WAVE
PROPAGATION
SNS 415
Nur ad-Din M. Salem
Radiation Mechanism
and Physics of Propagating Waves
Referring to Transmission Line, T.L., The total voltage and current waves
on the line can be written as
𝑽 𝒛 = 𝑽+
𝟎
−
𝑽
𝟎
𝒆−𝒋𝜷𝒛 + + 𝒆𝒋𝜷𝒛 = 𝑽𝟎+ 𝒆−𝒋𝜷𝒛 + 𝚪𝑳 𝒆𝒋𝜷𝒛
𝑽𝟎
and
𝑽𝟎+ −𝒋𝜷𝒛
𝑰 𝒛 =
𝒆
− 𝚪𝑳 𝒆𝒋𝜷𝒛
𝒁𝟎
For open-circuited T.L., 𝒁𝑳 = ∞ and 𝚪𝑳 =1. Thus
𝑽 𝒛 = 𝑽+𝟎 𝒆−𝒋𝜷𝒛 + 𝒆𝒋𝜷𝒛
𝑽+
𝟎
𝑰 𝒛 =
𝒆−𝒋𝜷𝒛 − 𝒆𝒋𝜷𝒛
𝒁𝟎
Radiation Mechanism
and Physics of Propagating Waves (Continue)
For open-circuited T.L., 𝒁𝑳 = ∞ and 𝚪𝑳 =1. Thus
𝑽 𝒛 = 𝑽+𝟎 𝒆−𝒋𝜷𝒛 + 𝒆𝒋𝜷𝒛
Z
Z=0
Load
New reference
𝒁 = −𝑙
𝑙
+
𝒋𝜷𝑙
−𝒋𝜷𝑙
𝑽 𝑙 = 𝑽+
𝒆
+
𝒆
=
𝟐𝑽
𝟎
𝟎 cos𝜷𝑙
𝑽 𝑙 = 𝟐𝑽+
𝟎 cos𝜷𝑙
Similarly,
𝑰 𝑙
=
−𝟐𝐣𝑽𝟎+
𝒁𝟎
sin𝜷𝑙 =
−𝟐𝐣𝑽𝟎+
𝒁𝟎
sin𝜷𝑙 =
𝟐𝑽+
𝟎
𝒁𝟎
sin𝜷𝑙
Radiation Mechanism
and Physics of Propagating Waves (Continue)
𝑰 𝑙
=
𝟐𝑽+
𝟎
𝒁𝟎
sin𝜷𝑙
𝑽 𝑙
𝟐𝑽+
𝟎
𝟐𝑽+
𝟎
⋯
𝑙
𝒁𝟎
⋯
3𝜆
4
𝜆
2
𝜆
4
0
= 𝟐𝑽+
𝟎 cos𝜷𝑙
⋯
𝑙
3𝜆
4
𝜆
2
𝜆
4
0
3𝜋
2
𝜋
𝜋
2
0
𝜷𝑙
𝜷𝑙
3𝜋
2
𝜋
𝜋
2
0
Current distribution along O.C. T.L
Voltage distribution along O.C. T.L
Radiation Mechanism
and Physics of Propagating Waves (Continue)
Radiation from T.L.
Small portion of EM energy
escapes from T.L. and radiates
Voltage Standing Waves
To
Source
𝚫
𝜆
2
𝚫 ≪ 𝝀 cancels radiation due to
opposite currents and opposite
polarity
O.C.
T.L.
𝑵𝒐𝒕 𝒂𝒍𝒍 𝒕𝒉𝒆 𝒇𝒐𝒓𝒘𝒂𝒓𝒅
𝒆𝒏𝒆𝒓𝒈𝒚 𝒊𝒔 𝒓𝒆𝒇𝒍𝒆𝒄𝒕𝒆𝒅
𝒃𝒚 𝒕𝒉𝒆 𝑶. 𝑪.
Radiation Mechanism
and Physics of Propagating Waves (Continue)
Enlargement of the O.C.
• Less effected by the
cancellation of radiation
Spreading the two wires
• Increasing of radiation
• This radiator is called dipole
• When the total length of two wires is a half –wavelength, the
antenna is called half-wave dipole.
•
𝜆
A dipole antenna is composed of a piece of quarter-wave T.L.
2
bent out and open-circuited.
𝜆
2
Radiation Mechanism
and Physics of Propagating Waves (Continue)
𝜆
2
Note that for dipole antenna:
𝜆
2
• 𝒁𝒊𝒏 𝑶. 𝑪 = −𝒋𝒁𝟎 𝐜𝐨𝐭 𝜷𝓵
• 𝒁𝒊𝒏 𝑶. 𝑪 ȁ𝜆 = 𝟎 ⟹ 𝐦𝐢𝐧. 𝐢𝐧𝐩𝐮𝐭 𝐢𝐦𝐩𝐞𝐝𝐚𝐧𝐜𝐞
4
• Has high current at input
Voltage
𝜆
2
Voltage
𝐼𝑚𝑎𝑥
𝒁𝒊𝒏 =0
Current
Current
𝜆
4
Min. 𝒁𝒊𝒏 ⟹Max. current at antenna feed point.
𝒁𝑳 =∞
Radiation Mechanism
and Physics of Propagating Waves (Continue)
+
-
+
-
-
+ve half cycle and -ve half cycle
I
+
+
I
Radiation Mechanism
and Physics of Propagating Waves (Continue)
𝝀
𝟐
is a resonant antennas:
• From the basic resonance theory a high Q resonant
circuit has a very narrow bandwidth. The same is
true for the resonant antenna. This type of radiator
has a narrow bandwidth.
• Multiple of half-wavelength of dipole antenna is
equivalent to multiples of quarter-wavelength T.L.
𝝀
• O.C. T.L. and resonance length T.L is equivalent to
𝟒
resonant antenna.
• The current distribution and radiation pattern differ
as the length of the antenna differ.
Radiation Mechanism
and Physics of Propagating Waves (Continue)
The current distribution and radiation pattern differ as the length of the antenna differ.
Maxwell’s equations with harmonically change of time
Maxwell’s equations with harmonics time dependance
Maxwell’s equations with time varying fields
‾
∂𝐵
(1) ∇ × 𝐸‾ = −
∂𝑡
(1.1a)
‾
∂𝐷
(2) ∇ × 𝐻‾ = ∂𝑡 + 𝐽‾𝑒
(3) ∇ ⋅ 𝐷‾ = 𝜌
⃗ =0
(4) ∇ ⋅ 𝐵
∂𝑓
(5) ∇ ⋅ 𝐽‾𝑒 = − ∂𝑡
From (1) to (4) are maxwell's equations in point form (differential form)
‾
∵ 𝐸 = 𝐸‾ (𝑥̃, 𝑦, 𝑧, 𝑡) = 𝐸‾ (𝑥, 𝑦, 𝑧)𝑒 𝑗𝜔𝑡
∂𝐸‾
= 𝑗𝜔𝐸⃗ (𝑥, 𝑦, 𝑧)𝑒 𝑗𝜔𝑡 = 𝑗𝜔𝐸‾ 𝑒 𝑗𝜔𝑡
∂𝑡
∂
∂2
∴
→ 𝑗𝜔
&
→ −𝜔2
∂𝑡
∂𝑡 2
(1.1b)
(1.1c)
(1.1d)
(1.1e)
Similarly, are for 𝐻‾.
∴ Maxarell, equations with harmonicallychange
(1) ∇ × 𝐸‾ = −𝑗𝜔𝐵‾ or ∇ × 𝐸‾ = −𝑗𝜔𝜇𝐻
(2) ∇ × 𝐻‾ = 𝑗𝜔𝐷‾ + 𝐽‾𝑒 or ∇ × 𝐻‾ = 𝑗𝜔 ∈ 𝐸‾ + 𝐽̅𝑒
(3) ∇ ⋅ 𝐷‾ = 𝜌
(4) ∇ ⋅ 𝐵‾ = 0
& ∇ ⋅ 𝐽‾ = −𝑗𝜔𝜌
(1.2a)
(1.2b)
(1.2c)
(1.2d)
(1.2e)
Vector and scalar potentials
(auxiliary potential equations for 𝐴 & 𝑉)
∇ ⋅ 𝐵‾ = 0
∵ ∇ ⋅ ∇ × 𝐴‾ = 0 ⇒ Div curl any vector = zero.
∴ 𝐵‾ = ∇ × 𝐴‾,
𝐴‾ is the megnete vector potentical
∵ ∇ × 𝐸‾ = −𝑗𝜔𝑀𝐻‾
∇ × 𝐸‾ + 𝑗𝜔∇ × 𝐴‾ = 𝐴˙
∇ × 𝐸‾ + 𝑗𝜔∇ × 𝐴‾ = 0
∇ × (𝐸‾ + 𝑗𝜔𝐴‾) = 0
∵ ∇𝑥∇𝑉 = 0 ⇒ (curl grad any scalor = zers)
& ∇𝑥 − ∇𝑉 = 0 (we choose −∇𝑉 = 𝐸‾ + 𝑗𝜔𝐴 )
∴ from (1.4) ⇒ 𝐸‾ = −𝑗𝜔𝐴‾ − ∇𝑉
where 𝑉 is called scalar potential
∵ ∇ × 𝐻‾ = 𝑗𝜔𝜖𝐸‾ + 𝐽‾
1
𝐻‾ = ∇ × 𝐴‾, 𝐸‾ = −𝑗𝜔𝐴‾ − ∇𝑉
𝜇
1
∴ ∇ × ∇ × 𝐴‾ = 𝑗𝜔 ∈ 𝐸‾ + 𝐽‾
𝜇
1
→ ∇ × ∇ × 𝐴‾ = 𝑗𝜔𝑡[−𝑗𝜔𝐴‾ − ∇𝑉] + 𝐽‾
𝜇
(1.3)
(1.4)
(1.5)
multiplied by 𝑀 both sides & substituting by
∇ 𝐴 × 𝐵‾ × 𝐴‾ = ∇(∇ ⋅ 𝐴‾) − (∇ ⋅ ∇)𝐴‾ = ∇(∇ ⋅ 𝐴‾) − ∇2 𝐴‾
∇(∇ ⋅ 𝐴‾) − ∇2 𝐴‾ = j𝜔2 𝜇0 𝜖0 𝐴‾[−𝑗𝜔𝜇0 𝜖0 ∇𝑉] + 𝜇0 𝐽‾
∇2 𝐴‾ + 𝜔2 𝜇0 𝜖0 𝐴‾ = ∇(∇ ⋅ 𝐴‾) + 𝑗𝜔𝜇0 𝜖0 ∇𝑉 − 𝜇0 𝐽‾
∇2 𝐴‾ + 𝜔2 𝜇0 𝑡0 𝐴 = −𝜇0 𝐽‾ + ∇(∇ ⋅ 𝐴‾ + 𝑗𝜔𝜇0 𝑍0 𝑉)
∇2 𝐴‾ + ⏟
𝜔2 𝜇0 𝜖0 𝐴‾ = −𝜇0 𝐽‾ + ∇(∇
⏟⋅ 𝐴‾ + 𝑗𝜔𝜇0 𝑡0 𝑉)
(1.6)
Lorents condition
𝐾𝑧2 =𝛽 2
2
𝛽 = 𝜔02
𝑘𝑧2 =
Mot 0 is the wave number or phase custant
𝜔
1
where 𝛽 = 𝜔0 √𝜇0 𝜖0 = 𝐶0 , 𝐶 = 𝜇 𝜖 = 3 × 108
√ 0 0
𝛽=
𝜔
𝑐
=
2𝜋𝑓0
𝑐
m/s
𝜇0 = 4𝜋 × 10−7 H/m
𝜖0 = 8.85 × 10−12 F/m
2𝜋
= 𝜆 , 𝜆0 = wave langth
𝑓0 = frequency
0
‾ 0 𝑡0 𝑉.
