ap2011 solutions 06

Solutions 6 – Antenna arrays Antennas and Propagation, Frühjahrssemester 2011

Antenna array (linear array, constant feed magnitude, isotropic radiators)

Problem 6.1 : A three-element array of isotropic sources has the phase and magnitude relationships shown in the figure below. The spacing between the elements is 2d l= . a) Find the array factor. b) Find the nulls.

-jd

y

z

d

-1

#1

#2

#3 +1

Method I : a)

( )cos cosAF 2 sin cosjkd jkde j e j kd jq q q+ -= - - + = - -

To normalize the array factor so that its maximum equals unity, the normalization factor has to be -3j.

( )( )

( )( )

13

13

AF 1 2 sin cos

1 2 sin cos

n kd q

p q

= +

= +

b)

( )

( )1

2 sin cos 1

1 5 13 7 11 19cos sin , , ,..., , ,

2 6 6 6 6 6 6

n

n nx

p q

p p p p p pp q -

= -

= - = - - - =

1

1 1

2 2

cos ,

99.5965

146.446

nn

x

x

x

qp

pq

pq

-æ ö÷ç ÷= ç ÷ç ÷è ø

= - =

= - =


Method II : a)

uniform array with /2x p= - :

3sin sin cos

2 2 2AF 1sin cossin

2 22

Ny pp q

py p q

é ù-ê úë û= =é ù-ê úë û

Here, the phase centre is in the physical center of the array (element #1) Using the trigonometric identities:

( )

( )

2 12

2 12

sin( ) sin cos cos sin

cos 1 cos2

sin 1 cos2

x y x y x y

x x

x x

+ = +

= +

= - the array factor can be written as

( ) ( )

( ) ( )

( )

( )

3 1 12 2 21 12 2

2 21 1 1 1 1 12 2 2 2 2 2

12

2 21 12 2

1 12 2

2

13

sin sin cos cos sinAF

sin sin

2 sin cos cos cos sin sinsin

3 cos sin

3 1 cos 1 cos

1 2 cos 1 2 cos cos 1 2 sin( cos )

AF 1 2 sin( cos )n

kd kd

kd

p

y y y y yy y

y y y y y yy

y y

y y

y q q

q

+= =

+ -=

= -

= + - -

= + = + - = +

= +

Same result as of Method I, where phase center was located at element #3.

b)

( )1

1

1

1,2,3...2cos ,

2 ,2 ,3 ,...

11 : cos 99.59

65

2 : cos 146.446

n

nnd N n N N N

n

n

lq x p

p-

-

-

=ìïé ù ï= - íê ú ï ¹ë û ïîé ù= - =ê úë ûé ù= - =ê úë û


Antenna array (linear array, constant feed magnitude, isotropic radiators)

Problem 6.2 : Design an ordinary end-fire uniform linear array with only one maximum so that its directivity is 20 dBi (above isotropic). The spacing between the elements is 4d l= , and its

length is much greater than the spacing. Determine the a) number of elements, b) overall length of the array (in wavelengths), c) approximate half-power beamwidth (in degrees), d) progressive phase between the elements (in degrees), e) amplitude level (compared to the maximum of the major lobe) of

the first minor lobe (in dB). a) see slide 5.28

( )

( ) ( )

0

0

4

20 100

100 4 4 1004

dD N

D dB

dN N N N

l

ll l

=

= =

= = = =

b)

( )1 99 24.754

L N dl

l= - = ⋅ =

c) see slide 5.18

( ) ( )1 11.391cos 1 cos 0.98228 10.799

HPBW 2 21.6

h

h

Ndl

qp

q

- -= - = =

= ⋅ =

d) see slide 5.15

90kdx = = e) For small values of y the normalized array factor can be written as (Slide 5.12)

( )sin2AF

2

n

N

N

y

y= , since ( )cos 1 coskd kd kdy q q= @ -

The maximum of the first minor lobe approximately occurs at

( ) ( )3

sin 1 cos2 2 2s sN N

kd kdp

y q=

or when


{ }1 3cos

2s d Nl p

q xp

- é ù= - ê úë û

At that point the magnitude of AFn reduces to

( )sin 22AF 0.2123

2

s

n

s

N

N

y

py= = =

which is in dB equal to

2AF 20 log 13.5dB

3n p

æ ö÷ç= = -÷ç ÷÷çè ø


Antenna array (square array, constant feed magnitude, isotropic radiators)

Problem 6.3 : Determine the azimuthal and elevation angles of the grating lobes for a 10 by 10 element uniform planar array when the spacing between the elements is λ. The maximum of the main beam is directed toward 0 060 , 90q f= = and the array is located on the xy-plane.

