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```									LECTURE 16: LINEAR ARRAY THEORY - PART II
(Linear arrays: Hansen-Woodyard end-fire array, directivity of a
linear array, linear array pattern characteristics – recapitulation;
3-D characteristics of an N-element linear array.)

1. Hansen-Woodyard end-fire array (HWEFA)
One of the shortcomings of end-fire arrays (EFA) is their

Fig. 6-11, pp.270, Balanis

1
To enhance the directivity of an end-fire array, Hansen and
Woodyard proposed that the phase shift of an ordinary EFA
β = ± kd
be increased as:
     2.94 
β = −  kd +        for maximum at θ = 0°       (16.1)
      N 
      2.94 
β = +  kd +        for maximum at θ = 180°      (16.2)
       N 
Conditions (16.1)–(16.2) are known as the Hansen – Woodyard
maximizing the directivity.
The normalized pattern AFn of a uniform linear array is:
N                    
sin  ( kd cosθ + β ) 
AFn ≈      2                           (16.3)
N
( kd cosθ + β )
2
N
if the argument ψ = ( kd cosθ + β ) is sufficiently small (see
2
previous lecture). We are looking for optimal β , which would
result in maximum directivity. Let:
β = − pd                      (16.4)
where d is the array spacing and p is the optimization parameter
 Nd
sin         ( k cosθ − p )

⇒ AFn =          2                   
Nd
( k cosθ − p )
2
Assume that Nd / 2 = q ; then
sin  q ( kd cosθ − p ) 
                   
⇒ AFn =                                    (16.5)
q ( kd cosθ − p )

2
sin Z
or AFn =         ; where Z = q ( kd cosθ − p )
Z
2     sin 2 Z
U (θ ) = AFn =                              (16.6)
Z2
2
 sin  q ( k − p )  
                   
U (θ = 0) =                                   (16.7)
 q(k − p) 
                     
2
U (θ )       z sin Z 
U n (θ ) =            =                          (16.8)
U (θ = 0)  sin z Z 
where:
z = q(k − p)
Z = q ( k cosθ − p ) , and
U n (θ ) - normalized power pattern with respect to θ = 0°.

Directivity at θ = 0°:
4π U (θ = 0)
D0 =                                  (16.9)
where Prad =   ∫∫ U n (θ )d Ω . To maximize the directivity, the
Ω
quantity U 0 = Prad / 4π will be minimized.
2π π              2
1           z sin Z 
U0 =
4π    ∫   ∫  sin z Z  sinθ dθ dθ         (16.10)
0   0          

2
1  z   sin  q ( k cosθ − p )  
2 π
                   sin θ dθ (16.11)
U0 =          ∫                     
2  sin z  0  q ( k cosθ − p ) 
                    

2
1  z   π cos 2 z − 1                1
U0 =              +        + Si(2 z )  =     g ( z ) (16.12)
2kq  sin z   2 2z                   2kq

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z
sin t
Here, Si ( z ) = ∫      dt . The minimum of g ( z ) occurs when
0
t
z = q ( k − p ) = −1.47              (16.13)
Nd
⇒        ( k − p ) = −1.47
2
Ndk Ndp
⇒         −        = −1.47, where dp = − β
2         2
N
⇒ ( dk + β ) = −1.47
2
2.94                  2.94 
β =−          − kd = −  kd +              (16.14)
N                     N 
Equation (16.14) gives Hansen-Woodyard condition for improved
directivity along θ = 0°. Similarly, for θ = 180° :
 2.94       
β = +         + kd               (16.15)
   N        
Usually, conditions (16.14) and (16.15) are approximated by:
      π 
β = ±  kd +                     (16.16)
      N
which is easier to remember and gives almost identical results
since the curve g ( z ) at its minimum is very flat.
Conditions (16.14)-(16.15), or (16.16), ensure minimum
beamwidth (maximum directivity) in the desired end-fire direction
but there is a trade-off in the side-lobe level, which is higher than
that of the ordinary EFA. Besides, conditions (16.14)-(16.15) have
to be complemented by additional requirements, which would
ensure low level of the radiation in the direction opposite to the
main lobe.

