chapter chapter 7 : antenna synthesis : antenna...

Chapter Chapter 7 7 : Antenna Synthesis: Antenna Synthesis

• Continuous sources vs. Discrete sources

• Schelkunoff polynomial method

• Fourier transform method

• Woodward-Lawson method

1

• Woodward-Lawson method

• Triangular, cosine and cosine-squared

amplitude distributions

Continuous sourcesContinuous sourcesRecall the array factor

If the number of elements increases in a fixed-length array,

the source approaches a continuous distribution.

∑=

− +==N

n

nj

n kdea1

)1( cos; βθψψAF

2

the source approaches a continuous distribution.

In the limit, the array factor becomes the space factor, i.e.,

)'()1( )'(zj

n

nj

nnezIea

φβ =−

The radiation characteristics of continuous sources can

be approximated by discrete-element arrays, i.e.,

∫−+=

2/

2/

)]'(cos'[')'(

l

l

zkzj

n dzezI nφθSF

Schelkunoff polynomial Schelkunoff polynomial

methodmethod

)cos( βθψ +==+= kdjjeejyxz

The array factor for an N-element, equally spaced, non-

uniform amplitude, and progressive phase excitation is given

by

∑=

− +==N

n

nj

n kdea1

)1( cos; βθψψAF

3

)cos( βθψ +==+= kdjjeejyxzLet

∑=

−− +++==N

n

N

N

n

n zazaaza1

1

21

1LAF

which is a polynomial of degree (N-1).


method (method (22))Thus

)())(( 121 −−−−= NN zzzzzza LAF

where z1,z2,…,zN-1 are the roots. The magnitude then

becomes

|||||||||| −−−= zzzzzza LAF

4

ψψψ ∠=∠== 1|||| zezzj

Note that

βθλπ

βθψ +=+= cos2

cos dkd

z is on a unit circle.

|||||||||| 121 −−−−= NN zzzzzza LAF


method (method (33))

VR=Visible Region

IR=Invisible Region

β = 0

5

β = 0



VR=Visible Region

IR=Invisible Region

β = π/4

6

β = π/4



7


method : Examplemethod : ExampleDesign a linear array with a spacing between the elements

of d=λ/4 such that it has zeros at θ=0,π/2,π. Determine the

number of elements, their excitation, and plot the derived

pattern.

8


method : Example patternmethod : Example pattern

−15

−10

−5

04−element array factor

9

0 20 40 60 80 100 120 140 160 180−50

−45

−40

−35

−30

−25

−20

θ [Degree]

|(A

F) n

| [d

B]

Fourier Transform MethodFourier Transform Method

The normalized space factor for a continuous line-source

distribution of length l can be given by

+

=→−=

==

−

−−

− ∫∫k

kk

dzezIdzezI

z

l

l

zjl

l

zkkj z

ξθθξ

θ ξθ

1

2/

2/

'2/

2/

)')cos(

coscos

')'(')'()(SF

10

+=→−= −

k

kkk z

z

ξθθξ 1coscos

where kz is the excitation phase constant of the source. If

I(z’)=I0/l,

−

−=

k

kkl

k

kkl

Iz

z

θ

θθ

cos2

cos2

sin

)( 0SF

Fourier Transform Method (Fourier Transform Method (22))

Since the current distribution extends only over -l/2≤z’≤ l/2,

∫∞

∞−== ')'()()( '

dzezIzjξξθ SFSF

∫∫∞

∞−

−∞

∞−

− == ξθπ

ξξπ

ξξdedezI

zjzj '' )(2

1)(

2

1)'( SFSF

The current distribution can then be given by

11

The approximate source distribution Ia (z’) is given by

∫∫ ∞−∞−== ξθ

πξξ

πdedezI )(

2)(

2)'( SFSF

≤≤−== ∫

∞

∞−

−

elsewhere0

2/'2/)(2

1)'(

)'(

'lzldezI

zI

zj

a

ξξπ

ξSF

∫−==2/

2/

' ')'()()(l

l

zj

aaa dzezIξξθ SFSFThus

Fourier Transform Method : Fourier Transform Method : ExampleExample 77..22

Determine the current distribution and the approximate radiation

pattern of a line source placed along the z-axis whose desired

radiation pattern is symmetrical about θ=π/2, and it is given by

≤≤

=elsewhere0

4/34/1)(

πθπθSF

12

elsewhere0

Fourier Transform Method : Fourier Transform Method :

ExampleExample


Linear ArrayLinear ArrayFor an odd number of elements, the array factor is given by