The main radiation occurs in the following direction:

0 00 0

0 0

3sin sin

60 , 90 2sin cos 0

q fq f

q f

ìïï =ï= = íï =ïïî

Grating lobes occur if the following conditions are fulfilled: (Slide 5.46)

sin cos 2

sin cos 2x x x

y y y

kd m

kd n

θ φ ξ πθ φ ξ π

Ψ = + = ±Ψ = + = ±

in the directions given by (Slide 5.47):

1

32tan for 0 360

n

mf f-

é ùê ú

ê ú= £ £ê úê úê úë û

(1)

1 1

32sin sin for 0 180

cos sin

nmq q

f f- -

é ùê úé ù ê ú= = £ £ê ú ê úê úë û ê ú

ê úë û

(2)

With dx = dy = λ follows kdx = kdy = 2π and the limits of m and n are given as:

( ) ( )0 0sin cos sin cos 2 sin cos 21,0,1

since: 1 sin cos 1

xkd mm

q f q f p q f p

q f

- = = üïï = -ýï- £ £ ïþ

( )0 03

sin sin sin sin 2 sin sin 22

1,02 3 3 2 3

1.866 sin sin 0.13392 2 2

ykd nn

q f q f p q f p

q f

æ ö üï÷ç ï- = - =÷ç ï÷çè ø ïï = -ýï+ - ïï- » - £ - £ » ïïþ This means that the main radiation lobe and the grating lobes can occur in the directions given by (1) and (2) for m = -1,0,1 and n = -1,0 (i.e., six combinations).


Case I: 0n = , 1m = - :

From (1):

140.8934 319.113

tan2 139.1066

f -ì- ïé ù ïï= - =ê ú íïê ú ë û ïïî

From (2):

( )

( )

1

11

sin 1.32281sin

cos sin 1.3228q

f

--

-

ì -ïæ ö- ïï÷ç= = í÷ç ÷çè ø ï +ïïî and

( )

( )

1

11

sin 1.32283 /2sin

sin sin 1.3228q

f

--

-

ì -ïæ ö ïï÷ç= = í÷ç ÷ç ïè ø +ïïî , respectively. This means that there is no solution for case I.

Case II: 0n = , 0m = ( main radiation) :

From (1):

( )190

tan270

f -ìïï= ¥ = íïïî

From (2): ( )1 0

sin0

q -=, meaning that the term "0/0 " can be anything, and

( )

( )

1

11

sin 0.866 60 or 1203 /2sin

sin sin 0.866 60 or 120q

f

--

-

ì =ïæ ö ïï÷ç= = í÷ç ÷ç ïè ø - = - -ïïî

Because of the limits of q (0 180q£ £ ), the main radiation occurs

at 90f = and 60q =

and (because of the symmetry)

at 90f = and 120q =

.

Case III: 0n = , 1m = :

From (1):

140.89343

tan2 220.89

f -ì ïé ù ï= =ê ú íïê ú ë û ïî

From (2):

( )

( )

1

11

sin 1.32281sin

cos sin 1.3228q

f

--

-

ìïæ ö ïï÷ç= = í÷ç ÷çè ø ï -ïïî and

( )

( )

1

11

sin 1.32283 /2sin

sin sin 1.3228q

f

--

-

ìïæ ö ïï÷ç= = í÷ç ÷ç ïè ø -ïïî , respectively.

This means that there is no solution in case III.

Case IV: 1n = - , 1m = - :

From (1):

17.633 /2 1

tan1 187.63

f -ì ïé ù- ï= =ê ú íï-ê ú ë û ïî

From (2):

( )

( )

1

11

sin 1.009 2701sin

cos sin 1.009 90q

f

--

-

ì - »ïæ ö- ïï÷ç= = í÷ç ÷çè ø ï »ïïî

and


( )

( )

1

11

sin 1.009 2703 /2 1sin

sin sin 1.009 90q

f

--

-

ì - »ïæ ö- ïï÷ç= = í÷ç ÷ç ïè ø »ïïî

The approximate “solution” of the arcsin-fuction means that there is no perfectly constructive interference of the waves originating from the respective array elements, but rather an “almost constructive interference”. As shown by the plots below, a large and significant “almost grating” lobe does exist.

