1

Chapter 4 : Linear Wire Antenna

• Infinitesimal Dipole

• Small Dipole

• Finite Length Dipole

• Half-Wavelength Dipole

• Linear Elements near or on Infinite Perfect

Conductors

2

Infinitesimal Dipole• Length l << λλλλ

• Used to represent capacitor-plate (top-hat-loaded) antennas

• Capacitive loading to maintain the uniform current

3

Radiated Field

• The current on the infinitesimal dipole is

assumed to be constant, i.e.,

• Recall that

• Since the dipole is infinitesimal, the following approximations hold:

0ˆ)'( Izz =I

')',','(4

),,( dlR

ezyxzyx

jkR

C

−

∫= JAπµ

0''' === zyx

constant|'| ==−= rrrRrr

'' dzdl =

4

Radiated Field (2)

jkrl

l

jkre

r

lIzdze

r

Izzyx

−

−

− == ∫ πµ

πµ

4ˆ'

4ˆ),,( 0

2/

2/

0A

0;sin;cos =−== φθ θθ AAAAA zzr

,)(ˆ

)(sin

1ˆ)sin(

sin

ˆ

∂∂

−∂∂

+

∂∂

−∂∂

+

∂∂

−∂∂

=×∇

θφ

φθθ

φθ

θθ

θ

φθ

φ

r

r

ArA

rr

rAr

A

r

AA

r

rA

Thus

Since A has only z component,

and

∂∂

−∂∂

=θµ

φ θrA

rArr

)(1ˆH

The magnetic field becomes:

5

Radiated Field (3)

jkr

r

ejkrr

lkIjH

HH

−

+=

==

11

4

sin

0

0

πθ

φ

θ

0

)(

111

4

sin

11

2

cos

2

0

2

0

=

−+=

+=

−

−

φ

θ πθ

η

πθ

η

E

ekrjkrr

lkIjE

ejkrr

lIE

jkr

jkr

r

Therefore,

Likewise, the electric field can be found to be

6

Power Density and

Radiation Resistance

• Poynting vector

)ˆˆ(2

1

)ˆ()ˆˆ(2

1)(

2

1

**

**

φφθ

φθ

θ

φθ

HEHEr

HEEr

r

r

−=

×+=×= HEW

+=

−=

232

2

0

32

22

0

)(

11

16

sincos||

)(

11

sin

8

krr

lIkjW

krj

r

lIWr

πθθ

η

θλ

η

θ

−==

⋅+=⋅=

∫ ∫

∫ ∫∫∫

3

2

02

0 0

2

2

0 0

2

)(

11

3sin

sinˆ)ˆˆ(

krj

lIddrW

ddrrWWrdP

r

r

S

λπ

ηφθθ

φθθθ

π π

π π

θsW

Outgoing Power

Components of

Poynting Vector

7

Power Density and

Radiation Resistance (2)

3

2

0

)(

1

3)

~~(2

kr

lIWW em λ

πηω −=−

)~~

(2 emrad WWjPP −+= ω

direction radialin All power; (reactive)

imaginary average-time)~~

(2

energy electric average-time~

energy magnetic average-time~

power radiated average-time

power

=−

=

=

=

=

em

e

m

rad

WW

W

W

P

P

ω

2

2

2

803

2

=

=λ

πλ

πη

llRr

Radiation Resistance

Complex Power

rrad RIlI

P2

0

2

0 ||2

1

3==

λπ

η

Reactive Power

8

Far Field (kr >> 1)• For kr >> 1, the fields can be approximated as

1

0

4

sin

4

sin

0

0

>>

===≅

≅

≅

−

−

kr

HHEE

er

lkIjH

er

lkIjE

rr

jkr

jkr

θφ

φ

θ

πθ

πθ

η

ηφ

θ ==H

EZw

space)-freefor (120 impedance intrinsic

impedance wave

Ω=

=

πη

Zw

Ratio of E and H:

