Chapter 4 : Linear Wire Antenna

• Infinitesimal Dipole
• Small Dipole
• Finite Length Dipole
• Half-Wavelength Dipole
• Linear Elements near or on Infinite Perfect
Conductors

1
 Infinitesimal Dipole
• Length l << λ
• Used to represent capacitor-plate (top-hat-
loaded) antennas
• Capacitive loading to maintain the uniform
current

2
 Radiated Field
• The current on the infinitesimal dipole is
assumed to be constant, i.e.,
I( z ' ) = zˆI 0
• Recall that
µ e − jkR
A ( x, y , z ) =
4π ∫C
J ( x' , y ' , z ' )
R
dl '

• Since the dipole is infinitesimal, the following

approximations hold:
x' = y ' = z ' = 0 dl ' = dz '
r r
R =| r − r ' |= r = constant
3
 Radiated Field (2)
Thus µI 0 − jkr l / 2 µI 0l − jkr
A( x, y, z ) = zˆ e ∫ dz ' = zˆ e
4πr − l / 2 4πr
Since A has only z component,
Ar = Az cos θ ; Aθ = − Az sin θ ; Aφ = 0
and
rˆ  ∂ ∂Aθ  θˆ  1 ∂Ar ∂ 
∇× A =  ( Aφ sin θ ) −  +  − (rAφ )
r sin θ  ∂θ ∂φ  r  sin θ ∂φ ∂r 
φˆ  ∂ ∂Ar 
+  (rAθ ) −  ,
r  ∂r ∂θ 
The magnetic field becomes:
1 ∂ ∂Ar 
H =φ ˆ
 (rAθ ) −
µr  ∂r ∂θ 
4
 Radiated Field (3)
Therefore, H r = Hθ = 0
kI 0l sin θ  1  − jkr
Hφ = j 1 + jkr  e
4πr  
Likewise, the electric field can be found to be
I 0l cos θ 1  − jkr
Er = η 1 + jkr  e
2πr 2  
kI 0l sin θ  1 1  − jkr
Eθ = jη 1+ − 2
e
4πr  jkr (kr ) 
Eφ = 0
5
 Power Density and
Radiation Resistance
1 1
• Poynting vector W = (E × H ) = (rˆEr + θˆEθ ) × (φˆH φ* )
*

2 2
1
= (rˆEθ H φ* − θˆEr H φ* )
2
2
η I 0l sin 2 θ  1 
Wr = 1 − j
Components of 8 λ r 2  (kr ) 3 
Poynting Vector k | I 0l |2 cos θ sin θ  1 
Wθ = jη 1 +
 ( kr ) 2 
16π 2 r 3  
2π π
P = ∫∫ W ⋅ ds = ∫ ∫ ( rˆWr + θˆWθ ) ⋅ rˆr 2 sin θdθdφ
0 0
S
Outgoing Power 2
2π π π I 0l  1 
=∫ ∫0 r
W r 2
sin θd θd φ = η 1 − j 3
0 3 λ  ( kr ) 
6
 Power Density and
Radiation Resistance (2)
P = power
Complex Power Prad = time - average radiated power
~
~ ~ Wm = time - average magnetic energy
P = Prad + j 2ω (Wm − We ) ~
We = time - average electric energy
~ ~
2ω (Wm − We ) = time - average imaginary
(reactive) power; All in radial direction
2 2 2
π I 0l 1 2π l 2 l 
Prad =η = | I 0 |2 Rr Rr = η   = 80π  
3 λ 2 3 λ  λ 
Radiation Resistance
2
~ ~ π I 0l 1
2ω (Wm − We ) = −η Reactive Power
3 λ (kr ) 3
7
 Far Field (kr >> 1)
• For kr >> 1, the fields can be approximated as
kI 0l sin θ − jkr 
Eθ ≅ jη e 
4πr

kI 0l sin θ − jkr 
Hφ ≅ j e kr >> 1 TEM Wave
4πr 
Er ≅ Eφ = H r = H θ = 0


Eθ
Ratio of E and H: Zw = =η
Hφ
Z w = wave impedance
η = intrinsic impedance (120π Ω for free - space)
8
 Directivity
Time-average power density:
2
1 1 η kI 0l sin 2 θ
Wav = Re(E × H ) = rˆ
*
| Eθ | = rˆ
2