Because of the limits of q (0 180q£ £ ), a grating lobe occurs

at 187.63f = and 90q =

.

Case V: 1n = - , 0m = :

From (1):

( )190

tan270

f -ìïï= ¥ = íïïî

From (2): ( )1 0

sin0

q -=, where the term "0/0 " can be anything, and

17.69 or 172.313 /2 1

sinsin 7.69 or 172.31

qf

-- -ìïæ ö- ï÷ç= = í÷ç ÷ç ïè ø ïî

Because of the limits of q (0 180q£ £ ), two grating lobes occur

at 270f = , 7.69q =

and

at 270f = , 172.31q =

.

Case VI: 1n = - , 1m = :

From (1):

17.63 352.373 /2 1

tan1 172.37

f -ì- ïé ù- ïï= =ê ú íïê ú ë û ïïî

From (2):

( )

( )

1

11

sin 1.009 901sin

cos sin 1.009 270q

f

--

-

ì »ïæ ö- ïï÷ç= = í÷ç ÷çè ø ï - »ïïî

and

( )

( )

1

11

sin 1.009 903 /2 1sin

sin sin 1.009 270q

f

--

-

ì »ïæ ö- ïï÷ç= = í÷ç ÷ç ïè ø - »ïïî

The approximate “solution” of the arcsin-fuction means that there is no perfectly constructive interference of the waves originating from the respective array elements, but rather an “almost constructive interference”. As shown by the plots below, a large and significant “almost grating” lobe does exist.

Because of the limits of q (0 180q£ £ ), a grating lobe occurs

at 357.37f = and 90q =

.


The 3D array factor is shown in the figure below in linear and logarithmic scale.

z

y

x

z

y

x

linear

dB

main beam(case II)

symmetricmain beam(case II)

grating lobe(case V)

grating lobe(case V)

grating lobe(case IV)

grating lobe(case VI)


Antenna array (non-linear array, constant feed magnitude, non-isotropic radiators)

Problem 6.4 : A corner reflector consists of two semi-infinite perfectly conducting planes at angle of 90 to each other. A /2l dipole is placed parallel to the intersection line of the two planes and at the distance d from the same line. The minimum distance to each plane from the dipole is equal. a) Determine the far-field if the input current of the dipole is 0I . b) Determine the nulls of the radiation pattern in the case /2d l= .

dFeed

a)

Thanks to image theory the corner reflector problem can be transformed into an array consisting of four elements:

r2

r2

r3 r1

d

d

Feed #1

Plate #1Image #2

d

dFeed #1

Plate #2

Image #2

Image #3

Image #4

d

d

(a) (b)

x

y

r


The distances from the four dipoles to the observation point r are

1

2

3

4

cos sin

sin sin

cos sin

sin sin

r r d

r r d

r r d

r r d

f q

f q

f q

f q

= -

= -

= +

= +

and the total electric field of the configuration is given by:

tot 1 2 3 4

sin cos sin sin sin cos sin sin0

jkd jkd jkd jkd

E E E E E

E e e e e

AE AF

q f q f q f q f- -

= + + +

é ù= - + -ë û= ⋅

with the single element pattern being the one of a /2l dipole:

( )00

cos cos2

2 sin

jkrj I eAE E

r

pqh

p q

-= = ,

and the array factor:

( ) ( )[ ]2 cos sin cos cos sin sinAF kd kdq f q f= - .

Thus, the total electric field is:

( )( ) ( )[ ]0

tot

cos cos2 2 cos sin cos cos sin sin

2 sin

jkrj I eE kd kd

r

pqh

q f q fp q

-= ⋅ ⋅ -

b)

Here, the zeros of the single element pattern and of the array factor have to be determined. For the array factor AF , the nulls are occurring for:

( ) ( )cos sin cos cos sin sin 0

cos cos 2 sin sin 02 2

kd kd

A B A BA B

q f q f- =

+ -- = ⋅ =

where:

sin cos sin cos

sin sin sin sin

A kd

B kd

q f p q f

q f p q f

= =

= =


This means that

sin 02

A B+= or sin 0

2A B-

= .

This gives

0A B+ = and therewith sin cos sin sin 0p q f p q f+ =

Thus, sin 0q = or cos sin 0f f+ = and thus 135 ,315f = .

0A B- = and therewith sin cos sin sin 0p q f p q f- =

Thus, sin 0q = or cos sin 0f f- = and thus 45 ,225f = .

