# Sol100318

```Solution to Exam in Antenna Theory
Time: 18 March 2010, at 8.00–13.00.
Location: Polacksbacken, Skrivsal
You may bring: Laboratory reports, pocket calculator, English dictionary, RådeWestergren: “Beta”,
Nordling-Österman: “Physics Handbook”, or comparable handbooks.
Six problems, maximum five points each, for a total maximum of 30 points.
1.
a) If the complex electric field is denoted E(r), find the corresponding instantaneous (timedependent) electric field E(r,t).
(1p)
b) The array factor of a N-element uniform array can be written
sin N2 ψ
,
AF =
sin 21 ψ
where ψ = kd cos θ + β is the progressive (total) phase shift. Specify the condition for β
for a
ii) end-fire-array;
iii) phased (or scanning) array.
(2p)
c) A half-wavelength dipole has the input impedance (73 + j42.5) Ω. What is the input
impedance of a quarter-wavelength monopole placed directly above an infinite perfect
electric conductor?
(1p)
d) A folded half-wavelength dipole has an input resistance of approximately
i) 50 Ω;
ii) 75 Ω;
iii) 150 Ω;
iv) 300 Ω;
v) 600 Ω
Solution
a) E(r,t) = Re{E(r)e jω t }
b)
i) Broadside array: ψ = kd cos 90◦ + β = 0 ⇒ β = 0
ψ = kd cos 0◦ + β = 0
⇒ β = −kd
ii) End-fire array:
ψ = kd cos 180◦ + β = 0 ⇒ β = kd
iii) Phased (or scanning) array: ψ = kd cos θ0 + β = 0 ⇒ β = −kd cos θ0
c) Zmonopole = 21 Zdipole = (36.5 + j21.25) Ω
d) Impedance of a folded half-wavelength dipole: iv) 300Ω
1
(1p)
2. Consider a very thin finite length dipole of length l which is symmetrically positioned about
the origin with its length directed along the z axis according to the figure. In the far-field region
the condition that the maximum phase error should be less than π /8 defines the inner boundary
of that region to be r = 2l 2 /λ . For r ≤ 2l 2 /λ , we are in the radiating near-field region and
the far-field approximation is not valid. By allowing a maximum
phase error of less than π /8,
p
show that the inner boundary of this region is at r = 0.62 l 3 /λ .
Hint: The vector potential is given by
e− jkR ′
µ
Ie (x′ , y′ , z′ )
dl .
A=
4π
R
Expand R, where the higher order terms become more important as the distance to the antenna
decreases. Note that r′ = z′ ẑ.
Z
z
R
l/2
r
r
θ
y
φ
x
−l/2
Solution
Apply the theorem of cosine to the triangle in the figure:
s
′ 2
z′
z
2
2
′2
′
R = r + z − 2rz cos θ or R = r 1 − 2 cos θ +
r
r
(1)
Expand the square root using
′ 2
√
z′
z
x x2 x3
4
(2)
1 + x = 1 + − + + O(x ) where x = −2 cos θ +
2 8 16
r
r

"
"
′ 2 #
′ 2 #2

z
z
1
z′
z′
1
R = r 1+
−2 cos θ +
−2 cos θ +
−

2
r
r
8
r
r

" #
"
′ 2 #3
z
z′ 4 
1
z′
(3)
+
+O
−2 cos θ +

16
r
r
r
" (
′ 3
′ 4 #
z′ 2 2
z
z
z′
1 z′ 2 1
−
cos θ − 4
cos θ +
= r 1 − cos θ +
r
2 r
8
r
r
r
"
#)
′ 4
′ 3
z
z
1
3
cos θ + O
(4)
− 8
16
r
r
" #)
(
′ 3
1 z′ 2
1
z′ 4
z′
z
= r 1 − cos θ +
(1 − cos2 θ ) +
cos θ (1 − cos2 θ ) + O
r
2 r
2 r
r
(5)
= r − z′ cos θ +
1
r
z′2
2
sin2 θ +
2
1
r2
z′3
2
cos θ sin2 θ + . . .
(6)
In the far-field region, the two first terms are used as the approximation for R and the third term
is the error. In this case, we consider the radiating near-field region and we have to approximate
R with the three first terms
1 z′2 2
′
sin θ
(7)
R ≈ r − z cos θ +
r 2
and the error is given by the fourth term
1 z′3
2
cos θ sin θ
r2 2
(8)
The maximum error is found when
∂ 1 z′3
z′3
2
θ
sin
θ
cos
sin θ (− sin2 θ + 2 cos2 θ ) = 0
=
∂ θ r2 2
2r2
(9)
We note that θ = 0 or 180◦ give no maximum because they make the fourth term equal to zero.
