field and wave electromagnetic

Field and Wave Electromagnetic

Chapter11

Antennas and Radiating Systems

Radio Technology Laboratory 2

Introduction (1)- Potential functions and

1-

for time harmonic cases

- The potential functions and are solutions of

non-homogeneous wave equations.

- For harmonic time dependence, the ph

A V

H A

E V j A

A V

μ

ω

= ∇×

= −∇ −

'

'

asor retarded potentials.

'4

1 2'

4

jkR

v

jkR

v

JA dv

R

V dv kR

e

e

μπ

ρ πω μεπε λ

−

−

=

= = =

∫

∫


Introduction (2)

0 Lorentz Gauge cf)

. 0

continuity equation

.

No need to evaluate the both integrals

VA

t

i e A j V

Jt

i e J j

με

ωμε

ρ

ωρ

∂⎛ ∇• + = →⎜ ∂⎜⎜ ∇• + =⎝

∂⎛ ∇• = − →⎜ ∂⎜⎜ ∇• = −⎝

∴


Introduction (3)-

1

- Three steps to determine electromagnetic fields from

a current distribution.

1. Determine from

2. Find from

3. Find from

E and H

E Hj

A J

H A

E H

ωε= ∇×


Radiation Fields of Elemental Dipoles* Elemental electric dipoles

Elemental oscillating electric dipole.

Short conducting wire of length terminated in two small conductive

spheres disks (capacitive loading)assume the

dl

Q

⇒current in the wire to be

uniform and to vary sinusoidally with time ( ) cos [ ]j ti t t e Ie ωω= = ℜ


- The current vanishes at the ends of the wire,

charge must be deposited there.

( )( ) ( ) Re[ ]

+ sign : for the charge on the upper end

j tdq ti t q t Qe

dtI j Q

Ior Q

j

ω

ω

ω

= ± =

= ±

= ± ⇒

sign : for the charge on the lower end

The pair of equal and opposite charges separated by

a short distance effectively form an electric dipole.

( )

This kind of oscillating di

p zQdl C m

−⇒

= ⋅⇒ pole is called a Hertzian dipole

The Elemental Electric Dipole (1)


0

0

- Electromagnetic fields of a Hertzian dipole

( )4

2 where = k =

c

cos - sin

j RIdl eA z

R

z R

βμπ

ω πβλ

θ θ θ

−

=

=

=




0

0

0 0

22

cos ( ) cos4

- sin ( )sin4

0

1 1[ ( ) ]

1 1 - sin [ ]

4 ( )

R

j R

R z

j R

z

R

j R

A RA A A

Idl eA A

R

Idl eA A

RA

AH A RA

R R

Idle

j R j R

θ φ

β

β

θ

φ

θ

β

θ φ

μθ θπμθ θπ

φμ μ θ

φ β θπ β β

−

−

−

= + +

⎡= =⎢

⎢⎢

= = −⎢⎢

=⎢⎢⎣

∂∂= ∇× = −

∂ ∂

= +



0

0

20 2 3

20 2 3

00

0

1

1 1 1 [ ( sin ) ( )]

sin

1 1 2cos [ ]

4 ( ) ( )

1 1 1 sin [ ]

4 ( ) ( )

0

where 120

j RR

j R

E Hj

R H RHj R R R

IdlE e

j R j R

IdlE e

j R j R j R

E

φ φ

β

βθ

φ

ωε

θ θωε θ θ

η β θπ β β

η β θπ β β β

μη πε

−

−

= ∇×

∂ ∂= −

∂ ∂

= − +

= − + +

=

= =


The Elemental Electric Dipole - Near Field (1)

2

2

- Near field

2 In the near zone, 1,

Consider only leading term.

