1

Lecture 8: Linear Wire Antennas – Dipoles and Monopoles(Small electric dipole antenna. Finite-length dipoles. Half-wavelengthdipole. Method of images - revision. Vertical infinitesimal dipole abovea conducting plane. Monopoles. Horizontal infinitesimal dipole above aconducting plane.)

The dipole and the monopole are two most widely used antennas forwireless mobile communication systems. Arrays of dipoles arecommonly used as base-station antennas in land-mobile systems. Themonopole is perhaps the most common antenna for portable equipment,such as cellular telephones, cordless telephones, automobiles, trains, etc.It has attractive features such as simple construction, relativelybroadband characteristics, small dimensions at high frequencies. Analternative to the monopole antenna for hand-held units is the loopantenna, the microstrip patch antenna, the spiral antennas, and others.

1. Small dipole

50 10l

λ λ< ≤ (8.1)

If one assumes that R r≈ and condition (8.1)holds, the maximum phase error in ( )Rβ thatcan occur is

max 182 10l

eβ π= = ≈ ,

at 0θ = . Reminder: a maximum total phaseerror of /8π is acceptable since it does notaffect substantially the integral solution for A.The assumption R r≈ will be made for both,the amplitude and the phase factors in thekernel of the VP integral.

z

/ 2l

/ 2l−

( )I z0

z'

0I

2

The current is a triangular function of 'z :

0

0

'1 , 0 ' / 2

/ 2( ')

'1 , / 2 ' 0

/ 2

zI z l

lI z

zI l z

l

⋅ − ≤ ≤ = ⋅ + − ≤ ≤

(8.2)

The VP integral is obtained as:0 / 2

0 0

/ 2 0

' 'ˆ 1 ' 1 '

4 / 2 / 2

lj R j R

l

z e z eA z I dz I dz

l R l R

β βµπ

− −

−

= ⋅ + + −

∫ ∫ (8.3)

The solution of (8.3) is particularly simple when it can be assumed thatR r≈ :

0

1ˆ

2 4

j reA z I l

r

βµπ

− = ⋅

(8.4)

The further away from the antenna the observation point is, the moreaccurate the expression in (8.4). Note that the result in (8.4) is exactlyone-half of the result obtained for A of an infinitesimal dipole, if 0Iwere the current uniformly distributed along the dipole. This is to beexpected because we made the same approximation for R, as in the caseof the infinitesimal dipole with a constant current distribution, and we

integrated a triangular function along l, whose average is obviously 0

12

I .

Therefore, we need not repeat all the calculations of the fieldcomponents, power and antenna parameters. We shall make use of ourknowledge of the infinitesimal dipole field. The far-field components ofthe small dipole are simply half those of the infinitesimal dipole:

0

0

sin8

sin8

0

j r

j r

r r

I l eE j

r

I l eH j

rE E H H

β

θ

β

ϕ

ϕ θ

βη θπ

β θπ

−

−

= = = =

, 1rβ (8.5)

3

The normalized field pattern is the same as that of the infinitesimaldipole: ( , ) sinE θ ϕ θ= (8.6)

The power pattern: 2( , ) sinU θ ϕ θ= (8.7)

-1

-0.5

0

0.5

1

-1 -0.5 0 0.5 1The beam solid angle:

22

0 0

3

0

sin sin

4 82 sin 2

3 3

A

A

d d

d

π π

π

θ θ θ ϕ

ππ θ θ π

Ω = ⋅

Ω = ⋅ = ⋅ =

∫ ∫

∫The directivity:

0

4 31.5

2A

Dπ= = =

Ω(8.8)

As expected, the directivity (and the beam solid angle, as well as theeffective aperture) is the same as those of the infinitesimal dipole,because the normalized patterns of both dipoles are the same.

sinθ2sin θ

0θ =

90θ =

4

The radiated power will be four times less than that of an infinitesimaldipole because the far fields are twice less:

2 2

0 014 3 12rad

I l I lπ πη ηλ λ

∆ ∆ Π = ⋅ =

(8.9)

As a result, the radiation resistance is also four times less compared tothat of the infinitesimal dipole:

2 2220

6r

l lR

π η πλ λ∆ ∆ = =

(8.10)

