Solutions Manual Antenna Theory - Analysis and Design 4th Edition Constantine A. Balanis

Solutions Manual
Antenna Theory - Analysis and Design 4th Edition

By
Constantine A. Balanis

( All Chapters Included - 100% Verified Solutions )

,P1: OTE/SPH P2: OTE
JWBS171-Sol-c02 JWBS171-Balanis March 4, 2016 19:56 Printer Name: Trim: 7in × 10in




CHAPTER 2
Solution Manual


Exact Approximate
( )( )
𝜋 𝜋 𝜋 𝜋
2.1. (a) dΩ = sin 𝜃 d𝜃 d𝜙 ΩA ≃ − −
3 4 3 6
60◦ 60◦ 𝜋∕3 𝜋∕3 ( )( )
𝜋 𝜋 𝜋2
ΩA = dΩ = sin 𝜃 d𝜃 d𝜙 ≃ =
∫45◦ ∫30◦ ∫𝜋∕4 ∫𝜋∕6 12 6 72

|𝜋∕3 |𝜋∕3 ΩA ≃ 0.13708 sterads
= (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
( ) ( )
𝜋 450 − 314.5585
ΩA = (0.366) = 0.09582 sterads × 100 = 43.06%
12 314.5585
{
0.09582 sterads
ΩA = ( )( )
180 180
0.09582 = 314.5585 (degrees)2
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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,P1: OTE/SPH P2: OTE
JWBS171-Sol-c02 JWBS171-Balanis March 4, 2016 19:56 Printer Name: Trim: 7in × 10in




2 SOLUTION MANUAL


4𝜋 4𝜋
(b) D0 = = = 131.1456 (dimensionless)
ΩA (sterads) 0.09582
= 10 log10 (131.1456) = 21.1775 dB

or

( )( )
180 180
4𝜋
D0 = 𝜋 𝜋 = 131.1456 (dimensionless) = 21.1775 dB
ΩA (degrees)2
{
131.1456 (dimensionless)
D0 =
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 1
=
[Re(E × H ∗ )] + [Re(E × Hej2𝜔t )]
2 2
1 E2 52
2.3. (a) W rad = Re[E × H ∗ ] = â r = â = 0.03315̂ar Watts∕m2
2 2𝜂 2(120𝜋) r
2𝜋 𝜋
(b) Prad = ∮ Wrad ds = ∫ (0.03315)(r2 sin 𝜃 d𝜃 d𝜙)
s 0 ∫0
2𝜋 𝜋
= (0.03315)(100)2 sin 𝜃 d𝜃 d𝜙
∫0 ∫0
𝜋
= 2𝜋(0.03315)(100)2 sin 𝜃 d𝜃 = 2𝜋(0.03315)(100)2 ⋅ (2)
∫0
= 4165.75 Watts

2.4. (a) U(𝜃) = cos 𝜃
U(𝜃h ) = 0.5 = cos 𝜃h ⇒ 𝜃h = cos−1 (0.5) = 60◦
2𝜋
⇒ Θh = 2(60◦ ) = 120◦ = rads.
3
U(𝜃n ) = 0 = cos 𝜃n ⇒ 𝜃n = cos−1 (0) = 90◦
⇒ Θn = 2(90◦ ) = 180◦ = 𝜋 rads.




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, P1: OTE/SPH P2: OTE
JWBS171-Sol-c02 JWBS171-Balanis March 4, 2016 19:56 Printer Name: Trim: 7in × 10in




SOLUTION MANUAL 3

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

(d) U(𝜃) = cos2 (2𝜃)
1
U(𝜃h ) = 0.5 = cos2 (2𝜃h ) ⇒ 𝜃h = 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

(e) U(𝜃) = cos(3𝜃)
1
U(𝜃h ) = cos(3𝜃h ) = 0.5 ⇒ 𝜃h = cos−1 (0.5) = 20◦
3
⇒ Θh = 2(20◦ ) = 40◦ = 0.698 rads
1
U(𝜃n ) = cos(3𝜃n ) = 0 ⇒ 𝜃n = cos−1 (0) = 30◦
3
⇒ Θn = 2(30◦ ) = 60◦ = 𝜋∕3 rads

(f) U(𝜃) = cos2 (3𝜃)
1
U(𝜃h ) = 0.5 = cos2 (3𝜃h ) ⇒ 𝜃h = cos−1 (0.5)1∕2 = 15◦
3
⇒ Θh = 2(15◦ ) = 30◦ = 𝜋∕6 rads
1
U(𝜃n ) = 0 = cos2 (3𝜃n ) ⇒ 𝜃n = cos−1 (0) = 30◦
3
⇒ Θn = 2(30◦ ) = 60◦ = 𝜋∕3 rads

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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