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Antennas

© Amanogawa, 2006 – Digital Maestro Series 1

Antennas

Antennas are transducers that transfer electromagnetic energy between a transmission line and free space.

Transmitting Antenna

I

I

Transmitter

Transmission Line Electromagnetic

Wave

Receiving Antenna

I

IReceiver

Transmission Line ElectromagneticWave

Antennas


Here are a few examples of common antennas:

Linear dipole fed by a two-wire line

Ground plane

Linear monopole fed by a single wire over a ground plane

Coaxial ground plane antenna

Parabolic (dish) antenna

Linear elements connected to outer conductor of the coaxial cable simulate the ground plane

Uda-Yagi dipole array

Passive elements

Loop dipole

Log−periodic array

Loop antenna Multiple loop antenna wound

around a ferrite core

Antennas


From a circuit point of view, a transmitting antenna behaves like an equivalent impedance that dissipates the power transmitted

The transmitter is equivalent to a generator.

I

I

Transmitter

Transmission Line

Transmitting Antenna

I

I

Transmitter

Transmission Line Electromagnetic

Wave

Zeq = Req + jXeq

P t R Ieq( ) =12

2

Vg

Zg

Antennas


A receiving antenna behaves like a generator with an internal impedance corresponding to the antenna equivalent impedance.

The receiver represents the load impedance that dissipates the time average power generated by the receiving antenna.

I

I

Transmission Line

Receiving Antenna

I

I

Receiver

Transmission Line

P t R Iin( ) =12

2

Veq

Zeq

ElectromagneticWave

ZinZR

Antennas


Antennas are in general reciprocal devices, which can be used both as transmitting and as receiving elements. This is how the antennas on cellular phones and walkie−talkies operate.

The basic principle of operation of an antenna is easily understood starting from a two−wire transmission line, terminated by an open circuit.

Vg

Zg

| I |

| I |Note: This is the return current on the second wire, not the reflected current already included in the standing wave pattern.

ZR → ∞ Open circuit

Antennas


Imagine to bend the end of the transmission line, forming a dipole antenna. Because of the change in geometry, there is now an abrupt change in the characteristic impedance at the transition point, where the current is still continuous. The dipole leaks electromagnetic energy into the surrounding space, therefore it reflects less power than the original open circuit ⇒ the standing wave pattern on the transmission line is modified

Z0Vg

Zg

| I |

| I |

| I0 |

| I0 |

Antennas


In the space surrounding the dipole we have an electric field. At zero frequency (d.c. bias), fixed electrostatic field lines connect the metal elements of the antenna, with circular symmetry.

E E

Antennas


At higher frequency, the current oscillates in the wires and the field emanating from the dipole changes periodically. The field lines propagate away from the dipole and form closed loops.

Antennas


The electromagnetic field emitted by an antenna obeys Maxwell’s equations

Under the assumption of uniform isotropic medium we have the wave equation:

Note that in the regions with electrical charges ρ

jj

E HH J E

ω µω ε

∇ × = −

∇ × = +

2

2

E H J EH J E

J H

j jj

ω µ ω µ ω µ εω ε

ω µ ε

∇ × ∇ × = − ∇ × =− +

∇ × ∇ × = ∇ × + ∇ ×

= ∇ × +

( )2 2E E E Eρ ε∇ × ∇ × = ∇∇ ⋅ − ∇ = ∇ − ∇

Antennas


In general, these wave equations are difficult to solve, because of the presence of the terms with current and charge. It is easier to use the magnetic vector potential and the electric scalar potential. The definition of the magnetic vector potential is

Note that since the divergence of the curl of a vector is equal to zero we always satisfy the zero divergence condition

We have also

B A= ∇ ×

( )B A 0∇ ⋅ = ∇ ⋅ ∇ × =

( )j jj E AE H A 0ωω µ ω∇ × = − = − ∇ × × =⇒ ∇ +

Antennas


We define the scalar potential φ first noticing that

and then choosing (with sign convention as in electrostatics)

Note that the magnetic vector potential is not uniquely defined, since for any arbitrary scalar field ψ

In order to uniquely define the magnetic vector potential, the standard approach is to use the Lorenz gauge