∑−=

==M

Mm

jm

meaψψθ )()( AFAF

For an even number of elements,

∑∑ −−

+ +==M

mj

m

mj

m eaea]2/)12[(

1]2/)12[()()( ψψψθ AFAF

14

where

−≤≤−+

≤≤−

=1

2

12

12

12

'

mMdm

Mmdm

zm

Mmmdzm ±±±== ,,2,1,0,'K

Elements’ locations

∑∑=−= m

m

Mm

m

1

βθψ += coskd

Even-number

Odd-number


Linear Array (Linear Array (22))For an odd number of elements, the excitation coefficients can

be obtained by

MmM

dedeT

ajmjm

T

Tm

≤≤−

== −

−

−

− ∫∫ ψψπ

ψψ ψπ

π

ψ )(2

1)(

1 2/

2/AFAF

15

MmM ≤≤−

where

For an even number of elements,

βθψ += coskd

≤≤

−≤≤−=

−−

−

+−

−

∫

∫Mmde

mMde

amj

mj

m

1)(2

1

1)(2

1

]2/)12[(

]2/)12[(

ψψπ

ψψπ

ψπ

π

ψπ

π

AF

AF


ExampleExampleSame as Example 7.2; non-zero only

2/cos2/ πβθψπ ≤+=≤− kd

4/34/ πθπ ≤≤

thus

thereforesin

πm

16

2

2sin

2

1

2

1 2/

2/ πψ

π

π

π

ψ

mdea

jm

m

== ∫−

−

0101.00588.0

010.00518.02170.0

0455.00895.03582.0

0496.00578.00.1

73

1062

951

840

==

−==−=

=−==

−===

±±

±±±

±±±

±±

aa

aaa

aaa

aaa

Result


ExampleExample

Quiz

A 5-element uniform linear array with a

spacing of λλλλ between elements is designed

to scan at θθθθ=ππππ/3. Assume that the array is

aligned along the z-axis.

a) Find the array factora) Find the array factor

b) Find the angle of the grating lobe.

c) Find the condition such that there exists no

grating lobe.

Quiz solution

0.6

0.7

0.8

0.9

1

| [d

B]

5−element array factor

d=λ

d=5λ/8

d=λ/2

0 20 40 60 80 100 120 140 160 1800

0.1

0.2

0.3

0.4

0.5

θ [Degree]

|(A

F) n

| [d

B]

Woodward-Lawson Method• Sampling the desired pattern at various discrete

locations.

• Use composing function of the forms:

mmmmmm NNbb φφψψ sin/)sin(or /)sin(

as the field of each pattern sample

• The synthesized pattern is represented by a finite

sum of composing functions.

• The total excitation is a sum of space harmonics.

Woodward-Lawson Method:

Line-source

2/'2/)'(cos'

lzlel

bzi mjkzm

m ≤≤−= − θ

M1

Let the source be represented by a sum of the following constant current source of length l.

Then the current source can be given by

∑−=

−=M

Mm

jkz

mmeb

lzI

θcos'1)'(

number) odd 12(for ,,2,1,0

number)even 2(for ,,2,1 where

+±±±=

±±±=

MMm

MMm

K

K

( )

( )m

m

mm kl

kl

bs

θθ

θθθ

coscos2

coscos2

sin

)(

−

−=

The field pattern of each current source is given by

Composing function


Line-source (2)For an odd number samples, the total pattern becomes

b can be obtained from the value at the sample points θ , i.e.,

( )

( )∑−= −

−=

M

Mmm

m

m kl

kl

b

θθ

θθθ

coscos2

coscos2

sin

)(SF

dmmb )(SF θθ ==

lkz lz

λπ =∆⇒=∆ = 2' '||

bm can be obtained from the value at the sample points θm, i.e.,

In order to satisfy the periodicity of 2π and faithfully reconstruct the desired pattern,


Line-source (3)

samples oddfor ,2,1,0,cos K±±=

=∆= ml

mmm

λθ

Thus the location of each sample is given by

,2,1,1212

=

−

=∆−

Kmmm λ

Therefore, M should be the closest integer to M=l/λ.

sampleseven for

,2,1,2

12

2

12

,2,1,22

cos

−−=

+=∆

+

=

=∆=

K

K

ml

mm

ml

m λθ


Example

5,,2,1,0),2.0(cos)(cos 11 ±±±==∆= −−Kmmmmθ

Same as Example 7.2; for l = 5λ.Since l = 5λ, M = 5 and ∆ = 0.2.

