Dipole Antenna / Hertzian dipole

Dipole Antenna / Hertzian dipole
January 2, 2013
1. The A
2. The B and H
3. The E
4. Far field approximation
5. The S , P rad and R rad
• The Hertzian dipole does not exists in real life, but the real antenna can be formed by taking
integration on Hertzian dipole.
Current element / dipole carry dynamic current
dl ≤ λ 10
, it is very small
i(t) = I 0 cos ωt , the current on the dipole
It is difficult to solve for the B and E directly.
Use Retarded Potential : V (x, y, z, t) −→ V ( )
x, y, z, t − r c , so the retarded current is
i(t) = I 0 cos ω ( t − c) r = I0 cos(ωt − ωr) = I c 0 cos(ωt − βr)
Retarded : time delay by r c or phase delay by βr 1

1 The retarded magnetic vector potential
In phasor form
A(r, t) = µ[i(t)]dl
4πr
ẑ = µI0 cos ( t − ) r dl
c ẑ
4πr
Ã(r) = µI 0dl
4πr e−jβr ẑ
( Recall ) Phasor transformation : V 0 cos (ωt + ϕ) = Re ( V 0 e j(ωt+ϕ)) = Re
⎛
⎞
⎝V 0 e
} {{ jϕ e
}
jωt ⎠
P hasor
For radiation by antenna, it is more convenient to select spherical coordinate system
( Recall ) The coordinate transform of rectangular to spherical V (x, y, z) → V (r, θ, ϕ)
Given vector ¯V = V xˆx + V y ŷ + V z ẑ , the vector representation in spherical coordinate is
ˆV = (V x sin θ cos ϕ + V y sin θ sin ϕ + V z cos θ) ˆr+(−V x sin ϕ + V y cos ϕ) ˆϕ+(V x cos θ cos ϕ + V y cos θ sin ϕ − V z sin θ) ˆθ
Since Ã(r) = µI 0
4πr dl e−jβr ẑ = A z ẑ only, A x = A y = 0 , thus
à = (A z cos θ) ˆr + (−A z sin θ) ˆθ
= µI 0dl
4πr cos θe−jβrˆr − µI 0dl
4πr sin θe−jβr ˆθ
( )
cos θ
= µI 0dl
4π
r
e−jβrˆr − sin θ
r
e−jβr ˆθ
• A(r, θ) is ϕ independent ⇐⇒
∂
∂ϕ = 0
• A = A rˆr + A θ ˆθ ⇐⇒ Aϕ = 0
2 B and H
Find B by B = ∇ × A
˜B = ∇ × Ã = ∣ ∣∣∣∣∣∣∣∣
∣
ˆr ˆθ ˆϕ ∣∣∣∣∣∣∣∣
r 2 sin θ r sin θ r
∂ ∂ ∂
∂r ∂θ ∂ϕ
A r rA θ r sin θA ϕ
Since A is ϕ independent ⇒ ∂
∂ϕ = 0 , and A ϕ = 0
ˆr ˆθ ˆϕ
∣ r
˜B 2 sin θ r sin θ r
=
∂ ∂
= 1 ∂ ∂ ∣∣∣∣ ( 1 ∂rAθ
0
r ∣
∂r ∂θ ˆϕ = ∂r ∂θ
A r rA θ
r ∂r
∣ A r rA θ 0 ∣
2
− ∂A )
r ˆϕ
∂θ

= µI 0dl
4πr
= 1 µI 0 dl
r 4π
(sin θjβ + e−jβr
(
− ∂ ∂r sin θe−jβr − ∂ )
cos θ
∂θ r e−jβr ˆϕ
)
sin θ ˆϕ = µI (
0dl
jβ
r
4π sin θe−jβr r + 1 )
ˆϕ
r 2
Thus, by H = B µ
˜H = I 0dl
4π sin θe−jβr ( jβ
r + 1 r 2 )
ˆϕ
• H r = H θ = 0
• H(r, θ) is ϕ independent ⇐⇒
∂
∂ϕ = 0
3 The E
Find E by ε ∂E
∂t = ∇ × H
In phasor form, εjωẼ = ∇ × ˜H, thus
Since H is ϕ independent ,
Ẽ = 1
jωε
∣
Ẽ = 1
jωε ∇ × ˜H = 1
jωε
∣
∂
∂ϕ = 0 , and H r = H θ = 0
∣
ˆr ˆθ ˆϕ ∣∣∣∣∣∣∣∣
(
r 2 sin θ r sin θ r
∂ ∂
= 1
0 jωε
∂r ∂θ
0 0 r sin θH ϕ
H ϕ = I (
0dl
jβ
4π sin θe−jβr r + 1 )
r 2
and thus
∂r sin θH ϕ
∂θ
∣
ˆr ˆθ ˆϕ ∣∣∣∣∣∣∣∣
r 2 sin θ r sin θ r
∂ ∂ ∂
∂r ∂θ ∂ϕ
H r rH θ r sin θH ϕ
ˆr
r 2 sin θ
∂r sin θH ϕ
∂θ
, so r sin θH ϕ = I 0dl
4π sin2 θe
(jβ −jβr + 1 )
r
= I 0dl
(jβ
4π 2 sin θ cos θe−jβr + 1 )
r
−
ˆθ
r sin θ
∂r sin θH ϕ
∂r
)
∂r sin θH ϕ
∂r
= I (
0dl
4π sin2 θ
[−jβe −jβr jβ + 1 ) ( )] −1
+ e −jβr = I 0dl
r
r 2 4π sin2 θ
[β 2 e −jβr − jβe−jβ
r
− e−jβr
r 2 ]
And therefore
ˆr ∂r sin θH ϕ
r 2 sin θ ∂θ
= I 0dl
2π cos θe−jβr ( jβ
r 2 + 1 r 3 )
ˆr
3