( ) 0φ∇ × ±∇ =

( ) ( )E A E Ajj ωω φ φ∇ × + = ∇ × −∇ − − ∇⇒ =

( )B A A ψ= ∇ × = ∇ × + ∇

jA 0ω µε φ∇ ⋅ + =

Antennas


From Maxwell’s equations

From vector calculus

( ) ( )

j

jj j

1H B J E

B J EA J A

ω εµ

µ ω µ εµ ω µ ε ω φ

∇ × = ∇ × = +

∇ × = +∇ × ∇ × = + − − ∇⇒

( ) ( ) j2 2AA A J A ω µ ε φµ ω µ ε∇ × ∇ × = − ∇ =∇ ∇ ⋅ − ∇+

( )j jAA ω ω µ φµ ε φ ε∇∇ ⋅ = − ⇒ ∇⋅ = − ∇

Lorenz Gauge

( ) ( ) 2∇ × ∇ × = ∇ ∇ ⋅ − ∇… … …

Antennas


Finally, the wave equation for the magnetic vector potential is For the electric field we have

The wave equation for the electric scalar potential is

2 2 2 2A A A A Jω µε β µ∇ + = ∇ + = −

( )

( )2 2

D E

A

Aj

j j jω µε φ

ρρ ω φερφ ω φ ωε

∇ ⋅ = ⇒ ∇⋅ = ∇ ⋅ − − ∇ =

∇ + = ∇ +∇ ⋅ = −−

2 2 2 2 ρφ ω µ ε φ φ β φε

∇ + = ∇ + = −

Antennas


The wave equations are inhomogenoeous Helmholtz equations, which apply to regions where currents and charges are not zero. We use the following system of coordinates for an antenna body

z

xy

rr '

r r '−

Radiating antenna body

Observation pointrr

J( ')( ')ρ

dV '

Antennas


The generals solutions for the wave equations are

The integrals are extended to all points over the antenna body where the sources (current density, charge) are not zero. The effect of each volume element of the antenna is to radiate a radial wave

( ) ( ) j r r

VJ r er dV

r r

''A '

4 '

βµπ

− −=

−∫∫∫

( ) ( ) j r r

Vr er dV

r r

''1 '4 '

βρφ

πε

− −=

−∫∫∫

j r rer r

'

'

β− −

−

Antennas


Infinitesimal Antenna

z

xy

r r r '= −

r ' 0=

Infinitesimal antenna body

Observation point

J(0)(0)ρ

dV '

S∆

z∆

I constant phasor= z λ∆ <<

Dielectric medium (ε , µ)

Antennas


The current flowing in the infinitesimal antenna is assumed to be constant and oriented along the z−axis

The solution of the wave equation for the magnetic vector potential simply becomes the evaluation of the integrand at the origin

( ) ( )( ) z

S r S

V r z

V

i

S zI J ' J 0 '

'J ' I

= ∆ ⋅ = ∆ ⋅ ∆ =

∆ ∆

∆

=

∆ ⋅

1H AI

A4 1E H

j r

zz e

ir

j

β µµπ

ω ε

−⇒

= ∇ ×∆ = = ∇ ×

Antennas


There is still a major mathematical step left. The curl operations must be expressed in terms of spherical coordinates

z

xy

r

θ

ϕiϕ

iϕ

iθ

ri

Azimuthal angle

Polar angle

Antennas


In spherical coordinates

( ) ( )

( ) ( )

( ) ( )

r

r

r

r

r

i r i r i

rr

A rA r A

A A ir

A r A ir r

r A A ir r

2

sin

1Asin

sin

1 sinsin

1 1sin

1

θ ϕ

θ ϕ

ϕ θ

ϕ θ

θ ϕ

θ

θ ϕθ

θ

θθ θ ϕ

θ ϕ

θ

∂ ∂ ∂∇ × =

∂ ∂ ∂

∂ ∂ = − ∂ ∂

∂ ∂ + − ∂ ∂

∂ ∂ + − ∂ ∂

Antennas


We had

For the fields we have

j r

z z r

j r

z ei i i i

r

j z ei

r r

ith

j

wI

A cos sin4

I 1A 1 sin4

β

θ

β

ϕ

µθ θ

π

µ βθ

π β

−

−

∆= = −

∆ ⇒ ∇ × = +

j rj z ei

r j rI1 1H A 1 sin

4

β

ϕβ

θµ π β

−∆ = ∇ × = +

Antennas


The general field expressions can be simplified for observation point at large distance from the infinitesimal antenna