So, we must have the maximum error for
√
θ = arctan(± 2)
(10)
The maximum phase error ≤ π8 , which gives
k
1 z′3
cos θ sin2 θ
r2 2
√
z′ =l/2, θ =arctan(± 2)
=
π l3
π
2π 1 1 2
√
= √
≤
2
λ r 3 3 12 3λ r2
8
We solve for r and obtain
r
l3
r ≥ 0.62
λ
(11)
(12)
which shows that the inner boundary of the radiating near-field region is
r
l3
r = 0.62
λ
(13)
QED 3
3. An infinitesimal horizontal electric dipole of length l and constant electric current I0 is placed
parallel to the y axis a height h = λ /2 above an infinite electric ground plane.
a) Find the spherical E- and H-field components radiated by the dipole in the far-zone.
b) Find the angles of all the nulls of the total field.
Solution
ψ̂
z
r̂
a) Introduce a new set of spherical coordinates (r, ψ , χ ),
where ψ̂ is given in the figure and χ̂ = r̂ × ψ̂ .
θ̂
r1
r
θ
kI0 l − jkr1
e
sin ψ ψ̂
4π r1
kI0 l − jkr2
Eψ2 ψ̂ = jη
e
sin ψ ψ̂
4π r2
p
p
sin ψ = 1 − cos2 ψ = 1 − |r̂ · ŷ|
q
= 1 − sin2 θ sin2 φ
Eψ1 ψ̂ = jη
r2
h
ψ
y
π −θ
x
h
φ
(1)
(2)
(3)
Theorem of cosine:
2 + h2 − 2rh cos θ
r12 = rq
r1 = r
h
1 − 2 hr cos θ +
r1 = r 1 − hr cos θ + 21
r1 ≈ r − h cos θ
h 2
r i
h 2
r
2 + h2 − 2rh cos(π − θ )
r22 = rq
2
√
r2 = r 1 + 2 hr cos θ + hr , but 1 + x ≈ 2x
h
i
2
2
r2 = r 1 + hr cos θ + 21 hr
, but hr ≪ hr
(4)
r2 ≈ r + h cos θ
Far-field approximation:
r1 ≈ r − h cos θ
r1 ≈ r + h cos θ
r1 ≈ r2 ≈ r
for phases
for amplitudes
(5)
(6)
q
h
i
kI0 l − jkr
= jη
1 − sin2 θ sin2 φ e jkh cos θ − e− jkh cos θ ψ̂ = [kh = π ]
E=
e
4π r
q
kI0 l − jkr
= jη
1 − sin2 θ sin2 φ [2 j sin(π cos θ )] ψ̂ ,
0 ≤ θ ≤ π /2
(7)
e
{z
}
4
π
r
|
{z
}|
AF
EF
q
kI0 l − jkr
1
1 − sin2 θ sin2 φ [2 j sin(π cos θ )] χ̂ ,
0 ≤ θ ≤ π /2
H = r̂ × E = j
e
η
4π r
(8)
Eψ1 ψ̂
+ Eψ2 ψ̂
b) Nulls of the AF:
AF = 2 j sin(π cos θ ) = 0
⇒ π cos θ = nπ ,
⇒
cos θ = n,
(9)
n = 0, ±1, ±2, . . .
n = −1, 0, 1
4
(10)
(11)
n = −1 : cos θ = −1 ⇒ θ = 180◦ > 90◦
n=0:
cos θ = 0
⇒ θ = 90◦
⇒ θ = 0◦
n=1:
cos θ = 1
⇒
not a null
(12)
Nulls of the EF:
⇒
⇒
1 − sin2 θ sin2 φ = 0
(13)
sin θ sin φ = 1
2
2
sin θ = ±1
⇒ θ = 90
◦
(14)
simultaneous with
simultaneous with
sin φ = ±1
φ = 90
◦
θ = 90◦ is already a null for the AF, so the EF does not introduce any new nulls.
Nulls of the total field for θ = 0◦ and 90◦ , regardsless of value of φ .
