ECE-D Antenna 1

8/14/2019 ECE-D Antenna 1

1/19

Antenna Theory

Radiation Mechanism

Dipole antenna-Fundamentals


2/19

Introduction

Antenna:

A radiating device

A Transducer

A matching element

Any wire or a structure can radiate EM waves,however, the shape and size should

compatible with local circuit, this study canbe extended as Antenna Theory


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Transmission line

A transmission line, of Z0 characteristics

impedance, is excited by a source

+

+

tapering


4/19

Vertical tapering

By vertical tapering, the electric fields willhave a curvature shape.


5/19

Give a sinusoidal signal

at t=T/4 at t=2T/4=T/2

at t=4T/4=T

/2


6/19

Dipole antenna

Fundamental antenna, the two arms of /4 length is placednear by, will act as dipole.


7/19

Current element

At a position, a dl length of wire is consideredand assumed that have uniform current flow.

Current element is a small current carried by

small length of wire, ie I.dl


8/19

Radiation of Small current element

Consider a wire of length L is carrying acurrent I, as

The Field at a distant point P, at a distance R

)sin(. tII m =

dl

P

R


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Retarded Vector Potential;

When current I is passing through the current element, thefield created a point P will phase shifted about R/c

So, to specify the field with respect to the present time, we canwrite the current term as

It is a retarded form,

can be compared with

A

dl

P

r

[ ]

=

C

rtII m sin

[ ] ( )rtII m = sin


10/19

The vector potential

IntegralPot.

4

4 == dv

r

J

r

IdA

z4

2/

2/dz

r

eIA

jkr

z

=

yI

r

( )rAz

the onlythe onlycomponentcomponent

x

z

zr

eIA

jkr

z 4

=

0

0

==

A

AIn spherical form

:Convert into Polar form = coszrAA =

sin

z

AA 0=

A


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Electric and Magnetic Fields

We can find E and H terms from;

By derivation, we will get:

j

AAjwE

)( +=

AH

=1

M a g n e tic Fie ld

C om po ne n ts

0=H

0=rH

+=

2

2kr

e

r

ejSin

IH

jkrjkr

E le ctric Fie ld

C om p on en ts

+

=

os

322 rk

ej

kr

eC

IE

-jkrjkr

r

322in

2

+=

rk

ej

kr

e

r

ejS

IE

-jkrjkrjkr

0=E

=w h e re


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consider the H

+=

2

2kr

e

r

ejSinIHjkrjkr

N e a r fie ld o r In d u ctio n fie ld

E a n d H are p h ase sh ifted b y 9

Fa r fie ld o r R a d ia tio n fie ld

N o p h a se sh ift b e tw e e n th e E a n d H fie ld

ri

NearField

FarField

We can find the position of riby

equatingboth the components


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Finding ri

+=

2

2kr

er

ejSinIHjkrjkr

A t ri ,the far field will change from near field both will be equale

,Or from the above equation real part is equale to imaginary

( ) ( )2

cossin

r

kr

r

krk=

;By approximation

21

rrk=

2/1 == kr ri= .0 159

If length of antenna is greater than , then

2

2l

ri =


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Induction/intermediateregion

Since kr value is much less than unity,the real term is almost vanish. Thisimaginary component creates a 90ophase shift between H and E.

Thus lead to polarization on thedirection of propagation, called Crossfield.

Thats if you expect a linearly polarizedantenna to receive within inductionfield, it cant

For example: An FM station transmits at100MHz, therefore: =3meter, so youcant get the signal before 0.47meter.

AM station at 1MHz, then ri=47 meter,


15/19

Far field

The kr is very greater than unity, thus

-jkrjes

lH

jkrje

l

inr2

I

r2

sinE

=H

EZw

C alled W a ve im p en d en ce

-In p h a se


16/19

Radiation pattern

Consider the far-field condition:

yI r

x

z

inr2

I

r2

sinE

s

l

H

l


17/19

Power density

The poynting vector:

The radical and transverse components

in ce th e tra n sv e rse p o w e r is n o t co n trib u tin g a t fa r fie ld s


18/19

Power in Real and Imaginary form says

Wm, We are magnetic and electric energy density,

that stored in current element (or antenna)

The real term is modified as

> >V a n ish e s a t k r 1

R ris radiation resistance


19/19

Problem

Find the Radiation resistance of adipole antenna about length of/50.

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ECE-D Antenna 1