Ga + jBa = (27.892 + j48.233) × 10−3 (c) When both slots are identical and λg ∕2 apart, their admittances add in parallel. That is, when the admittance of the second slot is transferred a distance of λg ∕2 to the input (where the other admittance is), they are identical; inspedances/admittnaces inside transmission lines repeat every λg ∕2. Yin = 2Ya = 2(Ga + jBa ) = 2(27.892 + j48.233) × 10−3 = (55.784 + j96.466) × 10−3 Zin =

1 1 = = 4.492 − j7.7685 Yin (55.784 + j96.466) × 10−3

Zin = 4.492 − j7.7685

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13

Solution Manual

13.1. To find the fields within the horn, we will use the cylindrical coordinate system of 𝜌, 𝜓, x of Fig. 13.2(a). Since the region within the horn is source-free, Maxwell’s equations ∇ × E = −j𝜔𝜇H

(1)

∇ × H = j𝜔𝜀E

(2)

must be satisfied. In terms of the cylindrical coordinates 𝜌, 𝜓, x, equations 1 and 2 reduces to j𝜔𝜀E𝜌 = j𝜔𝜀E𝜓 = j𝜔𝜀Ex = −j𝜔𝜇H𝜌 = −j𝜔𝜇H𝜓 = −j𝜔𝜇Hx =

1 𝜕Hx 𝜕H𝜓 − 𝜌 𝜕𝜓 𝜕x 𝜕H𝜌

(3)

𝜕Hx 𝜕𝜌

(4)

1 𝜕 1 𝜕H𝜌 (𝜌H𝜓 ) − 𝜌 𝜕𝜌 𝜌 𝜕𝜓

(5)

1 𝜕Ex 𝜕Ex − 𝜌 𝜕𝜓 𝜕x

(6)

𝜕x

𝜕E𝜌

−

𝜕Ex 𝜕𝜌

(7)

1 𝜕 1 𝜕E𝜌 (𝜌E𝜓 ) − 𝜌 𝜕𝜌 𝜌 𝜕𝜓

(8)

𝜕x

−

If we assume that the waveguide feeding the horn only supports the dominant TE10 -mode, the lowest order mode within the sectoral guide (horn) is that which is analogous to the TE10 -mode of the rectangular guide, with all the other modes attenuated in the transition region (throat) between the waveguide and the horn. Thus the dominant mode within the horn is one whose only non vanishing components are E𝜓 , H𝜌 , Hx . That is E𝜌 = Ex = H𝜓 = 0

(9)

Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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SOLUTION MANUAL

In addition 𝜕 =0 𝜕𝜓

(10)

Using (9) and (10), we can write (6) and (8) as j𝜔𝜇H𝜌 =

𝜕E𝜓

𝜕x 1 𝜕 (𝜌E𝜓 ) −j𝜔𝜇Hx = 𝜌 𝜕𝜌

(11) (12)

Substituting (11) and (12) into (4) we can write −𝜔2 𝜇𝜀E𝜓 =

𝜕 2 E𝜓 𝜕x2

+

] [ 𝜕 1 𝜕 (𝜌E𝜓 ) 𝜕𝜌 𝜌 𝜕𝜌

(13)

which when expanded can be written as 𝜕 2 E𝜓 𝜕𝜌2

( ) 2 1 𝜕E𝜓 𝜕 E𝜓 1 2 + + k − 2 E𝜓 = 0 + 𝜌 𝜕𝜌 𝜕x2 𝜌 where k2 = 𝜔2 𝜇𝜀

(14) (14a)

To solve (14), we make use of the method of separation of variables. We assume that E𝜓 (𝜌, X) = R(𝜌)X(x)

(15)

Substituting (15) into (14) leads to ( ) 1 𝜕R 1 𝜕2X 𝜕2 R 2 + R 2 + k − 2 RX = 0 X 2 +X 𝜌 𝜕𝜌 𝜕𝜌 𝜕x 𝜌

(16)

Dividing by RX and changing the partials to total derivation (16) reduces to 1 d2 X = −kx2 , kx2 = constant X dx2 [ ] ( 2 ) 1 1 d2 R 1 1 𝜕R 2 + + k − kx − 2 = 0 R d𝜌2 R 𝜌 𝜕𝜌 𝜌

(17) (18)

Multiplying (18) by 𝜌2 R reduces to 𝜌2

dR d2 R +𝜌 + [(k𝜌 𝜌)2 − 1]R = 0 d𝜌 d𝜌2 where k𝜌2 = k2 − kx2

(19) (19a)

Equation (19) is recognized as a special form (n = 1) of Bessel’s differential equation [equation V-1 of Appendix V] with a solution of R(𝜌) = AH1(2) (k𝜌 𝜌) + B1 H11 (k𝜌 𝜌)

(20)

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393

where A and B are constants. The Hankel functions of the first and second kind of order one (n = 1) were chosen as solutions because they represent traveling waves in the inward and outward, respectively, radial directions. The solution of (17) is of the form X(x) = C cos(kx x) + D sin(kx x) where C and D are constants. Using (20) and (21) we can write (15) as [ ] E𝜓 (𝜌, x) = AH1(2) (k𝜌 𝜌) + B1 H1(2) (k𝜌 𝜌) [C cos(kx x) + D sin(kx x)]

(21)

(22)

Applying the boundary conditions E𝜓 (𝜌, x = a∕2) = E𝜓 (𝜌, x = −a∕2) = 0 leads to

(

kx a 2

)

(

kx a 2

)

( ) =0 ⎫ 𝜋 ⎪ D = 0, kx = m ⇒ a ⎬ ) ( ) ( m = 1, 2, 3, … kx a kx a ⎪ − D sin =0 ⎭ C cos 2 2 C cos

+ D sin

(23)

(24)

(25)

Equation (22) can be rewritten as [ ] E𝜓 (𝜌, X) = Am cos(kx x) H1(2) (k𝜌 𝜌) + 𝛼m H1(1) (k𝜌 𝜌)

(26)

where Am and 𝛼m are constants for the modes m = 1, 3, 5, … The nonvanishing magnetic field components can be obtained from (11) and (12). That is H𝜌 (𝜌, X) =

[ ] k 1 𝜕E𝜓 = j x Am sin(kx x) H1(2) (k𝜌 𝜌) + 𝛼m H1(1) (k𝜌 𝜌) j𝜔𝜇 𝜕X 𝜔𝜇

Hx (𝜌, X) = −

k𝜌 [ ] 1 1 𝜕 (𝜌E𝜓 ) = j Am cos(kx x) H0(2) (k𝜌 𝜌) + 𝛼m H0(1) (k𝜌 𝜌) j𝜔𝜇 𝜌 𝜕𝜌 𝜔𝜇

(27) (28)

by using (V-18). If we consider only the lowest order mode (m = 1, kx = 𝜋∕a) and no reflected component [𝛼m H1(1) (k𝜌 𝜌) = 𝛼m H0(1) (k𝜌 𝜌) = 0], the fields within the horn can be written as E𝜌 = EX = H𝜓 = 0 ( ) 𝜋 E𝜓 (𝜌, X) = A1 cos x H1(2) (k𝜌 𝜌) a ( ) 𝜋 𝜋 A1 sin x H1(2) (k𝜌 𝜌) H𝜌 (𝜌, X) = j 𝜔𝜇a a ( ) k𝜌 𝜋 Hx (𝜌, X) = j A1 cos x H0(2) (k𝜌 𝜌) 𝜔𝜇 a [ ( )2 ]1∕2 f ≫f0 𝜋 2 2 1∕2 = k[1 − (fc ∕f )2 ]1∕2 ≃ k k𝜌 = [K − (𝜋∕2) ] = k 1 − ka 𝜔𝜇∕k𝜌2 = 𝜔𝜇∕k = 𝜂

(29a) (29b) (29c) (29d)

(29e) (29f)

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SOLUTION MANUAL

The cylindrical components E𝜓 (𝜌, X) and H𝜌 (𝜌, X) can be resolved to any point within the horn to their rectangular counter parts using (VII-7a) or (VII-7b). Thus Ez = −E𝜓 sin 𝜓,

Hz = H𝜌 cos 𝜓

(30)

Ey = E𝜓 cos 𝜓,

Hy = H𝜌 sin 𝜓

(31)

which for horns with small flares (𝜓 small) reduce to Ez = Hy = 0

(32)

Ey ≃ E𝜓 ,

(33)

Hz ≃ H𝜌

If the length of the horn is large, the Hankel functions can be approximated by their asymptotic expansions of (V-17), or √ √ 2 −j(k𝜌−𝜋∕4) (2) 2 −j(k𝜌−𝜋∕4) (2) , H0 (k𝜌) ≃ (34) e e H1 (k𝜌) ≃ j 𝜋k𝜌 𝜋k𝜌 where 𝜌 = (y2 + z2 )1∕2

(34a)

Choosing a new coordinate system (x′ , y′ , z′ ), as shown in Fig. 13.2(b) such that x′ = x,

y′ = y,

z′ = z − 𝜌1 ,

𝜌1 = 𝜌 cos 𝜓e

(35)

we can write 𝜌 of (34a) as ]1∕2 z′ =0 [ 𝜌 = (z2 + 𝜌2 )1∕2 = [(z′ + 𝜌1 )2 + y′2 ]1∕2 = 𝜌21 + y′2 ] [ ( )2 ( )4 1 y′ 1 y′ 𝜌 = 𝜌1 1 + − +⋯ 2 𝜌1 8 𝜌1

(36)

For narrow horns (y′ ≪ 𝜌1 ) [ ( )2 ] ⎧ 1 y′ ⎪𝜌 1 + 𝜌≃⎨ 1 2 𝜌1 ⎪𝜌 ⎩ 1

for phase terms for amplitude terms

Using (30)-(37), we can write (29a)-(29b) as Ez′ = Ex′ = Hy′ = 0 √ 2 −jk𝜌1 E1 = jA1 e , 𝜋k𝜌1 ( ′ )2 ) 𝜋 ′ −j 2k 𝜌y1 x e a ( ′ )2 ( ( 𝜂) ) 𝜋 ′ −j 2k 𝜌y1 𝜋 sin Hz′ (x′ , y′ ) ≃ jE1 x e ka a ( ′ )2 ) ( E 𝜋 ′ −j 2k 𝜌y1 Hx′ (x′ , y′ ) ≃ + 1 cos x e 𝜂 a

Ey′ (x′ , y′ ) ≃ E1 cos

(

(37)

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SOLUTION MANUAL

13.2. Using the geometry of Figure 13.2(b) 2𝜋 2𝜋 2𝜋 2𝜋 (a) 𝜙max = = k𝛿max = (𝜌 − 𝜌1 ) = (𝜌 − 𝜌e cos 𝜓e ) = 𝜌 (1 − cos 𝜓e ) 3 λ e λ e λ e ( 𝜓 )] [ 2𝜋 2𝜋 e = (10λ)(1 − cos 𝜓e ) = 2𝜋(10λ)(1 − cos 𝜓e ) = 2𝜋(10) 2 sin2 3 λ 2 ) ( 𝜓 𝜓 1 1 1 sin2 e = = 7.418◦ = ⇒ e = sin−1 √ 2 3(20) 60 2 20 𝜓e = 2(7.418◦ ) = 14.836◦ ,

2𝜓e = 29.672◦

b1 = 𝜌e sin 𝜓e = 10λ sin(14.836◦ ) = 2.56λ ⇒ b1 = 5.12λ 2 (c) 𝜌1 = 𝜌e cos 𝜓e = 10λ cos(14.836◦ ) = 9.667λ

(b)

[ ( ) ( )] 64a𝜌1 b1 b1 2 2 C +S DE = √ √ 𝜋λb1 2λ𝜌 2λ𝜌 1

1

b1

5.12λ = = 1.1644 √ 2λ(9.667λ) 2λ𝜌1 ( ) b 1 C2 √ = [C(1.1644)]2 = (0.738)2 = 0.5446 2λ𝜌1 ( ) b1 2 S = [S(1.1644)]2 = (0.591)2 = 0.3493 √ 2λ𝜌1 DE =

64(0.5λ)(9.667λ) (0.5446 + 0.3493) = 17.192 = 12.35 dB 𝜋(5.12)λ2

(d) GE = 𝜀t DE = 𝜀r 𝜀cd DE = 𝜀r (1)DE = (1 − |Γ|2 )DE = [1 − (0.2)2 ](17.192) = 0.96(17.192) = 16.50 = 12.18 dB ]1∕2 [ = 6.2458λ 13.3. 𝜌1 = 6λ, b1 = 3.47λ, a = 0.5λ ⇒ 𝜌e = 𝜌21 + (b1 ∕2)2 (a) s =

b21 8λ𝜌1

=

(3.47λ)2 = 0.25 8λ(6λ)

b1 sin 𝜃 = 0 λ For s = 1∕4 at 𝜃 = 0◦ . ] [ 1 + cos(0◦ ) E = 0 + 20 log10 = 0 + 0 = 0 dB 2 b 𝜃 = 10◦ : 1 sin 𝜃 = 3.47 sin(10◦ ) = 0.6. Thus at 𝜃 = 10◦ λ ] [ 1 + cos(10◦ ) E = −3.5 + 20 log10 = −3.5 − 0.066 = −3.566 dB 2 b 𝜃 = 20◦ : 1 sin 𝜃 = 3.47 sin(20◦ ) = 1.187 λ ] [ 1 + cos(20◦ ) E = −9.25 + 20 log10 = −9.25 − 0.266 = −9.516 dB 2 𝜃 = 0◦ :

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SOLUTION MANUAL

(b)

) ( )] [ ( 64𝜌1 a b b 1 1 DE = + S2 √ C2 √ 𝜋b1 λ 2λ𝜌 2λ𝜌 1

(13-19)

1

b1 3.47λ = 1.00 = √ √ 2λ𝜌1 2λ ⋅ 6λ C2 (1.00) = (0.77989)2 = 0.6082 S2 (1.00) = (0.43826)2 = 0.1921 64(6λ)(0.5λ) [0.6082 + 0.1921] = 14.095 = 11.49 dB 𝜋(3.47λ)(λ) √ √ b1 50 50 B= = 3.47 = 9.8179 λ 𝜌e ∕λ 6.2458

DE =

From Fig. 13.8 ⇒ GE = 81.32 Using (13.19c) DE =

81.32 a GE = 0.5 √ = 14.37 = 11.57 dB √ λ 50 50 𝜌e ∕λ 6.2458

13.4. 𝜌1 = 6λ, b1 = 6λ, a = 0.5λ ⇒ 𝜌e = [𝜌21 + (b1 ∕2)2 ]1∕2 = (62 + 32 )1∕2 λ 𝜌e = 6.708λ 2 2 (a) s = b1 ∕8λ𝜌1 = 6 ∕8(6) = 3∕4 ) ( b 1 + cos 𝜃 𝜃 = 0◦ : 1 sin 𝜃 = 0 ⇒ E = −3.3 + 20 log10 λ 2 𝜃=0◦ = −3.3 + 0 = −3.3 dB ) ( | b 1 + cos 𝜃 1 = 1.042 ⇒ E = −0.25 + 20 log10 sin 𝜃 || 𝜃 = 10◦ : λ 2 𝜃=10◦ |𝜃=10◦ = −0.25 − 0.066 = −0.316 dB ) ( b 1 + cos 𝜃 1 sin(20◦ ) = 2.052 ⇒ E = −3.25 + 20 log10 𝜃 = 20◦ : λ 2 𝜃=20◦ = −3.25 − 0.266 = −3.516 dB b1 6 = √ = 1.732 (b) √ 2λ𝜌1 12 Using Appendix IV ⇒ C2 (1.732) = (0.327)2 = 0.10693 S2 (1.732) = (0.51776)2 = 0.2681 [ ( ) ( )] 64a𝜌1 b1 b1 2 2 C Using (13-18) ⇒ DE = +S √ √ 𝜋b1 λ 2λ𝜌 2λ𝜌 1

1

64(0.5)6 [0.10693 + 0.2681] = 3.8197 = 5.82 dB 𝜋(6) √ √ b1 50 50 B= √ =6 = 16.38 𝜌 ∕λ 6.708 e λ

DE =

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SOLUTION MANUAL

397

From Fig. 13.8 ⇒ GE = 30.316 Using (13.19c) ⇒ DE =

13.5. b1 =

√ 2𝜋𝜌1

0.5(30.316) a GE =√ = 5.552 = 7.444 dB √ λ 50 50∕6.708 𝜌e ∕λ

Using (13-18a) 𝜌1 = 6λ ⇒ b1 = 3.46λ 𝜌1 = 10λ ⇒ b1 = 4.47λ 𝜌1 = 15λ ⇒ b1 = 5.48λ 𝜌1 = 20λ ⇒ b1 = 6.32λ 𝜌1 = 30λ ⇒ b1 = 7.75λ 𝜌1 = 50λ ⇒ b1 = 10.0λ 𝜌1 = 75λ ⇒ b1 = 12.25λ 𝜌1 = 100λ ⇒ b1 = 14.14λ

Using Figure 13.7 b1 ≃ 3.5λ b1 ≃ 4.5λ b1 ≃ 5.5λ b1 ≃ 6.5λ b1 ≃ 8.0λ b1 ≃ 10.3λ b1 ≃ 12.5λ b1 ≃ 14.5λ

13.6. 𝜌1 = 20λ, a = 0.5λ (a) From Figure 13.7 ⇒ b1 = 6.5λ ( ) b ∕2 6.5λ∕2 6.5 6.5 = tan−1 (0.1625) = 9.23◦ (b) tan 𝜓e = 1 = = ⇒ 𝜓e = tan−1 𝜌1 20λ 40 40 2𝜓e = 2(9.23◦ ) = 18.46◦ ) ( )] [ ( 64𝜌1 a b1 b1 2 2 (c) DE = +S (13-18) C √ √ 𝜋b1 λ 2λ𝜌 2λ𝜌 1

1

b1 6.5λ = 1.028 =√ √ 2λ𝜌1 2λ(20λ) Using the table in Appendix IV C2 (1.028) = (0.77)2 = 0.5929 S2 (1.028) = (0.44)2 = 0.1936 DE =

64(20)(0.5) (0.5929 + 0.1936) = 24.65 = 13.92 dB 𝜋(6.5)

which compares very well with the value from Fig. 13.7. (d) From Figure 13.6 ⇒ HPBW = 10◦ ]1∕2 [ (e) 𝜌e = 𝜌21 + (b1 ∕2)2 = [(20)2 + (6.5∕2)2 ]1∕2 λ = 20.26λ √ √ b 50 50 = 6.5 = 10.21 ⇒ GE = 81.6 (from Figure 13.9) B= 1 λ 𝜌e ∕λ 20.26 0.5(81.6) a G = 25.97 = 14.14 dB Using (13-20c) ⇒ DE = √ E = √ λ 50 50∕20.26 𝜌e ∕λ 2.286 λ = 0.8382λ 2.7273 1.016 λ = 0.3725λ b = 0.4 in = 1.016 cm ⇒ b = 2.7273

13.7. a = 0.9 in = 2.286 cm ⇒ a =

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SOLUTION MANUAL

30 × 109 = 2.7273 cm 11 × 109 D λ 30λ DE = 30 = 14.77 dB ⇒ E = = 35.79 ≃ 36 a 0.8382λ λDE ≃ 36 Using Figure 13.7 ⇒ 𝜌1 = 10λ, b1 ≃ 4.5λ, for a λ=

( 𝜓e = tan

−1

b1 ∕2 𝜌1

)

= tan−1 (0.225) = 12.68◦

2𝜓e = 25.36◦ √ √ 2λ𝜌1 = 2λ(10λ) = 20λ2 = 4.4721λ √ √ 𝜌e = 𝜌21 + (b1 ∕2)2 = (10)2 + (4.4721∕2)2 λ = 10.2470λ ( ) ) ( b1 ∕2 2.2361λ = 12.6044◦ = 0.2200 (rads) 𝜓e = tan−1 = tan−1 𝜌1 10λ √ √ √ b 50 50 (b) B = 1 = 4.4721 = 4.4721 4.8795 = 4.4721(2.2090) λ 𝜌e ∕λ 10.2470

13.8. (a) b1 ≃

√

B = 9.8787 ⇒ From Figure 13.8 ⇒ GE ≃ 81.5 DE =

81.5 a GE 81.5 = 0.7620 √ = 28.114 = 14.49 dB = 0.7620 √ λ 2.2090 50 50 𝜌e ∕λ 10.2470

λ=

c 30 × 109 = 3 cm = 1.1811 in = f 10 × 109

a=

0.9 0.4 λ = 0.7620λ, b = λ = 0.3387λ 1.1811 1.1811

(c) 𝜀ap = Aem =

Aem 2.237λ2 = = 0.6565 = 65.65% Ap 0.7620λ(4.4721λ) λ2 λ2 λ2 G0 = D0 = (28.114) = 2.2372λ2 4𝜋 4𝜋 4𝜋

2𝜋 (10.2470 − 10)λ = 1.5519 rads = 88.93◦ λ 13.9. To find the fields within an H-plane sectoral horn we can use a procedure similar to that of an E-plane sectoral horn of Problem 13.1. (d) Δ𝜙max = k(𝜌e − 𝜌1 ) =

For the H-plane horn, Maxwell’s equations reduce to j𝜔𝜀Ep = j𝜔𝜀E𝜓 = j𝜔𝜀Ey =

1 𝜕Hy 𝜕H𝜓 − 𝜌 𝜕𝜓 𝜕y 𝜕H𝜌 𝜕y

−

𝜕Hy 𝜕𝜌

1 𝜕 1 𝜕H𝜌 (𝜌H𝜓 ) − 𝜌 𝜕𝜌 𝜌 𝜕𝜓

(1) (2) (3)

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SOLUTION MANUAL

1 𝜕Ey 𝜕E𝜓 − 𝜌 𝜕𝜓 𝜕y

−j𝜔𝜇H𝜌 =

𝜕E𝜌

−j𝜔𝜇H𝜓 =

𝜕y

(4)

𝜕Ey

−

(5)

𝜕𝜌

1 𝜕 1 𝜕E𝜌 (𝜌E𝜓 ) − 𝜌 𝜕𝜌 𝜌 𝜕𝜓

−j𝜔𝜇Hy =

399

(6)

Using the geometry of Figure 13.9, the nonvanishing components which best match the TE10 -mode of the waveguide are Ey , H𝜌 , H𝜓 , or E𝜌 = E𝜓 = Hy = 0 and

𝜕 =0 𝜕y

(7)

Using (7) reduces (3) and (4) to −j𝜔𝜇H𝜌 = j𝜔𝜇H𝜓 =

1 𝜕Ey 𝜌 𝜕𝜓

(8)

𝜕Ey

(9)

𝜕𝜌

Substituting (8) and (9) into (3) leads to −𝜔2 𝜇𝜀Ey = 𝜌2

𝜕 2 Ey 𝜕𝜌2

+𝜌

1 𝜕 𝜌 𝜕𝜌 𝜕Ey 𝜕𝜌

(

+

𝜌

𝜕Ey 𝜕𝜌

𝜕 2 Ey 𝜕𝜓 2

) +

2 1 𝜕 Ey 𝜌2 𝜕𝜓 2

+ (𝜌k)2 Ey = 0

or

(10)

(11)

where k2 = 𝜔2 𝜇𝜀 Assuming (11) is separable, we can write Ey (𝜌, 𝜓) = R(𝜌)𝜓(𝜓)

(12)

and reduce (11) to 𝜌2

dR d2 R +𝜌 + [(k𝜌)2 − p2 ]R = 0 d𝜌 d𝜌2

𝜕2𝜓 + p2 𝜓 = 0 𝜕2 𝜓 2

(13) (14)

where p2 is a positive constant. For the horn, the desired solution of (13) and (14) are of the form [Equation (13) is Bessel’s equation, see (V-1)] R(𝜌) = AHp2 (k𝜌) + BHp(1) (k𝜌)

(15)

𝜓(𝜓) = C cos(p𝜓) + D sin(p𝜓)

(15a)

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SOLUTION MANUAL

where A,B,C,D are constants. Thus (12) can be written as [ ] Ey (𝜌, 𝜓) = AHp(2) (k𝜌) + BHp(1) (k𝜌) [C cos(p𝜓) + D sin(p𝜓)]

(16)

Referring to Figure (13.10(b)) and applying the boundary conditions of Ey (𝜌, 𝜓 = 𝜓h ) = Ey (𝜌, 𝜓 = −𝜓h ) = 0

(17)

we find that ( D = 0; p = m

𝜋 2𝜓h

) , m = 1, 3, 5, …

(18)

and write (16) as [ ] Ey (𝜌, 𝜓) = Bm cos(p𝜓) Hp(2) (k𝜌) + 𝛽m Hp(1) (k𝜌)

(19)

where Bm and 𝛽m are constants for the modes m = 1, 3, 5 ⋯. The nonvanishing magnetic field components H𝜌 and H𝜓 of (8) and (9) reduce to H𝜌 (𝜌, 𝜓) = − H𝜓 (𝜌, 𝜓) =

] [ p 1 1 1 𝜕Ey = −jBm sin(p𝜓) Hp(2) (k𝜌) + 𝛽m Hp(1) (k𝜌) j𝜔𝜇 𝜌 𝜕𝜓 𝜔𝜇 𝜌

] [ ′ ′ 1 𝜕Ey 𝜕 k = −jBm cos(p𝜓) Hp(2) (k𝜌) + 𝛽m Hp(1) (k𝜌) , = j𝜔𝜇 𝜕𝜌 𝜔𝜇 𝜕(k𝜌)

(20) (21)

The cylindrical components H𝜌 and H𝜓 can be resolved to their rectangular counter parts using (VII-7a) or (VII-7b). Thus Hz = H𝜌 cos 𝜓 − H𝜓 sin 𝜓

(22a)

Hx = H𝜌 sin 𝜓 + H𝜓 cos 𝜓

(22b)

which for small flare horns (𝜓 small) reduce to Hz (𝜌, 𝜓) ≃ H𝜌 (𝜌, 𝜓), Hx (𝜌, 𝜓) ≃ H𝜓 (𝜌, 𝜓)

(23)

Assuming the lowest order mode (m = 1) and only radially outward traveling waves, the fields within the horn can be written as E𝜌 = E𝜓 = Hy = 0

(24)

Ey (𝜌, 𝜓) = B1 cos(p𝜓)Hp(2) (k𝜌)

(25)

′

Hx (𝜌, 𝜓) = −jB1 𝜂 cos(p𝜓)Hp(2) (k𝜌) Hz (𝜌, 𝜓) = −jB1

p 1 sin(p𝜓)Hp(2) (k𝜌) 𝜔𝜇 𝜌

(26) (27)

Using the asymptotic form of the Hankel function as given by (V-17), we can write (25) as √ 2 p (28) cos(p𝜓)e−jk𝜌 Ey (𝜌, 𝜓) ≃ j B1 𝜋k𝜌 where 𝜌 = (z2 + x2 )1∕2

(28a)

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401

SOLUTION MANUAL

using a new coordinate system (x′ , y′ , z′ ), as shown in Figure 13.10(a) such that x′ = −x, y′ = −y, z′ = z − 𝜌2 , 𝜌2 = 𝜌h cos 𝜓h

(29)

We can write (28a) as 𝜌 = [z2 + x2 ]1∕2 = [(z′ + 𝜌2 )2 + x′2 ]1∕2

(29a)

Equation (29a) can be written, using the binomial expansion, at the aperture of the horn (z′ = 0) as [ 𝜌 = [𝜌2 + x ] 2

′2 1∕2

(

= 𝜌2 1 +

x′ 𝜌2

)2 ]1∕2

[ = 𝜌2

1 1+ 2

(

x′ 𝜌2

)2

1 − 8

(

x′ 𝜌2

]

)4

+⋯

(30)

For a small flare angle horns (x′ ≪ 𝜌2 ), (30) can be approximated by [ ( )2 ] ⎧ 1 x′ ⎪𝜌 1 + for phase terms 𝜌≃⎨ 2 2 𝜌2 ⎪𝜌 for amplitude terms ⎩ 2

(31)

Using (31) we can write (28) as ( Ey′ (x′ , 𝜓) ≃ E2 cos

𝜋 2𝜓h

)

( −j 2k

e

x′2 𝜌2

)

(32)

where √ p

E2 = j B1

2 −jk𝜌2 e 𝜋k𝜌2

(33)

The fields at the aperture of horn can be approximated by ( Ey′ (x′ )

= E2 cos

E Hx′ (x′ ) = − 2 cos 𝜂

) k ( x′2 ) 𝜋 ′ −j 2 𝜌2 x e a1 (

) k ( x′2 ) 𝜋 ′ −j 2 𝜌2 x e a1

√ √ 13.10. 𝜌2 = 6λ, a1 = 6λ, b = 0.25λ ⇒ 𝜌h = 𝜌1 2 + (a1 ∕2)2 = 36 + 9λ = 6.7082λ 4𝜋b𝜌2 (a) DH = {[C(u) − C(v)]2 + [S(u) − S(v)]2 } a1 λ ) (√ (√ ) λ𝜌2 a1 6 1 6 1 u= √ = 2.02 +√ +√ =√ a1 6 λ𝜌2 6 2 2 ) (√ (√ ) λ𝜌2 a1 6 1 6 1 v= √ = −1.44 −√ −√ =√ a 6 λ𝜌 6 2 2 2

(34)

(35)

(13-39)

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SOLUTION MANUAL

Using Appendix IV C(u) = C(2.02) = 0.5069, S(u) = S(2.02) = 0.3496 C(v) = C(−1.44) = −C(1.44) = −0.50396 S(v) = S(−1.44) = −S(1.44) = −0.70712 4𝜋(0.5)6 {(0.5069 + 0.50396)2 + (0.3496 + 0.70712)2 } = 6.718 = 8.27 dB 6 √ √ a1 50 50 A= =6 = 6(2.7301) = 16.38 ⇒ GH = 84.13 (from Fig. 13.15) λ 𝜌h ∕λ 6.7082