( )

( )

j r

r

j z ej r

ij r j r

ij r j r

2

2

I1E H4

1 12 cos

1 1sin 1

β

θ

βµω ε ε π

θβ β

θβ β

−∆= ∇ × =

× +

+ + +

( )r r

j r j r 21 1 21 1πββ λβ

>> >> ⇒ = >>

Antennas


At large distance we have the expressions for the Far Field

• At sufficient distance from the antenna, the radiated fields are perpendicular to each other and to the direction of propagation.

• The magnetic field and electric field are in phase and

These are also properties of uniform plane waves.

E H Hµ ηε

= =

j rj z ei

rI

H sin4

β

ϕβ

θπ

−∆≈

j rj z ei

rI

E sin4

β

θβµ θ

ε π

−∆≈

r2π λ>>

Antennas


However, there are significant differences with respect to a uniform plane wave:

• The surfaces of constant phase are spherical instead of planar, and the wave travels in the radial direction.

• The intensities of the fields are inversely proportional to the distance, therefore the field intensities decay while they are constant for a uniform plane wave.

• The field intensities are not constant on a given surface of constant phase. The intensity depends on the sine of the polar angle θ.

The radiated power density is

2*

22

1 1( ) Re E H2 2

Isin

2 4

r

r

P t i H

zi

r

ϕµε

βη θπ

= × =

∆ =

Antennas


The spherical wave resembles a plane wave locally in a small neighborhood of the point ( r, θ, ϕ ).

z( )P t

Eθ

Hϕ

rθJ

ϕ

Antennas


Radiation Patterns

Electric Field and Magnetic Field

z

x

or HEθ ϕ

θ

y

x

Plane containing the antenna

proportional to sinθ

Plane perpendicular to the antenna

omnidirectional or isotropic

Fixed r

Antennas


Time−average Power Flow (Poynting Vector)

z

x

( )P t

θ

Plane containing the antenna

proportional to sin2θ

Plane perpendicular to the antenna

omnidirectional or isotropic

y

x

Fixed r

Antennas


Total Radiated Power

The time−average power flow is not uniform on the spherical wave front. In order to obtain the total power radiated by the infinitesimal antenna, it is necessary to integrate over the sphere

Note: the total radiated power is independent of distance. Although the power density decreases with distance, the integral of the power over concentric spherical wave fronts remains constant.

2 20 0

22 2

2 30

4 3

sin ( )

I I42 sin2 4 3 4

totP d d r P t

z zr d

r

π π

π

π

ϕ θ θ

β βη πηπ θ θπ π

=

=

=

∆ ∆= =

∫ ∫

∫

Antennas


1 2tot totP P=

1totP

2totP

Antennas


The total radiated power is also the power delivered by the transmission line to the real part of the equivalent impedance seen at the input of the antenna

The equivalent resistance of the antenna is usually called radiation resistance. In free space

2 22 2I1 4 2 1 2I I

2 3 4 2 3

eq

tot eq

R

z zP R πη π πηλ π λ

∆ ∆ = = =

[ ] [ ]2

2820 01oo

eqozR πµη π

λη

ε= =

∆

= Ω =

⇒ Ω

Antennas


The total radiated power is also used to define the average power density emitted by the antenna. The average power density corresponds to the radiation of a hypothetical omnidirectional (isotropic) antenna, which is used as a reference to understand the directive properties of any antenna.

z

x

( , )P t θ

θ

aveP

Power radiation pattern of an omnidirectional average antenna

Power radiation pattern of the actual antenna

Antennas


The time−average power density is given by

The directive gain of the infinitesimal antenna is defined as

( )

Surface of wave front2

22

Total Radiated Powe

2

r

I1I12 3 44 4

ave

tot

P

zP zrr r

βη ηβπ ππ π

= =

∆ = = ∆ =

12 22

2

( , , ) I Isi( , )

3 sin2

n2 4 3 4ave

P t r z zP r r

Dθ β βη ηθθ

θ

πϕ

π

− ∆ ∆ = =

=

Antennas


The maximum value of the directive gain is called directivity of the antenna. For the infinitesimal antenna, the maximum of the directive gain occurs when the polar angle θ is 90°

The directivity gives a measure of how the actual antenna performs in the direction of maximum radiation, with respect to the ideal isotropic antenna which emits the average power in all directions.