5
(15)
(16)
4. A four-element uniform array has its elements placed along the z axis with distance d = λ /2
between them according to the figure below.
a) Derive the array factor and show that it can be written as
sin(2ψ )
, where ψ is the
sin(ψ /2)
progressive phase shift between the elements.
b) In order to obtain maximum radiation along the direction θ = 0◦ , where θ is measured
from the positive z axis, determine the progressive phase shift ψ .
c) Find all the nulls of the array factor.
z
d
y
d
x
d
Solution
z
θ
E∼
a)
r3
d
y
d
d
∑
Far-field approximation for phases:
3
r1 = r + d cos θ
2
1
r2 = r + d cos θ
2
1
r3 = r − d cos θ
2
3
r4 = r − d cos θ
2
and for amplitudes:
r
r2
r1
1
2 d cos θ
e− j(krn +βn ) e− jkr
AF
≈
rn
r
n=1
4
r4
3
2 d cos θ
r1 ≈ r2 ≈ r3 ≈ r4 ≈ r
AF = e− j 2 (kd cos θ +β ) + e− j 2 (kd cos θ +β ) + e j 2 (kd cos θ +β ) + e j 2 (kd cos θ +β )
3
1
1
− j 23 ψ
= [ψ = kp cos θ + β ] = e
− j 32 ψ
+e
+e
j 21 ψ
+e
j 23 ψ
(2)
(3)
(4)
(5)
(6)
(7)
(8)
j3ψ
(9)
3
AF e jψ − 1 = e− j 2 ψ e j4ψ − 1
(11)
=e
1 + e jψ + e j2ψ + e
− j 21 ψ
3
(1)
Use the expression for a geometric
series or form AF e jψ − AF, where AF e jψ =
3
ψ
ψ
ψ
ψ
ψ
−j
j2
j3
j4
j
e 2 e +e +e +e
.
3
(10)
AF e jψ − AF = e− j 2 ψ e j4ψ − 1
or
6
Solving for the AF gives
AF = e− j 2 ψ
3
=
j2ψ (e j2ψ − e− j2ψ )
3
3
e j4ψ − 1
sin ψ
− j 23 ψ e
=
e
= e− j 2 ψ e j 2 ψ
1
1
1
ψ
j
e −1
sin 12 ψ
e j 2 ψ (e j 2 ψ − e− j 2 ψ )
sin ψ
sin 12 ψ
Q.E.D.
(12)
(13)
b) For maximum along θ = 0◦ all sources must be in phase, i.e., ψ (θ = 0◦ ) = 0:
ψ = (kd cos θ + β )|θ =0◦ , d=λ /2 = 0
2π λ
cos 0◦ + β = 0 ⇒ β = −π
⇒
λ 2
⇒ ψ = π (cos θ − 1)
2ψ
=0
c) Null of the AF ⇒ AF = sin
sin ψ2
First we investigate the nominator:
sin[2π (cos θ − 1)] = 0
⇒
⇒
⇒
⇒
sin 2ψ = 0 and
(14)
(15)
(16)
sin ψ2 6= 0
(17)
n = 0, ±1, . . .
2π (cos θ − 1) = nπ ,
n
cos θ − 1 =
2
n 2=n
,
n = −4, −3, −2, −1, 0
cos θ = 1 + =
2
2
(18)
(19)
(20)
Then we insert the values of n and obtain the angle θ and simultaneously check the denominator:
⇒ θ = 180◦ ⇒ sin π2 (cos θ − 1) = 0 ⇒ Not a null!
n = −4 ⇒ cos θ = −1
n = −3 ⇒ cos θ = −1/2 ⇒ θ = 120◦ ⇒ sin π2 (cos θ − 1) 6= 0
OK!
π
◦
⇒ θ = 90
⇒ sin 2 (cos θ − 1) 6= 0
OK!
n = −2 ⇒ cos θ = 0
n = −1 ⇒ cos θ = 1/2
⇒ θ = 60◦ ⇒ sin π2 (cos θ − 1) 6= 0
OK!
⇒ θ = 0◦
⇒ sin π2 (cos θ − 1) = 0 ⇒ Not a null!
n = −0 ⇒ cos θ = 1
(21)
Nulls for θ = 60◦ , 90◦ , 120◦
7
5. Two identical constant current loops with radius a are placed a distance d apart according to
the figure below. Determine the smallest radius a and the smallest separation d so that nulls are
formed in the directions θ = 0◦ , 60◦ , 90◦ , 120◦ , and 180◦ , where θ is the angle measured from
the positive z axis.
z
a
y
d
a
x
Solution
Study the element factor(EF) and the array factor (AF) separately.