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a) For a maximum at θ = 0°:
2.94
ψ θ =0 = −
      2.94                             N
β = −  kd +            ⇒                            (16.17)
       N  θ = 0°                         2.94
ψ     θ =180
= −2kd −
N
In order to have a minimum of the pattern in the θ = 180
direction, one must ensure that:
| ψ |θ =180°  π ,                (16.18)
It is easier to remember Hansen-Woodyard conditions for
maximum directivity in the θ = 0° direction as:
2.94
| ψ |θ =0 =          π
N                     (16.19)
| ψ |θ =180°  π ,

b) For a maximum at θ = 180° :
2.94
ψ θ =180 =
2.94                             N
β = kd +              ⇒                        (16.20)
N θ =180°                     2.94
ψ θ =0 = 2kd +
N
In order to have a minimum of the pattern in the θ = 0
direction, one must ensure that
| ψ |θ = 0°  π ,               (16.21)
One can now summarize Hansen-Woodyard conditions for
maximum directivity in the θ = 180° direction as:
2.94 π
| ψ |θ =180 =       
N    N              (16.22)
| ψ |θ =0  π

If (16.18) and (16.21) are not observed, the radiation in the
opposite of the desired direction might even exceed the main beam
level. It is easy to show that the complimentary requirement of

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| ψ |= π at the opposite direction can be met, if the following
relation is observed:
 N −1 λ
d =                             (16.23)
 N 4
If N is large, d  λ / 4 . Thus, for a large uniform array, Hansen-
Woodyard condition can yield improved directivity, only if the
spacing between the array elements is approximately λ / 4 .

Solid line: d = λ / 4
Dotted line: d = λ / 2
N=10
      π 
β = −  kd + 
      N

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2.     Directivity of a linear array
2.1. Directivity of a BSA
2
 N            
 sin  kd cosθ               2
2
U (θ ) = AFn =       2          =  sin Z              (16.24)
 Z 
 N kd cosθ                 
 2
                 

U     U
D0 = 4π 0 = 0                                 (16.25)
where:
U av =
4π
θ = π / 2 in terms of AFn is unity:
U 0 = U max = U (θ = π / 2 ) = 1
1
⇒ D0 =                           (16.26)
U av
The radiation intensity averaged over all directions is calculated as:
2
N         
2π π    2                 π sin   kd cosθ 
1      sin Z              1       2          sin θ dθ
U av   =     ∫ ∫ Z2
4π 0 0
sin θ dθ dφ = ∫
20      N
kd cosθ
2
Change variable:
N                 N
Z=     kd cosθ ⇒ dZ = − kd cosθ dθ                    (16.27)
2                 2
Nkd
−
π         2                          2             2
1  sin Z                 1 2                    sin Z 
U av    = − ∫         d cosθ dθ = −                ∫               dZ (16.28)
20 Z                   2 N kd      Nkd           Z 
2

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Nkd
2              2
1             sin Z 
U av   =
Nkd       ∫  Z  dZ
Nkd        
(16.29)
−
2
2
 sin Z 
The function,           , is a relatively fast decaying function as Z
 Z 
increases. That is why, for large arrays, where Nkd / 2 is big
enough ( ≥ 20 ) , the integral (16.29) can be approximated by:
∞               2
1      sin Z        π
U av         ∫ 
Nkd −∞  Z 
 dZ =
Nkd
(16.30)

1     Nkd        d
D0 =           2N                  (16.31)
U av     π         λ
Substituting the length of the array L = ( N − 1) d in (16.31) yields:
 L  d 
D0 = 2 1 +                        (16.32)
 d  λ 

N
For a large array ( L  d ) :
D0  2 L / λ                       (16.33)