DH =

b G Using (13-40c) ⇒ DH = √ H = 0.25 λ 50 𝜌h ∕λ

(b) Using ⇒ t = 𝜃 = 30◦ :

a21 8λ𝜌2

=

62 8(6)

=

)

( 84.13

√ 50∕6.7082

= 7.704 = 8.867 dB

6 = 3∕4 8

a1 sin 𝜃1 = 6 sin(30◦ ) = 3 λ

(

1 + cos 30◦ 2 = −14.25 − 0.6022 = −14.8522

Using Figure 13.14 ⇒ E = −14.25 + 20 log10

)

) ( a1 1 + cos 45◦ sin 𝜃1 = 6 sin(45◦ ) = 4.24 ⇒ E = −24 + 20 log10 λ 2 = −25.3754 ) ( a 1 + cos 90◦ 𝜃 = 90◦ : 1 sin 𝜃1 = 6 sin(90◦ ) = 6 ⇒ E = −33.5 + 20 log10 λ 2 = −39.52

𝜃 = 45◦ :

13.11. a1 =

√ 3λ𝜌2

Using (13-41c) 𝜌2 = 6λ ⇒ a1 = 4.2426λ 𝜌2 = 8λ ⇒ a1 = 4.899λ 𝜌2 = 10λ ⇒ a1 = 5.477λ 𝜌2 = 15λ ⇒ a1 = 6.71λ 𝜌2 = 20λ ⇒ a1 = 7.746λ 𝜌2 = 30λ ⇒ a1 = 9.487λ 𝜌2 = 50λ ⇒ a1 = 12.247λ 𝜌2 = 75λ ⇒ a1 = 15.00λ 𝜌2 = 100λ ⇒ a1 = 17.32λ

Using Figure 13.16 a1 ≃ 4.25λ a1 ≃ 5.0λ a1 ≃ 5.5λ a1 ≃ 6.9λ a1 ≃ 7.75λ a1 ≃ 9.5λ a1 ≃ 12.4λ a1 ≃ 15.25λ a1 ≃ 17.6λ

13.12. H-plane Horn; f = 10 GHz ⇒ λ = 3 cm; D0 = 13.25 dB = 101.325 = 21.1349 a = 2.286 cm = 2.286∕3λ = 0.762λ; b = 1.016 cm = 1.016∕3λ = 0.33867λ For maximum directivity ⇒ a1 =

√ 3λ𝜌2

(13-41c)

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SOLUTION MANUAL

1 u= √ 2

(√ λ𝜌2 a1

a +√1 λ𝜌2

) ( ) √ √ 3λ𝜌2 1 1 + √ =√ √ √ + 3 = 1.633 3λ𝜌2 λ𝜌2 3 2

(√

) 1 =√ 2

403

λ𝜌2

(13-41a) ) (√ (√ ( ) ) √ √ λ𝜌2 3λ𝜌2 λ𝜌2 a 1 1 1 1 v= √ −√1 − √ =√ =√ √ √ √ − 3 = −0.816 a1 λ𝜌2 3λ𝜌2 λ𝜌2 3 2 2 2 (13-41b) C(u) = C(1.633) = 0.35172, S(u) = S(1.633) = 0.60929 C(v) = C(−0.816) = −C(0.816) = −0.72956 S(v) = S(−0.816) = −S(0.816) = −0.26381 Using (13-39) DH = =

4𝜋b𝜌2 [(0.35172 + 0.72956)2 + (0.60929 + 0.26381)2 ] a1 λ 4𝜋b𝜌2 (1.16916 + 0.76230) a1 λ

4𝜋(1.016)𝜌2 𝜌 3(21.1349) (1.93147) ⇒ 2 = (1.93147) = 2.5712 3a1 a1 4(1.016)𝜋 √ ⇒ 𝜌2 = 2.5712a1 = 2.5712 3λ𝜌2

21.1349 =

𝜌22 = (2.5712)2 (3λ𝜌2 ) ⇒ 𝜌2 = (2.5712)2 (3)λ = (2.5712)2 (9) = 59.5 cm √ √ (a) a1 = 3λ𝜌2 = 3(3)(59.5) = 23.14 cm (b) 𝜌2 = 59.5 cm ( ) ) ( a1 ∕2 11.57 = tan−1 (0.1945) = 11◦ (c) 𝜓h = tan−1 = tan−1 𝜌2 59.5 ⇒ 2𝜓h = 22◦ . 13.13. a = 0.9′′ = 2.286 cm, b = 0.4 in = 1.016 cm 30 × 109 = 2.7273 cm 11 × 109 a 2.286 b 1.016 = = 0.8382, = = 0.37253 λ 2.7273 λ 2.7273 D λ 16.3λ = 43.755 DH = 16.3 = 12.12 dB ⇒ H = b 0.37235λ f = 11 GHz ⇒ λ =

Using Figure 13.14 ⇒ 𝜌2 = 10λ, a1 = 5.5λ √ √ √ 13.14. (a) a1 ≃ 3λ𝜌2 = 3λ(10λ) = 30λ2 = 5.4772λ √ √ 𝜌h = 𝜌2 2 + (a1 ∕2)2 = (10)2 + (5.4772∕2)2 λ = 10.3682λ ( ) ) ( a1 ∕2 2.7386 = 15.3156◦ = 0.2739 (rads) 𝜓h = tan−1 = tan−1 𝜌2 10

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SOLUTION MANUAL

a A= 1 λ

(b)

√

√ 50 50 = 5.4772 = 5.4772(2.1960) = 12.0280 𝜌h ∕λ 10.3682

A = 12.0280 ⇒ From Figure 13.15 ⇒ GH ≃ 98.8 (98.8) 98.8 b GH = 0.3387 = 0.3387 √ = 15.2384 = 11.8294 dB √ λ 2.1960 50 50 𝜌h ∕λ 10.3682 Aem 1.2126λ2 = = = 0.6537 = 65.37% Ap 0.3387λ(5.4772λ)

DH =

𝜀ap

(c)

Aem = (d) Δ𝜙max

λ2 λ2 λ2 G0 ≃ D0 = (15.2384) = 1.2126λ2 4𝜋 4𝜋 4𝜋 2𝜋 = k(𝜌h − 𝜌2 ) = (10.3682 − 10)λ = 2𝜋(0.3682) λ = 2.3135 rads = 132.56◦

13.15. Referring to Figure 13.16 ( 𝜌e = 𝜌1 + 2

2

[( = b1

b1 2

𝜌e b1

(

)2

)2

⇒ 𝜌 1 = 𝜌e − 2

2

b1 2

[

)2

(

⇒ 𝜌1 = 𝜌e − 2

b1 2

)2 ]1∕2

]1∕2 1 − 4

Also 𝜌1 = pe +

b b 𝜌1 cot(𝜓e ) = pe + = pe + 𝜌 1 2 2 (b1 ∕2)

(

b b1

)

( ) b ⇒ pe = 𝜌1 1 − b1

Thus [( ) ]1∕2 ( ) ) 𝜌e 2 ( 1 ) b b − pe = 𝜌1 1 − = b1 1− b1 b1 4 b1 √ ( )2 𝜌e 1 − = (b1 − b) b1 4 (

In a similar manner, we can show that √ ( ph = (a1 − a) 13.16. 𝜌1 𝜌2 a1 b1

= 13.5′′ = 14.2′′ = 7.65′′ = 5.65′′

𝜌h a1

)2 −

1 4

= 34.49 cm ⎫ 𝜌e = [𝜌1 2 + (b1 ∕2)2 ]1∕2 = 13.7924′′ = 35.0327 cm = 36.07 cm ⎪ ⇒ ⎬ = 19.43 cm 𝜌h = [𝜌1 2 + (a1 ∕2)2 ]1∕2 = 14.7061′′ = 37.3536 cm ⎪ = 14.35 cm ⎭

a = 0.9′′ = 2.286 cm b = 0.4′′ = 1.016 cm

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SOLUTION MANUAL

√ ( (a)

pe = (b1 − b)

𝜌e b1

)2

√ 1 − = (5.65 − 0.4) 4

(

13.7924 5.65

)2 −

405

1 4

= 12.544′′ = 31.862 cm √ √ ( )2 ) ( 𝜌h 14.7061 2 1 1 − = (7.65 − 0.9) − ph = (a1 − a) a1 4 7.65 4 = 12.529′′ = 31.8246 cm Therefore pe ≃ ph , and the pyramidal horn is physically realizable. 30 × 109 = 3.658537 cm = 1.44′′ (b) f = 8.2 GHz ⇒ λ = 8.2 × 109 13.5 7.65 0.9 𝜌1 = λ = 9.375λ, a1 = λ = 5.3125λ, a = λ = 0.625λ 1.44 1.44 1.44 13.7925 λ = 9.578′′ 1.44 14.2 5.65 0.4 𝜌2 = λ = 9.861λ, b1 = λ = 3.9236λ, b = λ = 0.278λ 1.44 1.44 1.44 14.7061 λ = 10.213λ 𝜌h = 1.44 (√ (√ ) ) λ𝜌2 a1 9.861 1 5.3125 1 u= √ +√ +√ = √ = 1.6142 a1 λ𝜌2 9.861 2 5.3125 2 (√ (√ ) ) λ𝜌2 a1 9.861 1 1 5.3125 u= √ −√ = √ −√ = −0.7783 a1 λ𝜌 9.861 2 2 5.3125 𝜌e =

2

b1

3.9236 = 0.8835 = √ √ 2λ𝜌2 2 ⋅ (9.861) C(u) = C(1.6142) = 0.3595 ⎫ ⎪ S(u) = S(1.6142) = 0.6262 From Appendix IV C(v) = C(−0.7783) = −C(0.7783) = −0.7091 ⎬ ⎪ S(v) = S(−0.7783) = −S(0.7783) = −0.2326 ⎭ ( ) ⎫ b1 C √ = C(0.8835) = 0.7579 ⎪ 2λ𝜌2 ⎪ ( ) ⎬ From Appendix IV b1 ⎪ S √ = S(0.8835) = 0.3249 ⎪ 2λ𝜌2 ⎭ Using (13-50) or (13-50a) 8𝜋𝜌1 𝜌2 {[C(u) − C(v)]2 + [S(u) − S(v)]2 } Dp = a1 b1 =

{

( C2

√

b1

)

2λ𝜌1

8𝜋(13.5)(14.2) [(0.3595 + 0.7091)2 + (0.6262 + 0.2326)2 ] 7.65(5.65) × [(0.7579)2 + (0.3249)2 ] Dp = 142.456 = 21.54 dB

( + S2

√

b1 2λ𝜌1

)}

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SOLUTION MANUAL

[

(

Using (13-51) ⇒ Dp = 10 1.008 + log10 ⎫ (3.9236)2 = 0.205 ⎪ 8λ𝜌1 8(9.375) ⎪ ⎬ 2 a21 (5.3125) ⎪ = = 0.358 ⎪ t= 8λ𝜌2 8(9.861) ⎭ b21

s=

a1 b1 λ2

)] − (Le + Lh )

=

Using Figure 13.21 ⇒

Le = 0.5 dB Lh = 0.8 dB

Thus Dp = 10{1.008 + log10 [5.3125(3.9236)] − (0.5 + 0.8) = 23.2698 − 1.3 = 21.96 dB √ 50 50 = 5.3125 = 11.7546 ⇒ GH = 98.254 from Figure 13.15 𝜌h ∕λ 10.213 √ √ b1 50 50 B= = 3.9236 = 8.965 ⇒ GE = 85.325 from Figure 13.8 λ 𝜌e ∕λ 9.578 a A= 1 λ

√

GE GH (85.325)(98.254) = √ √ (10.1859)(2.285)(2.2126) 50 50 10.1859 𝜌e ∕λ 𝜌h ∕λ

Using (13-52e) Dp =

Dp = 162.794 = 22.14 dB f = 10.3 GHz: ⇒ λ =

30 × 109 = 2.9126 cm = 1.1467′′ 10.3 × 109

Using the same procedure as for f = 8.2 GHz, we have that 𝜌1 = 11.773λ, 𝜌2 = 12.383λ, 𝜌e = 12.0279λ, 𝜌h = 12.8247λ a1 = 6.671λ, b1 = 4.927λ, a = 0.7849λ, b = 0.3488λ u = 1.7135, v = −0.9675, √

b1

= w = 1.015

2λ𝜌1

C(u) = 0.32515, C(v) = −0.7750, C(w) = 0.780 S(u) = 0.5359, S(v) = −0.4063, S(w) = 0.440 Using (13-50) ⇒ Dp =

8𝜋(13.5)14.2 {(0.32515 + 0.7750)2 + (0.5359 + 0.4063)2 } (7.65)(5.65) × {(0.78)2 + (0.44)2 } = 187.56 = 22.73 dB

s=

b21 8λ𝜌1

= 0.2577 ⇒ Le = 0.9 : t =

a21 8λ𝜌2

= 0.4492 ⇒ Lh = 1.5

Using (13-51) ⇒ Dp = 10{1.008 + log10 [6.671(4.927)]} − (0.9 + 1.5) = 22.847 dB A = 6.671(1.97452) = 13.172 ⇒ GH = 98.673 B = 4.927(2.03887) = 10.0455 ⇒ GE = 81.550

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SOLUTION MANUAL

Using (13-52e) ⇒ Dp =

407

98.673(81.550) = 196.2329 = 22.93 dB 10.1859(1.97452) 2.03887

f = 12.4 GHz ⇒ λ =

30 × 109 = 2.41935 cm = 0.9525′′ 12.4 × 109

Using the same procedure as for f = 8.2 GHz, we have that 𝜌1 = 14.173λ,

𝜌2 = 14.908λ,

𝜌e = 14.48λ,

𝜌h = 15.4395λ

a1 = 8.0315λ,

b1 = 5.9318λ,

a = 0.9449λ,

b = 0.4199λ

u = 1.811,

v = −1.131,

b w = √ 1 = 1.114 2λ𝜌1

C(u) = 0.340,

C(v) = −0.745,

C(w) = 0.760

S(u) = 0.443,

S(v) = −0.560,

S(w) = 0.545

Using (13-50) ⇒ Dp =

8𝜋(13.5)(14.2) {(0.340 + 0.745)2 + (0.443 + 0.560)2 } 7.65(5.65)

× [(0.76)2 + (0.545)2 ] = 212.84 = 23.28 dB s=

b21 8λ𝜌1

= 0.31 ⇒ Le = 1.4; t =

a21 8λ𝜌2

= 0.541 ⇒ Lh = 2.1

Using (13-51) ⇒ Dp = 10{1.008 + log10 [8.0315(5.9318)]} − (1.4 + 2.1) = 23.36 dB A = 8.0318(1.79956) = 14.453 ⇒ GH = 95.1 B = 5.9318(1.858) = 11.0227 ⇒ GE = 80.6 Using (13-52e) ⇒ Dp =

95.1(80.6) = 225.063 = 23.52 dB (10.1859)(1.79956)(1.858)

In summary, all three equations yield nearly identical results. The computed directivities agree closely with those of a commercial unit. Using the computer program Horn Analysis, the following directivities are obtained: At 8.2 GHz → Dp = 21.7017 dB 10.3 GHz → Dp = 22.7914 dB 12.4 GHz → Dp = 23.3311 dB 13.17.

𝜌1 = 5.3′′ = 13.46 cm ⎫ ⎪ 𝜌2 = 6.2′′ = 15.748 cm ⎪ a1 = 3.09′′ = 7.85 cm ⎪ 𝜌e = [𝜌1 2 + (b1 ∕2)2 ]1∕2 = 5.428′′ = 13.786 cm b1 = 2.34′′ = 5.944 cm ⎬ 𝜌h = [𝜌1 2 + (a1 ∕2)2 ]1∕2 = 6.3896′′ = 16.2296 cm ⎪ a = 0.9′′ = 2.286 cm ⎪ b = 0.4′′ = 1.016 cm ⎪ ⎭ √ √ ( )2 ) ( 𝜌e 5.428 2 1 1 (a) pe = (b1 − b) − = (2.34 − 0.4) − = 4.394′′ = 11.16 cm b1 4 2.34 4

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SOLUTION MANUAL

√ √ ( )2 ) ( 𝜌h 6.3896 2 1 1 ph = (a1 − a) − = (3.09 − 0.9) − = 4.394′′ = 11.16 cm b1 4 3.09 4 Therefore pe = ph , and the pyramidal horn is physically realizable. 9 (b) f = 8.2 GHz ⇒ λ = 30 × 10 = 3.658537 cm = 1.44′′ 8.2 × 109 5.3 3.09 0.9 𝜌1 = λ = 3.68λ, a1 = λ = 2.1458λ, a = λ = 0.625λ, 1.44 1.44 1.44 5.428 λ = 3.764λ 𝜌e = 1.44 6.2 2.34 0.4 λ = 4.305λ, b1 = λ = 1.625λ, b = λ = 0.278λ, 𝜌2 = 1.44 1.44 1.44 6.3896 = 4.437λ 𝜌h = 1.44 ) ) (√ (√ λ𝜌2 a1 4.305 1 2.1458 1 = 1.415 +√ =√ +√ u= √ a1 λ𝜌2 4.305 2 2 2.1458 ) (√ (√ ) λ𝜌2 a1 4.305 1 2.1458 1 v= √ = −0.0476 −√ −√ =√ a1 λ𝜌 4.305 2 2 2.1458 2

b1

1.625 = 0.5989 =√ √ 2λ𝜌2 2(3.68) C(u) = C(1.415) = 0.5284 ⎫ ⎪ S(u) = S(1.415) = 0.711 From Appendix IV C(v) = C(−0.0476) = −C(0.0476) = −0.0476 ⎬ ⎪ S(v) = S(−0.0476) = −S(0.0476) = −0.00025 ⎭ C(0.5989) = 0.5801, S(0.5989) = 0.11004 Using (11-50) or (13-50a) 8𝜋𝜌1 𝜌2 Dp = {[C(u) − C(v)]2 + [S(u) − S(v)]2 } a1 b1 =

{

( C

2

√

b1

)

2λ𝜌1

( +S

2

b1

)}

√ 2λ𝜌1

8𝜋(5.3)(6.2) [(0.5284 + 0.0476)2 + (0.7111 + 0.00025)2 ][(0.5801)2 + (0.11004)2 ] (3.09)(2.34)

Dp = 33.36 = 15.23 dB [ Using (13-51) ⇒ Dp = 10 1.008 + log10

(

a1 b1 λ2

)] − (Le + Lh )

b1 2 (1.624)2 = = 0.0896 ⎫ ⎪ ⇒ Le = 0.1 dB 8λ𝜌1 8(3.679) ⎬ Using Figure 13.21 ⇒ L = 0.2 dB a1 (2.1457)2 h t= = = 0.1337 ⎪ ⎭ 8λ𝜌2 8(4.304)

s=

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SOLUTION MANUAL

409

Thus Dp = 10{1.008 + log10 [2.1458(1.6825)]} − (0.1 + 0.2) = 15.5 − 0.3 = 15.2 dB √ 50 50 = 2.1458 = 7.2037 ⇒ GH = 71.29 from Figure 13.15 𝜌h ∕h 4.437 √ √ b1 50 50 B= = 1.625 = 5.9158 ⇒ GE = 58.66 from Figure 13.8 λ 𝜌e ∕h 3.769 a A= 1 λ

√

Using (13-52e) (Computer Result ⇒ 15.2507 dB) Dp =

GE GH 58.66(71.29) = = 33.57 = 15.26 dB √ √ 10.1859(3.64275)(3.3573) 50 50 10.1859 𝜌e ∕λ 𝜌h ∕λ

f = 10.3 GHz: λ =

30 × 109 = 2.9126 cm = 1.1467′′ 10.3 × 109

Using the same procedure as for f = 8.2 GHz, we have that 𝜌1 = 4.622λ, 𝜌2 = 5.4068λ, 𝜌e = 4.7336λ, 𝜌h = 5.5722λ a1 = 2.695λ, b1 = 2.041λ, a = 0.7849λ, b = 0.3488λ u = 1.4296, v = −0.2095, w = √

b1

= 0.6713

2λ𝜌1

C(u) = 0.510, C(v) = −0.209,

C(w) = 0.6371

S(u) = 0.709, S(v) = −0.005,

S(w) = 0.1545

Using (13-50) ⇒ Dp =

8𝜋(5.3)6.2 {(0.510 + 0.209)2 + (0.709 + 0.005)2 } 3.09(2.34) × {(0.6371)2 + (0.1545)2 } = 50.4 = 17.02 dB

Dp = 17.02 dB s=

b1 2 a 2 = 0.1126 ⇒ Le = 0.2 : t = 1 = 0.1679 ⇒ Lh = 0.25 8λ𝜌1 8λ𝜌2

Using (13-51) ⇒ Dp = 10{1.008 + log10 [2.695(2.041)]} − (0.2 + 0.25) = 17.034 dB A = 2.695(2.9955) = 8.0729 ⇒ GH = 78.6 B = 2.041(3.25) = 6.633 ⇒ GE = 66.1 Using (13-52e) ⇒ Dp =

(

Computer Program Horn Analysis 17.0838 dB

66.1(78.6) = 52.39 = 17.19 dB 10.1859(2.9955)3.25

f = 12.4 GHz ⇒ λ =

30 × 109 = 2.41935 cm = 0.9525′′ 12.4 × 109

Using the same procedure as for f = 8.2 GHz, we have that 𝜌1 = 5.5643λ, 𝜌2 = 6.5092λ, 𝜌e = 5.6987λ, 𝜌h = 6.7082λ a1 = 3.2441λ, b1 = 2.4567λ, a = 0.9449λ, b = 0.4199λ

)

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SOLUTION MANUAL

b u = 1.455, v = −0.343, w = √ 1 = 0.7364 2λ𝜌1 C(u) = 0.490, C(v) = −0.339, C(w) = 0.685 S(u) = 0.705, S(v) = −0.0225, S(w) = 0.200 8𝜋(5.3)6.2 {(0.99 + 0.339)2 + (0.705 + 0.0225)2 } 3.09(2.34)

Using (13-50) ⇒ Dp =

× [(0.685)2 + (0.2)2 ] Dp = 70.75 = 18.497 dB s=

[Using computer program Horn Analysis = 18.5150 dB]

b1 2 = 0.136 ⇒ Le = 0.25; 8λ𝜌1

t=

a1 2 = 0.202 ⇒ Lh = 0.3 8λ𝜌2

Using (13-51) ⇒ Dp = 10{1.008 + log10 [3.2441(2.4567)]} − (0.25 + 0.3) = 18.544 dB A = 3.2441(2.7301) = 8.8568 ⇒ GH = 84.55 B = 2.4567(2.9621) = 7.2769 ⇒ GE = 69.65 Using (13-52e) ⇒ Dp =

69.65(84.55) = 71.492 = 18.54 dB 10.1859(2.7301)(2.9621)

In summary, all three equations [i.e. (13-50), (13-51), and (13-52e)] yield nearly identical results. The computed directivities agree closely with those measured of the commercially available unit of Figs. 13.22 and 13.23. 13.18. (a) 𝜀ap ≃ 50% (b) a1 = 4λ0 , b1 = 2.5λ0 D0 =

4𝜋 4𝜋 4𝜋 Aem = 2 (𝜀ap Ap ) = 2 (0.5)(4λ0 )(2.5λ0 ) = 20𝜋 = 62.83185 = 17.982 dB 2 λ0 λ0 λ0

(c) Aem = 𝜀ap Ap = 0.5(4λ0 )(2.5λ0 ) = 5λ2 0 PL = W i Aem (PLF) =

10 × 10−3 (5λ20 ) λ20

( ) 1 2

= 25 × 10−3 = 25 mWatts

13.19. G0 = 17.05 dB = 10 log10 G0 (dimensionless) ⇒ G0 (dim) = 101.705 = 50.7 30 × 109 = 2.7273 cm 11 × 109 2.286λ 1.016 a= = 0.8382λ, b = λ = 0.3725λ 2.7273 2.7273 f = 11 GHz. ⇒ λ =

The initial value of 𝜒, is taken, using (13-55), as 𝜒1 =

50.7 = 3.219127 √ 2𝜋 2𝜋

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SOLUTION MANUAL

411

which does not satisfy (13-54). After few iteration, it is found that 𝜒1 = 2.96795 is a more accurate value. Thus 𝜌e = 𝜒1 λ = 2.96795(2.7273) = 8.0945 cm = 3.1868′′ = 2.96795λ ( ) G 2 1 (50.7)2 ( 1 ) λ = 3.49156λ = 9.5225 cm = 3.749′′ 𝜌h = 0 3 λ= 𝜒1 2.96795 8𝜋 8𝜋 3 √ √ G 50.7 3 3 a1 = 0 λ= λ = 3.23646λ = 8.8268 cm = 3.475′′ 2𝜋 2𝜋𝜒1 2𝜋 2𝜋(2.96795) √ √ b1 = 2𝜒1 λ = 2(2.96795)λ = 2.43637λ = 6.6447 cm = 2.616′′ [( ) ]1∕2 𝜌e 2 1 pe = (b1 − b) − = 6.25263 cm = 2.46167′′ ⎫ ⎪ b1 4 ⎪ [( ) ]1∕2 ⎬ ⇒ 𝜌e ≃ 𝜌h ≃ 6.2526 cm 𝜌h 2 1 ⎪ ≃ 2.4617′′ ph = (a1 − a) − = 6.25269 cm = 2.46169′′ ⎪ a1 4 ⎭ a Using (13-52a) ⇒ A = 1 λ b Using (13-52b) ⇒ B = 1 λ

√ √

50 = 12.247 ⇒ GH = 98.921 from Figure 13.15 𝜌h ∕λ 50 = 9.9999 ⇒ GE = 81.518 from Figure 13.8 𝜌e ∕λ

Using (13-52e) Dp =

GE GH 81.518(98.921) = = 50.9695 = 17.07 dB √ √ √ √ 50 50 50 50 10.1859 10.1859 𝜌e ∕λ 𝜌h ∕λ 2.96795 3.49156

9λ, a = 0.5λ, 13.20. 𝜌1 = 𝜌2 =√ √ b = 0.22λ√ (a) a1 = 3λ𝜌2 = 3λ(9λ) = 3 3λ = 5.1962λ √ √ √ b1 = 2λ𝜌1 = 2λ(9λ) = 3 2λ = 4.2426λ

(b)

𝜌e =

√

b B= 1 λ

√ √ ( √ )2 √ √ √ 3 2 2 2 2 √ 𝜌1 + (b1 ∕2) = (9λ) + = (81 + 4.5)λ2 = 9.2466λ λ 2 √

50 4.2426λ = 𝜌e ∕λ λ

√

50 = 4.2426(2.3254) = 9.8749 9.2466

B = 9.8749 ⇒ GE ≃ 80 (from Fig. 13.8) 0.5(80) 40 a GE =√ = = 17.2 = 12.356 dB √ λ 2.32538 50 50 𝜌e ∕λ 9.2466 √ √ ( √ )2 √ √ √ ) ( √ 3 3 27 2 λ = 9.3675λ 81 + 𝜌h = 𝜌22 + (a1 ∕2)2 = √(9λ)2 + = λ 2 4

DE =

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SOLUTION MANUAL

a A= 1 λ

√

50 5.1962λ = 𝜌h ∕λ λ

√

50 = 5.1962(2.3103) = 12 9.3675

A = 12 ⇒ GH ≃ 95 (from Fig. 13.15) DH =

Dp =

b GH √ λ 50

0.22(95) 20.9 = √ = = 9.0463 = 9.5647 dB 2.3103 50 𝜌h ∕λ 9.3675

𝜋λ2 𝜋λ2 DE DH = (17.2)(9.0463) = 138.8815 = 21.4264 dB 32ab 32(0.5)(0.22λ) Dp = 138.8815 = 21.4264 dB

Alternate Method: t = a21 ∕8λ𝜌2 = 27λ2 ∕8λ(9λ) = 0.375 ⇒ LH = 1.15 (from Fig. 13.21) s = b21 ∕8λ𝜌1 = 18λ2 ∕8λ(9λ) = 0.25 ⇒ LE = 0.8 (from Fig. 13.21) [ ( )] a1 b1 Dp (dB) = 10 1.008 + log10 − (LE + LH ) = 10[1.008 + 1.343] − (1.15 + 0.8) λ2 Dp (dB) = 21.56 dB 13.21. λ = (a) (b) (c) (d)

30 × 109 = 3 cm 10 × 109 √ √ √ a1 ≃ 3λ𝜌 = 3λ(10λ) = 30λ2 = 5.477λ = 16.43 cm √ √ b1 ≃ 2λ𝜌 = 20λ2 = 4.472λ = 13.416 cm 4𝜋 4𝜋 G0 = 12 2 (a1 b1 ) = 12 2 (5.477λ)(4.472λ) = 153.89 = 21.87 dB λ λ 1 1 er ecd 𝜀ap = 1(1)𝜀ap = , 𝜀ap = = 50% 2 2 λ2 32 Aem = G = (153.89) = 110.2156 cm2 = 110.2156 × 10−4 m2 4𝜋 0 4𝜋 Prec = W i Aem = 10 × 10−6 × 110.2156 × 10−4 = 1,102.156 × 10−10 = 11.02156 × 10−8 Prec = 11.02156 × 10−8 = 0.1102156 𝜇Watts

13.22. The problem was solved using computer program. 𝜒 = 6.047 𝜌e = 37.0 cm 𝜌h = 40.8 cm a1 = 27.4 cm b1 = 21.3 cm 𝜌e = 𝜌h = 31.7 cm

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SOLUTION MANUAL

13.23. a = 0.5λ, b = 0.25λ, 2𝜓e = 43◦ , 2𝜓h = 50◦ √ (a) b1 ≃ 2λ𝜌1 ( ) b ∕2 −1 b1 ∕2 𝜓e = tan ⇒ 𝜌1 = 1 𝜌1 tan 𝜓e 𝜌1 =

