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).
Schelkunoff polynomial Schelkunoff polynomial
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
Schelkunoff polynomial Schelkunoff polynomial
method (method (33))
VR=Visible Region
IR=Invisible Region
β = 0
5
β = 0
Schelkunoff polynomial Schelkunoff polynomial
method (method (44))
VR=Visible Region
IR=Invisible Region
β = π/4
6
β = π/4
Schelkunoff polynomial Schelkunoff polynomial
method (method (55))
7
Schelkunoff polynomial Schelkunoff polynomial
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
Schelkunoff polynomial Schelkunoff polynomial
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
Fourier Transform Method : Fourier Transform Method :
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
Fourier Transform Method : Fourier Transform Method :
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
Fourier Transform Method : Fourier Transform Method :
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
Fourier Transform Method : Fourier Transform Method :
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
Woodward-Lawson Method:
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,
Woodward-Lawson Method:
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
Woodward-Lawson Method:
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)
Woodward-Lawson Method:
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.,
Woodward-Lawson Method:
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)'(
Woodward-Lawson Method:
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
Woodward-Lawson Method:
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)