TEM Wave

9

Directivity

2

22

02* sin

42ˆ||

2

1ˆ)Re(

2

1

r

lkIrErav

θπ

ηη θ ==×= HEW

22

2

2

02 |),,(|2

sin42

φθη

θπ

ηθ rE

rlkIWrU av ===

2

0max

42 πη lkI

U =

2

34 max

0 ==radP

UD π

πλ

πλ

8

3

4

2

0

2

== DAem

Time-average power density:

Radiation intensity:

Maximum radiation intensity:

Maximum directivity:

Maximum effective area:

10

Small Dipole

• Length λ/50 <λ/50 <λ/50 <λ/50 < l < λ/10λ/10λ/10λ/10

11

Radiated Field• The current on the small dipole is assumed to

be a triangular function, i.e.,

≤≤−+

≤≤−=

0'2/),'2

1(ˆ

2/'0),'2

1(ˆ

)'(

0

0

zlzl

Iz

lzzl

Iz

zI

−+

+=

∫

∫−

−

−

2/

00

0

2/0

')'2

1(

')'2

1(4

ˆ),,(

ljkR

l

jkR

dzR

ez

lI

dzR

ez

lIzzyx

πµ

A

where I0 is a constant. Vector potential becomes:

Approximating R ~ r yields the maximum phase error

kl/2 = π/10 for l=λ/10.

12

Radiated Field (2)

1

0

8

sin

8

sin

0

0

>>

===≅

≅

≅

−

−

kr

HHEE

er

lkIjH

er

lkIjE

rr

jkr

jkr

θφ

φ

θ

πθ

πθ

η

jkr

z er

lIzAzzyx

−==π

µ42

1ˆˆ),,( 0A

Using R~r:

The far-field can be given by

which is one-half of that for the infinitesimal dipole.

2

2

2

0

20||

2

==λ

πl

I

PR rad

rRadiation resistance:

13

Field Separation• For a very thin dipole, x’=y’=0, thus

)'cos'2()''2()(

)'()'()'()'(

222222

222222

zrzrzzzzyx

zzyxzzyyxxR

+−+=+−+++=

−++=−+−+−=

θ

θcos; where 2222rzzyxr =++=

14

15

Field Separation (2)

then,'cos'2

Let2

2

r

zrzx

+−=

θ

L+

+

+−= θθθθ 2

3

2

22

sincos2

'1sin

2

'1cos'

z

r

z

rzrR

Recall also the Taylor expansion:

L+++= 2

!2

)0('')0(')0()( x

fxffxf

L1682

1)1(32

2/1 xxxx +−+=+

which yields the same result:

)'cos'2(1

1 2

2zrz

rrR +−+= θ

Recall that

16

Far Field

θcos'zrR −=

2when

2

'sin

2

'1 2

max

22 π

θθ ==

r

zz

r

82

'2 π≤

r

zk

By retaining only the first two terms, i.e.,

λ

22lr ≥

λ

22when

Dr ≥

termphasefor 'ˆcos'

termamplitudefor

rrrrrR

rRr⋅−=−≈

≈

θ

The most significant neglected term has the maximum value

A maximum total phase error of π/8 is acceptable, thus

Far-field approximation

2/'2/ lzl ≤≤−

17

Radiating Near Field

+−= θθ 2

2

sin2

'1cos'

z

rzrR

[ ] 0cos2sinsin2

'sincos

2

'1 22

2

32

3

2=+−=

∂∂

θθθθθθ r

zz

r

[ ] 0cos2sin1

22 =+− =θθθθ

By retaining only the first three terms, i.e.,

)2(tan 1

1 ±= −θ

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.,

yields

18

Radiating Near Field (2)

83123

2

3

1

8sincos

2

'2

3

2

3

tan

2/'

2

2

3

1

πλ

πλπ

θθθθ

≤

=

=

−== r

l

r

l

r

kz

lz

≥

λ

32

33

2 lr

λ

3

62.0l

r ≥

which reduces to

or

If the maximum total phase error is allowed to be π/8,

Near-field regionλλ

32

62.02 D

rD

≥≥

19

Finite Length Dipole

• Length > λλλλ/10

• Current on the finite length dipole assuming

the wire is very thin

≤≤−

+

≤≤

−

=

0'2/,)'2

(sinˆ

2/'0,)'2

(sinˆ

)'(

0

0

zlzl

kIz

lzzl

kIz

zeI

where I0 is a constant. This distribution assumes

that the antenna is center-fed and the current

vanishes at the end points.