2. Finite-length infinitesimally thin dipoleA good approximation of the current distribution along the dipole’s

length is the sinusoidal one:

0

0

sin ' , 0 ' / 22

( ')

sin ' , / 2 ' 02

lI z z l

I zl

I z l z

β

β

− ≤ ≤ = + − ≤ ≤

(8.11)

It can be shown that the VP integral/ 2

/ 2

ˆ ( ') '4

l j R

l

eA z I z dz

R

βµπ

−

−

= ⋅ ∫ (8.12)

has an analytical (closed form) solution. Nevertheless, we shall follow astandard approach commonly used to calculate the far field. It is basedon the solution to the infinitesimal dipole field problem. The finite-length dipole is subdivided into an infinite number of infinitesimaldipoles of length 'dz . Each infinitesimal dipole produces the elementaryfar field described as:

( ')

( ')

sin '4

sin '4

0

j r

e z

j r

e z

r r

edE j I dz

r

edH j I dz

rdE dE dH dH

β

θ

β

ϕ

ϕ θ

ηβ θπ

β θπ

−

−

⋅

⋅

= = = =

(8.13)

5

Here, ( ')e zI denotes the current value of the current element at 'z . Using

the far-zone approximations:1 1

, for the amplitude factor

'cos , for the phase factorR rR r z θ−

(8.14)

the following approximation of the elementary far field is obtained:

'cos sin '4

j rj z

e

edE j I e dz

r

ββ θ

θ ηβ θπ

−

⋅ (8.15)

Using the superposition principle, the total far field is obtained as:/ 2 / 2

'cos( ')

/ 2 / 2

sin '4

l lj rj z

e zl l

eE dE j I e dz

r

ββ θ

θ θ ηβ θπ

−

− −

= ⋅ ⋅∫ ∫ (8.16)

The first factor

( ) sinj re

g jr

β

θ ηβ θ−

= (8.17)

is called the element factor. The element factor in this case is the farfield produced by an infinitesimal dipole of unit current element

1 (A m)I l⋅ = ⋅ . The second factor/ 2

'cos( ')

/ 2

( ) 'l

j ze z

l

f I e dzβ θθ−

= ∫ (8.18)

is called the space factor (or pattern factor, array factor). The patternfactor is dependent on the amplitude and phase distribution of the currentat the antenna (the source distribution in space).

The element factor is well known, and is the same for any currentelement, provided the angle θ is always associated with the current-element axis.

6

For the specific current distribution described by (8.11), the patternfactor is:

0'cos

0

/ 2

/ 2'cos

0

( ) sin ' '2

sin ' '2

j z

l

lj z

lf I z e dz

lz e dz

β θ

β θ

θ β

β

−

= ⋅ +

+ −

∫

∫(8.19)

The above integrals are solved having in mind that

( ) ( )2 2sin( ) sin coscx

c x ea b x e dx c a bx b a bx

b c⋅+ ⋅ = + − + +∫ (8.20)

The far field of the finite-length dipole is obtained as:

0

cos cos cos2 2( ) ( ) ,

2 sin

j r

l le

E g f j Ir

EH

β

θ

θϕ

β βθθ θ η

π θ

η

− − = ⋅ =

=

(8.21)

The amplitude pattern:

cos cos cos2 2( , )

sin

l l

E

β βθθ ϕ

θ

− = (8.22)

7

Patterns (in dB) for some dipole lengths l λ≤ :

8

The pattern of the dipole 1.25l λ=

9

The power pattern:2

cos cos cos2 2( , )

sin

l l

F

β βθθ ϕ

θ

− =

(8.23)

Note: The maximum of ( , )F θ ϕ is not necessarily unity, but for 2l λ<the major maximum is always at 90θ = .

The radiated powerFirst, the average power flux density is calculated as:

2

22 0

2 2

cos cos cos1 | | 2 2ˆ ˆ| |

2 8 sin

l lI

P r E rrθ

β βθη

η π θ

− = ⋅ = ⋅

(8.24)

The total radiated power is given by the integral:2

2

0 0

sinPds P r d dπ π

θ θ ϕΠ = = ⋅∫∫ ∫ ∫ (8.25)

2

20

0

cos cos cos| | 2 24 sin

l lI

dπ

β βθη θ

π θℑ

− Π = ∫ (8.26)

ℑ is solved in terms of the cosine and the sine integrals:

( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

1ln sin 2 2

21

cos ln 2 22

i i i

i i

C l C l l S l S l

l C l C l C l

β β β β β

β β β β

ℑ = + − + − +

+ + + −

(8.27)

Here:0.5772C is the Euler’s constant

cos cos( )

x

i

x

y yC x dy dy

y y

∞

∞

= = −∫ ∫ is the cosine integral

10

0

sin( )

x

i

yS x dy

y= ∫ is the sine integral.