To simplify Eq. IV let ⏟
∇ ⋅ 𝐴 = −𝑗𝜔𝜇
Lorents condition
∴ 𝑉=−
1
∇
𝐽𝜔𝜇0 𝜖0
(1.7)
⋅𝐴
∴ Eq. (1.6) becomes: ∇2 𝐴‾ + 𝜔2 𝑀 ∈ 𝐴‾ = −𝜇𝐽‾
(1.8)
The wave equations
∇ ‾
∈ ‾
‾ is the wave eqn.
If the region is free space
∇ ‾
,
‾
,
C
‾
In rectangular coordinate, the above vector wave eqn. will convert into scaler wave eqn.
as: ∇
∇
∇
,∇
∂
∂
Solution of 2nd order differential non-homogenous
‾
4
‾
∂
∂
∂
∂
wave equation is:
! "#$%
'(
&
&‾ is the distance from the source point Q the observation (or field) point, P.
z
)̀̅
&‾
&
)‾
)‾
&‾
R
)‾ )̀‾
|&‾ | |)‾ )̀‾|
ˆ
ˆ
ˆ
-ˆ
-ˆ
ˆ
/ˆ .
.
.
Method 1:
-/
.
-/ ˆ
-/
.
x
.
Given current distribution (J)
1‾2
1
∇
‾
-/
ˆ
Q . ̀ , ̀ , ̀/
&5
̅
P(x,y,z)
Radiating
antenna
body
-/
get |then1‾ |2 then 3‾ |2 at point P (free space)
then get 35 at point 6 from Maxwell eqn.
y
∇
⃗;
2
∴∇
3‾
1‾ /6
9
∈ 3‾ /6
9
‾/6
0 ⇒ free space is at point 6
1‾ 9 ∈ 3‾ at point ?
1
1‾
∇
Method 2:
∵ 3‾
9 ‾
∴ 3‾
∇( and from Lorentz condition ∇ ⋅ ‾
‾
9
H
#EF∈
∇.∇ ⋅ ‾/
and finding 1‾ using 1‾
3‾ ≅
9
In Far field
3J
0, 3K ≃
‾
9
K , 3M
H
∇
F
≃
9
9
∈ (;
(
9
∇⋅D‾
EFG
;
⃗
M
The fields from Infinitesimal dipole
"Ideal dipole, Hertzian dipole, short dipole with constant
current or elementary dipole”
Find EM fields generated by an electric current source ̅ of ideal dipole with const.
current placed on z-axis.
The characteristics of short dipole is: O ≪ Q very small length R ≪ Q very thin is not
practical antenna
S. /
S
constant along the dipole
S. / S
S. , T/ S. /! #EU
‾
The current density
ˆ
'V 'W‾ ⋅ 'O‾
⇒ S ‾ ⋅ 'W‾
‾'V
‾.'W‾ ⋅ 'O‾/
S'O‾
‾ ⋅ 'W‾/'O
.XYZ
[
∴ S. / S at O/2 ⩽ ⩽ O/2
ˆ
ˆ
&‾ )‾ )
ˆ ˆ
-/
ˆ
ˆ .
ˆ, as ⃗ → 0, &
∴ 'O‾
' ˆ
)
S ! #EU
∴ ‾
F_ [_ "#$J e/
e
c"e/
`aJ
Converting ‾
'
̂
̂
F_ [_ "#$J
e
`aJ
ˆH into spherical coordinate:
̂,
S O "#$J
!
cos i
4 )
S O "#$J
!
sin i
4 )
0 l
J )ˆ
K î
J
K
M
Using method:
3‾
and m‾
‾
9
∇
gives
(1) 1‾
n̂
9
‾
1
[_ ℓ #$
`a
pJ
∇.∇ ⋅ ‾/
H
Jq
r ! "s$J sin i
∴ 1M ≠ 0 but 1K 1J 0, 1‾ 1nn̂
9 S O "#$J
1
!
sin i u1
1M
v
9 )
4 )
(2) 3J
9
3K
3M
EF_ [_ e "#$J
!
cos
aJ q
0
i p1
S O "#$J
!
sin i u1
4 )
H
r
#$J
1
9 )
1
v
. )/
In far field ) >> Q (or ) > 10Q )
we neglect the terms proportional to
3J is inversely proportional to )
3K is inversely proportional to )
9 S O "#$J
!
sin i
4 )
S O "#$J
9
!
sin i
4 )
∴ 1M
3K
∵
then z
y
,z
|
{
120
H H
, , m)
Jq Jx
>> 1
F_ [_ "#$J
e
`aJ
∴ 1M
3K
9 S O "#$J
!
sin i
4 )
#}$[_ e "#$J
!
sin
`aJ
i
∴ ~3K ⊥ 1M € ⊥ propagationIIdirecton "r". This wave called TEM wave. The magnitude of
H
J
H
J
the field components is proportional to (i.e., E & H ∝ ).
3K .(/‚/
1M . /‚/
z.Ω/
and
3M
1K
z.Ω/
The same result of I and II in for field can be obtain from substituting the magnetic
vector potential ‾ in
3‾ ≅
9
where 3K
‾
9
K , 1M
#EF_ [_ e "#$J
!
sin
`aJ
3K
i
9
†
and 3J
0
9.120 / S O "#$J
!
sin i
4
3K &
930 S O "#$J
!
sin i
)
1n
S O "#$‰
!
sin i
4 )
∴ 1n&
3i
z
Analysis of radiation problem:
specify sources then required the field radiated by the source
synthesis problem:
•
specify radiation fields
then required to find
the sources
‾ (Magnetic vector potential)
≡A wxiliary potential vector is strictly mathematical tools.
Analysis
Œ̅
Œ̅
Radiated
fields
‹
3‹ &1
synthesis
One step analysis
‹
3‹ &1
̅
Two steps analysis, much easier and much simpler
‾ and 𝑯
‾ for short dipole with const. Current
𝑬
Radiation characteristics of short dipole with const. current
(Antenna parameters)
(1) Current distribution.
I(z)
𝐼𝑜
z
‾ and 𝑯
‾ fields in far field
(2) 𝑬
𝑗𝜂𝛽𝐼0 𝑙 −𝑗𝛽𝑟
𝐸𝜃
𝐸𝜃
𝑒
sin 𝜃 ,
= 𝜂 ⇒ 𝐻𝜙 =
4𝜋𝑟
𝐻𝜙
𝜂
𝑗𝛽𝐼0 𝑙 −𝑗𝛽𝑟
∴ 𝐻𝜙 = 4𝜋𝑟 𝑒
sin 𝜃.
𝐸𝜃 =
Ex. In far- Fall region 𝑅 ≈𝑟 ⩾ 10𝜆. For Mobile system 𝑓 = 9oo 𝑀𝐻𝑧
𝑐
3×108
+1𝑧 ⇒ 𝜆 = 𝑓 = 900×106 = 0.33 m = 33.33 cm
10𝜆 = 3.3 m ⇒ for field
(3) Radiation pattern: is a graphical representation of the radiation properties of the
antenna as a function of space coordinates 𝑓(𝜃, 𝜙)
Radiation pattern:
(I) field pattern |𝐸‾ |
(II) power pattern. |𝐸‾ |2
(I)
Field pattern (Field radiation pattern):
|𝐸𝜃 | =
𝜂𝛽𝐼 𝑙
( 𝜃 ) |sin⁡ 𝜃|
4𝜋𝑟
𝜃
0°
|𝐸𝜃 |𝑛
0
5°
10°
⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝐸𝜃 (r, 𝜃)
is a function of r & 𝜃
⇒ |𝐸𝜃 |𝑛 = |sin⁡ 𝜃| is the normalized field absolute.
……….
30°
……….. 45°
0.5
……….
0.707
60°
……….. 90°
180°
0.868
……….. 1
0
𝜃=0°
𝜃
𝑧
Figure-8
𝜃=−90°
𝜌
𝜃=90°
Vertical plane ≡ 𝜃 −⁡plane⁡≡ ⁡𝜌 − 𝑧 plane (or x-z if 𝜙=0, or
y-z if 𝜙 =π/2 plane) ≡ E-plane
E-plane: is a plane containing E-Field Vector
Horizontal plane (H-plane)
𝐻𝜙 = 𝑗
𝛽𝐼0 𝑙 −𝑗𝛽𝑟
𝛽𝐼0 𝑙
𝑒
sin⁡ 𝜃, |𝐻𝜙 | =
|sin⁡ 𝜃|
4𝜋𝑟
4𝜋𝑟
|𝐻𝜙 |𝑛 = |sin⁡ 𝜃|, 𝐻𝜙 (𝑟, 𝜃) is function f 𝑟 and 𝜃
|𝐻𝜙 |𝑛 is constant with 𝜙
H-plane: is the plane contain H-field vector.
𝜙=90o
𝜙
0°
5°
10° …
90° …
180° …
270° …
|𝐻𝜙 |𝑛
1
1
1
1
1
1
1
1
1
𝑦
1
(4) Radiated power
Radiated power density
1
𝑤
‾ = Re⁡{𝐸‾ 𝑋𝐻‾ ∗ }⁡𝑤/𝑚2
2
which gives the magnitude
and director of power density flow
and is called Poynting vector.