20

Radiated Field

0

'4

sin)'(

'4

sin)'(

===≅

≅

≅

−

−

θφ

φ

θ

πθ

πθ

η

dHdHdEdE

dzeR

zkIjdH

dzeR

zkIjdE

rr

jkRe

jkRe

The electric and magnetic field components in

the far field for the infinitesimal dipole dz’ are

given by

Using the far-field approximation yields

'4

sin)'( cos'dzee

r

zkIjdE

jkzjkre θθ π

θη −≅

21

Radiated Field (2)

444 3444 2144 344 21factor space

2/

2/

cos'

factorelement

2/

2/')'(

4

sin

== ∫∫ −

−

−

l

l

jkz

e

jkrl

ldzezI

r

kejdEE

θθθ π

θη

factor) (spacefactor)(element field total ×=

The total electric field can be obtained by

summing up contributions from all infinitesimal

dipoles, i.e.,

Thus, the electric field of the finite length dipole

can be given by

−+

+=

∫

∫−−

2/

0

cos'

0

2/

cos'0

')'2

(sin

')'2

(sin4

sin

ljkz

l

jkzjkr

dzezl

k

dzezl

kr

ekIjE

θ

θθ π

θη

22

Radiated Field (3)

−

≅−

θ

θ

πηθ

sin

2coscos

2cos

2

0

klkl

r

eIjE

jkr

)]cos()sin([)sin(22

γββγβαβα

γβα

α +−++

=+∫ xxe

dxxex

x

Using

Likewise, the magnetic field can be given by

yields

−

=≅−

θ

θ

πηθ

φsin

2coscos

2cos

2

0

klkl

r

eIj

EH

jkr

23

Current Distributions and

Radiation Pattern

24

Radiation Pattern for l=1.25λλλλ

25

Power Density and

Radiation Intensity

2

22

2

02

**

sin

2coscos

2cos

8

||ˆ||

2

1ˆ

)ˆˆRe(2

1)Re(

2

1

−

==

×=×=

θ

θ

πη

η

φθ

θ

φθ

klkl

r

IrEr

HEav HEW

Time-average power density:

Radiation intensity:2

2

2

02

sin

2coscos

2cos

8

||

−

==θ

θ

πη

klkl

IWrU av

26

Radiated Power

θθ

θ

πη

φθθ

π

π π

d

klkl

I

ddUUdPrad

∫

∫ ∫∫∫

−

=

=Ω=Ω

0

2

2

0

2

0 0

sin

2coscos

2cos

4

||

sin

−+++

− +−+=

)](2)2()2/ln()[cos(2

1

)](2)2()[sin(2

1)()ln(

4

|| 2

0

klCklCklCkl

klSklSklklCklCI

P

ii

iiirad πη

constant) s(Euler' 1566495772.0=C

Radiated power can be obtained by

which can be given by

where

(1)

27

Radiated Power (2)

∫∫ ∞

∞=−=

x

xi dy

y

ydy

y

yxC

coscos)(

∫=x

i dyy

yxS

0

sin)(

∫

−=

−+=

x

in

iin

dyy

yxC

xCxCxC

0

cos1)( where

)()ln()(

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

Cosine integral

Sine integral

0 5 10 15 20 25 30 35 40 45 50−5

−4

−3

−2

−1

0

1

2

3

4

5

Ci(x)

Si(x)

Cin

(x)