So, the radiated power can be written as:2

0| |4Iηπ

Π = ⋅ ℑ (8.28)

Radiation resistance:

20

2| | 2rRI

ηπ

Π= = ⋅ ℑ (8.29)

Directivity:

max max0 2

0 0

4 4

( , )sin

U FD

F d dπ ππ π

θ ϕ θ θ ϕ= =

Π∫ ∫

(8.30)

where:2

cos cos cos2 2( , )

sin

l l

F

β βθθ ϕ

θ

− =

is the power pattern (see (8.23) ).

Finally,

max0

2FD =

ℑ(8.31)

Input resistanceThe radiation resistance given in (8.29) corresponds to the radiated

power but it is not equal to the input resistance because the current at thedipole center (if its center is the feed point) is not necessarily of themaximum amplitude. If the dipole is lossless, the input power is equal tothe radiated one:

2 20| | | |

2 2in

in r

I IR R= (8.32)

According to the sinusoidal distribution assumed in (8.11), the current atthe center of the dipole ( ' 0z = ) is:

11

0 sin2in

lI I β =

(8.33)

2sin2

rin

RR

lβ⇒ =

(8.34)

3. Half-wavelength dipole

This is a classical and widely used thin wire antenna:2

lλ=

0

cos cos2

2 sin

j rI eE j

rE

H

β

θ

θϕ

π θη

π θ

η

− ⋅

=

(8.35)

Radiated power flow density:2

2 22 30 0

2 2 2 2

( ) normalized power pattern

cos cos| | | |2| | sin

2 8 sin 8

F

I IP E

r rθ

θ

π θη η η θπ θ π

−

= =

(8.36)

Radiation intensity:2

2 22 30 0

2 2

( ) normalized power pattern

cos cos| | | |2 sin8 sin 8

F

I IU r P

θ

π θη η θ

π θ π

−

= =

(8.37)

12

3-D power pattern (not in dB) of the half-wavelength dipole:

Radiated powerThe radiated power of the half-wavelength dipole is, of course, a

special case of the integral in (8.26).

22

0

0

220

0

cos cos| | 24 sin

| | 1 cos,

8

0.5772 ln(2 ) (2 ) 2.435i

Id

I ydy

y

C

π

π

π θη θ

π θ

ηπ

π πℑ

Π =

−Π =

ℑ = + −

∫

∫ (8.38)

2 20 02.435 | | 36.525 | |

8I I

ηπ

⇒ Π = = (8.39)

Radiation resistance

20

273

| |rRI

Π= Ω (8.40)

13

Directivity

max / 900

4 44 4 1.643

2.435

UUD θπ π == = = = =

Π Π ℑ(8.41)

Maximum effective area2

20 0.13

4eA Dλ λπ

= (8.42)

Input resistanceSince / 2l λ= ,

73in rR R= Ω (8.43)The imaginary part of the input impedance is approximately

42.5j+ Ω . To acquire maximum power transfer, this reactance has tobe removed by matching (that is shortening) the dipole:

• thick dipole 0.47l λ• thin dipole 0.48l λ

The input impedance of the dipole is very frequency sensitive; inother words, it depends strongly on the ratio /l λ . This is to be expectedfrom a resonant structure operating near the resonance, such as the half-wavelength dipole. It should be also kept in mind that the inputimpedance is influenced in a non-negligible way by the capacitanceassociated with the physical junction to the transmission line. Thestructure used to support the antenna, if any, can also influence the inputimpedance. That is why the curves that are given below describing theantenna impedance should be considered just representative of a typicalbehaviour.

14

Below, measurement results for the input impedance of a dipole aregiven.

Note the strong influence of the dipole diameter on its resonantproperties.