ˆθ ∂r sin θH ϕ
r sin θ ∂r
= I [
0dl β 2
4π sin θ e −jβr
r
− jβe−jβr
r 2
− e−jβr
r 3 ]
ˆθ
= 1
ωε
√ µ
Since η =
ϵ = β ωϵ
• E ϕ = 0
(
Ẽ = 1
jωε
ˆr
r 2 sin θ
∂r sin θH ϕ
∂θ
−
ˆθ
r sin θ
∂r sin θH ϕ
∂r
(
I 0 dl
β
2π cos θe−jβr r − j )
ˆr − 1
[
I 0 dl
β
2
2 r 3 ωε 4π sin θe−jβr jr − β r − 1 ]
ˆθ
2 jr 3
Ẽ = η I 0dl
2π cos θe−jβr ( 1
r 2 −
j )
ˆr + η I [
0dl
jβ
βr 3 4π sin θe−jβr r + 1 r − 2
)
j ]
ˆθ
βr 3
• E is ϕ independent ⇐⇒
∂
∂ϕ = 0
4 Far Field Approximations
Therefore the fields are
Ẽ = η I 0dl
2π cos θe−jβr ( 1
r 2 −
˜H = I 0dl
4π sin θe−jβr ( jβ
r + 1 r 2 )
ˆϕ
j )
ˆr + η I [
0dl
jβ
βr 3 4π sin θe−jβr r + 1 r − 2
When r is large ( far field ) , ignore all the term with power ≥ 2
j ]
ˆθ
βr 3
In time domain
˜H(r, θ) = I 0dl
4π sin θe−jβr jβ r ˆϕ
Ẽ(r, θ) = η I 0dl
4π sin θ jβ r ˆθ
H(r, θ, t) = − I 0βdl
4πr
sin θ sin (ωt − βr) ˆϕ Ẽ
= −η I 0βdl
4πr
sin θ sin (ωt − βr) ˆθ
5 Poynting Vector S , S avg , P rad and R rad
The Poynting vector S in time domain is
S = E × H = η I2 0β 2 dl 2
16π 2 r 2 sin2 θ sin 2 (ωt − βr) ˆr
The time-average Poynting vector in phasor domain is
˜S avg = 1 ] [Ẽ
2 Re × ˜H∗ = 1 2 Re [ ]
E θ Hϕ
∗ 1 ˆr =
2 η|H ϕ| 2ˆr = η I2 0β 2 dl 2
32π 2 r 2 sin2 θˆr
Thus the time-average radiated power is
4

( ) 2 2π
• β 2 = = 4π2
λ
• ´ 2π
dϕ = 2π
ϕ=0
‹
ˆ 2π ˆ π
( ηI
2
P rad = ˜S avg · d¯S =
0 β 2 dl 2
S
ϕ=0 θ=0 32π 2 r 2
λ 2
= ηI2 0β 2 dl 2
32π 2
ˆ 2π
• ´ π
θ=0 sin3 θdθ = − ´ π
θ=0 (1 − cos 2 θ) d cos θ = −
P rad =
ηI 2 0
Special case, in free space, η 0 ≈ 120π
ϕ=0
ˆ π
dϕ sin 3 θdθ
θ=0
[
] π
cos θ − cos3 θ
3
0
( ) 4π
2
dl 2
λ 2 2π · 4
32π 2 3 = ηI2 0π
3
( ) 2 dl
P rad = I040π 2 2 λ
) (r
sin 2 θ
2 sin θdθdϕ )
[
= − −2 − −2 ]
= 4 3 3
( ) 2 dl
λ
Since P = I 2 R (for static) P = 1 2 I2 0R ( for dynamic ), thus the fictitious resistance, the radiation
resistance R rad ( in free space ) is
Where the general radiation resistance
( ) 2 dl
R rad = 80π 2 λ
R rad = 2ηπ
3
( ) 2 dl
λ
5