In far-field 𝐸𝑟 = 𝐻𝑟 ≃ 0, 𝐸‾ and 𝐻‾ in general will be as:
𝐸‾ = 𝐸̅𝜃 𝜃ˆ + 𝐸‾𝜙 𝜙ˆ,⁡⁡⁡⁡⁡⁡𝐻‾ = 𝐻𝜃 𝜃ˆ + 𝐻𝜙 𝜙ˆ, ⁡𝜙ˆ. 𝜂ˆ
𝑟ˆ 𝜃ˆ
𝜙ˆ
1
1
‾ = Re⁡ {|0 𝐸𝜃 𝐸𝜙 |} = Re⁡{𝑟ˆ(𝐸𝜃 𝐻𝜙∗ − 𝐸𝜙 𝐻𝜃∗ )}
∴𝑊
2
2
0 𝐻𝜃∗ 𝐻𝜙∗
𝐸𝜙
𝐸𝜃
𝑟ˆ
𝐸𝜃 𝐸𝜃∗ 𝐸𝜙 𝐸𝜑∗
∵
= 𝜂⁡&
= −𝜂 ⇒ 𝑤
‾ = Re⁡ {
+
}
𝐻𝜙
𝐻𝜃
2
𝜂
𝜂
‾ = 𝑟 [|𝐸6 |2 + |𝐸𝜙 |2 ] w/m2
𝑊
2𝜂
and if 𝐸𝜃 and 𝐸𝜙 are in 𝑟𝑚𝑠 (root mean square), then
‾ = 𝑟ˆ 1 [|𝐸𝜃 |2 + |𝐸𝜙 |2 ] W/m2
𝑊
𝜂
𝑥
𝜙=0, 360
x-y Plane, Horizontal plane
H-Plane, 𝜙- plane, ϴ=90° plane
̅ for short dipole with const. Current oriented along Z-axis
∴𝑊
2
‾ = 𝛾ˆ 1 [|𝐸𝜃 |2 ] = 𝑟ˆ 1 (𝜂𝛽𝐼0 𝑙) sin2 ⁡ 𝜃
𝑊
2𝜂
2𝜂 4𝜋𝑟
𝜂
𝛽𝐼 𝑙 2
= 𝑟ˆ 32 ( 𝜋𝑟0 ) sin2 ⁡ 𝜃
Radiated power:
‾ ⋅ 𝑑𝑠‾,⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡𝑑𝑠‾
𝑃rad = ∬𝑠 𝑊
=
⁡= (𝑟𝑑𝜃)(𝑟 sin 𝜃𝑑𝜙)𝑟ˆ = 𝑟 2 sin𝜃⁡𝑑𝜃𝑑𝜙⁡𝑟ˆ⁡
⁡= 1
1 2𝜋 𝜋
2
∫0 ∫0 (|𝐸0 |2 + |𝐸𝜙 | ) 𝑟 2 sin𝜃⁡𝑑𝜃⁡𝑑𝜙
2𝜂
For short dipole 𝐸𝜙 = 0, |𝐸0 | =
𝜂𝛽𝐼0 𝑙
|sin⁡ 𝜃|
4𝜋𝑟
1 (𝐼0 𝑙𝛽)2 𝑟 2 2𝜋 𝜋
∴ Prad ⁡= 𝜂
∫ ∫ sin3 ⁡ 𝜃𝑑𝜃𝑑𝜙
2 16𝜋 2 𝑟 2 0
0
𝜋
1
2𝜋
= 𝜂(𝐼0 𝑙𝛽)2
∫ sin3 ⁡ 𝜃𝑑𝜃) → 4/3
2
16𝜋 2 0
𝜋
𝜋
where, ∫0 sin3 ⁡ 𝜃𝑑𝜃 = ∫0
sin
⏟ 2⁡ 𝜃
sin⁡ 𝜃(1−cos2 ⁡ 𝜃)
Let 𝑢 = cos⁡ 𝜃
𝑑𝑢 = −sin⁡ 𝜃𝑑𝜃
at 𝜃 = 0 → 𝑢 = 1
𝜃 = 𝜋 → 𝑢 = −1
Prad
1 (𝐼0 𝑙𝛽)2
⁡= 𝜂
2
6𝜋
𝜂(𝐼0 𝑙𝛽)2
⁡=
Watt
12𝜋
𝜋
𝜋
𝑑𝜃 = ∫0 sin⁡ 𝜃𝑑𝜃 − ∫0 sin⁡ 𝜃cos2 ⁡ 𝜃𝑑𝜃
−1
−cos⁡
𝜃|𝜋0 + ∫1 𝑢2 𝑑𝑢
⏟
1 3 −1
− (−1 − 1) + 𝑢 |
3
⏟
1
1
= 2 + (−1 − 1)
3
2
= 2− =
3
6 2 4
=
− = #
3 3 3
(5) The Radiation Resistance (𝑅rad )
Assume
is dissipated into an imaginary resistance called (𝑅rad ), then
Max. Current
⁡𝐼o
⁡𝑅rad
∴
for short dipole with const. current is
1
2
= |𝐼0 |2 ⁡⁡𝑅rad = 𝜂
∴ ⁡𝑅rad
⁡=
∴ 𝑅rad = 𝜂
𝑙2 𝛽2
𝜂 6𝜋 ⁡, ∵
𝛽=
(𝐼0 𝑙𝛽)2
12𝜋
2𝜋
𝜆
𝑙 2 2𝜋 2
( ) =
6𝜋 𝜆
𝑙 2
𝜆
𝑅rad = 80𝜋 2 ( )
Where 𝑅rad is used to measure the radiated power.
Since 𝑙 ≪ 𝜆(𝑙 ⩽ 𝜆/10) for short dipole, then if 𝑙 = 𝜆/10 ⇒ 𝑅rad ≈ 8Ω
As 𝑅rad ↓↓⁡⁡⁡⁡⇒
↓↓
∴ short dipole with con. current has low power radiation
Prob.1
Exer.1) A vertical wire 1 m long carries a current of 5 A of operating frequency 2MH𝑧
assuming that the wine is in free space. Calculat the strength of the radiated fueld
produced at distance 2 km and 25 km in the direction at right angles to the axis of the
wire.
Exer.2) Find the power radiated by a 50 cm dipole antenn operated at 30MHz with
current of 2 A. How much current would be needed to radiatef power of 5W?
1
Exer.3) Starting from Maxwell's equation, prove that 𝐸‾ = −𝑗𝜔𝐴‾ + 𝑗𝜔∈𝑀 ∇∇ ⋅ 𝐴‾ Where 𝐴‾ is
the magnetic vector potential.
Exer.4) Write down Maxwell is equations in time varying field and deduce these
equations in electrostatic and magnetostatic
Prob.2
Exer.4) On a dipole antenna of 1 m
Long and oriented along the 𝑧-axis? opercted at 10MHz, the current distribatan is:
5(1 − 2𝑧) 0 ⩽ 𝑧 ⩽ 1/2
1
𝐼(𝑧) = {
5(1 + 2𝑧) 0 ⩾ 𝑧 ⩾
2
a) Find the radiated electric field and compare it with the field of short dipole whose
current equals 5 A
Exer.5) Prove that 𝜂𝛽 = 𝜔𝜇
Note:
𝐼𝑚 𝛽(𝑙/2 − 𝑧)
𝑙
2𝑧
𝐼𝑚 𝛽 (1 − )
⏟ 2
𝑙
𝐼0
Antenna parameters applied on short dipole with const. current (Continue)
(6) Half power Beam Width (HPBW)
HPBW is the angular separation of the paints where the main beam of the field pattern
1
equals . On the power pattern these points are corresponding to one-half.
√2
HPBW for short dipole with const. current:
|𝐸𝜃 |𝑛 = |sin⁡ 𝜃|
𝜃1
1
√2
At 𝜃 = 𝜃1 , 𝜃2
|𝐸𝜃 |𝑛 = |sin 𝜃1,2 | =
𝜃1 = sin−1
1
√2
1
√2
= 45𝑜
𝜃2 = 180∘ − 45 = 135∘ ⁡(Where⁡sin⁡ 𝜃) = sin⁡(180 − 𝜃)
∴ 𝜃𝐻𝑃𝛽𝜔 = 𝜃2 − 𝜃1 = 135∘ − 45∘ = 90∘
1
√2
𝜃2
𝜃
Radiation power classification
We can classify radiation power into three Categories:
(1) Isotropic radiation pattern: (Ideal pattern)
•
•
•
•
It is hypothetical source which radiates equally in all directions (pattern does not
depends on 𝜃 or 𝜙 )
It is idea and not realizable
Used as a reference
̅ = 𝑊0 𝑟ˆ is its power density (𝑊0 = Prad = 𝑃𝑟𝑎𝑑2 )
𝑊
𝑆
4𝜋𝑟
where power radiated by isotropic source is
‾ 0 ⋅ ds̅ = ∫ 2𝜋 ∫ 𝜋 𝑊0 𝑟ˆ ⋅ 𝑟ˆ𝑟 2 sin⁡ 𝜃𝑑𝜃𝑑𝜙 = 2𝜋𝑊𝑜 𝑟 2 (−cos⁡ 𝜃|𝜋0 = 4𝜋𝑟 2 𝑊0
Prad = ∬𝑠 𝑊
0
0
𝑃rad
∴ 𝑊0 =
⏟
4𝜋𝑟 2
surface ares of sphere.
(2) Directional pattern: is directed in a given (or certain) direction
Ex. - dish antennas, Yagi-Uda array antenna, Radar.
(3) Omni-directional pattern:
It is a radiation pattern directed in one plane and is isotropically in the perpendicular plane.
Ex. Short dipole, λ⁄2 dipole antenna
omni-directional radiation pattern
Antenna parameters applied on short dipole with const. current (Continue)
‫قدرة الهوائي على توجيه المجال في اتجاه معين‬
(7) Directive Gain:
The directive gain for a non-isotropic source is equal to the ratio of its radiation density
in a given direction over that of an isotropic source (as a reference antenna)
𝐷𝑔⁡ =
𝑃𝑜𝑤𝑒𝑟⁡𝑑𝑒𝑛𝑠𝑖𝑡𝑦
𝑃𝑜𝑤𝑒𝑟⁡𝑑𝑒𝑛𝑠𝑖𝑡𝑦⁡𝑜𝑓⁡𝑖𝑠𝑜𝑡𝑟𝑜𝑝𝑖𝑐⁡𝑠𝑜𝑢𝑟𝑐𝑒⁡
2
1
1
‾ × 𝐻‾ ∗ } 2𝜂 [|𝐸𝜃 |2 + |𝐸𝜙 | ] 𝑊
Re⁡{𝐸
⁡⁡⁡⁡⁡⁡⁡⁡= 2
=
=
Prad⁡/4𝜋𝑟 2
𝑃rad /4𝜋𝑟 2
𝑊0
For short dipole with constant current:
‾
𝑊
𝑤0
∴ 𝐷𝑔
𝐷𝑔
𝐷𝑔
𝜂 𝛽 2 𝐼𝑜2 𝑙2 2
‾ 0 = 𝑟ˆ𝑤0
⁡= 𝑟ˆ
sin 𝜃, ⁡⁡⁡⁡𝑊
32 𝜋 2 𝑟 2
𝑃rad
𝜂(𝐼0 𝑙𝛽)2
⁡=
, ⁡𝑃
=
4𝜋𝑟 2 rad
12𝜋
2 2 2
𝜂𝛽 𝐼𝑜 𝑙
2
𝑊
2 𝑟 2 sin 𝜃
32𝜋
⁡=
=
⋅ (4𝜋𝑟 2 )
𝑊0
𝜂𝐼02 𝑙 2 𝛽2
12𝜋
4 × 12 2
⁡=
sin ⁡ 𝜃
32
3
⁡= sin2 ⁡ 𝜃
2
⁡= 1.5⁡sin2 ⁡ 𝜃
.