23max ( , ) sin 1Directivi .t2 2

y 5D πθ ϕ = = =

x

z

90°

aveP maxP

Antennas


The infinitesimal antenna is a suitable model to study the behavior of the elementary radiating element called Hertzian dipole.

Consider two small charge reservoirs, separated by a distance ∆z, which exchange mobile charge in the form of an oscillatory curent

+

−

( )I t

t

z∆

+

−

+

−+

−

+

− +

−

Antennas


The Hertzian dipole can be used as an elementary model for many natural charge oscillation phenomena. The radiated fields can be described by using the results of the infinitesimal antenna.

Assuming a sinusoidally varying charge flow between the reservoirs, the oscillating current is

current flowing out of reser

charge on reference resevoir rvoir

cos( ) ( ) ( ) Iphaso

or

o od dI t q t q t j qdt dt ω ω= = =⇒

Radiation pattern

oq

oI

Antennas


A short wire antenna has a triangular current distribution, since the current itself has to reach a null at the end the wires. The current can be made approximately uniform by adding capacitor plates.

The small capacitor plate antenna is equivalent to a Hertzian dipole and the radiated fields can also be described by using the results of the infinitesimal antenna. The short wire antenna can be described by the same results, if one uses an average current value giving the same integral of the current

max 2oI I=

ImaxIoz∆

Io

Antennas


Example − A Hertzian dipole is 1.0 mm long and it operates at the frequency of 1.0 GHz, with feeding current Io = 1.0 Ampéres. Find the total radiated power.

For a short dipole with triangular current distribution and maximum current Imax = 1.0 Ampére

8 9

22 3 2

Hertzian dipole3 10 10 0.3 m 300 mm

1 mm

4 2 1 2120 ( ) ( 1 10 )3 4 12 0.3

4.39 mW

o o

otot

I z

c fz

I zPη

λ

λλ

π η π ππλ π π

−

∆

= ≈ × = =∆ = ⇒

∆ = = ⋅

=

max 2 4.39 / 4 1.09 mWo totI I P= ⇒ = ≈

Antennas


Time−dependent fields - Consider the far−field approximation

( )

( )

( )

( )

2

I sinRe H Re

4

I sinRe cos( ) sin

I sins

I sinsin(

in

( )4

Re

( )

4

4

E

)

j t rj t

j t

j ze i e

r

zi j t r j t

E t

H t

zi t r

r

t

r

e

r

r

zi

rθ

ω βωϕ

ϕ

ϕ

ω

β θπ

β

β θη ω

θω β ω

β θω

π

π

β

β

π

β

−∆ = ≈

∆

≈ − + −

≈

=

≈ − −

∆−

∆

−

Antennas


Linear Antennas

Consider a dipole with wires of length comparable to the wavelength.

z∆

'θ

θ

'r

r

'iθiθ

'z

z

1L−

2L

Antennas


Because of its length, the current flowing in the antenna wire is a function of the coordinate z. To evaluate the far−field at an observation point, we divide the antenna into segments which can be considered as elementary infinitesimal antennas.

The electric field radiated by each element , in the far−field approximation, is

In far−field conditions we can use these additional approximations

'I' sin '

4 '

j rj z eE i

r

β

θβµ θ

ε π

−∆∆ =

r r z'

' 'cosθ θ

θ≈≈ −

Antennas


The lines r and r’ are nearly parallel under these assumptions.

z∆

'θ θ≈

θ

'r

r

'z

z2L

'cosz θThis length is neglected if

' 'cosr r z θ≈ −

Antennas


The electric field contributions due to each infinitesimal segment becomes

The total fields are obtained by integration of all the contributions

you cannotneglect here

'cos

4 'cosI

'4

j r j zj z e eE i

r z

β β θ

θβµ

ε π θπ

−∆∆ =

−

you can neglect here

sinθ

21

21

cos

cos

E sin I( )4

H sin I( )4

j r L j zL

j r L j zL

j ei z e dzr

j ei z e dzr

ββ θ

θ

ββ θ

ϕ

µ β θε π

β θπ

−

−

−

−

= ⋅

= ⋅

∫

∫

Antennas


Short Dipole

Consider a short symmetric dipole comprising two wires, each of length L << λ . Assume a triangular distribution of the phasor current on the wires