2 2η 2 4π r2
Radiation intensity:
2
η kI l r 2
U = r 2Wav = 0
sin 2 θ = | Eθ (r ,θ , φ ) |2
2 4π 2η 2
η kI 0l
Maximum radiation intensity: U max =
2 4π
U max 3
Maximum directivity: D0 = 4π =
Prad 2
λ2 3λ2
Maximum effective area: Aem = D0 =
4π 8π
9
 Small Dipole
• Length λ/50 < l < λ/10

10
 Radiated Field
• The current on the small dipole is assumed to
be a triangular function, i.e.,
 2
 zI 0 (1 − l z ' ),
ˆ 0 ≤ z' ≤ l / 2
I( z ' ) = 
 zˆI (1 + 2 z ' ), −l / 2 ≤ z ' ≤ 0
 0 l
where I0 is a constant. Vector potential becomes:
µ  0 2 e − jkR
A ( x, y, z ) = zˆ  ∫−l / 2 I 0 (1 + z ' ) dz '
4π  l R
l/2 2 e − jkR 
+∫ I 0 (1 − z ' ) dz ' 
0 l R 
Approximating R ~ r yields the maximum phase error
kl/2 = π/10 for l=λ/10.
11
 Radiated Field (2)
Using R~r: 1 µI 0l − jkr
A( x, y, z ) = zˆAz = zˆ e
2 4πr
which is one-half of that for the infinitesimal dipole.
The far-field can be given by
kI 0l sin θ − jkr 
Eθ ≅ jη e 
8πr

kI 0l sin θ − jkr 
Hφ ≅ j e kr >> 1
8πr 
Er ≅ Eφ = H r = H θ = 0


2
2 Prad 2 l 
Radiation resistance: Rr = = 20π  
λ
2
| I0 |
12
 Field Separation
• For a very thin dipole, x’=y’=0, thus
R = ( x − x' ) 2 + ( y − y ' ) 2 + ( z − z ' ) 2 = x 2 + y 2 + ( z − z ' ) 2
= ( x 2 + y 2 + z 2 ) + (−2 zz '+ z '2 ) = r 2 + (−2rz ' cos θ + z '2 )
where r 2 = x 2 + y 2 + z 2 ; z = r cos θ

13
14
 Field Separation (2)
Recall also the Taylor expansion:
f ' ' (0) 2
f ( x) = f (0) + f ' (0) x + x +L
2!
which yields the same result:
x x 2 x3
(1 + x)1/ 2 = 1+ − + L
2 8 16
Recall that
1
R = r 1 + 2 (−2rz ' cos θ + z '2 )
r
− 2rz ' cos θ + z '2
Let x = 2
, then
r
1  z '2  1  z '3

R = r − z ' cos θ +  sin θ  + 2  cos θ sin θ  + L
 2
  2

r 2  r  2 
15
 Far Field
By retaining only the first two terms, i.e.,
R = r − z ' cos θ
The most significant neglected term has the maximum
value
1  z '2  z ' 2
π
 sin 2 θ  = when θ =
r 2  max 2r 2
A maximum total phase error of π/8 is acceptable, thus
z '2 π − l / 2 ≤ z' ≤ l / 2 2l 2
k ≤ r≥
2r 8 λ
Far-field approximation
2D 2 R ≈ r for amplitude term
when r ≥ r
λ R ≈ r − r ' cos θ = r − rˆ ⋅ r ' for phase term
16
 Radiating Near Field
By retaining only the first three terms, i.e.,
1  z '2 2 
R = r − z ' cos θ +  sin θ 
r 2 
The most significant neglected term is the fourth term. In
order to find its maximum value, one can differentiate the
fourth term with respect to θ, and the result is set to 0, i.e.,
∂  1  z '3  z '3
∂θ
 2 cos θ sin 2
θ 
  = 2
sin [
θ − sin 2
θ + 2 cos 2
]θ =0
r  2   2r
yields

[− sin 2
θ + 2 cos θ ]
2
θ =θ1 =0 θ1 = tan −1 (± 2 )

17
 Radiating Near Field (2)
If the maximum total phase error is allowed to be π/8,

kz '3 π l 3  1  2  π  l3  π
cos θ sin 2
θ =    =  2  ≤
2r 2
z '=l / 2 λ 8r  3  3  12 3  λr  8
2
θ = tan −1 θ

which reduces to 2  l 3

r ≥
2
 
3 3λ
or
l3
r ≥ 0.62
λ
2D 2 D3
Near-field region ≥ r ≥ 0.62
λ λ
18
 Finite Length Dipole
• Length > λ/10
• Current on the finite length dipole assuming
the wire is very thin
  l 
 zI 0 sin k ( 2 − z ' ),
ˆ 0 ≤ z' ≤ l / 2
  
I e ( z' ) = 
 zˆI sin k ( l + z ' ), −l / 2 ≤ z ' ≤ 0
 0  2 


where I0 is a constant. This distribution assumes

that the antenna is center-fed and the current
vanishes at the end points.