( )sin , where 1- - 1

4 2j R

RR

Idl RH e j R

Rβ

φ

πβλ

βθ βπ

−

=

= +

Magnetic field intensity due to a current elementIdl is obtained using the Biot-Savart law in magnetostatics

'

' '0 0

2) ( ) ,

4 4 C

I Idl R dlcf dB A

R R

μ μπ π

×= = ∫


The Elemental Electric Dipole - Near Field (2)

30

30

30

2cos4

sin4

) , , ( 2cos sin )4

Note : the near field of an oscillating time-varying dipole

are the

R

R

E RE E

pE

R

pE

R

I pcf p zQdl Q E R

j R

θ

θ

θ

θπε

θπε

θ θ θω πε

= +

=

=

= = ± = +

quasi-static field

Electric field due to an elemental electric dipole of a moment in the e-direction

ρ


The Elemental Electric Dipole - Far Field (1)

0

- Far field

2 RIn the far zone, = 1,

Far zone leading terms of E and H fields are

( ) sin4

( ) sin , ( 0)4

j R

j R

R

R

Idl eH j

R

Idl eE j E

R

β

φ

β

θ

πβλ

β θπ

η β θπ

−

−

=

= =


The Elemental Electric Dipole - Far Field (2)

0

1. and are in space quardrature and in time phase.

2. : constant equal to the intrinsic impedance of

the medium.

3. The same propert

note E H

E

θ φ

θ

φ

η=Η

ies as those of a plane wave

(at very large distances from the dipole a spherical

wavefront closely resembles a plane wavefront)

4. The magnitude of the far-zone fields varies inversely

with the distance from the source.

5. The phase is a periodic function of R.

2 ( )

2

cR

f

π λλβ π

= =


The Elemental Magnetic Dipole (1)

* Elemental magnetic dipole

- Small filamentary loop of

radius b.

2

- Uniform time harmonic current

( ) cos

:

vector phasor magnetic moment

i t I t

m zI b zm

ω

π

=

= =



1

1 1

'0

1

( )

1

''0

1

' '

( )4

[1 ( )]

[(1 ) - ]4

) (- sin

j R

j R j R Rj R

j R

j R

I dlA e

R

e e e

e j R R

I dlA e j R j dl

R

cf dl x

β

β ββ

β

β

μπ

β

μ β βπ

φ

−

− − −−

−

−

=

=

− −

∴ = +

= +

∫

∫ ∫

' 'cos )y bdφ φ

0



'For every , there is another symmetrically

located differential current element on the

other side of the y-axis an equal amount

ˆof contribution to in the - .

But an equal amount of con

Idl

A x

⇒

tribution to

ˆin the opposite direction of

A

y



''0 2

12

2 2 21

'

12' 2 22

21

1' 2

sin (1 )

2

2 cos

cos sin sin

1 1 2 (1 sin sin ) (R )

1 2 (1 sin sin )

j RIbA j R e d

R

where R R b bR

R R

b bb

R R R R

b

R R

πβ

πμ φφ β φπ

ψ

ψ θ φ

θ φ

θ φ

−

−

−

−

= +

= + −

=

∴ = + −

−

∫

'1 (1 sin sin )

b

R Rθ φ+



20

2

02

202

202 3

0

202 3

0

(1 ) sin4

(1 ) sin4

1 1 sin [ ]

4 ( )

1 1 2cos [ ]

4 ( ) ( )

1 1 1 sin [ ]

4 ( ) ( )

j R

j R

j R

j RR

IbA j R e

Rm

j R eRj m

E ej R j R

j mH e

j R j R

j mH e

j R j R j R

β

β

βφ

β

θ

μφ β θ

μφ β θπωμ β θπ β βωμ β θπη β βωμ β θπη β β β

−

−

−

−

−

= +

= +

∴ = +

= − +

= − + + j Rβ



0

0

0

- Far field

( ) sin4

( ) sin4

j R

j R

m eE

R

m eH

R

β

φ

β

θ

ωμ β θπωμ β θπη

−

−

=

= −



0

H.W 11-1, 11-2, 11-3, 11-4, 11-5

) let ( , ) : Electric and magnetic field of the electric dipole.

( , ) : Electric and magnetic field of the magnetic dipole.

e e

m m

e

cf E H

E H

E Hη= 00

00 0

0

, ( )

i.e Hertzian electric dipole and elemental

magnetic dipole are dual devices.

mm e

EEH

H

if Idl j m where

φ

θ

ηη

ωμβ β ω μ εη

= − =

= = =


Antenna Patterns and Antenna Parameters (1)

* Antenna patterns and antenna parameters

- Antenna pattern (radiation pattern) : the relative far-zone field

strength versus direction at a fixed distance from an antenna.