𝑤𝑜
Power density pattern of short dipole as an example
. 𝑊(𝜃, ∅)
𝑫𝒈(𝜽, 𝝓) =
Isotopic antenna
power density
𝐖⁡(𝜽⁡,𝝓)
(8) Directivity "Max directive gain”
𝐷0 =
𝑊max
𝑊0
It is the value of the directive gain in the direction of its maximum value.
𝐷0
𝑊max
𝑊max
=
𝑊0
𝑃rad /4𝜋𝑟 2
4𝜋𝑟 2 𝑊max
⁡=
𝑃rad
⁡=
Notes:
•
For short dipole 𝐷0 = 𝐷𝑔 |
= 1.5
𝜃=90
∘
where 𝑊max = 𝑊|𝜃=90
•
The directivity of short dipole 𝐷0 = 1.5 calculated in 𝑑𝐵 :
𝐷o ⁡𝑑𝐵 = 10log⁡ 1.5 = 1.76𝑑𝐵
•
𝐷o of isotropic 𝐷𝑔 = 𝐷0 = 𝑊0 = 1
𝑊
0
𝐷o ⁡dB⁡|isotrapic = 10log⁡ 1 = 0 dB
•
Directivity of Dish antenna is from:
30 dB (corresponding to 1000) to 40 dB (corresponding to 10000).
•
By Increasing 𝐷0, the power concentration will increase and both angle and HPBW will decrease.
(9) Efficiency of Antenna
𝑒
𝑃owor radiated by antenna
× 100
input power
𝑃rad
⁡=
× 100
𝑃in
1 2
𝑃rad
2 𝐼0 ⁡𝑅rad
⁡=
=
× 100
1 2
1 2
𝑃rad + 𝑃loss
𝐼
⁡𝑅
+
𝐼
⁡𝑅
2 0 rad 2 0 Less
1 2
2 𝐼0 𝑅rad
⁡=
× 100
1 2
(𝑅
)
𝐼
+
𝑅
rad
Loss
2 0
𝑅rad
⁡=
× 100
𝑅rad + 𝑅Loss
⁡=
𝑙 2
𝜆
Ex. Since 𝑅rad = 80𝜋 2 ( ) for short dipole, and for 𝑙 =
𝑅rad |𝑙= 𝜆 =
10
𝑙 2
80𝜋 2 (𝜆)
= 8Ω
8
𝜆
,
10
so
and if 𝑅Loss = 2Ω, 𝑒 = 10 × 100 = 80%
(10) Gain "Antenna Gain" (Directivity is reduced by losses)
Directive Gain is a gain of directional properties while Gain is the ratio of the power
density in a given direction to the power density that would be obtained if the power
accepted (inputted) by the antenna were radiated isotopically.
𝑊(𝜃, 𝜙)
4𝜋𝑟 2 𝑊(𝜃, 𝜙)
=
Pin ⁡/4𝜋𝑟 2
Pin
2
1
2
|𝐸𝜃 | + |𝐸𝜙 | ] 𝑃
[
2𝜂
rad
𝐺(𝜃, 𝜙) =
⋅
Pin /4𝜋𝑟 2
𝑃rad
2
2
1 [|𝐸𝜃 | + |𝐸𝜙 | ] Prad
𝐺=
⋅
2𝜂 𝑃rad /4𝜋𝑟 2
Pin
𝑃rad
𝐺 = 𝐷𝑔 ⋅
𝑃in
𝐺 = 𝐷𝑔 ⋅ 𝑒⁡
(Gain = Directivegain × efficiency)
at max. ⇒ (Max. gain) 𝐺0 = 𝐷0 . e⁡⁡⁡⁡⁡
𝐺=
Max. gain = (dircetivity) ×(efficiency)
.
Pin =10 W
. 8W
Antenna is a passive element and gain is for the loss of the power due to concentrating of
a power in a certain direction.
Ex. Find the power density in 𝐖𝐦−𝟐 at distance of 𝟑𝟎⁡𝐤𝐦 from an antenna
that is radiating 5⁡kW with a directivity of 37⁡dB. What directivity is required if the
antenna power density is 2.5⁡mW⁡m−2 ?
Solv:
𝑤 =? ? , 𝑟 = 30⁡km, 𝑃rad = 5 × 103 ⁡W, 𝐷0𝑎𝐵 = 37𝑑𝐵 = 10log⁡ 𝐷0
𝑊
𝑊
𝐷0 =
=
(4𝜋)(30 × 103 ) = 5011.87
2
Prad⁡/4𝜋𝑟
5 × 103
5011.87(5 × 103 )
𝑊=
⁡= 2.22 × 10−3 ⁡W/m2
4𝜋(30,000)2
⁡= 2.22⁡mW/m2
2
If 𝑤 = 2.5mw/m ⇒ 𝐷0 = 37.5⁡dB.
Do = 5011.87
Prob.3
Exer. For short dipole with triangle current (in Exer.4 in Prob.2) extended along the 𝑧-axis, find
(i) the radiation patter in free space
(ii) The radiated power
(iii) The radiation resistance
(𝐼𝑉)The⁡𝐻𝑃𝐵𝑊
(v) The Directive gain
(Vi) the grin
(vii) The efficacy.
(viii) The directivity
For short dipole with triangle current Find:
1) Current distributed
z axis
2𝑧
𝐼0 (1 − ) 0 ⩽ 𝑧 ⩽ 𝑙/2
𝑙
𝐼(𝑧) = {
2𝑧
𝐼0 (1 + ) −𝑙/2 ⩽ 𝑧 ⩽ 0
𝑙
𝑙/2
2) Field 𝑬𝜽 & 𝑯𝝓
Since 𝐴𝑧 =
𝜇0 𝐼0 𝑙 𝑒 −𝑗𝛽𝑟
,
8𝜋
𝑟
then 𝐴𝜃 = − 𝐴𝑧 sin 𝜃 =
𝑰
.𝒐
− 𝜇0 𝐼0 𝑙 𝑒 −𝑗𝛽𝑟
sin 𝜃
8𝜋
𝑟
x axis
𝐴𝜙 = 0 = 𝐴𝑟 = 0
In Far Field:
𝐸̅ = −𝑗𝜔 𝐴̅
𝐸𝑟 = 0,
𝐸𝜙 = 0
Eθ = −jωAθ
𝑗𝜂𝛽𝐼0 𝑙 −𝑗𝛽𝑟
𝐸𝜃 =
𝑒
sin 𝜃
8𝜋𝑟
𝐸𝜃
𝛽𝐼0 𝑙 −𝑗𝛽𝑟
𝐻𝜙 =
=𝑗
𝑒
sin 𝜃
𝜂
8𝜋𝑟
I(z)
−𝑙/2
3) Radiation Pattern:
𝑧 𝜃=0°
𝜃
𝑰𝒐
z
−𝑙/2
ȁ𝐸𝜃 ȁ
𝜃=−90°
𝑙/2
ห𝐻𝜙 ห
𝜌
𝜃=90°
𝑦
𝜙
𝑥
𝜙=0, 360
(4) Radiation power:
y axis
𝑃rad =
𝛽 2 𝜂I02 𝑙 2
48𝜋
(5) Radiation resistance
𝑙 2
𝑅rad = 20𝜋 2 ( )
𝜆
3
3
2
(6) 𝐷𝑔 = 2 sin 𝜃, 𝐷0 = 2 (same as short dipole with const. current)
(7) 𝛉𝐇𝐏𝐁𝐖 = 90∘ (same as short dipole with const. current)
(8) 𝑮, 𝒆, 𝐃 ⇒ 𝐺 = 𝐷𝑔 𝑒 (same as short dipole with const. current)
Dipole Antennas of finite lengths (Practical dipoles)
The ideal short dipole (of const. current) was discussed. It is not practical dipole
since its current is not equal zero at the ends of its length.
Once the current distribution is known, all the other parameters of antenna (such
as radiation resistance, radiated power, directivity, ..., etc.) can be obtained.
1) Current Distribution
In practical dipole of varying length, the current distribution can be generally
expressed ass
𝐼 sin 𝛽(𝑙/2 − 𝑧) 0 ⩽ 𝑧 ⩽ 𝑙/2
𝐼(𝑧) = { 𝑚
𝐼𝑚 sin 𝛽(𝑙/2 + 𝑧) −𝑙/2 ⩽ 𝑧 ⩽ 0
where the dipole is placed symmetrically along the 𝑧-axis, 𝐼m is the maximum current,
𝛽𝑙
the current is zero at the ends, and 𝐼0 = 𝐼𝑚 sin 2 = 𝐼(0) is the current at the feeding
point.
Special case: For the practical short dipole 𝑙 << 𝜆
𝑙
∵ 𝐼(𝑧) = 𝐼m Sin 𝛽 (2 − 𝑧)
0 ⩽ 𝑧 ⩽ 𝑙/2
2𝜋 𝑙
,
𝜆 𝜆
≪ 1 and since sin 𝑥 = 𝑥 for 𝑥 << 1
𝑙
𝐼𝑚 𝛽 ( − 𝑧) 0 ⩽ 𝑧 ⩽ 𝑙/2
2
∴ 𝐼(𝑧) = {
𝑙
𝑙
Im 𝛽 ( + 𝑧) − ⩽ 𝑧 ⩽ 0
2
2
𝑙
2𝑧
𝐼𝑚 𝛽 (1 − ) 0 ⩽ 𝑧 ⩽ 𝑙/2
2
𝑙
={
𝑙
2𝑧
𝐼𝑚 𝛽 (1 + ) −𝑙/2 ⩽ 𝑧 ⩽ 0
2
𝑙
𝛽=
∵ 𝐼0 = 𝐼(0) = 𝐼𝑚𝛽
∴
𝐼0 = 𝐼𝑚𝛽
𝑙
2
𝑙
2
2𝑧
𝐼0 (1 − ) 0 ⩽ 𝑧 ⩽ 𝑙/2
𝑙
𝐼(𝑧) = {
2𝑧
𝐼0 (1 + ) −𝑙/2 ⩽ 𝑧 ⩽ 0
𝑙
Thus, it is the same practical short dipole with triangle current.
Example Dipole of length 7.5 m is connected to a generator of variable frequency 𝑓 =
1𝑀𝐻𝑧 ∶ 40 𝑀𝐻𝑧.