m θθθθm bmm θθθθm bm

0 90 10 90 1

1 78.46 1 -1 101.54 1

2 66.42 1 -2 113.58 1

3 53.13 1 -3 126.87 1

4 36.87 0 -4 143.13 0

5 0 0 -5 180 0

Woodward-Lawson Method: Example (2)

Composing functions for line-source (l = 5λ)


Linear array

( )

( )m

m

mm kdN

kdN

bf

θθ

θθθ

coscos2

sin

coscos2

sin

)(

−

−=

The pattern of each sample (uniform array) can be written as (assuming l = Nd)

Composing function

2

( )

( )∑−= −

−

=M

Mmm

m

m kdN

kdN

b

θθ

θθθ

coscos2

sin

coscos2

sin

)(AF

dmmb )(AF θθ ==

For an odd number elements, the array factor becomes

bm can be obtained from the value at the sample points θm, i.e.,


Linear array (2)

,2,1,2

12

2

12

=

−=∆

−Km

Nd

mm λ

samples oddfor ,2,1,0,cos K±±=

=∆= mNd

mmm

λθ

The location of each sample is given by

sampleseven for

,2,1,2

12

2

12

22cos

−−=

+=∆

+=

KmNd

mm

Ndm λ

θ

The normalized excitation coefficient of each element is given by

∑−=

′−=M

Mm

zjk

mnmneb

Nza

θcos1)'(


Example

Element number

n

Position

z’n

Coefficient

an

±1 ±0.25λλλλ 0.5696

Same as Example 7.2; for N=10 and d = λ/2.

The coefficients can be found to be

±1 ±0.25λλλλ 0.5696

±2 ± 0.75λλλλ -0.0345

±3 ± 1.25λλλλ -0.1001

±4 ± 1.75λλλλ 0.1108

±5 ± 2.25λλλλ -0.0460

To obtain the normalized amplitude pattern of unity at θ=π/2,

the array factor has be divided by ∑ = 4998.0na


Example

0.8

1

1.2

1.4

No

rma

lize

d m

ag

nitu

de

Line−source

Linear Array (N=10,d=λ/2)

0 20 40 60 80 100 120 140 160 1800

0.2

0.4

0.6

0.8

θ [Degree]

No

rma

lize

d m

ag

nitu

de

Triangular, cosine, cosine

squared distributions

Mutual CouplingMutual Coupling

• Consider two antennas

2221212

2121111

IZIZV

IZIZV

+=

+=

=

2

1

2221

1211

2

1

I

I

ZZ

ZZ

V

V

2112 ZZ =If reciprocal,

32

Equivalent Circuit

Two-port NetworkT-Network Equivalent Circuit

Mutual Coupling: 2 antennas

01

111

2 =

=I

I

VZ

02

112

1=

=I

I

VZ

01

221

2=

=I

I

VZ

02

222

1=

=I

I

VZ

2112 ZZ =

for reciprocal networks

:, 2211 ZZ

Input impedance02=I

021=I

I

impedancepoint driving

active

21

=ddZ

1

21211

1

11

I

IZZ

I

VZ d +==

Input impedance

2

12122

2

22

I

IZZ

I

VZ d +==

2

12

1

21 on depends ;on depends :Note

I

IZ

I

IZ dd

Mutual Coupling: 2 antennas (2)

• As I1 and I2 change, the driving point impedance changes.

• In a uniform array, the phase of I1 and I2

is changed to scan the beam.

• As the beam is scanned, the driving port • As the beam is scanned, the driving port impedance in each antenna changes.

• In general, Z11,Z12=Z21,Z22 can be calculated using numerical techniques.

• For some special cases, they can be calculated analytically.

Mutual Coupling: N antennas

=

NNNNN

N

N

N I

I

I

ZZZ

ZZZ

ZZZ

V

V

V

M

L

MOMM

L

L

M

2

1

21

22221

11211

2

1 IZV =

jkIj

iij

k

I

VZ

≠=

=,0

In an N-element uniform array β)1( −= njeIIIn an N-element uniform array β)1(

0

−= nj

n eII

∑=

−

−

=

+++==

N

n

nj

n

Nj

N

j

d

eZ

eZeZZI

VZ

1

)1(

1

)1(

11211

1

11

β

ββL

changes. as changes :Note 1 βdZ

Mutual Coupling: 2 dipoles

Mutual Coupling: 2 dipoles (2)

chapter chapter 7 : antenna synthesis : antenna...

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