Lecture 9 Types of Antenna

1

Antenna is a structure designed for radiating and receiving EM energy in a prescribed manner

Far field region ( the distance where the receiving antenna is located far enough for the transmitter to appear as a point source)

The shape or pattern of the radiated field is independent of r in the far field.

Normalized power function or normalized radiation intensity

2

22Lr

max

( , , )( , )nP rP P

Directivity is the overall ability of an antenna to direct radiated power in a given direction.

An antenna’s pattern solid angle:

Total radiated power can be written as

Antenna efficiency e is measured as

3

( , )p nP d

maxmax max

( , ) 4( , )( , )

n

n ave p

PD DP

2max .rad pP r P

.rad rad

rad diss rad diss

P Re P P R R

If the current distribution of a radiating element is known, we can calculate radiated fields.

In general, the analysis of the radiation characteristics of an antenna follows the three steps below:

1. Determine the vector magnetic potential from known of assumed current on the antenna.

2. Find the magnetic field intensity from .

3. Find the electric field intensity from .

A

J

H

A

E

H

4

From the point form of Gauss’s law for magnetic field,

0B

Define

therefore

we can express as

where Jd = current density at the point source (driving point) R = distance from the point source to the observation point (m)

B A

( ) 0A

A

0

4dJA dVR

5

From here we can determine , then find in free space.

We can then find the electric field from

The time-averaged radiated power is

B

H

0 00 .rE a H

The subscript “0” represents the observation point.

0 01( , , ) Re( )2P r E H

W/m2.

6

1. Hertzian dipole (electric dipole)

2. Small loop antenna (magnetic dipole)

3. Dipole antenna

7

A short line of current that is short compared to the operating wavelength. This thin, conducting wire of a length dl carries a time-harmonic current

0( ) cos( )i t I t A

and in a phasor form 0jI I e A.

8

The current density at the source seen by the observationpoint is

A differential volume of this current element is dV = Sdz.

.j Rd z

IJ e aS

9

Therefore

Then

Where at the observation point.

For short dipole, R r, thus we can write

Conversion into the spherical coordinate gives

.j Rd zJ dV Ie dza

/ 20

0/ 24

j Rl

zl

eA I dzaR

0A A

00 .4

j r

zeA Il ar

cos sin .z ra a a 10

Therefore

We can then calculate for 0.B

00 (cos sin ).4

j r

reA Il a ar

11

Multiply 2 to both nominator and denominator, so we have

We are interested in the fields at distances very far from the antenna, which is in the region where

1r

20

0 2

1 sin .4

j r jA Il e ar r

2r or 2R

12

Under a far-field condition, we could neglect

and

Then

and

Finally, W/m2.

2

1( )R

3

1 .( )R

0 sin

4

j rIl eH j ar

00 00 sin .4

j r

rIl eE a H j ar

2 2 2

202 2( , ) sin

32r

I lP r a

r

13

Since the current along the short Hertzian dipole is uniform,we refer the power dissipated in the radial distance Rrad to I,

2maxrad pP r P

2

max

( , )( , ) sinnP rP rP

2 2 8sin sin sin 3p d d d

2 2 2

240 .I l

2

2rad

radI RP or 2 280 ( )rad

lR m.14

15

91cos(2 10 )i t

16

a) Pmax at r = 100 m

b) What is the time-averaged power density at P (100, /4, /2)?

c) Radiation resistance

17

Assume a <<

A complicate derivation brings to

18

0

0sin

4

j rIS eH ar

0 sin4

j rIS eE ar

If the loop contains N-loop coil then S = Na2

Longer than Hertizian dipole therefore they can generate higher radiation resistance and efficiency.

19

Divide the dipole into small elements of Hertzian dipole. Then find and .

Figure of dipole

H

E

20

H

E

The current on the two halves are Symmetrical and go to zero at the ends.

We can write Where

Assume = 0 for simplicity.

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( , ) ( ) cosi z t I z t

0

0

sin( ( )); 02 2( )sin( ( )); 02 2

j

j

L LI e z zI z

L LI e z z

From

22

sin4

j rI dl edH j ar

0 / 20

/ 2 0

sin( ( )) sin ' sin( ( )) sin '4 2 2

j r j rL

L

I e L e LH j a z dz z dz

R R

In far field but since small differences can be critical.

, 'r R j r j Re e

We can write

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cosR r z

( cos )j R j r ze e

0 / 2cos cos0

/ 2 0

sin( ( )) sin( ( ))4 2 2

sinL

j z j z

L

j rI L LH j a e z dz e z dz

e

From

In our case

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2 2cos( ) sin( ) cos( )ax

ax ee c bx dx a c bx b c bxa b

, cos , ,2lx z a j c b

0cos( cos ) cos( )2 2

2 sin

j rl l

I eH j ar

0 0rE a H Ha

202

15( , ) ( ) rIP r F ar

where

25

2

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

l l

F

max

( )( ) ( )nFP F

20

max max2

15 ( )IP Fr

1. Find Pn(), calculate F() over the full range of for length L in terms of wavelength then find Fmax (this step requires Matlab)

2. Find p

3. Dmax (Directivity)

4. Rrad

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2

max

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

l l

dF

4p

max30 ( ) pF

Link to Matlab file

27

28

2L

2 2L

22

02 2

cos ( cos )15 2( , )sin

rIP r ar

20

max 2

15IPr

2

2max

cos ( cos )( ) 2( ) ( ) sinnFP F

Using Matlab, we get

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p = 7.658

Dmax = 1.64

Rrad = 73.2

This is much higher than that of the Hertzian dipole.