Sketch the current distribution at a) 𝑓 = 1MHz b) 𝑓 = 20MHz c) 𝑓 = 40MHz
Solution
a) 𝑓 = 1MHz ⇒ 𝜆 =
𝑐
𝑓
=
3×108
3×106
= 300 m ⇒
𝜆
ℓ
ℓ≪𝜆
ℓ<
⇒
⇒ = 0.025
10
7.5 m < 300 m
𝜆
7.5 < 30
→ ∴ short dipole with triangle current
I (z)
𝑰𝒐
z
−𝑙/2
𝑐
𝑙/2
3×108
b) 𝑓 = 20MHz ⇒ 𝜆 = 𝑓 = 20×106 = 15 m
ℓ
⇒𝜆=
7.5
15
𝜆
= 0.5 ⇒ the current distribution is of 2 dipole antenna
I (z)
𝑰𝒐
z
−𝑙/2
𝑐
3 × 108
𝑐) 𝑓 = 30 ⇒ 𝜆 = =
= 10
𝑓 30 × 106
𝑙 = 7.5𝑚, 𝜆 = 10
𝑙
3𝜆
= 0.75 ⇒ 𝑙 =
𝜆
4
𝜆
2
<𝑙<𝜆⇒
0.5 <
𝑙
<1
𝜆
𝑙/2
Other cases
3×108
𝑙
d) 𝑓 = 40MHz ⇒ 𝜆 = 40×106 = 7.5 m ⇒ 𝜆 = 1 ⇒ 𝑙 = 𝜆
I (z)
𝑰𝒐
z
−𝑙/2
𝑙/2
I (z)
𝑙
𝜆
𝑰𝒐
=1.5
3
𝑙= 𝜆
2
−𝑙/2
𝑙
=2
𝜆
𝑙 =2𝜆
z
𝑙/2
2) Field of dipole antenna of different length
“Field of practical dipole antenna of finite length”
𝐸‾ = 𝐸0 𝜃ˆ
𝛽𝑙
𝑙
𝑗𝜂𝐼𝑚 −𝑗𝛽𝑟 cos ( 2 cos 𝜃) − cos (𝛽 ⁄2)
𝐸𝜃 =
𝑒
[
]
2𝜋𝑟
sin 𝜃
is the field of dipole antenna of length 𝑙 placed symmetrically along z-axis. 𝐼𝑚 is the max.
current of the current distribution given by
𝐼 sin 𝛽(𝑙/2 − 𝑧) 0 ⩽ 𝑧 ⩽ 𝑙/2
𝐼(𝑧) = { 𝑚
𝐼𝑚 sin 𝛽(𝑙/2 + 𝑧) −𝑙/2 ⩽ 𝑧 ⩽ 0
𝜂 = 120𝜋 in free space
𝛽𝑙
𝑙
𝑒 −𝑗𝛽𝑟 cos ( 2 cos 𝜃) − cos (𝛽 ⁄2)
𝐸𝜃 = 𝑗60𝐼𝑚
[
]
𝑟
sin 𝜃
Example:
For 𝜆⁄2 dipole antenna placed along the z-axis, determine
(i) radiation pattern, E-and-H planes
(ii) radiated power (iii) radiation resistance
(iv)Directivity (v) HPBW (vi) relation between gain, Dg and e
solution
𝜆
(i) 𝑙 = 2 , 𝛽𝑙 =
∴ 𝐸𝜃
2𝜋
𝜆
1 𝜆
⋅ 2 ⋅ 2 = 𝜋/2
𝜋
𝑒 −𝑗𝛽𝑟 cos (2 cos 𝜃) − cos (𝜋/2)
= 𝑗60𝐼𝑚
[
]
𝑟
sin 𝜃
= 𝑗60𝐼𝑚
𝑒 −𝑗𝛽𝑟 cos (𝜋/2cos 𝜃)
𝑟
sin 𝜃
∴Radiation pattern:
ȁEθ ȁn =
20
cos (𝜋/2cos 𝜃)
sin 𝜃
0
0
10
30
40
50
60
70
80
90
ȁEθ ȁn
0
0.0239 0.0946 0.2089 0.359 0.532 0.0.707 0.859 0.963 1
𝜃
E-plane
ȁ𝐸𝜃 ȁ
1
2𝜋
𝜋
More directive than short dipole
2
(ii) 𝑃rad = 2𝜂 ∫0 ∫0 ȁ𝐸𝜃 ȁ2 + ห𝐸𝜙 ห 𝑟 2 sin 𝜃𝑑𝜃𝑑𝜙
60𝐼𝑚 cos (π/2cos 𝜃)
∵ ȁ𝐸𝜃 ȁ =
∣
𝑟
sin 𝜃
2
1 2𝜋 𝜋 3600𝐼𝑚
cos 2 (𝜋/2cos 𝜃) 2
𝑃rad =
∫0 ∫0
𝑟 sin 𝜃𝑑𝜃𝑑𝜃
2𝜂
𝑟2
sin2 𝜃
2
2
3600𝐼𝑚
𝜋 cos (𝜋/2cos 𝜃)
=
(2𝜋)
∫⏟
𝑑𝜃
⏟0
2𝜂
sin 𝜃
solved numerically and equals to 1.22
∵ 𝜂 = 120𝜋
2
∴ 𝑃rad = 30𝐼𝑚
∗ 1.22
(iii) 𝑅rad =? ?
1
2
2
∵ Prad = 𝐼𝑚
𝑅rad ⇒ 𝑅rad =
2𝑃rad
2
𝐼𝑚
∴ 𝑅rad = 2(30𝐼𝑚 2 ∗ 1.22)/𝐼𝑚2
≃ 73.2Ω
∴ as 𝑅rad increases, 𝑃rad increases.
(iv) 𝐷0 =? ?
1
1 3600 Im2 cos2 (𝜋/cos 𝜃)
2
2]
[ȁ𝐸
ȁ
+
ȁ𝐸𝜙ȁ
]
𝑊 2𝜂 0
2𝜂 [ 𝑟 2
sin2 𝜃
∵ 𝐷𝑔 =
=
=
2 ∗ 1 ⋅ 22
𝑊0
Prad /4𝜋𝑟 2
30𝐼𝑚
𝐷0 |𝜆⁄2 > 𝐷0 |
4𝜋𝑟 2
cos 2 (𝜋/2cos 𝜃)
= 1.64
Sort dipole
sin2 𝜃
∴ 𝐷0 = 𝐷𝑔 ห
max
= 𝐷𝑔 ห
cos (𝜋2 cos 𝜃)
|
sin 𝜃
∘
(v) HPBW ⇒ ȁ𝐸𝜃 ȁ𝑛 = |
𝜃𝐻𝑃𝐵𝜔 = 𝜃2 − 𝜃1 = 78
(vi) 𝑒 =
𝑃rad
𝜌in
if 𝑅𝐿 = 2Ω̇
× 100 =
𝜃=90𝑜
𝑃rad
𝑃rad +𝑃loss
=
× 100
1
√2
= 1.64
𝐷0 ȁ𝜆 > 𝐷0 ȁ𝑠ℎ𝑜𝑟𝑡 𝑑𝑖𝑝𝑜𝑙𝑒
2
𝑒=
𝑅rad
𝑅rad +𝑅𝐿
× 100 =
73.2
× 100
73.2 + 2
= 97.3% which is greater than short dipol (𝑒ȁ 𝑙 = 𝜆 = 80%)
𝜆 10
𝐺
𝐺
𝑊
𝑃𝑟𝑎𝑑
∗
2
𝑃in /4𝜋𝑟 𝑃rad
𝑊
𝑃𝑟𝑎𝑑
=
∗
2
𝑃𝑟𝑎𝑑 /4𝜋𝑟
𝑃in
= 𝐷𝑔 ∗ 𝑒
=
for max 𝐺 = 𝐺0 at 𝐷𝑔 = 𝐷0
∴ 𝐺0 = 𝐷0 𝑒
Monopole antenna
A conductor wire antenna of length above an infinity large ground plane (perfect
conductor of infinite conductivity, i.e., = ∞) forms a monopole antenna (see
fig.1) when fed at the base the resulting
fields are identical to the dipole’s.
2
2
Equivalent
to
~
~
Infinite ground plane
“Perfect conductor”
=∞
2
Image
Fig:1
Comarigon between dipole and monople
Comparizon
dipole
monople
=
Construction
(1)
=
2
|
I (z)
Current
Distribution
(2)
2
I
=
I
|
z
/2
I
Y
z "=0
"
2
/2
=
/2 Z
0
H-plane
z "
Radiation
Pattern
(3)
4
$
#
s
E-plane
x
∅
H-plane
#
"=' 2
$
Y
s
x
∅
()
(4)
=rad
(5)
OP
6
OD
7
#
, /012 cos '/2cos "
= *60
.
3
:
sin "
#
but exists from
0 ⩽ " ⩽ '/2 and for
0 ⩽ < ⩽ 2'
=>?@
A
E E
=
∫D ∫D |() | > GHI )@)@J
B
=rad
A
E E/
=
∫D ∫D |() | > GHI )@)@J
B
=rad |mono = A/ KLMN|dipole
1
| |
2T #
QR =
Prad /4'-
1
| |
2T #
QY =
Prod
2 /4'-
OD =1.643
OD =2*1.643=3.289
" HPBW=78
HPBW
8
The same
"HPBW
" HPBW=
\]^
= 39
Radiation
Resistance
9
Efficiency
10
Gain
11
You have to complete this table
Monopole antenna
𝑙
A conductor wire antenna of length above an infinity large ground plane (perfect
2
conductor of infinite conductivity, i.e., 𝜎 = ∞) forms a monopole antenna (see
̅ fields are identical to the dipole’s.