2.1 Directivity of ordinary EFA
Consider an EFA with maximum radiation at θ = 0 , i.e.
β = −kd .
2
 N                    
 sin  kd ( cosθ − 1)                2
2     2                      =  sin Z 
U (θ ) = AFn       =                                        (16.34)
N                   Z 
      kd ( cosθ − 1) 
 2
                      
N
where: Z =     kd ( cosθ − 1) .
2

8
2π π        2                 π           2
P       1       sin Z               1  sin Z 
4π ∫ ∫  Z 
U av = rad =                  sin θ dθ dφ = ∫           sin θ dθ
4π        0 0
2 0 Z 
Again, change of variables is used:
N                       N
Z = kd cosθ ⇒ dZ = − kd cosθ dθ                    (16.35)
2                        2
π          2                   − Nkd           2
1  sin Z                 1 2          sin Z 
U av = − ∫ 
2 Nkd ∫  Z 
 d cosθ = −                       dZ
2 0 Z                            0
Nkd      2
1       sin Z 
Nkd ∫  Z 
U av =                   dZ              (16.36)
0
If ( Nkd ) is sufficiently large, the above integral can be
approximated as:
∞           2
1  sin Z             1 π
Nkd ∫  Z 
U av =                  dZ =                 (16.37)
0
Nkd 2
1     2 Nkd      d
⇒ D0           =       = 4N             (16.38)
U av     π        λ
It is seen that the directivity of an EFA is approximately twice as
large as the directivity of the BSA (compare (16.38) and (16.31)).
Another (equivalent) expression can be derived for D0 in terms of
the array length L = ( N − 1) d :
 L  d 
D0 = 4 1 +                          (16.39)
 d  λ 
For large arrays, the following approximation holds:
D0 = 4 L / λ if L  d                     (16.40)

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2.2 Directivity of HW EFA
If the radiation has its maximum at θ = 0°, then the minimum
of U av was obtained as (16.12):
2
min    1 2  Z min          π cos(2Z min ) − 1                
U av =                        +                 + Si(2 Z min )  (16.41)
2k Nd  sin Z min    2      2 Z min                     
where:
π
Z min = −1.47  −
2
2
min    1  π  π 2                   0.878
⇒ U av =                + − 1.8515 =             (16.42)
Nkd  2   2 π                 Nkd
1         Nkd            d 
D0 = min =             = 1.789  4 N        (16.43)
U av       0.878               λ 
From (16.43), one can see that using HW conditions leads to
improvement of the directivity of the EFA with a factor of 1.789.
Equation (16.43) can be expressed via the length L of the array as:
  L  d                 L 
D0 = 1.789  4  1 +    = 1.789  4       (16.44)
   d  λ                 λ 

Example: Given a linear uniform array of N isotropic elements
(N=10), find the directivity D0 if:
a) β = 0 (BSA)
b) β = −kd (Ordinary EFA)
π
c) β = − kd −      (Hansen-Woodyard EFA)
N
In all cases, d = λ / 4 .

a) BSA
d
D0  2 N   = 5 ( 6.999 dB )
λ

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b) Ordinary EFA
d
D0  4 N   = 10 (10 dB )
λ

c) HW EFA
  d 
D0  1.789  4 N    = 17.89 (12.53 dB )
  λ 

3. Pattern characteristics of linear uniform arrays - recapitulation

NULLS ( AFn = 0 ):
 n λ
θ n = arccos  ±   , where n = 1, 2,3,4,... and n ≠ N , 2 N ,3 N ,...
 N d

MAXIMA ( AFn = 1):
 mλ 
θ n = arccos  ±    , where m = 0,1, 2,3,...
   d 

HALF-POWER POINTS:
 1.391λ          πd
θ h  arccos  ±       , where    1
   π Nd          λ

HALF-POWER BEAMWIDTH:
π          1.391λ   π d
θ h = 2  − arccos          ,    1
2          π Nd   λ