ρe

E-plane

b1 ∕2 b1 ∕2 = = 2.53665(b1 ∕2) tan(21.5◦ ) 0.39391 ρh

𝜌1 = 1.26932b1

a1/2

b21 = 2λ𝜌1 = 2λ(1.26932b1 ) = 2.53865b1 λ

ψh

H-plane

ρ2

b1 = 2.53865λ √ a1 ≃ 3λ𝜌2 ( ) a ∕2 −1 a1 ∕2 𝜓h = tan ⇒ 𝜌2 = 1 2 tan 𝜓h 𝜌 𝜌2 =

b1/2

ψe ρ1

a1 ∕2 a1 ∕2 = = 2.14451(a1 ∕2) tan(25◦ ) 0.46631 = 1.07225a1

a21

= 3λ𝜌2 = 3λ(1.07225a1 ) = 3.21676λa1 a1 = 3.21676λ

(b) 𝜌1 = 1.26932b1 = 1.26932(2.53865λ) = 3.22236λ 𝜌1 = 3.22236λ 𝜌2 = 1.07225a1 = 1.07225(3.21676λ) = 3.44917λ 𝜌2 = 3.44917λ √ √ 𝜌e = 𝜌21 + (b1 ∕2)2 = λ (3.22236)2 + (1.26932)2 = 3.46335λ √ √ 𝜌h = 𝜌22 + (a1 ∕2)2 = λ (3.44917)2 + (1.60838)2 = 3.80574λ [( (c) 𝜌e = (b1 − b)

𝜌e b1

)2

]1∕2 1 − 4

[( = (2.53865 − 0.25)λ

3.46335 2.53865

)2

1 − 4

]1∕2

= 2.28865λ(1.26932) 𝜌e = 2.90503λ [( 𝜌h = (a1 − a)

𝜌h a1

)2

]1∕2 1 − 4

= 2.71676λ(1.07225) 𝜌h = 2.91304λ

[( = (3.21676 − 0.5)λ

3.80574 3.21676

)2 −

1 4

]

413

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SOLUTION MANUAL

13.24. dm =

√

3lλ ⇒ l =

√

L2 + (dm ∕2)2

√ Using dm = 3lλ dm = 15.03λ dm = 12.29λ dm = 9.54λ dm = 7.81λ dm = 6.79λ dm = 5.58λ dm = 5.01λ dm = 4.37λ dm = 3.62λ dm = 2.66λ dm = 2.03λ dm = 1.64λ

From Figure 13.26 L = 75λ ⇒ dm = 14.5λ ⇒ l = 73.35λ L = 50λ ⇒ dm = 12λ ⇒ l = 50.36λ L = 30λ ⇒ dm = 9.4λ ⇒ l = 30.37λ L = 20λ ⇒ dm = 7.6λ ⇒ l = 20.36λ L = 15λ ⇒ dm = 6.7λ ⇒ l = 15.37λ L = 10λ ⇒ dm = 5.6λ ⇒ l = 10.38λ L = 8λ ⇒ dm = 5λ ⇒ l = 8.38λ L = 6λ ⇒ dm = 4.2λ ⇒ l = 6.36λ L = 4λ ⇒ dm = 3.5λ ⇒ l = 4.37λ L = 2λ ⇒ dm = 2.5λ ⇒ l = 2.36λ L = λ ⇒ dm = 1.9λ ⇒ l = 1.38λ L = 0.5λ ⇒ dm = 1.5λ ⇒ l = 0.9λ

13.25. L = 19.5′′ , dm = 15′′ , d = 2.875′′ √ (a) From Figure 13.24 l = L2 + (dm ∕2)2 = 20.89′′ The optimum gain will occur when, according to (13-59) dm =

√

2 2 3lλ ⇒ dm = 3lλ ⇒ λ = dm ∕3l = (15)2 ∕[3(20.89)] = 3.5898′′ = 9.118 cm

c 3 × 1010 = = 3.29 × 109 Hz = 3.29 GHz λ 9.118 19.5′′ 15′′ L= λ = 5.432λ, d = λ = 4.1785λ m 3.5898′′ 3.5898′′ f =

For these two L and dm , from (Fig. 13-26) ⇒ Dc ≃ 19.5 dB d2 3 According to (13-58d) and (13-58c), s = m = 8λl 8 L(s) ≅ (0.7853 − 0.3976s + 13.112s2 + 3.901s3 ) = 2.6858 dB Using these, we can compute the directivity from (13-58). ( Thus Dc = 10 log10

𝜋dm λ

)2 − 2.9 = 10 log10 [𝜋(4.1785)]2 − 2.53

= 22.363 − 2.6858 = 20.68 dB which agrees closely with the value obtained from Fig. 13.26 f = 2.5 GHz → λ =

c 30 × 109 = 12 cm = 4.7244′′ = f 2.5 × 109

From Fig. 13-26 From Fig. 13-26 ↓ 19.5 15 L= λ = 4.1275λ, dm = λ = 3.175λ, Dc ≅18.5 dB 4.7244 4.7244

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SOLUTION MANUAL

415

Using equations 2 dm

(13-58d) ⇒ s =

8λl

=

(15)2 = 0.284975 8(4.7244)(20.89)

(13-58c) ⇒ L(s) = [0.7853 − 0.3976(s) + 13.112s2 + 3.901s3 ] = 1.8271 (13-58) ⇒ Dc = 10 log10 [𝜋(3.175)]2 − 1.8271 = 19.978 − 1.8271 Dc = 18.15 dB f = 5 GHz ⇒ λ =

30 × 109 = 6 cm = 2.3622′′ 5 × 109

Using Fig. 13.26 L=

15 19.5 λ = 8.255λ, dm = λ = 6.35λ, Dc ≃ 20 dB (From Fig 13-26) 2.3622 2.3622

Using equations (13-58d) ⇒ s =

2 dm

8λl

=

(15)2 = 0.56995 8(2.3622)(20.89)

(13-58c) ⇒ L(s) = [0.7853 − 0.3976s + 13.112s2 + 3.901s3 ] = 5.5403 (13-58) ⇒ Dc = 10 log10 [𝜋(6.35)]2 − 5.5403 = 25.998 − 5.5403 = 20.46 dB (b) The cut off frequency of the dominant TE11 -mode of a circular waveguide is given by ′ 𝜒11 ′ is the first zero of the derivative of the Bessel function where 𝜒11 (fc )11 = √ ′ ) = 0], and it is equal 2𝜋a 𝜇𝜀 of the first kind of order one [i.e. J1′ (𝜒11 ′ to 𝜒11 = 1.841 Thus (fc )11 =

1.841 1.841 1.841c √ = √ = 𝜋d 2𝜋(d∕2) 𝜇𝜀 𝜋d 𝜇𝜀

For d = 2.875′′ = 7.3025 cm (fc )11 =

1.841(30 × 109 ) = 2.4074 × 109 = 2.4074 GHz 𝜋(7.3025)

13.26. Conical horn; l =√ 10λ √ (a) dm = 2am ≃ 3λl = 3λ(10λ) = 5.477λ ⇒ dm = 5.477λ, am = 2.7385λ 𝜓c = sin−1 (am ∕l) = sin−1 (2.7385∕10) = sin−1 (0.27385) = 15.8935◦ 𝜓c = 15.8935◦ ⇒ 2𝜓c = 31.787◦ (b)

s=

2 dm

8lλ

=

(5.477λ)2 29.997529 = = 0.375 8λ(10λ) 80

L(s) = −10 log10 (𝜀ap ) ≃ (0.7853 − 0.3976s + 13.112s2 + 3.901s3 )

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SOLUTION MANUAL

L(s) = −10 log10 (𝜀ap ) = 2.6858 ⇒ 𝜀ap = 10−2.6858∕10 = 0.5388 𝜀ap = 0.5388 = 53.88% ( )2 C − L(s) λ ) ( ) ( 𝜋dm 2 2𝜋a 2 − L(s) = 10 log10 − L(s) = 10 log10 λ λ ) ( 5.477λ𝜋 2 = 10 log10 − (2.6858) λ

(c)

Dc (dB) = 10 log10

= 10 log10 (17.2065)2 − 2.6858 = 10(2.47139) − 2.6858 = 24.7139 − 2.6858 Dc (dB) = 22.06 dB ⇒ Dc = 102.206 = 160.69 Dc = 160.89 = 22.06 dB √

13.27. dm =

3lλ ⇒ s =

2 dm

8λl

=

3 . [From (13-58c) ⇒ L(s = 3∕8) ≃ 2.6858 dB] 8

( )2 ( )2 C C − 2.6858 ⇒ 10 log10 = 25.286 λ λ ( )2 ( )2 C C = 2.5286 ⇒ = 102.5286 = 337.754 log10 λ λ

From (13-58) ⇒ Dc = 22.6 = 10 log10

𝜋dm √ C 18.378λ = = 377.754 = 18.378 ⇒ dm = λ λ 𝜋 18.378λ = 5.85λ dm = 𝜋 30 × 109 = 2.7273 cm 11 × 109 dm = 5.85λ = 5.85(2.7273) = 15.9545 cm = 6.281 in. λ=

Since √ d2 (15.9545)2 3lλ ⇒ l = m = = 31.111 cm = 12.248 in. 3λ 3(2.7273) √ √ L = l2 − (dm ∕2)2 = (31.111)2 − (15.9545∕2)2 = 30.071 cm = 11.839 in. ) ) ( ( 15.9545 −1 dm ∕2 −1 𝜓c = tan = tan = 14.857◦ ⇒ 2𝜓c = 29.71◦ L 2(30.071)

dm =

13.28. (a) Using Fig. 13.26, for a Dc = 20 dB dm ≃ 4.4λ

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(b) Using (13-59) dm = √ (c) L =

√ d2 (4.4λ)2 3lλ ⇒ l = m = = 6.45λ 3λ 3λ

l ≃ 6.45λ √ (l)2 − (dm ∕2)2 = (6.45λ)2 − (4.4∕2λ)2 = 6.06λ

L = 6.06λ

From Fig. 13.26 ⇒ L ≃ 6λ (

(d) Total Flare angle = 2𝜓c = 2 tan−1

dm ∕2 L

)

( = 2 tan−1

4.4∕2 6.06

)

2𝜓c = 2 tan−1 (0.363) = 2(19.95) = 39.90◦ 2𝜓c = 39.90◦ 2 dm

(4.4λ)2 = 0.375 8λl 8λ(6.45λ) L(s) = (0.753 − 0.3976s + 13.112s2 + 3.901s3 )|s=0.375 = 2.6858

(e) s =

=

2.6858 = −10 log10 (𝜀ap ) log10 𝜀ap = −0.26858 𝜀ap = 10−0.26858 = 0.5380 = 53.88% 𝜀ap = 53.88% The expected value for 𝜀ap for an optimum gain horn is about 50%. [ ( ) ] C 4𝜋 − L(s) 13.29. Dc (dB) = 10 log10 𝜀ap 2 (𝜋a2 ) = 10 log10 λ λ d2 30 × 109 s= m, λ= = 3 cm 8λl 10 × 109 √ 3lλ 3 = For optimum directivity, dm ≃ 3lλ. Thus s = 8lλ 8 3 For s = : 8 [ ( )2 ( )2 ] ( ) 3 3 3 + 13.112 + 3.901 L(s) ≃ 0.7853 − 0.3976 8 8 8 L(s) ≃ 2.6858 dB

ψc

dm

(13-58d) (13-58c)

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Dc (dB) = 20 = 10 log10

( )2 C − 2.6858 λ

22.6858 = 10 log10 (C∕λ)2 , (C∕λ)2 = 102.6858 = 185.6 = dm dm dm 𝜓c

(

2𝜋a λ

(

)2 =

√ 185.6 = λ = 4.3365λ = 4.3365(3) = 13 cm 𝜋 = 13 cm = 5.122 inches √ 2 2 = 3lλ ⇒ dm = 3lλ ⇒ l = dm ∕3λ = (13)2 ∕3(3) = 18.78 cm ) ( 6.5 = sin−1 (0.346) = 20.25◦ = sin−1 18.78

dm 𝜋 λ

)2

13.30. 𝜓c = 25◦ d ∕2 (a) sin 𝜓c = m l For optimum directivity √ 3lλ √ √ √ 3lλ∕2 3lλ 1 3λ = = sin 𝜓c = l 2l 2 l dm =

1 3λ 3λ 3λ 3λ = = ⇒l= 2 ◦ 2 4 l 4(sin 25 ) 4(0.42262)2 4 sin 𝜓c √ √ √ l = 4.199λ ⇒ dm = 3lλ = 3(4.199)λ2 = 12.597λ

sin2 𝜓c =

dm = 3.5493λ (b)

2 s = dm ∕8λl = (3.5493λ)2 ∕8λ(4.199λ) = 0.375

L(s) ≃ 0.7853 − 0.3976s + 13.112s2 + 3.901s3 = 2.6858 L(s) = 2.6858 = −10 log10 (𝜀ap ) 𝜀ap = 10−0.26858 = 0.51145 = 53.88% 𝜀ap ≃ 53.88% (c) The expected 𝜀ap for maximum horn directivity is ≃ 50%. So this design closely matches the expected value. ) ( ( )2 𝜋dm 2 C (d) Dc (dB) = 10 log10 − L(s) = 10 log10 − 2.6858 λ λ = 10 log10 [𝜋(3.5493)]2 − 2.6858 = 10 log10 (11.15045)2 − 2.6858 Dc = 20.946 − 2.6858 = 18.26 dB = 101.826 = 66.99 Dc = 66.99 = 18.26 dB

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13.31. A = 20𝜋λ2 Using (13-53) ( ) 1 4𝜋 1 4𝜋 (a1 b1 ) = (Area) G0 = 2 2 2 λ λ2 ⏟⏟⏟ ⏟⏟⏟ 𝜀ap 𝜀ap 1 4𝜋 (a) G0 = (20𝜋λ2 ) = 394.784 = 25.964 dB 2 λ2 1 4𝜋 1 (b) G0 ≃ (20𝜋λ2 ) = (80𝜋 2 ) = 40𝜋 2 = 394.784 2 λ2 2 G0 = 394.784 = 25.964 dB Alternate: dm =

√ d2 3lλ 3lλ, s = m = = 3∕8 8λl 8λl

s = 3∕8, L(s)|s=3∕8 = (0.753 − 0.3976s + 13.112s2 + 3.901s3 )|s=3∕8 = 2.6858 L(s) = −10 log10 (𝜀ap ) = 2.6858 ⇒ 𝜀ap = 10−0.26858 = 0.5388 ( ) 4𝜋 G0 = 0.5388 2 (20𝜋λ2 ) = 0.5388(80𝜋 2 ) = 425.42 λ G0 = 425.42 = 26.29 dB 13.32. (a) Aem = (1 − |Γ|2 )

(0.03)2 λ2 D0 = 0.99 75 = 5.317764 × 10−3 m2 4𝜋 4𝜋

(f = 10 GHz → λ = 0.03 m)

Aem = 0.005317764 m2

(b) Pmax = Aem Wi = (1 × 10−6 Watts/m2 )(0.005317764) = 5.317 × 10−9 Watts Pmax = 5.317764 × 10−9 Watts 13.33. f = 10.3 GHz ⇒ λ0 = 30 × 109 ∕(10.3 × 109 ) = 2.9126 cm 2.9126 2.9126 (a) λ0 ∕4 < d < λ0 ∕2 ⇒ λ0 < d < λ0 ⇒ 0.72185 cm < d < 1.4563 cm 4 2 (b) W < λ0 ∕10 = 0.29126 cm (c) t ≤ W∕10 = 0.29126∕10 = 0.029126 cm 13.34. (a) Using the Fig. P13.34(a) [see next page] E-plane: 4.54λ, H-plane: 5.2λ (b) Using the Fig. P13.24(b) [see next page] E-plane: 5.06λ, H-plane: 4.5λ

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Phase center d (wavelengths)

6 5 4 3

ψe

a

d

ρe = 5λ , a = 0.7λ E-plane

b 2

ρe

1 0

5

H-plane

10 Flare angle ψe (degrees)

15

20

(a) E-plane sectoral horn

Phase center d (wavelengths)

6 5 4 3

a d

ψh

ρh = 5λ , a = 0.7λ H-plane

b 2 1 0

E-plane

ρh

5

10

15

20

Figure P13.34 Phase center location, as a function of flare angle, for E- and H-plane sectoral horns. (Source: Y. Y. Hu, “A Method of Determining Phase Centers and its Applications to Electromagnetic Horns,” Journal of Franklin Institute, vol. 271, pp. 31–39, January 1961)

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14

Solution Manual

14.1. By using (14-1), the effective dielectric constant is equal to 𝜀reff =

Using (14-19),

[ ( )]−1∕2 1 10 + 1 10 − 1 1 + 12 = 6.8568 + 2 2 1.2

W = 1.2 h 120𝜋 Zc = √ 𝜀reff [1.2 + 1.393 + 0.667 ln(1.2 + 1.444)] =

120 ⋅ 𝜋 = 44.415 ohms 2.6185(3.24152)

[ ( )]−1∕2 1 6.8 + 1 6.8 − 1 1 + 12 = 4.8667 + 2 2 1.5 At low frequencies, the characteristic impedance can be found by using (14-19), if W > 1. h

14.2. 𝜀reff =

120𝜋 Zc (f = 0) = √ 4.8667[1.5 + 1.393 + 0.667 ln(1.5 + 1.444)] = 170.889∕(3.6132) = 47.296 ohms 14.3. W = 0.4λ0 , h = 0.05λ0 , f = 10 GHz, 𝜀r = 2.25 ⇒ W∕h = 0.4∕0.05 = 8 > 1 𝜀reff =

] [ ( )] 𝜀r + 1 𝜀 r − 1 [ h 2.25 + 1 2.25 − 1 0.05 −1∕2 1 + 12 = 1 + 12 + + 2 2 W 2 2 0.4

𝜀reff = 1.625 + 0.625(2.5)−1∕2 = 1.625 + 0.625(0.63246) = 2.02 𝜀reff = 2.02 Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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120𝜋 [ ( )] √ W W 𝜀reff + 1.393 + 0.667 ln + 1.444 h h 120𝜋 = √ [ ( )] 0.4 0.4 2.02 0.05 + 1.393 + 0.667 ln 0.05 + 1.444

Zc =

120𝜋 1 1 = 24.3557 ⇒ Yc = = = 0.04106 Zc = √ Z 24.3557 c 2.02(10.89067) [ ( )2 ] [ ] 0.4λ 1 W 1 2𝜋 0 1 − (k0 h) = 0.5λ0 1− (a) G1 = 120λ0 24 120λ0 24 λ0 [ ] 1 0.4 0.4 1 − (0.1𝜋)2 = [1 − 0.00411] = 3.32 × 10−3 120 24 120 [ ( )] 0.4λ0 W 2𝜋 B1 = [1 − 0.636 ln(k0 h)] = + 0.05λ0 1 − 0.636 ln 120λ0 120λ0 λ0 =

0.4 [1 − 0.636 ln(0.1𝜋)] 120 0.4 0.4 [1 − 0.636(−1.15786)] = [1 + 0.7364] = 5.79 × 10−3 = 120 120

=

Y1 = G1 + jB1 = (3.32 + j5.79) × 10−3 Yin = Y1 + Yc = (3.32 + j5.79) × 10−3 + 41.06 × 10−3 = (44.38 + j5.79) × 10−3 Yin = (44.38 + j5.79) × 10−3 : Capacitive (b) Place an inductor in parallel. BL =

1 1 1 1 = = = 5.79 × 10−3 ⇒ L = 𝜔L 2𝜋fL 2𝜋fBL 2𝜋(10 × 109 )(5.79 × 10−3 ) L = 2.75 × 10−9 Henries 1 1 = = 22.533 ′ Yin 44.38 × 10−3 ′ = 22.533 Zin

′ ′ = Gin = 44.38 × 10−3 ⇒ Zin = (c) Yin

30 × 109 14.4. fr = 2 GHz, 𝜀r = 10.2, h = 0.05 inches = 0.127 cm ⇒ λ0 = = 15 cm 2 × 109 √ √ 𝜈0 2 2 30 × 109 = = 7.5(0.422577) = 3.169 cm (a) W = 2fr 𝜀r + 1 2(2 × 109 ) 10.2 + 1 ] [ ] 𝜀 + 1 𝜀r − 1 [ h −1∕2 10.2 + 1 10.2 − 1 0.127 −1∕2 1 + 12 1 + 12 𝜀reff = r = + + 2 2 w 2 2 3.169 = 5.6 + 4.6[1 + 12(0.04)]−1∕2 𝜀reff = 5.6 + 4.6(0.82174) = 5.6 + 3.78 = 9.38 [ [ ] ] 3.169 + 0.264 (𝜀reff + 0.3) Wh + 0.264 (9.38 + 0.3) 0.127 ΔL = 0.412 [ [ ] = 0.412 ] h 9.38 − 0.258 3.169 + 0.8 (𝜀reff − 0.258) Wh + 0.8 0.127

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423

(9.68)(25.21676) ΔL = 0.412 = 0.4281 ⇒ ΔL = 0.4281(0.127) = 0.05437 cm h (9.122)(25.75276) L=

30 × 109 7.5 λ − 2ΔL = − 2(0.05437) = √ − 0.10874 √ 2 9 2(2 × 10 ) 9.38 9.38

L = 2.448837 − 0.10874 = 2.34 cm L = 2.34 cm, W = 3.169 cm (b) 0.48λ < L < 0.49λ ⇒ 2.35 cm < L < 2.40 cm λ 30 × 109 30 × 109 λ= √ 0 = = = 4.89767 cm √ 9 𝜀reff 2 × 109 9.38 2 × 10 (3.06268) (c) G1 =

[ ( [ ] )2 ] 1 3.169 W 1 2𝜋 1 − (k0 h)2 = 1− 0.127 120λ0 24 120(15) 24 7.5 3.169 [1 − 0.0017597] = 120(15) G1 = 1.7597 × 10−3 Rin =

1 1 = = 284.139 2G1 2(1.7597 × 10−3 )

(d) 150 = 284.139 cos

2

(

√ ) 𝜋 L 150 2.34 −1 y ⇒ y0 = cos = cos−1 (0.72657) L 0 𝜋 284.139 𝜋

2.34 (0.75748) = 0.5642 cm 𝜋 y0 = 0.5642 cm

y0 =

14.5. fr = 9 GHz ⇒ λ0 = 30 × 109 ∕9 × 109 = 3.333 cm, 𝜀r = 2.56 √ √ 𝜈 2 2 30 × 109 30 √ (a) W = 0 0.56179 = = = 2fr 𝜀r + 1 2(9 × 109 ) 2.56 + 1 18 1.2492 λ = 0.37476λ0 3.333 0 ( ) 2 k0 W ⎤ ⎡ sin cos 𝜃 ( )2 ⎥ 𝜋⎢ 2 2𝜋W 1 ⎥ sin3 𝜃 d𝜃 ⎢ D0 = , I1 = ∫0 ⎢ ⎥ λ0 I1 cos 𝜃 ⎥ ⎢ ⎦ ⎣ [ ] sin(X) = −2 + cos(x) + XSi (X) + X 2𝜋 X = k0 W = (0.37476λ0 ) = 2.35471 rads λ0 sin(X) sin(2.35471) cos(2.35471) = 0.706056, Si (2.35471) ≃ 1.725, = = 0.30082 X 2.35471 I1 = [−2 − 0.706056 + 2.35471(1.725) + 0.30082] = 1.656637 [ ] 2𝜋 1 D0 = (0.37476λ0 ) = 3.34687 = 5.24639 dB λ0 1.656637 W = 1.2492 cm =

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(b) D2 =

2 1 + g12 D0

g12 ≪1

≃ 2D0 = 2(3.34687) = 6.6937 = 8.25669 dB

30 × 109 = 3 cm 14.6. f = 10 GHz, 𝜀r = 4, h = 0.25 cm ⇒ λ0 = 10 × 109√ √ √ 𝜈 2 2 30 × 109 3 2 = = = 0.9487 cm (a) W = 0 2fr 𝜀r + 1 2(10 × 109 ) 4 + 1 2 5 ] [ ] 𝜀 + 1 𝜀r − 1 [ h 4+1 4−1 0.25 −1∕2 1 + 12 = 1 + 12 (b) 𝜀reff = r + + 2 2 W 2 2 0.9487 𝜀reff = 2.5 + 1.5(0.4902) = 2.5 + 0.7352 = 3.2352 3 1 30 × 109 = = = 0.8339 √ √ √ 2(1.7987) 9 2fr 𝜀reff 𝜇0 𝜀0 2(10 × 10 ) 3.2352 [ ] ) ( W (𝜀reff + 0.3) + 0.264 + 0.264 (3.2352 + 0.3) 0.9487 0.25 ΔL h (d) = 0.412 ( ) [ ] = 0.412 W h (3.2352 − 0.258) 0.9487 + 0.8 (𝜀reff − 0.258) + 0.8 0.25 h (c) Leff =

(3.5352)(4.0588) ΔL = 0.412 = 0.4321 h (2.9772)(4.5948) ΔL = 0.4321(0.25) = 0.1080 L = Leff − 2ΔL = 0.8339 − 2(0.1080) = 0.6178 14.7. W = 1.6046 cm, fr = 4.6 × 109 Hz, h = 0.45 cm, 𝜀r = 6.8 v0 30 × 109 (a) L = = 1.2505 cm √ √ = 2fr 𝜀r 2(4.6 × 109 ) 6.8 ) ( ) 𝜀r + 1 𝜀 r − 1 ( h −1∕2 6.8 + 1 6.8 − 1 0.45 −1∕2 1 + 12 1 + 12 = + + (b) 𝜀reff = 2 2 W 2 2 1.6046 𝜀reff = 3.9 + 2.9(0.4786) = 5.288 ( ) ( ) W 1.6046 (𝜀reff + 0.3) + 0.264 + 0.264 0.45 (5.288 + 0.3) ΔL h = 0.412 ( ) ( ) = 0.412 W h (5.288 − 0.268) 1.6046 + 0.8 (𝜀reff − 0.268) + 0.8 0.45 h ΔL = 0.4015 h ΔL = 0.4015 h = 0.4015(0.45) = 0.1807 cm Le = L + 2ΔL = 1.2505 + 2(0.1807) = 1.6119 cm fr′ =

𝜈0 30 × 109 = 4.046 × 109 , because of fringing = √ √ 2Le 𝜀reff 2(1.6119) 5.288

14.8. (a) Using (14-6), the width of the patch is 30 W= 2(1.6) 𝜀reff

√

2 = 3.962 cm 10.2 + 1 [ ( )] 0.127 −1∕2 10.2 + 1 10.2 − 1 1 + 12 = = 9.51 + 2 2 3.962

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(

) 3.962 + 0.264 (9.51 + 0.3) 0.127 ΔL = (0.127)(0.412) ( ) = 0.05455 cm (9.51 − 0.3) 3.962 + 0.8 0.127 30 λ − 2(0.05455) = 2.931 cm L = − 2ΔL = √ 2 2(1.6) 9.51 (b) Using (14-12) and (14-12a) I1 sin(X) , I1 = −2 + cos(X) + XSi (X) + , X = k0 W X 120𝜋 2 2𝜋 2𝜋 X= W= (3.962) = (0.4226)𝜋 = 1.3277 λ0 18.75

G1 =

I1 = −2 + cos(1.3277) + (1.3277)(1.204348) +

sin(1.3277) = 0.57075 1.3277

G1 = 0.57075∕(120𝜋 2 ) = 4.81916 × 10−4 Siemens Resonant input impedance, Rin = Zin =

1 2G1

= 1.0375 × 103 = 1037.5 ohms

Direct numerical calculation from (14-12) results into G1 = 4.819021 × 10−4 , Rin =

1 = 1037.56 ohms 2G1

(c) Numerical calculation from (14-18a) G12 = 3.92904 × 10−4 1 1 = 571.56 ohms = 2(G1 + G2 ) 2(4.819 + 3.929) × 10−4 ( ) 𝜋 y0 (e) Rin (y = y0 ) = Rin (y = 0) cos2 L ( ) 𝜋 2 75 = 571.56 cos y0 2.931 y0 = 1.1197 cm (0.4408 inch) √ 30 2 14.9. (a) W = = 7.412 cm 2(1.6) 2.2 + 1. [ ( )] 0.1575 −1∕2 2.2 + 1 2.2 − 1 1 + 12 𝜀reff = = 2.1356 + 2 2 7.412 (d) Rin =

2.1356 + 0.3 ΔL = (0.1575)(0.412) ⋅ 2.1356 − 0.258 L=

7.412 + 0.264 0.1575 7.412 + 0.8 0.1575

= 0.0832 cm

λ 30 − 2(0.0832) = 6.2487 cm − 2ΔL = √ 2 2(1.6) 2.1356

∴ The patch is a realistic dimension for the roof of a personal car.