fig.1) when fed at the base the resulting 𝐸̅ 𝑎𝑛𝑑𝐻
𝐼(𝑧)
𝐼(𝑧)
𝑙⁄
2
Equivalent
𝑙⁄
2
to
~
~
Infinite ground plane
“Perfect conductor”
𝝈=∞
𝑙⁄
2
𝐼(𝑧)
Image
Fig:1
𝝀
𝝀
Comarigon between 𝟐 dipole and 𝟒 monople
𝝀
Comparizon
𝟐
𝑙⁄
𝟐
𝟒
monople
𝝀⁄
𝟒
𝝀⁄
𝟒
𝑙 = 𝜆⁄4
𝑙 = 𝜆⁄2
Construction
(1)
−𝑙⁄
𝟐
Current
Distribution
(2)
𝝀
dipole
𝜆⁄
2
𝝀⁄
𝟒
𝐼𝑚
I
I (z)
ȁ𝐼(𝑧)ȁ
𝐼0 = 𝐼𝑚
I
z
−𝑙/2
𝑙/2
𝐼𝑚
𝜆⁄
2
𝝀⁄
𝟒
I
𝐼0 = 𝐼𝑚
𝑙/2 Z
0
𝝀⁄
𝟒
H-plane
z 𝜃
Radiation
Pattern
(3)
𝐻𝜙
𝐸𝜃
Y
s
E-plane𝑰𝒐
x
∅
H-plane
z 𝜃=0
𝜃
𝐻𝜙
𝜃 = 𝜋⁄2
Y
𝐸𝜃
s
x
∅
𝑬𝜽
(4)
𝐼𝑚 −𝑗𝛽𝛾 cos 𝜋/2cos 𝜃
𝐸𝜃 = 𝑗60
𝑒
[
]
𝑟
sin 𝜃
The same 𝐸𝜃 but exists from
0 ⩽ 𝜃 ⩽ 𝜋/2 and for
0 ⩽ 𝜙 ⩽ 2𝜋
𝑷rad
(5)
𝑫𝒈
6
𝑫𝟎
7
𝑷𝒓𝒂𝒅
𝟏 𝟐𝝅 𝝅
=
∫ ∫ (ȁ𝑬𝜽 ȁ𝟐 )𝒓𝟐 𝐬𝐢𝐧 𝜽𝒅𝜽𝒅𝝓
𝟐𝜼 𝟎 𝟎
𝑷rad
𝟏 𝟐𝝅 𝝅/𝟐
ȁ𝑬𝜽 ȁ𝟐 𝒓𝟐 𝐬𝐢𝐧 𝜽𝒅𝜽𝒅𝝓
=
∫ ∫
𝟐𝜼 𝟎 𝟎
𝑷rad ȁmono = 𝟏/𝟐𝐏𝐫𝐚𝐝ȁdipole
1
(ȁ𝐸 ȁ2 )
2𝜂 𝜃
𝐷𝑔 =
Prad /4𝜋𝑟 2
1
(ȁ𝐸 ȁ2 )
2𝜂 𝜃
𝐷𝑔 =
Prod2
/4𝜋𝑟 2
2
𝑫𝟎 =1.643
𝑫𝟎 =2*1.643=3.289
𝜃 HPBW=780
HPBW
8
𝜃 HPBW=
780
2
= 390
𝜃HPBW
Radiation
Resistance
9
Efficiency
10
Gain
11
You have to complete this table
Small loop antenna
•
Loop antenna has many conficuirationts:1) rectangle, 2) square,3) triangle 4) ellipse 5) circle.
•
Circular or rectangular loop is the most popular because of their simplicity in analysis and
constructions.
•
Loop antenna has two categories: Electricity small and Electricity large:
1 Electricity small loop: the overall length (number of turns N times circumference, c) is less than
one-tenth of wave length (NC <
𝜆
).
10
2 Electricity large loop: the circumference approaches one free space wavelength (C~𝜆).
Electricity small loop
Electricity large loop
z
Radiation
z
𝐸∅
𝐸∅
y
y
x
x
Loop placed at x-y plane
C~𝜆
Loop placed at x-y plane
𝜆
NC < 10
Configuration
Rectangular loop
Square loop
Circular loop
Small square loop of side 𝒍
Small square loop of side 𝑙 is composed of four short dipoles constant current 𝐼0
placed horizontally in x-y pane.
Construction:
𝑍
𝑰𝒐
4
1
3
𝒚
𝑙
𝑙
𝒙
2
The far fields of small square loop:
The magnetic vector potential for each short dipole segment takes the form
𝜇0 I0 𝑙 ̅ −𝑗𝛽𝑅
𝐴‾ =
4𝜋𝑅
𝑒
where
𝐴‾ = (𝐴𝑥1 − 𝐴𝑥3 )𝑥ˆ + (𝐴𝑦2 − 𝐴𝑦4 )𝑦ˆ
⋮
𝜔𝜇0 𝛽𝐼0 𝑙 2 −𝑗𝛽𝑟
∴ 𝐸𝜙 =
𝑒
sin 𝜃 , ∵ 𝜔𝜇0 = 𝜂𝛽
4𝜋𝑟
𝜂𝛽2 𝐼0 𝑙 2 −𝑗𝛽𝑟
∴ 𝐸𝜙 = 4𝜋𝑟 𝑒
sin 𝜃
𝐻𝜃 =
and
−𝐸𝜙
𝜂
The radiation patterns:
𝐸𝜙 =
ȁ𝐸∅ ȁ𝑛
0
1
𝜂𝛽2 𝐼0 𝑙 2
4𝜋𝑟
10 20
1 1
30
1
𝑒 −𝑗𝛽𝑟 sin 𝜃 is plotted in 𝜙-plane
40 50
1 1
…………………..
𝐻𝜃 = −
𝜃
𝐸𝜙
𝜂
=−
0
𝛽2 I0 𝑙 2
10
ȁ𝐻𝜃 ȁ𝑛
4𝜋𝑟
20
𝑒 −𝑗𝛽𝑟 sin 𝜃 is plotted in θ-Plane
……………………..
You have to calculate it
z 𝜽
𝑬𝝓
𝑯𝜽
Y
𝝆
E- plane
x
H-plane
𝝓
𝜙 − plane
𝜃 − plane
The radiated power:
Prad =
1
2𝜂
2𝜋
𝜋
2
∫0 ∫0 [ȁ𝐸𝜃 ȁ2 + |𝐸𝜙 | ] ⋅ 𝑟 2 sin𝜃 𝑑𝜃𝑑𝜙
𝜂2 𝛽 4 𝐼02 𝑙 4 2
sin 𝜃
|𝐸𝜙 | =
16𝜋 2 𝑟 2
𝜂𝛽 4 𝐼02 𝑙 4
∴ 𝑃𝑟𝑎𝑑 =
Watt
2
12𝜋
The radiation resistance:
𝑅𝑟𝑎𝑑 =
∴ 𝑅𝑟𝑎𝑑
2𝑃𝑟𝑎𝑑
ȁ𝐼0 ȁ2
=
𝜂𝛽 4 𝑙 4
6𝜋
4
, the area of a square loop = 𝑆 = 𝑙2, 𝛽4 = (2𝜋𝜆) , and
𝜂 = 120𝜋.
(120𝜋)(2)4 (𝜋)4 𝑆 2 20(16)𝜋 4 𝑆 4
𝑆2
4
=
=
= 320𝜋 ( 4 )
(6𝜋)(𝜆4 )
𝜆4
𝜆
Note: For circular loop s = 𝜋𝑟02
While for square loop s = 𝒍𝟐
Notes:
• All the other radiation characteristics (patterns) of small loop antenna (like direction gain, directivity,
gain, …. etc.) can be obtained.
• Loop antenna with small circumference (small loop) have small radiation resistance (𝑅𝑟𝑎𝑑 < 𝑅𝑙𝑜𝑆𝑆 ). They are
very poor radiators, and they are seldom employed for transmission in radio communication, they are used
in receiving mode, such as portable radios and pagers because its 𝑅𝑟𝑎𝑑 < 𝑅𝑙𝑜𝑆𝑆 .
• The radiation resistance can be increased by increasing (electricity) its circumference and/or the number
turns. Another way is to insert, within its circumference, a ferrite core of high permeability, 𝜇𝑟 , which
raise the magnetic field and hence the radiation resistance. This form called ferrite loop.
Helical antenna
Construction:
𝑟𝑜
Helix axis
𝑟𝑜
𝑳
𝑺
𝛼
𝐺𝑁𝐷 𝑝𝑙𝑎𝑛𝑒
L
S
S
𝛼
C
𝐶𝑜𝑎𝑥𝑖𝑎𝑙 𝑙𝑖𝑛𝑒
ONE turn helix
C
Parameters of helix:
𝑺
(1) 𝑳 =Length of one turn,
(2) Total length= N𝑳,
𝑺
(3) 𝑵 = no. of turns,
(4) 𝒉 = hight= axial length= NS,
𝑺
(5) 𝑺 = spacing between turns,
)6) 𝑪 = circumference of a helix,
𝑺
(7) 𝑟0 = helix radius,
(8) α pitch angle: is the angle formed by a line tangent to the helix wire and plane perpendicular to the helix axis.
𝑠
Relation of one helix = 𝑐 = 2𝜋𝑟𝑜 , 𝐿 = √𝐶 2 + 𝑠 2 , and 𝑡𝑎𝑛 𝛼 =
𝑐
If α = 0
The winding is flattened and the helix reduces to a loop antenna of N turns.
If α = 90°
The helix reduces to a linear wire.
Helical antenna “Modes of operations”
1. Normal mode: The radiation is normal to the
helix axis “broadside”.
*Occurs when L<<𝜆, 𝑠 ≪ 𝜆 and C<< 𝜆
*Used in mobile applications.
2. Axial mode: The radiation is to the axial direction
“end fire”.
*Occurs when L> 𝜆, S > 𝜆 and C > 𝜆
*Used in satellite applications
*Higher gain and higher bandwidth
Analysis of normal mode helical antenna
For normal mode operation the radiation from one helix can be considered as a radiation from
a small circular loop of radius 𝑟0 and short dipole length (S).
𝐸̅𝐻 =
𝐸̅𝐶 +
𝐸̅𝐷
Small Short Dipole
Circular
loop
Helical
𝑟0
S
𝛼 = 0 (circular loop)
̅𝐷
where 𝐸
=
𝑗𝜂𝛽𝐼0 𝑆
4𝜋𝑟
.
ˆ
𝑒 −𝑗𝛽𝑟 sin 𝜃 𝜃
α=900 (Short dipole)
𝐼0
and 𝑙 = 𝑆
short dipole
For small square loop of side length (𝑙):
2 𝐼 𝑙 2 𝑒 −𝑗𝛽𝑟
0
sin 𝜃
4𝜋
𝑟
𝜂𝛽
𝐸‾ =
𝜙ˆ,
𝑙 2 =area of a square loop
𝐼0
Small loop
Note that the relation between square loop field and short dipole is
𝐸𝜙 (Square loop) =
𝛽𝑙
𝑗
𝐸𝜃 (dipole) = −𝑗𝛽𝑙𝐸𝜃 (dipole)
For small circular loop of radius 𝒓𝟎 :
Replacing the square loop area, 𝑙 2 , by the circle loop area,𝜋𝑟02 gives:
𝐸‾𝐶
∴ 𝐸‾𝐻
𝜂𝛽 2 𝐼0 (𝜋𝑟02 ) −𝑗𝛽𝑟
𝑒
sin 𝜃 𝜙ˆ
4𝜋𝑟
𝑗𝜂𝛽𝐼0 𝑆 −𝑗𝛽𝑟
𝜂𝛽 2 𝐼0 (𝜋𝑟02 ) −𝑗𝛽𝑟
=
𝑒
sin 𝜃 𝜃ˆ +
𝑒
sin 𝜃 𝜙ˆ
4𝜋𝑟
4𝜋𝑟
=
In phasor domain 𝐸̅𝐻 has the form
𝐸‾𝐻 = |
𝜂𝛽𝐼0 𝑆
𝜂𝛽 2 𝐼0 (𝜋𝑟02 )
sin 𝜃| ∠(90 − 𝛽𝑟) 𝜃ˆ + |
sin 𝜃| ∠(−𝛽𝑟) 𝜙ˆ
4𝜋𝑟
4𝜋𝑟
The E-field of helical antenna has 𝐸𝐻𝜃 and 𝐸𝐻𝜙 components with not same magnitude and it has a 𝜋/2
phase difference. This Field form an Elliptical polarization.