EF: The electric field from a constant current loop is
Eφ = η
kaI0 − jkr
e
J1 (ka sin θ )
2r
(1)
The Bessel function J1 (ka sin θ ) has zeros for ka sin θ = 0, 3.8317, ...
ka sin θ = 0
⇒
ka sin θ = 3.8317
θ = 0◦ or θ = 180◦
3.8317 3.8317λ
⇒ a=
=
k sin θ
2π sin θ
(2)
(3)
Since the elements are fed with equal phases (and amplitudes), the AF must have a maximum in the broadside direction θ = 90◦ and not a null. Therefore, the null for θ = 90◦
must come from the EF, so
a=
3.8317
= 0.61λ
2π sin 90◦
(4)
AF We must choose d so that AF(θ = 60◦ = AF(θ = 120◦ ) = 0
z
r1
r
Far-field approximation:
r2
x
d/2
θ _d
cos θ
2
d/2
_d cos θ
2
AF = e
r1 = r − d2 cos θ
r2 = r + d2 cos θ
y
r1 ≈ r2 ≈ r
jk d2 cos θ
+e
AF(θ = 60 ) = 0
◦
− jk d2 cos θ
⇒
⇒
⇒
for phases
(5)
for amplitudes
(6)
kd
= 2 cos
cos θ
2
πd
πd
◦
cos
cos 60 = cos
=0
λ
2λ
πd
π
= (2n + 1) ,
n = 0, ±1, ±2, . . .
2λ
2
n = 0, ±1, ±2, . . .
d = (2n + 1)λ ,
8
(7)
(8)
(9)
(10)
Choose d = λ as the smallest distance.
2π λ
cos θ = 2 cos(π cos θ )
AF = 2 cos
λ 2
(11)
We have to check the zero at θ = 120◦ :
π
AF(θ = 120◦ ) = 2 cos(π cos 120◦ ) = 2 cos −
=0
2
OK!
(12)
For zeros along θ = 0◦ , θ = 60◦ , θ = 90◦ , θ = 120◦ , and θ = 180◦ , we choose the radius
of the loops to be a = 0.61λ and the spacing between the loops as d = λ .
9
6. Design a linear array of isotropic elements placed along the z axis such that the nulls of the
array factor occur at θ = 60◦ , θ = 90◦ , and θ = 120◦ . Assume that the elements are spaced a
distance d = λ /4 apart and that β = 45◦ .
a) Sketch and label the visible region on the unit circle.
b) Find the required number of elements.
c) Determine the excitation coefficients.
N
Hint: The array factor of an N-element linear array is given by AF =
∑ an ej(n−1)ψ , where
n=1
ψ = kd cos θ + β . Use the representation z = ejψ .
Solution
a) The visible region:
AF =
N
N
n=1
n=1
∑ an e j(n−1)ψ = ∑ anzn−1 = a1 + a2 z + · · · + aN zN−1
= (z − z1 )(z − z2 ) · · · · · (z − zN−1 )
ψ = kd cos θ + β =
θ = 0◦
⇒
θ = 90◦
⇒
θ = 180◦
⇒
z = e jψ
⇒
(1)
π π
2π 5λ
cos θ − = (5 cos θ − 1)
λ 8
4
4
(2)
π
(5 − 1) = π
4
π
π
ψ = (0 − 1) = −
4
4
3π
π
ψ = (−5 − 1) = −
4
2
ψ=
|z| = 1
⇒
(3)
(4)
(5)
unit circle
(6)
When θ varies from 0◦ to 180◦ , ψ varies from π to −3π /2
ψ = − 32π , θ = 180◦
Visible region
(the whole unit circle and
ψ
ψ = π , θ = 0◦
ψ = − π4 , θ = 90◦
b) Nulls:
θ1 = 0◦
⇒
ψ1 = π
θ2 = 90◦
⇒
ψ2 = −
θ3 = 180◦
⇒
π
4
3π
ψ3 = −
2
⇒
⇒
⇒
10
z1 = e jψ1 = e jπ = −1
1
z2 = e jψ2 = e− jπ /4 = √ (1 − j)
2
z3 = e jψ3 = e− j
3π
2
=j
(7)
(8)
(9)
AF = (z − z1 )(z − z2 )(z − z3 ) = a1 + a2 z + a3 z2 + a4 z3
⇒
(10)
4 elements (a1 , a2 , a3 , a4 ) are needed
(11)
c) Excitation coefficients:
(z − z1 )(z − z2 )(z − z3 ) = . . .
= z3 − z2 (z1 + z2 + z3 ) +z (z1 z2 + z2 z3 + z3 z1 ) −z1 z2 z3
{z
} |
{z
} | {z }
|
−a3
a2
(12)
a1
1
1
a1 = −z1 z2 z3 = −(−1) √ (1 − j) j = √ (1 + j)
(13)
2
2
√
1
1
a2 = z1 z2 + z2 z3 + z3 z1 = (−1) √ (1 − j) + √ (1 − j) j + j(−1) = j
2−1
2
2
(14)
1
1
1
(15)
a3 = −(z1 + z2 + z3 ) = − −1 + √ − j √ + j = 1 − √ (1 − j)
2
2
2
a4 = 1
(16)
11
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