28

Radiation Resistance and

Input Resistance

rinin R

IR

I

2

||

2

|| 2

0

2

=

−+++

− +−+==

)](2)2()2/ln()[cos(2

1

)](2)2()[sin(2

1)()ln(

2||

22

0

klCklCklCkl

klSklSklklCklCI

PR

ii

iiirad

r πη

r

in

in RI

IR

2

0

=

Radiation resistance becomes

Input resistance can be given by

Since

For a dipole of length l,

=2

sin0

klII in

=

2sin 2 kl

RR r

in

Input Resistance

assuming loss-less

29

Directivity

Q

F

P

UD

rad

maxmax0

)(24

θπ ==

−+++

− +−+=

)](2)2()2/ln()[cos(2

1

)](2)2()[sin(2

1)()ln(

klCklCklCkl

klSklSklklCklCQ

ii

iii

Directivity is given by

and

where 2

sin

2coscos

2cos

)(

−

=θ

θθ

klkl

F

30

Radiation resistance, input

resistance and directivity

31

Half-wavelength Dipole

≅−

θ

θπ

πηθ

sin

cos2

cos

2

0

r

eIjE

jkr

=≅−

θ

θπ

πηθ

φsin

cos2

cos

2

0

r

eIj

EH

jkr

θπ

ηθ

θπ

πη 3

22

2

0

2

22

2

0 sin8

||

sin

cos2

cos

8

||

r

I

r

IWav ≅

=

Let l = λ/2, then

θπ

ηθ

θπ

πη 3

2

2

0

2

2

2

02 sin8

||

sin

cos2

cos

8

|| IIWrU av ≅

==

Power density

Radiation Intensity

32

Half-wavelength Dipole (2)

435.2)2()2ln()2( ≅−+= πππ iin CCC

5.4273 jZ in +=

Radiated Power

Radiation Resistance

)2(8

||cos1

4

||

sin

cos2

cos

4

||

2

02

0

2

0

0

2

2

0

ππ

ηπ

η

θθ

θπ

πη

π

π

in

rad

CI

dyy

yI

dI

P

∫

∫

=

−=

=

643.1435.2

44 max

0 ≅==radP

UD π

73)2(4||

22

0

≅== ππη

inrad

r CI

PR

Input Impedance

Directivity

where

(2)

33

Wire antennas near or on infinite

perfect conductor

34

Image Theory

PMCon tanˆ0 0H =×→= nH

NOTE: The fields

obtained are valid

only in the top

half-plane.

PECon tanˆ0 0E =×→= nE

35

Vertical Electric Dipole

1

1

0 sin4

1

θπ

ηθr

lekIjE

jkrd

−

=

2

2

0

2

2

0

sin4

sin4

2

2

θπ

η

θπ

ηθ

r

lekIj

r

lekIjRE

jkr

jkr

v

r

−

−

=

=Direct Component

Reflected Component

36

Vertical Electric Dipole (2)

rrr ≅≅ 21

θθ cos]cos2[ 222

1 hrrhhrr −≅−+=

In far field:

θθ cos]cos2[ 222

2 hrrhhrr +≅++=Phase term

Amplitude term

Parallel ray

approximation

37

Vertical Electric Dipole (3)

00

0)]coscos(2[sin4

0

<=

≥≅−

z

zkhr

lekIjE

jkr

θθπ

ηθTotal Electric Field

12

lobes ofnumber +≅λh

38

Vertical Electric Dipole (4)

)cos(cossin2

ˆ||2

1ˆ)Re(

2

1 22

2

0

2

2* θθλ

ηη θ kh

lI

rrErav ==×= HEW

)cos(cossin2

||2

22

2

022

θθλ

ηη θ kh

lIE

rU ==

2

0max

2 λη lI

U =

+−=

== ∫∫ ∫

32

2

0

2/

0

2

0

2/

0

)2(

)2sin(

)2(

)2cos(

3

1

sin2sin

kh

kh

kh

khlI

dUddUPrad

λπη

θθπφθθππ π

Power Density

Radiation Intensity

Maximum Radiation Intensity

Radiated Power

(3)