- E-plane pattern = the magnitude of the normalized field strength

(with respect to the peak value) versus for a constant

- H-plane pattern = the magnitude of the normalized field strength

versus

θ φ

for πφ θ =2



e.g. Hertzian dipole

θ φ

φ θ π= /2



- Antenna parameters

(1) Width of main beam = beamwidth (3dB beamwidth)

(2) Sidelobe levels unwanted radiation (first sidelobe : -40dB)

(3) Directivity

- Directive gain in terms of radiation

⇒

2

intensity.

- Radiation intensity = the time-average power per unit solid angle

: (W/sr) cf) sr : steradian (solid angle)

Radiation intensity (W/sr)

- Total time-average pavU R∴ = P

ower radiated

(W)

differential solid angle sin

r avP ds Ud

d d dθ θ φ

= = Ω

Ω = =∫ ∫iP



- Directive gain

( , ) = the ratio of the radiation intensity

in the direction ( , ) to the

average radiation intensity

( , ) 4 (

/ 4

D

r

G

U U

P

θ φθ φ

θ φ π θπ

= =

max max

2

max2 2

0 0

, )

- Directivity = the maximum directive gain of an antenna

4 (Dimensionless)

4

( , ) sin

av r

Ud

U UD

U P

E

E d dπ π

φ

π

π

θ φ θ θ φ

Ω

= =

=

∫

∫ ∫



22 2

02 2

22 2 2

02

22

2 2

0 0

. : Hertzian dipole

1 1 ( ) sin

2 2 32

( ) U sin

32

4 ( , ) 4 sin 3 ( , ) sin

2(sin )sin

av

av

D

Ex

Idle E H E H

R

IdlR

UG

Ud d d

D

θ φ

π π

η β θπ

η β θπ

π θ φ π θθ φ θθ θ θ φ

∗

−

= ℜ × = =

= =

∴ = = =Ω

=

∫ ∫ ∫

P

P

( , ) 1.5 1.762DG dBπ φ = →



p- Power gain G = the ratio of its maximum radiation intensity

to the radiation intensity of a

lossless isotropic source with

i

max

the same input power P

total input power ( : ohmic loss power)

4

- Radiation efficiency

- Radiation resistan

i r l l

pi

p rr

i

P P P P

UG

P

G P

D P

π

η

= +

=

= =

22

ce = the value of a hypothetical resistance

21

2

rr m r r

m

PP I R R

I= ∴ =


Thin Linear Antennas (1)* Thin linear antennas

( ) sin ( ),

sin ( ), 0

sin ( ), 0.

m

m

m

I z I h z

I h z z

I h z z

β

ββ

= −

− >⎧= ⎨ + <⎩


Thin Linear Antennas (2)

'

0 0'

1' 2 2 2

'

-00

- The far-field contribution from the differential current element is

( ) sin .4

( 2 cos ) cos

1 1

sin sin ( )

4

j R

j R j zm

Idz

Idz edE dH j

R

R R z Rz R z

R RI

E H j e h z eR

β

θ φ

β βθ φ

η η β θπ

θ θ

η β θη βπ

−

= =

= + − ≅ −

≅

= = − cos

cos

sin ( ) : even function

cos( cos ) sin( cos )

h

h

j z

dz

h z

e z j z

θ

β θ

β

β θ β θ

−

−

= +

∫

Odd functioneven function


Thin Linear Antennas (3)-0

0 0

-

' '

sin sin ( )cos( cos )

260

( ),

cos( cos ) cos where ( )

sin

) sin( ) ?

1, , sin( ),

hj Rm

j Rm

Ax

Ax Ax

IE H j e h z z dz

Rj I

e FR

h hF

cf e Bx C dx

u e u e v Bx C vA

βθ φ

β

η β θη β β θπ

θ

β θ βθθ

= = −

=

−=

+ =

= = = +

∫

∫cos( )

sin( ) Ax

B Bx C

I e Bx C dx

= +

= +∫

H.W



2

2

2 2

2 2

1sin( ) cos( )

1sin( ) [ cos( ) sin( )

1sin( ) cos( )