Input resistance of dipole antenna

15

Input reactance of a dipole antenna

One can calculate the input resistance as a function of /l λ usingequations (8.29) and (8.34). These equations, however, are valid onlyfor infinitesimally thin dipoles. Besides, it is important to compute theexact reactance, too. In practice, dipoles are most often tubular, and theyhave some finite diameter d. General-purpose numerical methods (suchas the Method of Moments or FDTD) are used to calculate the antennaimpedance. When finite-thickness wire antennas are to be analyzed andno assumption is made for the current distribution along the wire, theMoM is applied to the classical Pocklington’s equation or to its variation,the Hallen’s equation. A classical method producing closed formsolutions for the self-impedance and the mutual impedance of straight-wire antennas is the induced emf method, which will be discussed later.The induced emf method does assume sinusoidal current distribution.

16

4. Method of images – revision

magnetic conductor+

-iJ

+

-oJ + -

oJ oMoM

+ -iJ

iMiM

electric conductor

oM

+

-iJ

+

-oJ

+-iJ

+ -oJ oM

iMiM

17

5. Vertical electric current element above perfect conductor

conductor

actual source

image source

Pdirect

reflected

z

1θ

2θ

1r

2r

1 1( , )ε µ

h

h

r

θ

The field at the observation point P is a superposition of the fields of theactual source and the image source, both radiating in a homogeneousmedium of constitutive parameters 1 1( , )ε µ . The actual source is acurrent element 0( )I l∆ (infinitesimal dipole).

1

2

0 11

0 22

( ) sin4

( ) sin4

j rd

j rr

eE j I l

r

eE j I l

r

β

θ

β

θ

ηβ θπ

ηβ θπ

−

−

∆

∆

= ⋅

= ⋅(8.44)

Expressing the distances 1| |r and 2| |r in terms of | |r and h (using thecosine theorem) gives:

2 21

2 22

2 cos

2 cos( )

r r h rh

r r h rh

θ

π θ

= + −

= + − −(8.45)

We shall use the binomial expansion of 1r and 2r to obtainapproximations of the amplitude and the phase terms, which wouldsimplify the evaluation of the total far field and the VP integral.

18

For the amplitude term:

1 2

1 1 1r r r

(8.46)

For the phase term, we shall use a second order approximation (see alsothe geometrical interpretation below).

1

2

cos

cos

r r h

r r h

θθ

−+

(8.47)

x

y

z

h

h θ

2 cosh θ

σ = ∞

1r

2r

r

The total far field is:d rE E Eθ θ θ= + (8.48)

( ) ( )cos cos0( )sin

4j r h j r hI l

E j e er

β θ β θθ ηβ θ

π− − − +∆ = ⋅ + (8.49)

( ) ( )0

( )( )

sin 2cos cos , 04

0 , 0

j r

fg

eE j I l h z

r

E z

β

θ

θθ

θ

ηβ θ β θπ

−

∆ ⋅ ≥

= <

(8.50)

Again, it should be noted that the far field expression can bedecomposed into two factors: the field of the elementary source ( )g θand the pattern factor (also array factor) ( )f θ .

19

The normalized power pattern is:

( ) 2( ) sin cos cosF hθ θ β θ= ⋅ (8.51)

Elevation plane patterns of a vertical infinitesimal electric dipole fordifferent height above a perfectly conducting plane.

20

As the vertical dipole is moved further away from the infinite conducting(ground) plane, more and more lobes are introduced in the power pattern.This effect is called scalloping of the pattern. The number of lobes is

2nint 1

hn

λ = +

21

Total radiated power2 / 2

2 2

0 0

1| | sin

2Pds E r d d

π π

θ θ θ ϕη

Π = =∫∫ ∫ ∫/ 2

2 2

0

| | sinE r dπ

θπ θ θη

Π = ∫ (8.52)

( )/ 2

2 2 2 20

0

( ) sin cos cosI l h dπ

ηβ θ β θ θ∆Π = ⋅∫( )

( )( )

( )

2

02 3

cos 2 sin 213 2 2

h hI l

h h

β βπηλ β β∆ Π = − +

(8.53)

• As 0hβ → , the radiated power of the vertical dipole approachestwice the value of the radiated power of a dipole of the same lengthin free space.

• As hβ → ∞ , the radiated power of both dipoles becomes the same.

