Lecture Notes_Antenna Arrays
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Hon Tat Hui Antenna Arrays
NUS/ECE EE6832
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Antenna Arrays1 Introduction
1. They can provide the capability of a steerable beam (radiation direction change) as in smart antennas.
2. They can provide a high gain (array gain) by using simple antenna elements.
3. They provide a diversity gain in multipath signal reception.
4. They enable array signal processing.
Antenna arrays are becoming increasingly important in wireless communications. Advantages of using antenna arrays:
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Hon Tat Hui Antenna Arrays
NUS/ECE EE6832
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2 Far-Field Expression of An Antenna Array
Observation point P(x,y,z) at infinity
x
y
zr1
r2
rN
r0
ri
1
Ni
21, 2,, i,, N
= position vectors of the antenna elements
r1, r2,, ri,, rN= distances of the antenna
elements from the observation point
r0 = distance of the origin from the observation pointAn arbitrary antenna array
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The sum of the far fields E radiated from the array is:
( ) ( )( )( )
1
where is the far field of the th antena given by:
, ,
, the component of the radiation pattern , t
i
N
ii
ijkr
i i i i i
i
i
i
f f w K e
ff
=
=
= + ==
E EE
E a a
he component of the radiation pattern the weighting factor of the excitation source a constant accounting for the path loss
i
i
wK
==
(1)
(2)
(for all ) antenna element located at the origin.i( ) ( )Note that , and , are obtained with the thi if f i
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For identical antenna elements, the element radiation patterns are the same and independent of i. Hence,
( ) ( )1
, , iN
jkri i
if f w K e
== + E a a
Furthermore, we assume that Ki is approximately constant such that
( )0 0 ,i i rr r r r = = a
1 2 NK K K K= = = ="We can also write
(3)
(4)
(5)
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Then
( ) ( ) ( )( ) ( ) ( )
0
0
,
1, ,
, , ,
i r
Njkjkr
ii
jkrarray
Ke f f w e
Ke f f f
=
= + = +
aE a aa a (6)
The result in (6) is known as the Principle of Pattern Multiplication, which states that the array pattern is the product of the antenna element pattern multiplied with an array factor .( ),arrayf Hereafter, we focus on the study of the array factor.
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( ) ( ),1
1
, i r
i
Njk
array iiN
jbi
i
f w e
w e
=
=
=
=
a
Putting in the matrix form, we have( ), Tarrayf = w b
1
2
N
ww
w
= w #
( )( )
( )
11
22
,
,
,
r
r
N N r
jkjb
jkjb
jb jk
eee e
e e
= =
a
a
a
b # #
(7)
(8)
(9)
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Eq. (8) is an important expression which tells us that the array radiation pattern (array factor) can be obtained from a given array element weight vector w and vice versa. If we consider the weight vector as an N dimensionalfunction w(xi,yi,zi), i =1, 2, , N, then the array factor and the weight function is related by a certain type of transformation operation . That is,
( ) ( )( ), , ,arrayf w x y z = Similarly, the weight function w(x,y,z) can be determined from a given required array factor function.
( ) ( )( )1, , ,arrayw x y z f = (11)
(10)
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3 Uniform Liner Arrays (ULAs)
Far field observation point
d
r rN-1
y
xDipole NDipole 1
Dipoles are parallel to the zdirection
An N-element uniform antenna array with an element separation d and at the plane of = /2
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Uniform linear arrays (ULAs) mean that the array elements are same as each other and they are aligned along a straight line with equal element separations. If the excitation currents have the same amplitude (I = 1) but the phase difference between adjacent elements is (the progressive phase difference), then the weight vector is:
( )
1
22
3
1
j
j
j NN
IwIewIew
w Ie
= =
w##
(12)
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The array factor (AF) for this array specified on the plane = /2 is:
( ) ( ) ( )
( )
1 1 cos
1 1
( cos ) 2( cos ) ( 1)( cos )
1( 1) 2
1
AF 2,
1
sin2
sin2
where cos and 0 , 2
i
N Nj i j i djb
array ii i
j kd j kd j N kd
N j Nj n
n
f w e I e e
e e e
Ne e
kd
= =
+ + +
=
= = = == + + + +
= = = +
"
(13)
(14)
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( )nsin1 2AFsin
2
N
=
The normalized array factor is:
The relation between |AFn|, , d, and is shown graphically on next page. Note that |AFn| is a period function of , which is in turn a function of . The angle is in the real space and its range is 0 to 2. However, is not in the real space and its range can be greater than or smaller than 0 to 2, leading to the problem of grating lobes or not achieving the maximum values of the |AFn| expression.