39

Vertical Electric Dipole (5)1

32

max0

)2(

)2sin(

)2(

)2cos(

3

124

−

+−==

kh

kh

kh

kh

P

UD

rad

π

+−

==32

2

2

0 )2(

)2sin(

)2(

)2cos(

3

12

||

2

kh

kh

kh

khl

I

PR rad

r λπη

Directivity

Radiation Resistance

40

λλλλ/4 Monopole

25.215.36

)5.4273(2

1

(dipole)2

1

j

j

ZZ inin

+=

+=

=

41

Fields due to y-directed dipole

r

leIy

jkr

πµ

4ˆ 0

−

=A

φθπ

µφθθ sincos

4sincos 0

r

leIAA

jkr

y

−

==

φπ

µφφ cos

4cos 0

r

leIAA

jkr

y

−

==

φπ

µωω

φθπ

µωω

φφ

θθ

cos4

sincos4

0

0

r

leIjAjE

r

leIjAjE

jkr

jkr

−

−

−=−≅

−=−≅

Vector Potential A

Recall that )()( Aj

Ajrrrr

⋅∇∇−−=ωµε

ωE

Far-field Electric Field

42

Fields due to y-directed dipole (2)

ψψζ

ψζψζψψψζψζψ

cosˆsinˆˆ

sinˆsincosˆcoscosˆˆ

cosˆsinsinˆcossinˆˆ

xz

yxz

yxzr

+−=

−+=

++=

ψπ

µωψ sin

4

0

r

leIjE

jkr−

≅

then

Far-field Electric Field

Introduce a “new” spherical coordinate system (r,ψ,ζ) such that

ψπη

ωµζ sin

4

0

r

leIjH

jkr−

≅Far-field Magnetic

Field

(This can be obtained by letting (x,y,z)->(z,x,y) and (ψ,ζ)->(θ,φ))

43

Horizontal Electric Dipole

ψπ

ηψ sin4 1

01

r

lekIjE

jkrd

−

=ψ

πη

ψπ

ηψ

sin4

sin4

2

0

2

0

2

2

r

lekIj

r

lekIjRE

jkr

jkr

h

r

−

−

−=

=Direct Component

Reflected Component

44

Horizontal Electric Dipole (2)

rrr ≅≅ 21

θθ cos]cos2[ 222

1 hrrhhrr −≅−+=θθ cos]cos2[ 222

2 hrrhhrr +≅++=

In far field:

Phase term

Amplitude term

φθψ sinsinˆˆcos =⋅= ry

45

Horizontal Electric Dipole (3)

)]cossin(2[sinsin14

220 θφθπ

ηψ khjr

lekIjE

jkr

−≅−

Total Electric Field

λh2

lobes ofnumber ≅

46

Horizontal Electric Dipole (4)

)cos(sin)sinsin1(2

ˆ 222

2

0

2θφθ

λη

khlI

rrav −=W

)cos(cos)sinsin1(2

||2

222

2

022

θφθλ

ηη θ kh

lIE

rU −==

+−−=

== ∫∫ ∫

32

2

0

2/

0

2

0

2/

0

)2(

)2sin(

)2(

)2cos(

2

)2sin(

3

2

2

sin2sin

kh

kh

kh

kh

kh

khlI

dUddUPrad

λπ

η

θθπφθθππ π

Power Density

Radiation Intensity

Radiated Power

+−−

==32

2

2

0 )2(

)2sin(

)2(

)2cos(

2

)2sin(

3

2

||

2

kh

kh

kh

kh

kh

khl

I

PR rad

r λπη

Radiation Resistance

47

Horizontal Electric Dipole (5)

>

≤

=

2/)(

4

2/)(

)(sin4 2

0

π

π

khkhR

khkhR

kh

D

223220

15

322

15

8

3

2

3

2

=

+−

=→

λλπ

ηλπ

λπη

hlhlR

kh

r

Directivity

For small kh

Maximum Radiation Intensity

==>

=≤

=

1)cossin( and 0,2/2

0,2/)(sin2

max

2

0

2

2

0

max

θφπλ

η

θπλ

η

khkhlI

khkhlI

U

48

Horizontal Electric Dipole (6)

+−−=

32 )2(

)2sin(

)2(

)2cos(

2

)2sin(

3

2)(

kh

kh

kh

kh

kh

khkhR

21

220

0

)sin(5.7)(

15

8

3

2

3

2)(sin4

=

+−−→

=kh

khkhkhD

khFor small kh

where