Wire Antennas: Maxwell’s equations

E1

Maxwell’s equations govern radiation variables:

( ) ( )

( )D

B

(ε is permittivity; εo = 8.8542 × 10-12 farads/m for vacuum) (µ is permeability; µo = 4π • 10-7 henries/m) (1 Tesla = 1 Weber m-2 = 104 gauss)

Lec 09.6- 1 1/11/01

1­m v field electric E

1­m a field magnetic H

2 ­m Coulombs nt displaceme electric E ε =

(Teslas) density flux magnetic H µ =

Maxwell’s Equations: Dynamics and Statics

E2

t BE ∂ ∂

t DJH ∂ ∂

+=

D

0B =

Maxwell’s equations

→

→

→

→

⎟ ⎠ ⎞

⎜ ⎝ ⎛ ∂

θ ∂

∂ ∂∆∇++∆∇

1ˆ r 1ˆ

r r;zzyyxx

( )=

( )= ρ -3

0=

0=

Statics

Lec 09.6- 2 1/11/01

− = × ∇

× ∇

ρ = • ∇

• ∇

φ∂ φ +

θ∂ θ + ∂ ∂ ∂ ∂ ∂ ∂

sin r

2 ­m a J

C m

Static Solutions to Maxwell’s Equations

rpq

vq volume of source

p

0EsinceE =×∇

0BsinceAB =•∇

dv r4

1 qv q

pq

q p ∫

ρ

πε =φ

dv r J

4A

qv q pq

qp ∫π

µ=

E3

Maxwell’s equations govern variables: ( )( )

Lec 09.6- 3 1/11/01

φ−∇ =

× ∇ =

p point at potential tic electrosta is volts

p at potential vector the is

1­

1­

m a field magnetic H

m v field electric E

Dynamic Solutions to Maxwell’s Equations

E4

0BsinceAB =•∇

( ) ( )

q

q pq p qv pq

J t r c A t ( l4 r

−µ= π ∫

ms1031c 18 oo

−×≅=

pq

q

jkr q

p qv pq

J eA 4 r

− µ= π ∫

λπ=ω=εµω= 2ck oowhere propagation constant

( ) ( ) ( ) ( )rq →→→

( )E H j= from Faraday’s law Lec 09.6- 4

1/11/01

× ∇ =

dv static so ution, delayed)

light) of (velocity ε µ

Sinusoidal steady state: dv

r E r B r A J : method Solution

where − ∇ × ωε

S = E × H

d << λ/2π

E5

( ) ( )( )hereW

dr,2

me >>

>>π λ

<<

−θπ

π jkro

o inˆr 2 , E j e

4 r Ω=εµ≡ηλπ= 377,2 ooo

characteristic impedance of free space −θ

≅ φ π

jkro inˆH e

z

d

Q(t)

-Q(t)

θ

φ

y

x

θ, E

φ, Hˆ

r ˆ

Io

Lec 09.6- 5 1/11/01

Elementary Dipole Antenna (Hertzian Dipole)

W power, active store strongest are fields c quasistati the and

r field" near " In

>> λ ≅ θ η kI dsFar field:

k where

kI ds4 jr

Elementary Dipole Antenna (Continued)

E

2r

π λ

<<

π λ

≅ 2

r π λ

>> 2

r

"r 2

λ≥

HES ∗ ×∆

( ) ( ) ( ) ( ) ( ) ( )

2 1

×=

==

E6 Lec 09.6- 6

1/11/01

field" near "

D" aperture of field far D2

vector" Poynting " m W 2 ­

2 ­

2 ­

m W t H t E t S and

m W S Re t S density power average where

Elementary Dipole Antenna (Continued)

E7

z

θ

y

S

for a short dipole( ) 2

oo r22

r λ

θη =

Watts3

ˆSR2 1P

2 o

4et λη

π =Ω•= ∫ π

Total power transmitted

HES ∗ ×∆

( ) ( ) ( ) ( ) ( ) ( )

2 1

×=

==

Lec 09.6- 7 1/11/01

sin d I t S

d I d r

vector" Poynting " m W 2 ­

2 ­

2 ­

m W t H t E t S and

m W S Re t S density power average where

Equivalent Circuit for Short Dipole Antenna

Rr = “Radiation resistance”

Reactance

+

– V–

Reactance is capacitive for a short dipole antennaand inductive for a small loop antenna

jX

z

+ – d 0

Io

I(z) deff

V–

E8

→oI

→oI ( )∫ ∞

∞−

=→ effoo π

=

2 eff

2 effo

r 2

ot 3RI

2 1P

λ πη

==

Example: λ = 300m(1 MHz), deff = 1m ⇒ Rr ≅ 0.01Ω such a mismatch requires transformers or hi-Q resonators for a good coupling to receivers or transmitters

( )2 effor d

3 2R ληπ

= ohms for a short dipole antenna

Lec 09.6- 8 1/11/01

dz z I d I d I λ << d Since

2 d d Here

d I

Wire Antennas of Arbitrary Shape Finding self-consistent solution is difficult

(matches all boundary conditions) Approximate solutions are often adequate

r( )

p– +

I( ) θ( )

oI

V

Approach

( )-1)

( ) fffffarfield EHA →→→2)

AAAA A d)(sine)(I)(ˆ r4

jkE )(jkr

L ff θθ

π η

≅ −∫3) G1

Lec 09.6- 9 1/11/01

r I guess to reasoning line TEM Use

r I

Estimating Current Distribution I( )on Wire Antennas

~ → coax H( )

E( )

E

H

σ = ∞

Most energy stored within 1–2 wire radii.