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(b) From (14-12), with numerical calculation G1 = 1.57259 × 10−3 , Rin =

1 = 317.95 ohms 2G1

(c) Using (14-18a), G12 = 4.58053 × 10−4 1 1 (d) Rin = = = 246.23 ohms 2(G1 + G12 ) 2(1.57259 × 10−3 + 4.58053 × 10−4 ) ( ) (e) Rin (y = y0 ) = Rin (y = 0) cos2 𝜋 y0 L ( ) 𝜋 75 = 246.23 cos2 y0 6.2487 y0 = 1.9615 cm (0.7722 inch) 14.10. f0 = 10 GHz, 𝜀r = 10.2, h = 0.05 in = 0.127 cm 30 W= 2(10)

√

2 = 0.634 cm 10.2 + 1

10.2 + 1 10.2 − 1 W 0.634 + [1 + 12(4.992)−1 ]−1∕2 , = = 4.992 2 2 h 0.127 (8.093 + 0.3) (4.992 + 0.264) = 8.093ΔL = (0.127)(0.412) = 0.0509 cm (8.093 − 0.258) (4.992 + 0.8)

𝜀reff =

L=

W = 0.634 cm = 0.2496 in 30 1 − 2(0.0509) = 0.4255 cm, √ L = 0.4255 cm = 0.1675 in 2(10) 8.093

Yc’ β

Y1

Yin

~ Y2

Y1

I

14.11. Y1 = G1 + jB1 ̃ 2 + jB̃ 2 Ỹ 2 = G Resonance, Im (Yin ) = 0 ⇒ Yin = Gin Yin = Y1 + Ỹ 2 ,

where Ỹ 2 = Yc Ỹ 2 = Yc

Y1 + jYc tan 𝛽l Yc + jY1 tan 𝛽l G1 + j(B1 + Yc tan 𝛽l) (Yc − B1 tan 𝛽l) + jG1 tan 𝛽l

Gin = Y1 + Yc

G1 + j(B1 + Yc tan 𝛽l) (Yc − B1 tan 𝛽l) + jG1 tan 𝛽l

(1) (2)

Gin [(Yc − B1 tan 𝛽l) + jG1 tan 𝛽l] = (G1 + jB1 )[(Yc − B1 tan 𝛽l) + jG tan 𝛽l] + Yc [G1 + j(B1 + Yc tan 𝛽l)]

(3)

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427

Separating into real and imaginary parts. Gin (Yc − B1 tan 𝛽l) = G1 (Yc − B1 tan 𝛽l) − B1 G tan 𝛽l + Yc G1 jGin G1 tan 𝛽l = j[B1 (Yc − B1 tan 𝛽l) + G21 tan 𝛽l + Yc (B1 + Yc tan 𝛽l)]

(4) (5)

Rearranging (4) Gin (Yc − B1 tan 𝛽l) = 2G1 (Yc − B1 tan 𝛽l) ⇒ Gin = 2G1

(6)

Substituting this result into (5) 2G21 tan 𝛽l = B1 (Yc − B1 tan 𝛽l) + G21 tan 𝛽l + Yc (B1 + Yc tan 𝛽l) ] [ (tan 𝛽l) G21 + B21 − Yc2 = 2Yc B1 , tan 𝛽l =

2Yc B1 B21

+ G21 − Yc2

(7)

̃ z = G1 , since Re (Y1 ) = G1 From Yin = Y1 + Ỹ 2 , and Gin = 2G1 , we can see G Also, since imaginary part of Yin = 0 at resonance Yin = Gin + jBin , Bin = B1 + B̃ 2 = 0, ∴ Ỹ 2 = G1 − jB1

Im (Yin ) = 0, Bin = 0 B̃ 2 = −B1

Total input admittance is, Yin = Y1 + Ỹ 2 = 2G1 14.12.

2Ym

2Ym

X

Y1 – Ym

Y1 – Ym

Yin l1

l2

This circuit has been “derived” by considering the equivalent circuit admittance for a symmetric 2-port junction. Y11

Y12

V1

Y21

Y22

V2

=

I1

Y12

I2

Y11 – Y12

Y11 – Y12

For a symmetric voltage, Node X of the Fig. P14-12 sees an open circuit. Hence, the new circuit becomes

Y1 – Ym

Yin

l1

Y1 – Ym

l2

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SOLUTION MANUAL

For an antisymmetric voltage, node X sees a zero voltage, a grounded circuit. Thus, the new circuit becomes

2Ym

Y1 – Ym Y1 – Ym

2Ym

Yin l1

Y1 + Ym

l2

Y1 + Ym

Yin l1

l2

If we assume Bm to be neglible, since its effect should be to change the resonant frequency slightly, then

Rin |Y=Y0 ≃

cos2

(

𝜋 y L 0

)

2(G ± Gm )

where + = odd mode − = even mode 14.13. Y1

Yin

Yc’ β l1

~ ~ Y 1 Y2

Yc’ β

Y1

l2

We wish to find the impedance at resonance l = l 1 + l2 , For transmission line theory Yin = Ỹ 1 + Ỹ 2 , where Ỹ 1 = Yc

G cos 𝛽l1 + j(B cos 𝛽l1 + Yc sin 𝛽l1 ) (Yc cos 𝛽l − B sin 𝛽l1 ) + jG sin 𝛽l1

(l1 = y0 )

Ỹ 2 = Yc

G cos 𝛽l2 + j(B cos 𝛽l2 + Yc sin 𝛽l2 ) (Yc cos 𝛽l2 − B sin 𝛽l2 ) + jG sin 𝛽l2

(l2 = l − y0 )

At resonance, the input impedance at an arbitray feed point is real. By transforming the slot admittances G1 + jB1 to the common point and adding them together, the input impedance at resonance is found as [ ] ( ) G2 + B2 ( ) B ) ( 1 2𝜋 1 1 1 2 𝜋 2 𝜋 sin sin cos y + y − y Rin (y = y0 ) = 2(G1 ± G12 ) L 0 L 0 Yc L 0 Yc2 From Example 14.2. W = 1.186 cm, h = 0.1588 cm, ⇒ Zc = 26.0146 ohms, λ = 3 cm.

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SOLUTION MANUAL

429

Using (14-12) and (14-12a) G1 = 0.00157 Siemens. Using (14-8b), B1 = 5.6 × 10−3 . ) ( ∴ G21 + B21 ∕Yc2 = [(1.57 × 10−3 )2 + (5.6 × 10−3 )2 ]∕[1.47 × 10−3 ] = 0.02289 ≪ 1 ∴ B1 ∕Yc = 5.6 × 10−3 ∕(0.03844) = 0.14568 < 1 ( ) 𝜋 1 Rin (y = y0 ) = cos2 y0 2(G1 ± G12 ) L

14.14. W = 0.634 cm, L = 0.4255 cm, h = 0.127 cm, f = 10 GHz 1 = 284.751 (a) Using (14-8a) ⇒ G1 = 0.00175592, Rin = 2G1 ( ) 𝜋 y (b) 50 = 284.751 cos2 L 0 ) √ ( 50 𝜋 y0 = = 0.4190 cos L 1037.30084 𝜋 y = cos−1 (0.4190) = 1.13845. ∴ y0 = 0.15419 cm L 0 14.15. 𝜀r = 2.2, W0 = 0.2984 cm, h = 0.1575 cm, L = 0.9068 cm W0 0.2984 h = 0.5278 = = 1.8946 h 0.1575 W0 2.2 + 1 2.2 − 1 + [1 + 12(0.5278)]−1∕2 = 1.6 + 0.6[0.3693] 2 2 𝜀reff = 1.8216 √ 120𝜋∕ 𝜀reff Zc = [ ] W0 W0 + 1.393 + 0.667 ln + 1.444 h h √ 120𝜋∕ 1.8216 = 1.8946 + 1.393 + 0.667 ln[3.3386]

𝜀reff =

120𝜋∕(1.3497) = 68.264 4.0917 ( ) 𝜋 68.264 = 152.44 cos2 y0 L ( ) 𝜋 68.264 cos2 y = = 0.4478 L 0 152.44 √ L 0.9068 (0.8377) = 0.2418 cm (0.0952 in) y0 = cos−1 ( 0.4478) = 𝜋 𝜋 Zc =

14.16. Zc = 50 ohms 𝜀r = 2.2, 𝜀reff =

[ ]−1∕2 2.2 + 1 2.2 − 1 h + 1 + 12 2 2 W0 120 ⋅ 𝜋

50 = √ [ ( W W 𝜀reff h0 + 1.393 + 0.667 ln h0 ∴ W0 = 0.4933 cm

| | | )] | | + 1.444 | |h=0.1588 cm

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SOLUTION MANUAL

228.3508 ohms 50 ohms

εr = 2.2 W = 1.186 cm W0 = 0.4933 cm y0 = 0.3126 cm L = 0.906 cm

14.17. (a) 75 ohms 75 = 228.3508 cos2 ( cos

(

𝜋 y 0.906 0

)

) √ 75 𝜋 y0 = 0.906 228.3508

𝜋 y = cos−1 (0.57309) = 0.9605, y0 = 0.277 cm 0.906 0 Using (14-19), we can find the width of microstrip feed line by iterative method. 75 =

[ √ 𝜀reff

120𝜋 ( )] W W + 1.393 + 0.667 ln + 1.444 0.1588 0.1588

∴ Dimension of line: Width = 0.2546 cm (b) 100 ohms

100 = 228.3508 cos y0 =

2

(

) ( ) √ 𝜋 𝜋 100 y , cos y = 0.906 0 0.906 0 228.3508

0.906 0.906 cos−1 (0.661757) = (0.847636) = 0.2444 cm 𝜋 𝜋

We can find the dimension of width W of microstrip feed line by iterative procedure as Part (a) in this problem. 100 =

√

𝜀reff

[

120𝜋 ( )] , W W + 1.393 + 0.667 ln + 1.444 0.1588 0.1588

if

W >1 h

or ] [ ⎛100 = √60 ln 8h + W , if W ≤ 1 W 4h h ⎜ 𝜀reff ∴ Dimension of ⎜ {[ } W = 0.14285 cm ] ] [ ⎜ 𝜀 + 1 𝜀r − 1 h −1∕2 W 2 ⎜𝜀reff = r 1 + 12 + 0.04 1 − + , ⎝ 2 2 W h

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SOLUTION MANUAL

14.18. Using (14-31), 𝜀r = 2.2, L = 0.906 cm, W = 1.186 cm m=0 0 0 0

n 0 0 0

p 1 2 3

0 0 0

⋮ 1 1 1

0 1 2

11.1622 GHz → ②nd 14.0465 GHz → ③nd 20.3822 GHz → ⑤th

0

⋮ 2

0

22.3245 GHz

:

fr 8.52698 GHz → ①st 17.0540 GHz → ④th 25.5809 GHz

) TM001 , TM010 , TM011 , TM002 , … 8.5269 GHz, 11.16 GHz, 14.047 GHz, 17.0540 GHz

( ⋮ 14.19. (a) TMz

2 1 𝜕 Az , 𝜔𝜇𝜀 𝜕x𝜕z 2 1 𝜕 Az , Ey = −j 𝜔𝜇𝜀 𝜕y𝜕z ( 2 ) 1 𝜕 2 + 𝛽 Az , Ez = −j 𝜔𝜇𝜀 𝜕z2

Ex = −j

1 𝜕Az 𝜇 𝜕y 1 𝜕Az Hy = − 𝜇 𝜕x

Z

Hx =

Hz = 0

h

X

εr W

L

Az = [C1 cos(𝛽x x) + D1 sin(𝛽x x)] ⋅ [C2 cos(𝛽y y) + D2 sin(𝛽y y)]x × [C3 cos(𝛽z z) + D3 sin(𝛽z z)] Boundary Conditions: At x = 0 and x = L, Hy = 0 At y = 0 and y = W, Hx = 0

}

PMC walls } At z = 0 and z = h, Ex = Ey = 0 PEC walls Applying the boundary conditions to the walls leads to Az = Amnp cos(𝛽x x) cos(𝛽y y) cos(𝛽z z) ( p𝜋 m, n, p = 0, 1, 2, … m𝜋 n𝜋 and 𝛽x = , 𝛽y = , 𝛽z = , m=n=p≠0 L W h Thus

( ) ( p𝜋 ) ( ) ( ) ( p𝜋 ) m𝜋 m𝜋 1 n𝜋 sin x cos y sin z 𝜔𝜇𝜀 L h L W h ( ) ( p𝜋 ) ( ) ( ) ( p𝜋 ) n𝜋 m𝜋 1 n𝜋 cos x sin y sin z Ey = −j 𝜔𝜇𝜀 W h L W h [ ( ) ] ) ( ) ( p𝜋 ) ( p𝜋 2 1 n𝜋 m𝜋 2 + 𝛽 cos − x cos y cos z Ez = −j 𝜔𝜇𝜀 h L W h

Ex = −j

Y

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SOLUTION MANUAL

( ) ( ) ( ) ( p𝜋 ) m𝜋 1 n𝜋 n𝜋 cos x sin y cos z 𝜇 W L W h ( ) ( ) ( ) ( p𝜋 ) m𝜋 1 m𝜋 n𝜋 sin x cos y cos z Hy = 𝜇 L L W h

Hx = −

Hz = 0 √ (b)

( fr )TM mnp

1 = √ 2𝜋 𝜇𝜀

(

m𝜋 L

)2

( +

n𝜋 W

)2 +

( p𝜋 )2 h

(c) if L > W > h, lowest mode will be TMz100 and ( fr )TM 100 = (d) ( fr )TM 100 =

1 √ 2L 𝜇𝜀

1 √

2L 𝜇𝜀

14.20. W = 2 cm, L = 5 cm, h = 0.1568 cm, 𝜀r = 2.2 (a) Second-order Dominant mode for L = 5 cm > L∕2 > W = 2 cm is TMx020 ; m = 0, n = 2, p = 0 [See Fig. 14.16(c), (14-35)] ( ) 𝜈 1 2𝜋 1 =√ 0 (b) ( fr )mnp = ( fr )020 = =√ √ L 2𝜋 𝜇𝜀reff 𝜇𝜀reff Le 𝜀reff Leff e [ ] [ ] −1∕2 𝜀 + 1 𝜀r − 1 h 0.1568 −1∕2 2.2 + 1 2.2 − 1 1 + 12 1 + (12) 𝜀reff = r = + + 2 2 W 2 2 2 3.2 1.2 𝜀reff = + (1 + 0.9408)−1∕2 = 1.6 + 0.6(0.71781) = 2.03 2 2 ) ( 2 + 0.264 (2.03 + 0.3) (2.33)(13.0191) ΔL 0.1568 = 0.412 = 0.412(1.2629) ( ) = 0.412 2 h 1.772(13.5551) (2.03 − 0.258) + 0.8 0.1568 ΔL = 0.5203 ⇒ ΔL = 0.5203 h = 0.5203(0.1568) = 0.08159 cm h Le = L + 2ΔL = 5 + 2(0.08159) = 5.163 cm 30 × 109 = 4.07822 GHz , ( fr )020 = √ 2.03(5.163)

λ0 =

30 × 109 = 7.356 cm 4.07822 × 109

[ ( [ ] )2 ] 2𝜋 1 2 W 1 2 1 − (k0 h) = (c) G1 = 1− 0.1568 120λ0 24 120(7.356) 24 7.356 = 2.2657(0.99925) G1 = 2.264 × 10−3 , Rin =

(d) Rin =

G12 = 0.24921 × 10−3

1 1 = = 220.85 2G1 2(2.264 × 10−3 ) 1 1 = 248.16 = 2(G1 − G12 ) 2(2.264 − 0.24921) × 10−3

The minus (−) sign is used because for the TM020 mode the voltage (field) distribution is even (symmetric). See Figure 14.16(c).

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SOLUTION MANUAL

14.21. (a) TMy 2 1 𝜕 Ay 𝜔𝜇𝜀 𝜕x𝜕y ( 2 ) 1 𝜕 2 + 𝛽 Ey = −j Ay 𝜔𝜇𝜀 𝜕y2 2 1 𝜕 Ay Ez = −j 𝜔𝜇𝜀 𝜕y𝜕z

Ex = −j

Hx = −

1 𝜕Ay 𝜇 𝜕z

y

Hy = 0 Hz =

h

1 𝜕Ay 𝜇 𝜕x

z

x

L

εr W

Ay = [C1 cos(𝛽x x) + D1 sin(𝛽x x)][C2 cos(𝛽y y) + D2 sin(𝛽y y)][C3 cos(𝛽z z) + D3 sin(𝛽z z)] Boundary Condition At x = 0 and x = W, Hz = 0 → PMC walls At y = 0 and y = h,

Ex = Ez = 0 → PEC walls

At z = 0 and z = L,

Hx = 0 → PMC walls

Applying the boundary conditions to the walls leads to Ay = Amnp cos(𝛽x x) cos(𝛽y y) cos(𝛽z z) and 𝛽x =

m𝜋 , W

𝛽y =

n𝜋 , h

𝛽z =

p𝜋 L

(

m, n, p = 0, 1, 2, … m=n=p≠0

)

Thus ( ) ( p𝜋 ) ( ) ( ) ( p𝜋 ) m𝜋 m𝜋 1 n𝜋 sin x sin y cos z 𝜔𝜇𝜀 W L W h L [ ( ) ] ) ( ) ( p𝜋 ) ( 1 n𝜋 2 n𝜋 m𝜋 + 𝛽 2 cos − x cos y cos z Ey = −j 𝜔𝜇𝜀 h W h L ( ) ( p𝜋 ) ( ) ( ) ( p𝜋 ) n𝜋 m𝜋 1 n𝜋 cos Ez = −j x sin y sin z 𝜔𝜇𝜀 h L W h L ( ) ( ) ( ) ( p𝜋 ) m𝜋 1 p𝜋 n𝜋 cos x cos y sin z Hx = − 𝜇 L W h L Ex = −j

Hy = 0 Hz = −

1 𝜇

(

√ (b)

( fr )TM mnp

=

1 √

2𝜋 𝜇𝜀

) ( ) ( ) ( p𝜋 ) m𝜋 m𝜋 n𝜋 sin x cos y cos z W W h L

(

m𝜋 W

)2

( +

n𝜋 h

(c) if L > W > h, lowest mode will be y 1 (d) ( fr )TM √ 001 = 2L 𝜇𝜀

)2 +

y TM001

( p𝜋 )2 L

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SOLUTION MANUAL

14.22. The array factor of 2 elements of zero phase difference, placed along the y-direction and separated by Le , is (AF)2 = ej = ej

k0 Le 2

cos 𝛾

k0 Le 2

sin 𝜃 sin 𝜙

(AF)2 = 2 cos

(

+ e−j

k0 Le 2

+ e−j

cos 𝛾 k0 Le 2

sin 𝜃 sin 𝜙

)

k0 Le sin 𝜃 sin 𝜙 2

14.23. At the two radiating edges (x = 0 and x = L) Ez (x = L) = E0 ,

Hy (x = L) = 0

Ez (x = 0) = −E0 ,

Hy (x = 0) = 0

We have constant distribution of Ez and zero distribution of Hy . z

n^ 2

Slot 2 y

W

x

L

Slot 1 n^ 1

Thus, on slot 1 (x = L) M 1 = −̂n1 × [Ez (x = L)̂az ] = −̂ax × E0 â z = E0 â y , J 1 = n̂ 1 × [Hy (x = L)̂ay ] = 0,

M 1 = E0 â y

J1 = 0

similarly, on slot 2, we have M 2 = −̂n2 × [−E0 â z ] = E0 â y J2 = 0 Since M 1 and M 2 are parallel to the PEC ground plane, their images below the plane are identical to those above the plane. In addition, M 1 and M 2 are constant over the apertures. Thus, removing the ground plane is equal to doubling the height of the slots. To find the radiated field by the two slots, first integrate over slot 1, and then use array factor to combine the contribution from both slots. (a) On slot 1, keeping in mind that the height is doubled M 1 = E0 â y , J 1 = 0

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SOLUTION MANUAL

435

z M2

y

M2

M1 M1

x

then L𝜃 =

∫∫

My cos 𝜃 sin 𝜙e+jk0 r

= E0 cos 𝜃 sin 𝜙

W 2

ejk0 (y

ds′

′ sin 𝜃⋅sin 𝜙+z′ cos 𝜃)

2

L𝜃 = hWE0 cos 𝜃 sin 𝜙

L𝜙 =

h

∫− W ∫−h

[

Y=

′ cos 𝜓

sin Y sin Z Y Z

dz′ dy′

]

k0 W sin 𝜃 sin 𝜙, Z = k0 h cos 𝜃 2 ∫∫

My cos 𝜙ejk0 r

= E0 cos 𝜙

W 2

′ cos 𝜓

h

∫− W ∫−h

ejk0 (y

ds′ ′ sin 𝜃 sin 𝜙+z′ cos 𝜃)

2

[

L𝜙 = hWE0 cos 𝜙

sin Y sin Z Y Z

dz′ dy′

]

thus ) k0 e−jk0 r ( L𝜙 + 𝜂0 N𝜃 0 4𝜋r ] [ hWE sin Y sin Z 0 k0 −jk0 r cos 𝜙 E𝜃′ ≈ −j e 4𝜋r Y Z

E𝜃 ≈ −j

k0 e−jk0 r (L𝜃 − 𝜂0 N𝜙 ) 4𝜋r ] hWE0 k0 −jk r [ sin Y sin Z E𝜙′ ≈ +j e 0 cos 𝜃 sin 𝜙 4𝜋r Y Z E𝜙 ≈ +j

(b) The array factor of 2 elements of zero phase difference, placed along the x-direction and separated by Le , is (AF)2 = ej = ej ( (c) (AF)2 = 2 cos

)

k0 Le sin 𝜃 cos 𝜙 2

k0 Le 2

cos 𝛾

k0 Le 2

sin 𝜃 sin 𝜙

+ e−j

k0 Le 2

+ e−j

cos 𝛾 k0 Le 2

sin 𝜃 sin 𝜙

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SOLUTION MANUAL

Finally, the field radiated by both slots is Et = E′ (AF)2 that is ] k0 W(E0 h) −jk r [ sin Y sin Z e 0 cos 𝜙 cos X 2𝜋r Y Z ] k0 W(E0 h) −jk r [ sin Y sin Z =j e 0 cos 𝜃 sin 𝜙 cos X 2𝜋r Y Z k0 Le sin 𝜃 cos 𝜙 = 2 k W = 0 sin 𝜃 sin 𝜙 2 = k0 h cos 𝜃

E𝜃 = −j E𝜙 X Y Z

14.24. At the two radiating edges (z = 0 & z = L) Ey (z = L) = E0 ,

Hx (z = L) = 0

Ey (z = 0) = −E0 ,

Hx (z = 0) = 0

y n^ 2 Slot 2 x L W Slot 1

z n^ 1

(a) Thus, on slot 1 (z = L) M 1 = −̂n1 × [Ey (z = L)̂ay ] = −̂az × E0 â y = E0 â x J1 = 0 Similarly, on slot 2, we have M 2 = E0 â x J2 = 0 L𝜃 =

∫∫

Mx cos 𝜃 cos 𝜙ejk0 r

= E0 cos 𝜃 cos 𝜙

′ cos 𝜓

ds′ = E0 cos 𝜃 cos 𝜙

W 2

h

∫− W ∫−h

ejk0 (r

′ cos 𝜓)

dy dz

2

W 2

h

∫− W ∫−h 2

ejk0 (x

′ sin 𝜃 cos 𝜙+y′ sin 𝜃 sin 𝜙)

dy dx

] [ k W sin X sin Y , X = 0 sin 𝜃 cos 𝜙, Y = k0 h sin 𝜃 sin 𝜙 L𝜃 = hWE0 cos 𝜃 cos 𝜙 X Y 2 ] [ ′ sin X sin Y L𝜙 = −Mx sin 𝜙ejk0 r cos 𝜓 ds′ = hWE0 sin 𝜙 ∫∫ X Y

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SOLUTION MANUAL

Thus

437

] hWE0 k0 −jk r [ sin X sin Y e 0 sin 𝜙 4𝜋r X Y ] [ hWE0 k0 −jk r sin X sin Y E𝜙′ ≈ +j e 0 cos 𝜃 cos 𝜙 4𝜋r X Y E𝜃′ ≈ +j

(AF)2 = ej

(b)

k0 Le 2

cos 𝜃

(

= 2 cos

+ e−j

k0 Le 2

k0 Le cos 𝜃 2

cos 𝜃

)

(c) Finally, the field radiated by both slots is Et = E′ (AF)2 ( ) ] k0 Le k0 hWE0 −jk r [ sin X sin Y 0 cos sin 𝜙 E𝜃 = j e ⋅ cos 𝜃 2𝜋r X Y 2 ( ) ] k0 Le k0 hWE0 −jk r [ sin X sin Y 0 cos cos 𝜃 cos 𝜙 E𝜙 = j e ⋅ cos 𝜃 2𝜋r X Y 2 k W X = 0 sin 𝜃 cos 𝜙 2 Y = k0 h sin 𝜃 sin 𝜙 14.25. W = 1.186 cm, L = 0.906 cm, h = 0.1588 cm 𝜀r = 2.2, f = 10 GHz. 𝜀reff =

( ) 0.1588 −1∕2 2.2 + 1 2.2 − 1 1 + 12 = 1.972 + 2 2 1.186

Effective length: Le = L + 2ΔL =

λ 30 = = 1.068 cm √ 2 2(10) 1.972 λ0 =

√ ΘE ≃ 2 sin−1 √ −1

ΘH ≃ 2 sin

7.03λ0 2 4(3Le2 + h2 )𝜋 2

c 30 = = 3 cm f0 10

≃ 2 sin−1 (0.4649) = 2(0.4835) = 0.9670 rads = 55.41◦

1 ≃ 2 sin−1 2 + k0 W

√

1 2+

2𝜋 (1.186) 3

= 2 sin−1 (0.472247)

ΘH = 2(0.4918) = 0.9837 rads = 56.36◦ 4𝜋 4𝜋 = = 13.21 = 11.21 dB ΘE ΘH (0.9670)(0.9837) 22.181 22.181 = = 11.66 = 10.67 dB (b) D0 = 2 (0.9670)2 + (0.9837)2 ΘE + Θ2H The D0 s obtained are high because the beamwidth obtained using (14-58) and (14-59) are smaller than those obtained using the Matlab program Microstrip which are ΘE = 88◦ and ΘH = 76◦ . Using these values for the two beamwidths, the respective directivities are D0 (Kraus) = 6.1682 = 7.9 dB and D0 (T − P) = 5.38 = 7.3 dB, which are more representative for a microstrip antenna. (a) D0 =

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SOLUTION MANUAL

14.26. g12 =

G12 3.92904 × 10−4 2 2 = = 0.8153, DAF = = 1.101746 −4 G1 1 + g 1 + 0.8153 4.819021 × 10 12

Using (14-53) and (14-53a) I1 = 0.57074 ) ( ) ( 1 2𝜋 ⋅ 3.962 2 = (1.32768)2 (1.752) = 3.0885 = 4.897 dB D0 = 18.75 0.57074 ∴ D2 = D0 DAF = (3.0885)(1.101746) = 3.4027 = 5.32 dB Using (14-55a), I2 = 1.62558 Finally using (14-55) ( D2 =

14.27. g12 =

2𝜋W λ0

)2

𝜋 = (1.32768)2 (𝜋∕1.62558) = 3.4066 = 5.32 dB I2

G12 4.58053 × 10−4 2 = = 0.2912878, DAF = = 1.5488486 G1 1 + g12 1.57259 × 10−3

Using (14-53) and (14-53a), ( D0 =

I1 = 1.8625,

2𝜋 ⋅ 7.412 18.75

)2

1 = 3.3123 = 5.20 dB 1.8625

∴ D2 = D0 DAF = (3.3123)(1.5488486) = 5.130 = 7.1 dB Using (14-55a), I2 = 6.4152. Finally using (14-55) ( D2 =

2𝜋W λ0

)2

𝜋 = I2

(

2𝜋(7.412) 18.75

)2

𝜋 = 3.0211 = 4.8016 dB 6.4152

14.28. Using the equivalence principle the cavity can be modeled as a circular loop antenna of radius ae with a magnetic current of Im = â 𝜙 2V0 cos 𝜙′ z

r

θ θ' = π 2

ψ ϕ'

x

a^r R y

dl' = aedϕ'

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SOLUTION MANUAL

where at 𝜙′ = 0, V0 = hE0 J1 (kae ) E𝜃 ≃ −j

ke−jkr (L + 𝜂N𝜃 ) 4𝜋r 𝜙

E𝜙 ≃ +j

ke−jkr (L − 𝜂N𝜙 ) but N𝜃 = N𝜙 = 0 4𝜋r 𝜃

∴ Our task then is to find L𝜙 and L𝜃 L𝜃 =

∫ ∫S

M𝜙 cos 𝜃 sin(𝜙 − 𝜙′ )ejkr

′ cos 𝜓

ds′

where r′ cos 𝜓 = 𝜌′ sin 𝜃 cos(𝜙 − 𝜙′ ) = ae sin 𝜃 cos(𝜙 − 𝜙′ ) 𝜌′ = ae because the current source exist only at 𝜌′ = ae and ds′ = 𝜌′ d𝜌′ d𝜙′ = ae d𝜙′ 2𝜋

L𝜃 =

∫0 2𝜋

=

∫0

M𝜙 cos 𝜃 sin(𝜙 − 𝜙′ )ejkae sin 𝜃 cos(𝜙−𝜙 ) ae d𝜙′ ′

′

2V0 a cos 𝜙′ cos 𝜃 sin(𝜙 − 𝜙′ )ejX cos(𝜙−𝜙 ) d𝜙′

= 2V0 (a) cos 𝜃

2𝜋

∫0 ′

where X = kae sin 𝜃, use cos 𝜙′ =

′

cos 𝜙′ sin(𝜙 − 𝜙′ )ejX cos(𝜙−𝜙 ) d𝜙′ ′

′

ej𝜙 + e−j𝜙 , 2

sin(𝜙 − 𝜙′ ) =

′

ej(𝜙−𝜙 ) − e−j(𝜙−𝜙 ) 2j

Substituting L𝜃 = 2V0 ae cos 𝜃 =

2𝜋

∫0

′ ′ ′ ′ 1 j𝜙′ (e + e−j𝜙 )(ej(𝜙−𝜙 ) − e−j(𝜙−𝜙 ) )ejX cos(𝜙−𝜙 ) d𝜙′ 4j

V0 ae cos 𝜃 2𝜋 j𝜙′ j(𝜙−𝜙′ ) ′ ′ ′ ′ ′ ′ [e e − ej𝜙 e−j(𝜙−𝜙 ) + e−j𝜙 ej(𝜙−𝜙 ) − e−j𝜙 −j(𝜙−𝜙 ) ] ∫0 2j ′

. e+jX cos(𝜙−𝜙 ) d𝜙′ V0 ae cos 𝜃 2𝜋 j𝜙 ′ ′ ′ [e − ej(2𝜙 −𝜙) + e−j(2𝜙 −𝜙) − e−j𝜙 ]ejX cos(𝜙−𝜙 ) d𝜙′ ∫0 2j [ 2𝜋 2𝜋 V0 ae cos 𝜃 ′ L𝜃 = ej𝜙 ① ′ d𝜙′ − ②′ ejX cos(𝜙−𝜙 ) d𝜙′ ∫ ∫0 jX cos(𝜙−𝜙 ) j2𝜙 −j𝜙 2j e e 0 ] 2𝜋 2𝜋 −j2𝜙′ j𝜙 ′ −j𝜙 ′ + e e ③ d𝜙 − e ④ d𝜙 ∫0 ∫0 ejX cos(𝜙−𝜙′ ) ejX cos(𝜙−𝜙′ ) =

