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

Show that the directivity D of an antenna may be written
( ) ( )
( ) ( )
∫∫ Ω
= ∗
∗
π
θ φ θ φ
π
θ φ θ φ
4
2
max max 2
, ,
4
1
, ,
r d
Z
E E
r
Z
E E
D
Solution:
av U
D U max (θ ,φ )
=
, 2
max max U(θ ,φ ) = S(θ ,φ ) r , = ∫∫ Ω
π
θ φ
π 4
( , )
4
U 1 U d av
U(θ ,φ ) = S(θ ,φ )r2 , ( ) ( )
Z
S E E θ φ θ φ
(θ ,φ ) , ,
∗
=
Therefore
( ) ( )
( ) ( )
∫∫ Ω
= ∗
∗
π
θ φ θ φ
π
θ φ θ φ
4
2
max max 2
, ,
4
1
, ,
r d
Z
E E
r
Z
E E
D q.e.d.
Note that r2 = area/steradian, so U = Sr2 or (watts/steradian) = (watts/meter2) × meter2
2-7-2. Approximate directivities.
Calculate the approximate directivity from the half-power beam widths of a unidirectional
antenna if the normalized power pattern is given by: (a) Pn = cos θ, (b) Pn = cos2 θ, (c) Pn
= cos3 θ, and (d) Pn = cosn θ. In all cases these patterns are unidirectional (+z direction)
with Pn having a value only for zenith angles 0° ≤ θ ≤ 90° and Pn = 0 for 90° ≤ θ ≤ 180°.
The patterns are independent of the azimuth angle φ.
Solution:
(a) 1 o o
HP θ = 2 cos − (0.5) = 2 × 60 = 120 , 278
(120)
40,000
2 D = = (ans.)
(b) 1 o o
HP θ = 2cos− ( 0.5) = 2× 45 = 90 , 4.94
(90)
40,000
2 D = = (ans.)
2
(c) 1 3 o o
HP θ = 2cos− ( 0.5) = 2× 37.47 = 74.93 , 7.3
(75)
40,000
2 D = = (ans.)
2-7-2. continued
(d) 2cos 1( 0.5)
HP
θ = − n , (cos 1( 0.5))2
10,000
n D − = (ans.)
*2-7-3. Approximate directivities.
Calculate the approximate directivities from the half-power beam widths of the three
unidirectional antennas having power patterns as follows:
P(θ,φ) = Pm sin θ sin2 φ
P(θ,φ) = Pm sin θ sin3 φ
P(θ,φ) = Pm sin2 θ sin3 φ
P(θ,φ) has a value only for 0 ≤ θ ≤ π and 0 ≤ φ ≤ π and is zero elsewhere.
Solution:
To find D using approximate relations,
we first must find the half-power beamwidths.
= 90 −θ
2
HPBW or
2
θ = 90 − HPBW
For sin θ pattern,
2
1
2
sin sin 90 HPBW = 



θ =  − ,



− − 
2
sin 1
2
90 HPBW 1 , 90
2
sin 1
2
HPBW 1 − 



− −  , ∴HPBW =120o
For sin2 θ pattern,
2
1
2
sin2 sin2 90 HPBW = 



θ =  − ,
2
1
2
sin 90 HPBW = 



 − , ∴HPBW = 90o
For sin3 θ pattern,
2
1
2
sin3 sin3 90 HPBW = 



θ =  − ,
3
3 2
1
2
sin 90 HPBW = 



 − , ∴HPBW = 74.9o
*2-7-3. continued
Thus,
3.70
(120)(90)
3.82 40,000
(120)(90)
41,253sq. deg. 41,253
HP HP
= = = ≅ =
θ φ
D (ans.)
4.45
(120)(74.9)
4.59 40,000
(120)(74.9)
= 41,253 = ≅ = (ans.)
5.93
(90)(74.9)
6.12 40,000
(90)(74.9)
= 41,253 = ≅ = (ans.)
*2-7-4. Directivity and gain.
(a) Estimate the directivity of an antenna with θHP = 2°, φHP = 1°, and (b) find the gain of
this antenna if efficiency k = 0.5.
Solution:
(a) 4
HP HP
2.0 10
(2)(1)
= 40,000 = 40,000 = ×
θ φ
D or 43.0 dB (ans.)
(b) G = kD = 0.5(2.0×104 ) = 1.0×104 or 40.0 dB (ans.)
2-9-1. Directivity and apertures.
Show that the directivity of an antenna may be expressed as
( ) ( )
∫∫ ( ) ( )
∫∫ ∫∫
∗
∗
=
Ap
Ap Ap
E x y E x y dxdy
E x y dxdy E x y dxdy
D
, ,
4 , ,
λ2
π
where E(x, y) is the aperture field distribution.
Solution: If the field over the aperture is uniform, the directivity is a maximum (= Dm)
and the power radiated is P′ . For an actual aperture distribution, the directivity is D and
the power radiated is P. Equating effective powers
for P(θ,φ) = sin θ sin2φ
for P(θ,φ) = sin θ sin3φ
for P(θ,φ) = sin2 θ sin3φ
4
D P′ = D P m , ( ) ( ) ∫∫
∗ =
′