Radiation resistance

( )( )

( )( )

2

2 320

cos 2 sin 22 12

| | 3 2 2r

h hlR

I h h

β βπηλ β β∆ Π = = − +

(8.54)

• As 0hβ → , the radiation resistance of the vertical dipoleapproaches twice the value of the radiation resistance of a dipole ofthe same length in free space:

1, 0

2mp dpin inR R hβ= = (8.55)

• As hβ → ∞ , the radiation resistance of both dipoles becomes thesame.

22

Radiation intensity

( )22

2 2 2 20| |sin cos cos

2 2E I l

U r P r hθ η θ β θη λ

∆ = = =

(8.56)

The maximum of ( )U θ occurs at / 2θ π= (except for hβ → ∞ ):

0max 2

I lU

ηλ∆ =

(8.57)

This value is 4 times greater than maxU of a free-space dipole.

Maximum directivity

( )( )

( )( )

max0

2 3

24

cos 2 sin 213 2 2

UD

h h

h h

π β ββ β

= =Π − +

(8.58)

If 0hβ = , 0 3D = , which is twice the max. directivity of a free-space

current element ( 0 1.5idD = ) The maximum of 0D occurs when2.881hβ = ( 0.4585h λ= ). Then, 0 / 2.8816.566 hD β == .

23

6. MonopolesA monopole is a dipole that has been divided into half at its center

feed point and fed against a ground plane. It is normally / 4λ long (aquarter-wavelength monopole), but it might by shorter when spacerestrictions dictate shorter lengths. Then, the monopole is a smallmonopole whose counterpart is the small dipole (see Section 1, thisLecture). Its current has linear distribution with its maximum at the feedpoint, and its null at the monopole’s edge.

The vertical monopole is extensively used for AM broadcasting(f=500 to 1500 kHz, λ =200 to 600 m), because it is the shortest mostefficient antenna at these frequencies, as well as because verticallypolarized waves suffer less attenuation at close to the groundpropagation. Vertical monopoles are widely used as base-stationantennas in mobile communications, too.

Monopoles at base stations and radiobroadcast stations are supportedby suitable towers and guy wires. The guy wires must be separated intoshort enough ( /8λ≤ ) pieces, which are insulated from each other tosuppress any parasitic currents. Special care is taken for propergrounding of the monopole. Usually multiple radial wire rods, each0.25 0.35λ− long, are buried at the monopole base in the ground tosimulate perfect ground plane, so that the pattern approximates closelythe theoretical one, i.e. the pattern of the / 2λ -dipole. Losses in theground plane cause undesirable deformation of the pattern as shownbelow for an infinitesimal dipole above an imperfect ground plane.

24

Monopole fed against alarge solid ground plane

Practical monopole with radialwires to simulate perfect ground

l

Several important conclusions follow from the image theory and thediscussion in Section 5.

• The field distribution in the upper half-space is the same as that ofthe respective free-space dipole

• The currents and the charges on a monopole are the same as on theupper half of its dipole counterpart, but the terminal voltage is onlyhalf that of the dipole. The input impedance of a monopole istherefore only half that of the respective dipole:

12

mp dpin inZ Z= (8.59)

(See also (8.55).)• The total radiated power of a monopole is half the power radiated

by its dipole counterpart, since it radiates in half-space (but its fieldis the same). As a result, the beam solid angle of the monopole ishalf that of the respective dipole and its directivity is twice thedirectivity of the dipole.

0 0

4 42

12

mp dpmp

dpAA

D Dπ π= = =

Ω Ω(8.60)

25

The quarter-wavelength monopole

This is a straight wire of length / 4l λ= mounted over a groundplane. From the discussion above, it can be expected that the quarter-wavelength monopole should be very similar to the half-wavelengthdipole in the hemisphere above the ground plane.

• Its radiation pattern is the same as that of a free-space / 2λ -dipole,only that it is non-zero only for 0 90θ< ≤ (above ground).

• The field expressions are the same as those of the / 2λ -dipole.• The radiated power of the / 4λ -monopole is half that of the / 2λ -

dipole.• The radiation resistance of the / 4λ -monopole is half that of the

/ 2λ -dipole: ( )0.5 0.5 73 42.5 36.5 21.25,mp dpin inZ Z j j= = + = + Ω .