① ②

2𝜋

∫0

′

ej𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = ej𝜙 2𝜋 ⋅ J0 (X)

2𝜋

∫0

′

2𝜋

′

e−j𝜙 ejX cos(𝜙−𝜙 ) ej2𝜙 d𝜙′ = ej𝜙

= −ej𝜙

𝜙−2𝜋

∫𝜙

ej(−2)u ejX cos u du,

∫0

′

′

ej2(𝜙 −𝜙) ejX cos(𝜙−𝜙 ) d𝜙′

⟨u = 𝜙 − 𝜙′ , du = −d𝜙′ ⟩

439

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SOLUTION MANUAL

[

2𝜋

= ej𝜙

ej(−2)u ejX cos u du = ej𝜙

∫0 j𝜙

2

]

2𝜋 J (X) j−n n

= −2𝜋ej𝜙 J−2 (X) n=−2

j𝜙

= −2𝜋e (−1) J2 (X) = −2𝜋e J2 (X) ③

2𝜋

′

∫0

= −e−j𝜙 [ = e−j𝜙 ④

∴ L𝜃 = = L𝜃 =

𝜙−2𝜋

∫𝜙−0

′

2𝜋

ej2u ejX cos u du

∫0

= −e−j𝜙 (2𝜋)J2 (X) n=2 ′

e−j𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = e−j𝜙 2𝜋J0 (X)

∫0

V0 ae cos 𝜃 j𝜙 [e 2𝜋J0 (X) + 2𝜋ej𝜙 J2 (X) − 2𝜋e−j𝜙 J2 (X) − 2𝜋e−j𝜙 J0 (X)] 2j 2𝜋V0 ae cos 𝜃 ′ [J0 (X)(ej𝜙 − e−j𝜙 ) + J2 (X)(ej𝜙 − e−j𝜙 )] 2j 2𝜋V0 ae cos 𝜃 2j sin 𝜙[J0 (X) + J2 (X)] = 2𝜋V0 ae cos 𝜃 sin 𝜙[J0 (X) + J2 (X)] 2j e−jkr V a cos 𝜃 sin 𝜙[J0 (kae sin 𝜃) + J2 (kae sin 𝜃)] 2r 0 e 2𝜋

∫0

2V0 ae cos 𝜙′ cos(𝜙 − 𝜙′ )ejkae sin 𝜃 cos(𝜙−𝜙 ) d𝜙′ ′

2𝜋

= 2V0 ae

∫0 2𝜋

= 2V0 ae

∫0 [

L𝜙 =

]

′

ej2(𝜙−𝜙 ) ejX cos(𝜙−𝜙 ) d𝜙′

∫0

ej2u ⋅ ejX cos u du = e−j𝜙

2𝜋 J (X) j−n n

2𝜋

∴ E𝜙 ≃ jk L𝜙 =

2𝜋

′

e−j2𝜙 ej𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = e−j𝜙

′

cos 𝜙′ cos(𝜙 − 𝜙′ )ejX cos(𝜙−𝜙 ) d𝜙′ ′

′

′

′

ej𝜙 + e−j𝜙 ej(𝜙−𝜙 ) + e−j(𝜙−𝜙 ) jX cos(𝜙−𝜙′ ) ′ d𝜙 e 2 2

2𝜋 2𝜋 V0 ae ′ ej𝜙 ① ′ d𝜙′ + ej2𝜙 −j𝜙 ② d𝜙′ ∫ ∫0 jX cos(𝜙−𝜙 ) 2 e ejX cos(𝜙−𝜙′ ) 0 ] 2𝜋 2𝜋 −j2𝜙′ j𝜙 ′ −j𝜙 ′ + e e ③ d𝜙 + e ④ d𝜙 ∫0 ∫0 ejX cos(𝜙−𝜙′ ) ejX cos(𝜙−𝜙′ )

① ④ ② ③

2𝜋

′

ej𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = 2𝜋ej𝜙 J0 (X)

∫0 2𝜋

∫0 2𝜋

′

′

′

′

e−j𝜙 ej2𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = −2𝜋ej𝜙 J2 (X)

∫0 2𝜋

∫0

′

e−j𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = 2𝜋e−j𝜙 J0 (X)

ej𝜙 e−j2𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = −2𝜋e−j𝜙 J2 (X)

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SOLUTION MANUAL

441

V0 ae [2𝜋J0 (X)ej𝜙 + 2𝜋J0 (X)e−j𝜙 − 2𝜋J2 (X)ej𝜙 − 2𝜋J2 (X)e−j𝜙 ] 2 L𝜙 = 2𝜋V0 ae cos 𝜙[J0 (X) − J2 (X)] L𝜙 =

∴ E𝜃 ≃

−jke−jkr 2𝜋V0 ae cos 𝜙[J0 (X) − J2 (X)] 4𝜋r

E𝜃 ≃ −jk

e−jkr V a cos 𝜙[J0 (ka sin 𝜃) − J2 (ka sin 𝜃)] 2r 0 e

The E𝜃 and E𝜙 components of the far-field are given as

E𝜃 ≃ −j E𝜙 ≃ jk

kae V0 e−jkr ′ } {cos 𝜙J02 2r

ae V0 e−jkr {cos 𝜃 sin 𝜙J02 } 2r

′ J02 = J0 (kae sin 𝜃) − J2 (kae sin 𝜃)

J02 = J0 (kae sin 𝜃) + J2 (kae sin 𝜃) 14.29. Az = Bmnp Jm (k𝜌 𝜌)[A2 cos(m𝜙) + B2 sin(m𝜙)] cos(kz z) Boundary Conditions: H𝜙 (𝜌 = a, 0 ≤ 𝜙 ≤ 𝜋, 0 ≤ z ≤ h) = 0 H𝜌 (0 ≤ 𝜌 ≤ a, 𝜙 = 0, or 𝜙 = 𝜋, 0 ≤ z ≤ h) = 0

H𝜌 (𝜙 = 0) =

1 1 𝜕Az 11 = B J (k 𝜌)[−A2 sin(m𝜙) m + mB2 ⋅ cos(m𝜙)] cos kz z 𝜇 𝜌 𝜕𝜙 𝜇 𝜌 mnp m 𝜌 ∴ B2 = 0,

H𝜌 (𝜙 = 𝜋) =

11 B J (k 𝜌)[−A2 sin(m𝜙) m] cos kz z = 0 ⇒ m𝜋 = sin−1 (0) = q𝜋 𝜇 𝜌 mnp m 𝜌 m = q = 1, 2, 3, 4, …

H𝜙 (𝜌 = a) = −

1 𝜕Az 1 = − Bmnp Jm′ (k𝜌 a)[A2 cos(m𝜙)] cos(hz z) = 0. 𝜇 𝜕𝜌 𝜇

Jm′ (k𝜌 a) = 0,

′ k𝜌 = 𝜒mn ∕a

From the Boundary conditions E𝜌 (0 ≤ 𝜌 ≤ a, 0 ≤ 𝜙 ≤ 𝜋, z = 0) = E𝜌 (0 ≤ 𝜌 ≤ a, 0 ≤ 𝜙 ≤ 𝜋, z = h) = 0 E𝜙 (0 ≤ 𝜌 ≤ a, 0 ≤ 𝜙 ≤ 𝜋, z = 0) = E𝜙 (0 ≤ 𝜌 ≤ a, 0 ≤ 𝜙 ≤ 𝜋, z = h) = 0 kz =

p𝜋 . h

∴ m = 1, 2, 3, 4, … , n = 1, 2, 3, 4, … , p = 0, 1, 2, 3, 4, …

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SOLUTION MANUAL

Therefore the resonant frequencies for the TMzmnp modes can be written √ √ √ 2𝜋fr 𝜇𝜀 = 𝜔r 𝜇𝜀 = 𝛽r = 𝛽𝜌2 + 𝛽z2 (k𝜌 = 𝛽r , kz = 𝛽z ) √ ( ′ )2 ( ) 𝜒mn p𝜋 2 1 + , m = 1, 2, 3, … fr = √ a n 2𝜋 𝜇𝜀 n = 1, 2, 3, … p = 0, 1, 2, 3, … 14.30. The solution to this problem is identical to that of Problem 14-27 with exceptions of m. TMz :

m𝜋 = sin−1 (0) = q𝜋, q = 1, 2, 3, 4, … , m = 2, 4, 6, … 2 √ ( ′ )2 ( ) 𝜒mn p𝜋 2 1 + fr = √ a n 2𝜋 𝜇𝜀 m = 2, 4, 6, 8, … n = 1, 2, 3, … p = 0, 1, 2, 3, …

14.31. Without considering the feed point, we can use cavity model. The solution to this problem is identical to that of Problem 14-29, with exceptions of m, TMzmnp . H𝜌 (𝜙 = 𝜙0 ) =

11 B J (k 𝜌)[−A2 m sin(m𝜙0 )] cos kz z = 0, 𝜇 𝜌 mnp m 𝜌

∴ m𝜙0 = sin−1 (0) = q𝜋 q𝜋 ∴ m= , q = 1, 2, 3, 4, 5, … √ 𝜙0 ( ′ )2 ( ) 𝜒mn p𝜋 2 q𝜋 1 + , m= , q = 1, 2, 3, … fr = √ a n 𝜙0 2𝜋 𝜇𝜀 n = 1, 2, 3, … p = 0, 1, 2, 3, … 14.32. If we use the cavity model, Az can be written as Az (𝜌, 𝜙, z) = [A1 Jm (k𝜌 𝜌) + B1 Ym (k𝜌 𝜌)][C2 cos m𝜙 + D2 sin m𝜙][C3 cos kz z] Applying the boundary condition leads to H𝜙 (𝜌 = a, 0 ≤ 𝜙 < 2𝜋, 0 ≤ z ≤ h) = −

1 𝜕Az 𝜇 𝜕𝜌

1 = − [A1 Jm′ (k𝜌 a) + B1 Ym′ (k𝜌 a][C2 cos m𝜙 + D2 sin m𝜙][c3 cos kz z] = 0 𝜇 H𝜙 (𝜌 = b, 0 ≤ 𝜙 ≤ 2𝜋, 0 ≤ z ≤ h) 1 = − [A1 Jm′ (k𝜌 b) + B1 Ym′ (k𝜌 b)][C2 cos m𝜙 + D2 sin m𝜙][c3 cos kz z] = 0. 𝜇 E𝜙 (a ≤ 𝜌 ≤ b, 0 ≤ 𝜙 ≤ 2𝜋, z = 0, z = h) = 0 E𝜙 = −j

2 1 1 𝜕 Az ⇒ 𝜔𝜇𝜀 𝜌 𝜕𝜙𝜕z

kz =

p𝜋 , p = 0, 1, 2, … h

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SOLUTION MANUAL

][

[

Jm′ (k𝜌 a) Ym′ (k𝜌 a)

A1

Jm′ (k𝜌 b) Ym′ (k𝜌 b)

B1

] =

[ ] 0

443

∴ Jm′ (k𝜌 a)Ym′ (k𝜌 b) − Jm′ (k𝜌 b)Ym′ (k𝜌 a) = 0

0

To find the resonant frequency for TMzmnp mode, we must solve the following equation. Jm′ (k𝜌 a)Ym′ (k𝜌 b) − Jm′ (k𝜌 b)Ym′ (k𝜌 a) = 0 kmn must satisfy the characteristic equation. 14.33. The solution to this problem is identical to that of Problem 14-32 with following exceptions:

Ez = −j

1 𝜔𝜇𝜀

(

𝜕2 + k2 𝜕Z 2

) Az = −j

k𝜌2 𝜔𝜇𝜀

[A1 Jm (k𝜌 𝜌) + B1 Ym (k𝜌 𝜌)]

× [C2 cos(m𝜙) + D2 sin m𝜙][C3 cos kz z]. Ez (𝜙 = 0) = −j

k𝜌2 𝜔𝜇𝜀

[A1 Jm (k𝜌 𝜌) + B1 Ym (k𝜌 𝜌)][C2 (1) + D2 (0)]C3 cos kz z = 0. ⇒ C2 = 0

Ez (𝜙 = 𝜙0 ) = −j

k𝜌2 𝜔𝜇𝜀

[A1 Jm (k𝜌 𝜌) + B1 Ym (k𝜌 𝜌)][D2 sin(m𝜙0 )]C3 cos kz z = 0 ⇒ sin(m𝜙0 ) = 0

m𝜙0 = sin−1 (0) = q𝜋 ⇒ m =

q𝜋 , q = 1, 2, 3, 4, … 𝜙0

Jm′ (k𝜌 a)Ym′ (k𝜌 b) − Jm′ (k𝜌 b)Ym′ (k𝜌 a) = 0 14.34. The designed center frequency = 1.6 GHz. The dielectric constant of the substrate = 10.2 (i.e., RT∕duroid) h = 0.127 cm. Using (14-69a) F=

8.791 = 1.7203555 √ 1.6 10.2

Therefore using (14-69) F 1.7203555 = = 1.701525 cm [ ( ) ]}1∕2 {1 + 0.0222555}1∕2 𝜋F 2h ln + 1.7726 1+ 𝜋𝜀r F 2h

a= {

Grad = 8.084 × 10−4 Gc = 7.326 × 10−4 Gd = 4 × 10−4 Gt = Grad + Gc + Gd = 19.409 × 10−4 Rin (𝜌′ = ae ) = 515.221 ohms 𝜌0 ≃ 0.42 cm

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SOLUTION MANUAL

14.35. fr = 1.6 GHz, 𝜀r = 2.2, h = 0.1575 cm Using (14-69a) F=

8.791 = 3.704 √ 1.6 2.2

Therefore using (14-69) a= (

1 √

14.36. ( fr )mno =

2𝜋 𝜇𝜀

′ 𝜒mn

a (

) , 𝜀r = 10.2, h = 0.127 cm )

) 3.0542 ′ , 𝜒21 = 3.0542 a a 2𝜋 𝜇𝜀 2𝜋 𝜇𝜀 ) ) ( ( 30 × 109 3.0542 3.0542 1 = = √ √ a a 2𝜋 𝜇0 𝜀0 𝜀r 2𝜋 10.2

(a) ( fr )210 = (b) ( fr )210

3.704 = 3.587 cm {1 + (0.0123)[ln(36.944) + 1.7726]}1∕2

1 √

30 × 109 a= √ 2𝜋 10.2

′ 𝜒21

(

1 √

=

3.0542 9 × 108

(

) = 5.0734 cm

a = 5.0734 cm 14.37. (a) For a circular microstrip operating in the TMzmn0 mode, the resonant frequency for the mn mode is given by, negelecting fringing ( ′ ) 𝜒mn 1 (14-65) ( fr )mn0 = √ a 2𝜋 𝜇𝜀 ′ = 3.0542 For 𝜀 = 𝜀r 𝜀0 and mn = 21 ⇒ 𝜒21

(14-64)

Thus z

9 ( fr )TM 210 = 1.9 × 10 =

1 2𝜋 𝜀r 𝜀0 𝜇0 √

1.9 × 109 = a=

(

3.0542 a

) =

1 √ √ 2𝜋 𝜀r 𝜀0 𝜇0

(

3.0542 a

)

) ( 30 × 109 3.0542 √ a 2𝜋 10.2

30(3.0542) = 2.40317 cm √ (1.9)2𝜋 10.2

{

]}1∕2 [ ( ) 𝜋a 2h + 1.7726 ln (b) ae = a 1 + 𝜋a𝜀r 2h { [ ( ) ]}1∕2 2(0.127) 3.0542𝜋 = 2.40317 1 + ln + 1.7726 𝜋(2.40317)10.2 2(0.127)

(14-67)

= 2.40317{1 + 0.003364[ln(37.7758) + 1.7726]}1∕2 ae = 2.40317{1 + 0.003364[3.63167 + 1.7726]}1∕2 = 2.40317{1 + 0.01818}1∕2

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SOLUTION MANUAL

445

ae = 2.40317(1.00905) = 2.42492 ae = 2.42492 cm ( 1 (c) ( fr )TM √ √ 210 (with fringing) = 2𝜋 𝜀r 𝜇0 𝜀0

′ 𝜒21

)

ae

=

) ( 30 × 109 3.0542 √ 2𝜋 10.2 2.42492

( fr )TM 210 (with fringing) = 1.88296 GHz 14.38. Desired frequency: f = 900 MHz ⇒ λ =

1 m = 0.333 m 3

𝜀r = 10.2, h = 0.127 cm: TMz210 (a) Resonant frequency; From equation (14-63a), (k𝜌 )2 + (kz )2 = 𝜔2r 𝜇𝜀

(

′ k𝜌 = 𝜒mn ∕a, kz =

∴ TMz210 → (m = 2, n = 1, p = 0).

p𝜋 ) n

(𝜒mn ∕a)2 = 𝜔2r 𝜇𝜀 = (2𝜋frmno )2 𝜇𝜀 ( ′ ) ) ( 𝜒mn 3.0542 1 1 , ( fr )210 = (fr )mn0 = √ √ √ a a 2𝜋 𝜇0 𝜀0 𝜀r 2𝜋 𝜇𝜀 (fr )210 =

3.0542 √ c 2𝜋a 𝜀r

(c = 3 × 108 m∕s)

(b) Neglect fringing. a=

3.0542c 3.0542 = 0.057 m √ √ = 210 2𝜋( fr ) 𝜀r 2𝜋(3) 10.2

a = 5.7 cm (c) E𝜃 ≃ −j

ke−jkr (L + 𝜂N𝜃 ) 4𝜋r 𝜙

ke−jkr (L − 𝜂N𝜃 ) 4𝜋r 𝜃 Using the equivalent principle the cavity can be modeled as a circular loop antenna of radius a with magnetic current for general mode. E𝜙 ≃ +j

Im = 2V0 cos(m𝜙′ ) Similar procedure like Problem 14-23, L =

L𝜃 =

∬s

M𝜙 cos 𝜃 sin(𝜙 − 𝜙′ )ejkr

= 2V0 a cos 𝜃

2𝜋

∫0

∬s

′ cos 𝜓

ds′

I m e+jkr

′ cos 𝜓

ds′

X = ka sin 𝜃 ′

cos n𝜙′ sin(𝜙 − 𝜙′ )e+jX cos(𝜙−𝜙 ) d𝜙′

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L𝜃 = 2V0 cos 𝜃 =a

2𝜋

∫0

′ ′ ′ ′ 1 jm𝜙′ + e−jm𝜙 )(ej(𝜙−𝜙 ) − e−j(𝜙−𝜙 ) )ejX cos(𝜙−𝜙 ) d𝜙′ (e 4j

V0 cos 𝜃 2𝜋 j(m𝜙′ +𝜙−𝜙′ ′ ′ ′ ′ [e ) − ejm𝜙 −j(𝜙−𝜙 ) + e−jm𝜙 +j(𝜙−𝜙 ) ∫0 2j jX cos(𝜙−𝜙′ )

−e−j(m𝜙 +𝜙−𝜙 )]e d𝜙′ { 2𝜋 [ j(m−1)𝜙′ j𝜙 ′ ′ V0 cos 𝜃 e ej(m+1)𝜙 e−j𝜙 e−j(m+1)𝜙 +j𝜙 e =a − + ∫0 2j ① ② ③ ] ′ e−j(m−1)𝜙 e−j𝜙 jX cos(𝜙−𝜙′ ) ′ − d𝜙 e ④ ′

①ej𝜙

2𝜋

′

2𝜋

= ejm𝜙

′

′

′

= e−jm𝜙

2𝜋

′

∫0

′

′

2𝜋

j

J (X), used Jm (X) = −(m−1) m−1

∫0

′

j−n 2𝜋 jnu jX cos u e e du 2𝜋 ∫0

V0 cos 𝜃 jm𝜙 m−1 [e (j) Jm−1 (X) − ejm𝜙 (j)m+1 Jm+1 (X) j

= −j2𝜋aV0 cos 𝜃[sin m𝜙(j)m Jm−1 (X) + sin m𝜙(j)m Jm+1 (X)] ke−jkr [−j2𝜋V0 cos 𝜃 sin m𝜙(Jm−1 (X) + Jm+1 (X))](j)m 4𝜋r

= (j)m

ke−jkr [V0 a cos 𝜃 sin m𝜙(Jm−1 (X) + Jm+1 (X))] 2r 2𝜋

L𝜙 = 2V0 a

∫0 2𝜋

L𝜙 = 2V0 a

′

∫0

X = ka sin 𝜃

′

e+j(m−1)(𝜙−𝜙 ) ejX cos(𝜙−𝜙 ) d𝜙′

+ e−jm𝜙 (j)m+1 Jm+1 (X) − e−jm𝜙 (j)m−1 Jm−1 (X)]

∴ E𝜙 = +j

′

ej(m+1)(𝜙−𝜙 ) ejX cos(𝜙−𝜙 ) d𝜙′

2𝜋

′

e−j(m−1)𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = e−jm𝜙

= e−jm𝜙

∴ L𝜃 = a𝜋

′

ej(m+1)(𝜙−𝜙 ) ejX cos(𝜙 −𝜙) d𝜙′

∫0

2𝜋 Jm+1 (X) j−(m+1)

2𝜋

∫0

2𝜋

J(1−m) (X) j−(1−m)

2𝜋

′

e−j(m+1)𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = e−jm𝜙

∫0

′

2𝜋 Jm+1 (X) j−(m+1)

2𝜋

④e−j𝜙

′

ej(m+1)𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = e+jm𝜙

∫0

′

ej(m−1)(𝜙 −𝜙) ejX cos(𝜙−𝜙 ) d𝜙′

∫0

ej(1−m)(𝜙−𝜙 ) ejX cos(𝜙−𝜙 ) d𝜙′ = ejm𝜙

∫0

2𝜋

= ejm𝜙 ③ej𝜙

2𝜋

′

ej(m−1)𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = ejm𝜙

∫0

②e−j𝜙

′

cos m𝜙′ cos(𝜙 − 𝜙′ )ejka sin 𝜃 cos(𝜙−𝜙 ) d𝜙′ ′

cos m𝜙′ cos(𝜙 − 𝜙′ )ejka sin 𝜃 cos(𝜙−𝜙 ) d𝜙′ ′

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447

2𝜋

′ ′ ′ ′ 1 jm𝜙′ + e−jm𝜙 ][ej(𝜙−𝜙 ) + e−j(𝜙−𝜙 ) ]ejX cos(𝜙−𝜙 ) d𝜙′ [e 4 ] [ V a 2𝜋 ej(m−1)𝜙′ ej𝜙 e−j(m+1)𝜙′ ej𝜙 ej(m+1)𝜙′ e−j𝜙 e−j(m−1)𝜙′ e−j𝜙 = 0 + + + 2 ∫0 ① ② ③ ④

L𝜙 = 2V0 a

∫0

′

⋅ ejX cos(𝜙−𝜙 ) d𝜙′ ①:

2𝜋

∫0

′

= ejm𝜙 ②

′

∫0

′

∫0

2𝜋

′

′

′

′

ej(m+1)(𝜙−𝜙 ) ejX cos(𝜙−𝜙 ) d𝜙′

∫0

2𝜋 Jm+1 (X) j−(m+1)

2𝜋

′

2𝜋

′

e−j𝜙 e−j(m−1)𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = e−jm𝜙

∫0

′

ej(m+1)(𝜙 −𝜙) ejX cos(𝜙−𝜙 ) d𝜙′

2𝜋 Jm+1 (X) j−(m+1)

e−j𝜙 ej(m+1)𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = ejm𝜙 = ejm𝜙

④

2𝜋

′

ej𝜙 e−j(m+1)𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = e−jm𝜙

2𝜋

′

2𝜋

2𝜋

∫0

′

ej(m−1)(𝜙 −𝜙) ejX cos(𝜙 −𝜙) d𝜙′

∫0

Jm−1 (X) j−(m−1)

= e−jm𝜙 ③

2𝜋

′

ej𝜙 ej(m−1)𝜙 ejX cos(𝜙−𝜙 ) d𝜙′ = ejm𝜙

∫0

= e−jm𝜙

′

′

ej(m−1)(𝜙−𝜙 ) ejX cos(𝜙−𝜙 ) d𝜙′ 2𝜋

Jm−1 (X). j−(m−1)

V0 a [2𝜋{(ejm𝜙 + e−jm𝜙 )(j)m−1 jm−1 (X) + ejm𝜙 + e−jm𝜙 )(j)m+1 jm+1 (X)}] 2 = V0 a𝜋[2 cos m𝜙(j)m (−j)Jm−1 (X) + 2 cos m𝜙(j)m (j)Jm+1 (X)]

∴ L𝜙 =

∴ L𝜙 = 2𝜋V0 a cos m𝜙[(−j)(j)m ][Jm−1 (X) − Jm+1 (X)] As similar procedure like L𝜃 . L𝜙 = 2𝜋V0 a(−j)(j)m cos m𝜙[Jm−1 (X) − Jm+1 (X)] ∴ E𝜃 = −jk

e−jkr (2𝜋V0 a(−j)(j)m cos m𝜙[Jm−1 (X) − Jm+1 (X)] 4𝜋r

ke−jkr [V0 a cos m𝜙(Jm−1 (X) − Jm+1 (X))] 2r (X = ka sin 𝜃)

= −(j)m

For m = 2, the far field is E𝜃 =

ke−jkr [V0 a cos 2𝜙(J1 (X) − J3 (X))] 2r

E𝜙 = −

ke−jkr [V0 a cos 𝜃 sin 2𝜙(J1 (X) + J3 (X))] 2r

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(d) E- and H-Plane Field Pattern of Circular Patch (TMz210 mode) normalized azimuthal (x − y plane) amplitude pattern (dB) 0 30

30

60

60

90

90 –30 dB –20 dB 120

120 –10 dB 0 dB

150

150

180 E-Plane H-Plane

(e) U = |E𝜙 |2 + |E𝜃 |2 at 𝜃 = 90◦ . U ∝ (cos 2𝜙)2 at 𝜃 = 90◦

90 y 120

0 dB

60

–20 dB

150

30

180

x

210

330

240

300 270

Fig. P14.38

0

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(f) Directivity TMz210 mode D0 = D𝜃 + D𝜙 D𝜃 = D𝜙 =

4𝜋U𝜃 max = 4.3739 (Prad )𝜃 + (Prad )𝜙 4𝜋U𝜙 max (Prad )𝜃 + (Prad )𝜙

= 1.4866

∴ D0 = D𝜃 + D𝜙 = 4.3739 + 1.4866 = 5.8605 D0 = 7.6 dB 4.2012 8 √ c (c = 3 × 10 m∕s) 2𝜋a 𝜀r 4.2012 1 4.2012c = 0.0698 m = 6.98 cm (b) a = √ = √ 2𝜋(3) 10.2 2𝜋( fr )310 𝜀r (c) Far-zone field. From Part (c) in Problem 14-38

14.39. (a) ( fr )310 =

E𝜃 = (−j)3 =j

ke−jkr {V0 a cos 3𝜙[J2 (X) − J4 (X)]}, X = ka sin 𝜃 2r

ke−jkr {V0 a cos 3𝜙[J2 (ka sin 𝜃) − J4 (ka sin 𝜃)]} 2r

E𝜙 = −j

ke−jkr {V0 a sin 3𝜙 cos 𝜃[J2 (ka sin 𝜃) + J4 (ka sin 𝜃)]} 2r

(d) See the Fig. P14-39. E- and H-Plane Field Pattern of Circular Patch (TMz310 mode) 0 30

30

60

60

90

90 –30 dB –20 dB 120

120 –10 dB 150

0 dB 180 E-Plane H-Plane

Fig. P14.39(d)

150

449

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(e) See Fig. P14.39(e) 90 y 120

0 dB

60

–20 dB

150

30

180

x

210

0

330

240

300 270

Fig. P14.39(e)

(f) D0 = D𝜃 + D𝜙 D𝜃 = 5.98148 D𝜙 = 1.111 D0 = 7.092 = 8.5 dB 14.40. (a) TMz410

( fr )410 =

5.3175 √ c 2𝜋a 𝜀r

(c = 3 × 108 m∕s)

5.3175 = 0.0883 m, a = 8.83 cm √ 2𝜋(3) 10.2 ke−jkr (c) E𝜃 = −(j)4 [V0 a cos 4𝜙(J3 (ka sin 𝜃) − J5 (ka sin 𝜃))] 2r

(b) a =

(j)4 = 1 E𝜙 = (j)4

ke−jkr {V0 a cos 𝜃 sin 4𝜙[J3 (ka sin 𝜃) + J5 (ka sin 𝜃)]} 2r

(j)4 = 1 (d) E- and H-Plane Field Pattern of Circular Patch (TMz410 mode) [see Fig. P14.40(d)]

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0 30

30

60

60

90

90 E-Plane H-Plane

–30 dB –20 dB

120

120 –10 dB 0 dB

150

150

180

Fig. P14.40(d)

(e) see Fig. P14.40e. 90 y 120

0 dB

60

–20 dB

150

30

180

x

210

330

240

300 270

Fig. P14.40(e)

(f) D0 = D𝜃 + D𝜙 = 7.0929 + 0.853 = 7.945 = 9 dB 14.41. f = 10 GHz, ⇒ λ = 3 cm, d = 0.1 cm 𝜀r = 2.2, h = 0.1575 cm

0

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√ Z jXp = j √0 tan( 𝜀r k0 h) 𝜀r (√ ) 120𝜋 2𝜋 tan = j√ 2.2 0.1575 3 2.2 jXp = j254.167(0.53245) = j135.33 Form K. R. Carver & J. W. Mink, “Microstrip Antenna Technology” IEEE Trans. AP, Vol 29, pp. 2–24, 1981. 14.42.