[ sin( ) cos( )]

cos , , -

Ax Ax

Ax Ax Ax

Ax Ax

Ax

Be Bx C e Bx C dx

A AB

e Bx C e Bx C B e Bx C dxA A

B Be Bx C e Bx C I

A A A

eI A Bx C B Bx C

A Blet A jk B k C khθ

= + − +

= + − + + +

= + − + −

∴ = + − ++= = =

∫

∫



- Half wave dipole

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

4 4 2 ( ) sin sin

cos( cos )60 2 sin

cos( cos )2

2 sin

j Rm

j Rm

F

j IE e

R

jIH e

R

βθ

βφ

λ π λ πβ θ θλθ

θ θπ θ

θ

π θ

π θ

−

−

−= =

⎧ ⎫⎪ ⎪

∴ = ⎨ ⎬⎪ ⎪⎩ ⎭⎧ ⎫⎪ ⎪

= ⎨ ⎬⎪ ⎪⎩ ⎭


Thin Linear Antennas (6)2

2

2

2

2 2 2

0 0 0

2

0

cos( cos )151 22 sin

cos ( cos )2sin 30

sin

cos ( cos )2 1.218 obtained by numerical solution

sin

mav

r av m

IE H

R

P R d d I d

d

θ φ

π π π

π

π θ

π θ

π θθ θ φ θ

θπ θ

θθ

∗

⎧ ⎫⎪ ⎪

= = ⎨ ⎬⎪ ⎪⎩ ⎭

= =

= ⇒

∫ ∫ ∫

∫

P

P

using Simpson's law



2

2 0 2max

max

2 73.1

15 (90 )

4 60 1.64

36.54

rr

m

av m

r

PR

I

U R I

UD

P

ππ

∴ = = Ω

= =

= = =

P



cos0

- Effective antenna length

Assume a center-feed linear dipole

and a general phasor current distribution ( )

30 {sin ( ) }

(0) : the input curr

hj R j z

h

I z

jE H e I z e dz

RI

β β θθ φη β θ−

−= = ∫

0

cos

ent at the feed point of the antenna

30 (0) ( )

sin where ( ) ( )

(0)

the effective length of the transmitting antenna

j Re

h j ze h

j IE H e l

R

l I z e dz

βθ φ

β θ

η β θ

θθ

−

−

⇒ = =

=Ι

⇒

∫



1 ( ) ( )

2 (0)

The length of an equivalent linear antenna with

a uniform current (0) such that it radiates the

same

h

e hl I z dz

I

I

πθ−

= =

⇒

∫

far-zone field in the plane2

πθ =



Ex 11-6) Assume a sinusoidal current distribution on a center-fed, thin

straight half-wave dipole. Find its effective lenth.

What is its maximum value?

cos4

4

( ) sin sin ( )4

cos( cos )2 2 ( )sin

j ze

e

l z e dz

l

λβ θ

λλθ θ β

π θθ

β θ

−= −

⎡ ⎤⎢ ⎥

= ⎢ ⎥⎢ ⎥⎣ ⎦

∫



2 maximum value of ( ) ( )

2 2

note

1 ( )

(0)

Valid only for relatively short antennas having

a current maximum at the fee

e e

h

e h

l l

l I z dzI

π λ λθβ π

−

= = = <

=

⇒

∫

d point



( )

Negative sign is to conform with the convention that the

electric potential increases in a direction opposite to

that of the electric field

oce

i

Vl

Eθ = −

⇒

The effective length of an antenna

for receiving is the same as that

for transmitting.

⇒

Receiving Antennas and Reciprocity

39

Circuit relation for Radiation into Free Space

40

Effective Area of Receiving Antenna

41

Effective Area for Hertzian Dipole

42

z

Δ z

-h

h

0

r'

For matched termination:

⎟⎟⎠

⎞⎜⎜⎝

⎛=⋅= 2

2

1EAAaPP eerR η

)(θθ fr

eZIaE

jkr−

=

θθ sin23)( =f

322

ηλl

jZ =

θθ 2sin)23()( =g

2

6

4⎟⎠⎞

⎜⎝⎛=λ

πη lRr

)6/4()/(8

)sin(

8 2

22

πληθ

l

El

R

VP

r

OCR ==

πλθθλ

π 4)()sin(

8

3 22 gAe ==

cf) Normalization g( , )d =4θ φ πΩ∫

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