19
 Radiated Field
The electric and magnetic field components in
the far field for the infinitesimal dipole dz’ are
given by kI e ( z ' ) sin θ − jkR
dEθ ≅ jη e dz '
4πR
kI e ( z ' ) sin θ − jkR
dH φ ≅ j e dz '
4πR
dEr ≅ dEφ = dH r = dH θ = 0
Using the far-field approximation yields
kI e ( z ' ) sin θ − jkr jkz 'cosθ
dEθ ≅ jη e e dz '
4πr

20
 Radiated Field (2)
The total electric field can be obtained by
summing up contributions from all infinitesimal
dipoles, i.e., − jkr
l/2 sin θ  l / 2
ke 
Eθ = ∫ dEθ = jη
 ∫−l / 2
jkz 'cosθ
I ( z ' ) e dz '

π 3 144424443
e
−l / 2 4 r
144244
element factor space factor

total field = (element factor) × (space factor)

Thus, the electric field of the finite length dipole
can be given by
kI 0e − jkr sin θ  0  l  jkz 'cosθ
Eθ = jη ∫−l / 2 sin k ( + z ' ) e dz '
4πr   2 
l/2  l  jkz 'cosθ 
+∫ sin k ( − z ' ) e dz '
0
 2  
21
 Radiated Field (3)
Using
αx
e
∫ sin( βx + γ )dx = α 2 + β 2 [α sin( βx + γ ) − β cos(βx + γ )]
αx
e

yields   kl   kl  
− jkr  cos cos θ  − cos  
    2 
I 0e 2
Eθ ≅ jη
2πr  sin θ 
 

Likewise, the magnetic field can be given by

  kl   kl  
− jkr  cos cos θ  − cos  
    2 
Eθ I 0e 2
Hφ ≅ = j
η 2πr  sin θ 
 

22
 Current Distributions and
Radiation Pattern

23
 Radiation Pattern for l=1.25λ

24
 Power Density and
Radiation Intensity
Time-average power density:
1 1
Wav = Re(E × H * ) = Re(θˆEθ × φˆH φ* )
2 2
2
  kl   kl  
2  cos cos θ  − cos  
| Eθ |2 = rˆη 02 2     2 
1 | I | 2
= rˆ
2η 8π r  sin θ 
 
Radiation intensity:
2
  kl   kl  
2  cos cos θ  − cos  
U = r 2Wav = η 0 2     2 
|I | 2
8π  sin θ 
 

25
 Radiated Power
Radiated power can be obtained by
2π π
Prad = ∫∫ UdΩ = ∫ ∫ U sin θdθdφ
Ω 0 0
2
  kl   kl 
| I 0 |2 cos 2 cos θ  − cos 2 
π     
=η
4π ∫
0 sin θ
dθ

which can be given by

| I 0 |2
 1
Prad =η C + ln(kl ) − Ci (kl ) + sin(kl )[ Si (2kl ) − 2 Si (kl )]
4π  2 (1)
1 
+ cos(kl )[C + ln(kl / 2) + Ci (2kl ) − 2Ci (kl )]
2 
where
C = 0.5772156649 (Euler's constant)
26
 Radiated Power (2)
∞ cos y x cos y
Ci ( x ) = − ∫ dy = ∫ dy Cosine integral
x y ∞ y
x sin y
Si ( x) = ∫ dy Sine integral
0 y 5

Ci(x) is related to Cin(x) by 4

Cin ( x) = C + ln( x) − Ci ( x)
2

 1 − cos y 
x 0

where Cin ( x) = ∫  dy −1

0
 y  −2
Ci(x)
−3
Si(x)

−4 Cin(x)