(15)
where is a constant to make the largest value of |AFn| equal to one. Note that is not necessarily equal to N.
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The relation between AFn,, d, and
|AFn( )|
= kd cos +
kd
kdcos
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3.1 Properties of the normalized array factor AFn:1. |AFn| is a periodic function of , with a period of 2.
This is because|AFn( + 2)| = |AFn()|.
2. As cos() = cos(-), |AFn| is symmetric about the line of the array, i.e., = 0 & . Hence it is enough to know |AFn| for 0 .
3. The maximum values of |AFn| occur when (see Supplementary Notes):
1 ( cos ) , 0,1,2,2 2
kd m m = + = = "1
max main beam directions cos ( 2 )2m
d
= =
(16)
(17)
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Note that there may be more than one angles maxcorresponding to the same value of m because cos-1(x) is a multi-value function. If there are more than one maximum angles max, the second and the subsequent maximum angles give rise to the phenomenon of grating lobes. The condition for grating lobes to occur is that d (disregarding the value of ) as shown below:
Main lobe 1st grating lobe 2nd grating lobe1st grating lobe2nd grating lobe
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Main lobe 1st grating lobe1st grating lobe|AFn( )|
kd
=kdcos
Visible region
2nd grating lobe2nd grating lobe
(1) When d 0.5, no grating lobes can be formed for whatever value of . (2) When d , grating lobe(s) is (are) formed for whatever value of . (3) When 0.5
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4. There are other angles corresponding to the maximum values for the minor lobes (minor beams) but these angles cannot be found from the formula in no. 3 above.
5. When and d are fixed, it is possible that can never be equal to 2m. In that case, the maximum values of |AFn| cannot be determined by the formula in no. 3.
6. The main beam directions max are not related to N. They are functions of and d only.
7. The nulls of |AFn| occur when:1,2,3,
, 2 ,2 ,3 ,
nnN n N N N
== "
"1
null2 null directions cos ( )
2n
d N
= = (18)
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8. The null directions null are dependent on N.9. The larger the number N, the closer is the first null (n = 1)
to the first maximum (m = 0). This means a narrower main beam and an increase in the directivity or gain of the array.
10.The angle for the main beam direction (m = 0) can be controlled by varying or d.
Note that there may be more than one angles nullcorresponding to a single value of n because cos-1(x) is a multi-value function.
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Example 1A uniform linear array consists of 10 half-wave dipoles with an inter-element separation d = /4 and equal current amplitude. Find the excitation current phase difference such that the main beam direction is at 60 (max = 60).Solutions
( )( ) ( )
1maxmain beam dirction 60 cos 22
2 2 cos 60 0.5
2 45 360 315 , when 14
md
m
m m
= = = = =
= = + = =
d = /4, max = 60, N = 10
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Other values of corresponding to other values of m are outside the range of 0 2 and are not included.
n
sin 5 cos1 2AF
1sin cos2 2
10
+ = + =
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3.2 Phased (Scanning) ArraysIt was mentioned earlier that by controlling the values of dor , the maximum radiation direction of an array can be arbitrarily pointed to any direction. In practice, the element separation d is usually fixed while the excitation current phase between elements is controlled electronically. The current amplitudes of the all the elements are assumed to be the same. This kind of steerable direction arrays is called uniform phased scanning arrays. To accomplish this, the excitation current phase must be adjusted so that:
0 0cos 0, coskd kd == + = = (19)
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For a phased scanning array, the length L = (N-1)d of the array (where N is the number of elements) can be determined from the graph on next page. The graph shows the relation between the array length, the half-power beamwidth, and the maximum radiation direction. The half-power beamwidth is an alternative way to specify the gain of the array. Once the half-power beamwidth and the maximum radiation direction are specified, the number of elements required to design such an array can be calculated.