λ impedance transformer, Z0 = 50 ohms, 𝜀r = 2.2 4 RL = 100 ohms, W0

W1

Z0 Z1 Z1 = 70.7

Z0 = 50

RL = 100 Ω

Z1 =

√

Z0 RL ∴

Z1 =

√ 50(100) = 70.7106

W0 = 0.4891 cm W1 = 0.28 cm 14.43. Two section binomial transformer. Using(9-37) 𝜌n = 2−N

RL − Z0 N! RL + Z0 (N − n)!n!

for N = 2, RL = 100, Z0 = 50,

RL − Z0 100 − 50 1 = = RL + Z0 100 + 50 3

1 2! 1 = . 3 2! ⋅ 0! 12 1 2! 1 = n = 1; 𝜌1 = 2−2 3 (1!)(1!) 6 n = 0; 𝜌0 = 2−2

n = 0 → 𝜌0 =

Z − Z0 1 ⇒, = 1 12 Z1 + Z0

n = 1 → 𝜌1 =

1 Z2 − Z1 ⇒, = 6 Z2 + Z1 W0 = 0.4891 cm W2 = 0.2091 cm W1 = 0.3790 cm

Z1 = 1.182 Z0 = 59.09 Z2 = 1.399 Z1 = 82.73

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453

14.44. Using (9-42) and (9-49), we obtain Γ = e−jN𝜃

ZL − Z0 TN (sec 𝜃m cos 𝜃) ZL + Z0 TN (sec 𝜃m ) 100 Ω

W0 Z0 = 50 Ω

W1

W2

Z1 = 59.09 Z2 = 82.73

W0 = 0.4891 cm

W2 = 0.2091 cm

T2 (sec 𝜃m cos 𝜃) = 2(sec 𝜃m cos 𝜃)2 − 1 = 2 sec2 𝜃m (1 + cos 2𝜃) − 1 Let the maximum tolerable value of 𝜌 be 𝜌m = 0.05. 𝜌m =

| ZL − Z0 1 1 | = 0.05, ∴ TN (sec 𝜃m )|N=2 = = 6.67 | (ZL + Z0 )TN (sec 𝜃m ) |N=2 3 0.05 T2 (sec 𝜃m ) = 2 sec2 𝜃m − 1 = 6.67

and hence sec 𝜃m = 1.96 and 𝜃m = 1.04 rad. 2𝜌0 cos 2𝜃 + 𝜌1 = 𝜌m T2 (sec 𝜃m cos 𝜃) = 𝜌m sec2 𝜃m cos 2𝜃 + 𝜌m (sec2 𝜃m − 1) ∴ 𝜌0 = 12 𝜌m sec2 𝜃m = 0.096 𝜌1 = 𝜌m (sec2 𝜃m − 1) = 0.142 Thus, the impedances Z1 and Z2 are given by Z1 =

1 + 𝜌0 1 + 0.096 ⋅ Z0 = Z = 1.21Z0 = 60.5 1 − 𝜌0 1 − 0.096 0

Z2 =

1 + 𝜌1 1 + 0.142 ⋅Z = Z = 1.62Z0 = 81 1 − 𝜌1 0 1 − 0.142 0

By applying the iterative procedure in (14-19b), we can find the appropriate value of width. Z1 = 60.5 ⇒ W1 = 0.3691 cm Z2 = 81.0 ⇒ W2 = 0.2181 cm 14.45. The input impedance of a λ∕2 dipole is Zd = 73 + j42.5 A folded dipole is a step-up impedance transformer whose input impedance is given by (9-26) which, for a λ∕2 length, reduces to (9-27), which gives two equal parallel paths of

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SOLUTION MANUAL

currents. The technique can be extended to N parallel paths/currents leading to an input impedance of Zin = N 2 Zd . In contrast to the folded dipole with N parallel current paths, the folded slot consists of N parallel slots, and the folded slot is a step-down impedance transformer where the impedance is equal to 1 Zin (slot) = 2 Zs (slot) N 𝜂2 where Zs = according to Babinet’s Principle. From Example 12.6, 4Zd Zs = 362.95 − j211.31 Therefore the input impedance of the folded slot is equal to Zin (slot) =

1 1 Zs = (362.95 − j211.31) = 90.738 − j52.828 2 4 (2) Zin (slot) = 90.738 − j52.828

14.46. For a cubic resonator of Figure 14.57, using the Matlab computer program DRA Analysis Design, we get the following: ———————————————————————————————————————————————————————– (a) 𝜀r = 8.9 Input ———————————————————————————————————————————————————————– Select the desired geometry: 1. Cubic resonator 2. Cylindrical 3. Hemicylindrical 4. Hemispherical Selected DRA: 1 Enter the length a (in cm): 1 Enter the width b (in cm): 1 Enter the height c (in cm): 0.3 Enter the relative permittivity, er: 89 ———————————————————————————————————————————————————————– Output ———————————————————————————————————————————————————————– The first five modes for a cubic resonator are: fTE(1,1,0) = 3.48 GHz fTM(1,0,1) = 3.09 GHz fTM(1,0,1) = fTE(1,1,1) = 3.48 GHz fTM(0,1,1) = 3.09 GHz fTM(0,1,1) = fTE(2,1,0) = 4.43 GHz fTM(1,1,1) = 3.48 GHz fTE(1,1,0) = fTE(1,2,0) = 4.43 GHz fTM(2,0,1) = 4.14 GHz fTE(1,1,1) = fTE(2,1,1) = 4.43 GHz fTM(0,2,1) = 4.14 GHz fTM(1,1,1) = » ———————————————————————————————————————————————————————– (b) 𝜀r = 89 Input ———————————————————————————————————————————————————————– Select the desired geometry: 1. Cubic resonator 2. Cylindrical 3. Hemicylindrical 4. Hemispherical Selected DRA: 1

3.09 3.09 3.48 3.48 3.48

GHz GHz GHz GHz GHz

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SOLUTION MANUAL

Enter the length a (in cm): 1 Enter the width b (in cm): 1 Enter the height c (in cm): 0.3 Enter the relative permittivity, er: 8.9 ———————————————————————————————————————————————————————– Output ———————————————————————————————————————————————————————– The first five modes for a cubic resonator are: fTE(1,1,0) = 10.99 GHz fTM(1,0,1) = 9.77 GHz fTM(1,0,1) = 9.77 GHz fTE(1,1,1) = 10.99 GHz fTM(0,1,1) = 9.77 GHz fTM(0,1,1) = 9.77 GHz fTE(2,1,0) = 14.02 GHz fTM(1,1,1) = 10.99 GHz fTE(1,1,0) = 10.99 GHz fTE(1,2,0) = 14.02 GHz fTM(2,0,1) = 13.09 GHz fTE(1,1,1) = 10.99 GHz fTE(2,1,1) = 14.02 GHz fTM(0,2,1) = 13.09 GHz fTM(1,1,1) = 10.99 GHz »

14.47. For a hemicylindrical resonator of Figure 14.59, using the Matlab computer program DRA Analysis Design, we get the following: ———————————————————————————————————————————————————————– (a) 𝜀r = 8.9 Input ———————————————————————————————————————————————————————– Select the desired geometry: 1. Cubic resonator 2. Cylindrical 3. Hemicylindrical 4. Hemispherical Selected DRA: 3 Enter the radius a (in cm): 0.3 Enter the length h (in cm): 1 Enter the relative permittivity, er: 8.9 ———————————————————————————————————————————————————————– Output ———————————————————————————————————————————————————————– The first five modes for a Hemicylindrical resonator are: fTE(0,1,1) = 13.07 GHz fTM(1,1,1) = 10.14 GHz fTM(1,1,1) fTE(0,1,2) = 14.88 GHz fTM(1,1,2) = 12.38 GHz fTM(1,1,2) fTE(0,1,3) = 17.96 GHz fTM(1,1,3) = 15.95 GHz fTM(0,1,1) fTE(1,1,1) = 20.60 GHz fTM(2,1,1) = 16.49 GHz fTM(0,1,2) fTE(0,1,4) = 21.78 GHz fTM(2,1,2) = 17.95 GHz fTM(1,1,3) » ———————————————————————————————————————————————————————– (b) 𝜀r = 89 Input ———————————————————————————————————————————————————————– Select the desired geometry: 1. Cubic resonator 2. Cylindrical 3. Hemicylindrical 4. Hemispherical Selected DRA: 3 Enter the radius a (in cm): 0.3 Enter the length h (in cm): 1 Enter the relative permittivity, er: 89

= = = = =

10.14 12.38 13.07 14.88 15.95

GHz GHz GHz GHz GHz

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———————————————————————————————————————————————————————– Output ———————————————————————————————————————————————————————– The first five modes for a Hemicylindrical resonator are: fTE(0,1,1) = 4.13 GHz fTM(1,1,1) = 3.21 GHz fTM(1,1,1) fTE(0,1,2) = 4.71 GHz fTM(1,1,2) = 3.92 GHz fTM(1,1,2) fTE(0,1,3) = 5.68 GHz fTM(1,1,3) = 5.04 GHz fTE(0,1,1) fTE(1,1,1) = 6.51 GHz fTM(2,1,1) = 5.21 GHz fTE(0,1,2) fTE(0,1,4) = 6.89 GHz fTM(2,1,2) = 5.68 GHz fTM(1,1,3) »

= = = = =

3.21 3.92 4.13 4.71 5.04

GHz GHz GHz GHz GHz

14.48. For a hemispherical resonator of Figure 14.60, using the Matlab computer program DRA Analysis Design, we get the following ———————————————————————————————————————————————————————– (a) 𝜀r = 8.9: Input ———————————————————————————————————————————————————————– Select the desired geometry: 1. Cubic resonator 2. Cylindrical 3. Hemicylindrical 4. Hemispherical Selected DRA: 4 Enter the radius a (in cm): 0.3 Enter the relative permittivity, er: 8.9 ———————————————————————————————————————————————————————– Output ———————————————————————————————————————————————————————– The dominant mode for a hemicylindrical resonator is: fTE(1,1,1) = 14.6389 GHz (degenerate; even, odd) » ———————————————————————————————————————————————————————– (b) 𝜀r = 89: Input ———————————————————————————————————————————————————————– Select the desired geometry: 1. Cubic resonator 2. Cylindrical 3. Hemicylindrical 4. Hemispherical Selected DRA: 4 Enter the radius a (in cm): 0.3 Enter the relative permittivity, er: 89 ———————————————————————————————————————————————————————– Output ———————————————————————————————————————————————————————– The dominant mode for a hemicylindrical resonator is: fTE(1,1,1) = 4.6292 GHz (degenerate; even, odd) » ———————————————————————————————————————————————————————– 14-49 Input ———————————————————————————————————————————————————————– Select the desired geometry:

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1. Cubic resonator 2. Cylindrical 3. Hemicylindrical 4. Hemispherical Selected DRA: 2 Select one of the following: 1. Analysis 2. Design Selected one: 2 Select one of the following modes: 1. Transverse Electric (TE 01d) 2. Transverse Magnetic (TM 01d) 3. Hybrid (HE 11d) Selected mode: 2 Enter the fractional bandwidth (in percent):2.887 Enter VSWR:3 Enter the resonant frequency (in GHz):10 ———————————————————————————————————————————————————————– (a)Q(specified)= 39.9966 ———————————————————————————————————————————————————————– (b)Your dielectric constant should be greater than 27.9000 ———————————————————————————————————————————————————————– (c)Enter your dielectric constant:38 ———————————————————————————————————————————————————————– Output ———————————————————————————————————————————————————————– (d) a(cm)= 0.4158 h(cm) = 0.1650 ———————————————————————————————————————————————————————– » ———————————————————————————————————————————————————————– 14-50 Input ———————————————————————————————————————————————————————– Select the desired geometry: 1. Cubic resonator 2. Cylindrical 3. Hemicylindrical 4. Hemispherical Selected DRA: 2 Selected one of the following: 1. Analysis 2. Design Selected one:2 Selected one of the following modes: 1. Transverse Electric (TE 01d) 2. Transverse Magnetic (TM 01d) 3. Hybrid (HE 11d) Selected mode:3 Enter the fractional bandwidth (in percent):2.887 Enter VSWR:3 Enter the resonant frequency (in GHz):10 ———————————————————————————————————————————————————————–

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(a)Q(specified)= 39.9966 ———————————————————————————————————————————————————————– (b)The dielectric constant should be within the range of: 35.8423 < er < 95.1567 ———————————————————————————————————————————————————————– (c)Enter your dielectric constant:38 ———————————————————————————————————————————————————————– Output ———————————————————————————————————————————————————————– (d) a(cm)= 0.2688 h(cm) = 0.1723 ———————————————————————————————————————————————————————– »

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Solution Manual

15.1. Array Factor: AF = C[ejks cos 𝜓1 − e−jks cos 𝜓2 ] (For far-zone: 𝜓1 = 𝜓2 = 𝜓) AF = 2j sin(ks cos 𝜓) cos 𝜓 = â x ⋅ â r aˆx

ψ

s

15.2. (a) For all three corner reflectors (𝛼 = 60◦ , 45◦ , 30◦ ), the geometrical coordinate system shown in the adjacent figure is used. The sources will be numbered so that the feed will be #1. The images are designated as #2, #3, … (in a counter clock wise rotation), as shown in Figure 15.4(b) for the 90◦ corner reflector. y

s z

ϕ x Source

Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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𝜶 = 60◦ Using the source arrangement of Figure 15.3(b) ( E(r, 𝜃, 𝜙) =

e−jkr1 e−jkr2 e−jkr3 e−jkr4 e−jkr5 e−jkr6 + + + + + r1 r2 r3 r4 r5 r6

) f (𝜃, 𝜙)

for the main source (#1) and the five images. For far-field observations: For phase terms r1 ≃ r − s cos 𝜓1 = r − s(̂ax ⋅ â r ) = r − s sin 𝜃 cos 𝜙 r2 ≃ r − s cos 𝜓2 = r − s[(0.5̂ax + 0.866̂ay ) ⋅ â r ] = r − s(0.5 sin 𝜃 cos 𝜙 + 0.866 sin 𝜃 sin 𝜙) r3 ≃ r − s cos 𝜓3 = r − s[(−0.5̂ax + 0.866̂ay ) ⋅ â r ] = r − s(−0.5 sin 𝜃 cos 𝜙 + 0.866 sin 𝜃 sin 𝜙) r4 ≃ r − s cos 𝜓4 = r − s(−̂ax ⋅ â r ) = r + s sin 𝜃 cos 𝜙 r5 ≃ r − s cos 𝜓5 = r − s[(−0.5̂ax − 0.866̂ay ) ⋅ â r ] = r + s(0.5 sin 𝜃 cos 𝜙 + 0.866 sin 𝜃 sin 𝜙) r6 ≃ r − s cos 𝜓6 = r − s[(0.5̂ax − 0.866̂ay ) ⋅ â r ] = r + s(−0.5 sin 𝜃 cos 𝜙 + 0.866 sin 𝜃 sin 𝜙) where â r = â x sin 𝜃 cos 𝜙 + â y sin 𝜃 sin 𝜙 + â z cos 𝜃 For amplitude terms r1 ≃ r 2 ≃ r 3 ≃ r 4 ≃ r 5 ≃ r 6 ≃ r Making these substitutions and combining terms (first with fourth, second with fifth, and third with sixth), we can write that E = f (𝜃, 𝜙)

e−jkr 2{sin(ks sin 𝜃 cos 𝜙) r

− sin[ks(0.5 sin 𝜃 cos 𝜙 + 0.866 sin 𝜃 sin 𝜙)]} Using the identities of sin(x ± y) = sin x cos y ± cos x sin y reduces the field to { E =2 E0

(

) ks sin(ks sin 𝜃 cos 𝜙) − 2 sin sin 𝜃 cos 𝜙 cos 2

where E0 = f (𝜃, 𝜙)

e−jkr r

(√

)}

3 ks sin 𝜃 sin 𝜙 2

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SOLUTION MANUAL

Utilizing the identity of sin 2x = 2 sin x cos x we can write in the final form that (√ )] [ ( ) ) ( 3 E ks ks = 4 sin sin 𝜃 cos 𝜙 cos sin 𝜃 cos 𝜙 − cos ks sin 𝜃 sin 𝜙 E0 2 2 2 𝜶 = 45◦ Using the source arrangement of Figure 15.3(c) ( E=

e−jkr1 e−jkr2 e−jkr3 e−jkr4 e−jkr5 e−jkr6 e−jkr7 e−jkr8 + + + + + + + r1 r2 r3 r4 r5 r6 r7 r8

) f (𝜃, 𝜙)

For far-field observations: For phase terms r1 = r − s cos 𝜓1 = r − s(̂ax ⋅ â r ) = r − s sin 𝜃 cos 𝜙 [ ] 1 s r2 = r − s cos 𝜓2 = r − s √ (̂ax + â y ) ⋅ â r = r − √ (sin 𝜃 cos 𝜙 + sin 𝜃 sin 𝜙) 2 2 r3 = r − s cos 𝜓3 = r − s(̂ay ⋅ â r ) = r − s sin 𝜃 sin 𝜙 [ ] 1 s r4 = r − s cos 𝜓4 = r − s √ (−̂ax + â y ) ⋅ â r = r + √ (sin 𝜃 cos 𝜙 − sin 𝜃 sin 𝜙) 2 2 r5 = r − s cos 𝜓5 = r − s(−̂ax ⋅ â r ) = r + s sin 𝜃 cos 𝜙 [ ] 1 s r6 = r − s cos 𝜓6 = r − s √ (−̂ax − â y ) ⋅ â r = r + √ (sin 𝜃 cos 𝜙 + sin 𝜃 sin 𝜙) 2 2 r7 = r − s cos 𝜓7 = r − s(−̂ay ⋅ â r ) = r + s sin 𝜃 sin 𝜙 [ ] 1 s r8 = r − s cos 𝜓8 = r − s √ (̂ax − â y ) ⋅ â r = r − √ (sin 𝜃 cos 𝜙 − sin 𝜃 sin 𝜙) 2 2 For amplitude terms r 1 ≃ r2 ≃ r3 ≃ r4 ≃ r5 ≃ r6 ≃ r7 ≃ r8 ≃ r Making these substitutions and combining terms (first with fifth, second with sixth, third with seventh, fourth with eighth), we can write { E =2 E0

cos(ks sin 𝜃 cos 𝜙) + cos(ks sin 𝜃 sin 𝜙)

− 2 cos

(

) } ( ) ks ks sin 𝜃 sin 𝜙 √ sin 𝜃 cos 𝜙 cos 2 2

where E0 = f (𝜃, 𝜙)

e−jkr r

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SOLUTION MANUAL

𝜶 = 30◦ The procedure for this reflector follows those of the others. Using the geometry of Figure 15.3(d) ( −jkr e 1 e−jkr2 e−jkr3 e−jkr4 e−jkr5 e−jkr6 e−jkr7 e−jkr8 E= + + + + + + + r1 r2 r3 r4 r5 r6 r7 r8 ) −jkr −jkr −jkr −jkr e 9 e 10 e 11 e 12 + + + + f (𝜃, 𝜙) r9 r10 r11 r12 For far field observations: For phase terms r1 = r − s cos 𝜓1 = r − s(̂ax ⋅ â r ) = r − s sin 𝜃 cos 𝜙 ] [ √ s √ 1 r2 = r − s cos 𝜓2 = r − s ( 3̂ax + â y ) ⋅ â r = r − ( 3 sin 𝜃 cos 𝜙 + sin 𝜃 sin 𝜙) 2 2 ] [ √ √ s 1 r3 = r − s cos 𝜓3 = r − s (̂ax + 3̂ay ) ⋅ â r = r − (sin 𝜃 cos 𝜙 + 3 sin 𝜃 sin 𝜙) 2 2 r4 = r − s cos 𝜓4 = r − s(̂ay ⋅ â r ) = r − s sin 𝜃 sin 𝜙 ] [ √ √ s 1 r5 = r − s cos 𝜓5 = r − s (−̂ax + 3̂ay ) ⋅ â r = r − (− sin 𝜃 cos 𝜙 + 3 sin 𝜃 sin 𝜙) 2 2 ] [ s √ 1 √ r6 = r − s cos 𝜓6 = r − s (− 3̂ax + â y ) ⋅ â r = r − (− 3 sin 𝜃 cos 𝜙 + sin 𝜃 sin 𝜙) 2 2 r7 = r − s cos 𝜓7 = r − s(−̂ax ⋅ â r ) = r + s sin 𝜃 cos 𝜙 ] [ s √ 1 √ r8 = r − s cos 𝜓8 = r − s (− 3̂ax − â y ) ⋅ â r = r + ( 3 sin 𝜃 cos 𝜙 + sin 𝜃 sin 𝜙) 2 2 ] [ √ √ s 1 r9 = r − s cos 𝜓9 = r − s (−̂ax − 3̂ay ) ⋅ â r = r + (sin 𝜃 cos 𝜙 + 3 sin 𝜃 sin 𝜙) 2 2 r10 = r − s cos 𝜓10 = r − s(−̂ay ⋅ â r ) = r + sin 𝜃 sin 𝜙 ] [ √ √ s 1 r11 = r − s cos 𝜓11 = r − s (̂ax − 3̂ay ) ⋅ â r = r + (− sin 𝜃 cos 𝜙 + 3 sin 𝜃 sin 𝜙) 2 2 ] [ √ s √ 1 r12 = r − s cos 𝜓12 = r − s ( 3̂ax − â y ) ⋅ â r = r + (− 3 sin 𝜃 cos 𝜙 + sin 𝜃 sin 𝜙) 2 2 For amplitude terms r1 ≃ r2 ≃ r3 ≃ r4 ≃ r5 ≃ r6 ≃ r7 ≃ r8 ≃ r9 ≃ r10 ≃ r11 ≃ r12 ≃ r Making these substitutions and combining terms (first with seventh, second with eighth, third with ninth, fourth with tenth, fifth with eleventh, sixth with twelfth), we can write { (√ ) ( ) 3 1 E = 2 cos(ks sin 𝜃 cos 𝜙) − 2 cos ks sin 𝜃 cos 𝜙 cos ks sin 𝜃 sin 𝜙 E0 2 2 (√ )} ( ) 3 1 − cos(ks sin 𝜃 sin 𝜙) + 2 cos ks sin 𝜃 cos 𝜙 cos ⋅ ks sin 𝜃 sin 𝜙 2 2 where E0 = f (𝜃, 𝜙)

e−jkr r

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463

Relative Field Strength (│E/E0│)

5.0 4.0 3.0

α = 60°

2.0 1.0 0

0

2.0

4.0 6.0 Feed-to-Vertex Spacing (s/λ)

7.0

10.0

Relative Field Strength (│E/E0│)

8.0

6.0

4.0

α = 45° 2.0

Relative Field Strength (│E/E0│)

0

0

2.0

4.0 6.0 Feed-to-Vertex Spacing (s/λ)

7.0

10.0

8.0

6.0

α = 30° 4.0

2.0

0

0

2.0

4.0 6.0 Feed-to-Vertex Spacing (s/λ )

8.0

10.0

Figure P15.2 Relative field strengths along the axis (𝜃 = 90◦ , 𝜙 = 0◦ ) for 𝛼 = 60◦ , 45◦ , 30◦ corner reflectors as a function of feed-to-vertex spacing

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SOLUTION MANUAL

15.3. (a) For a corner reflector with an included angle 𝛼 = 36 degrees, it is possible to replace the antenna with a system of images. The system of images comprises ten sources spaced equally about a circle of radius s, with alternating polarities. The field determined from this system of images is valid for −18◦ ≤ 𝜙 ≤ 18◦ . y #4

#3

#5

#2 36°

#6

#1 x

#7

#10 #8

#9

(x1 , y1 ) = (s, 0) (x2 , y2 ) = (s cos 36◦ , s sin 36◦ ) (x3 , y3 ) = (s cos 72◦ , s sin 72◦ ) (x4 , y4 ) = (s cos 108◦ , s sin 108◦ ) = (−s cos 72◦ , s sin 72◦ ) (x5 , y5 ) = (s cos 144◦ , s sin 144◦ ) = (−s cos 36◦ , s sin 36◦ ) (x6 , y6 ) = (s cos 180◦ , s sin 180◦ ) = (−s, 0) (x7 , y7 ) = (s cos 216◦ , s sin 216◦ ) = (−s cos 36◦ , −s sin 36◦ ) (x8 , y8 ) = (s cos 252◦ , s sin 252◦ ) = (−s cos 72◦ , −s sin 72◦ ) (x9 , y9 ) = (s cos 288◦ , s sin 288◦ ) = (s cos 72◦ , −s sin 72◦ ) (x10 , y10 ) = (s cos 324◦ , s sin 324◦ ) = (s cos 36◦ , −s sin 36◦ ) E1 = E0 ejksu ◦

E2 = −E0 ejksu cos 36 ejksv sin 36 ◦

E3 = E0 ejksu cos 72 ejksv sin 72

◦

◦

◦

E4 = −E0 e−jksu cos 72 ejksv sin 72 ◦

E5 = E0 e−jksu cos 36 ejksv sin 36

◦

◦

E6 = −E0 e−jksu ◦

◦

E7 = E0 e−jksu cos 36 e−jksv sin 36 ◦

◦

E8 = −E0 e−jksu cos 72 e−jksv sin 72 ◦

◦

E9 = E0 ejksu cos 72 e−jksv sin 72 ◦

◦

E10 = −E0 ejksu cos 36 e−jksv sin 36

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SOLUTION MANUAL

E=

10 ∑

◦

◦

◦

En = E0 [ejksu − ejksu cos 36 ejksv sin 36 + ejksu cos 72 ejksv sin 72

465

◦

n=1 ◦

◦

◦

◦

◦

◦

− e−jksu cos 72 ejksv sin 72 + e−jksu cos 36 ejksv sin 36 − e−jksu + e−jksu cos 36 e−jksv sin 36 ◦

◦

◦

◦

◦

◦

− e−jksu cos 72 e−jksv sin 72 + ejksu cos 72 e−jksv sin 72 − ejksu cos 36 e−jksv sin 36 ] ◦

= E0 [j2 sin(X) − j2 sin(X cos 36◦ )ejY sin 36 + j2 sin(X cos 72◦ )ejY sin 72 ◦

◦

◦

− j2 sin(X cos 36◦ )e−jY sin 36 + j2 sin(X cos 72◦ )e−jY sin 72 ] = E0 [j2 sin(X) − j4 sin(X cos 36◦ ) cos(Y sin 36◦ ) + j4 sin(X cos 72◦ ) cos(Y sin 72◦ )] = jE0 ⋅ 2[sin(X) − 2 sin(X cos 36◦ ) cos(Y sin 36◦ ) + 2 sin(X cos 72◦ ) cos(Y sin 72◦ )] X = ksu = ks sin 𝜃 cos 𝜙

where;

Y = ksv = ks sin 𝜃 sin 𝜙 neglecting the “j”, we can write the array factor as AF(𝜃, 𝜙) = 2[sin(X) − 2 sin(X cos 36◦ ) cos(Y sin 36◦ ) + 2 sin(X cos 72◦ ) cos(Y sin 72◦ )] (b) See the plot below for computation of the relative field strength. 10 9

Relative Field Strength

8 7 6 5 4 3 2 1 0

0

1

2 3 4 5 6 7 8 Feed-to-Vertex Spacing (wavelengths)

9

10

(c) AF(𝜃 = 90◦ , 𝜙 = 0◦ ) = 2[sin(ks) − 2 sin(ks cos 36◦ ) + 2 sin(ks cos 72◦ )] dAF = 2[cos(ks) − 2 cos 36◦ cos(ks cos 36◦ ) + 2 cos 72◦ cos(ks cos 72◦ )] d(ks)

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d(AF) corresponding to a maximum in the relative field strength d(ks)

The first zero of occurs for

ks = 6.415

or

s = 1.021λ

The relative field strength is given by )| | ( |AF 𝜃 = 𝜋 , 𝜙 = 0 | = 7.482 | | 2 | | 5 4 3 2 DAF/d(ks)

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0

1

(d)

2 3 4 5 6 7 8 Feed-to-vertex spacing (wavelengths)

9

10

90 1 120

60 0.8 0.6

150

30 0.4 0.2

180

0

210

330 Normalized power pattern

240

300 270

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SOLUTION MANUAL

( )[ ( ) (√ )] | ks 15.4. AF = 4 sin ks sin 𝜃 cos 𝜙 cos ks sin 𝜃 cos 𝜙 − cos 3⋅ ⋅ sin 𝜃 sin 𝜙 || 𝜃= 𝜋2 2 2 2 | ◦ )[ ( ) (√ ] )] ( )[ ( ) ks ks ks ks ks cos − cos cos −1 AF = 4 sin 3 ⋅ 0 = 4 sin 2 2 2 2 2 ] ( )[ ( ) ks ks cos −1 AF = 4 sin 2 2 ] ( )[ ( ) ks ks cos −1 =0 AF = 4 sin 2 2 (

(

ks 2

)

ks = sin−1 (0) = m𝜋, m = 0, 1, 2, … . 2 2m𝜋 2m𝜋 s= = = mλ, m = 0, 1, 2, … k 2𝜋∕λ ( ) ( ) ks ks ks − 1 = 0 ⇒ cos =1⇒ cos = cos−1 (1) = 2m𝜋, 2 2 2 4m𝜋 4m𝜋 s= = = 2mλ, m = 1, 2, … k 2𝜋∕λ sin