I( )AI

[ ] 2 32

om 2

oe r 1JmH

4 1E

4 1W ∝µ=ε= −

like-

G2 Lec 09.6- 10

1/11/01

me, W , W ,

TEM is wires thin on I,V Therefore

Examples: Current Distributions and Patterns I(z)

Io

z

d θ

E(θ)

D = λ/2

Short dipole, d << λ/2

λ/2 z

Io ↑↓

“Half-wave dipole”

“Full-wave antenna”

Rr ≅ 73Ω, jX= 0

( )jkr o

ff k kˆE

in 2 2

−η≅ θ

π θ ⎡ ⎤ ⎢ ⎥⎣ ⎦

A A for center-fed wire of length A

2ifcos

2cos λ

=⎟⎠ ⎞

⎜⎝ ⎛ θπ

= A

G3 Lec 09.6- 11

1/11/01

j I e cos cos cos

2 r sθ −

Examples: Current Distributions and Patterns

Vee (broadband)

↑ ↓Io

+=

Traveling wave

Radiation decay

+

Standing wave

↑↓ Io

Long-wire antenna

I(z)

Io↑↓ 2Io

z

Current transformer, jX ≠ 0

G4 Lec 09.6- 12

1/11/01

Mirrors, Image Charges and Currents We can replace planar mirrors (σ = ∞) with image charges and currents;E, H solution unchanged.

(σ = ∞)

J

Sources

Image chargesand currentsJ

Say 1 MHz (λ = 300m)D ≅ 1m (short dipole); D << λ Deff ≅ 2D/2 = D; X is capacitive

+

+

–

–

Io

⊥(σ = ∞),H//(σ = ∞), matching boundary conditions. Also, the uniqueness theoremsays any solution is the valid one.

H

Image current

Antenna pattern

Ground plane

D Image current

≅ D

G5 Lec 09.6- 13

1/11/01

Note: Anti-symmetry of image charges and currents guarantees E

Antenna Arrays

1 3

2 4

A1Io ↑

A2Io ↑

AiIo ↑

z y x

θ i

B e-jϕ i

ri

p E

ϕ i = 2π ri/λ

N identically oriented antenna elements

J1

o i Nj j ( , )

p i i 1

E E( , )e ϕ − ϕ

= ≅ ∑

0zyx ato===

2

i

),(ji

factor element

2 o

2 i,EE),(G ∑•=∝

Lec 09.6- 14 1/11/01

A e θ φ θ φ I from

factor array

e A φ θ ϕ − φ θ φ θ

Examples of Antenna Arrays

+ –

Full-wavelength antenna:

←λ /2→

Array factor Element factor

Pattern

↑↓ i(t)

i(z)

t = 0

⇒ × =

Array

J2

1 2 3 N

Linear array:

0 z2 z3 zi

zi cos θ θ

z …

ije φ−

0;2Let

i

iii

α

= λ π

+α=ϕ

Lec 09.6- 15 1/11/01

current element ith of phase is

z at 0 cos z = θ

Linear Array Example:Half-Wave Dipole Plus Reflector

Antenna pattern

λ/4

Image current

Reflector >> λ/2y

x

Io

z

λ/2

=

y

Element factor

z

x null

null

×

y

x z

Arrayfactor

⇒

J3 Lec 09.6- 16

1/11/01

Io

λ/2

λ/2

x

y z

To cancel y-direction lobes, drive two duplicate antennas in phase, λ/2 apart in y-direction ⇒

x z

y

Used in 1927 to discover galactic radiation at 27 MHz while seeking

∝ 72

y

x z

∝ 62

∝ 12

J4 Lec 09.6- 17

1/11/01

Linear Array Example: Jansky Antenna

radio interference on AT&T transatlantic radio telephone circuits.

= 49(x – pol.)

= 36(z – pol.) = 1(x – pol.)

Genetic Algorithms for Designing Wire Antennas

3. Need vector description to represent each possible antenna.

5. Metrics and vector descriptions favoring simplicity bias the design accordingly.

1. Need performance metric, e.g. target gain or pattern plus cost function.

2. Need software tool to compute that metric for wire antennas.

4.

especially if the antenna is not too reactive.Yields rats-nest configurations that perform well. e.g.

Ground plane

J5 Lec 09.6- 18

1/11/01

Need genetric algorithm to randomly vary vector so its metric can be computed. The algorithm hill-climbs efficiently toward optimum design,