y
d Ant. NAnt. 1
L
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Maxim
um radiation direction
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Example 2Design a uniform linear phased scanning array whose maximum radiation direction is in 30 ( = 30). The desired half-power beamwidth is 2 while the element separation is d = /4. Determine the excitation current phase , the length of the array L, and the number of elements N in the array.SolutionsSince the array is uniform, the current amplitude is same for all elements. The excitation current phase is found from:
02 cos cos30 1.36 77.94
4o okd rad = = = =
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To find the length L of the array, we use the graph on page 24. From that graph with HPBW=2 and maximum radiation direction =30:
(L+d)/=50Therefore with d = /4,
L = 49.75
49.751 1 2000.25
LNd
= + = + =
The number of elements is then:
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4 Circular Arrays
( )cossin cos
n n
n
r aa
==
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4.1 Advantages of Circular Arrays
1. Unlike linear arrays, circular arrays can provide a 2D angular scan, both horizontal and vertical scans.
2. Unlike 2D planar arrays, circular arrays are basically 1D linear arrays but in a circular form.
3. Unlike linear arrays, a circular array can scan horizontally for 360 with no distortions near the end-fire directions.
4. Unlike linear arrays, distortions in the array pattern of a circular array due to mutual coupling effect are same for each element and this makes it easier to deal with the mutual coupling effect.
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( )[ ] ( )[ ]( )[ ]
( )[ ]
1 1 2 2sin cos sin cos
sin cos
sin cos
1
AFN N
n n
j ka j ka
j ka
Nj ka
n
e e
e
e
+ +
+
+=
= ++ +="
4.2 Array FactorFor a uniform circular array with N elements and an equal excitation current amplitude I0 and a current phase of n (reference to the central point of the array) for the nth element (and n=2n/N), the array factor is:
(For a detailed derivation of the AF above, see ref. [1])
(20)
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Note that in the above expression, AF is a sum of Ncomplex exponentials. The magnitude of each complex exponential is 1. Hence the maximum value of the magnitude of |AF| is the addition of magnitudes of the complex exponentials, i.e., N. The maximum radiation direction (max, max) is therefore achieved when:
( )( )( )
max max 1
max max 2
max max
sin cos 2 2sin cos 4 2
sin cos 2 2
0,1,2,N
ka N qka N q
ka qq
+ = + = + = =
#
"
(21)
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Hence in a circular array, we first choose a desired maximum radiation direction (max, max). Then the excitation phase n for each element is determined according to the above formula. The n so determined may not be equally increasing from one element to the next. This is different from the case of a linear array.
One excitation method to achieve the above maximum radiation direction is:
( )[ )
max max2 sin cos 2 , 1,2, , is choosen to make 0,2n
n
q ka n N n Nq
= =
" (22)
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Example 3A uniform circular array with a radius a = 0.5 and the number of elements N = 8. The maximum radiation direction of the array factor AF is at (60, 30). What should be the excitation phases n for the elements?SolutionsUsing the formula for n , we have:
12 22 sin cos
2 3 6 82 2.633.66
q
= = =
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22 42 sin cos
2 3 6 82 1.364.92
q
= = =
32 62 sin cos
2 3 6 80 0.700.70
q =
= +=
42 82 sin cos
2 3 6 80 2.362.36
q =
= +=
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52 102 sin cos
2 3 6 80 2.632.63
q =
= +=
62 122 sin cos
2 3 6 80 1.361.36
q =
= +=
72 142 sin cos
2 3 6 82 0.705.58
q
= = =
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82 162 sin cos
2 3 6 82 2.363.92
q
= = =
Azimuth pattern ( = 60) Vertical pattern ( = 30)
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Note that in the above discussion on circular arrays, we have only derived the array factor AF. The array pattern of any circular array with practical antenna elements must be obtained by multiplying the array factor with the actually element radiation pattern. For example, if the elements are half-wave dipole antennas, then the array pattern F() is:
( )[ ]( )[ ] ( )[ ]sin cos
1
cos 2 cos( ) AF
sincos 2 cos
sinn n
Nj ka
n
F
e
+
=
=
= (23)
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References:
1. C. A. Balanis, Antenna Theory, Analysis and Design, John Wiley & Sons, Inc., New Jersey, 2005.
2. W. L. Stutzman and G. A. Thiele, Antenna Theory and Design, Wiley, New York, 1998.
3. David K. Cheng, Field and Wave Electromagnetic, Addison-Wesley Pub. Co., New York, 1989.
4. John D. Kraus, Antennas, McGraw-Hill, New York, 1988.5. Fawwaz T. Ulaby, Applied Electromagnetics, Prentice-Hall, Inc.,
New Jersey, 2007.