−5
0 5 10 15 20 25 30 35 40 45 50

27
 Radiation Resistance and
Input Resistance
Radiation resistance becomes
2P η  1
Rr = rad2 = C + ln( kl ) − C ( kl ) + sin(kl )[ Si (2kl ) − 2 Si (kl )]
2π 
i
| I0 | 2
1 
+ cos(kl )[C + ln(kl / 2) + Ci (2kl ) − 2Ci (kl )]
2 
2 2
| I | | I |
Since in R = 0 R assuming loss-less
in r 2
2 2 I 
Input resistance can be given by Rin =  0  Rr
 kl   I in 
For a dipole of length l, I in = I 0 sin  
2
Rr
Rin =
Input Resistance 2  kl 
sin  
2
28
 Directivity
Directivity is given by
U max 2 F (θ ) max
D0 = 4π =
Prad Q
where 2
  kl   kl  
 cos 2 cos θ  − cos 2  
F (θ ) =     
 sin θ 
 
and
 1
Q = C + ln(kl ) − Ci (kl ) + sin(kl )[ Si (2kl ) − 2 Si (kl )]
 2
1 
+ cos(kl )[C + ln(kl / 2) + Ci (2kl ) − 2Ci (kl )]
2 
29
 Radiation resistance, input
resistance and directivity

30
 Half-wavelength Dipole
Let l = λ/2, then
 π   π 
− jkr  cos cos θ   − jkr  cos cos θ  
Eθ ≅ jη
I 0e
  2  Hφ ≅
Eθ
= j
I 0e
 
2 
2πr  sin θ  η 2πr  sin θ 
   
2
 π 
2  cos cos θ   2
Power density Wav = η 2 2     ≅ η 0 sin 3 θ
| I0 | 2 | I |
8π r  sin θ  8π 2 r 2
 
2
Radiation Intensity  π 
2  cos cos θ   2
U = r 2Wav = η 0 2     ≅ η 0 sin 3 θ
| I | 2 | I |
8π  sin θ  8π 2
 
31
 Half-wavelength Dipole (2)
π 
Radiated Power cos  cos θ 
2
| I 0 |2 π 2  dθ
Prad = η
4π ∫0 sin θ (2)
| I 0 |2 2π  1 − cos y 
2
| I0 |
=η
4π ∫0  y dy =η 8π Cin (2π )
where Cin (2π ) = C + ln(2π ) − Ci (2π ) ≅ 2.435
U max 4
Directivity D0 = 4π = ≅ 1.643
Prad 2.435
2 Prad η
Radiation Resistance Rr = = Cin (2π ) ≅ 73
| I0 | 2
4π
Input Impedance Z in = 73 + j 42.5
32
 Wire antennas near or on infinite
perfect conductor

33
 Image Theory

Etan = 0 → nˆ × E = 0
on PEC

H tan = 0 → nˆ × H = 0
on PMC

NOTE: The fields

obtained are valid
only in the top
half-plane.
34
 Vertical Electric Dipole

Reflected Component
− jkr2
kI le
Direct Component Eθr = jRvη 0 sin θ 2
− jkr1 4πr2
kI le
Eθd = jη 0 sin θ1
4πr1 kI 0le − jkr2
= jη sin θ 2
4πr2
35
 Vertical Electric Dipole (2)
In far field:

Parallel ray
approximation

r1 = [r 2 + h 2 − 2rh cos θ ]2 ≅ r − h cos θ

Phase term
r2 = [r 2 + h 2 + 2rh cos θ ]2 ≅ r + h cos θ
r1 ≅ r2 ≅ r Amplitude term
36
 Vertical Electric Dipole (3)
kI 0le − jkr
Total Electric Eθ ≅ jη sin θ [2 cos(kh cos θ )] z ≥ 0
Field 4πr
=0 z<0
2h
number of lobes ≅ +1
λ

37
 Vertical Electric Dipole (4)
Power Density 2
1 1 η I 0l
Wav = Re(E × H ) = r
*
ˆ | Eθ | = r 2
2
ˆ sin 2 θ cos 2 (kh cos θ )
2 2η 2r λ
Radiation Intensity 2
r2 2 η I 0l
U= | Eθ | = sin 2 θ cos 2 (kh cos θ )
2η 2 λ
2
Maximum Radiation Intensity η I 0l
U max =
2 λ
Radiated Power
2π π /2 π /2
Prad = ∫ ∫ U sin θdθdφ = 2π ∫ U sin θdθ
0 0 0
2 (3)
I 0l  1 cos(2kh) sin(2kh) 
= πη  − +
λ  3 (2kh) 2
(2kh) 3 
38
 Vertical Electric Dipole (5)
−1
Directivity U max  1 cos(2kh) sin(2kh) 
D0 = 4π = 2 − 2
+ 3 
Prad  3 ( 2 kh ) ( 2 kh ) 
Radiation Resistance 2
2 Prad l  1 cos(2kh) sin(2kh) 
Rr = = 2πη    3 − (2kh) 2 + (2kh) 3 
λ 
2
| I0 |  