Antenna polarization
Antenna polarization is defined by polarization of the EM wave which is defined as the direction at
which the Electric field vector 𝐸̅ is aligned (traced).
Linear polarization
Elliptical polarization
” Only one component of 𝐸̅ field vector”
“The two components of 𝐸̅ field vector has
ȁ𝐸𝜃 ȁ ≠ |𝐸𝜙 | but ∡𝐸𝜃 − ∡𝐸∅ = ± 𝜋⁄2
Vertically (V)
Horizontally (H)
(𝐸̅ ⊥ 𝐸𝑎𝑟𝑡ℎ)
(𝐸̅ ∥𝐸𝑎𝑟𝑡ℎ)
Condition of circular polarization for helical antenna
ȁ𝐸𝜃 ȁ = |𝐸𝜙 | and ∡ 𝐸𝜃 − ∡𝐸𝜙 = 𝜋/2
𝜂𝛽𝐼0 𝑆
𝜂𝛽 2 𝐼0 (𝜋𝑟02 )
∴|
sin 𝜃| = |
sin 𝜃|
4𝜋𝑟
4𝜋𝑟
2𝜋
∴ 𝑆 = 𝛽(𝜋𝑟02 ),
𝛽=
,
𝐶 = 2𝜋𝑟0
𝜆
2𝜋 2 𝑟02 (2𝜋𝑟0 )(𝜋𝑟0 ) (2𝜋𝑟0 ) 𝑐 𝑐 2
∴𝑆=
=
=
=
𝜆
𝜆
𝜆 2 2𝜆
∴ 𝐶 = √2𝑺𝜆 is the condition for circular polarization
Circular polarization
The two components of
̅
𝐸 field vector has same
magnitudeȁ𝐸𝜃 ȁ = |𝐸𝜙 |
and the phase difference
∡𝐸𝜃 − ∡𝐸∅ = ± 𝜋⁄2
Ex. Design a ten-turn helical antenna which at 500MHz operates in the normal mode. The spacing
between turn is 𝜆/40. It is desired that the antenna possesses circular polarization. Determine
i) circumference of the helix
ii) length of a single turn
iii) Overall length of the entire helix
iv) pitch angle in degrees.
Solving:
300×106
N=10 turns, F= 500MHz⇒ 𝜆= 500×106=0.6m
𝑆=
𝜆
0.6
=
= 0.015 m
40 40
𝜆
i) 𝐶 = √2𝑆𝜆 = √2 ( ) 𝜆 = √2(0.015)(0.6) = 0.134 m
40
ii) 𝐿 = √𝑆 2 + 𝐶 2 = √(0.015)2 + (0.134)2 = 
iii) 𝑁𝐿 = 10𝐿 =
𝑠
iv) 𝛼 = tan−1 𝑐
Antenna Array
The dipole antenna is very simple antenna is used when nearly omni directional pattern is required but
its gain is low.
When higher directivity or gain is required, several dipoles (or other elementary radiators) can be
arranged to form array and a directive beam of radiation can be obtained. A more directive beam means
that the antenna will also have a higher gain (𝐺 = 𝐷𝑔 ⋅ 𝑒)
Array Analysis
Consider an array (shown in fig.1) consists of N identical antenna element with the same operation but
excited with relative amplitude 𝑐𝑖 and phase 𝛼𝑖 for 𝒍̇ -𝒕𝒉 antenna
Z
.2
.1
d
d
.3 .
4
0
𝑐3 𝑒 𝑗𝛼3
𝑐1 𝑒 𝑗𝛼1
𝑐2 𝑒 𝑗𝛼2
.
5
… N.
𝑐𝑁 𝑒 𝑗𝛼𝑁
Y
*Excitation of short dipole and 𝝀⁄𝟐 dipole:
∵ 𝐸𝜃 =
𝑗𝜂𝛽𝐼0 𝑙 −𝑗𝛽𝑁 2
𝑒
sin⁡ 𝜃
4𝜋𝑟
short dipole
𝐸𝜃 can be put in the general form:
𝐸𝑖 = 𝑐𝑖𝑒 𝑗𝛼𝑖 𝑓(𝜃, 𝜙) ⋅
𝑒 −𝑗𝛽𝑅𝑖
,
𝑅𝑖
where 𝑓(𝜃, 𝜙) =
𝑗𝜂𝛽𝑙
sin⁡ 𝜃
4𝜋
and
𝐼0 = 𝑐𝑖 𝑒 𝑗𝛼𝑖 = 𝑐𝑖 ∠𝛼𝑖 is the excitation current
where 𝑓(𝜃, 𝜙) does not affected by dipole position similarly for 𝜆⁄2 dipole antenna :
𝑒 −𝑗𝑟𝑟 cos⁡(𝜋/2cos⁡ 𝜃)
𝑟
sin⁡ 𝜃
can take the form:
∵ 𝐸𝜃 = 𝑗602 𝐼𝑚
𝐸𝑖 = 𝐶𝑖 𝑒 𝑗𝛼𝑖 𝑓(𝜃, 𝜙) ⋅ 𝑒 −𝑗𝛽𝑅𝑖 ,𝐼𝑚 = 𝑐𝑖 𝑒 𝑗𝛼𝑖 = 𝑐1 ∠𝛼𝑖
𝑓(𝜃, 𝜙)describe the electric field 𝑅𝑖 radiation pattern of elementary antenna used in array, 𝐶 ≡ Current
amplitute, 𝑐 ≡ Current amplitude,𝛼 ≡ current phase.
∴ All types of Antennas have the same form but differ in 𝑓(𝜃, 𝜙)
∴ Total Electric ̅̅̅
𝐸𝑡 :
̅̅̅
̅̅̅
̅̅̅
𝐸𝑡 = 𝐸1 + 𝐸2 + ̅̅̅
𝐸3 + ⋯ + 𝐸𝑁
Where 𝐸1 = 𝐸𝑐 ,𝑒 𝑗𝛼1 ⁡𝑓(𝜃, 𝜙) = 𝑗60
cos⁡(𝜋/2cos⁡ 𝜃)
sin⁡ 𝜃
𝑒 −𝑗𝛽𝑅1
𝐸1 = 𝑐1 𝑒 𝑓(𝜃, 𝜙)
𝑅1
−𝑗𝛽𝑅2
𝑒
𝐸2 = 𝐶2 𝑒 𝑗𝛼2 𝑓(𝜃, 𝜙)
𝑅2
𝑒 −𝑗𝛽𝑅𝑁
𝐸𝑁 = 𝐶𝑁 𝑒 𝑗𝛼𝑁 𝑓(𝜃, 𝜙)
𝑅𝑁
𝑗𝛼1
In far field region
•
phase terms: 𝑅1 = 𝑟, 𝑅2 = 𝑟 − 𝑑cos⁡ 𝜓, 𝑅3 = 𝑟 − 2𝑑cos4, 𝑅4 = 𝑟 − 3𝑑cos⁡ 𝜓
•
Magnitude terms: 𝑅 ≈ 𝑅 ≈ 𝑅 ≈ ⋯ 𝑅 = 𝑟.
1
1
1
2
1
3
1
𝑁
1
𝑒 −𝑗𝛽𝑟
𝑒 −𝑗𝛽𝑟 𝑗𝛽𝑑cos⁡ 𝜓
+ 𝑐2 𝑒 𝑗𝛼2 𝑓(𝜃, 𝜙)
𝑒
𝑟
𝑟
⁡+𝑐3 𝑒 𝑗𝛼3 𝑓(𝜃, 𝜙)𝑒 −𝑗𝛽𝑟 𝑗𝛽𝛽𝑑𝛼0 𝜓
𝑒 −𝑗𝛽𝑟 𝑗2𝛽𝑑cos⁡ 𝜓
𝑒 −𝑗𝛽𝑟 𝑗(𝑁−1)𝛽𝑑 𝑐𝑜𝑠4
𝑔
+𝑐3 𝑒 𝑗𝛼3 𝑓(𝜃, 𝜙)
𝑒
+ ⋯ 𝑐𝑁 𝑒 𝑗𝛼𝑁 𝑓(𝜃, 𝜙)
𝑒
𝑟
𝑟
∴ 𝐸𝑡 = 𝑐1 𝑒 𝑗𝛼1 𝑓(𝜃, 𝜙)
Uniform Array
It is an array of equi-spaced elements that are fed with currents of equal magnitude and having a
progressive phase shift along the array.