=0⇒

∴ s = mλ,

m = 0, 1, …

m = 1, 2, …

15.5. From (15-14) we can write for r′ = r0 , 𝜃 = 𝜃0 2f = r0 (1 + cos 𝜃0 ) Using Figure 15.10 and the definition of the sine function sin 𝜃0 =

d∕2 d∕2 d ⇒ r0 = = r0 sin 𝜃0 2 sin 𝜃0

When substituting it above, we can write 2f = r0 (1 + cos 𝜃0 ) =

d d 1 + cos 𝜃0 (1 + cos 𝜃0 ) ⇒ f = 2 sin 𝜃0 4 sin 𝜃0

Using the trigonometric identity (See Appendix VI-1) ( ) 1 + cos 𝛼 𝛼 = cot 2 sin 𝛼 reduces the above equation to ( ) ( ) 𝜃0 𝜃0 f 1 d ⇒ = cot f = cot 4 2 d 4 2 ( ) 𝜃0 f 1 = cot d 4 2 ) ( ◦ f 1 1 1 90 = cot(45◦ ) = = 0.25 = cot (a) 𝜃0 = 90◦ : d 4 2 4 4 ) ( f 1 1 180◦ = cot(90◦ ) = 0 = cot 𝜃0 = 180◦ : d 4 2 4 ◦ ◦ 90 < 𝜃0 < 180 : 0.25 > f ∕d > 0

15.6. From (15-25) ⇒

𝜙=0

467

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SOLUTION MANUAL

f 1 = cot(0) = ∞ d 4 ( ◦) f 1 1 90 = cot(45◦ ) = 0.25 = cot 𝜃0 = 90◦ : d 4 2 4 0◦ < 𝜃0 < 90◦ : ∞ > f ∕d > 0.25

(b) 𝜃0 = 0◦ :

15.7. The far-field region is commonly taken to exist at distances greater than 2D2 ∕λ from the antenna, λ being the wavelength, D: dimension of diameter. R ≥ 2D2 ∕λ,

D = 10 meters,

f = 2 GHz → λ = 0.15 meters

At 2 GHz ⇒ R ≥ 2(100)∕(0.15) = 1333.3 meters f = 4 GHz → λ = 0.075 meters. At 4 GHz ⇒ R ≥ 2(100)∕(0.075) = 2666.66 meters. 15.8. From (2-110) Aem =

λ2 4𝜋 D ⇒ D = 2 Aem 4𝜋 λ

Using (12-40) D=

4𝜋 4𝜋 4𝜋 2 4𝜋 Aem = 2 𝜀ap Ap = 2 (1)(𝜋a2 ) = 2 2 λ λ λ λ

( )2 ( )2 d 𝜋d = 2 λ

15.9. On the surface of the reflector, the current density is given by (15-29), which can be written using (15-32) and (15-32a), as √ Js = 2

√

𝜀 [̂n × (Ŝ i × Ei )] = 2 𝜇

𝜀 C 𝜇 1

′ √ e−jkr Gf (𝜃 ′ , 𝜙′ ) ′ [̂n × (̂a′r × ê i )] r

The polarization of the source can be written as â y = â ′r sin 𝜃 ′ sin 𝜙′ + â 𝜃 ′ cos 𝜃 ′ sin 𝜙′ + â ′𝜙 cos 𝜙′ Referring to Figure 15.13, the unit vector ê i can be written as ê i =

â ′r × (̂ay × â ′r ) |̂a′r × (̂ay × â ′r )|

which by using the vector identity A × (B × C) = (A ⋅ C)B − (A ⋅ B)C can be expressed as ê i =

(̂a′r ⋅ â ′r )̂ay − (̂a′r ⋅ â y )̂a′r |(̂a′r ⋅ â ′r )̂ay − (̂a′r ⋅ â y )̂a′r |

=

â y − (̂a′r ⋅ â y )̂a′r |̂ay − (̂a′r ⋅ â y )̂a′r |

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SOLUTION MANUAL

Using the transformation of (VII-12) or (VII-12a), we can write the numerator as â y − (̂a′r ⋅ â y )̂a′r = â ′𝜃 cos 𝜃 ′ sin 𝜙′ + â ′𝜙 cos 𝜙′ and the denominator as √ |̂ay − (̂a′r

⋅ â y )̂a′r |

=

1 − sin2 𝜃 ′ sin2 𝜙′

Thus ê i =

â y ⋅ (̂a′r ⋅ â y )̂a′r |̂ay ⋅ (̂a′r ⋅ â y )̂a′r |

=

â ′𝜃 cos 𝜃 ′ sin 𝜙′ + â ′𝜙 cos 𝜙′ √ 1 − sin2 𝜃 ′ ⋅ sin2 𝜙′

and √ Js = 2

𝜀 C 𝜇 1

√ ′ ′ √ √ e−jkr e−jkr 𝜀 Gf (𝜃 ′ , 𝜙′ ) ′ [̂n × (̂ar × ê i )] = 2 ⋅ C1 ⋅ Gf (𝜃 ′ , 𝜙′ ) ′ u r 𝜇 r

where u = n̂ × (̂a′r × ê i ) = (̂n × ê i )̂a′r − (̂n × â ′r )̂ei . Eventually u reduces to (15-34) by using (15-18) and (VII-12) and (VII-12a). 15.10. E = (̂ax + â y sin 𝜙 cos 𝜙)f (r, 𝜃, 𝜙) (a) Cross Pol = sin 𝜙 cos 𝜙 Maximum:

𝜙 = 0 ⇒ zero 𝜙 = 90◦ ⇒ zero 𝜙 = 180◦ ⇒ zero 1 sin(2𝜙) 2 𝜙 = 45◦ ⇒ 0.5

(b) Cross Pol = sin 𝜙 cos 𝜙 = Maximum:

𝜙 = 135◦ ⇒ 0.5 (c)

(̂ax + â y sin 𝜙 cos 𝜙) f (r, 𝜃, 𝜙) E = (̂ax + â y sin 𝜙 cos 𝜙)f (r, 𝜃, 𝜙) = √ 2 2 1 + sin 𝜙 cos 𝜙 â x + â y sin 𝜙 cos 𝜙 E = â w f (r, 𝜃, 𝜙), â w = √ 1 + sin2 𝜙 cos2 𝜙 PLF = |̂ax ⋅ â w |2 =

(d) PLF = |̂ay ⋅ â w |2 =

| 1 1 | = = −3 dB | 2 2 ◦ 2 1 + sin 𝜙 cos 𝜙 |𝜙=45 sin2 𝜙 cos2 𝜙 || 0.25 = = 0.2 = −7 dB | 2 2 ◦ 1.25 | 𝜙=45 1 + sin 𝜙 cos 𝜙

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SOLUTION MANUAL

(̂ax + â y sin 𝜙 cos 𝜙) â a = √ 1 + sin2 𝜙 cos2 𝜙

(e)

|2 |⎛ | â + â sin 𝜙 cos 𝜙 ⎞ ⎛ â + â sin 𝜙 cos 𝜙 ⎞| |⎜ x ⎟ ⎜ x ⎟|| y y PLF = ||⎜ √ ⎟ ⋅ ⎜√ ⎟| | |⎜ |⎝ 1 + sin2 𝜙 cos2 𝜙 ⎟⎠ ⎜⎝ 1 + sin2 𝜙 cos2 𝜙 ⎟⎠| | | | 1 + sin2 𝜙 ⋅ cos2 𝜙 |2 | | =| | = 1 = 0 dB | 1 + sin2 𝜙 ⋅ cos2 𝜙 | | | 𝜃0

2𝜋

15.11.

∫0

∫0

J s ejkr ⋅̂ar ds′ = 2 ′

√

𝜀 C I 𝜇 1

where 𝜃0

2𝜋

I=

∫0

√ Gf (𝜃 ′ , 𝜙′ )

∫0

r′

× e−jkr [1−sin 𝜃 ′

(

× [̂n × (−̂az × ê r )](r′ )2 sin 𝜃 ′ sec

𝜃′ 2

′ sin 𝜃 cos(𝜙′ −𝜙)−cos 𝜃 ′ ⋅cos 𝜃]

) d𝜃 ′ d𝜙′

By using (15-37a) [ ( ′ )] ( ′) 𝜃 𝜃 â 𝜃 ⋅ u = â 𝜃 ⋅ −̂az (̂n ⋅ ê r ) − ê r cos = (̂n ⋅ ê r ) sin 𝜃 − â 𝜃 ⋅ ê r cos 2 2 [ ( ′ )] ( ′) 𝜃 𝜃 â 𝜙 ⋅ u = â 𝜙 ⋅ −̂az (̂n ⋅ ê r ) − ê r cos = −̂a𝜙 ⋅ ê r cos 2 2 Toward 𝜃 = 𝜋 equation (15-49) reduces to (15-51). This is accomplished by making the following substitutions. The exponential of (15-49b) can be written as r′ [1 − sin 𝜃 ′ sin 𝜃 cos(𝜙′ − 𝜙) − cos 𝜃 ′ cos 𝜃]𝜃=𝜋 = r′ [1 + cos 𝜃 ′ ] = 2f by using (15-14). Also [

(

â 𝜃 ⋅ u|𝜃=𝜋 = (̂n ⋅ ê r ) sin 𝜃 − â 𝜃 ⋅ ê r cos ( â 𝜙 ⋅ u|𝜃=𝜋 = −̂a𝜙 ⋅ ê r cos

𝜃′ 2

)

𝜃′ 2

(

)] 𝜃=𝜋

= −̂a𝜃 ⋅ ê r cos

𝜃′ 2

)

| | | |𝜃=𝜋

| | | |𝜃=𝜋

Assuming azimuthal symmetry, we can write I as ( ′) ( ′) √ 𝜃 𝜃 −jk(2f ) ′ ′ = −2𝜋 Gf (𝜃 )e (̂a𝜃 ⋅ ê r + â 𝜙 ⋅ ê r )|𝜃=𝜋 r sec cos sin(𝜃 ′ )d𝜃 ′ ∫0 2 2 √ 𝜃0 = −2𝜋 (̂a𝜃 ⋅ ê r + â 𝜙 ⋅ ê r )|𝜃=𝜋 Gf (𝜃 ′ )e−j2kf r′ sin 𝜃 ′ d𝜃 ′ ∫0 𝜃0

I|𝜃=𝜋

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SOLUTION MANUAL

( Since

r′

=f

sec2

𝜃′ 2

) , as given by (15-14a), we can write I as

I|𝜃=𝜋 = −2𝜋fe−j2kf ( Since sin 𝜃 ′ = sin

𝜃′ 𝜃′ + 2 2

∫0

[(̂a𝜃 ⋅ ê r + â 𝜙 ⋅ ê r )]𝜃=𝜋

)

( = 2 sin

−j2kf

I|𝜃=𝜋 = −4𝜋fe

𝜃0

𝜃0

∫0

𝜃′ 2

)

( cos

𝜃′ 2

√ Gf (𝜃 ′ )

sin 𝜃 ′ d𝜃 cos2 (𝜃 ′ ∕2)

)

( ′) √ 𝜃 ′ [(̂a𝜃 ⋅ ê r + â 𝜙 ⋅ ê r )]𝜃=𝜋 Gf (𝜃 ) tan d𝜃 ′ 2

Assuming the cross polarized field is small (̂a𝜃 ⋅ ê r + â 𝜙 ⋅ ê r )𝜃=𝜋 ≃ −1 Thus I|𝜃=𝜋 = −4𝜋fe−j2kf

𝜃0

∫0

( ′) √ 𝜃 Gf (𝜃 ′ ) tan d𝜃 ′ 2

and ( ′) [√ ]1∕2 𝜃0 √ 𝜔𝜇e−jkr 𝜃 𝜀 Pt (4𝜋fe−j2kf ) Gf (𝜃 ′ ) tan d𝜃 ′ ∫0 2𝜋r 𝜇 2𝜋 2 ( ′) [√ ]1∕2 𝜃0 √ 2𝜔𝜇f 𝜃 𝜀 Pt −jk(r+2f ) ′ Gf (𝜃 ) tan E(r, 𝜃 = 𝜋) = −j e d𝜃 ′ ∫0 r 𝜇 2𝜋 2

E(r, 𝜃 = 𝜋) = −j

which is used to form (15-52a), (15-53), and (15-54). 15.12.

f = 3 GHz, λ = 0.1 m, d = 1 meter. ( )2 𝜋d D0 = 𝜀ap , where 𝜀ap = aperture efficiency λ (a) uniform illumination, 𝜀ap = 1 ( D0 =

𝜋⋅1 0.1

)2 = 986.96 = 29.94 dB

(b) 𝜀ap = 𝜀t 𝜀s = (0.8)(0.85) = 0.68 ) 𝜋⋅1 2 (0.68) = 671.13 0.1 D0 = 671.13 = 28.27 dB (

D0 =

total aperature efficiency: 𝜀ap = 0.68 = −1.6749 dB

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SOLUTION MANUAL

15.13. f ∕d = 0.4284 (a) Using (15-24) ( ) ⎡ 1 f ⎤ [ ] ⎢ 2 d ⎥ 0.5(0.4284) −1 ⎢ −1 ⎥ 𝜃0 = tan = tan = tan−1 (1.7698) = 60.53◦ 2 − 1∕16 ⎢ ( f )2 ⎥ (0.4284) 1 ⎥ ⎢ − ⎣ d 16 ⎦ Θ0 = 2𝜃0 = 121.06◦ (b) Using (15-55) ( ′) | 𝜃0 √ |2 𝜃 ′| ′ | 𝜀ap = cot G (𝜃 ) ⋅ tan d𝜃 f |∫ | 2 | 0 | ( ) ( ) ( ) |2 𝜃0 || 𝜃0 √ 𝜃′ 𝜃′ ′| = cot 2 2 cos tan d𝜃 | 2 || ∫0 2 2 | ( ′ )]𝜃0 |2 ( ′) ( ) ( ) |[ |2 𝜃0 || 𝜃0 𝜃 𝜃 | ′| 2 𝜃0 | −2 cos = 2 cot 2 sin = 2 cot d𝜃 | | | | 2 || ∫0 2 2 || 2 | 0 | ( )]2 ( )]2 ( )[ ( )[ 𝜃0 𝜃0 𝜃0 𝜃0 𝜀ap = 2 cot 2 = 8 cot 2 2 − 2 cos 1 − cos 2 2 2 2 ) [ ( )] ( 2 60.53◦ 60.53◦ 1 − cos 𝜀ap (𝜃0 = 60.53◦ ) = 8 cot 2 = 8⋅(2.937)(0.018576) 2 2 𝜀ap (𝜃0 = 60.53◦ ) = 0.43644 ⇒ 𝜀ap = 43.644% (

2

(

𝜋d λ

)2

𝜃0 2

)

3 × 108 = 3 × 10−2 = 0.03 meters 10 × 109 ( )2 𝜋(42.672) = 8.715 × 106 = 69.4 dB d = 42.672 meters, D0 = 0.43644 0.03

(c) D0 = 𝜀ap

, f = 10 GHz ⇒ λ =

(d) Using (15-65) D ≥ D0

( 1−

m2 2

)2

[ ( ) ]2 1 𝜋 2 = 1− = 0.9618 ⇒ D ≥ 0.9618D0 2 16

D ≥ 0.9618(8.715 × 106 ) = 8.3822 × 106 = 69.234 dB 15.14. f ∕d = 0.38

[

0.5(0.38) (a) Using (15-24) ⇒ 𝜃0 = (0.38)2 − 1∕16 From Figure 15.20 ⇒ Gf = cosn 𝜃 ′ = cos2 𝜃 ′ tan−1

] = 66.68◦

(b) For Gf = cos2 𝜃 ′ ⇒ 𝜀ap (𝜃0 = 66.68◦ ) ≃ 0.83 f ∕d = 0.38 = 10λ∕d ⇒ d = 10λ∕(0.38) = 26.3158λ [ ]2 ( )2 𝜋(26.3158)λ 𝜋d D= 𝜀ap = 0.83 = 5672.98 = 37.538 dB λ λ

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SOLUTION MANUAL

473

(c) From Figure 15.23, the field strength of the cos2 (𝜃 ′ ) pattern along 𝜃 ′ = 𝜃0 = 66.68◦ is 8 dB down from the maximum. { 3.428 cos4 (𝜃 ′ ∕2), 0 ≤ 𝜃 ′ ≤ 𝜋∕2 3 × 108 f = 1 GHz ⇒ λ = = 0.3 m 15.15. Gf = 109 0 Elsewhere ( ) ⎡ 1 f ⎤ 1 ⎤ ⎡ (0.5) ⎢ 2 d ⎥ ⎥ ⎢ 2 ⎥ = tan−1 ⎢ = tan−1 (1.3333) = 53.13◦ (a) 𝜃0 tan−1 ⎢ ( )2 2 − 1∕16 ⎥ ⎢ f ⎥ (0.5) 1 ⎥ ⎢ ⎢ ⎥ − ⎦ ⎣ ⎣ d 16 ⎦ 2𝜃0 = 2(53.13◦ ) = 106.26◦ ( (b) 𝜀ap = cot 2

=

=

= 𝜀ap =

(c)

𝜃0 2

( ′ )|2 ) | 𝜃0 √ 𝜃 | | Gf (𝜃 ′ ) tan | | |∫0 2 || |

( ′) |2 | 𝜃 √ | | ( ) | 𝜃0 ( ′ ) sin | 2 𝜃 | 2 𝜃0 | ′ cot G0 cos4 | ( ′ ) d𝜃 | | 2 ||∫0 2 𝜃 | cos | | 2 | | ( ′) ( ′) ( ) | 𝜃0 2 | 𝜃0 𝜃 𝜃 | | cot 2 cos G0 | sin d𝜃 ′ | |∫0 | 2 2 2 | | ( ) | 𝜃0 ( ) 2 | G0 1 𝜃 2 𝜃0 | ′ ′| 2 𝜃0 G0 cot sin 𝜃 d𝜃 | = cot |[− cos 𝜃 ′ ]00 |2 | | | ∫ 2 | 0 2 4 2 | ) ( ◦ 53.13 3.428 3.428 2 | − cos(53.13◦ ) + 1|2 = cot 2 (2) | − 0.6 + 1|2 4 2 4 𝜀ap = 0.5486 = 54.86%

D0 = 𝜀ap

(

𝜋d λ

)2

[ = 0.5486

10𝜋 0.3

]2

= 0.5486(104.71976)2 = 6,016.07 = 37.79 dB

15.16. f ∕d = 0.433, d = 10 m, f = 10 GHz, Gf (𝜃 ′ ) = 2.667 cos2 (𝜃 ′ ∕2) | 1 (f ) | | | | | 1 | | | | (0.433) | | | | 2 d 2 −1 −1 | | | | (a) 2𝜃0 = 2 tan | ( )2 | = 2 tan | (0.433)2 − 1∕16 | | | f | | 1 | | | | − | | d | | 16 | | = 2 tan−1 |1.73215| = 2(60◦ ) = 120◦ 2𝜃0 = 120◦ ( ′) ) | 𝜃0 √ |2 𝜃 | | Gf (𝜃 ′ ) tan d𝜃 ′ | | | |∫0 2 | | √ ) ( ′) ( ) ( | 𝜃0 |2 | | 𝜃 𝜃′ 2 𝜃0 ′| 2 | cos = cot 2.667 | tan d𝜃 | ∫ 2 2 2 | 0 | | | (

(b) 𝜀ap = cot 2

𝜃0 2

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SOLUTION MANUAL

𝜀ap

𝜀ap

( ′) | |2 𝜃 | | ( ′ ) sin ( ) | 𝜃0 | 2 𝜃 2 𝜃0 | ′| = 2.667 cot cos | ( ′ ) d𝜃 | | 2 ||∫0 2 𝜃 | cos | | 2 | | ( ) ( ) ( )| 2 𝜃0 𝜃0 | 𝜃′ 𝜃 ′ || = 2.667 cot 2 sin d |2 | 2 || ∫0 2 2 || ( ′ )|𝜃0 2 ( )|2 ( )| ( )| 𝜃0 | 𝜃 | || 2 𝜃0 | 2 𝜃0 | = 2.667(4) cot |− cos | | = 2.667(4) cot |1 − cos | | | | 2 | 2 |0 | 2 | 2 ||

𝜀ap (𝜃0 = 60◦ ) = 2.667(4) cot 2 (30◦ ) |1 − cos(30◦ )|2 = 2.667(4)(1.73205)2 [1 − (0.866)]2 = 2.667(4)(3)(0.13397)2 = 0.57445 𝜀ap (𝜃0 = 60◦ ) = 0.57455 = 57.455% PL = Aem W i ( ) (10𝜋)2 (λ)2 [ 𝜋d ]2 λ2 λ2 10𝜋 2 100𝜋 2 Aem = 𝜀ap = 𝜀ap = 𝜀ap D0 = = 𝜀ap 4𝜋 4𝜋 λ 4𝜋 λ 4𝜋 4𝜋

(c)

Aem = 25𝜋𝜀ap = 78.53982(0.57455) = 45.12505 m2 ) ( 100 2 Aem = 45.12505 m2 = 45.125 = 451, 250.5148 cm2 1 PL = 451, 250.5148(10−6 ) = 0.45125 Watts PL = 0.45125 Watts { 15.17. f ∕d = 0.433, d = 10 m, Gf =

G0 cos2 (𝜃 ′ ∕2),

0◦ ≤ 𝜃 ′ ≤ 90◦

0,

Elsewhere

1 ⎤ ⎡ (0.433) ⎥ ⎢ 2 −1 ◦ ◦ (a) 𝜃0 = tan ⎢ ⎥ = 60 ⇒ 2𝜃0 = 120 2 ⎢ (0.433) − 1∕16 ⎥ ⎦ ⎣ 𝜋∕2

2𝜋

(b)

∫0

∫0

Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′ d𝜙′ = 2𝜋 ⇒

𝜋∕2

G0

∫0

𝜋∕2

G0

∫0

𝜋∕2

∫0

𝜋∕2

∫0

Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′ = 4𝜋

Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′ = 2

G0 𝜋∕2 (1 + cos 𝜃 ′ ) sin 𝜃 ′ d𝜃 ′ 2 ∫0 [ ]𝜋∕2 ] G G [ 1 cos2 𝜃 ′ = 0 − cos 𝜃 ′ − = 0 1+ 2 2 2 2 0

cos2 (𝜃 ′ ∕2) sin 𝜃 ′ d𝜃 ′ =

cos2 (𝜃 ′ ∕2) sin 𝜃 ′ d𝜃 ′ = G0 =

3G0 =4 2 8 = 2.667◦ 3

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SOLUTION MANUAL

( ′) ) | 𝜃0 √ |2 𝜃 | ′| ′ = Gf (𝜃 ) tan d𝜃 | | | |∫0 2 | | ( ) | 𝜃0 √ ( ′) |2 ′ ∕2) | | sin(𝜃 𝜃 2 𝜃0 | ′ = cot G0 cos2 d𝜃 || | ′ 2 |∫0 2 cos(𝜃 ∕2) | | | ( ) | 𝜋∕2 |2 𝜃0 | | = G0 cot 2 sin(𝜃 ′ ∕2) d𝜃 ′ | | | 2 ||∫0 | ( ′ ) ( ′ )|2 ( ) | 𝜋∕2 𝜃0 | 𝜃 𝜃 | = G0 cot 2 2 sin d | | | ∫ 2 | 0 2 2 || ( ) 𝜃 |2 2 𝜃0 | = G0 cot |[−2 cos(𝜃 ′ ∕2)]00 | | | 2 (

(c) 𝜀ap

𝜀ap

cot 2

𝜃0 2

= G0 cot 2 (30◦ )|(−2 cos(30◦ ) + 2)|2 = G0 (1.732)2 | − 2(0.866) + 2|2 ( ) 8 [0.26795]2 = 0.57434 = 57.434% = (1.732)2 3 𝜀ap (𝜃0 = 60◦ ) = 0.57434 = 57.434%

15.18. d = 10 m, 2𝜃0 = 120◦ { ′

Gf (𝜃 ) = 𝜋∕2

2𝜋

(a)

∫0

∫0

[

∫0

]𝜋∕2

′

cot 2

=

=

=

Elsewhere 𝜋∕2

∫0

Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′

0

𝜋∕2

∫0

[1 + cos 𝜃 ′ ] sin 𝜃 ′ d𝜃 ′ = 4𝜋

] [ 3𝜋 1 = = 𝜋G0 1 + G = 4𝜋 ⇒ G0 = 8∕3 2 2 0

) | 𝜃0 √ |2 | ′ ′| Gf (𝜃) tan(𝜃 ∕2) d𝜃 | | | |∫0 | | ( ) | 𝜃0 2 𝜃0 | sin(𝜃 ′ ∕2) ′ || G0 cot 2 cos(𝜃 ′ ∕2) d𝜃 | | | 2 ||∫0 cos(𝜃 ′ ∕2) | ( ′) ( ) | 𝜃0 2 | 𝜃0 | 𝜃 | G0 cot 2 sin d𝜃 ′ | | | 2 ||∫0 2 | ( ) ( )| ( )| 2 |2 |𝜃0 || 𝜃0 | 2 𝜃0 | ′ 2 𝜃0 | | G0 cot + 1| |2(− cos(𝜃 ∕2)| | = 4G0 cot |− cos | | | | 2 | 2 | 2 |0 | | (

(b) 𝜀ap =

0,

cos2 (𝜃 ′ ∕2) sin(𝜃 ′ )d𝜃 ′ = G0 𝜋

cos2 𝜃 ′ − cos 𝜃 − 2

𝜋G0

0◦ ≤ 𝜃 ′ ≤ 90◦

Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′ d𝜙′ = 2𝜋 𝜋∕2

= 2𝜋G0

G0 cos2 (𝜃 ′ ∕2),

𝜃0 2

𝜀ap = 4G0 cot 2 (30◦ )[1 − cos(30◦ )]2

475

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SOLUTION MANUAL

( ) 8 (1.732)2 (1 − 0.866) = 0.57434 = 57.434% 𝜀ap = 4 3 { G0 cos2 (𝜃 ′ ), 0◦ ≤ 𝜃 ′ ≤ 90◦ ′ Gf (𝜃 ) = 0, Elsewhere (c) Using (15-59a), 𝜃0 = 60◦ , n = 2 { 𝜀ap (n = 2) = 24

( 2

sin

𝜃0 2

)

[

(

+ ln cos

𝜃0 2

)]}2

( cot

2

𝜃0 2

)

= 24{sin2 (30◦ ) + ln[cos(30◦ )}2 cot 2 (30◦ ) = 24{(0.5)2 + ln(0.866)}2 (1.7321)2 = 24[0.25 − 0.1438]2 (3) = 25(0.1062)2 (3) 𝜀ap (n = 2) = 0.8114 = 81.14% 1. The cos2 (𝜃 ′ ∕2) has lower overall 𝜀ap ; 57.434% vs. 81.14% for cos2 (𝜃 ′ ). r cos2 (𝜃 ′ ∕2) has higher taper efficiency (𝜀 ) because its pattern is more uniform across t the reflector. r cos2 (𝜃 ′ ∕2) has lower spillover efficiency (𝜀 ) because its pattern is much more intense s than that of cos2 (𝜃 ′ ) outside the bounds of the reflector. However the product of the two efficiencies (𝜀t and 𝜀s ) is lower for the cos2 (𝜃 ′ ∕2) than that for the cos2 (𝜃 ′ )[𝜀ap = 𝜀t 𝜀s ]. ( ′) 2𝜋 𝜋∕2 ′ 4 𝜃 15.19. (a) G (𝜃 )dΩ = G0 cos sin 𝜃 ′ d𝜃 ′ d𝜙′ ∫ ∫Ω f ∫0 ∫0 2 ( ′) 𝜋∕2 4 𝜃 = 2𝜋G0 cos sin 𝜃 ′ d𝜃 ′ = 4𝜋 ∫0 2 Thus (

𝜋∕2

G0

cos

∫0

4

𝜃′ 2

) sin 𝜃 d𝜃 = G0 ′

′

𝜋∕2

∫0

( 2 ⋅ cos

4

𝜃′ 2

)

( sin

𝜃′ 2

)

( cos

𝜃′ 2

) d𝜃 ′ = 2

or ( 𝜋∕2

G0

∫0

( cos5

𝜃′ 2

)

( sin

𝜃′ 2

) d𝜃 ′ = G0

𝜋∕2

∫0

cos5

𝜃′ 2

) ( ( ′ )) 𝜃 d cos 2 =1 −1∕2

( ′ )] )]5 [ ]𝜋∕2 [ 6 ′ cos (𝜃 ∕2) 𝜃 1 G0 d cos =− = G0 cos ∫0 2 6 2 0 ] [ ( ) 𝜋 3 3 − cos(0) = −3 ⇒ G0 = G0 cos6 = 3.4286 = 4 1 − cos6 (𝜋∕4) 1 − 0.125 𝜋∕2 [

(

𝜃′ 2

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SOLUTION MANUAL

( ′) ) | 𝜃0 √ |2 𝜃 | ′| ′ = Gf (𝜃 ) tan d𝜃 | | | |∫0 2 | | ( ′) ( ) | 𝜃0 ( ′) |2 √ 𝜃0 | 𝜃 𝜃 | = cot 2 G0 cos2 tan d𝜃 ′ | | | 2 ||∫0 2 2 | (

(b) 𝜀ap

477

cot 2

𝜃0 2

( = G0 cot 2

𝜃0 2

( ′) ( ′) ) | 𝜃0 |2 𝜃 𝜃 | | cos sin d𝜃 ′ | | | |∫0 2 2 | |

)| 𝜃0 |2 | |1 sin 𝜃 ′ d𝜃 ′ | | | | 2 ∫0 | | ] ( )( )[ ]2 ( )[ 1 − cos 𝜃0 2 𝜃0 𝜃0 1 |𝜃0 = G0 cot 2 = G0 cot 2 − cos 𝜃 ′ | |0 2 4 2 2 (

= G0 cot 2

( = G0 cot

2

𝜃0 2

𝜃0 2

)

( 4

sin

𝜃0 2

)

) 𝜃0 ( ) ( ) ( ) 2 4 𝜃0 2 𝜃0 2 𝜃0 sin = G0 cos = G sin ( ) 0 2 2 2 𝜃0 sin2 2 (

cos2

𝜀ap

[

(

= G0 cos 𝜀ap = G0

[

𝜃0 2

)