This result was to-be-expected given the geometry of the problem: the dipole polarized parallel to the corner mirror metal plates shall indeed give a radiation null in the direction of the plates (that is, φ=45° and φ=315°).

For the element AE , the nulls are occurring for:

( )cos cos2 0

sin

pq

q= .

For 0q = the expression has the form 00

, and thus the rule of l'Hospital has to be

applied to check if a null of the radiation pattern is occuring at 0q =

( )0 0

0

sin cos sin( ) (0) 2 2lim lim 0( ) cos(0)f fg gq q

q

p pq qq

q q

=

¢= = =

¢.

This result was to-be-expected given the geometry of the problem: the dipole has a null in the direction θ=0°, so does a planar group of four parallel dipoles.


Antenna array (linear array, non-constant feed magnitude, dipole radiators)

Problem 6.5 : Assume an antenna array composed of 3 infinitesimal horizontal dipoles positioned along the z -axis, as shown in figure below. The currents on the dipoles are constant along the wire and have the following time dependence:

D1: ( ) ( )1 0 sinI t I tw= D2: ( ) ( )2 02 cosI t I tw= D3: ( ) ( )3 0 sinI t I tw= -

a) Calculate the array factor ( , )AF dq and the complete far field ( , )FFE d rq, of

this antenna array in function of d ,q and r . b) Which angle q maximizes theAF , having distance d of half a wavelength? c) Calculate distance d for which the array radiates in end-fire direction. a)

The phase difference between D2 and D1 is 90 . As well the phase difference between D3 and D2 is 90 . So we have an array with uniform spacing d and progressive phase x ( 90= ).

The far field radiation of one single infinitesimal dipole is located in the origin and oriented like shown in figure is

0 cos4

jkrI leE j

rq h qp

-=

The excitations can be written as


( )

( )

( )

1 0

2 0

3 0

cos 90

2 cos

cos 90

I I t

I I t

I I t

w

w

w

= -

=

= - -

The sum of the far field contributions of the 3 dipoles can be written as

( cos ) ( cos )0 2 2

( cos ) ( cos )0 2 2

cos 24

cos 24

j kr kd j kr kdjkr

j kd j kdjkr

I lE j e e e

r

I lj e e e

r

p pq q

q

p pq q

h qp

h qp

- - + - + +-

- - +-

é ù= ê + - ú

ê úë ûé ù

= ê + - úê úë û

cos cos2AF 2j jkd jkde e ep

q q- -é ù= + -ê úë û

( )AF 2 2 sin coskd q= +

The total field becomes

( )( )0 cos 2 2 sin cos4

jkrFF

I lE j e kd

rh q q

p-= +

b) The maximum of the AF (AFmax = 4) appears if the argument of the sine function is

cos 2 , 0, 1, 2, ......2

kd n where np

q p= + =

solving for q we get

( )max

2 2 12 2arccos arccos arccos 22

n nn

kd

p pp pq

p

æ ö æ ö+ +÷ ÷ç ç÷ ÷ç ç÷ ÷ç ç= = = +÷ ÷ç ç÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

A real solution occurs only for arguments of the arcos whose modulus is smaller or equal to one. Therefore the only possible n is 0n = .

max1 1

arccos 2 arccos 602 2

nqæ ö æ ö÷ ÷ç ç= + = + = ÷ ÷ç ç÷ ÷÷ ÷ç çè ø è ø

This array factor is shown in the following figure (note: direction for positive z / θ=0° pointing horizontally in the picture)


1

2

3

4

30

210

60

240

90

270

120

300

150

330

180 0

c)

The endfire direction of the given array is either 0q = or 180q = . In order to achieve radiation in the endfire direction the argument of the cosine function has to be maximized.

( )AF( , ) 2 2 sin cos 4

for 0 :

cos 22

, where 0,1, 2,....4

d kd

kd kd n

d n n

q qq

pq p

ll

= + ==

= = +

= + =

( )AF( , ) 2 2 sin cos 4

for 180 :

cos 2 22 2

, where 1, 2,....4

d kd

kd kd n kd n

d n n

q qq

p pq p p

ll

= + ==

= - = + = - -

= - =

The following picture shows the AF for d = λ / 4. (note: direction for positive z / θ=0° pointing horizontally in the picture)

1

2

3

4

30

210

60

240

90

270

120

300

150

330

180 0