MINOR LOBE MAXIMA:
 λ  2s + 1                              πd
θ s  arccos  ±           , where s = 1, 2,3,... and    1
  2d  N                                 λ

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FIRST-NULL BEAMWIDTH (FNBW):
π          λ 
θ n = 2  − arccos     
2          Nd  

FIRST SIDE LOBE BEAMWIDH (FSLBW):
π          3λ   π d
θ s = 2  − arccos        ,   1
 2         2 Nd   λ


B. Ordinary end-fire array

NULLS ( AFn = 0 ):
    n λ
θ n = arccos 1 −     , where n = 1, 2,3,... and n ≠ N , 2 N ,3 N ,...
 N d

MAXIMA ( AFn = 1):
 mλ 
θ n = arccos 1 −    , where m = 0,1,2,3,...
    d 

HALF-POWER POINTS:
 1.391λ           πd
θ h = arccos 1 −       , where    1
    π Nd          λ

HALF-POWER BEAMWIDTH:
 1.391λ  π d
θ h = 2arccos 1 −        ,  1
    π Nd  λ


MINOR LOBE MAXIMA:
 ( 2 s + 1) λ                             πd
θ s = arccos 1 −            , where s = 1, 2,3,... and    1
     2 Nd                                 λ

12
FIRST-NULL BEAMWIDTH:
    λ 
θ n = 2arccos 1 −   
 Nd 

FIRST SIDE LOBE BEAMWIDH:
    3λ  π d
θ s = 2arccos 1 −    ,    1
 2 Nd  λ

C. Hansen-Woodyard end-fire array

NULLS:
                λ 
θ n = arccos 1 + (1 − 2n )        , where n = 1, 2,... and n ≠ N , 2 N ,...
              2 Nd 


MINOR LOBE MAXIMA:
    sλ                             πd
θ s = arccos 1 −     , where s = 1, 2,3,... and    1
 Nd                                λ

SECONDARY MAXIMA:
                     λ                            πd
θ m = arccos 1 + [1 − (2m + 1)]       , where m = 1, 2,... and    1
                   2 Nd                            λ

HALF-POWER POINTS:
            λ           πd
θ h = arccos  1 − 0.1398     , where     1, N -large
            Nd          λ

HALF-POWER BEAMWIDTH:
           λ           πd
θ h = 2arccos 1 − 0.1398     , where     1, N -Large
           Nd           λ

13
FIRST-NULL BEAMWIDTH:
    λ 
θ n = 2arccos 1 −   
 2 Nd 

4. 3-D characteristics of a linear array
In the previous considerations, it was always assumed that the
linear-array elements are located along the z-axis, thus, creating a
problem, symmetrical around the z-axis. If the array axis has an
arbitrary orientation, the array factor can be expressed as:
N                                    N
j ( n −1)( kd cos γ + β )
AF = ∑ an e                                = ∑ an e ( ) ,
j n −1 ψ
(16.45)
n =1                                  n =1
where an is the excitation amplitude and ψ = kd cos γ + β .
The angle γ is subtended between the array axis and the
radius-vector to the observation point. Thus, if the array axis is
ˆ
along the unit vector a :
a = sin θ a cos φa x + sin θ a sin φa y + cosθ a z
ˆ                  ˆ                  ˆ          ˆ (16.46)
and the radius – vector to the observation point is:
r = sin θ cos φ x + sin θ sin φ y + cosθ z
ˆ                ˆ                ˆ        ˆ     (16.47)
the angle γ can be found from:
cos γ = a ⋅ r
ˆ ˆ
= sin θ cos φ sin θ a cos φa x + sin θ sin φ sin θ a sin φa y + cosθ cosθ a z
ˆ                              ˆ               ˆ
⇒ cos γ = sin θ sin θ a cos (φ − φa ) + cosθ cosθ a            (16.48)
If a = z (θ a = 0°) , then cos γ = cosθ , γ = θ .
ˆ ˆ

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