( sin

]2

1 sin(𝜃0 ) 2

𝜃0 2

)]2

G0 3.4286 2 sin2 (𝜃0 ) = sin 𝜃0 = 0.85715 sin2 𝜃0 4 4

=

𝜀ap = 0.85715 sin2 𝜃0 (c) 𝜃0 = 90◦ for maximum aperture efficiency. The total subtended angle is equal to Θ0 = 2𝜃0 = 180◦ . The maximum aperture efficiency is equal to 𝜀ap |max = 85.715% [ ] 0.5(0.25) = 90◦ 15.20. (a) 𝜃0 = tan−1 (0.25)2 − 1∕16 From (15-59b) ⇒ 𝜀ap = 40{sin4 (45◦ ) + ln(cos(45◦ ))}2 cot(45◦ ) = 0.3730 [ ]2 8𝜋 = 65446.4434 (b) D = 0.373 3 × 108 ∕(5 × 109 ) 90◦

(c) 𝜀s =

∫0 90◦

∫0

Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′ = 1. Gf (𝜃 ) sin 𝜃 d𝜃 ′

′

′

Because the dish has such a large included angle, surrounds the feed pattern completely. 𝜀ap 0.373 (d) 𝜀s 𝜀t = 𝜀ap ⇒ 𝜀t = = = 0.373 𝜀s 1

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SOLUTION MANUAL

15.21. d = 10 meters, 2𝜃0 = 80◦ , Gf = G0 cos8 (𝜃 ′ ), 0◦ ≤ 𝜃 ′ ≤ 90◦ {

[ ( )] 1 − cos4 (𝜃0 ) 𝜃0 (1 − cos 𝜃0 )3 − 2 ln cos − (a) 𝜀ap (n = 8, 𝜃0 = 40 ) = 18 4 2 3 }2 ( ) 𝜃0 1 cot 2 − sin2 𝜃0 2 2 { [ ( ◦ )] (1 − cos 40◦ )3 1 − cos4 (40◦ ) 40 − = 18 − 2 ln cos 4 2 3 }2 ( ◦) 40 1 − sin2 (40◦ ) cot 2 2 2 { }2 1 − 0.34436 = 18 − 2(−0.0622) − 0.00427 − 0.20659 7.54863 4 ◦

𝜀ap (n = 8) = 18(0.07743)2 (7.54863) = 0.81465 = 81.465% 𝜀ap = 0.81465 = 81.465% | (b) 1. G(n) = 2(n + 1)||n=8 = 2(8 + 1) = 18 0 |n=8 2. 𝜀t [Fig. 15.20(b)] ≃ 0.90 = 90% 3. 𝜀s [Fig. 15.20(b)] ≃ 0.91 = 91% (c) f = 10 GHz ⇒ λ = 3 cm = 0.03 m ( D0 =

𝜋d λ

)2

(

𝜀ap =

10𝜋 0.03

)2 (0.81465) = 1,096,622.71(0.81465)

D0 = 893,363.69 = 59.51 dB 15.22. The integral to find 𝜀ap or 𝜀t directly is difficult. Best way is to find 𝜀ap from directivity and then divide by 𝜀spillover . ( D0 =

𝜋d λ

)2

[ 𝜀ap = 5.42 × 106

30◦

𝜀s =

𝜋(10) (3 × 108 ∕25 × 109 )

G0 ∫0

cos10 (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′

90◦ G0 ∫0

cos10 (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′

⇒

∫

]2 𝜀ap = 5.42 × 106 ⇒ 𝜀ap = 0.7909

cos10 𝜃 ′ sin 𝜃 ′ d𝜃 ′ =

◦

𝜀s = 𝜀t =

𝜀ap 𝜀s

− cos11 (𝜃 ′ )|30 0

◦ − cos11 (𝜃 ′ )|90 0

=

0.2055 − 1 = 0.7945 −1

since all other efficiencies are 100%. 𝜀t =

0.7909 = 0.9954 0.7945

− cosm+1 𝜃 ′ m+1

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SOLUTION MANUAL

3 × 108 1 1 = × 10−1 = 0.0333 = meter 9 3 30 9 × 10 ( ) ⎡ 1 f ⎤ ⎡ 1 (0.536) ⎤ ⎡ ⎤ ⎢ 2 d ⎥ ⎢ ⎢ ⎥ ⎥ 0.268 2 −1 ⎥ = tan−1 ⎢ (a) 𝜃0 = tan−1 ⎢ ( )2 ⎥ = tan ⎢ ⎥ 1 1 ⎢ f ⎥ 1 ⎥ ⎢ (0.536)2 − ⎢ (0.536)2 − ⎥ ⎥ ⎢ − ⎣ ⎣ 16 ⎦ 16 ⎦ ⎣ d 16 ⎦

15.23. f = 9 GHz ⇒ λ =

𝜃0 = tan−1 (1.1922) = 50◦ { 𝜀ap (n = 2) = 24

( sin2

𝜃0 2

)

[ ( )]}2 ( ) 𝜃0 𝜃0 cot 2 + ln cos 2 2

= 24{sin2 (25) + ln[cos(25◦ )]}2 cot 2 (25) 𝜀ap (n = 2) = 24{(0.4226)2 + ln[0.9063]}2 (2.1445) = 0.7099 𝜀ap = 0.71 ( D0 =

𝜋d λ

)2

( 𝜀ap =

𝜋(10) 1∕30

)2 (0.71) = [300𝜋]2 (0.71) = (942.478)2 0.71

D0 = 8.882644 × 105 (0.71) = 6.306677 × 105 = 57.998 ≃ 58 dB (b) Aem =

λ2 1 D = 4𝜋 0 4𝜋

= 𝜀ap Aphysical Aem =

(

1 30

)2

(6.306677 × 105 ) = 55.763 m2 ( 2) ) ( 𝜋d 𝜋 ⋅ 100 = 55.763 m2 = 0.71 = 0.71 4 4

PT ⇒ PT = Aem Wi = 55.763(10 × 10−6 ) Wi

PT = 557.633 × 10−6 = 0.557633 × 10−3 Watts PT = 557.633 𝜇 Watts = 0.557633 m Watts ( )2 𝜋(10)2 d 𝜋d2 Aphysical = 𝜋r2 = 𝜋 = = = 78.5398 m2 2 4 4 15.24. (a) D ≃ 𝜀ap

(

𝜋d λ

)2 ,

λ=

3 × 108 = 0.06 m 5 × 109

( ) d = 𝜋d = 3𝜋 meters 2 ( )2 ) ( )2 ( )2 ( 𝜋d C 3𝜋 2 3𝜋 D ≃ 𝜀ap = 𝜀ap = 0.75 = 0.75 = 0.75(50𝜋)2 λ λ 0.06 3∕50 C = 2𝜋a = 2𝜋

D ≃ 0.75(2500𝜋 2 ) ≃ 18,505.51 ≃ 42.673 dB (0.06)2 λ2 D0 = (18,505.51) = 5.3 m2 4𝜋 4𝜋 PT = Aem Wi = 5.3(10 × 10−6 ) = 53 × 10−6 Watts

(b) Aem =

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SOLUTION MANUAL

(c) PT = 53 × 10−6 (1 − |Γ|2 ) = 53 × 10−6 (1 − |0.2|2 ) = 50.88 × 10−6 Watts PT = 50.88 × 10−6 Watts 30 × 109 = 3 cm = 0.03 m 15.25. 𝜃0 = 60◦ , λ = 9 ( 10)× 10 ( ◦) 𝜃0 f 1 1 1 1 60 = cot(30◦ ) = (1.732) = 0.433 = cot = cot d 4 2 4 2 4 4 d = f ∕0.433 = 5∕0.433 = 11.547 meters d = 11.547∕0.03 = 384.9λ ⇒ a = 192.45λ 29.2 29.2 = = 0.15◦ a∕λ 192.45 (b) Sidelobe Level = −17.6 dB ( )2 𝜋d (c) Directivity = = [𝜋(384.9)]2 = (1,209.199)2 = 1,462,162.253 λ D0 = 1,462,162.253 = 61.65 dB (a) HPBW ≃

41.253 = 1,833,466.67 = 62.63 dB 0.15(0.15) 72,815 72,815 72,815 = = 1,618,111.11 = D0 (Tai & Pereira) = 2 2 2 2(0.15)2 Θ1d + Θ2d 2(𝜃d )

(d) D0 (Kraus) =

(e)

2 e−(4𝜋𝜎∕λ)

= 62.09 dB = Loss Factor (LF) 𝜎 = 0.64 × 10−3 m∕0.03 m = 21.33 × 10−3 λ −3 )2

LF = e−(4𝜋(21.33)×10

2

= e−(0.268) = e−0.07187

LF(dB) = 20 log10 e−0.07187 = 20(−0.07187) log10 e = 20(−0.07187)(0.43429) = (−0.07187)(8.686) LF(dB) = 0.624 dB 15.26. f ∕d = 0.357, d = 10 meters, Gf (𝜃 ′ ) = cos2 (𝜃 ′ ∕2) 𝜃0

(a) 𝜀s =

∫0 ∫0

𝜋

70◦

Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′ = Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′

∫0 ∫0

𝜋

cos2 (𝜃 ′ ∕2) sin 𝜃 ′ d𝜃 ′

cos2 (𝜃 ′ ∕2) sin 𝜃 ′ d𝜃 ′

1 1 ⎤ ⎤ ⎡ ⎡ (f ∕d) (0.357) ⎥ ⎥ ⎢ ⎢ 2 2 −1 = tan−1 [2.7483] = 70◦ 𝜃0 = tan−1 ⎢ ( ) ⎥ = tan ⎢ 2 − 1∕16 ⎥ 1 (0.357) ⎥ ⎥ ⎢ (f ∕d)2 − ⎢ ⎦ ⎣ ⎣ 16 ⎦ ) ( 𝜃0 𝜃0 1 + cos 𝜃 ′ N= cos2 (𝜃 ′ ∕2) sin 𝜃 ′ d𝜃 ′ = sin 𝜃 ′ d𝜃 ′ ∫0 ∫0 2 { 𝜃0 } 1 ′ ′ ′ ′ = (sin 𝜃 + cos 𝜃 sin 𝜃 )d𝜃 2 ∫0 {[ ]𝜃 } [ ] cos2 𝜃 ′ 0 1 1 1 1 ′ − cos 𝜃 − N= = − cos 𝜃0 + cos2 𝜃0 − 1 − 2 2 2 2 2 𝜃0 =70◦ 0

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SOLUTION MANUAL

481

[ ] 1 1 1 0.3420 + (0.1170) − 1.5 = − (−1.0995) = 0.5498 2 2 2 [ ] 𝜋∕2 1 1 1.5 cos2 (𝜃 ′ ∕2) sin 𝜃 ′ d𝜃 ′ = − cos 𝜃0 + cos2 𝜃0 − 1.5 = = 0.75 D= ∫0 2 2 2 𝜃0 =90◦

N=−

𝜀s =

N 0.5498 = = 0.7330 = 73.30% D 0.75

( ′) |2 | 𝜃0 √ | ′| ′ ) tan 𝜃 G (𝜃 d𝜃 | | ( ) |∫ ( ) f | 2 𝜃0 | 0 | = 2 cot 2 𝜃0 N 2 (b) 𝜀t = 2 cot 𝜃0 2 2 D Gf (𝜃 ′ ) sin 𝜃 ′ d𝜃 ′ ∫0 √ ( ′) ( ′) ( ′) |2 | 𝜃0 √ |2 || 70◦ | 𝜃 𝜃 𝜃 | | N=| Gf (𝜃 ′ ) tan cos2 d𝜃 ′ | = || tan d𝜃 ′ || |∫0 | ∫ 2 2 2 | | 0 | | | | ( ′) 2 | | 𝜃 | | ( ′) ( ′ ) sin ( ′) 2 70◦ | | | 70◦ | 2 𝜃 𝜃 𝜃 | | ′| ′| =| cos cos tan d𝜃 | = | ( ′ ) d𝜃 | | | |∫0 |∫0 2 2 2 𝜃 | | | | cos | | 2 | | ( ′ )]70◦ |2 ( ′) ( ′ ) ( ′ )|2 | [ 70◦ |2 | | 70◦ 𝜃 𝜃 𝜃 𝜃 | | | | | | 2 − cos N=| sin sin = d𝜃 ′ | = |2 d | | | | | |∫0 | ∫0 | | 2 2 2 2 0 | | | | | | N = |2[− cos(35◦ ) + 1]|2 = (0.3617)2 = 0.1308 ( ′) 𝜃0 70◦ ′ ′ ′ 2 𝜃 D= Gf (𝜃 ) sin(𝜃 ) d𝜃 = cos sin 𝜃 ′ d𝜃 ′ = 0.5498 ∫0 ∫0 2 ( ) ) ( 𝜃0 N 0.1308 0.1308 𝜀t = 2 cot 2 = 2 cot 2 (35◦ ) = 4.0792 2 D 0.5498 0.5498 𝜀t = 0.9706 = 97.06% (c) 𝜀ap = 𝜀s 𝜀t = 0.7330(0.9706) = 0.7115 = 71.15% (

𝜋d (d) D0 = λ

)2

3 × 108 𝜀ap , λ = = 0.03 m, D0 = 10 × 109

(

𝜋(10) 0.03

)2 (0.7115) = 780.23 × 103

D0 = 780.23 × 103 = 58.92 dB 2

2

(e) D = D0 e−(4𝜋𝜎∕λ) = 780.23 × 103 e−(4𝜋∕100) = 780.23 × 103 e−(0.1257) = 780.23 × 103 (0.9843) = 767.98 × 103 = 58.85 dB [ 15.27.

A(𝜌′ )

= A0 1 −

(

𝜌′ a

)2 ] , a = 50λ

(a) According to Table 7.2 𝜀ap = 0.75 = 75%

2

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SOLUTION MANUAL

( )2 ) 2𝜋(50λ) 2𝜋a 2 = 0.75 = 0.75(100𝜋)2 λ λ D0 = 0.75(98,696.044) = 74,022.033 = 10 log10 (74,022.033) (

(b) D0 = 0.75

D0 = 74,022.033 = 48.694 dB (c) HPBW ≃

36.4 36.4 = = 0.728 degrees a∕λ 50 HPBW = 0.728 degrees

(d) D0 (Kraus) =

41,283 41,283 = = 1.8868494(41,283) 2 (Θ1d ) (0.728)2

(2-27)

D0 (Kraus) = 77,894.8 = 48.915 dB D0 (T&P) =

72,815 72,815 72,815 = ( 2 )= 2 2(0.728)2 + Θ2d 2 Θ1d

Θ21d

(2-30b)

D0 (T&P) = 68,695.47 = 48.369 dB ΔD = |D0 (Kraus) − D0 | = |48.915 − 48.694| = 0.221 dB ΔD = |D0 (T&P) − D0 | = |48.369 − 48.694| = 0.325 dB Because the aperture has only one main lobe with very small HPBW, then either Kraus’ or T&P’ formulas are applicable. ( ) ⎡ 1 f ⎤ 1 ⎤ ⎡ (0.43) ⎢ 2 d ⎥ ⎥ ⎢ 2 ⎥ = tan−1 ⎢ 15.28. (a) f ∕d = 0.43 ⇒ 𝜃0 = tan−1 ⎢ ( )2 2 − 1∕16 ⎥ ⎢ f ⎥ (0.43) 1 ⎥ ⎥ ⎢ ⎢ − ⎦ ⎣ ⎣ d ⎦ 16 𝜃0 = tan−1 (1.757) = 60.347◦ (b)

𝜀s = 100%

(c)

𝜀t = 100%

(d)

𝜀ap

because the entire surface(of the ) paraboloidal reflector is illuminated 𝜃′ 4 uniformly using the sec feed pattern. 2 = 𝜀s 𝜀t = 100% since both(𝜀s & ) 𝜀t are ) for a feed with a pattern of ( 100% 𝜃0 𝜃′ 2 4 Gf = cot sec 2 2

( (e) D0 =

because the entire power radiated by the feed, using the specified feed pattern, is captured by the reflector (is within the subtotal angle of the reflector)

𝜋d λ

)2

𝜀ap

(15-54)

3 × 108 = 0.3 × 10−1 = 0.03 meters = 3 cm 10 × 109 [ ]2 ) ( 𝜋(10) 10𝜋 2 D0 = (1) = = 1,096,622.71 = 60.4 dB 0.03 0.03 λ=

D0 = 1,096,622.71 = 60.4 dB

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SOLUTION MANUAL

( (f) PL = Aem

Wi

=

[ 2 ] ) (3) λ2 i D W = (1,096,622.71) 10 × 10−6 4𝜋 0 4𝜋 PL = 7.854 Watts

15.29. From Figure 15.28. (a) f ∕d = 0.50 ⇒ a1 = 1.34λ, b1 = 0.94λ with 𝜌1 and 𝜌2 of any value. (b) f ∕d = 0.75 ⇒ a1 = 1.77λ(𝜌2 = 2λ), 1.83λ(𝜌2 = 3λ, 4λ, ∞) ⇒ b1 = 1.22λ(𝜌1 = 2λ), 1.26λ(𝜌1 = 3λ, 4λ, ∞) (c) f ∕d = 1.0 ⇒ a1 = 2.14λ(𝜌2 = 2λ), 2.27(𝜌2 = 3λ), 2.37λ(𝜌2 = 4λ), 2.4λ(𝜌2 = ∞) ⇒ b1 = 1.49λ(𝜌1 = 2λ), 1.57λ(𝜌1 = 3λ), 1.64λ(𝜌2 = 4λ, ∞) [ ( ′ )2 ] 𝜌 15.30. Aperture electric field distribution = 1 − a (a) According to Table 7.2 𝜀ap = 75% = 0.75 4𝜋 4𝜋 4𝜋 Aem = 2 [𝜀ap Aphysical ] = 2 [0.75(𝜋(20λ)2 )] 2 λ ( ) λ λ 3 2 𝜋(400) = 1,200 𝜋 = 11,843.525 = 4𝜋 4 D0 = 11,843.525 = 40.735 dB

(b) D0 =

4𝜋 4𝜋 4𝜋 2 a2 2 𝜀 A = 𝜀 (𝜋a ) = 𝜀ap ap physical ap λ2 λ2 λ2 ( 2 2) ) ( ( )2 2𝜋a 2 C 4𝜋 a = 𝜀ap = 𝜀 = 𝜀 ap ap λ λ λ2

D0 =

483

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16

Solution Manual

16.1. There is AWGN noise in the system. === Noise Information === Mean: 0.000000, Variance: 0.100000 === DOA Estimations === Theta(degrees) SOI −0.046 16.2. There is AWGN noise in the system. === Noise Information === Mean: 0.000000, Variance: 0.100000 === DOA Estimations === Theta(degrees) SOI −0.071 SNOI1 60.034 16.3. There is AWGN noise in the system. === Noise Information === Mean: 0.000000, Variance: 0.100000 === DOA Estimations === Theta(degrees) Phi(degrees) SOI 19.956 90.021 16.4. There is AWGN noise in the system. === Noise Information === Mean: 0.000000, Variance: 0.100000

Antenna Theory: Analysis and Design, Fourth Edition. Constantine A. Balanis. © 2016 John Wiley & Sons, Inc. Published 2016 by John Wiley & Sons, Inc. Companion Website: www.wiley.com/go/antennatheory4e

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SOLUTION MANUAL

=== DOA Estimations === SOI SNOI1 SNOI2

Theta(degrees) 19.987 59.940 45.006

Phi(degrees) 90.053 180.026 269.988

16.5. The output y(t) due to the desired signal SOI p(t) is y(t) = Pej𝜔0 t (ẇ 1 + ẇ 2 )

(1)

In order to get the desire signal in the output, we get (ẇ 1 + ẇ 2 ) = 1

(2)

Since the phase shift of the interfering signal is k(λ∕4) sin(45◦ ) = due to the interfering signal SNOI n(t) is √

y(t) = Nej(𝜔0 t−

2𝜋∕4)

ẇ 1 + Nej(𝜔0 t+

√

2𝜋∕4)

ẇ 2 = Nej𝜔0 t [e−j

√ 2𝜋∕4

√ 2𝜋∕4, the output y(t) √

ẇ 1 + ej

2𝜋∕4

ẇ 2 ]

(3)

In order to get rid of the interfering signal, we get e−j

√ 2𝜋∕4

√ 2𝜋∕4

ẇ 1 + ej

ẇ 2 = 0

(4)

Solving (2) and (4) leads to ẇ 1 = 0.5 − j0.2478 ẇ 2 = 0.5 + j0.2478

(5)

The amplitude pattern of the two-element linear array without mutual coupling is shown in Fig. P16.5.

100

Amplitude

10−1 Without coupling With coupling With coupling (Normalized)

10−2

10−3

10−4 0

Figure P16.5

10

20

30 40 50 60 Observation angle (degrees)

70

80

90

Amplitude pattern of SOI at 𝜃0 = 0◦ , and SNOI at 𝜃0 = 45◦

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487

16.6. The output y(t) due to the desired signal SOI p(t) is y(t) = Pej𝜔0 t (ẇ 1 + ẇ 2 )

(1)

In order to get the desire signal in the output, we get (ẇ 1 + ẇ 2 ) = 1

(2) √ 3𝜋∕4, the output y(t)

Since the phase shift of the interfering signal is k(λ∕4) sin(60◦ ) = due to the interfering signal SNOI n(t) is √

y(t) = Nej(𝜔0 t−

3𝜋∕4)

ẇ 1 + Nej(𝜔0 t+

√

3𝜋∕4)

ẇ 2 = Nej𝜔0 t [e−j

√ 3𝜋∕4

√

ẇ 1 + ej

3𝜋∕4

ẇ 2 ]

(3)

In order to reject the interfering signal, we get e−j

√

3𝜋∕4

√

ẇ 1 + ej

3𝜋∕4

ẇ 2 = 0

(4)

Solving (2) and (4) leads to ẇ 1 = 0.5 − j0.1068 ẇ 2 = 0.5 + j0.1068

(5)

The amplitude pattern of the two-element linear array without mutual coupling is shown in Fig. P16.6. 100

Amplitude

10−1 Without coupling With coupling With coupling (Normalized)

10−2

10−3

10−4 0

10

20

30 40 50 60 Observation angle (degrees)

70

80

90

Figure P16.6 Amplitude pattern of SOI at 𝜃0 = 0◦ , and SNOI at 𝜃0 = 60◦

16.7. Using w̃̇ 1 = ẇ 1 w̃̇ 2 = ẇ 2

( (

c21 c22 −j c22 c11 − c12 c21 c22 c11 − c12 c21 c12 c11 +j c22 c11 − c12 c21 c22 c11 − c12 c21

) ) (1)

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SOLUTION MANUAL

and the coupling coefficients c11 = c22 = 2.37 + j0.340 c12 = c21 = −0.130 − j0.0517

(2)

by using ẇ 1 and ẇ 2 from Problem 16.5, we have w̃̇ 1 = 0.1977 − j0.1194 w̃̇ 2 = 0.2292 + j0.0618

(3)

The amplitude pattern of the two-element linear array with mutual coupling is shown in Fig. P16.5. 16.8. Using w̃̇ 1 = ẇ 1 w̃̇ 2 = ẇ 2

( (

c21 c22 −j c22 c11 − c12 c21 c22 c11 − c12 c21 c12 c11 +j c22 c11 − c12 c21 c22 c11 − c12 c21

) ) (1)

and the coupling coefficients c11 = c22 = 2.37 + j0.340 c12 = c21 = −0.130 − j0.0517

(2)

by using ẇ 1 and ẇ 2 from Problem 16.6, we have w̃̇ 1 = 0.2025 − j0.0613 w̃̇ 2 = 0.2174 + j0.0029

(3)

The amplitude pattern of the two-element linear array with mutual coupling is shown in Fig. P16.6. 16.9. There is no noise in the system. Linear Array Beamforming Pattern (N = 10, d = 0.5λ, SOI θ = 0°) 0 30

30

60

90

0

60

–10

–20

–30

dB

–30

–20

–10

Figure P16.9 Beamformed pattern with SOI at 𝜃0 = 0◦

0

90

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=== Weights (Amplitude) of Each Element === # 1 2 3 4 5 6 7 8 9 10

Exaxt Value 0.099996 0.099996 0.099996 0.099996 0.099996 0.099996 0.099996 0.099996 0.099996 0.099996

Normalized 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000

=== Beta (Phase in degrees) of Each Element === # 1 2 3 4 5 6 7 8 9 10

Exaxt Value 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

Normalized 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000 0.000000

16.10. There is no noise in the system. Linear Array Beamforming Pattern (N = 10, d = 0.5λ, SOI θ = 30°) 0 30

30

60

90

0

60

–10

–20

Figure P16.10

–30

dB

–30

Exaxt Value 0.099996 0.099996 0.099996

–10

Beamformed pattern with SOI at 𝜃0 = 30◦

=== Weights (Amplitude) of Each Element === # 1 2 3

–20

Normalized 1.000000 1.000000 1.000000

0

90

489

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4 5 6 7 8 9 10

0.099996 0.099996 0.099996 0.099996 0.099996 0.099996 0.099996

1.000000 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000

=== Beta (Phase in degrees) of Each Element === 1 2 3 4 5 6 7 8 9 10

Exaxt Value 0.000000 −90.000000 −180.000000 −270.000000 −360.000000 −450.000000 −540.000000 −630.000000 −720.000000 −810.000000

Normalized 0.000000 270.000000 180.000000 90.000000 0.000000 270.000000 180.000000 90.000000 0.000000 270.000000

16.11. There is no noise in the system. Linear Array Beamforming Pattern (N = 10, d = 0.5λ, SOI θ = 0°) 0 30

30

60

90

0

60

–10

Figure P16.11

–20

–30

dB

–30

Exaxt Value 0.092007 0.091550 0.110347 0.110727 0.092007 0.091550 0.110347 0.110727 0.092007 0.091550

–10

0

90

Beamformed pattern with SOI at 𝜃0 = 0◦ and SNOI at 𝜃1 = 30◦

=== Weights (Amplitude) of Each Element === 1 2 3 4 5 6 7 8 9 10

–20

Normalized 1.000000 0.995028 1.199337 1.203465 1.000000 0.995028 1.199337 1.203465 1.000000 0.955028

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=== Beta (Phase in degrees) of Each Element === 1 2 3 4 5 6 7 8 9 10

Exaxt Value −5.950321 5.832412 5.063294 −4.715329 −5.950321 5.832412 5.063294 −4.715329 −5.950321 5.832412

Normalized 0.000000 11.782733 11.013615 1.234992 360.000000 11.782733 11.013615 1.234992 360.000000 11.782733

16.12. There is no noise in the system. Linear Array Beamforming Pattern (N = 10, d = 0.5λ, SOI θ = 30°) 0 30

30

60

90

0

60

–10

–20

–30

dB

–30

–20

–10

0

90

Figure P16.12 Beamformed pattern with SOI at 𝜃0 = 30◦ and SNOI at 𝜃1 = 60◦

=== Weights (Amplitude) of Each Element === 1 2 3 4 5 6 7 8 9 10

Exaxt Value 0.101620 0.092685 0.089463 0.096229 0.104277 0.104575 0.096814 0.089631 0.092208 0.101116

Normalized 1.000000 0.912075 0.880365 0.946950 1.026147 1.029076 0.952708 0.882021 0.907374 0.995033

=== Beta (Phase in degrees) of Each Element === 1 2

Exaxt Value −4.093956 −94.060928

Normalized 0.000000 270.033028

491

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3 4 5 6 7 8 9 10

−178.815401 −265.211353 −357.431501 −452.332727 −544.854537 −631.585717 −716.191601 −805.778643

185.278555 98.882603 6.662455 271.761229 179.239419 92.508239 7.902355 278.315313

16.13. There is no noise in the system. ϕ = 270° ϕ = 180°

ϕ = 0° ϕ = 90° Figure P16.13 Beamformed pattern with SOI at 𝜃0 = 0◦

16.14. There is no noise in the system. ϕ = 180° ϕ = 270°

ϕ = 90° ϕ = 0° Figure P16.14

Beamformed pattern with SOI at 𝜃0 = 30◦ , 𝜙0 = 45◦

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16.15. There is no noise in the system. ϕ = 180°

ϕ = 270°

ϕ = 90°

ϕ = 0°

Figure P16.15 Beamformed pattern with SOI at 𝜃0 = 0◦ , and SNOI at 𝜃1 = 30◦ , 𝜙1 = 45◦

16.16. There is no noise in the system. ϕ = 180°

ϕ = 270°

ϕ = 0°

ϕ = 90° ◦

Figure P16.16.1 Beamformed pattern with SOI at 𝜃0 = 30 , 𝜙0 = 45◦ , and SNOI at 𝜃1 = 60◦ , 𝜙1 = 45◦ Beamforming pattern vs θ at ϕ = 45°

0

Beamforming pattern (dB)

−50 −100 −150 −200 −250 −300 −350

0

20

Figure P16.16.2

40

60 80 100 120 Elevation (degrees)

140

160

180

Beamforming pattern at elevation plane of 𝜙 = 45◦

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Alternate solution to 5.25 is on Next Page

494

Alternate Solution� (a) Loop placed along the xz plane: z

z

X

A small electric loop is equivalent to an infinitesimal magnetic dipole placed normal to the surface of the loop. The fields of a magnetic dipole can be obtained by using duality, from those of an electric dipole. When an electric infinitesimal dipole is placed parallel to the y axis, the fields in the far-zone region are given by (from Example 4.5 pg. 199), -jwµlole-jkr cos()sin¢ Eo � 41rr -jwµlole-jkr cos¢ 47fT jwµlole-jkr cos¢ Ho::::::: 47rrJT

E �

H �

jwµI0 ze-jkr cos()sin¢ 47rrJT

Er '.::: 0, Hr '.::: 0 When a magnetic dipole is placed above a PEC ground plane, the image is in the same direction as that of the source. Hence, the array factor is 2cos(kh cose). Using duality and incorporating the array factor, the far-zone radiated fields of the magnetic dipole are given by, ze-Jkr -jw1d · cos¢ [2cos(kh cos())] Eo ::::::: m 47rrJT -jwµI ze-jkr m cosesin¢ [2cos(khcose)J E �

41rw

Ho :::::::

jwµI ze-1kr cos()sin¢ [2cos(kh cos())] m 41rrJ2 r