39
 λ/4 Monopole

1
Z in = Z in (dipole)
2
1
= (73 + j 42.5)
2
= 36.5 + j 21.25

40
 Fields due to y-directed dipole
µI 0le − jkr
Vector Potential A A = yˆ
4πr
µI 0le − jkr
Aθ = Ay cos θ sin φ = cos θ sin φ
4πr
− jkr
µI 0le
Aφ = Ay cos φ = cos φ
4πr
r r j r
Recall that E(r ) = − jωA − ∇(∇ ⋅ A)
ωµε
µI 0le − jkr
Eθ ≅ − jωAθ = − jω cos θ sin φ
Far-field Electric Field 4πr
µI 0le − jkr
Eφ ≅ − jωAφ = − jω cos φ
4πr
41
 Fields due to y-directed dipole (2)
Introduce a “new” spherical coordinate system (r,ψ,ζ) such that
rˆ = zˆ sinψ cos ζ + xˆ sinψ sin ζ + yˆ cosψ
ψˆ = zˆ cosψ cos ζ + xˆ cosψ sin ζ − yˆ sinψ
ζˆ = − zˆ sinψ + xˆ cosψ
(This can be obtained by letting (x,y,z)->(z,x,y) and (ψ,ζ)->(θ,φ))
then
− jkr
µI 0le
Far-field Electric Field Eψ ≅ jω sinψ
4πr
− jkr
ωµI 0le
Far-field Magnetic Hζ ≅ j sinψ
Field 4πηr
42
 Horizontal Electric Dipole

Reflected Component
− jkr2
kI le
Direct Component Eψr = jRhη 0 sinψ
− jkr1 4πr2
kI le
Eψd = jη 0 sinψ kI 0le − jkr2
4πr1 = − jη sinψ
4πr2
43
 Horizontal Electric Dipole (2)
In far field:

r1 = [r 2 + h 2 − 2rh cos θ ]2 ≅ r − h cos θ Phase term

r2 = [r 2 + h 2 + 2rh cos θ ]2 ≅ r + h cos θ
r1 ≅ r2 ≅ r Amplitude term
cosψ = yˆ ⋅ rˆ = sin θ sin φ
44
 Horizontal Electric Dipole (3)
Total Electric kI 0le − jkr
Eψ ≅ jη 1 − sin 2 θ sin 2 φ [2 j sin(kh cos θ )]
Field 4πr

2h
number of lobes ≅
λ

45
 Horizontal Electric Dipole (4)
Power Density 2
η I 0l
Wav = rˆ 2 (1 − sin 2 θ sin 2 φ ) sin 2 (kh cos θ )
2r λ
Radiation Intensity
2
r2 η I l
U= | Eθ |2 = 0
(1 − sin 2 θ sin 2 φ ) cos 2 (kh cos θ )
2η 2 λ
Radiated Power
2π π /2 π /2
Prad = ∫ ∫ U sin θdθdφ = 2π ∫ U sin θdθ
0 0 0
2
π I 0l  2 sin(2kh) cos(2kh) sin(2kh) 
=η  − − +
2 λ 3 2kh (2kh) 2
(2kh) 3 
Radiation Resistance
2
2 Prad    2 sin(2kh) cos(2kh) sin(2kh) 
l
Rr = = πη    − − +
| I0 | 2
 λ  3 2kh (2kh) 2
(2kh) 3 
46
 Horizontal Electric Dipole (5)
For small kh
 l   2 2 8  2πh  
2 2 2 2
kh →0
32π 3  l   h 
Rr = πη    − +    =η    
 λ   3 3 15  λ   15  λ   λ 
Maximum Radiation Intensity
η I 0l 2 2
 sin (kh) kh ≤ π / 2, θ = 0
2 λ
U max =  2
η I l
 2 λ kh > π / 2, φ = 0 and sin( kh cos θ max ) = 1
0


Directivity  4 sin 2 (kh)
 kh ≤ π / 2
 R (kh)
D0 = 
 4 kh > π / 2
 R (kh)
47
 Horizontal Electric Dipole (6)
 2 sin(2kh) cos(2kh) sin(2kh) 
where R(kh) =  − − 2
+ 3 
 3 2 kh ( 2 kh ) ( 2 kh ) 
For small kh −1 2
kh →0
2 2 8   sin(kh) 
D0 = 4 sin 2 (kh)  − + (kh) 2  = 7.5 
 3 3 15   kh 

48