The integral in the field expressions becomes

( )( )

max

max

1 0I( )

1 0I z L z

zI z L z

− ≥= + <

cosmax

since short

1

dipolefor a

cos

2max 1

1

2I( ) I( )2

z

L LjL L

j

z Lz e dz z dz

z L L

I

e β

β θ

θ

πβ βλ

−≈

−

= ⋅ =

≈

≈ =

⇒

∫ ∫

Antennas


The final expression for far−fields of the short dipole are similar to the expressions for the Hertzian dipole where the average of the triangular current distribution is used

average current

max

max

max

E sin 24 2

sin4

H sin4

j r

j r

j r

j e Ii Lr

j I L eir

j I L eir

zβθ

βθ

βϕ

µ β θε π

µ β θε π

β θπ

−

−

−

= ⋅ ⋅

=

∆

=

Antennas


Half−wavelength dipole

Consider a symmetric linear antenna with total length λ/2 and assume a current phasor distribution on the wires which is approximately sinusoidal

The integral in the field expressions is

maxI( ) cos( )z I zβ=

( )4

cos maxmax 2

4

2 coscos cos2sin

j z II z e dzλ

β θ

λ

π θββ θ−

= ∫

Antennas


We obtain the far−field expressions

and the time−average Poynting vector

max

max

cosE cos2 sin 2

cosH cos2 sin 2

j r

j r

j e Iir

j e Iir

βθ

βϕ

µ π θε π θ

π θπ θ

−

−

=

=

22max

2 2 2cos( ) cos28 sin

rIP t ir

µ π θε π θ

=

Antennas


The total radiated power is obtained after integration of the time−average Poynting vector

The integral above cannot be solved analytically, but the value is found numerically or from published tables. The equivalent resistance of the half−wave dipole antenna in air is then

( )

2.4376

22max 0

2max

1 cos1 12 4

1 0.1939782

eq

tot

R

uP I duu

I

πµε π

µε

≈

− =

= ⋅

∫

( 2) 0.193978 73.07eqR µλε

= ⋅ ≈ Ω

Antennas


The direction of maximum radiation strength is obtained again for polar angle θ =90° ande we obtain the directivity

The directivity of the half−wavelength dipole is marginally better than the directivity for a Hertzian dipole (D = 1.5).

The real improvement is in the much larger radiation resistance, which is now comparable to the characteristic impedance of typical transmission line.

2max2 2

2 2max2 2

( , ,90 ) 8 1.64114 2.4376

8tot

IP t r rDP r I

r

µε π

µπεπ

°= = ≈

⋅

Antennas


From the linear antenna applet

Radiation Pattern for E and H Power Radiation Pattern

Antennas


For short dipoles of length 0.0005 λ to 0.05 λ


Antennas



Antennas


For general symmetric linear antennas with two wires of length L, it is convenient to express the current distribution on the wires as

The integral in the field expressions is now

( ) ( ) maxI sinz I L zβ= −

( )

( ) ( )

cosmax

max2

sin

2 cos cos cossin

Lj z

LI L z e dz

I L L

β θβ

β θ ββ θ

−− =

= −

∫

Antennas


The field expressions become

( ) ( )

( ) ( )

21

21

cos

max

cos

max

E sin I( )4

cos cos cos2 sin

H sin I( )4

cos cos cos2 sin

j r L j zL

j r

j r L j zL

j r

j ei z e dzr

j I ei L Lr

j ei z e dzr

j I ei L Lr

ββ θ

θ

βθ

ββ θ

ϕ

βϕ

µ β θε π

µ β θ βε π θ

β θπ

β θ βπ θ

−

−

−

−

−

−

= ⋅

= −

= ⋅

= −

∫

∫

Antennas


Examples of long wire antennas


Antennas



antennas - amanogawaamanogawa.com/archive/docs/antennas1.pdfantennas are in general reciprocal...

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