∴ 𝐶1 = 𝐶2 = 𝐶3 = ⋯ = 𝐶𝑁 and 𝛼1 = 0, 𝛼2 = 𝛼, 𝛼3 = 2𝛼, 𝛼4 = 3α, … 𝛼N = (𝑁 − 1)𝛼
∴ 𝐸𝑡 = 𝐶𝑓(𝜃, 𝜙)
𝑒 −𝑗𝛽𝑟
𝑒 −𝑗𝛽𝑟 𝑗𝛽𝑑cos⁡ 𝜓
+ 𝐶𝑒 𝑗𝛼 𝑓(𝜃, 𝜙)
𝑒
⁡𝛼𝑁 ⁡(𝑁 − 1)𝛼
𝑟
𝑟
𝑒 −𝑗𝛽𝑟 𝑗2𝛽𝑑cos⁡ 𝜓𝑟
𝑒
+⋯
𝑟
𝑒 −𝑗𝛽𝑟 𝑗(𝑁−1)𝛽𝑑cos⁡ 𝜓
+𝑐𝑒 𝑗(𝑁−1)𝛼 𝑓(𝜃, 𝜙)
𝑒
𝑟
𝑒 −𝑗𝛽𝑟
= 𝑐𝑓(𝜃, 𝜙)
[1 + 𝑒 𝑗𝛼 𝑒 𝑗(𝛽𝑑cos⁡ 𝜓) + 𝑒 𝑗2𝛼 𝑒 𝑗2𝛽𝑑cos⁡ 𝜑 + 𝑒 𝑗(𝑁−1)𝛼 𝑒 𝑗(𝑁−1)𝛽𝑑cos⁡ 𝜓 ] + ⋯
𝑟
𝑒 −𝑗𝛽𝑟
= 𝐶𝑓(𝜃, 𝜙)
[1 + 𝑒 𝑗(𝛼+𝛽𝑑cos⁡ 𝜑) + 𝑒 𝑗2(𝛼+𝛽𝑑cos⁡ 𝜓)
𝑟
= (𝑓(𝜃, 𝜙)𝑒 −𝑗𝛽𝑟 + 𝑒 𝑗(𝑁−1)(𝛼+𝛽𝑑cos⁡ 𝜓) ]
𝑒 −𝑗𝛽𝑟
=
𝐶𝑓(𝜃, 𝜙)
[1
⏟+ 𝑒 𝑗𝑢 + 𝑒 𝑗2𝑢 + 𝑒 𝑗3𝑢 + ⋯ 𝑒 𝑗(𝑁−1)𝑢 ]
1
⏟
𝑟
Array⁡Factor⁡(AF)
+𝑐𝑒 𝑗2𝛼 𝑓(𝜃, 𝜙)
Field of one ⁡antenna⁡element
Where 𝑢 = 𝛼 + 𝛽𝑑Cos⁡ 𝜓, eq.1 represents the principle of pattern multiplication
and 𝐴𝐹 = 1 + 𝑒 𝑗𝑢 + 𝑒 𝑗2𝑢 + 𝑒 j3𝑢 + ⋯ + 𝑒 𝐽(𝑁−1)𝑢 → (2)
To put the summation of AF in closed form:
multiplying both sides of eq. (1) by 𝑒 𝑗𝑢
∴ 𝑒 𝑗𝑢 𝐴𝐹 = 𝑒 𝑗𝑢 + 𝑒 𝑗2𝑢 + 𝑒 𝑗3𝑢 + 𝑒 𝑗4𝑢 + ⋯ 𝑒 𝑗𝑁𝑢 → (3)
𝐸𝑞(3) − 𝐸𝑞 (2) gives
𝑒 𝑗𝑢 𝐴𝐹 − 𝐴𝐹 = −1 + 𝑒 𝑗𝑁𝑢
𝐴𝐹(𝑒 𝑗𝑢 − 1) = −1 + 𝑒 𝑗𝑁𝑢
𝑗𝑢
∴ 𝐴𝐹
𝑒 −1
⁡= 𝑗𝑢
=
𝑒 −1
𝑒𝑗
𝑁𝑢
𝑁𝑢
𝑗
2 |𝑒 2
𝑗𝑢
Magnitude
𝑢
𝑒 2 [𝑒 2 − 𝑒 −𝑗2 ]
𝑁𝑢
2)
=
𝑢
2𝑗sin⁡ (2)
𝑢
sin⁡ (𝑁 2)
𝑢
𝑗𝑁−1
2 ⁡⁡⁡⁡⁡⁡⁡⁡⁡⁡
𝑒
⏟
𝑢
Phase
⏟sin⁡ (2)
2𝑗sin⁡ (
𝑢
𝑒 𝑗(𝑁−1)2 ⁡⁡⁡⁡⁡⁡
𝑗𝑢
− 𝑒 −𝑗𝑁𝑢 |
Principle of pattern multiplication of eq.1:
The field pattern of array of certain
element can be obtained by multiplying the
field of single element by the array factor
𝑁𝑢
sin ( 2 )
∴ ⁡ |𝐴𝐹| =
𝑢
sin (2)
Main lobe “major lobe”
|𝐴𝐹|
. . . . .
−
4𝜋
𝑁
−
2𝜋
𝑁
0
. . . . .
2𝜋
𝑁
𝑢 = 𝛼 + 𝛽𝑑 cos 𝜓
4𝜋
𝑁
HPBW
1st side lobe
FNBW
Side (miner) lobes
2st side lobe
A sketch of |𝐴𝐹| versus 𝑢 is shown in Fig.2
Note that:
*FNBW≡ First Null Beam Width
*Increasing the number of elements N⇒ decreasing the beam width and increasing 𝐷0
|𝐴𝐹| in polar coordinate
N
Back lobe
Fig. 3. |𝐴𝐹| in polar coordinate corresponding to the sketch of Fig. 2
Note that:
*The max. of main lobe occurs at 𝑢 = 0
|𝐴𝐹|𝑢=0
𝑢
𝑁
𝑁𝑢
sin⁡ (𝑁 )
Cos⁡ ( )
2
2
2 =𝑁
= lim
lim
𝑢 = 𝑢→0
1
𝑢→0
sin⁡ (2)
Cos⁡(𝑢/2)
2
The zeros (nulls) are given by
𝑁𝑢
sin⁡ 2
𝑁𝑢
|𝐴𝐹| = 0 =
𝑢 ⇒ sin⁡ ( 2 ) = 0
sin⁡ 2
𝑁𝑢
= ±𝑛𝜋, 𝑛 = 1,2,3, … , 𝑁
2
2𝑛𝜋
,𝑢=±
, 𝑛 = 1,2,3, … , 𝑁
∴⁡
𝑁
The secondary maxima (side lobe) occur approximately midway between the nulls:
The amplitude of the first side lobe is
3𝜋
3𝜋
sin⁡ ( )
3𝜋
1
2 |=
𝑁
∴ |𝐴𝐹|𝑢 =
=|
|=|
𝑁 3𝜋
3𝜋
3𝜋
𝑁
sin⁡ (
)
sin⁡ ( )
|sin⁡ 2𝑁|
2 𝑁
2𝑁
3𝜋
When 2>N>3𝜋 ⇒N> 2 ⇒ N>4.7
𝑆𝑖𝑛⁡𝑥 ≈ 𝑥⁡𝑖𝑓⁡𝑥 < 1
∴ |𝐴𝐹|
𝑢=
3𝜋 2𝑁
=
𝑁 3𝜋
*The ratio between the first sidelobe to the main Lobe: Ratio =
2𝑁
/𝑁
3𝜋
2𝑁
2
= 3𝜋𝑁 = 3𝜋 = 0.212
This ratio is independent of 𝑁 as long as 𝑁 is large enough. Thus, it is not possible to reduce this side
lobe radiation relative to the main beam (below 0.212), no matter how many elements we put into the
array. On the other hand, it is possible to decrease the width of the main beam by increasing the
number of elements N.
Two modes of uniform array
(1) Board-Side array “normal mode “: The broad side array is a uniform array with max. at 𝜓 =
⁡𝜋⁄2(normal to the axis of array)
∴ 𝑢 = 𝛼 + 𝛽𝑑 cos 𝜓 = 𝛼 + 𝛽𝑑cos⁡ 𝜋⁄2 = 𝛼
and max. occurs at 𝑢 = 0 (in |𝐴𝐹| versus 𝑢, the maxi at 𝑢 = 0 )
𝑟
z
𝜓
.. . ..
𝜓 = 900
y
𝜋
2
∴ 0 = 𝛼 + 𝛽𝑑 cos ⇒ 𝛼 = 0⁡⁡𝛼 = 0 zero phase current
In broad side array;
N
The max. radiation of the signal antenna
element should be directed toward 𝜓 = 900
𝑢
∴ 𝑈 = 𝛽𝑑cos⁡ 𝜓 orcos⁡ 𝜓 𝛽𝑑
First null occurs at 𝒖 =
𝟐𝝅
𝑵
⇒𝝍=
𝑢 = 𝛽𝑑cos𝜓
2𝜋
𝜋
= 𝛽𝑑cos⁡ ( 2 + Δ𝜓)
𝑁
2𝜋
− 𝑁 = −𝛽𝑑sin⁡ Δ𝜓 (Chose-ve
2𝜋
2𝜋
∴ sin⁡ Δ𝜓 = 𝑁𝛽𝑑 , 𝛽 = 𝜆
−2𝜋 0 2𝜋
𝑁
𝑁
𝑢 = 𝛽𝑑 cos 𝜓
∴±
𝜆
𝑁𝑑
For 𝑁 is very large= Δ𝜓 < 1
∴ Δ𝜓 = 𝜆/𝑁𝑑
sin⁡ Δ𝜓 =
sign)
𝛥𝜓 𝛥𝜓
𝜓
𝜋
2
𝜋
+ ⁡Δ𝜓
2
∴ The width of the main beam width of brood side array (𝑀𝐵𝑊𝑏𝑠 )
is 𝑀𝐵𝑊𝑏𝑠 = 2Δ𝜓 =
2𝜆
𝑁𝑑
z
(2) End fire array: is uniform array with max.at “Axial mode”
𝜓=0
𝜓
𝑢 = 𝛼 + 𝛽𝑑 cos 𝜓
N
∴ 𝑢 = ⁡𝛼 + 𝛽𝑑 and the max. occurs at u=0
.
. ...
0=⁡𝛼 + 𝛽𝑑
−2𝜋 0 2𝜋
𝑁
𝑁
∴ 𝜶 = −𝜷𝒅
Phase current excitation
N
∴ 𝒖 = ⁡𝜷𝒅 𝐜𝐨𝐬 𝝍 − 𝜷𝒅
𝑢 = ⁡𝛽𝑑(cos⁡𝜓 − 1
2
2𝜋
±
𝑁
2𝜋
2𝑁𝛽𝑑
sin⁡ (
Δ𝜓
⁡= 𝛽𝑑(cos⁡ Δ𝜓 − 1)
⁡= sin2 ⁡
Δ𝜓
( choose - ve )
2
Δ𝜓
2𝜋
𝜆
) ⁡= √
=√
2
2𝑁𝛽𝑑
2𝑁𝑑
∴
Δ𝜓
2
𝜆
𝜆
2𝜆
⁡= √
⇒ Δ𝜓 = 2√
=√
2𝑁𝑑
2𝑁𝑑
𝑁𝑑
2𝜆
∴ 𝑀𝐵𝑊𝑏𝑠 = 2Δ𝜓 = 2√
𝑁𝑑
𝜓
0
at 𝑢 = 𝑁𝜋 ⇒ 𝜓 = Δ𝜓
Δ𝜓 Δ𝜓
+
) ⁡= cos⁡(Δ𝜓)
2
2
Δ𝜓
Δ𝜓
cos2 ⁡
− sin2 ⁡
⁡= cos(Δ𝜓) ∗
2
2
Δ𝜓
Δ𝜓
cos2 ⁡
+ sin2 ⁡
⁡= cos 0 = 1 ∗∗
2
2
Δ𝜓 ⁡
∗∗ − ∗= 2sin2 ⁡
= 1 − cos⁡(Δ𝜓)
2
Δ𝜓
= 2sin2 ⁡
= 1 − cos⁡(Δ𝜓)
Δ𝜓
2
−2sin2 ⁡
= cos⁡(Δ𝜓) − 1
2
Δ𝜓
−2sin2 ⁡
= cos⁡(Δ𝜓) − 1
2
cos⁡ (
Is the main beam width of end fire array
b
Comparing a with b⇒an array of the same length (same Nd) end fire has a broader width of main beam
than does the broadside array.
𝜓=0
y
Problems
P.1 Design a ten-turn helical antenna which at 500 MHz operates in the normal mode. The spacing
𝜆
between turns is 40. It is desired that the antenna possesses circular polarization. Determine.
(a) circumference of the helix
(b) length of a single turn
(c) overall length of the entire helix
(d) pitch angle in degrees.
P.2 An array of 10 isotropic elements is placed along the x-axis a distance 𝜆⁄4 apart. Assume uniform
distribution.
find the (1) progressive phase (in degrees),
(2) the ratio of amplitudes of the main beam to the first side lobe in dB and the (3) main beam for
(a) broad side array
(b) end-fire array