• The directivity of the / 4λ -monopole is:

0 02 2 1.643 3.286mp dpD D= = ⋅ =

Some approximate formulas for rapid calculations of the input resistanceof a dipole and the respective monopole:

Let, for dipole

2

2 , for monopole

l lG

lG l

β πλ

β πλ

= =

= =

Then,

• if 04

Gπ< <

2

2

20 ,dipole

10 ,monopole

in

in

R G

R G

=

=

26

• if4 2

Gπ π< <

2.5

2.5

24.7 ,dipole

12.35 ,monopole

in

in

R G

R G

=

=

• if 22

Gπ < <

4.17

4.17

11.14 ,dipole

5.57 ,monopole

in

in

R G

R G

=

=

7. Horizontal current element above a perfectly conducting planeThe analysis is analogous to that of a vertical current element above a

ground plane. The difference arises in the element factor ( )g θ becauseof the horizontal orientation of the current element. Let’s assume thatthe current element is oriented along the y-axis, and the angle between rand the dipole’s axis (y-axis) is ψ .

x

y

z

h

h θ

2 cosh θ

σ = ∞

1r

2r

r

P

ϕ

ψ

27

( ) ( ) ( )d rE P E P E P= + (8.61)1

01

( ) sin4

j rd e

E j I lr

β

ψ ηβ ψπ

−

∆= (8.62)

2

02

( ) sin4

j rr e

E j I lr

β

ψ ηβ ψπ

−

∆= − (8.63)

One can express the angle ψ in terms of ( , )θ ϕ .ˆ ˆ ˆ ˆ ˆcos ( sin cos sin sin ˆ cos )y r y x y zψ θ ϕ θ ϕ θ= ⋅ = ⋅ + +

2 2

cos sin sin

sin 1 sin sin

ψ θ ϕ

ψ θ ϕ

⇒ =

⇒ = −(8.64)

The far-field approximations are:

1 2

1

2

1 1 1, for the amplitude term

cosfor the phase term

cos

r r r

r r h

r r h

θθ

= =

− +

Substituting the far-field approximations and equations (8.62), (8.63),(8.64) in the total field expression (8.61) yields:

( )2 20

array factor ( , )element factor ( , )

( , ) ( ) 1 sin sin 2 sin cos4

j r

fg

eE j I l j h

r

β

ψ

θ ϕθ ϕ

θ ϕ ηβ θ ϕ β θπ

−

∆= − (8.65)

28

The normalized pattern

( ) ( )2 2 2( , ) 1 sin sin sin cosF hθ ϕ θ ϕ β θ= − ⋅ (8.66)

As the height increases beyond a wavelength (h λ> ), scalloping appearswith the number of lobes being:

nint 2h

nλ

=

(8.67)

90ϕ =

90ϕ =

29

Following a procedure similar to that of the vertical dipole, the radiatedpower and the radiation resistance of the horizontal dipole can be found.

( ) ( )( )

( )( )

( )

2

02 3

sin 2 cos 2 sin 222 3 2 2 2

R h

h h hI l

h h h

β

β β βπ ηλ β β β∆ Π = − − +

(8.68)

( )2

r

lR R hπη β

λ∆ = ⋅

(8.69)

By expanding the sine and the cosine functions into series, it can beshown that for small values of ( )hβ the following approximation holds:

22

/ 0

3215h

hR β

πλ→

(8.70)

It is also obvious that if 0h = , then 0rR = and 0Π = . This is to beexpected because the dipole is short-circuited by the ground plane.

Radiation intensity

( ) ( )22

2 2 2 20| | 1 sin sin sin cos2 2r I l

U E hψη θ ϕ β θ

η λ∆ = = −

(8.71)

The maximum value of (8.71) depends on whether ( )hβ is less than/ 2π or greater:

• If2

hπβ ≤

4h

λ ≤

( )2

20max / 0

sin2

I lU h θ

η βλ =

∆ =

(8.72)

• If2

hπβ >

4h

λ >

2

0/ arccos , 0max 22 h

I lU πθ ϕ

β

ηλ

= =

∆ =

(8.73)

Maximum directivity

For small hβ ,( ) 2

0

sin7.5

hD

h

ββ

=